III Schramm--Loewner Evolutions (Full)

Part III — Schramm–Loewner Evolutions

Based on lectures by J. Miller

Notes taken by Dexter Chua

Lent 2018

These notes are not endorsed by the lecturers, and I have modified them (often

significantly) after lectures. They are nowhere near accurate representations of what

was actually lectured, and in particular, all errors are almost surely mine.

Schramm–Loewner Evolution (SLE) is a family of random curves in the plane, indexed

by a parameter

κ ≥

0. These non-crossing curves are the fundamental tool used to

describe the scaling limits of a host of natural probabilistic processes in two dimensions,

such as critical percolation interfaces and random spanning trees. Their introduction

by Oded Schramm in 1999 was a milestone of modern probability theory.

The course will focus on the definition and basic properties of SLE. The key ideas are

conformal invariance and a certain spatial Markov property, which make it possible

to use Itˆo calculus for the analysis. In particular we will show that, almost surely, for

κ ≤

4 the curves are simple, for 4

≤ κ <

8 they have double points but are non-crossing,

and for

κ ≥

8 they are space-filling. We will then explore the properties of the curves

for a number of special values of

(locality, restriction properties) which will allow us

to relate the curves to other conformally invariant structures.

The fundamentals of conformal mapping will be needed, though most of this will be

developed as required. A basic familiarity with Brownian motion and Itˆo calculus will

be assumed but recalled.

Contents

0 Introduction

1 Conformal transformations

1.1 Conformal transformations

1.2 Brownian motion and harmonic functions

1.3 Distortion estimates for conformal maps

1.4 Half-plane capacity

2 Loewner’s theorem

2.1 Key estimates

2.2 Schramm–Loewner evolution

3 Review of stochastic calculus

4 Phases of SLE

5 Scaling limit of critical percolation

6 Scaling limit of self-avoiding walks

7 The Gaussian free field

0 Introduction

Schramm–Loewner evolution is a random curve in a complex domain

(which

we will often take to be

for convenience), parametrized by a positive real

number

. This was introduced by Schramm in 1999 to describe various scaling

limits that arise in probability theory and statistical physics.

Recall that if we take a random walk on the integer lattice

, and take the

scaling limit as the grid size tends to 0, we converge towards a Brownian motion.

There are other discrete models that admit a scaling limit, and the limit is often

something that is not Brownian motion.

Schramm showed that if the scaling limit satisfies certain conformal invariance

properties, which many models do, then it must be

SLE

for some

. If we can

identify exactly which

it belongs to, then this will completely determine what

the scaling limit is, and often conveys a lot of extra information such as the

critical exponents. This has been done for several discrete models:

– The scaling limit of loop-erased random walk is SLE

– The scaling limit of hexagonal percolation is SLE

(see chapter 5)

–

It is conjectured that the scaling limit of self-avoiding random walk is

SLE

8/3

(see chapter 6)

– The “level sets” of a Gaussian free field are SLE

’s (see chapter7).

In this course, we will prove some basic properties of

SLE

, and then establish

the last three results/conjectures.

1 Conformal transformations

1.1 Conformal transformations

Definition

(Conformal map)

Let

U, V

be domains in

. We say a holomorphic

function f : U → V is conformal if it is a bijection.

We will write

for the open unit disk, and

for the upper half plane. An

important theorem about conformal maps is the following:

Theorem

(Riemann mapping theorem)

Let

be a simply connected domain

with

U 6

and

z ∈ U

be any point. Then there exists a unique conformal

transformation f : D → U such that f(0) = z, and f

(0) is real and positive.

We shall not prove this theorem, as it is a standard result. An immediate

corollary is that any two simply connected domains that are distinct from

are

conformally equivalent.

Example.

Take

. Then for

z ∈ D

, the map promised by the Riemann

mapping theorem is

f(w) =

w + z

1 + ¯zw

In general, every conformal transformation f : D → D is of the form

f(w) = λ

w + z

1 + ¯zw

for some |λ| = 1 and z ∈ D.

Example. The map f : H → D given by

f(z) =

z − i

z + i

is a conformal transformation, with inverse

g(w) =

i(1 + w)

1 − w

In general, the conformal transformations H → H consist of maps of the form

f(z) =

az + b

cz + d

with a, b, c, d ∈ R and ad − bc 6= 0.

Example. For t ≥ 0, we let H

= H \ [0, 2

√

ti]. The map H

→ H given by

z 7→

+ 4t

is a conformal transformation. Observe that this map satisfies

(z) − z| = |

+ 4t − z| → 0 as z → ∞.

So g

(z) ∼ z for large z.

Observe also that the family g

(z) satisfies the ODE

∂g

∂t

(z)

, g

(z) = z.

We can think of these functions

as being generated by the curve

(

) = 2

√

where for each

, the function

is the conformal transformation that sends

H \ γ([0, t]) to H (satisfying |g

(z) − z| → ∞).

Given the set of functions

, we can recover the curve

as follows — for

each

z ∈ H

, we can ask what is the minimum

such that

is not defined on

. By ODE theorems, there is a solution up till the point when

(

) = 0, in

which case the denominator of the right hand side blows up. Call this time

(

We then see that for each

, there is a unique

such that

(

) =

, and we have

γ(t) = z.

More generally, suppose

is any simple (i.e. non-self-intersecting) curve

starting from 0. Then for each

, we can let

be the unique conformal

transformation that maps

H\γ

([0

, t

]) to

with

(

)

−z| → ∞

as above (we will

later see such a map exists). Then Loewner’s theorem says there is a continuous,

real-valued function W such that

∂g

∂t

(z) − W

, g

(z) = z.

This is the chordal Loewner equation. We can turn this around — given a

function

, what is the corresponding curve

(

)? If

= 0, then

(

) = 2

√

More excitingly, if we take

√

κB

, where

is a standard Brownian motion,

and interpret this equation as a stochastic differential equation, we obtain

SLE

1.2 Brownian motion and harmonic functions

Recall that a function

is holomorphic iff it satisfies the Cauchy–

Riemann equations

∂u

∂x

∂v

∂y

∂u

∂y

= −

∂v

∂x

This in particular implies that both u and v are harmonic.

Definition

(Harmonic function)

A function

→ R

is harmonic if it is

and

∆f =



∂

∂x

+ ··· +

∂

∂x



f = 0.

Indeed, we can calculate

∂

∂x

∂

∂y

∂

∂x

∂v

∂y

∂

∂y



−

∂v

∂x



= 0.

In Advanced Probability, we saw that Brownian motion is very closely related

to harmonic functions.

Definition

(Complex Brownian motion)

We say a process

is a

complex Brownian motion if (B

, B

) is a standard Brownian motion in R

Recall some results from Advanced Probability:

Theorem.

Let

be a harmonic function on a bounded domain

which is

continuous on

. For

z ∈ D

, let

be the law of a complex Brownian motion

starting from z, and let τ be the first hitting time of D. Then

u(z) = E

[u(B

)].

Corollary

(Mean value property)

is a harmonic function, then, whenever

it makes sense, we have

u(z) =

2π

u(z + re

iθ

) dθ.

Corollary

(Maximum principle)

Let

be harmonic in a domain

. If

attains

its maximum at an interior point in D, then u is constant.

Corollary

(Maximum modulus principle)

Let

be a domain and let

D → C

be holomorphic. If

|f|

attains its maximum in the interior of

, then

is constant.

Proof.

Observe that if

is holomorphic, then

log |f|

is harmonic and attains its

maximum in the interior of

. It then follows that

|f|

is constant (if

|f|

vanishes

somewhere, then consider

log |f

for some large

, and do some patching if

necessary). It is then a standard result that a holomorphic function of constant

modulus is constant.

The following lemma should be familiar from the old complex analysis days

as well:

Lemma

(Schwarz lemma)

Let

D → D

be a holomorphic map with

(0) = 0.

Then

(

)

| ≤ |z|

for all

z ∈ D

. If

(

)

|z|

for some non-zero

z ∈ D

, then

f(w) = λw for some λ ∈ C with |λ| = 1.

Proof. Consider the map

g(z) =

(

f(z)/z z 6= 0

(0) z = 0

Then one sees that

is holomorphic and

(

)

| ≤

1 for all

z ∈ ∂D

, hence for all

z ∈ D

by the maximum modulus principle. If

(

)

for some

∈ D \{

}

then g must be constant, so f is linear.

1.3 Distortion estimates for conformal maps

We write

for the collection of conformal transformations

D → D

, where

is any simply connected domain with 0

∈ D

and

D 6

, with

(0) = 0 and

(0) = 1. Thus, it must be of the form

f(z) = z +

∞

n=2

Theorem

(Koebe-1/4 theorem)

f ∈ U

and 0

< r ≤

1, then

, r/

⊆

f(rD).

This says f cannot look like this:

By scaling it suffices to prove this for the case

= 1. This follows from the

following result:

Theorem. If f ∈ U, then |a

| ≤ 2.

The proof of this proposition will involve quite some work. So let us just

conclude the theorem from this.

Proof of Koebe-1/4 theorem.

Suppose

D → D

is in

, and

6∈ D

. We shall

show that |z

| ≥

. Consider the function

f(z) =

f(z)

− f(z)

Since

is composition of conformal transformations, it is itself conformal, and a

direct computation shows

f ∈ U. Moreover, if

f(z) = z + a

+ ··· ,

then

f(z) = z +





+ ··· .

So we obtain the bounds



≤ 2.

By the triangle inequality, we must have |z

−1

| ≤ 4, hence |z

| ≥

The 1/4 theorem bounds the distortion in terms of the value of

(0). Con-

versely, if we know how much the distortion is, we can use the theorem to derive

bounds on f

(0).

Corollary.

Let

be domains and

z ∈ D

˜z ∈

. If

D →

is a conformal

transformation with f(z) = ˜z, then

≤ |f

(z)| ≤

where d = dist(z, ∂D) and

d = dist(˜z, ∂

D).

Proof.

By translation, scaling and rotation, we may assume that

˜z

= 0,

d = 1 and f

(0) = 1. Then we have

D = f (D) ⊇ f(D) ⊇ B(0, 1/4).

d ≥

, as desired. The other bound follows by considering f

−1

We now proceed to prove the theorem. We first obtain a bound on area,

using IA Vector Calculus:

Proposition. Let f ∈ U. Then

area(f(D)) = π

∞

n=1

n|a

Proof.

In the ideal world, we will have something that helps us directly compute

area

(

)). However, for the derivation to work, we need to talk about what

does to the boundary of

, but that is not necessarily well-defined. So we

compute area(f (rD)) for r < 1 and then take the limit r → 1.

So fix

r ∈

1), and define the curve

(

) =

(

iθ

) for

θ ∈

]. Then we

can compute

¯z dz =

(x − iy)(dx + i dy)

(x − iy) dx + (ix + y) dy

f(rD)

2i dx dy

= area(f(rD)),

using Green’s theorem. We can also compute the left-hand integral directly as

¯z dz =

2π

f(re

iθ

(re

iθ

)ire

iθ

dθ

2π

∞

n=1

¯a

−iθn

∞

n=1

n−1

iθ(n−1)

ire

iθ

dθ

= π

∞

n=1

Definition

(Compact hull)

A compact hull is a connected compact set

K ⊆ C

with more than one point such that C \ K is connected.

We want to obtain a similar area estimate for compact hulls. By the Riemann

mapping theorem, if

is a compact hull, then there exists a unique conformal

transformation

C \

C \ K

that fixes infinity and has positive derivative

∞

, i.e.

lim

z→∞

(

)

/z >

0. Let

be the set of compact hulls containing 0 with

lim

z→∞

(

)

= 1. If

K ∈ H

, then the Laurent expansion of the corresponding

F (z) = z + b

∞

n=1

Observe there is a correspondence between the space of such

and

by sending

F to 1/F (1/z), and vice versa.

Proposition. If K ∈ H, then

area(K) = π

1 −

∞

n=1

n|b

Observe that the area is always non-negative. So in particular, we obtain the

bound

∞

n=1

n|b

≤ 1.

In particular, |b

| ≤ 1. This will be how we ultimately bound |a

Proof.

The proof is essentially the same as last time. Let

r >

1, and let

(

) (or, if you wish,

C \ F

(

C \ r

)), and

(

) =

(

iθ

). As in the

previous proposition, we have

area(K

) =

¯z dz

2π

F (re

iθ

(re

iθ

)ire

iθ

dθ

= π

−

∞

n=1

n|b

−2n

Then take the limit as r → 1.

By the correspondence we previously noted, this gives us some estimates on

the coefficients of any

f ∈ U

. It turns out applying this estimate directly is not

good enough. Instead, we want to take the square root of f(z

Lemma.

Let

f ∈ U

. Then there exists an odd function

h ∈ U

with

(

)

f(z

Proof. Note that f(0) = 0 by assumption, so taking the square root can poten-

tially be problematic, since 0 is a branch point. To get rid of the problem, define

the function

f(z) =

(

f(z)

z 6= 0

(0) z = 0

Then

is non-zero and conformal in

, and so there is a function

with

g(z)

f(z). We then set

h(z) = zg(z

Then

is odd and

(

)

(

). It is also immediate that

(0) = 0 and

(0) = 1. We need to show

is injective on

. If

, z

∈ D

with

(

) =

(

then

g(z

) = z

g(z

). (∗)

By squaring, we know

f(z

) = z

f(z

)

Thus,

(

) =

(

) and so

. But then (

∗

) implies

. So

injective, and hence h ∈ U.

We can now prove the desired theorem.

Proof of theorem. We can Taylor expand

h(z) = z + c

+ c

+ ··· .

Then comparing h(z)

= f(z

) implies

Setting

(

) =

h(1/z)

, we find that the

−1

coefficient of

−

, and as we

previously noted, this must be ≤ 1.

1.4 Half-plane capacity

Definition

(Compact

-hull)

A set

A ⊆ H

is called a compact

-hull if

is compact,

H ∩

and

H \ A

is simply connected. We write

for the

collection of compact H-hulls.

A large part of the course will be about studying families of compact

-hulls,

and in particular random families of compact

-hulls. In this chapter, we will

establish some basic results about compact

-hulls, in preparation for Loewner’s

theorem in the next chapter. We begin with the following important proposition:

Proposition.

For each

A ∈ Q

, there exists a unique conformal transformation

: H \ A → H with |g

(z) − z| → 0 as z → ∞.

This conformal transformation

will be key to understanding compact

H-hulls. The proof of this requires the following theorem:

Theorem

(Schwarz reflection principle)

Let

D ⊆ H

be a simply connected

domain, and let

D → H

be a conformal transformation which is bounded on

bounded sets and sends

R ∩D

. Then

extends by reflection to a conformal

transformation on

∗

= D ∪ {¯z : z ∈ D} = D ∪

by setting φ(¯z) = φ(z).

Proof of proposition.

The Riemann mapping theorem implies that there exists

a conformal transformation

H \ A → H

. Then

(

)

→ ∞

z → ∞

. By the

Schwarz reflection principle, extend

to a conformal transformation defined on

C \ (A ∪

A).

By Laurent expanding g at ∞, we can write

g(z) =

n=−∞

−N

Since

maps the real line to the real line, all

must be real. Moreover, by

injectivity, considering large z shows that N = 1. In other words, we can write

g(z) = b

−1

z + b

∞

n=1

with b

−1

6= 0. We can then define

(z) =

g(z) − b

−1

Since

and

−1

are both real, this is still a conformal transformation, and

(z) − z| → 0 as z → ∞.

To show uniqueness, suppose

, g

are two such functions. Then

◦ g

−1

H → H

is such a function for

∅

. Thus, it suffices to show that if

H → H

is a conformal mapping such that

(

)

− z →

0 as

z → ∞

, then in fact

But we can expand g(z) − z as

g(z) − z =

∞

n=1

and this has to be holomorphic at 0. So c

= 0 for all n, and we are done.

Definition

(Half-plane capacity)

Let

A ∈ Q

. Then the half-plane capacity of

A is defined to be

hcap(A) = lim

z→∞

z(g

(z) − z).

Thus, we have

(z) = z +

hcap(A)

∞

n=2

To justify the name “capacity”,

hcap

(

) had better be non-negative and

increasing in

. These are both true, as we will soon prove. However, let us first

look at some examples.

Example. z 7→

√

+ 4t

is the unique conformal transformation

√

]

→ H

with |

√

+ 4z − z| → 0 as z → ∞. We can expand

+ 4t = z +

+ ··· .

Thus, we know that

hcap

([0

√

]) = 2

. This justifies our previous funny

parametrization of [0,

√

ti].

Example.

The map

z 7→ z

maps

D → H

. Again, we have



z +

− z



→

as z → ∞, and so hcap(H ∩

D) = 1.

Proposition.

(i) Scaling: If r > 0 and A ∈ Q, then hcap(rA) = r

hcap(A).

(ii)

Translation invariance: If

x ∈ R

and

a ∈ Q

, then

hcap

(

) =

hcap

(

(iii)

Monotonicity: If

A ∈ Q

are such that

A ⊆

. Then

hcap

(

)

≤ hcap

(

Proof.

(i) We have g

(z) = rg

(z/r).

(ii) Observe g

A+x

(z) = g

(z − x) + x.

(iii) We can write

= g

(

A\A)

◦ g

Thus, expanding out tells us

hcap(

A) = hcap(A) + hcap(g

(

A \ A)).

So the desired result follows if we know that the half-plane capacity is

non-negative, which we will prove next.

Observe that these results together imply that if

A ∈ Q

and

A ⊆ r

(

D ∩ H

then

hcap(A) ≤ hcap(r(

D ∩ H)) ≤ r

hcap(

D ∩ H) = r

So we know that hcap(A) ≤ diam(A)

Compared to the above proofs, it seems much less straightforward to prove

non-negativity of the half-plane capacity. Our strategy for doing so is to relate

the half-plane capacity to Brownian motion! This is something we will see a lot

in this course.

Proposition.

Let

A ∈ Q

and

be complex Brownian motion. Define the

stopping time

τ = inf{t ≥ 0 : B

6∈ H \A}.

Then

(i) For all z ∈ H \ A, we have

im(z − g

(z)) = E

[im(B

)]

(ii)

hcap(A) = lim

y→∞

y E

[im(B

)].

In particular, hcap(A) ≥ 0.

(iii) If A ⊆

D ∩ H, then

hcap(A) =

iθ

[im(B

)] sin θ dθ.

Proof.

(i)

Let

(

) =

(

z − g

(

)). Since

z − g

(

) is holomorphic, we know

harmonic. Moreover,

is continuous and bounded, as it

→

0 at infinity.

These are exactly the conditions needed to solve the Dirichlet problem

using Brownian motion.

Since

(

)) = 0 when

z ∈ ∂

(

H \ A

), we know

(

) =

(

−

)). So the result follows.

(ii) We have

hcap(A) = lim

z→∞

z(g

(z) − z)

= lim

y→∞

(iy)(g

(iy) − iy)

= lim

y→∞

y im(iy − g

(iy))

= lim

y→∞

y E

[im(B

)]

where we use the fact that

hcap

(

) is real, so we can take the limit of the

real part instead.

(iii) See example sheet.

In some sense, what we used above is that Brownian motion is conformally

invariant, where we used a conformal mapping to transfer between our knowledge

of Brownian motion on

to that on

H \ D

. Informally, this says the conformal

image of a Brownian motion is a Brownian motion.

We have in fact seen this before. We know that rotations preserve Brownian

motion, and so does scaling, except when we scale we need to scale our time

as well. In general, a conformal mapping is locally a rotation and scaling. So

we would indeed expect that the conformal image of a Brownian motion is a

Brownian motion, up to some time change. Since we are performing different

transformations all over the domain, the time change will be random, but that

is still not hard to formalize. For most purposes, the key point is that the image

of the path is unchanged.

Theorem.

Let

D ⊆ C

be domains, and

D →

a conformal transforma-

tion. Let

be Brownian motions starting from

z ∈ D, ˜z ∈

respectively,

with f (z) = ˜z. Let

τ = inf{t ≥ 0 : B

6∈ D}

˜τ = inf{t ≥ 0 :

6∈

Set

σ(t) = inf



s ≥ 0 :

dr = t



= f(B

σ(t)

Then (B

: t < ˜τ

) has the same distribution as (

: t < ˜τ).

Proof. See Stochastic Calculus.

Example.

We can use conformal invariance to deduce the first exit distribution

of a Brownian motion from different domains. First of all, observe that if we

take our domain to be

and start our Brownian motion from the origin, the

first exit distribution is uniform. Thus, applying a conformal transformation, we

see that the exit distribution starting from any point z ∈ D is

f(e

iθ

) =

2π



1 − |z|

iθ

− z|



Similarly, on H, starting from z = x + iy, the exit distribution is

f(u) =

(x − u)

+ y

Note that if x = 0, y = 1, then this is just

f(u) =



+ 1



This is the Cauchy distribution!

2 Loewner’s theorem

2.1 Key estimates

Before we prove Loewner’s theorem, we establish some key identities and esti-

mates. As before, a useful thing to do is to translate everything in terms of

Brownian motion.

Proposition. Let A ∈ Q and B be a complex Brownian motion. Set

τ = inf{t ≥ 0 : B

6∈ H \A}.

Then

– If x > Rad(A), then

(x) = lim

y→∞

πy



− P

∈ [x, ∞)]



– If x < −Rad(A), then

(x) = lim

y→∞

πy



∈ (−∞, x]] −



Proof. First consider the case A = ∅ and, by symmetry, x > 0. Then

lim

y→∞

πy



− P

∈ [x, ∞)]



= lim

y→∞

πy P

∈ [0, x)]

= lim

y→∞

πy

π(s

+ y

)

= x,

where the first equality follows from the fact that Brownian motion exits through

the positive reals with probability

; the second equality follows from the

previously computed exit distribution; and the last follows from dominated

convergence.

Now suppose A 6= ∅. We will use conformal invariance to reduce this to the

case above. We write g

= u

+ iv

. We let

σ = inf{t > 0 : B

6∈ H}.

Then we know

∈ [x, ∞)] = P

(iy)

∈ [g

(x), ∞)]

= P

(iy)

∈ [g

(x) − u

(iy), ∞)].

Since

(

)

− z →

0 as

z → ∞

, it follows that

(iy)

→

1 and

(

)

→

0 as

y → ∞. So we have



(iy)



∈ [g

(x) − u

(iy), ∞)



− P



∈ [g

(x), ∞)





= o(y

−1

)

as y → ∞. Combining with the case A = ∅, the the proposition follows.

Corollary. If A ∈ Q, Rad(A) ≤ 1, then

x ≤ g

(x) ≤ x +

if x > 1

x +

≤ g

(x) ≤ x if x < −1.

Moreover, for all A ∈ Q, we have

(z) − z| ≤ 3 Rad(A).

Proof.

Exercise on the first example sheet. Note that

z 7→ z

sends

H \

This corollary gives us a “zeroth order” bound on

(

)

− z

. We can obtain

a better first-order bound as follows:

Proposition.

There is a constant

c >

0 so that for every

A ∈ Q

and

|z| >

2 Rad(A), we have



(z) −



z +

hcap(A)





≤ c

Rad(A) · hcap(A)

|z|

Proof. Performing a scaling if necessary, we can assume that Rad(A) = 1. Let

h(z) = z +

hcap(A)

− g

(z).

We aim to control the imaginary part of this, and then use the Cauchy–Riemann

equations to control the real part as well. We let

v(z) = im(h(z)) = im(z − g

(z)) =

im(z)

|z|

hcap(A).

Let B be a complex Brownian motion, and let

σ = inf{t ≥ 0 : B

6∈ H \

τ = inf{t ≥ 0 : B

6∈ H}.

Let

(

z, e

iθ

) be the density with respect to the Lebesgue measure at

iθ

for

Then by the strong Markov property at the time σ, we have

im(z − g

(z)) =

iθ

[im(B

)]p(z, e

iθ

) dθ.

In the first example sheet Q3, we show that

p(z, e

iθ

) =

im(z)

|z|

sin θ



1 + O



|z|



. (∗)

We also have

hcap(A) =

iθ

[im(B

)] sin θ dθ.

|v(z)| =



im(z − g

(z)) −

im(z)

|z|

hcap(A)



iθ

[im(B

)]p(z, e

iθ

) dθ −

im(z)

|z|

iθ

[im(B

)] sin θ dθ



By applying (∗), we get

|v(z)| ≤ c

c hcap(A) im(z)

|z|

where c is a constant.

Recall that

is harmonic as it is the imaginary part of a holomorphic function.

By example sheet 1 Q9, we have

|∂

v(z)| ≤

c hcap(A)

|z|

, |∂

v(z)| ≤

c hcap(A)

|z|

By the Cauchy–Riemann equations,

(

)) satsifies the same bounds. So we

know that

(z)| ≤

c hcap(A)

|z|

Then

h(iy) =

∞

(is) ds,

since h(iy) → 0 as y → ∞. Taking absolute values, we have

|h(iy) =

∞

(is)| ds

≤ c hcap(A)

∞

−3

≤ c

hcap(A)y

−z

To get the desired bound for a general

, integrate

along the boundary of the

circle of radius |z| to get

|h(z)| = |h(re

iθ

)| ≤

c hcap(A)

|z|

+ h(iz).

The following is a very useful fact about Brownian motion:

Theorem

(Beurling estimate)

There exists a constant

c >

0 so that the

following holds. Let

be a complex Brownian motion, and

A ⊆

be connected,

0 ∈ A, and A ∩ ∂

D 6= ∅. Then for z ∈ D, we have

[B[0, τ] ∩A = ∅] ≤ c|z|

1/2

where τ = inf{t ≥ 0 : B

6∈ D}.

We will not prove this, since it is quite tricky to prove. The worst case

behaviour is obtained when A = [−i, 0].

To the theorem, we need one more estimate, which uses the Beurling estimate.

Proposition.

There exists a constant

c >

0 so that the following is true:

Suppose A,

A ∈ Q with A ⊆

A and

A \ A is coonnected. Then

diam(g

(

A \ A)) ≤ c

(

(dr)

1/2

d ≤ r

Rad(

A) d > r

where

d = diam(

A \ A), r = sup{im(z) : z ∈

A}.

Proof. By scaling, we can assume that r = 1.

– If d ≥ 1, then the result follows since

(z) − z| ≤ 3Rad(A),

and so

diam(g

(

A \ A)) ≤ diam(A) + 6Rad(A) ≤ 8 diam(

A).

–

d <

1, fix

z ∈ H

so that

(

z, d

)

⊇

A \ A

. It then suffices to bound

the size of g

(U) (or, to be precise, g

(U \ A)).

Let B be a complex Brownian motion starting from iy with y ≥ 2. Let

τ = inf{t ≥ 0 : B

6∈ H \A}.

For B[0, τ ] to reach U , it must

(i)

Reach

(

1) without leaving

H \ A

, which occurs with probability

at most c/y for some constant c, by example sheet.

(ii)

It must then hit

before leaving

H \ A

. By the Beurling estimate,

this occurs with probability ≤ cd

1/2

Combining the two, we see that

lim sup

y→∞

[B[0, τ] ∩U 6= ∅] ≤ cd

1/2

By the conformal invariance of Brownian motion, if

inf{t ≥

0 :

6∈

H}, this implies

lim sup

y→∞

y P

[B[0, σ] ∩ g

(

A \ A) 6= ∅] ≤ cd

1/2

Since g

(

A \ A) is connected, by Q10 of example sheet 1, we have

diam(g

(

A \ A)) ≤ cd

1/2

We can finally get to the key content of Loewner’s theorem. An important

definition is the following:

Definition.

Suppose (

)

t≥0

is a family of compact

-hulls. We say that (

)

(i) non-decreasing if s ≤ t implies A

⊆ A

(ii)

locally growing if for all

T >

0 and

ε >

0, there exists

δ >

0 such that

whenever 0 ≤ s ≤ t ≤ s + δ ≤ T , we have

diam(g

\ A

)) ≤ ε.

This is a continuity condition.

(iii) parametrized by half-plane capacity if hcap(A

) = 2t for all t ≥ 0.

We write

be the set of families of compact

-hulls which satisfy (i) to (iii).

We write A

for the set of such families defined on [0, T ].

Example.

is is a simple curve in

starting from 0 and

, t

]. This

clearly satisfies (i), and the previous proposition tells us this satsifies (ii). On the

first example sheet, we show that we can reparametrize

so that

hcap

(

) = 2

for all t ≥ 0. Upon doing so, we have A

∈ A.

Theorem.

Suppose that (

)

∈ A

. Let

. Then there exists a continuous

function U : [0, ∞) → R so that

∂

(z) =

(z) − U

, g

(z) = z.

This is known as the chordal Loewner ODE , and

is called the driving

function.

Proof.

First note that since the hulls are locally growing, the intersection

s≥t

(

) consists of exactly one point. Call this point

. Again by the

locally growing property, U

is in fact a continuous function in t.

Recall that if A ∈ Q, then

(z) = z +

hcap(A)

+ O



hcap(A)Rad(A)

|z|



If x ∈ R, then as g

A+x

(z) − x = g

(z − x), we have

(z) = g

(z + x) − x = z +

hcap(z)

z + x

+ O



hcap(A)Rad(A + x)

|z + x|



. (∗)

Fix ε > 0. For 0 ≤ s ≤ t, let

s,t

= g

◦ g

−1

Note that

hcap(g

t+ε

\ A

)) = 2ε.

Apply (

∗

) with

(

t+ε

\ A

−U

, and use that

Rad

(

) =

Rad(A − U

) ≤ diam(A) to see that

(z) = g

t,t+ε

(z) = z +

2ε

z − U

+ 2ε diam(g

t+ε

\ A

))O



|z − U



t+ε

(z) − g

(z) = (g

t,t+ε

− g

t,t

) ◦ g

(z)

2ε

(z) − U

+ 2ε diam(g

t+ε

\ A

))O



(z) − U



Dividing both sides by

and taking the limit as

ε →

0, the desired result follows

since diam(g

t+ε

\ A

)) → 0.

We can ask about the reverse procedure — given a continuous function

can we find a collection

∈ A

whose driving function is

? We can simply let

be the solution to this differential equation. We can then let

= H \ domain(g

On the example sheet, we show that this is indeed a family in

, and

are

indeed conformal transformations.

2.2 Schramm–Loewner evolution

We now take

to be a random process adapted to

(

s ≤ t

). We can

then define

(

) as the solution to the Loewner equation for each

, and define

accordingly. We then obtain a random family A

in A.

Definition

(Conformal Markov property)

We say that (

) satisfy the confor-

mal Markov property if

(i) Given F

, (g

t+s

) − U

)

s≥0

= (A

)

s≥0

(ii) Scale-invariance: (rA

t/r

)

t≥0

= (A

Here (i) is the Markov property, and the second part is the conformal part,

since the only conformal maps H → H that fix 0 and ∞ are the rescalings.

Schramm was interested in such processes because they provide us with

potential scaling limits of random processes, such as self-avoiding walk or perco-

lation. If we start with a process that has the Markov property, then the scaling

limit, if exists, ought to be scale invariant, and thus have the conformal Markov

property. Schramm classified the list of all possible such scaling limits:

Theorem

(Schramm)

If (

) satisfy the conformal Markov property, then there

exists κ ≥ 0 so that U

√

κB

, where B is a standard Brownian motion.

Proof.

The first property is exactly the same thing as saying that given

, we

have

t+s

− U

)

s≥0

= (U

is a continuous process with stationary, independent increments. This

implies there exists κ ≥ 0 and a ∈ R such that

√

κB

+ at,

where B is a standard Brownian motion. Then the second part says

(rU

t/r

)

t≥0

= (U

)

t≥0

So U satisfies Brownian scaling. Plugging this in, we know

√

κB

t/r

+ at/r

= r

√

κB

+ at.

Since r

√

κB

t/r

= r

√

κB

, we must have a = 0.

This finally gives us the definition of SLE.

Definition

(Schramm–Loewner evolution)

For

κ >

SLE

is the random

family of hulls encoded by

√

κB

, where

is a standard Brownian motion.

When κ = 0, we have U

= 0. This corresponds to the curve γ(t) = 2

√

ti.

Usually, when talking about SLE

, we are instead thinking about a curve.

Theorem

(Rhode–Schramm, 2005)

If (

) is an

SLE

with flow

and driving

function

, then

−1

H → A

extends to a map on

for all

t ≥

0 almost surely.

Moreover, if we set

(

) =

−1

(

), then

H \ A

is the unbounded component of

H \ γ([0, t]).

We will not prove this. From now on,

SLE

will often refer to this curve

instead.

In the remainder of the course, we will first discuss some properties of

SLE

and then “prove” that the scaling limit of certain objects are

SLE

’s. We will not

be providing anything near a full proof of the identification of the scaling limit.

Instead, we will look at certain random processes, and based on the properties of

the random process, we deduce that the scaling limit “ought to” have a certain

analogous property. We will then show that for a very particular choice of

the curve

SLE

has that property. We can then deduce that the scaling limit is

SLE

for this κ.

What we will do is to prove properly that the

SLE

does indeed have

the desired property. What we will not do is to prove that the scaling limit

exists, satisfies the conformal Markov property and actually satisfies the desired

properties (though the latter two should be clear once we can prove the scaling

limit exists).

3 Review of stochastic calculus

Before we start studying SLE, we do some review of stochastic calculus. In

stochastic calculus, the basic object is a continuous semi-martingale. We write

these as

= M

+ A

where

is a continuous local martingale and

is continuous with bounded

variation.

The important concepts we will review are

(i) Stochastic integrals

(ii) Quadratic variation

(iii) Itˆo’s formula

(iv) L´evy characterization of Brownian motion

(v) Stochastic differential equations

The general setting is that we have a probability space (Ω

, F, P

) with a

continuous time filtration

, which satisfies the “usual conditions”, namely

contains all P-null sets, and F

is right continuous, i.e.

s>t

Stochastic integral

is a continuous semi-martingale, and

is a previsible process,

we set

where the first integral is the Itˆo integral, and the second is the Lebesgue–Stieltjes

integral. The first term is a continuous local martingale, and the second is a

continuous bounded variation process.

The Itˆo integral is defined in the same spirit as the Riemann integral. The

key thing that makes it work is that there is “extra cancellation” in the definition,

from the fact that we are integrating against a martingale, and so makes the

integral converge even though the process we are integrating against is not of

bounded variation.

Quadratic variation

If M is a continuous local martingale, then the quadratic variation is

[M]

= lim

n→∞

te−1

k=0

(k+1)2

−n

− M

−n

)

and is the unique continuous non-decreasing process of bounded variation such

that

−

[

]

is a continuous local martingale. Applying the same definition

to a bounded variation process always gives zero, so if

is a semi-martingale, it

makes sense to define

[X]

= [M + A]

= [M]

Also, we have

( · )

d[M]

where the integral on the right is the Lebesgue–Stieltjes integral.

Itˆo’s formula

Itˆo’s formula is the stochastic calculus’ analogue of the fundamental theorem of

calculus. It takes a bit of work to prove Itˆo’s formula, but it is easy to understand

what the intuition is. If f ∈ C

, then we have

f(t) = f(0) +

k=1

(f(t

) − f(t

k−1

))

for any partition 0 =

< ··· < t

. If we were to do a regular fundamental

theorem of calculus, then we can perform a Taylor approximation

f(t) = f(0) +

k=1

k−1

)



− t

k−1

) + o(t

− t

k−1

)



→ f (0) +

(s) ds

as max |t

− t

k−1

| → 0.

In Itˆo’s formula, we want to do the same thing but with Brownian motion.

Suppose B is a Brownian motion. Then

f(B

) = f(B

) +

k=1



f(B

) − f(B

k−1

)



When we do a Taylor expansion of this, we have to be a bit more careful than

before, as we cannot ignore all the higher order terms. The point is that the

quadratic variation of B

is non-zero. We write this as

f(0)+

k=1

k−1

)(B

−B

k−1

)(B

−B

k−1

)

+o((B

k−1

)

Taking the limit, and using that E[(B

− B

k−1

)

] = t

− t

k−1

, we get

f(B

) = f(0) +

) dB

) ds.

More generally, if

is a continuous semi-martingale, and

f ∈

1,2

× R), so that it is C

in the first variable and C

in the second, then

f(t, X

) = f(0, X

) +

∂

f(s, X

) ds +

∂

f(s, X

) dA

∂

f(s, X

) dM

∂

f(s, X

) d[M]

L´evy characterization of Brownian motion

Theorem

(L´evy characterization)

Let

is a continuous local martingale with

[M]

= t for t ≥ 0, then M

is a standard Brownian motion.

Proof sketch.

Use Itˆo’s formula with the exponential moment generating function

iθM

+θ

/2[M]

Stochastic differential equations

Suppose (Ω

, F, P

) and (

) are as before, and (

) is a Brownian motion adapted

to (

). We say a process

satisfies satisfies the stochastic differential equation

= b(X

) dt + σ(X

) dB

for functions b, σ iff

b(X

) ds +

σ(X

) dB

+ X

for all

t ≥

0. At the end of the stochastic calculus course, we will see that there

is a unique solution to this equation provided b, σ are Lipschitz functions.

This in particular implies we can solve

∂

(z) =

(z) − U

, g

(z) = z

for U

√

κB

, where B

is a standard Brownian motion.

4 Phases of SLE

We now study some basic properties of

SLE

. Our first goal will be to prove the

following theorem:

Theorem. SLE

is a simple curve if κ ≤ 4, and is self-intersecting if κ > 4.

When proving this theorem, we will encounter the Bessel stochastic differential

equation, and so we shall begin by understanding this SDE and the associated

Bessel process.

Definition

(Square Bessel process)

Let

= (

, . . . , B

) be a

-dimensional

standard Brownian motion. Then

= kX

= (B

)

+ (B

)

+ ··· + (B

)

is a square Bessel process of dimension d.

The square Bessel process satisfies a stochastic differential equation, which

we can obtain from Itˆo’s formula. Itˆo’s formula tells us

= 2B

+ ··· + 2B

+ d · dt.

For reasons that will become clear soon, we define

+ ··· + B

1/2

Then tautologically, we have

= 2Z

1/2

+ d · dt.

Observe that

is a continuous local martingale, and the quadratic variation is

equal to

[Y ]

)

+ ··· + (B

)

ds = t.

By the L´evy characterization,

is a standard Brownian motion. So we can

write

Lemma.

= 2Z

1/2

+ d · dt.

where

B is a standard Brownian motion.

This is called the square Bessel stochastic differential equation of dimension

Ultimately, we are interested in the Bessel process, i.e. the square root of the

square Bessel process.

Definition

(Bessel process)

The Bessel process of dimension

, written BES

= Z

1/2

Applying Itˆo’s formula, we have

−1/2

−

−3/2

d[Z]

The square Bessel SDE lets us understand the first term, and we can calculate

[Z]

= 4Z

Simplifying, this becomes

Lemma.



d − 1



−1

dt + d

This is the Bessel stochastic differential equation of dimension d.

Observe that we can make sense of this equation for any

d ∈ R

, not necessarily

integral. However, the equation only makes sense when

= 0, and

−1

fails

to be Lipschitz as

→

0. To avoid this problem, we say this is defined up to

when it hits 0.

Proposition. Let d ∈ R, and U

a BES

(i) If d < 2, then U

hits 0 almost surely.

(ii) If d ≥ 2, then U

doesn’t hit 0 almost surely.

This agrees with what we expect from the recurrence of Brownian motion,

for integral

. In particular, for

= 2, a Brownian motion gets arbitrarily close

to 0 infinitely often, so we expect that if we are just a bit lower than

= 2, we

would hit 0 almost surely.

Proof. The proof is similar to the proof of recurrence of Brownian motion. For

a ∈ R

, we define

= inf{t ≥ 0 : U

= a}.

We then consider

[

< τ

], and take the limit

a →

0 and

b → ∞

. To do so, we

claim

Claim. U

2−d

is a continuous local martingale.

To see this, we simply compute using Itˆo’s formula to get

2−d

= (2 − d)U

1−d

(2 − d)(1 − d)U

−d

d[U]

= (2 − d)U

1−d

(2 − d)(d − 1)

1−d

dt +

(2 − d)(1 − d)U

−d

= (2 − d)U

1−d

Therefore

is a continuous local martingale. Since

t∧τ

∧τ

is bounded, it is a

true martingale, and so optional stopping tells us

2−d

= E[U

2−d

∧τ

] = a

2−d

P[τ

< τ

] + b

2−d

P[τ

< τ

– If d < 2, we set a = 0, and then

2−d

= b

2−d

P[τ

< τ

Dividing both sides as b, we find that





2−d

= P[τ

< τ

Taking the limit b → ∞, we see that U

hits 0 almost surely.

– If d > 2, then we have

P[τ

< τ

] =





2−d

−





2−d

P[τ

< τ

] → 0

as a → 0 for any b and U

> 0. So we are done in this case.

–

= 2, then our martingale is just constant, and this analysis is useless.

In this case, we consider

log U

and perform the same analysis to obtain

the desired conclusion.

Now take an SLE

curve. For x ∈ R, consider the process

= g

(x) − U

= g

(x) − g

(γ(t)).

Essentially by definition,

starts being in

when

= 0 (formally, we need

to take limits, since x is not in H). We thus define

= inf{t ≥ 0 : V

= 0}.

This is then the time

cuts

off from

∞

. We thus want to understand

[

< ∞

We can calculate

= d(g

(x) − U

) =

(x) − U

dt −

√

κ dB

dt −

√

κ dB

This looks almost like the Bessel SDE, but we need to do some rescaling to write

this as



√



2/κ

√

dt + d

= −B

So we get that

√

κ ∼ BES

, with

= 1 + 4

/κ

. Thus, our previous result

implies

P[τ

< ∞] =

(

1 κ > 4

0 κ ≤ 4

Theorem. SLE

is a simple curve if κ ≤ 4, and is self-intersecting if κ > 4.

Proof.

κ ≤

4, consider the probability that

(

) hits

([0

, n

]). This is

equivalently the probability that

(

) hits

∂A

. This is bounded above by

the probably that

(

))

− U

hits

∂H

. But by the conformal Markov

property, g

(γ(t + n)) −U

is an SLE

. So this probability is 0.

κ >

4, we want to reverse the above argument, but we need to be a bit

careful in taking limits. We have

lim

n→∞

P[γ(t) ∈ [−n, n] for some t] = 1.

On the other hand, for any fixed n, we have

lim

m→∞

P[g

) − U

⊇ [−n, n]] = 1.

The probability that SLE

self-intersects is

≥ P[g

) − U

⊇ [−n, n] and g

(γ(t + m)) −U

∈ [−n, n] for some t].

By the conformal Markov property,

(

))

− U

is another

SLE

given

. So this factors as

P[g

) − U

⊇ [−n, n]]P[γ(t) ∈ [−n, n] for some t].

Since this is true for all

and

, we can take the limit

m → ∞

, then the limit

n → ∞ to obtain the desired result.

It turns out there are two further regimes when κ > 4.

Theorem. If κ ≥ 8, then SLE

is space-filling, but not if κ ∈ (4, 8).

This will be shown in the second example sheet. Here we prove a weaker

result:

Proposition. SLE

fills ∂H iff κ ≥ 8.

To phrase this in terms of what we already have, we fix

κ >

4. Then for any

0 < x < y, we want to understand the probability

g(x, y) = P[τ

= τ

where we assume 0

< x < y

. If

, this means the curve

cuts

and

from

∞

ft the same time, and hence doesn’t hit [

x, y

]. Conversely, if

[

] = 0,

then with probability one,

hits (

x, y

), and this is true for all

and

. Thus,

we want to show that g(x, y) > 0 for κ ∈ (4, 8), and g(x, y) = 0 for κ ≥ 8.

To simplify notation, observe that

(

x, y

) =

, y/x

) as

SLE

is scale-

invariant. So we may assume x = 1. Moreover, we know

g(1, r) → 0 as r → ∞,

since P[τ

< t] → 1 as t → ∞, but P[τ

< t] → 0 as r → ∞ for t fixed.

Definition (Equivalent events). We say two events A, B are equivalent if

P[A \ B] = P[B \ A] = 0,

Proposition. For r > 1, the event {τ

= τ

} is equivalent to the event

sup

t<τ

− V

< ∞. (∗)

Proof.

If (

∗

) happens, then we cannot have that

< τ

, or else the supremum

is infinite. So (

∗

)

⊆ {τ

}

. The prove the proposition, we have to show that



= τ

, sup

t<τ

− V

= ∞



= 0.

For M > 0, we define

= inf



t ≥ 0 :

− V

≥ M



It then suffices to show that

≡ P



= τ



sup

t<τ

− V

≥ M



= P[τ

= τ

| σ

< τ

] → 0 as M → ∞.

But at time

, we have

= (

+ 1)

, and

if these are cut off at

the same time. Thus, the conformal Markov property implies

= g(1, M + 1).

So we are done.

So we need to show that



sup

t<τ

− V

< ∞



(

> 0 κ ∈ (4, 8)

= 0 k ≥ 8

We will prove this using stochastic calculus techniques. Since ratios are hard, we

instead consider

= log



− V



We can then compute



− d



)



d − 1



− V

)



dt −

, Z

= log(r −1).

Here we can already see why κ = 8 is special, since it corresponds to d =

When faced with such complex equations, it is usually wise to perform a

time change. We define σ(t) by the equation

σ(0) = 0, dt =

dσ(t)

σ(t)

)

So the map

t 7→

= −

σ(t)

is a continuous local martingale, with

[

−

σ(t)

)

ds = t.

So by the L´evy characterization, we know that

is a Brownian motion.

Now let

= Z

σ( t)

Then we have



− d





d − 1



σ(t)

− V

σ(t)

dt + d

In integral form, we get



− d



t +

d − 1

σ(s)

− V

σ(s)

≥



− d



Now if κ ≥ 8, then d = 1 +

≤

. So we have

≥

So we find that

sup

≥

+ sup

= +∞.

So we find that

sup

t<τ

− V

= ∞.

So g(x, y) = 0 for all 0 < x < y if κ ≥ 8.

Now if κ ∈ (4, 8), we pick ε > 0 and set

r = 1 +

Then we have

log

(

r −

1) =

log

(

ε/

2). We will show that for small

, we have

, r

)

0. In fact, it is always positive, which is to be shown on the example

sheet.

We let

τ = inf{t > 0 :

= log ε}.

Then

t∧τ



− d



t ∧ τ +

d − 1

t∧τ

σ(s)

− V

σ(s)

≤

t∧τ



− d



t ∧ τ +

d − 1

t∧τ

≤

t∧τ



− d



t ∧ τ −



d − 1



(t ∧ τ)ε

t +

t∧τ



− d





d − 1





(t ∧ τ).

We let

∗

t +



− d +

d − 1



We then have shown that

∗

t∧τ

≥

t∧τ

We assume that

ε >

0 is such that



− d +

d−1



0. Then

∗

is a Brownian

motion with a negative drift, starting from

log

, so (by the second example

sheet), we have



sup

t≥0

∗

< log ε



> 0.

So we know that



sup

t≥0

< log ε



> 0.

So we have



1, 1 +



> 0.

This concludes the proof of the theorem.

SLE on other domains

So far, we only defined

SLE

in the upper half plane

from 0 to

∞

. But we

can define SLE in really any simply connected domain. If

D ⊆ C

is a simply

connected domain, and

x, y ∈ ∂D

are distinct, then there exists a conformal

transformation ϕ : H → D with ϕ(0) = x and ϕ(∞) = y.

SLE

from

is defined by setting it to be

(

˜γ

), where

˜γ

is an

SLE

from 0 to

∞

. On the second example sheet, we check that this

definition is well-defined, i.e. it doesn’t depend on the choice of the conformal

transformation.

5 Scaling limit of critical percolation

The reason we care about SLE is that they come from scaling limits of certain

discrete models. In this chapter, we will “show” that

SLE

corresponds to the

scaling limit of critical percolation on the hexagonal lattice.

Let

D ⊆ C

be simply connected, and

x, y ∈ ∂D

distinct. Pick a hexagonal

lattice in

with hexagons of size

such that

and

are lattice points. We

perform critical percolation on the hexagonal lattice, i.e. for each hexagon, we

colour it black or white with probability

. We enforce the condition that

the hexagons that intersect the clockwise arc of

∂D

from

to all be black,

and those along the counterclockwise arc must be white.

Then there exists a unique interface

that connects

with the property

that the black hexagons on its left and white on its right. It was conjectured

(and now proved by Smirnov) that the limit of the law of

exists in distribution

and is conformally invariant.

This means that if

is another simply connected domain, and

˜x, ˜y ∈ ∂

are

distinct, then

D →

is a conformal transformation with

(

) =

˜x

(

) =

˜y

then

(

) is equal in distribution of the scaling limit of percolation in

from

˜x

to ˜y.

Also, percolation also satisfies a natural Markov property: if we condition on

up to a given time

, then the rest of

is a percolation exploration in the

remaining domain. The reason for this is very simple — the path only determines

the colours of the hexagons right next to it, and the others are still randomly

distributed independently.

If we assume these two properties, then the limiting path

satisfies the

conformal Markov property, and so must be an

SLE

. So the question is, what

is κ?

To figure out this

, we observe that the scaling limit of percolation has a

locality property, and we will later see that

SLE

is the only SLE that satisfies

this locality property.

To explain the locality property, fix a simply-connected domain

(for

simplicity), and assume that 0

∈ ∂D

. Fixing a point

y ∈ ∂D

, we perform the

percolation exploration as before. Then the resulting path would look exactly

the same as if we performed percolation on

(with black boundary on

and

white boundary on

), up to the point we hit

∂D \ ∂H

. In other words,

doesn’t feel the boundary conditions until it hits the boundary. This is known

as locality.

It should then be true that the corresponding

SLE

should have an analogous

property. To be precise, we want to find a value of

so that the following is

true: If

is an

SLE

from 0 to

∞

, run up until it first hits

∂D \ ∂H

, then

(

) is a (stopped)

SLE

from 0 to

∞

where

D → H

is a conformal

transformation with

(0) = 0,

(

) =

∞

. This is the locality property. We will

show that locality holds precisely for κ = 6.

Suppose that (A

) ∈ A has Loewner driving function U

. We define

= ψ(A

Then

is a family of compact

-hulls that are non-decreasing, locally growing,

and

∅

. However, in general, this is not going to be parametrized by capacity.

On the second example sheet, we show that this has half plane capacity

˜a(t) = hcap(

) =

γ(ψ

))

ds,

which should be somewhat believable, given that hcap(rA) = r

hcap(A).

has a “driving function”

given by

= ψ

), ψ

= ˜g

◦ ψ ◦ g

−1

, g

= g

We then have

∂

˜g

(z) =

∂

˜a

˜g

(z) −

, ˜g

(z),

To see this, simply recall that if A

is γ([0, t]) for a curve γ, then U

= g

(γ(t)).

To understand

, it is convenient to know something about ψ

Proposition. The maps (ψ

) satisfy

∂

(z) = 2



(ψ

))

(z) − ψ

)

−

)

z − U



In particular, at z = U

, we have

∂

) = lim

z→U

∂

(z) = −3ψ

Proof. These are essentially basic calculus computations.

To get Loewner’s equation in the right shape, we perform a time change

σ(t) = inf



u ≥ 0 :

(ψ

))

ds = t



Then we have

∂

˜g

σ( t)

(z) =

˜g

σ(t)

−

σ(t)

, ˜g

(z) = z.

It then remains to try to understand d

σ(t)

. Note that so far what we said

works for general

. If we put

√

κB

, where

is a standard Brownian

motion, then Itˆo’s formula tells us before the time change, we have

= dψ

)



∂

) +

)



dt +

√

κψ

) dB

κ − 6

) dt +

√

κψ

) dB

After the time change, we get

σ(t)

κ − 6

σ(t)

)

σ(t)

)

dt +

√

κ d

where

σ( t)

) dB

is a standard Brownian motion by the L´evy characterization and the definition

of σ(t).

The point is that when κ = 6, the drift term goes away, and so

σ(t)

√

So (

σ(t)

) is an SLE

. Thus, we have proved that

Theorem.

is an

SLE

, then

(

) is an

SLE

up until hitting

(

∂D \ ∂H

)

if and only if κ = 6.

So κ = 6 is the only possible SLE

which could be the limit of percolation.

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

7 The Gaussian free field

We end by discussing the Gaussian free field, which we can think of as a two-

dimensional analogue of Brownian motion, i.e. a random surface. We will show

that the level curves of the Gaussian free field are SLE

To define the Gaussian free field, we have to do some analysis.

Notation.

– C

∞

is the space of infinitely differentiable functions on C

– C

∞

is the space of functions in C

∞

with compact support.

– If D is a domain, C

∞

(D) is the functions in C

∞

supported in D.

Definition

(Dirichlet inner product)

Let

f, g ∈ C

∞

. The Dirichlet inner

product of f, g is

(f, g)

∇

2π

∇f(x) · ∇g(x) dx.

This defines an inner product function on

∞

. If

D ⊆ C

is a non-trivial

simply-connected domain, i.e. not ∅ or C, we can define

Definition

(

))

We write

(

) for the Hilbert space completion of

∞

(D) with respect to ( ·, ·)

∇

Elements of

(

) can be thought of as functions well-defined up to a null

set. These functions need not be continuous (i.e. need not have a continuous

representatives), and in particular need not be genuinely differentiable, but they

have “weak derivatives”.

We will need the following key properties of H

(D):

Proposition.

(i)

Conformal invariance: Suppose

D →

is a conformal transformation,

and f, g ∈ C

∞

(D). Then

(f, g)

∇

= (f ◦ ϕ

−1

, g ◦ ϕ

−1

)

∇

In other words, the Dirichlet inner product is conformally invariant.

In other words,

∗

(

)

→ H

(

) given by

f 7→ f ◦ ϕ

−1

is an

isomorphism of Hilbert spaces.

(ii)

Inclusion: Suppose

U ⊆ D

is open. If

f ∈ C

∞

(

), then

f ∈ C

∞

(

Therefore the inclusion map

(

)

→ H

(

) is well-defined and

associates

(

) with a subspace of

(

). We write the image as

supp

(U).

(iii) Orthogonal decomposition: If U ⊆ D, let

harm

(U) = {f ∈ H

(D) : f is harmonic on U}.

Then

(D) = H

supp

(U) ⊕ H

harm

(U)

is an orthogonal decomposition of

(

). This is going to translate to a

Markov property of the Gaussian free field.

Proof.

Conformal invariance is a routine calculation, and inclusion does not

require proof. To prove orthogonality, suppose

f ∈ H

supp

(

) and

g ∈ H

harm

(

Then

(f, g) =

2π

∇f(x) · ∇g(x) dx = −

2π

f(x)∆g(x) dx = 0.

since f is supported on U and ∆g is supported outside of U.

To prove that they span, suppose

f ∈ H

(

), and

the orthogonal projec-

tion of

onto

supp

(

). Let

f −f

. We want to show that

is harmonic.

It would be a straightforward manipulation if we can take ∆, but there is no

guarantee that f

is smooth.

We shall show that

is weakly harmonic, and then it is a standard analysis

result (which is also on the example sheet) that

is in fact genuinely harmonic.

Suppose ϕ ∈ C

∞

(U). Then since g

⊥ H

supp

(U), we have

0 = (g

, ϕ) =

2π

∇g

(x) · ∇ϕ(x) dx = −

2π

(x)∆ϕ(x) dx.

This implies g

is C

∞

on U and harmonic.

We will define the Gaussian free field to be a Gaussian taking values in

To make this precise, we first understand how Gaussian random variables on

work.

Observe that if

, . . . , α

are iid

1) variables, and

, . . . , e

is a stan-

dard basis of R

, then

h = α

+ ··· + α

is a standard n-dimensional Gaussian random variable.

If x =

∈ R

, then

(h, x) =

j=1

∼ N (0, kxk).

Moreover, if

x, y ∈ R

, then (

h, x

) and (

h, y

) are jointly Gaussian with covariance

(x, y).

Thus, we can associate with

a family of Gaussian random variables (

h, x

)

indexed by

x ∈ R

with mean zero and covariance given by the inner product

on R

. This is an example of a Gaussian Hilbert space.

We now just do the same for the infinite-dimensional vector space

(

One can show that this is separable, and so we can pick an orthonormal basis

). Then the Gaussian free field h on D is defined by

h =

∞

j=1

where the α

’s are iid N(0, 1) random variables. Thus, if f ∈ H

(D), then

(h, f)

∇

∼ N (0, kfk

∇

More generally, if

f, g ∈ H

(

), then (

h, f

)

∇

and (

h, g

)

∇

are jointly Gaussian

with covariance (

f, g

)

∇

. Thus, the Gaussian free field is a family of Gaussian

variables (

H, f

)

∇

indexed by

f ∈ H

(

) with mean zero and covariance (

·, ·

)

∇

We can’t actually quite make this definition, because the sum

∞

j=1

does not converge. So

is not really a function, but a distribution. However,

this difference usually does not matter.

We can translate the properties of

into analogous properties of the

Gaussian free field.

Proposition.

(i)

D →

is a conformal transformation and

is a Gaussian free field

on D, then h ◦ ϕ

−1

is a Gaussian free field on

(ii)

Markov property: If

U ⊆ D

is open, then we can write

with

and

independent where

is a Gaussian free field on

and

harmonic on U.

Proof.

(i) Clear.

(ii)

Take

to be the projection onto

supp

(

). This works since we can take

the orthonormal basis (

) to be the union of an orthonormal basis of

supp

(U) plus an orthonormal basis of H

harm

(U).

Often, we would rather think about the

inner product of

with something

else. Observe that integration by parts tells us

(h, f)

∇

−1

2π

(h, ∆f)

Thus, we would be happy if we can invert ∆, which we can by IB methods.

Recall that

∆(−log |x − y|) = −2πδ(y − x),

where ∆ acts on

, and so

−log |x − y|

is a Green’s function for ∆. Given a

domain

, we wish to obtain a version of the Green’s function that vanishes on

the boundary. To do so, we solve ∆

= 0 on

with the boundary conditions

(y) = −log |x − y| if y ∈ ∂D.

We can then set

G(x, y) = −log |x − y| −

(y).

With this definition, we can define

∆

−1

ϕ(x) = −

2π

G(x, y)ϕ(y) dy.

Then ∆∆

−1

ϕ(x) = ϕ(x), and so

(h, ϕ) ≡ (h, ϕ)

= −2π(h, ∆

−1

ϕ)

∇

Then (h, ϕ) is a mean-zero Gaussian with variance

(2π)

k∆

−1

ϕk

∇

= (2π)

(∆

−1

ϕ, ∆

−1

ϕ)

∇

= −2π(∆

−1

ϕ, ∆∆

−1

ϕ)

= (−2π∆

−1

ϕ, ϕ)

ϕ(x)G(x, y)ϕ(y) dx dy.

More generally, if ϕ, ψ ∈ C

∞

(D), then

cov((h, ϕ), (h, ψ)) =

ϕ(x)G(x, y)ψ(y) dx dy.

On the upper half plane, we have a very explicit Green’s function

G(x, y) = G

(x, y) = −log |x − y| + log |x − ¯y|.

It is not hard to show that the Green’s function is in fact conformally invariant:

Proposition.

Let

be domains in

and

is a conformal transformation

D →

D. Then G

(x, y) = G

(ϕ(x), ϕ(y)).

One way to prove this is to use the conformal invariance of Brownian motion,

but a direct computation suffices as well.

Schramm and Sheffield showed that the level sets of

, i.e.

(

) = 0

}

are

SLE

’s. It takes some work to make this precise, since

is not a genuine

function, but the formal statement is as follows:

Theorem

(Schramm–Sheffield)

Let

. Let

γ ∼ SLE

from 0 to

∞

Let

its Loewner evolution with driving function

√

κB

= 2

, and set

= g

− U

. Fix W ⊆ H open and let

τ = inf{t ≥ 0 : γ(t) ∈ W }.

Let

be a Gaussian free field on

λ >

0, and

𝒽

be the unique harmonic

function on

with boundary values

and

−λ

. Explicitly, it is

given by

𝒽 = λ −

2λ

arg( ·).

Then

h + 𝒽

= (h + 𝒽) ◦ f

t∧τ

where both sides are restricted to W .

A few words should be said about why we should think of this as saying the

level curves of

are

SLE

’s. First, observe that since

𝒽

is harmonic, adding

𝒽

doesn’t change how

pairs with other functions in

(

). All it does is to

change the boundary conditions of

. So we should think of

𝒽

as a version

of the Gaussian free field with boundary conditions

(h + 𝒽)(x) = sgn(x)λ.

It is comforting to know the fact that the boundary conditions of

is well-defined

as an element of L

Similarly, (

𝒽

)

◦f

t∧τ

is a Gaussian free field with the boundary condition

that it takes

to the right of

γ ∼ SLE

and

−λ

to the left of it. Thus, we can

think of this as taking value 0 along

. The theorem says this has the same

distribution just

𝒽

. So we interpret this as saying the Gaussian free field has

the same distribution as a Gaussian free field forced to vanish along an SLE

What we stated here is a simplified version of the theorem by Schramm and

Sheffield, because we are not going to deal with what happens Γ hits W .

The proof is not difficult. The difficult part is to realize that this is the thing

we want to prove.

Proof. We want to show that if ϕ ∈ C

∞

(W ),

((h + 𝒽) ◦ f

t∧τ

, ϕ)

= (h + 𝒽, ϕ).

In other words, writing

(ϕ) = (𝒽 ◦ f

, ϕ), σ

(ϕ) =

ϕ(x)G

(x), f

(y))ϕ(y) dx dy,

we want to show that

((h + 𝒽) ◦ f

t∧τ

, ϕ) ∼ N(m

(ϕ), σ

(ϕ)).

This is the same as proving that

iθ((h+𝒽)◦f

t∧τ

,ϕ)

= exp



iθm

(ϕ) −

(ϕ)



Let F

= σ(U

: s ≤ t) be the filtration of U

. Then

iθ((h+𝒽)◦f

t∧τ

,ϕ)



t∧τ

= E

iθ(h◦f

t∧τ

,ϕ)

| F

t∧τ

iθm

t∧τ

(ϕ)

= exp



iθm

t∧τ

(ϕ) −

t∧τ

(ϕ)



If we knew that

exp



iθm

(ϕ) −

(ϕ)



is a martingale, then taking the expectation of the above equation yields the

desired results.

Note that this looks exactly like the form of an exponential martingale, which

in particular is a martingale. So it suffices to show that

(

) is a martingale

with

(ϕ)]

= σ

(ϕ) − σ

(ϕ).

To check that m

(ϕ) is a martingale, we expand it as

𝒽 ◦ f

(z) = λ −

2λ

arg(f

(z)) = λ −

2λ

im(log(g

(z) − U

)).

So it suffices to check that

log

(

)

−U

) is a martingale. We apply Itˆo’s formula

to get

d log(g

(z)−U

) =

(z) − U

dt−

(z) − U

−

κ/2

(z) − U

)

dt,

and so this is a continuous local martingale when

= 4. Since

(

) is bounded,

it is a genuine martingale.

We then compute the derivative of the quadratic variation

d[m

(ϕ)]

ϕ(x) im



(x) − U





(y) − U



ϕ(y) dx dy dt.

To finish the proof, we need to show that d

(

) takes the same form. Recall

that the Green’s function can be written as

(x, y) = −log |x − y| + log |x − ¯y| = −Re(log(x − y) − log(x − ¯y)).

Since we have

log(f

(x) − f

(y)) = log(g

(x) − g

(y)),

we can compute

d log(g

(x) − g

(y)) =

(x) − g

(y)



(x) − U

−

(y) − U



−2

(x) − U

)(g

(y) − U

)

dt.

Similarly, we have

d log(g

(x) − g

(y)) =

−2

(x) − U

)(g

(y) − U

)

dt.

So we have

(x, y) = −im



(x) − U





(y) − U



dt.

This is exactly what we wanted it to be.