Energy of flows on Z 2 percolation clusters


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1 Energy of flows on Z 2 percolaton clusters Chrstopher Hoffman 1,2 Abstract We show that f p > p c (Z 2 ), then the unque nfnte percolaton cluster supports a nonzero flow f wth fnte q energy for all q > 2. Ths extends the work of Grmmett, Kesten, and Zhang and Levn and Peres n dmensons d 3. As an applcaton of our technques we exhbt a graph that has transent percolaton clusters, but does not admt exponental ntersecton tals. Ths answers a queston asked by Benjamn, Pemantle, and Peres. Keywords : percolaton, energy, electrcal networks, exponental ntersecton tals. Subject classfcaton : Prmary: 60J45; Secondary: 60J10, 60J65, 60J15, 60K35. 1 Department of Mathematcs, Unversty of Maryland, College Park, MD, Research supported n part by an NSF postdoctoral fellowshp 1
2 1 Introducton Consder Bernoull bond percolaton on Z d wth parameter p. Recall that ths s the ndependent process on Z d whch retans an edge wth probablty p and deletes an edge wth probablty 1 p. For all d > 1, there exsts a crtcal parameter p c (Z d ) < 1 such that f p < p c, then a.s. there s no connected component wth nfntely many edges. If p > p c, then a.s. there s a unque connected component wth nfntely many edges [1]. Ths component s called the unque nfnte cluster. Kesten proved that p c (Z 2 ) = 1/2 [9]. The phase where p > p c s called supercrtcal Bernoull percolaton. We consder only the supercrtcal case. Let G = G(E, V ) be the graph wth vertex set V and edge set E. Consder each undrected edge on G as two drected edges, one n each drecton. Let vw be the drected edge from v to w. A flow f on G wth source v 0 s an nonnegatve edge functon such that the net flow out of any vertex v v 0 s zero: w f(v 0 w) w f(wv 0 ) = 0. The strength of a flow f from the orgn s the amount flowng from 0: w f(0w) w f(w0). A multsource flow f on G s the sum (or ntegral) of flows on G. The q energy of a flow (or multsource flow) f on G s E q (f) = e E f(e) q. The energy of a flow s the 2 energy of the flow. Note that f a multsource flow on G has fnte q energy then t s the sum (or ntegral) of flows wth fnte q energy. Grmmett, Kesten and Zhang proved that f d 3 and p > p c (Z d ), then smple random walk on the nfnte percolaton cluster, C (Z d, p), s a.s. transent [6]. They proved ths result by constructng a flow on the percolaton cluster wth fnte energy, whch s equvalent to transence of smple random walk on the cluster. Grmmett, Kesten, and Zhang showed that there exsts a tree n C (Z d, p) where the branches bfurcate at farly regular ntervals. Ths tree gves rse to a flow n a natural way. It s then easy to bound the energy of ths flow. Benjamn, Pemantle and Peres [3] gave an alternatve proof of Grmmett, Kesten and Zhang s result. They constructed unpredctable processes on Z. They used them to create a measure on the collecton of paths n Z d whch emanate from 0. For d 3 the measure µ they created has exponental ntersecton tals. That s, there exsts C and a θ < 1 such that for 2
3 all n µ µ { (ϕ, ψ) : ϕ ψ n } Cθ n, where ϕ ψ s the number of edges n the ntersecton of ϕ and ψ. We say that a graph G admts exponental ntersecton tals f there exsts a measure µ on paths on G wth exponental ntersecton tals. Then they proved that any graph that admts exponental ntersecton tal has transent percolaton clusters for some p < 1. They asked whether the converse s true. We show that t s not. Levn and Peres adapted the approach of Benjamn et al. to show that for d 3 these flows have fnte q energy a.s. for q > d/(d 1) [10]. Ths result s optmal snce Z d also supports flows of fnte q energy f and only f q > d/(d 1) [11]. In ths paper we wll use the method of Grmmett, Kesten, and Zhang to extend Levn and Peres s result to d = 2. Theorem 1.1 For every p > p c (Z 2 ) and q > 2 there exsts a flow on C (Z 2, p) wth fnte q energy a.s. Levn and Peres also proved a result usng a generalzaton of the energy of a flow. For any d 2 and α > 0, let H d,α (u) = u d/(d 1) /[log(1 + u 1 )] α for u > 0 and H d,α (0) = 0. Levn and Peres proved that f d 3 then C (Z d, p) supports a flow of fnte H d,α energy for any α > 2. (That s e H d,α (f(e)) <.) Hoffman and Mossel sharpened ths result by showng that C (Z d, p) supports a flow of fnte H d,α energy for d 3 and α > 1 [8]. Ths last result s optmal because there exsts a flow of fnte H d,α energy on Z d f and only f α > 1 [12]. We wll extend ths result of Levn and Peres to Z 2. Theorem 1.2 For every p > 1/2 and α > 2 there exsts a flow on C (Z 2, p) wth fnte H 2,α energy a.s. We are unable to prove a verson of Hoffman and Mossel s results for d = 2, but we conjecture that t s true. Conjecture 1.1 For every p > 1/2 and α > 1 there exsts a flow on C (Z 2, p) wth fnte H 2,α energy a.s. 3
4 Ths would requre a new approach as the Grmmett et al. method cannot be extended to reach ths concluson and the Benjamn et al. approach says nothng about graphs, lke Z 2, that do not admt exponental ntersecton tals. Häggström and Mossel used Benjamn, Pemantle and Peres s approach to extend the Grmmett, Kesten, and Zhang result n another way. They defned two classes of subgraphs of Z 3 and showed these subgraphs admt exponental ntersecton tals. Thus smple random walk on the nfnte percolaton cluster of those subgraphs s transent [7]. On certan subgraphs of Z 3 where Häggström and Mossel showed that smple random walk on the nfnte percolaton cluster s transent t s clear that the Grmmett, Kesten, and Zhang approach wll not work. We gve an example of a graph where the Grmmett et al. approach works but the Benjamn et al. approach does not. More specfcally we prove the followng. Theorem 1.3 There exsts a graph G whch has transent percolaton clusters but does not admt exponental ntersecton tals. To do ths we construct a graph G whch s the drect product of a tree and Z. The subset of G whch projects onto a branch of the tree s a subgraph of Z 2. On the percolaton cluster restrcted to that subset we construct a flow of fnte H 2,3 energy. Integratng these flows over all branches gves a multsoure flow of fnte energy on the percolaton cluster of G. Thus smple random walk on the nfnte percolaton cluster of G s transent. Then we show that ths graph does not support exponental ntersecton tals. 2 Flows on C (Z 2, p) We proceed usng the method of Grmmett, Kesten, and Zhang. We wll fnd a tree n C (Z 2, p). The branches of ths tree splt nto two at farly regular ntervals. Wth ths tree we wll assocate a flow. Snce we have control on how often the branches of the tree bfurcate we wll be able bound the q energy of the flow. Ths method allows us to show that there exst flows on C (Z 2, p) of fnte q energy for all q > 2. Frst we ntroduce some notaton. A path P s a sequence of open drected edges where the end of one edge s the begnnng of the next edge. A path P connects x and 4
5 y f x and y are endponts of edges n P. We wrte x y f there exsts an open path from x to y. We wrte x f x s part of the unque nfnte cluster. If x y then we let D(x, y) be the length of the shortest open path from x to y. We use the taxcab metrc on Z 2, (x, y), (x, y ) = x x + y y. Let β > 2, ρ 1, and b be constants whch wll be defned later. Let a = 20ρb. Let X k () = (β k, (β/2) k ) for all, 2 k 1 < 2 k 1, and k 1. Defne X k () =.5(X k+1 (2) + X k+1 (2 1)). The tree n the percolaton cluster that we wll construct wll have the followng propertes. For each and k there wll be one branch gong from near X k () to near X k (). Ths branch wll then bfurcate near X k+1 (2). One of the branches goes toward X k+1 (2), whle the other branch goes toward X k+1 (2 1). Defne L(u, v) to be the elements n Z 2 whch are wthn 2 of the lne segment jonng u and v. Let B(k) be all (x, y) Z 2 such that x + y k. Defne T k () = B(ak) + ( L(X k (), X k ()) L(X k+1 (2), X k+1 (2 1)) ), where + represents Mnkowsk addton. Lemma 2.1 There exsts a functon K(a) such that T k () T k (j) = for all j and k > K(a). We also have that T k () T k+1 (j) = for all j 2 or j 2 1 and k > K(a). Proof: If T k () T k (j) s not empty the there exsts a pont n L(X k (), X k ()) L(X k+1 (2), X k+1 (2 1)) whch s wthn 2ak of a pont n L(X k (j), X k (j)) L(X k+1 (2j), X k+1 (2j 1)). 5
6 Wthout loss of generalty ths mples ether X k (), X k (j) 2ak or X k+1 (2), X k+1 (2j 1) 2ak. Those two dstances are at least (β/2) k. So the frst condton s satsfed f (β/2) K(a) > 2aK(a). The second condton s also satsfed wth the same choce of K(a) for the same reason. Our next goal s to gve a suffcent condton for there to be an open path from a pont near X k () to a pont near X k+1 (2). Our condton wll also mply that the path les entrely nsde T k () and has length bounded by C X k (), X k+1 (2). Fnd y 1 = y 1 (k, ) through y t = y t (k, ) such that 1. y 1 = X k (), 2. y t X k () bk y u, y u+1 8bk, for all 1 u t 1 4. y u L(X k (), X k ()), for all 1 u t 1 and 5. t β k+1 /4bk. Defne E u = E u (k, ) to be the event that 1. there exsts z 1 y u + B(bk) such that z 1, and 2. any two ponts z 2 (y u + B(bk)) and z 3 (y u+1 + B(bk)) whch are connected have D(z 2, z 3 ) ak/2. Ths mples that the shortest path connectng z 2 and z 3 les n y u + B(ak) T k (). If all the E u hold then there s a path from near X k () to near X k () nsde T k (). In a smlar manner we can fnd y 1 through y t such that 1. y 1 = X k+1 (2 1), 2. y t = X k+1 (2), 3. 4bk y u, y u+1 8bk, for all 1 u t 1 4. y u L(X k+1 (2 1), X k+1 (2)), for all 1 u t 1, 6
7 5. t 5ρβ k+1 /bk, and 6. there exst u and u such that y u = y u. Defne E u = E u(k, ) to be the event that 1. there exsts z 1 y u + B(bk) such that z 1, and 2. any two ponts n z 2 (y u + B(bk)) and z 3 (y u+1 + B(bk)) whch are connected have D(z 2, z 3 ) ak/2. Ths mples that the shortest path connectng z 2 and z 3 les n y u + B(ak) T k (). Defne E k () = ( u E u (k, )) ( u E u(k, )). In ths next lemma we show that f E k () holds for one k and then there exst a path from near X k () to near X k+1 (2) and a path from near X k () to near X k+1 (2 1). We also show that f E k () holds for all k suffcently large and all then there exsts a tree wth the desred propertes n C (Z 2, p). Ths mples that there exsts a flow of fnte q energy for all q > 1 + log 2 β. Lemma 2.2 If there exsts a K such that E k () holds for all k K and then there exsts a flow on C (Z 2, p) wth fnte q energy for all q > 1 + log 2 β. Proof: It causes no loss of generalty to assume that K > K(a). We construct a multsource flow that has a source near each X K (). We wll show that the multsource flow has fnte energy. Thus there exsts a flow wth fnte energy. Snce E k () holds for each k K and t s possble to pck ponts p k () such that p k () X k () + B(bk) and p k (). For each k and t s possble to pck ponts z u y u + B(bk) such that z u. It s also possble to select z u y u + B(bk) such that z u. We can also requre that there exst u and u such that z u = z u, z 1 = p k (), z 1 = p k+1 (2 1), and z t = p k+1 (2). The second condton n the defnton of E u mples that z u z u+1. Snce all the E u and E u hold we can pece together these paths to form paths P k () and P k() such that 1. P k () whch connects p k () and p k+1 (2) 2. P k() whch connects p k () and p k+1 (2 1) 3. P k (), P k() T k (), and 7
8 4. P k (), P k() (ak/2)(β k+1 /4bk) = 5ρβ k+1 /2. The unon of the P k () and the P k() does not necessarly form a tree. However t s easy to remove branches so that t does form a tree. Instead of dong ths we use the P k () and the P k() to defne a flow drectly. For each edge e assgn t mass f(e) = k, 1 2 k ( IPk ()(e) + I P k ()(e) ). Lemma 2.1, the fact that K K(a), and condton 3 on the P k () mples that f e P k () P k() and k > K(a) then f(e) 4(2 k ). If f(e) q + f(e) q < e P k () e P k () then f(e) q < s fnte. Thus the followng calculaton shows f has fnte q energy. f(e) q + (4(2 k )) q 5ρβ k+1 e P k () e P k () f(e) q C2 kq β k C2 k(q log 2 β) 2 k+1 2C2 k(q log 2 β 1) <. Now we show that wth probablty one there exsts a K so that E k () holds for all k > K and all. To do ths we need to bound the probabltes of E u and the E u. Ths requres one two theorems. The frst follows from the work of Russo [13], Seymour and Welsh [14], and Kesten [9]. Theorem 2.1 Gven p >.5 there exsts C 1 and α 1 so that P(B(k) ) < C 1 2 α 1k. The second s due to Antal and Psztora [2]. Theorem 2.2 [2] Let p >.5. Then there exsts ρ = ρ(p) [1, ), and constants C 2, and δ > 0 such that for all y P(0 y, D(0, y) > ρ y ) < C 2 2 δ y. 8
9 Lemma 2.3 There exsts an α > 0 such that for all k P(E u (k, )) > 1 C 3 2 αbk. Proof: By Theorem 2.1 the probablty that condton 1 n the defnton of E u (k, ) does not hold s less than C 1 2 α1bk. If condton 2 s not true then there exst two ponts that are connected but the dstance between them s large. Theorem 2.2 mples that P ( z 2 and z 3 such that D(z 2, z 3 ) > ak/2 = 10ρbk ρ z 2 z 3 ) < (bk) 2 C 2 2 2δbk. Thus P(E u (k, )) > 1 C 1 2 α1bk (bk) 2 C 2 2 2δbk > 1 C 3 2 αbk for an approprate choce of C 3 and α. We also get the same bound for P(E u). Lemma 2.4 For p >.5 there exsts b, C 4, and α > 1 such that for all k and P(E k ()) 1 C 4 2 α k. Proof: Choose b large enough so that α = αb log 2 β > 1. P(E k ()) 1 u P(E u (k, ) C ) u P(E u(k, ) C ) 1 5ρβ k+1 C 3 2 αbk 1 C 4 2 k log 2 β αbk 1 C 4 2 α k. Proof of Theorem 1.1: Gven p >.5 and q > 2 choose β so that β > 2 and q > 1 + log 2 β. The prevous lemma mples that lm n P(,k>n E k ()) = 1. Thus the Borel Cantell lemma and Lemma 2.2 there exsts a flow of fnte q energy wth probablty 1. 9
10 3 A Refnement In ths secton we wll be concerned wth a generalzaton of the energy of a flow. For any functon H we defne the H energy of a flow f as E H (f) = e H(f(e)). We wll be workng wth a specal class of functons. For any α > 0, let H 2,α (u) = u 2 /[log(1 + u 1 )] α for u > 0 and H 2,α (0) = 0. The man result of ths secton s that for d = 2 there exsts a flow of fnte H 2,α energy on C (Z 2, p) for all α > 2. The proof s very smlar to the prevous secton. Let β > 1, ρ 1, and b be constants whch wll be defned later. Let a = 20ρb. Defne t k = 2 k k β. Let X k () = (t k, (t k )/2 k ) for 2 k and k 0. Then defne T k () as n secton 2. Lemma 3.1 For each β > 1 there exsts a functon K(a) such that T k () T k (j) = (1) for all j and k > K(a). We also have that T k () T k+1 (j) = (2) for all j 2 or j 2 1 and k > K(a). Proof: If T k () T k (j) s not empty the there exsts a pont n L(X k (), X k ()) L(X k+1 (2), X k+1 (2 1)) whch s wthn 2ak of a pont n L(X k (j), X k (j)) L(X k+1 (2j), X k+1 (2j 1)). Wthout loss of generalty ths mples ether X k (), X k (j) 2ak or X k+1 (2), X k+1 (2j 1) 2ak. Those two dstances are at least k β. So the frst condton s satsfed f K(a) β > 2aK(a). Ths can be acheved whenever β > 1. The second condton s also satsfed wth the same choce of K(a) for the same reason. 10
11 Pck the y 1,..., y t, y 1,..., y t as n secton 2. Ths can be done wth t t k+1 /4bk < 2 k 1 (k + 1) β /bk. Defne E u, E u, and E k () the same as n the prevous secton. Lemma 3.2 If E k () holds for all k K and all then there exsts a flow on C (Z 2, p) wth fnte H 2,α energy for all α > 1 + β. Proof: Agan we create a multsource flow wth fnte energy. We assume that K > K(a). Snce E k () holds for each k and t s possble to pck ponts p k () such that p k () X k () + B(bk) and p k (). We can also defne z u and z u as n the prevous secton. Snce E k () hold then there exsts paths P k () and P k() such that 1. P k () whch connects p k () and p k+1 (2) 2. P k() whch connects p k () and p k+1 (2 1) 3. P k (), P k() T k (), and 4. P k (), P k() (ak/2)(2 k 1 (k + 1) β /bk) < 5ρ2 k (k + 1) β. For each edge e assgn t mass f(e) = k, 1 2 k ( IPk ()(e) + I P k ()(e) ). Lemma 3.1 mples that f e P k () P k() and k > K(a) then If, e P k () f(e) 4(2 k ). (3) H 2,α (f(e)) + H 2,α (f(e)) < e P k () then H 2,α (f(e)) <. Thus the followng calculaton shows f has fnte H 2,α energy. e P k () H 2,α (f(e)) + H 2,α (f(e)) (4(2 k )) 2 e P k () [log(1 + (4(2 k )) 1 )] α 10ρ2k (k + 1) β 11 C(2 k )k β [ log(4(2 k ))] α
12 < C (2 k )k β [k 2] α 2k+1 C k β α Proof of Theorem 1.2: Gven p >.5 and α > 2 choose β so that β > 1 and α > 1 + β. Choose b so that Lemma 2.4 mples that lm n P(,k>n E k ()) = 1. Thus by Lemma 3.2 and the BorelCantell lemma there exsts a flow of fnte H 2,α energy wth probablty 1. 4 A graph wth transent percolaton clusters that do not admt exponental ntersecton tals In ths secton we wll construct a graph G. Usng the results of the prevous secton we are able to show that G has transent percolaton clusters. Then we wll show that G does not admt exponental ntersecton tals. Let b = 2 2 /5. Before we defne G we defne T = {(x, y) x N, y {0, 1} N, y = 0 for all such that b x} to be the tree whose branches splt at dstance b from the root. Thus a vertex (x, y) s dstance x from the root. The vertex (x, y) s connected to the vertex (x + 1, y). If x = b for some then the vertex (x, y) s also connected to the vertex (x + 1, ỹ), where ỹ = 1 and ỹ j = y j for all j. At dstance d from the root of T there are O(log d) 5 branches. Let G = {(x, y, z) (x, y) T, z Z, and z.6x}. Theorem 4.1 For all p > 1/2 smple random walk on the nfnte percolaton clusters on G s a.s. transent. Proof: We use the constructon n the prevous secton wth β = 1.5. To each y {0, 1} N corresponds G y = {(x, y, z) G y = y for all such that x < b and y = 0 else}, 12
13 a subset of G whch looks lke a wedge of Z 2. By the prevous secton we have wth postve probablty exhbted a flow, f y, on the nfnte percolaton cluster of G y. Ths flow has fnte H 2,3 energy. Defne the multsource flow F (e) = f y (e)dm, where m s (1/2, 1/2) product measure on {0, 1} N. Denote by T k () all branches (x, y, z) such that (x, z) T k (). Now we calculate the energy of F. e T k () F (e) 2 = ( e T k () = ( e T k () ) 2 f y (e)dm B e f y (e)dm where B e = (y y = y for all b < x) for e = (x, y, z). The last equalty s true because the defnton of f y the nequalty mples that f y (e) = 0 for all y / B e. Jensen s nequalty generates ( Snce m(b e ) < C(log e ) 5 e T k () X ) 2 ( ) fdm (m(x)) (f) 2 dm. X F (e) 2 e T k () e T k () C k 5 k 5 <. C C k 2 C f k 5 y (e) 2 dm B e C k 5 e T k () f y (e) 2 dm ) 2 f y (e) 2 dm (4(2 k )) 2 10ρ2 k (k + 1) 3 (4) Lne (4) s true by equaton (3) and condton 4 n the proof of Lemma 3.2. Thus f p >.5 nfnte percolaton clusters on G support flows of fnte energy and are transent. 13
14 Now we show that there s no measure on paths on G wth exponental ntersecton tals. We do ths as follows. Frst we defne a functon f(ψ, ϕ). We show that f µ s a measure on paths wth exponental ntersecton tals then fd(µ µ) s fnte. Then we show that for any measure on paths on G that the ntegral must be nfnte. Thus the graph G does not admt exponental ntersecton tals. It causes no loss of generalty to assume that µ s supported on transent paths. Let ψ ϕ be the number of edges that ψ and ϕ have n common. Let χ e (ψ, ϕ) be the event that e ψ ϕ. Defne f(ψ, ϕ) = If e ψ let l(e) be the largest number such that ψ l(e) = e. Let ψ be the path gven by edges ψ +1, ψ +2,... Notce that ψ ϕ =1 5. f e χ e (ψ, ϕ)( ψ ϕ ψ l(e) ϕ ) 5. (5) Lemma 4.1 If µ s a measure wth exponental ntersecton tals then fd(µ µ) <. Proof: Because µ has exponental ntersecton tals there exsts constants C and θ < 1 such that µ µ( ψ ϕ ) Cθ. Thus fd(µ µ) = = ψ ϕ =1 5 =1 5 d(µ µ) χ ψ ϕ d(µ µ) 5 µ µ( ψ ϕ ) 5 Cθ <. Ths next lemma s the man tool we wll use to show that fd(µ µ) = for any measure µ on paths n G. 14
15 Lemma 4.2 There exsts C such that for any µ and e G E( ψ ϕ ψ l(e) ϕ e ψ ϕ) > C log e. Proof: Let E n (e) be the set of all edges at dstance n from e. Snce e = ϕ l(e) and µ s supported on transent paths then there exsts a k > l(e) so that ϕ k E n (e). We can rewrte ths as P(e ψ \ ψ l(e) e ψ) 1. e E n(e) The expected number of ntersectons of ψ \ ψ l(e) and ϕ n E n (e) s E( ψ ϕ E n (e) ψ l(e) ϕ E n (e) e ψ ϕ) P(e ψ \ ψ l(e) e ψ) 2 e E n(e) 1/ E n (e). The last nequalty s true because of the CauchySchwartz nequalty. Let e = (x, y, z), e = (x, y, z ) and e E n (e). Defne e = x + z. Then.5 e < x e. If n e /10 then.3x < x < 1.2x. As any nterval (j, 4j) has at most one b there can be at most 3 possble values for z. For each x and z there are at most two possble values for y. Thus for any e G and n e /10 we have that E n (e) 12n. Puttng these two facts together we get that E( ψ ϕ ψ l(e) ϕ e ψ ϕ) > e /10 n=1 1/12n > C log e. Theorem 4.2 G does not admt exponental ntersecton tals. Proof: We argue by contradcton. If there were a measure µ wth exponental ntersecton tals then Lemma 4.1 says that fd(µ µ) <. The followng calculaton shows that ths ntegral must be nfnte. fd(µ µ) χ e (ψ, ϕ)( ψ ϕ ψ l(e) ϕ ) 5 d(µ µ) (6) e χ e (ψ, ϕ)e( ψ ϕ ψ l(e) ϕ e ψ ϕ) 5 d(µ µ) (7) e χ e (ψ, ϕ)(c log j) 5 d(µ µ) (8) e e =j 15
16 j j j j (C log j) 5 (µ µ)(e ψ, e ϕ) e =j (C log j) 5 µ(e ψ) 2 e =j (C log j) 5 1 C j(log j) 5 (9) C j Lne (6) comes from lne (5). Lne (7) follows from Jensen s nequalty. Lne (8) follows from Lemma 4.2. Lne (9) s true because the number of e such that e = s bounded by C (log ) 5. Thus µ does not have exponental ntersecton tals. Proof of Theorem 1.3: Theorem 1.3 s a combnaton of Theorems 4.1 and 4.2. References [1] M. Azenman, H. Kesten, C. Newman (1987) Unqueness of the nfnte cluster and contnuty of connectvty functons for short and long range percolaton. Comm. Math. Phys. 111, no. 4, [2] P. Antal and A. Psztora (1996). On the chemcal dstance n supercrtcal Bernoull percolaton. Ann. Probab [3] I. Benjamn, R. Pemantle and Y. Peres (1998). Unpredctable paths and percolaton. Ann. Probab., to appear. [4] P. G. Doyle and E. J. Snell (1984). Random walks and electrcal networks. Carus Math. Monographs 22, Math. Assoc. Amer., Washngton, D. C. [5] G.R. Grmmett (1989). Percolaton. Sprnger, New York. [6] G. R. Grmmett, H. Kesten and Y. Zhang (1993). Random walk on the nfnte cluster of the percolaton model. Probab. Th. Rel. Felds 96, [7] O. Häggström and E. Mossel (1998). Nearestneghbor walks wth low predctablty profle and percolaton n 2 + ɛ dmensons. Ann. Probab., to appear. 16
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