Return Probabilities of a Simple Random Walk on Percolation Clusters Deborah Heicklen 1;2, Christopher Homan 3;4 Abstract Benjamini and Schramm [6] sh

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1 Return Probabilities of a Simple Random Walk on Percolation Clusters Deborah Heicklen 1;2, Christopher Homan 3;4 Abstract Benjamini and Schramm [6] showed that the probability that a simple random walk on the innite percolation cluster in Z d returns to the origin at time t is greater than or equal to Ct?d=2. In this paper we show that this probability is less than or equal to C 0 t?d=2 (log t) 3d2 =2+d. We use this result to show that the uniform spanning forest on innite percolation clusters in Z d ; d 5 is a.s. supported on graphs with innitely many connected components. Keywords : percolation, simple random walk, return times, spanning forests. Subject classication : Primary: 60J45; Secondary: 60J10, 60J65, 60J15, 60K35. 1 Department of Mathematics, University of California, Berkeley, California, Research supported in part by an NSF postdoctoral fellowship 3 Department of Mathematics, University of Maryland, College Park, MD, Research supported in part by an NSF postdoctoral fellowship 1

2 1 Introduction Consider Bernoulli bond percolation on Z d with parameter p. Recall that this is the independent process on Z d which retains an edge with probability p and deletes an edge with probability 1? p. For all d > 1, there exists a critical parameter p c < 1 that depends on d, such that if p > p c, then a.s. there is a unique connected component with innitely many edges [7]. This component is called the innite cluster. The phase where p > p c is called supercritical Bernoulli percolation. Supercritical Bernoulli percolation can be viewed as a random perturbation of the original graph. For more on this, see [11]. It is then natural to ask what properties persist through this random perturbation. In particular, we focus on the properties of a simple random walk f(t)g on the innite cluster. A simple random walk on the innite cluster means that each neighbor of (t) in the cluster is equally likely to be (t + 1). Rayleigh's monotonicity principle says that if a simple random walk on a graph G is recurrent then a simple random walk on any subgraph of G is recurrent. This implies that a simple random walk on the innite percolation cluster in Z 2 is recurrent. Grimmet, Kesten, and Zhang proved that in Z d ; d 3 a simple random walk on the innite percolation cluster, C 1 (Z d ; p), is a.s. transient [12]. Haggstrom and Mossel used a method of Benjamini, Pemantle, and Peres to show that a large class of subgraphs of Z d have transient percolation clusters [15] [5]. Although on subgraphs transient percolation clusters corresponds well with transience, the methods in these papers tell us nothing about the probability that a simple random walk on the percolation cluster returns to the origin at time t. They also do not answer the question of whether two independent simple random walks on the percolation cluster will intersect innitely often a.s. The probability that a simple random walk on Z d returns to the origin at time t decays on the order of Ct?d=2. Benjamini and Schramm proved that the probability that a simple random walk returns to the origin at time t on a percolation cluster with parameter p is decreasing in p [6]. This implies that this return probability is greater than or equal to Ct?d=2. In this paper we get a bound in the other direction. We show that this probability is bounded above by C 0 t?d=2 (log t) d2 =2 : This result allows us to show that another property of the simple random walk persists after supercritical Bernoulli percolation. Namely, in Z d ; d 5, two simple random walks 2

3 intersect nitely often a.s. [18]. In section 7 we use the bound on the return probability to show that this property also holds on the innite percolation cluster in Z d ; d 5. Together with results of [4], [3] and [17], this in turn shows that for Z d ; d 5, the uniform spanning forest (USF) on percolation clusters is supported on innitely many components. We dene the wired, free and uniform spanning forests in section 7. This corresponds with a result of Pemantle that the USF in Z d ; d 4, is supported on trees, while in Z d ; d 5 the USF is supported on graphs that have innitely many connected components [19]. Benjamini and Schramm's result proves that two independent simple random walks on C 1 (Z d ; p); d 4; have innite expected number of intersections. If this could be extended to show that two simple random walks have innite intersections almost surely then this would prove that the USF on C 1 (Z d ; p); d 4; is supported on a tree a.s. We write [j 1 ; k 1 ][j 2 ; k 2 ]:::[j d ; k d ] for the subgraph of Z d that includes all vertices (v 1 ; :::; v d ) such that l i v i k i for all i. The edges of this graph are those that have both their endpoints in this set of vertices. For any graph G we write ng for the graph that is generated by replacing each edge of G with n edges in parallel. We say a distinguished vertex of ng is a vertex that corresponds with one in G. For any graph G we write V (G) for the vertices of G. The outline of the return time probability proof is as follows. We want to show that there exist subgraphs of C 1 (Z d ; p) where we can calculate the return times. In particular we will show that for some > 0 with high probability there exists a embedding of n[2 n ; 2 n ] d in C 1 (Z d ; p) centered around the origin. This embedding will look like a wiggly grid sitting inside the percolation cluster. It is easy to bound the return times of simple random walk on this subgraph. We will then show that there exists a graph which extends the embedding of n[2 n ; 2 n ] d and spans the percolation cluster intersected with some nite ball where we can still bound the return probabilities. This subgraph will look like the wiggly grid with bushes attached to various vertices in the grid. The bushes ensure that the subgraph spans the nite section of the percolation cluster. They also do not increase the return time probabilities signicantly. We then use a result of Benjamini and Schramm to show that the return probabilities on the percolation cluster are bounded by those on the subgraph. Thus we obtain bounds for the return probabilities on the percolation cluster. 3

4 Section 4 shows that the probability that there exists a embedding of n[2 n ; 2 n ] d centered near the origin inside C 1 (Z d ; p) is increasing to 1 exponentially in n. The rst step is to show that there exists an odd integer such that given that x and y are connected, the conditional probability that there exists a path P that connects x and y such that jp j = jx? yj is approaching 1 as jx? yj approaches 1. This is done in sections 2 and 3. Then in section 4 we show how to put these paths together to form a embedding of n[2 n ; 2 n ] d. In section 5 we show how to extend this grid to a spanning subgraph that we can still calculate the return probabilities on. In section 6 we show that this allows us to bound the return probabilities on the percolation cluster. In section 7 we show that the bounds on the return times show that the USF on C 1 (Z d ; p); d 5 is supported on graphs with with innitely many connected components. 2 Approximate Pathlengths in the Percolation In this section we show that if two points x and y are connected in the innite cluster and far apart, then with high probability there exists a path between them in the cluster of approximately any length in a certain interval. Namely, for any length q 2 [1:5; 36], there is a path of length qjx? yj jx? yj. These paths are a rst approximation for the sides of the grid we construct in section 4. In section 3 we construct paths of length exactly 5jx? yj. These will be the sides of the grid. First we introduce some notation. Let! 2 f0; 1g Zd be a realization of percolation. An open edge e is one such that!(e) = 1. An open path P is a connected set of open edges with no cycles such that two vertices have degree 1. These are the endpoints of P. An open path P connects x and y if x and y are endpoints of edges in P. We also say P is a path from x to y. We write x y if there exists an open path from x to y. We write x 1 if x is part of the unique innite cluster. If x y then we let D(x; y) be the distance of the shortest open path from x to y. A path P is in a set of vertices V if the endpoints of every edge in P are in V. We write V (G) for the vertices of a graph. We say that G 0 spans G if V (G) V (G 0 ): Throughout the next two sections we prove that there exists an odd integer, such that if x y and jx? yj is large, then with high probability there exists a path P which 4

5 connects x and y such that jp j = 5jx? yj. We show that conditioned on x y, the probability that there does not exist such a path decreases exponentially in jx? yj. We use the taxicab metric on Z d, j(x 1 ; :::; x d ); (y 1 ; :::; y d )j = jx 1? y 1 j + ::: + jx d? y d j: Dene L(u; v) to be the elements in Z d which are within d=2 of the line segment joining u and v. Let B k be all (x 1 ; :::; x d ) 2 Z d such that jx 1 j+:::+jx d j k and B k (x) = x+b k. Also let B (x; y) = B jx?yj (x) [ B jx?yj (y). All constants A i and C i may depend on d and p. There are a number of places in the following sections where in order to be precise in choosing our parameters, it would be necessary to use the greatest integer function. We dene our parameters without this as the calculations are already intensive. Similar estimates hold if the parameters are chosen using the greatest integer function when necessary. We will show that for any q 2 (1:5jx? yj; 36jx? yj) there exists a path P lying in a certain region, that connects x and y region and jp j is approximately q. Then in the next section we will use an inductive argument to show that there exists a path of exactly length 5jx? yj. There are three main tools that we will use to get a path of approximately the right length. The rst is the following result of Antal and Pisztora. Theorem 2.1 [1] Let p > p c (d). Then there exists = (p; d) 2 [1; 1); and constant C 1 such that for all y P(D(0; y) > jyj j 0 y) < e?c1jyj : The second is the method of Grimmet, Kesten, and Zhang [12]. This method was used to show the existence of a tree in C 1 (Z d ; p) with certain branching properties. This tree was used to show that for d 3, a simple random walk on C 1 (Z d ; p) is transient a.s. The method works just as well for Z 2 [16]. This method requires the following standard facts about the innite percolation cluster. Lemma 2.1 Given p > p c (d) there exists C 2 so that P(B k 6 1) < e?c2k : Proof: For d 3 a proof of this can be found in [13]. For d = 2 this can be proven by combining results in [1] and [16]. ] 5

6 Remark 2.2 Are there better sources for this? The third is a combinination two theorems, one of Chayes, Chayes and Newman, and the other of Barsky, Grimmett and Newman, which gives the following result. Theorem 2.2 [8], [2] Given p > p c (d) there exists C 3 so that P(0 6 1 j k ) < e?c3k : Combining these last two theorems we get the following lemma. Lemma 2.3 Let S be a connected set with diam(s) = n. Then there exists C 17 that P(S 6 1) e?c 17p n : such Proof: Since diam(s) = n it is possible to nd x i, 1 i p n=4, such that x i 2 S and B p n(x i ) \ B p n(x j ) = ; for all i 6= j. Then P(S 6 1) P(6 9x i such that x i 1) P(6 9x i such that x p n(x i )) + P(x i 6 1 j x p n(x i )) P(x i p n(x i )) p n=4 + e?c 3 P(x i 6 1) p n=4 + e?c 3 e?c 17p n : p n p n ] Is the lemma true if we replace p n with n? Is there a standard reference for this? Now we will use a method similar to the one introduced by Grimmet, Kesten, and Zhang to show that innite percolation clusters in Z d ; d 3 are a.s. transient. We will use this method to show that for any two points x; y 2 Z d the probability that there exists a reasonably short and direct path from near x to near y is going to 1 exponentially in jx? yj. Dene E r (z) = to be the event that there exists w 2 B r (z) 1. such that w 1 and 6

7 2. any point w 0 2 (B 2r (z)) C such that w 0 w satises D(w; w 0 ) jw? w 0 j: Notice that E r (z) = is not an increasing event. This makes some of the future arguments more technical since the FKG inequality will not apply. Lemma 2.4 There exists A 4 ; C 4 > 0 such that for all r P(E r (z)) > 1? A 4 e?c4r : Proof: By Lemma 2.1 the probability that line 1 does not hold is less than e?c2r. Theorem 2.1 implies that P (9 z 2 w such that jz 2? wj = j, D(z 2 ; w) jz 2? wj) < 2dj d?1 e?c1j : Thus P(E r (z)) > 1? e?c2r? 2dj d?1 e?c 1j > 1? A 4 e?c 4r j2r for an appropriate choice of A 4 and C 4. ] Dene F r (z) to be the event that 1. z 1 and 2. any point w 0 2 B 2r (z) C such that z w 0 satises D(z; w 0 ) jz? wj: Let k = Fix = 1=2k 2, and > 1 an odd integer from Theorem 2.1. We dene sets in Z d, called wedges, around x and y in which the various constructed paths from x to y will lie. The wedges allow enough room to construct paths of many dierent lengths, while ensuring that when the paths are concatenated, they do not intersect except near x or y. In the following denition, 6 xz; xy means the angle between the line segments xz and xy. Dene and W (x; y) = fz 2 Z d j z 2 B :1=k 2(x; y) or (6 xz; xy 15 and 6 yz; yx 15 )g W 0 (x; y) = fz 2 Z d j z 2 B :1=k 2(x; y) or (6 xz; xy 30 and 6 yz; yx 30 )g: 7

8 Denition 2.5 A pair of points x and y in Z d is (; ;!) good, written x ' y, if there exists an open path P 0 from x to y such that for all q 2 [1:5; 36], there exists open path P! such that 1. jp j 2 [(q? )jx? yj; (q + )jx? yj], 2. P W (x; y), 3. P j B1=4 (x;y) = P 0 j B1=4 (x;y). Now we show that most pairs of vertices that are connected and far apart are good pairs. Lemma 2.6 There exists A 5 ; C 5 > 0 such that P(x ' y j x y) > 1? A 5 e?c5jx?yj : Proof: The idea is as follows. By lemma 2.1, if x y, then it is likely that there is a path in the percolation cluster whose length is bounded above by jx? yj. We show that it is possible to pick a path, P 0 inside the wedge from x to y with length less than 1:5jx? yj. Moreover we show it is possible to pick a dierent path so that the length falls within any specied range. To do this, we create P 0 and a disjoint collection of possible detours that are contained in the dened wedges. We lay down a series of evenly spaced points, called markers, throughout the line segment between x and y and the possible detours. Then we show that with high probability there exists points in the balls around the markers such that for any two adjacent points w and w 0, there is a path whose length is bounded above by jw? w 0 j. This follows from lemma 2.4. We construct P 0 by concatenating the paths between the markers that lie along the line segment between x and y. By concatenating as many of the detours as necessary, we create paths between x and y of varying lengths. Each detour is in a neighborhood of three sides of a rectangle that have a xed width and height. If we include a detour, we do not include the path between the markers lying on the line segment between x and y, which is the fourth side of the rectangle. The width of each detour is relatively small compared to the height. We show that by the way the parameters are chosen, there are enough detours to ensure that the path is long enough. On the other hand, each detour 8

9 adds only a small amount. Hence it is possible to add the right number of detours so that the length of the path falls within the specied range. Let m = 4 be the number of markers on the short side of each rectangle and M = be the number of markers on the long side of each rectangle. Dene # = 60= to be the total number of possible detours. Let n = = be the number of markers on the line segment between x and y, = jx? yj=n be the size of the increments between the markers and v = (y? x)=jy? xj. Finally let v 0 be a unit vector perpendicular to v. We dene the markers equally spaced throughout the region in which the path is permitted. Let z 1 i;0 = x + iv 0 i n; zi;j 2 = x + (n=4 + 2mj)v + iv 0 0 i M; 0 j #; zi;j 3 = x + (n=4 + 2mj + i)v + Mv 0 i m; 0 j #; zi;j 4 = x + (n=4 + (2m + 1)j)v + iv 0 0 i M; 0 j #: We will show that if E =4 (zi;j) l holds for all i; j and l then x and y are a good pair. First we show that if the events E =4 (zi;0) 1 hold for all i then there is a path P 0 from B =4 (x) to B =4 (y) inside L(x; y) + B that has jp 0 j < 1:5jx? yj. To see this, let w i 2 B =4 (zi;0) 1 be the point in the denition of E =4 (zi;0). 1 Then jp 0 j = i i i D(w i ; w i?1 ) jw i? w i?1 j 2(=4) + i jz 1 i;0? z 1 i?1;0j =2jx? yj + jx? yj: Furthermore P 0 lies inside L(x; y) + B. This is because the maximum that P 0 can travel away from the line between x and y is bounded by max i D(w i ; w i?1 ) 2 + =4 : If F =4 (x) and F =4 (y) also hold then there is a path P 0 from x to y inside W (x; y) such that jp 0 j < 1:5jx? yj. 9

10 Now we show that if in addition the events E =4 (z l i;j) hold for all i; j; l then there are paths from near x to near y inside L(x; y) + B (+M ) of many dierent lengths. Suppose in addition to the events E =4 (z 1 i;0), F =4 (x), and F =4 (y) holding that for a xed j the events E =4 (z 2 i;j), E =4 (z 3 i;j), and E =4 (z 4 i;j) hold for all i. Then there is a path P j such that 2M < jp j j < 3M, P j \ P 0 = ; and the only vertices shared by P j and P 0 are the endpoints of P j. We can form a new path from x to y by cutting out a piece of P 0 and attaching P j. We say this path takes detour j. By this construction, the possible detours occur in the middle half of the line segment between x and y. This is because 2m# 2(4)(60=) < n=2. Furthermore, the maximum height of the detour marker points is M jx? yj < jx?yj 4 tan 15. This implies that P j B + (L(z0;j; 2 zm;j) 2 [ L(z0;j; 3 zm;j) 3 [ L(z0;j; 4 zm;j)) 4 W (x; y): These containment conditions imply that the P j are all disjoint. Now suppose E =4 (zi;j) l holds for all i; j; l and F =4 (x) and F =4 (y) also hold. Then for any i 2 [0; :::; #] we can form a path that takes exactly i detours. Since 2M < jp t j for each t a path that takes i detours has length at least 2iM. Thus there is a path that takes # detours which has length at least 2M# 2jx? yj(3002 )(60) jx? yj: Adding one extra detour adds at most length 4M 4jx? yj jx? yj: to the length of a path. Thus if E =4 (zi;j) l holds for all i; j; l and F =4 (x) and F =4 (y) also hold then x and y are a good pair. Now we calculate the probability that all of this happens. Notice that n + 3#M is the total number of markers. Also notice that P(x y) P(x 1) 2. P (x 6' y j x y) < P [i;j;l (E =4 (zi;j)) l c j x y + 2P (F=4 (x)) c j x y < (n + 3#M)P (E=4 (0)) c =P(x y) + 2P(x 6 1 j x y) + 2P(x 1; (F =4 (x)) c j x y) 10

11 < (n + 3#M)A 4 e?c 4=4 =P(x 1) 2 + 2e?C3jx?yj + 2A 4 e?c 4=4 =P (x 1) 2 < (n + 3#M + 2)A 4 =P(x 1) 2 e?c 4=4 + 2e?C 3jx?yj < (n + 3#M + 2)A 4 =P(x 1) 2 e?c4jx?yj=4n + 2e?C 3jx?yj < A 5 e?c5jx?yj : Thus the lemma is true for some appropriately chosen A 5 and C 5. ] Corollary 2.7 There exists A 6 and C 6 so that P(9y 2 B r (x) C such that x y, x 6' y) < A 6 e?c 6r and P(9y; z 2 B r (x)such that y z, y 6' z; jy? zj > r=2k 2 ) < A 6 e?c 6r : 3 Exact Pathlengths in the Percolation This section is devoted to showing that with high probability we can nd a path in the percolation cluster from x to y of length exactly 5jx? yj. Denition 3.1 A pair x and y in Z d is said to be (;!) very good, written x = y, if there exists a path P 0 from x to y satisfying the following property. For all q 2 (3jx? yj; 10jx? yj) such that q + jx? yj = 0 mod 2 there exists P! such that 1. jp j = q, 2. P W 0 (x; y); and 3. P j B1=16 (x;y) = P 0 j B1=16 (x;y). The proof that most pairs that are connected are very good is inductive. The main idea in the inductive step is contained in this next lemma. The three lemmas after that will allow us to get the induction started. Lemma 3.2 Suppose x 0 ; y 0 2 W (x; y) such that 3jx? yj < jx? 8 x0 j; jy? y 0 j < 5 jx? yj. 8 Suppose 1 jx? yj < jx 0? y 0 j < 1 jx? k 2 20 x0 j and similarly for jy? y 0 j. If x ' x 0, y ' y 0 and x 0 = y 0, then x = y. 11

12 Proof: The basic is idea is that given q we nd paths P 1 ; P 2 ; and P 3 as follows. The denition of a good pair will allow us to choose P 1 so that P 1 connects x and x 0, P 1 2 W (x; x 0 ), and jp 1 j a little less than :5q. It also allows us to choose P 2 so that P 2 connects y and y 0, P 2 2 W (y; y 0 ), and jp 2 j a little less than :5q. Then since x 0 = y 0 we can choose P 3 so that P 3 connects x 0 and y 0, P 3 2 W (x 0 ; y 0 ), and jp 3 j = q? jp 1 j? jp 2 j. Then form a path P by concatenating P 1 ; P 3 and P 2 we get a path from x to y of length q. The problem is that P may have loops. These can only occur in B jx?x 0 j=10k 2(x0 ) [ B jx 0?y 0 j=10k 2(x0 ) B jx 0?y 0 j=16(x 0 ) or B jy?y 0 j=10k 2(y0 ) [ B jx 0?y 0 j=10k 2(y0 ) B jx 0?y 0 j=16(y 0 ): Let q 1 be the length of the loops near x 0 and q 2 be the length of the loops near y 0. Notice that q 1 + q 2 < 2jx? yj. Now modify P 1 to nd P 0 1 with jp1j 0 2 (:5(q + q 1 + q 2 )? 1:6jx 0? y 0 j? jx? yj; :5(q + q 1 + q 2 )? 1:6jx 0? y 0 j + jx? yj): This is possible because :5(q + q 1 + q 2 )? 1:6jx 0? y 0 j? jx? yj > (1:5? )jx? yj? :1jx? x 0 j > 2:4jx? x 0 j? 1:6jx? x 0 j? :1jx? x 0 j > 1:5jx? x 0 j and :5(q + q 1 + q 2 )? 1:6jx 0? y 0 j + jx? yj < 6jx? yj + jx? yj < 36jx? x 0 j: Modify P 2 to nd P 0 2 with jp 2 j in the same interval. Choose P 3 so that it connects x 0 and y 0, P 3 2 W (x 0 ; y 0 ). Finally nd jp3j 0 = q + q 1 + q 2? jp1j 0? jp2j. 0 This can be done because q + q 1 + q 2? jp1j 0? jp2j 0 2 (3:2jx 0? y 0 j? 2jx? yj; 3:2jx 0? y 0 j + 2jx? yj) (3jx 0? y 0 j; 10jx 0? y 0 j): 12

13 Since both ends of P 1 ; P 2 ; and P 3 are the same as the ends of P 0 1; P 0 2; and P 0 3 this modication does not change the concatenation in B 1=10 (x; y) or change the loops. Thus concatenating P 0 1; P 0 3; and P 0 2 and then removing the loops near x 0 and y 0 gives a path P from x to y of length q. Finally P W (x; x 0 ) [ W 0 (x 0 ; y 0 ) [ W (y; y 0 ) W 0 (x; y). same argument gives us the following. ]The Corollary 3.3 Suppose x 0 ; y 0 2 W (x; y) such that 3 jx? yj < jx? 8 x0 j; jy? y 0 j < 5 jx? yj, 8 x 0 = y 0 and P 0 is the base path between them. Also suppose 1 jx? yj < jx 0? y 0 j < k 2 1 jx? 20 x0 j and similarly for jy? y 0 j. If there exists x 00 2 V (P 0 ) \ B :01jx 0?y 0 j(x 0 ) and y 00 2 V (P 0 ) \ B :01jx 0?y 0 j(y 0 ) such that x ' x 00 and y ' y 00 then x = y. The main result of this section is that P(x = y j x y) is converging to 1 exponentially in jx? yj. Before we show this, we show in the next two lemmas that it is not converging to 0 exponentially in jx? yj. This step is necessary to prove the base case of the induction. Denition 3.4 B 2k n(x) is very good if there exists y; z 2 B 2k n(x) such that y; z 1, y = z and jy? zj = k n. Lemma 3.5 If there exists an A 20 and a C 20 such that P(B 2k n+1(x) is very good j B 2k n(x) is very good) > 1? A 20 e?c 20k n ; then there exist an N and A 12 such that if jx? yj = k n for n N then P(x = y j x y) > A 12 jx? yj?2d : Proof: Notice that jw?w 0 j=k n ;w;w 0 2B 2k n P(w = w 0 ) P(B 2k n is very good ): Since P(w = w 0 ) is the same for each of these pairs, this implies that P(w = w 0 ) P(B 2k n is very good )=(2k nd ) 2. Since P(B 2k n+1(x) is very good j B 2k n(x) is very good) > 1? A 20 e?c 20k n ; 13

14 there exist N and C so that P(B 2k n(x) is very good for all n > N) > C: By the previous remark, this implies there exists N such that if jx? yj = k n for some n > N P(x = y j x y) P(x = y) P(B 2k n is very good)=(2k nd ) 2 A 12 jx? yj?2d : Lemma 3.6 There exist A 12 such that if jx? yj = k n for some n then P(x = y j x y) > A 12 jx? yj?2d : ] Proof: To prove this we show that there are A 20 and C 20 such that P(B 2k n+1(x) is very good j B 2k n(x) is very good) > 1? A 20 e?c 20k n : This implies the result by the previous lemma. The idea of the proof is as follows. It suces to nd a pair of points y; z 2 B 2k n+1(x) such that jy? zj = k n+1 and y = z. Since B 2k n(x) is very good, we rst nd points y 0 and z 0 in B 2k n(x) such that jy 0? z 0 j = k n and y 0 = z 0. Now we need to nd points y and z that are the right distance from y 0, z 0 and each other, and such that y 0 ' y, z 0 ' z. This would allow us to apply lemma 3.2. Technical considerations force us to add another step to the process. We replace y 0 and z 0 with points y 00 and z 00 near y 0 and z 0 respectively, which lie on the path between y 0 and z 0 and such that y ' y 00 and z ' z 00. Once we choose y, we need to choose z satisfying the hypotheses of lemma 3.2 and such that jy? zj = k n+1. Thus we choose z lying k n+1(y) and so that it is roughly colinear with y, y 0 and z 0. In order to do this, we nd z 1 and z 2 near this area, but on either side of B k n+1(y) such that z 1 ' z 2. The path joining them will intersect B k n+1(y). Let the intersection point be z. Once we nd z, we now choose y 00 and z 00 to satisfy the hypotheses in lemma

15 If B 2k n(x) is very good then there exist y 0 ; z 0 2 B 2k n(x) such that y 0 ; z 0 1, y 0 = z 0 and jy 0? z 0 j = k n. Let P be a path connecting y 0 and z 0. Let v be a unit vector pointing from y 0 to z 0. If there exist y; y 00 ; z; z 00 such that 1. y 2 B k n(y 0? (k n+1 =2)v), 2. z 2 B 6k n(y + (k n+1 )v), 3. jy? zj = k n+1 ; 4. y 00 2 V (P ) \ B :01jy 0?z 0 j(y 0 ), 5. z 00 2 V (P ) \ B :01jy 0?z 0 j(z 0 ), 6. y ' y 00, and 7. z ' z 00. Then corollary 3.3 shows that y = z. Since y; z 1 we have that B 2k n+1(x) is very good. Thus by the FKG inequality P(B 2k n+1(x) is not very good j B 2k n(x) is very good) P(6 9y; y 00 such that y 2 B k n(y 0? (k n+1 =2)v), y 00 2 V (P ) \ B :01jy 0?z 0 j(y 0 ), y ' y 00, and y; y 00 1 j B 2k n(x) is very good) + P(6 9z; z 00 such that z 2 B 6k n(y + (k n+1 )v), z 00 2 V (P ) \ B :01jy 0?z 0 j(z 0 ), z ' z 00 jz? yj = k n+1 and z; z 00 1 j B 2k n(x) is very good) P(6 9y; y 00 ) + P(6 9z; z 00 ): Suppose there exists y 2 B k n(y 0? (k n+1 =2)v) and such that y 1. If there exists y 00 2 V (P ) \ B :01jy 0?z 0 j(y 0 ) such that y 00 1 then y y 00. If there do not exist a; b 2 B 2k n+1(x) such that a b, a 6' b and ja? bj > k n?1 then y ' y 00. Additionally suppose there exist z 1 2 B k n =(y + (k n+1? 2k n =)v) and z 2 2 B k n =(y + (k n+1 + 2k n =)v) such that z 1 ; z 2 1. Then z 1 z 2 and z 1 ' z 2. Thus there is a z such that jy? zj = k n+1, z 1 and jz? (y + (k n+1 )v)j < 6k n. If there exists z 00 2 V (P ) \ B :01jy 0?z 0 j(z 0 ) such that z 00 1 then z z 00 and z ' z 00. Thus if all this holds then there exists y; y 00 ; z and z

16 Now we bound the probability that this happens. P(B 2k n+1(x) is not very good j B 2k n(x) is very good) < 3P(B k n = 6 1) + 2P(P \ B jy 0?z 0 j=100(y 0 ) 1) + P(6 9a; b 2 B 2k n+1(x) such that a b, a 6' b and ja? bj > k n?1 ) < 3e?C 2k n = + 2e?C 17k n=2 =100 + e?c 6k n?1 = < A 20 e?c 20k n=2 : ] The base case of the induction is proved in the next lemma. Lemma 3.7 For any d there exists an N = N(d) and C 8 (N) such that for all n N and all 0 1 e?c 3k n+1 + 2e?C 6k n+1 P(x 1) 2 + (e?c 2k n + e?c 8k n + e?c 3k n ) :0016k2+ < e?c 8k n+1+ and P(x 6 = y j jx? yj = k N and x y) < e?c 8k N : Proof: Fix d. By lemma 3.6 the last equation is satised if e?c 8k N = 1?C=k Nd. This denes C 8 in terms of N. Notice as N! 1, C 8! 0. By choosing N suciently large, we can force C 8 to be much smaller than the other constants in the rst equation. This implies that the terms with C 8 in them are the dominant terms of the inequality. Since e?:0016c 8k n+2+ = e?1:6c 8k n+1+ < e?c 8k n+1+ the inequality is true for N suciently large. ] Lemma 3.8 There exist A 8 and C 8 such that P(x = y j x y) > 1? A 8 e?c 8jx?yj : Proof: The proof is by induction. We will show that this is true with A 8 = 1 if jx? yj is suciently large. This implies the result. The previous lemma establishes the base 16

17 case, which is n = N. Given that the induction hypothesis is true for k n we will show that it is true for all x; y such that k n+1 jx? yj k n+2. This implies the result for all jx? yj k N +1. The idea for the proof is as follows. We start with two points x and y that are connected and jx? yj is approximately k n+1. We lay down about k disjoint copies of B 2k n in the middle of W 0 (x; y). Using the induction hypothesis, we show that with high probability at least one of these balls contains a pair satisfying the hypotheses of lemma 3.2. By applying this lemma to the pairs x; x 0, x 0 ; y 0, and y 0 ; y, we obtain that with high probability x = y. We claim that if there exist x 0 and y 0 satisfying the following conditions then x and y are a very good pair. 1. x 0 ; y 0 2 W (x; y) 5 2. jx? yj > jx? 8 x0 j > 1 jx? yj, jx? yj > jy? 8 y0 j > 1 jx? yj, 4 4. jx 0? y 0 j = k n, 5. x 0 = y 0, 6. x 0 ; y 0 1, 7. x; y 1, 8. all w such that jx? wj > 1 jx? yj either w 6 x or x ' w, and 4 9. all w such that jy? wj > 1 jx? yj either w 6 y or y ' w. 4 Lines 6 and 7 imply that x; y; x 0 and y 0 are all connected. Combining this with lines 8 and 2 shows that x ' x 0. Similarly line 3 and 9 show that y ' y 0. Thus we have satised the conditions of lemma 3.2 and x and y are a very good pair. Now we bound the conditional probabilities given x y that lines 1-6 are not satised, and that line 7, 8, or 9 is not satised. To make the rst bound let v be the unit vector parallel to x?y and v 0 be a unit vector perpendicular to x?y. Let = log k (jx?yj=k n+1 ). Dene x i;j = (x+y)=2+5k n iv+5k n jv 0 for jij; jjj :04k 1+=2 : This implies that x i;j 2 W (x; y), 5 8 jx? yj > jx i;j? xj > :25jx? yj, 17

18 6 and 5jx?yj > jx 8 i;j?yj > :25jx?yj for all i; j and jx i;j?x i 0 ;j 0j > 4kn for all i; j 6= i 0 ; j 0. Now the events that there exist x 0 and y 0 in B 2k n(x i;j ) satisfying lines 1-5 are independent. P(6 9x 0 ; y 0 satisfying lines 1 through 6 j x y) P(6 9x 0 ; y 0 satisfying lines 1-6) (1) P(6 9x 0 ; y 0 2 B 2k N (x i;j ) satisfying lines 1-6) (2) i;j P(6 9x 0 ; y 0 2 B 2k N (x i;j ) satisfying lines 1-6) :0016k2+ (3) [P(6 9x 0 ; y 0 2 B 2k N (x i;j ) such that jx? yj = k n and x 0 y 0 ) + (4) P(x 0 = y 0 j x 0 y 0 ) + P(x j x 0 y 0 )] :0016k2+ (5) [P(B k N (x i;j ) 6 1) + e?c 8k n + e?c 3k n ] :0016k2+ (6) [e?c 2k n + e?c 8k n + e?c 3k n ] :0016k2+ (7) Line 1 follows from the FKG inequality. Line 3 follows from lemma 2.1, the induction hypothesis, and lemma 2.2. P(line 7 does not hold j x y) = P(x; y 6 1 j x y) e?c 3k n+1 by lemma 2.2. Corollary 2.7 tells us that P(lines 8 or 9 do not hold j x y) 2P(lines 8 does not hold)=p(x y) Putting this all together gives P(lines 1-9 hold j x y) 2e?C 6k n+1 P(x 1) 2 : > 1? (e?c 2k n + e?c 8k n + e?c 3k n ) :0016k2+? e?c 3k n+1? 2e?C 6k n+1 > 1? e?c 8k n+1+ > 1? e?c 8jx?yj : P(x 1) 2 The next to last line is by lemma 3.7 and the last line is by the choice of. ] 18

19 4 Building the grid 6 In this section we rst show that with high probability there exists corner points, called n hubs, in the percolation cluster. Each corner point has 2d paths emanating from it. Then we show that we can glue these points together with paths of exactly the right length to form a large nite grid centered near 0. This is possible because with high probability, the points at the end of adjacent paths of two respective n hubs form a very good pair. Remember we write [j 1 ; k 1 ] [j 2 ; k 2 ] ::: [j d ; k d ] for the subgraph of Z d that includes all vertices (v 1 ; :::; v d ) such that j i v i k i for all i. We will show that with high probability there exists a copy of n[2 n ; 2 n ] d in! with center near the origin. Let v 1 ; :::; v 2d be the unit vectors in each of the 2d directions in Z d Denition 4.1 Given! and. A point x 2 Z d is an n hub if there exists paths P 1 ; :::; P 2d such that 1. P i connects x n (x), 2. V (P i ) \ V (P j ) = x if i 6= j, 3. jp i j 2n, and 4. V (P i ) n B n=2 (x) fz j v i ; xz 20g: Lemma 4.2 There exists A 9 > 0 and J P(x is a k j hub for all j > J ) > A 9 : Proof: We prove the lemma by showing that there exists A; C > 0 so that P(x is a k j+1 hub j x is a k j hub and x 1) 1? Ae?Ckj=2 : The true rate of decay is exponential in k j. This will be implicit in a later lemma. Suppose x is a k j hub and x 1. If there exist a disjoint collection of paths P 0 i such that 1. P 0 i fz j 6 v i ; xz 30g n B k j =2(x); 2. V (P 0 i ) \ V (P i ) 6= ;, 19

20 3. jp 0 ij 1:6k j+1, 4. V (P 0 i ) n B k j+1 =2(x) fz j 6 v i ; xz 20g; and 5. V (P 0 i ) k j+1(x) 6= ; then x is a k j+1 hub. Conditions 2 and 5 above and condition 1 from the denition of an n hub show that a subset of P i [ P 0 i satises condition 1 in the denition of an n + 1 hub. Condition 1 above and condition 2 in the denition of n hub imply condition 2 in the denition of n + 1 hub. Condition 3 above and condition 3 from the denition of an n hub show it satises condition 3 in the denition. Condition 4 above shows it satises condition 4 in the denition. If there exists y i 2 B k j(x + k j+1 v i ) n B k j+1(x) and z i 2 P i n B 2k j =3(x) such that such that y i ; z i 1 and y i ' z i then there exists an appropriate P 0 i. Conditions 1,2, and 4 are satised because y i ' z i. Condition 3 is satised because y i ' z i and jy i? z i j < (1 + 1=k)k j+1. Condition 5 is satised because y i 2 (B k j+1(x)) c, z i 2 P i, and y i ' z i. P(6 9P 0 i j x is a k j hub and x 1) < P(6 9y i ; z i j x is a k j hub and x 1) < P(6 9y i ; z i ) < P((B k j(x + k j+1 v i ) n B k j+1(x)) 6 1) + P(P i n B 2k j =3(x) 6 1) + P(9y; z 2 B 2k j+1(x) with y z, jy? zj k j?2 =2, y 6' z) < e?c 2k j =2 + e?c 172k j=2 =3 + A 6 e?c 6k j < Ae?Ckj=2 which proves the lemma. ] Lemma 4.3 There exists C 10 > 0 such that P(9x 2 B n such that x is an n hub) 1? e?c 10n : Proof: The proof is by induction. For the base case we will choose a J and C 10 such that for all n k J+1, 20

21 P(9x 2 B n such that x is an n hub) 1? e?c 10n : This is possible by the previous lemma, which states that there exists A 9 that P(x is a k j hub for all j > J) > A 9 : and J such The ergodic thoerem implies that for any j as J! 1 P(9x 2 B k J such that x is a k j hub for all j > J)! 1: Choose J so that for all j > J P(6 9x 2 B k J such that x is a k j hub for all j > J) k < :5: (8) Choose C 10 so that 1. for all n k J+1, P( 9 an n hub in B n ) 1? e?c 10n, and 2. e?c 3k j =P(h is a k j hub) + 2de?C 2=2k j + A 6 e?c 6k j :5e?C 10k j+2 for all j > J. The rst condition implies the base case. Suppose the induction hypothesis has been proven for k j. We will now show that it is true for all k j+1 n k j+2. If there is an h 2 B n which is a k j hub, h 1, B k j(h + v i n) n B n (h) 1 for all i, and there are no y; z 2 B 2n with y z, y 6' z, and jy? zj > n=k 2 then h is an n hub. This follows from the argument in the previous lemma. Let = log k (n=k j ). Notice that n = k j+, and 1 2. Find x 1 ; :::; x 2k so that B k j(x i ) B n=3 for each i and B k j(x i ) \ B k j(x j ) = ; for all i 6= j. This can be done like in lemma 3.8. The events that there exists a k j hub in B k j(x i ) are independent. So P(6 9h 2 B n such that h is a k j hub) P(6 9h 2 B k j such that h is a k j hub) 2k P(6 9h 2 B k j such that h is a k j hub) k (e?c 10k j ) k :5(e?C 10k j+ ): 21

22 By lemma 2.2 we have that P(h 6 1 j h is a k j hub) e?c 3k j =P(h is a k j hub): Thus P(6 9h 2 B n such that h is an n hub) < P(6 9h 2 B n such that h is an k j hub) + P(h 6 1 j h is an k j hub) + 2dP((B k j(h + v i n) n B n (h)) 6 1) + P(9y; z 2 B 2n such that y z, jy? zj > k j and y 6' z) < :5(e?C 10k j+ ) + e?c 3k j =P(h is a k j hub) + 2de?C 2=2k j + A 6 e?c 6k j < :5(e?C10n ) + :5(e?C 10k j+2 ) < e?c10n : ] For an = (d) which will be xed later dene the graph n[?2 n ; 2 n ] d to be the n grid. Remember a distinguished vertex of ng is a vertex that corresponds with one in G. In this case, they are the corner points of the n grid. We say there exists a copy of the n grid around 0 in! if there exists an embedding of the n grid in! with the distinguished vertex corresponding to (v 1 ; :::; v d ) in B n=100 ((n=5)(v 1 ; :::; v d )). Lemma 4.4 If 1. for each v = (v 1 ; :::; v d ) 2 [?2 n ; 2 n ] d there exists an x v 2 B n=100 ((n=5)v) which is an n=100 hub and x v 1 and 2. each pair z; z 0 2 [?n2 n ; n2 n ] d such that jz? z 0 j > n=4 and z z 0 is a very good pair then there is a copy of the n grid around 0 in!. 22

23 Proof: Given a v pick the distinguished vertex corresponding to v to be an n=100 hub in B n=100 ((n=5)v). The fact that distinguished vertices are n=100 hubs gives us the start and end of the path between any two neighboring distinguished vertices. The ends of these two paths form a very good pair. Thus we can connect them with a path of the right length so that when we remove loops the distance between any two distinguished vertices is n. It is easy to check that these paths are disjoint. ] Lemma 4.5 There exist ; C 11 > 0 and A 11 such that P(9 an n grid around 0 in!) > 1? A 11 e?c11n : Proof: By lemma 4.3 the probability that there are not the desired n hubs is less than (2 n+1 + 1) d (e?c10n=100 ). The probability that they are not all connected to the innite cluster is less than (2 n+1 + 1) d e?c3n=100 =P (x 1) 2. By lemma 3.8 the probability that the desired pairs are not very good is less than (2n2 n ) 2d (A 8 e?c8n=4 ). Thus the constants can be chosen. ] This xes for the rest of the paper. 5 Constructing the spanning subgraph A connected graph G is a good n graph around 0 if 1. there is a subgraph of G which is a copy of the n grid around 0 2. each v which is connected to the n grid is connected by a path of length at most nd, and 3. for no v does there exist disjoint paths in G that connect v to the n grid. Notice that the third condition implies that any walk in G which leaves the n grid at vertex x must return to the n grid at vertex x. Given a vertex v in the n grid we dene the bush attached to v as follows. It contains any vertex v 0 in G such that all paths from the n grid to v 0 go through v. It also contains all vertices which are connected to v by a single edge. 23

24 Any walk in G generates a unique walk in the n grid. Thus it generates a unique nearest neighbor walk in Z d. This property will be used heavily in the next section. Dene C(x) to be the open cluster containing x. Dene C n (x) = C(x) \ [?n2 n ; n2 n ] d : Lemma 5.1 If C 1 (Z d ; p), 2. for each v = (v 1 ; :::; v d ) 2 [?2 n ; 2 n ] d there exists an x v 2 B n=100 ((n=5)v) which is an n hub and x v 1 and 3. each pair z; z 0 2 [?n2 n ; n2 n ] d such that jz? z 0 j > n=4 and z z 0 is a very good pair. then there is a good n graph around 0 that spans C n (0). Proof: By lemma 4.4 the last two conditions imply that there is a copy of the n grid around 0. We will inductively show that we can attach each vertex of C n (0) to the n grid in a way that is consistent with the denition of a good graph. Suppose x is a vertex in C n (0) but not in the n grid. Then by the third condition there is a path from x to the n grid of length at most nd. This path can be chosen to intersect the n grid in only one vertex. The union of this path and the n grid satises the conditions of a good graph. Now we want to show that if we have an extension E of the n grid that satises the conditions of a good graph then we can extend it to include at least one more vertex so that the new extension E 0 still satises the conditions of a good graph. Let x be a vertex in C n (0) but which is not in E. By the second property there exists a path, P, connecting x to the n grid with length at most nd. It is clear that x [ V (E) V (P [ E). The problem is P [E might not satisfy the conditions of a good graph. We will now pare down P [ E to get a new graph E 0 which spans E [ P and also satises the conditions of a good graph. Starting at the n grid move on P towards x. Stop at the rst vertex v in V (E). From v there are two disjoint paths to the n grid. Remove the edge touching v in the longer of the two paths. If they are both the same length then remove either 24

25 one. Now start at v and move along P toward x and repeat this procedure until you get to x. Call the resulting graph E 0. Now we claim that E 0 is a good n graph. Suppose there is a vertex y with two disjoint paths to the n grid. Then at least one edge of one of the paths is in P and one edge is in E. Thus there is a vertex in V (E) \ V (P ) connected to one edge in E 0 \ E and one edge in P \ E 0. But whenever this situation arose we removed one of the two edges. Thus there is no such vertex. Also by the way we chose to remove edges the distance from any vertex in V (E [ P ) to the n grid does not increase. Thus the distance from any point in V(E [ P ) is at most nd. Thus the resulting graph is a good graph. Thus we can construct succesive extensions until we get a good n graph that spans C n (0). ] Lemma 5.2 P(9 a good n graph around 0 that spans C n (0) j 0 2 C 1 (Z d ; p)) > 1? A 11 e?c11n : Proof: The proof is the same as the proof of lemma 4.5. ] 6 Bounding the return probabilities on C 1 (Z d ; p) First we bound the return probabilities of a continous time random walk on a good n graph. Then we apply a result of Benjamini and Schramm to get bounds for the continuous time random walk on C 1 (Z d ; p). Then we bound the return probabilities of simple random walk on C 1 (Z d ; p). The continuous time random walk is dened as follows. Let Z v be a continuous time Markov chain on the vertices, with Z v (0) = v and jumping rate one accross each edge in G. The times of the jumps are determined by a family of Poisson processes with mean 1. To bound the return probabilities on a good n graph G, we consider the projection of a walk on G to a walk on the distinguished vertices of the n grid and then the projection onto Z d. A walk, P, is the sequence vertices visited by Z v and P t is the sequence of vertices visited up to time t. In the case that G is a good n graph we dene proj(p t ) to be the projection of the walk P to a nearest neighbor walk on Z d. First we eliminate every element of the sequence P t which is not a distinguished vertex. Then we replace consecutive occurrences of the same distinguished vertex with one occurrence 25

26 of that vertex. Finally we obtain the sequence proj(p t ) by mapping this sequence of distinguished vertices into Z d in the canonical way. We now show that with high probability, jproj(p t )j is large. We do this in three steps. First we show that the number of vertices in P t is large. Then we bound the amount of time the random walk spends in the bushes. To do this we introduce an intermediate sequence S(P t ). After eliminating every element of the sequence P t which is not on the n grid, we then obtain the sequence S(P t ) by replacing consecutive occurrences of the same vertex on the n grid with one occurence of that vertex. Finally we estimate how often the random walk hits the distinguished vertices. Lemma 6.1 There exists C 12 such that for any good n graph G and vertex v P(jproj(P t )j < t=(8n 2 )(4nd) 3d ) < C 12 t?d=2 : Proof: First we show that with high probability jp t j is large. For any vertex v the random variable that tells how long the walk will remain at v is a Poisson process with mean equal to one over the number of edges in G that end at v. Thus it is dominated by a Poisson process with mean 1. Let S t=2 be the sum of t=2 Poisson random variables with mean 1. Thus P(jP t j < t=2) < P(S t=2 > t). The right hand side is decreasing exponentially in t so it is bounded by Ct?d=2. Next we show that with high probability the projection of P t onto the n grid is large. The cover time EC(G) of a graph G is is dened as the maximum, over all vertices v of G, of the expected number of steps before a simple random walk, starting at v, has visited all the vertices of G. Thus for any graph G 0 and any pair of vertices v 0 and v 00 the probability that simple random walk started at v 0 has not reached v 00 by time 2EC(G 0 ) is less than.5. This implies that the probability that simple random walk started at v 0 has not reached v 00 by time cec(g 0 ) is less than 2?c=2. The cover time for a graph with N vertices is bounded by N 3 [10]. For any vertex v in the n grid, the bush attached to v has at most (2nd) d vertices. Thus the cover time for the bush attached to v is bounded by (2nd) 3d. Let v be the number of steps that a simple random walk started at v takes before it hits one of the two neighboring vertices in the n grid. The argument above implies that P( v > c(2nd) 3d ) < 2?c=2 for all v and c. >From this we bound P(jS(P t )j < t=4(2nd) 3d j jp t j > t=2): 26

27 It is less than the probability that the sum of t=4(2nd) 3d copies of v have sum at least t=2. This is also decreasing exponentially in t so it is bounded by C 0 t?d=2. Finally we estimate P(jproj(P t )j < t=(8n 2 )(2nd) 3d j js(p t )j t=4(2nd) 3d ): Let v be a distinguished vertex and dene Y to be the number of steps that a simple random walk (in the n grid) starting at v takes before it hits a neighboring distinguished vertex. An argument similar to one in the previous paragraph implies P(Y > cn 2 ) < 2?c=2. We also have that P(jproj(P t )j < t=(8n 2 )(2nd) 3d j js(p t )j t=4(2nd) 3d ) is less than the probability that the sum of t=(8n 2 )(2nd) 3d independent copies of Y is less than 4(2nd) 3d. This probability is decreasing exponentially in t so it is bounded by C 00 t?d=2. Combining these three estimates gives us the lemma. ] The same estimates as in the rst paragraph of the proof of the previous lemma give us the following lemma. Lemma 6.2 There exists C 12 ; C 18 > 1 such that P(jP t j C 18 t) 2?C12t : Let P t be the last element in P t. Now we show that given that the number of steps in the projection is high, the return probability is low. To prove this we use the estimate for the return probability of simple random walk on Z d. Lemma 6.3 There exists C 16 such that for all n and t such that t 2 n P( P t = 0 j C 18 t > jproj(p t )j t=(8n 2 )(2nd) 3d ) < C 16 t?d=2 (8n 2 ) d=2 (2nd) 3d2 =2 : Proof: The distribution that proj(p t ) generates on paths on (2 n ; 2 n ) d given that proj(p t ) = i is the same as the distribution simple random walk generates on paths of length i. Also P t = 0 implies that proj(p t ) is one of the distinguished vertices neighboring 0. There is a natural measure preserving projection from simple random walk paths on Z d to simple random walk paths on (2 n ; 2 n ) d. The only way that a path projects to one ending at 0 or a neighbor of 0 is if the path on Z d ends at 0 or a neighbor of 0 or if the path ever goes outside of (2 n ; 2 n ) d. For i C 18 t the probability that the 27

28 simple random walk leaves (2 n ; 2 n ) d is less than or equal to C2?2n =C 18. The return probabilities for simple random walk on Z d give us that for each i t=(8n 2 )(2nd) 3d P( P t = 0 j jproj(p t )j = i) < C 0 t?d=2 (8n 2 ) d=2 (2nd) 3d2 =2 : Combining these two estimates proves the lemma. ] Combining lemmas 6.1, 6.2, and 6.3 gives us the following. Lemma 6.4 There exists a C 13 such that for any good n graph G, any v 2 G, and any t 2 n p G t (v; v) C 13 t?d=2 (8n 2 ) d=2 (2nd) 3d2 =2 : Now we are ready to apply the following theorem of Benjamini and Schramm. Theorem 6.1 [6] If G 0 G and V (G) = V (G 0 ) then v2v (G) p G0 t (v; v) Lemma 6.5 There exists C 14 > 0 such that v2v (G) p G t (v; v:) E(p! t (0; 0) j 0 2 C 1 (Z d ; p)) < C 14 t?d=2 (log t) 3d2 =2+d : Proof: The idea for this lemma is as follows. We divide the percolation cluster into pieces, restricting our attention to the innite cluster. We intersect C 1 (Z d ; p) \ B m with evenly spaced boxes that cover most of this region, but are separated by a distance 2(C 18 t + nd). The term nd is there so that when we construct a good n graph around this box, it will not intersect the adjacent box. The C 18 t term is there so that with high probability, a random walk that starts in one of the good n graphs will not intersect an adjacent good n graph. By dividing the cluster this way, it is possible to analyze the behavior of the random walk on each piece separately. With high probability it is possible to nd a good n graph around these boxes. Then we are able to apply Benjamini and Schramm's result to each piece and average. Given t choose n to be the smallest integer such that 2 n =5 3C 18 t 1+d=2 and 2A 11 e?c11n t?d=2. This implies that there exists a C such that n C log t for all t. Let D be the density of C 1 (Z d ; p). 28

29 Divide B m \ C 1 (Z d ; p) into two regions, R m 1 and R m 2 as follows. Let B = fz 2 Z d j z i 2 [?2 n =5 + C 18 t; 2 n =5? C 18 t] 8ig: Then dene R m 1 = (B + 2(5nd + 2 n =5)Z d ) \ C 1 (Z d ; p) \ B m ; and R m 2 = (B m \ C 1 (Z d ; p)) n R m 1. Dene For any vertex v let G n (v)! be a good n graph centered around v if one exists. G m = [ v22(5nd+2 n =5)Z dg n(v) \ B m : Notice that G m is the disjoint union of good n graphs so we have the same bounds for p Gm t (v; v) as we had for any good n graph. Let C 1 denote C 1 (Z d ; p). Also dene B 0 = fz 2 Z d j z i 2 [?2 n =5? 2nd; 2 n =5 + 2nd]g and R m 3 = (B 0 + 2(5nd + 2 n =5)Z d ) \ B m \ C 1 (Z d ; p): Notice that V (G m ) R m 3. By the ergodic theorem, (D)E(p! t (0; 0) j 0 2 C 1 (Z d ; p)) = lim m 1 jb m j v2c 1\B m p! t (v; v): (9) We split up the sum on the right hand side into two parts and estimate each one. v2c 1\B m p! t (v; v) = v2r m 1 p! t (v; v) + v2r m 2 1 p! t (v; v) A : (10) We estimate the last term on the right hand side by calculating the number of vertices in the region R m 2. p! t (v; v) jr m 2 j v2r m 2 29 jb m jd 2C 18t + 5nd 2 n =5

30 3C 18 t jb m jd 3C 18 t 1+d=2 jb m jt?d=2 : (11) In order to estimate the rst term, notice that any point in R m 1 from (R m 3 ) c. Furthermore, v2r m 1 p! t (v; v) = ((p! t (v; v) j jp t j < C 18 t)p(jp t j < C 18 t) + v2r m 1 (p! t (v; v) j jp t j C 18 t)p(jp t j C 18 t)) v2r m 1 is at least C 18 t away ((p Rm 3 t (v; v) j jp t j < C 18 t)p(jp t j < C 18 t)) + C 12 jb m jt?d=2 :(12) Let V m be the vertex set of R m 3 nv (G m ). Applying the result of Benjamini and Schramm to the graph R m 3 and the spanning subgraph G m [ V m yields the following estimate. v2r m 1 ((p Rm 3 t (v; v) j jp t j < C 18 t)p(jp t j < C 18 t) v2r m 3 v2r m 3 v2g m p Gm p Rm 3 t (v; v) p Gm [V m t (v; v) t (v; v) + jv m j (13) Putting together lines 9 through 13 and then applying lemma 6.4 yields the following. (D)E(p! t (0; 0) j 0 2 C 1 (Z d ; p)) lim m 1 jb m j lim m 1 jb m j lim m 1 jb m j v2r m 1 \Gm p Gm 1 t (v; v) + jv m j + C 12 jb m jt?d=2 + jb m jt?d=2 A v2g m C 13 t?d=2 (8n 2 ) d=2 (2nd) d2=2 C 14 t?d=2 (log t) 3d2 =2+d :! + 3C 12 t?d=2 v2g m C 13 t?d=2 (8(C log t) 2 ) d=2 (C log t) 3d2=2! + 3C 12 (t?d=2 ) Dene P G t (v; v) to be the probability that simple random walk on G returns to v after t steps. 30 ]