In particular,
P
n η
1 l
b
S
n
= ω; η
2 m,n
b
S
n
= λ
o
≍ P ¦
η
1 l
b
S
n
= ω
© P
n η
2 m,n
b
S
n
= 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
= ω
η
2
b
S
n
= λ
© = P
¦ η
1
b
Y [0, b ξ
]
= ω
R
η
2
b
Y [0, b ξ
]
= 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
]
= ω
R
η
2
b
Y [0, b ξ
]
= 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
]
= 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
= ω
© .
4.2 The separation lemma
Throughout this section S will be a random walk and b S will be an independent infinite LERW. Let
F
k
denote the σ-algebra generated by
{S
n
: n ≤ σ
k
} ∪ {b S
n
: n ≤ b
σ
k
}. 1054
For positive integers j and k, let A
k
be the event A
k
= {S[1, σ
k
] ∩ b S[0,
b σ
k
] = ;}, D
k
be the random variable D
k
= k
−1
min {distSσ
k
, b S[0,
b σ
k
], distb S
b σ
k
, S[0, σ
k
]}, and T
k j
be the integer valued random variable T
k j
= min{l ≥ k : D
l
≥ 2
− j
}. The goal of this section is to prove the following separation lemma which states that, conditioned
on the event A
k
that the random walk S and the infinite LERW b S do not intersect up to the circle of
radius k, the probability that they are further than some fixed distance apart from each other at the circle of radius k D
k
≥ c
1
is bounded from below by a constant c
2
0.
Theorem 4.7 Separation Lemma. There exist constants c
1
, c
2
0 such that for all k,
P D
k
≥ c
1
A
k
≥ c
2
. The proof of Theorem 4.7 depends on two lemmas. Lemma 4.8 roughly states that the probability
that S and b S stay close together without intersecting each other is very small. More precisely, the
probability that T
j −1
≥ 1 + c j
2
2
− j
T
j
and that the paths don’t intersect is less than 2
−β j
2
. Lemma 4.9 states that if S and b
S are separated, then there is a substantial probability that they stay separated and don’t intersect. To wit, if
{T
j
k} and A
T
j
hold, then the probability that A
2k
and {D
2k
≥ 2
− j
} hold is greater than 2
−α j
. The proof of the separation lemma then combines the two lemmas to show that
P ¦
T
j −1
≤ 1 + c j
2
2
− j
T
j
A
2k
© ≥ 1 − 2
α j−β j
2
. Since
∞
Y
j=1
1 + c j
2
2
− j
∞, and
∞
Y
j= j
1 − 2
α j−β j
2
0, then conditioned on A
2k
, there is a probability bounded below that S and b S separate to some fixed
distance before leaving the ball of radius 2k no matter how close the two paths were upon leaving the ball of radius k.
Lemma 4.8. For all c 0, there exists β = βc 0 and j
c such that for all j ≥ j and all k,
P §
T
k j
−1
≥ 1 + c j
2
2
− j
T
k j
; A
2k
F
T
k j
ª
1
T
k j
≤ 3k
2 ≤ 2
−β j
2
.
1055
Proof. We let j be such that for all j
≥ j , c j
2
2
− j
12. Since k is fixed we will write T
j
for T
k j
from now on. We suppose that b S[0,
b σT
j
] and S[0, σT
j
] are any paths such that T
j
≤
3k 2
holds. We also assume that D
T
j
2
− j+1
or else there is nothing to prove.
Now consider K := b S[0,
b σ1 + c j
2
2
− j
T
j
] and let ρ = inf{n ≥ σT
j
: distS
n
, K ≤ 2
− j+1
S
n
}. Notice that even though we assume that D
T
j
2
− j+1
, ρ is not necessarily equal to σT
j
. If
ρ σ1+4·2
− j
T
j
then this means that T
j −1
1+4·2
− j
T
j
. However, if ρ ≤ σ1+4·2
− j
T
j
, then by the Beurling estimates for random walk Theorem 2.3, there exists c
′
1 such that
P ¦
S[ ρ, σ1 + 8 · 2
− j
T
j
] ∩ K = ; ©
≤ c
′
. The same estimate will hold starting at T
j
+ 8k2
− j
, k = 0, 1, . . . , ⌊c j
2
8⌋. Therefore,
P
n T
j −1
≥ 1 + c j
2
2
− j
T
j
; A
2k
F
T
j
o
1 T
j
≤ 3k
2
≤ P
n T
j −1
≥ 1 + c j
2
2
− j
T
j
; A
T
j
+c j
2
2
− j
F
T
j
o
1 T
j
≤ 3k
2 ≤ c
′ ⌊c j
2
8⌋
= 2
−β j
2
.
Lemma 4.9. There exists α ∞ and c 0 such that for all j and k,
P §
A
2k
; D
2k
≥ c2
− j
F
T
k j
ª ≥ c2
−α j
1A
T
k j
. Proof. Since k is fixed, we will omit the superscript k from now on. Let z
1
= Sσ
T
j
and z
2
= b S
b σ
T
j
. Without loss of generality, we may assume that T
j
2k or else there is nothing to prove and also that argz
2
argz
1
. Note that z
1
= z
2
= T
j
and k ≤ T
j
≤ 2k. Suppose that A
T
j
holds. By definition of T
j
, there exists c 0 and half-wedges
W
1
= {z : 1 − c2
− j
T
j
≤ |z| ≤ 1 + c2
− j
T
j
, −c2
− j
≤ argz − argz
1
≤ c2
− j
} and
W
2
= {z : 1 − c2
− j
T
j
≤ |z| ≤ 1 + c2
− j
T
j
, −c2
− j
≤ argz − argz
2
≤ c2
− j
} such that S[0,
σ
T
j
] ∩ W
2
= ;, b S[0,
b σ
T
j
] ∩ W
1
= ;, and distW
1
, W
2
≥ c2
− j
T
j
. Using Lemma 2.4 and Proposition 3.5, it is easy to verify that there exists a global constant c
′
such that
P
z
1
¦ Sσ
W
1
≥ 1 + c2
− j
T
j
© ≥ c
′
, and
P
z
2
¦ bSbσ
W
2
≥ 1 + c2
− j
T
j
© ≥ c
′
. 1056
Now consider the half-wedges W
′ 1
= {z : T
j
≤ |z| ≤ 2k, − 3c
2 2
− j
≤ argz − argz
1
≤ π
6 }
and W
′ 2
= {z : T
j
≤ |z| ≤ 2k, π
6 ≤ argz − argz
1
≤ 3c
2 2
− j
}. Applying Lemma 2.4 and Corollary 3.7 to W
′ 1
and W
′ 2
, one obtains that for any z
′ 1
∈ ∂ W
1
such that z
′ 1
≥ 1 + c2
− j
T
j
P
z
′ 1
§ Sσ
W
′ 1
= 2k ª
≥ c
′
2
−α j
, and for any z
′ 2
∈ ∂ W
2
such that z
′ 2
≥ 1 + c2
− j
T
j
P
z
′ 2
§ b
S b
σ
W
′ 2
= 2k ª
≥ c
′
2
−α j
, The result then follows since W
′ 1
and W
′ 2
are distance c2
− j
T
j
apart and S and b S are independent.
Proof of Theorem 4.7. We again fix k and let T
j
= T
k 2
j
. Let s =
∞
Y
j=1
1 + c j
2
2
− j
where c is chosen so that s ≤ 32. We also let j
be such that for j ≥ j
, 2
−β j
2
+α j
1, where α and β = βc are as in Lemmas 4.8 and 4.9.
To prove the theorem, it suffices to show that for all m,
P ¦
D
k
≥ c
1
A
k
; 2
−m
≤ D
k 2
2
−m+1
© ≥ c
2
. By Lemma 4.9, it is enough to find a constant c
′ 2
such that
P ¦
T
j
≤ 3k4 A
k
; 2
−m
≤ D
k 2
2
−m+1
© ≥ c
′ 2
. In fact, we will show that
P ¦
T
j
≤ ks2 A
k
; 2
−m
≤ D
k 2
2
−m+1
© ≥ c
′ 2
. Let
B
m
= {2
−m
≤ D
k 2
2
−m+1
}, and
C
j
= {T
j −1
≤ 1 + c j
2
2
− j
T
j
}.
1057
Then,
P ¦
T
j
≤ 3k4 A
k
; B
m
© ≥ P
m
\
j= j +1
C
j
A
k
; B
m
=
m
Y
j= j +1
P
C
j
A
k
; B
m
;
m
\
l= j+1
C
l
=
m
Y
j= j +1
P
C
j
A
k
; B
m
; A
T
j
;
m
\
l= j+1
C
l
=
m
Y
j= j +1
1 − P
C
c j
; A
k
B
m
; A
T
j
; T
m l= j+1
C
l
P A
k
B
m
; A
T
j
; T
m l= j+1
C
l
. However,
B
m
∩ A
T
j
∩
m
\
l= j+1
C
l
∈ F
T
j
,
B
m
∩ A
T
j
∩
m
\
l= j+1
C
l
⊂ {T
j
≤ 3k
2 },
and B
m
∩ A
T
j
∩
m
\
l= j+1
C
l
⊂ A
T
j
. Therefore, by Lemmas 4.8 and 4.9,
P ¦
T
j
≤ 3k4 A
k
; B
m
© ≥
m
Y
j= j +1
1 − 2
−β j
2
+α j
≥
∞
Y
j= j +1
1 − 2
−β j
2
+α j
= c
2
0.
Using the same techniques, one can prove a “reverse” separation lemma. Let e S be a random walk
started uniformly on the circle ∂ B
n
and conditioned to hit 0 before leaving B
n
. Let X be the time reversal of b
S
n
so that X is also a process from ∂ B
n
to 0. As before, for k ≤ n, let
e A
k
= {e S[0,
e σ
k
] ∩ X [0, b σ
k
] = ;}, e
D
k
= k
−1
min {diste
S e
σ
k
, X [0, b
σ
k
], distX b
σ
k
, e S[0,
e σ
k
]}. Then,
Theorem 4.10 Reverse Separation Lemma. There exists c
1
, c
2
0 such that
P ¦
e D
k
≥ c
1
e A
k
© ≥ c
2
.
1058
5 The growth exponent
5.1 Introduction
