Up to constant independence of the initial and terminal parts of a LERW path

4 Some results for loop-erased random walks

4.1 Up to constant independence of the initial and terminal parts of a LERW path

For this section only, we no longer restrict our random walks to be two-dimensional. When it is necessary to specify what dimension we are in, we will denote the dimension by d. Although we have avoided using it up to now, it will be convenient to use “big-O” notation in this section. Recall that f n = Oan if there exists C ∞ such that f n ≤ C an. Here, C can depend on the dimension but on no other quantity. We will also write f n = [1 + Oan] gn, if there exists C ∞ such that 1 − C an ≤ f n gn ≤ 1 + C an. Recall that for a natural number l, Ω l denotes the set of paths ω = [0, ω 1 , . . . , ω k ] such that ω j ∈ B l , j = 0, 1, . . . , k − 1 and ω k ∈ ∂ B l . Given a set K such that B l ⊂ K, and such that P σ K ∞ = 1, we define µ l,K on Ω l to be the measure obtained by running a random walk up to the first exit time σ K of K, loop-erasing and restricting to B l . More precisely, for ω ∈ Ω l , µ l,K ω = P ¦ b S K [0, b σ l ] = ω © . If B l ⊂ K 1 and B l ⊂ K 2 are such that P ¦ σ K i ∞ © = 1 for either i = 1 or i = 2, we define a measure µ l,K 1 ,K 2 on Ω l as follows. Let X denote random walk conditioned to leave K 1 before K 2 as long as this has positive probability; if not, µ l,K 1 ,K 2 is not defined. Then for ω ∈ Ω l , we let µ l,K 1 ,K 2 ω = P ¦ b X K 1 [0, b σ l ] = ω © . This is the measure on Ω l obtained by running X up to σ X K 1 , loop-erasing and restricting to B l . Note that µ l,K is equal to µ l,K,Λ . In this section, we establish some relations between the measures defined above. In fact we will show that for n ≥ 4 and any K 1 and K 2 such that B nl ⊂ K 1 and B nl ⊂ K 2 , µ l,K 1 ,K 2 ω = h 1 + O 1 log n i µ l,K 1 ∩K 2 ω d = 2; ” 1 + On −1 — µ l,K 1 ∩K 2 ω d ≥ 3. Proposition 4.2 µ l,K 1 ω = h 1 + O 1 log n i µ l,K 2 ω d = 2; ” 1 + On 2 −d — µ l,K 2 ω d ≥ 3. Proposition 4.4 1044 This implies that if B 4l ⊂ K 1 and B 4l ⊂ K 2 then µ l,K 1 ,K 2 ω ≍ µ l,K 1 ∩K 2 ω µ l,K 1 ω ≍ µ l,K 2 ω recall that the symbol ≍ means that each side is bounded by a constant multiple of the other side, the constant depending on the random walk S and on nothing else. We use these facts to prove that for a LERW b S n , η 1 l b S n and η 2 4l,n b S n see the definitions in section 2.3 are independent up to constants Proposition 4.6. Lemma 4.1. Let d = 2. For ω ∈ Ω l and y ∈ ∂ B l , P y σ nl ξ ω = O 1 log n . Proof. Let y be such that P y σ nl ξ ω = max y ∈∂ B l P y σ nl ξ ω . We will show that P y σ nl ξ ω ≤ C log n which will clearly imply the result for all y ∈ ∂ B l . P y σ nl ξ ω = P y σ 2l ξ ω X z ∈∂ B 2l P z σ nl ξ ω P y S σ nl = z σ 2l ξ ω . By part 1 of Lemma 3.2, there exists c 0 such that P y σ 2l ξ ω ≤ 1 − c. Furthermore, for z ∈ ∂ B 2l , P z σ nl ξ ω = P z σ nl ξ l + X y ∈∂ B l P z S ξ l = y; ξ l σ nl P y σ nl ξ ω ≤ P z σ nl ξ l + P y σ nl ξ ω . By [16, Proposition 6.4.1], P z σ nl ξ l ≍ log n −1 . Therefore, P y σ nl ξ ω ≤ 1 − c ” Clog n −1 + P y σ nl ξ ω — which gives the desired result. 1045 Proposition 4.2. Suppose that n ≥ 4, K 1 and K 2 are such that B nl ⊂ K 1 and B nl ⊂ K 2 , and that ω = [0, ω 1 , . . . , ω k ] ∈ Ω l . Then, µ l,K 1 ,K 2 ω = h 1 + O 1 log n i µ l,K 1 ∩K 2 ω d = 2; ” 1 + On −1 — µ l,K 1 ∩K 2 ω d ≥ 3. In particular, if B 4l ⊂ K 1 and B 4l ⊂ K 2 then µ l,K 1 ,K 2 ω ≍ µ l,K 1 ∩K 2 ω. Proof. Let K = K 1 ∩ K 2 . Let X be a random walk conditioned to exit K 1 before K 2 . Then by formula 5, µ l,K ω = pωG K ωP ω k σ K ξ ω , and µ l,K 1 ,K 2 ω = p X ωG X K ωP ω k ¦ σ K 1 ξ ω σ K 1 σ K 2 © . By Lemma 2.2, G X K ω = G K ω. Furthermore, if we let hz = P z ¦ σ K 1 σ K 2 © , then by 3, p X ω = h ω k h0 p ω. The function h is harmonic in B nl and ω k ∈ B l . Therefore, by the difference estimates for harmonic functions [16, Theorem 6.3.8], h ω k = ” 1 + On −1 — h0, and thus, p X ω = ” 1 + On −1 — p ω. Hence, it suffices to show that P ω k ¦ σ K 1 ξ ω ∧ σ K 2 © = h 1 + O 1 log n i P ω k ¦ σ K 1 σ K 2 © P ω k σ K ξ ω d = 2; ” 1 + On −1 — P ω k ¦ σ K 1 σ K 2 © P ω k σ K ξ ω d ≥ 3. Let y ∈ ω be such that P y ¦ σ K 1 σ K 2 © = max y ∈ω P y ¦ σ K 1 σ K 2 © . Then P y ¦ σ K 1 σ K 2 © = P y ¦ σ K 1 σ K 2 σ K ξ ω © P y σ K ξ ω + P y ¦ σ K 1 σ K 2 ξ ω σ K © P y ξ ω σ K ≤ P y ¦ σ K 1 σ K 2 σ K ξ ω © P y σ K ξ ω + P y ¦ σ K 1 σ K 2 © P y ξ ω σ K . Therefore, P y ¦ σ K 1 σ K 2 © ≤ P y ¦ σ K 1 σ K 2 σ K ξ ω © = P y ¦ σ K 1 σ K 2 ∧ ξ ω © P y σ K ξ ω , 1046 and hence P ω k ¦ σ K 1 σ K 2 © ≤ max y ∈ω P y ¦ σ K 1 σ K 2 ∧ ξ ω © P y σ K ξ ω . 12 A similar argument shows that P ω k ¦ σ K 1 σ K 2 © ≥ min y ∈ω P y ¦ σ K 1 σ K 2 ∧ ξ ω © P y σ K ξ ω . 13 Now let y be any point on the path ω. Then since B nl is a subset of both K 1 and K 2 , P y ¦ σ K 1 σ K 2 ∧ ξ ω © = P y σ nl ξ ω X z ∈∂ B nl P z ¦ σ K 1 σ K 2 ∧ ξ ω © P y S σ nl = z σ nl ξ ω , and P y σ K ξ ω = P y σ nl ξ ω X z ∈∂ B nl P z σ K ξ ω P y S σ nl = z σ nl ξ ω . However [10, Lemma 2.1.2], P y S σ nl = z σ nl ξ ω = h 1 + O log n n i P S σ nl = z d = 2; ” 1 + On −1 — P S σ nl = z d ≥ 3; Let y 1 be such that P y 1 ¦ σ K 1 σ K 2 ∧ ξ ω © P y 1 σ K ξ ω = max y ∈ω P y ¦ σ K 1 σ K 2 ∧ ξ ω © P y σ K ξ ω , and y 2 be such that P y 2 ¦ σ K 1 σ K 2 ∧ ξ ω © P y 2 σ K ξ ω = min y ∈ω P y ¦ σ K 1 σ K 2 ∧ ξ ω © P y σ K ξ ω . Then by 12 and 13, if d = 2, P ω k ¦ σ K 1 σ K 2 ∧ ξ ω © P ω k ¦ σ K 1 σ K 2 © P ω k σ K ξ ω ≤ P y 1 ¦ σ K 1 σ K 2 ∧ ξ ω © P y 2 σ K ξ ω P y 1 σ K ξ ω P y 2 ¦ σ K 1 σ K 2 ∧ ξ ω © ≤ 1 + C log nn 2 1 − C log nn 2 ≤ 1 + C ′ log n n . The lower bound and the case d ≥ 3 follows in the same way. 1047 We now define a measure on unrooted loops in Λ. See [16, Chapter 9] for more details. A rooted loop η = [η , η 1 , . . . , η k ] is a path in Λ such that η = η k ; η is called the root of the loop. We say that two rooted loops η and η ′ are equivalent if η ′ = [η j , η j+1 , . . . , η k −1 , η , . . . , η j ] for some j. We call the equivalence classes under this relation unrooted loops. We will denote by e η the unrooted loop corresponding to the rooted loop η. Recall the notation p η := k Y i=1 p η i −1 , η i = P η S i = η i , i = 0, . . . k . Notice that this does not depend on the root of η and therefore p e η is well defined for unrooted loops e η. We define a measure m on the set of unrooted loops as follows. Given an unrooted loop e η, let α e η be the number of distinct rooted representatives of e η. Then we define m e η = α e ηp e η | e η| , where | e η| denotes the number of steps of a representative of e η. Any two representatives of e η have the same number of steps so that m is well defined. Recall the definition of G K ω given in 4. The following lemma allows us to express G K ω in terms of the unrooted loop measure. Lemma 4.3. G K ω = exp m e η : e η ⊂ K; e η ∩ ω 6= ; . Proof. We will first show that for any z ∈ Λ, G K z = exp{m e η : e η ⊂ K; z ∈ e η}. 14 Let ρ = P z ξ z σ K . Then by the strong Markov property for random walk, G K z = 1 + ρG K z, and thus, G K z = 1 1 − ρ = e − ln1−ρ = exp    ∞ X j=1 ρ j j    . Given an unrooted loop e η, let κ e η be the largest integer m that divides eη = n such that η k = η k+n m for all k ≤ n − nm. Then κ e η = eη α eη and therefore m e η = p η κ e η . Let β e η = β z e η := { j : η j = z} 1048 be the number of times that e η hits z. Then the number of representatives of e η that are rooted at z is β e ηκ e η. Hence, ρ j = X η rooted at z η⊂K βη= j p η = X e η unrooted z ∈ e η, e η⊂K β e η= j X η rooted at z p η = X e η unrooted z ∈ e η, e η⊂K β e η= j β e ηp e η κ e η = j · mz ∈ e η; e η ⊂ K; β e η = j. Therefore, ∞ X j=1 ρ j j = ∞ X j=1 mz ∈ e η; e η ⊂ K; β e η = j = mz ∈ e η; e η ⊂ K, which proves 14. Let E j , j = 0, . . . , k, be the set of unrooted loops e η such that ω j ∈ e η and e η ⊂ K \ {ω , . . . , ω j −1 }. Then the E j are disjoint and their union is the set of all unrooted loops e η such that e η ∩ ω 6= ; and e η ⊂ K. This observation along with 14 finishes the proof. Proposition 4.4. Suppose that n ≥ 4, K 1 and K 2 are such that B nl ⊂ K 1 and B nl ⊂ K 2 , and that ω ∈ Ω l . Then µ l,K 1 ω = h 1 + O 1 log n i µ l,K 2 ω d = 2; ” 1 + On 2 −d — µ l,K 2 ω d ≥ 3. In particular, if B 4l ⊂ K 1 and B 4l ⊂ K 2 then µ l,K 1 ω ≍ µ l,K 2 ω. Proof. By Formula 5, for any ω ∈ Ω l , µ l,K i ω = pωG K i ωP ω k ¦ σ K i ξ ω © i = 1, 2. Let en = log n −1 if d = 2 and en = n 2 −d if d ≥ 3. Let ω ′ = [ω ′ , . . . , ω ′ k ′ ] be any other path in Ω l . We will show that P ω k ¦ σ K 1 ξ ω © P ω k ¦ σ K 2 ξ ω © ≤ [1 + Cen] P ω ′ k′ ¦ σ K 1 ξ ω ′ © P ω ′ k′ ¦ σ K 2 ξ ω ′ © 15 and G K 1 ω G K 2 ω ≤ [1 + C en] G K 1 ω ′ G K 2 ω ′ . 16 1049 For this will imply that µ l,K 1 ω ′ µ l,K 2 ω ′ ≤ [1 + C en] µ l,K 1 ω µ l,K 2 ω and 1 = X ω ′ ∈Ω l µ l,K 1 ω ′ = X ω ′ ∈Ω l µ l,K 1 ω ′ µ l,K 2 ω ′ µ l,K 2 ω ′ ≤ [1 + C en] µ l,K 1 ω µ l,K 2 ω X ω ′ ∈Ω l µ l,K 2 ω ′ = [1 + C en] µ l,K 1 ω µ l,K 2 ω . One then gets the other bound by reversing the roles of K 1 and K 2 . We first show 15. Since B nl ⊂ K 1 P ω k ¦ σ K 1 ξ ω © = P ω k σ nl ξ ω X z ∈∂ B nl P z ¦ σ K 1 ξ ω © P ω k S σ nl = z σ nl ξ ω . If d ≥ 3, then [16, Proposition 6.4.2] for z ∈ ∂ B nl , P z ¦ σ K 1 ξ ω © ≥ P z ξ l = ∞ ≥ 1 − C n 2 −d . Therefore, if d ≥ 3, ” 1 − C n 2 −d — P ω k σ nl ξ ω ≤ P ω k ¦ σ K 1 ξ ω © ≤ P ω k σ nl ξ ω . One gets a similar formula with K 2 replacing K 1 and ω ′ replacing ω, from which 15 follows for the case d ≥ 3. To prove 15 for the case d = 2, we first note that [10, Lemma 2.1.2] P ω k S σ nl = z σ nl ξ ω = 1 + O log n n P S σ nl = z . Furthermore, for z ∈ ∂ B nl , P z ¦ σ K 1 ξ ω © = P z ¦ σ K 1 ξ l © + X y ∈∂ i B l P y ¦ σ K 1 ξ ω © P z ¦ S ξ l = y; ξ l σ K 1 © . By applying Lemma 4.1 and [10, Lemma 2.1.2] again we get that for y ∈ ∂ i B l , P y ¦ σ K 1 ξ ω © = P y σ nl ξ ω X w ∈∂ B nl P w ¦ σ K 1 ξ ω © P y S σ nl = z σ nl ξ ω ≤ C log n X w ∈∂ B nl P w ¦ σ K 1 ξ ω © P S σ nl = w ≤ C log n P ω k ¦ σ K 1 ξ ω © P ω k σ nl ξ ω . 1050 Thus, P z ¦ σ K 1 ξ ω © ≤ P z ¦ σ K 1 ξ l © + C log n P ω k ¦ σ K 1 ξ ω © P ω k σ nl ξ ω P z ¦ ξ l σ K 1 © ≤ P z ¦ σ K 1 ξ l © + C log n P ω k ¦ σ K 1 ξ ω © P ω k σ nl ξ ω . Therefore, P ω k ¦ σ K 1 ξ ω © = P ω k σ nl ξ ω X z ∈∂ B nl P z ¦ σ K 1 ξ ω © P ω k S σ nl = z σ nl ξ ω ≤ 1 + C log n n P ω k σ nl ξ ω X z ∈∂ B nl P z ¦ σ K 1 ξ ω © P S σ nl = z ≤ 1 + C log n n P ω k σ nl ξ ω X z ∈∂ B nl P z ¦ σ K 1 ξ l © P S σ nl = z + C log n P ω k ¦ σ K 1 ξ ω © , and hence, P ω k ¦ σ K 1 ξ ω © ≤ 1 + C log n P ω k σ nl ξ ω X z ∈∂ B nl P z ¦ σ K 1 ξ l © P S σ nl = z , with a similar lower bound. We get similar bounds with ω ′ replacing ω and K 2 replacing K 1 from which 15 follows. We now show 16. By Lemma 4.3, G K 1 ωG K 2 ω ′ G K 2 ωG K 1 ω ′ = exp {m e η ∩ ω 6= ;; e η ∩ ω ′ = ;; e η ⊂ K 1 ; e η ∩ Λ \ K 2 6= ; + m e η ∩ ω = ;; e η ∩ ω ′ 6= ;; e η ∩ Λ \ K 1 6= ;; e η ⊂ K 2 − m e η ∩ ω 6= ;; e η ∩ ω ′ = ;; e η ∩ Λ \ K 1 6= ;; e η ⊂ K 2 − m e η ∩ ω = ;; e η ∩ ω ′ 6= ;; e η ⊂ K 1 ; e η ∩ Λ \ K 2 6= ;} ≤ exp{m e η ∩ ω ′ = ;; e η ∩ Λ \ K 2 6= ; + m e η ∩ ω = ;; e η ∩ Λ \ K 1 6= ;}. We also get a similar lower bound by exchanging the roles of K 1 and K 2 . Therefore, it suffices to show that there exists C ∞ such that for all l and all ω ∈ Ω l , m e η ∩ B l 6= ;; e η ∩ ω = ;; e η ∩ Λ \ B nl 6= ; ≤ C en. Given an unrooted loop e η with representative η = [η , . . . , η k ], let e η = min{ η i : i = 0, . . . , k}. 1051 Let e η ∗ be such that eη ∗ = eη and such that arg e η ∗ = min{argη i : η i = eη }. Suppose first that d = 2. Then for j ≤ l2 and z ∈ ∂ B j , m e η ∩ ω = ;; e η ∩ Λ \ B nl 6= ;; ⌈ e η ⌉ = j; e η ∗ = z ≤ P z ¦ σ 2 j ξ j −1 © ‚ max y ∈∂ B 2 j P y σ l ξ ω Œ × max v ∈∂ B l P v ¦ σ nl σ j −1 © max w ∈∂ B nl P w ¦ ξ z ξ j −1 © ≤ C 1 j 2 j l 1 2 log l − log j − 1 log nl − log l 1 j ≤ C l − 1 2 log n −1 j − 3 2 log l j , where the exponent 1 2 comes from the Beurling estimates Theorem 2.3. If l 2 j ≤ l, then m e η ∩ ω = ;, e η ∩ Λ \ B nl 6= ;, ⌈ e η ⌉ = j, e η ∗ = z ≤ P z ¦ σ nl ξ j −1 © max w ∈∂ B nl P y ¦ ξ z ξ j −1 © ≤ C log j − log j − 1 log nl − log l j −1 ≤ C j −2 log n −1 . Therefore, m e η ∩ B l 6= ;; e η ∩ ω = ;; e η ∩ Λ \ B nl 6= ; = l 2 X j=1 X z ∈∂ B j m e η ∩ ω = ;; e η ∩ Λ \ B nl 6= ;; e η = j; e η ∗ = z + l X j=l 2 X z ∈∂ B j m e η ∩ ω = ;; e η ∩ Λ \ B nl 6= ;; e η = j; e η ∗ = z ≤ C log n   l − 1 2 l 2 X j=1 j − 1 2 log l j + l X j=l 2 1 j    ≤ C log n . 1052 The case d ≥ 3 is easier. In this case, m e η ∩ ω = ;, e η ∩ Λ \ B nl 6= ;, ⌈ e η ⌉ = j ≤ ‚ max z ∈∂ B j P z ¦ σ nl ξ j −1 © Œ max w ∈∂ B nl P w ¦ ξ j ∞ © ≤ C ‚ j 2 −d − j − 1 2 −d nl 2 −d − l 2 −d Œ ‚ nl 2 −d l 2 −d Œ ≤ C ‚ j 1 −d l 2 −d − nl 2 −d Œ n 2 −d ≤ C n 2 −d l d −2 j 1 −d . Thus, m e η ∩ B l 6= ;; e η ∩ ω = ;; e η ∩ Λ \ B nl 6= ; = l X j=1 m e η ∩ ω = ;, e η ∩ Λ \ B nl 6= ;, ⌈ e η ⌉ = j ≤ C n 2 −d l d −2 l X j=1 j 1 −d ≤ C n 2 −d . Corollary 4.5. Recall that b S denotes an infinite LERW. Suppose that n ≥ 4, K is such that B nl ⊂ K, and ω ∈ Ω l . Then, P ¦ b S[0, b σ l ] = ω © = h 1 + O 1 log n i P ¦ b S K [0, b σ l ] = ω © d = 2; ” 1 + On 2 −d — P ¦ b S K [0, b σ l ] = ω © d ≥ 3. In particular, P ¦ b S[0, b σ l ] = ω © ≍ P ¦ b S 4l [0, b σ l ] = ω © . Proof. This follows immediately from Proposition 4.4 and the definition of the infinite LERW b S: P ¦ b S[0, b σ l ] = ω © = lim m →∞ P ¦ b S m [0, b σ l ] = ω © = lim m →∞ µ l,B m ω. We conclude this section with the proof that η 1 and η 2 are independent up to constants for the LERW b S n . Proposition 4.6. Let 4l ≤ m ≤ n. Then for any ω ∈ Ω l , λ ∈ e Ω m,n , P n η 1 l € b S n Š = ω; η 2 m,n € b S n Š = λ o =    ” 1 + Ologm l −1 — P ¦ η 1 l € b S n Š = ω © P n η 2 m,n € b S n Š = λ o d = 2; ” 1 + O l m — P ¦ η 1 l € b S n Š = ω © P n η 2 m,n € b S n Š = λ o d ≥ 3. 1053 In particular, P n η 1 l € b S n Š = ω; η 2 m,n € b S n Š = λ o ≍ P ¦ η 1 l € b S n Š = ω © P n η 2 m,n € b S n Š = λ o , i.e. η 1 and η 2 are “independent up to constants”. Proof. We fix l, m and n throughout and let η 1 = η 1 l , η 2 = η 2 m,n . Let X be a random walk started at 0 conditioned to leave B n before returning to 0. Then b X and b S n have the same distribution. Let Y be a random walk started on ∂ B n according to harmonic measure from 0 and conditioned to hit 0 before returning to ∂ B n . By reversing paths, for all z ∈ ∂ B n , P S ξ ∧ σ n = z = P z S ξ ∧ σ n = 0 . Therefore, X and Y R the time-reversal of Y have the same distribution. Recall that one obtains the same distribution on LERW by erasing loops from random walks forwards or backwards. Therefore, if ω and λ are as above, P ¦ η 1 € b S n Š = ω η 2 € b S n Š = λ © = P ¦ η 1 € b Y [0, b ξ ] Š = ω R η 2 € b Y [0, b ξ ] Š = λ R © . Now let Z be a random walk starting at λ , conditioned to hit 0 before leaving B n \ λ. Then by the domain Markov property for LERW Lemma 2.4, P ¦ η 1 € b Y [0, b ξ ] Š = ω R η 2 € b Y [0, b ξ ] Š = λ R © = P ¦ η 1 b Z[0, b ξ ] = ω R © . However, by again reversing paths as above, and noting that the loop-erasure of a random walk starting at 0 and conditioned to avoid 0 after the first step has the same distribution as the loop- erasure of an unconditioned random walk, P ¦ η 1 € b Z[0, b ξ ] Š = ω R © = µ l,Λ \{λ },K ω, where K = B n \ {λ 1 , . . . , λ k }. Let k = m l, ek = log k −1 if d = 2 and ek = k −1 if d ≥ 3. Since k ≥ 4, B m ⊂ K and B m ⊂ Λ \ {λ }, we can apply Propositions 4.2 and 4.4 to conclude that µ l,Λ \{λ },K ω = [1 + Oek]µ l,B n \λ ω = [1 + Oek]µ l,n ω = [1 + Oek] P ¦ η 1 € b S n Š = ω © .

4.2 The separation lemma