III Schramm--Loewner Evolutions - Scaling limit of self-avoiding walks}\label{sec:saw

6Scaling limit of self-avoiding walks}\label{sec:saw

III Schramm--Loewner Evolutions

6 Scaling limit of self-avoiding walks

We next think about self-avoiding walk (SAW).

Definition

(Self-avoiding walk)

Let

= (

V, E

) be a graph with uniformly

bounded degree, and pick

x ∈ V

and

n ∈ N

. The self-avoiding in

starting

from

of length

is the uniform measure on simple paths in

starting from

of length n.

Self-avoiding walk tends to be difficult to understand, since they usually

don’t have the Markov property. However, in the scaling limit, it tends to be

more well-behaved. Restricting to the case of

, it has been shown that

for d ≥ 5, the scaling limit is just Brownian motion (Hara and Slade).

= 4, the same is conjectured to be true. In

= 3, there is no conjecture

for the continuous object which should describe its scaling limit. In this course,

we are only interested in

= 2. In this case, the self-avoiding walk is conjectured

to converge to SLE

8/3

(Lawler–Schramm–Werner).

In percolation, the property that allowed us to conclude that the scaling limit

was

SLE

is the locality property. In self-avoiding walks, the special property

is restriction. This is a very simple concept — if

= (

, E

) is a subgraph

with

x ∈ V

, then the self-avoiding walk on

conditioned to stay in

is a self-avoiding walk on

. Indeed, uniform measures always restricts to the

uniform measure on a smaller subset. It turns out

SLE

8/3

is the only

SLE

which

satisfies a continuum version of restriction. We will only show one direction,

namely that

SLE

8/3

satisfies the restriction property, but the other direction is

also known.

Brownian excursion

When studying the restriction property, we will encounter another type of process,

known as Brownian excursion. Fix a simply connected domain

D ⊆ C

and

points

x, y ∈ ∂D

distinct. Then roughly speaking, a Brownian excursion is a

Brownian motion starting at

conditioned to leave

∂D

. Of course, we

cannot interpret these words literally, since we want to condition on a zero

probability event.

To make this rigorous, we first use conformal invariance to say we only have

to define this for H with x = 0, y = ∞.

To construct it in

, we start with a complex Brownian motion

with

= 0 and

ε >

0. We then condition

on the event that

hits

R 

0 before hitting 0. This is a positive probability event. We then take limits

R → ∞

and

ε →

0. On the second example sheet, we will show that this makes

sense, and the result is called Brownian excursion.

It turns out the limiting object is pretty simple. It is given by

B = (

)

where

is a standard Brownian motion and

∼ BES

, and the two are

independent.

The key property about Brownian excursion we will need is the following:

Proposition.

Suppose

be a compact

-hull and

is as usual. If

x ∈ R \A

then

[

B[0, ∞) ∩ A = ∅] = g

(x).

Proof.

This is a straightforward computation. Take

iε

with

ε >

0, and

let B

be a Brownian motion. Define

= inf{t ≥ 0 : im(B

) = R}.

Then the desired probability is

lim

ε→0

lim

R→∞

[B[0, σ

] ∩A = ∅ | B[0, σ

] ∩R = ∅].

By Bayes’ theorem, this is equal to

lim

ε→0

lim

R→∞

[B[0, σ

] ∩(A ∪ R) = ∅]

P[B[0, σ

] ∩ R = ∅]

We understand the numerator and denominator separately. The gambler’s ruin

estimate says the denominator is just

ε/R

, and to bound the numerator, recall

that for z ∈ H \ A, we have

(z) − z| ≤ 3Rad(A).

Thus, using conformal invariance, we can bound

(z)

[B[0, σ

R+3Rad(A)

] ∩ R = ∅] ≤ P

[B[0, σ

] ∩ (A ∪ R) = ∅]

≤ P

(z)

[B[0, σ

R−3Rad(A)

] ∩ R = ∅].

So we get

im(g

(z))

R + 3Rad(A)

≤ numerator ≤

im(g

(z))

R − 3Rad(Z)

Combining, we find that the desired probability is

lim

ε→0

im(g

(x + iε))

= g

(x).

The restriction property

We now return to understanding the restriction property. We assume that

κ ≤

since this is the range of κ-values so that SLE

is simple.

Recall that Q is the set of all compact H-hulls. We define

= {A ∈ Q :

A ∩ (−∞, 0] = ∅}

−

= {A ∈ Q :

A ∩ [0, ∞) = ∅}

For A ∈ Q

= Q

∪ Q

−

, we define ψ

: H → A → H by

(z) = g

(z) − g

(0).

This is the version of

that fixes 0, which is what we need. This is the unique

conformal transformation with

(0) = 0, lim

z→∞

(z)

= 1.

We will need the following fact about SLE:

Fact. SLE

is transient, i.e. if γ is an SLE

in H from 0 to ∞, then

lim

t→∞

γ(t) = ∞ almost surely.

This is not very difficult to prove, but the proof is uninteresting and we have

limited time. Since

SLE

is simple for

κ ≤

4 and is also transient, it follows that

for all A ∈ Q

0 < P[γ[0, ∞) ∩ A = ∅] < 1.

This is useful because we want to condition on this event. Write

= {γ[0, ∞) ∩ A = ∅}.

Definition

(Restriction property)

We say an

SLE

satisfies the restriction

property if whenever

is an

SLE

, for any

A ∈ Q

, the law of

(

) conditional

on V

is that of an SLE

curve (for the same κ).

Observe that the law of

is determined by the probabilities

A 7→ P

[

] for

all A ∈ Q

Lemma. Suppose there exists α > 0 so that

P[V

] = (ψ

(0))

for all A ∈ Q

, then SLE

satisfies restriction.

Proof.

Suppose the assertion in the lemma is true. Suppose that

A, B ∈ Q

Then we have that

P[ψ

(γ[0, ∞)) ∩ B = ∅ | V

] =

P[γ[0, ∞) ∩ (ψ

−1

(B) ∪ A) = ∅]

P[γ[0, ∞) ∩ A = ∅]

(ψ

−1

(B)∪A)

(0))

(ψ

(0))

(ψ

(0))

(ψ

(0))

(ψ

(0))

= (ψ

(0))

= P[V

where we used that

−1

(B)∪A

= ψ

◦ ψ

So the law of ψ

(γ) given V

is the law of γ.

We now have to show that

SLE

8/3

satisfies the condition in the lemma. Let

be the filtration of U

√

κB

. Then

= P[V

| F

]

is a bounded martingale with

= P[V

]. Also,

→ 1

by the martingale convergence theorem. Also, if we define the stopping time

τ = inf{t ≥ 0 : γ(t) ∈ A},

then we get

= P[V

| F

] = P[V

| F

{t<τ}

= P[V

(A)−g

(0)

{t<τ}

by the conformal Markov property.

Observe that if

is another bounded

-martingale with the property that

→ 1

as t → ∞, then M

for all t ≥ 0, since

= E[1

| F

] =

Given what we were aiming for, we consider

= (ψ

(A)−g

(0)

(0))

{t<τ}

Lemma. M

t∧τ

is a continuous martingale if

κ =

, α =

These numbers are just what comes out when we do the computations.

Proof.

Recall that we showed that

is a probability involving Brownian ex-

cursion, and in particular is bounded in [0

1]. So the same is true for

, and

hence

t∧τ

. So it suffices to show that

t∧τ

is a continuous local martingale.

Observe that

t∧τ

= (ψ

t∧τ

(A)−g

t∧τ

(0)

(0))

So if we define

= (ψ

(A)−g

(0)

(0))

then it suffices to show that

is a continuous local martingale by optional

stopping. We write

= ˜g

◦ ψ

◦ g

−1

where ˜g

= g

(γ(0,t])

. We then have

= (ψ

))

In the example sheet, we show that

∂

) =

)

2ψ

)

−

000

By Itˆo’s formula, we get

= αN



(α −1)κ + 1

)



−



000

)



+ αN

)

√

κ dB

Picking

ensures the second d

term vanishes, and then setting

kills

the first dt term as well, and we are done.

We next want to establish that M

→ 1

as t → ∞.

Recall that

is the probability that a Brownian excursion on

H \ γ

, t

]

from

(

) to infinity does not hit

. So we would expect this to be true, since

, as

increases, the tip of the SLE gets further and further away from

and so it is difficult for a Brownian excursion to hit

; conversely on

, it

eventually gets to A, and then we are dead.

By scaling, we can assume that

sup{im(ω) : ω ∈ A} = 1.

It is convenient to define the stopping times

σ(r) = inf{t ≥ 0 : im(γ(t)) = r}.

Note that σ

< ∞ almost surely for all r > 0 since SLE

8/3

is transient.

Lemma. M

t∧τ

→ 1 on V

as t → ∞.

Proof.

Let

be a Brownian excursion in

H \ γ

, σ

] from

(

) to

∞

. Let

be a complex Brownian motion, and

= inf{t ≥ 0 : im(B

) = R}, z = γ(σ

) + iε.

Then the probability that

B hits A is

1 − ψ

) = lim

ε→0

lim

R→∞

[B[0, τ

] ⊆ H \ γ[0, σ

], B[0, τ

] ∩ A 6= ∅]

[B[0, τ

] ⊆ H \ γ[0, σ

]]

, (∗)

We will show that this expression is

≤ Cr

−1/2

for some constant

C >

0. Then we

know that

∧τ

→

1 as

r → ∞

. This is convergence along a subsequence,

but since we already know that M

t∧τ

converges this is enough.

We first tackle the denominator, which we want to bound from below. The

idea is to bound the probability that the Brownian motion reaches the lime

(

) =

+ 1 without hitting

R ∪ γ

, σ

]. Afterwards, the gambler’s ruin

estimate tells us the probability of reaching

(

) =

without going below the

im(z) = r line is

R−r

In fact, we shall consider the box

= [

−

(

) of side length 2 centered

(

). Let

be the first time

leaves

, and we want this to leave via the

top edge

. By symmetry, if we started right at

(

), then the probability of

leaving at

is exactly

. Thus, if we are at

(

) +

iε

, then the probability

of leaving via ` is >

What we would want to show is that

[B(η) ∈ ` | B[0, η] ∩γ[0, σ

] = ∅] >

. (†)

We then have the estimate

denominator ≥

· P

[B[0, η] ∩ γ[0, σ

] = ∅] ·

R − r

Intuitively, (

†

) must be true, because

, σ

] lies below

(

) =

, and so if

, η

] doesn’t hit

, σ

], then it is more likely to go upwards. To make this

rigorous, we write

< P

[B(η) ∈ `]

= P

[B(η) ∈ ` | B[0, η] ∩γ[0, σ

] = ∅] P[B[0, η] ∩ γ[0, σ

] = ∅]

+ P

[B(η) ∈ ` | B[0, η] ∩γ[0, σ

] 6= ∅] P[B[0, η] ∩ γ[0, σ

] 6= ∅]

To prove (

†

), it suffices to observe that the first factor of the second term is is

≤

, which follows from the strong Markov property, since

[

(

)

∈ `

]

≤

whenever im(w) ≤ r, which in particular is the case when w ∈ γ[0, σ

To bound the numerator, we use the strong Markov property and the Beurling

estimate to get

[B hits A without hitting R ∪γ[0, σ

]] ≤ P

[B[0, η] ∩ γ[0, σ

]] · Cr

−1/2

Combining, we know the numerator in (∗) is

≤

C · r

−1/2

· P[B[0, η] ∩ γ[0, σ

] = ∅].

These together give the result.

Lemma. M

t∧τ

→ 0 as t → ∞ on V

This has a “shorter” proof, because we outsource a lot of the work to the

second example sheet.

Proof.

By the example sheet, we may assume that

is bounded by a smooth,

simple curve β : (0, 1) → H.

Note that γ(τ) = β(s) for some s ∈ (0, 1). We need to show that

lim

t→τ

) = 0.

For m ∈ N, let

= inf



t ≥ 0 : |γ(t) − β(s)| =



Since β is smooth, there exists δ > 0 so that

` = [β(s), β(s) + δn] ⊆ A,

where n is the unit inward pointing normal at β(s). Let

= g

(`) − U

Note that a Brownian motion starting from a point on

has a uniformly positive

chance of exiting

H \ γ

, t

] on the left side of

, t

] and on the right side as

well.

On the second example sheet, we see that this implies that

⊆ {w : im(w) ≥ a|Re(w)|}

for some

a >

0, using the conformal invariance of Brownian motion. Intuitively,

this is because after applying

− U

, we have uniformly positive probability of

exiting via the positive or real axis, and so we cannot be too far away in one

directionn.

Again by the second example sheet, the Brownian excursion in

from 0 to

∞ hits L

with probability → 1 as m → ∞.

We thus conclude

Theorem. SLE

8/3

satisfies the restriction property. Moreover, if

γ ∼ SLE

8/3

then

P[γ[0, ∞) ∩ A = ∅] = (ψ

(0))

5/8

There is a rather surprising consequence of this computation. Take

, . . . , γ

to be independent SLE

8/3

’s. Then we have

P[γ

[0, ∞) ∩ A = ∅ for all j] = (ψ

(0))

Note that this is the same as the probability that the hull of

, . . . , γ

des not

intersect

, where the hull is the union of the

’s together with the bounded

components of H \

In the same manner, if

, . . . ,

are independent Brownian excursions,

then

[0, ∞) ∩ A = ∅ for all j] = (ψ

(0))

Thus, the hull of γ

, . . . , γ

has the same distribution as the hull of

, . . . ,

Moreover, if we take a boundary point of the hull of, say,

, . . . , γ

, then

we would expect it to belong to just one of the

SLE

8/3

’s. So the boundary of

the hull of

, . . . , γ

looks locally like an

SLE

8/3

, and the same can be said for

the hull fo

, . . . ,

. Thus, we conclude that the boundary of a Brownian

excursion looks “locally” like SLE

8/3