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
κ
(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
D
(which
we will often take to be
H
for convenience), parametrized by a positive real
number
κ
. 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
Z
d
, 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
κ
for some
κ
. If we can
identify exactly which
κ
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
2
The scaling limit of hexagonal percolation is SLE
6
(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
4
’s (see chapter7).
In this course, we will prove some basic properties of
SLE
κ
, and then establish
the last three results/conjectures.
1 Conformal transformations
1.1 Conformal transformations
Definition
(Conformal map)
.
Let
U, V
be domains in
C
. We say a holomorphic
function f : U V is conformal if it is a bijection.
We will write
D
for the open unit disk, and
H
for the upper half plane. An
important theorem about conformal maps is the following:
Theorem
(Riemann mapping theorem)
.
Let
U
be a simply connected domain
with
U 6
=
C
and
z U
be any point. Then there exists a unique conformal
transformation f : D U such that f(0) = z, and f
0
(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
C
are
conformally equivalent.
Example.
Take
U
=
D
. 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
t
= H \ [0, 2
ti]. The map H
t
H given by
z 7→
p
z
2
+ 4t
is a conformal transformation. Observe that this map satisfies
|g
t
(z) z| = |
p
z
2
+ 4t z| 0 as z .
So g
t
(z) z for large z.
Observe also that the family g
t
(z) satisfies the ODE
g
t
=
2
g
t
(z)
, g
0
(z) = z.
We can think of these functions
g
t
as being generated by the curve
γ
(
t
) = 2
ti
,
where for each
t
, the function
g
t
is the conformal transformation that sends
H \ γ([0, t]) to H (satisfying |g
t
(z) z| ).
Given the set of functions
g
t
, we can recover the curve
γ
as follows for
each
z H
, we can ask what is the minimum
t
such that
g
t
is not defined on
z
. By ODE theorems, there is a solution up till the point when
g
t
(
z
) = 0, in
which case the denominator of the right hand side blows up. Call this time
τ
(
z
).
We then see that for each
t
, there is a unique
z
such that
τ
(
z
) =
t
, and we have
γ(t) = z.
More generally, suppose
γ
is any simple (i.e. non-self-intersecting) curve
in
H
starting from 0. Then for each
t
, we can let
g
t
be the unique conformal
transformation that maps
H\γ
([0
, t
]) to
H
with
|g
t
(
z
)
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
t
=
2
g
t
(z) W
t
, g
0
(z) = z.
This is the chordal Loewner equation. We can turn this around given a
function
W
t
, what is the corresponding curve
γ
(
t
)? If
W
= 0, then
γ
(
t
) = 2
ti
.
More excitingly, if we take
W
t
=
κB
t
, where
B
t
is a standard Brownian motion,
and interpret this equation as a stochastic differential equation, we obtain
SLE
κ
.
1.2 Brownian motion and harmonic functions
Recall that a function
f
=
u
+
iv
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
f
:
R
k
R
is harmonic if it is
C
2
and
f =
2
x
2
1
+ ··· +
2
x
2
k
f = 0.
Indeed, we can calculate
2
u
x
2
+
2
u
y
2
=
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
B
=
B
1
+
iB
2
is a
complex Brownian motion if (B
1
, B
2
) is a standard Brownian motion in R
2
.
Recall some results from Advanced Probability:
Theorem.
Let
u
be a harmonic function on a bounded domain
D
which is
continuous on
¯
D
. For
z D
, let
P
z
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
z
[u(B
τ
)].
Corollary
(Mean value property)
.
If
u
is a harmonic function, then, whenever
it makes sense, we have
u(z) =
1
2π
Z
2π
0
u(z + re
) dθ.
Corollary
(Maximum principle)
.
Let
u
be harmonic in a domain
D
. If
u
attains
its maximum at an interior point in D, then u is constant.
Corollary
(Maximum modulus principle)
.
Let
D
be a domain and let
f
:
D C
be holomorphic. If
|f|
attains its maximum in the interior of
D
, then
f
is constant.
Proof.
Observe that if
f
is holomorphic, then
log |f|
is harmonic and attains its
maximum in the interior of
D
. It then follows that
|f|
is constant (if
|f|
vanishes
somewhere, then consider
log |f
+
M|
for some large
M
, 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
f
:
D D
be a holomorphic map with
f
(0) = 0.
Then
|f
(
z
)
| |z|
for all
z D
. If
|f
(
z
)
|
=
|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
f
0
(0) z = 0
.
Then one sees that
g
is holomorphic and
|g
(
z
)
|
1 for all
z D
, hence for all
z D
by the maximum modulus principle. If
|f
(
z
0
)
|
=
|z
0
|
for some
z
0
D \{
0
}
,
then g must be constant, so f is linear.
1.3 Distortion estimates for conformal maps
We write
U
for the collection of conformal transformations
f
:
D D
, where
D
is any simply connected domain with 0
D
and
D 6
=
C
, with
f
(0) = 0 and
f
0
(0) = 1. Thus, it must be of the form
f(z) = z +
X
n=2
a
n
z
n
.
Theorem
(Koebe-1/4 theorem)
.
If
f U
and 0
< r
1, then
B
(0
, r/
4)
f(rD).
This says f cannot look like this:
f
By scaling it suffices to prove this for the case
r
= 1. This follows from the
following result:
Theorem. If f U, then |a
2
| 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
f
:
D D
is in
U
, and
z
0
6∈ D
. We shall
show that |z
0
|
1
4
. Consider the function
˜
f(z) =
z
0
f(z)
z
0
f(z)
.
Since
˜
f
is composition of conformal transformations, it is itself conformal, and a
direct computation shows
˜
f U. Moreover, if
f(z) = z + a
2
z
2
+ ··· ,
then
˜
f(z) = z +
a
2
+
1
z
0
z
2
+ ··· .
So we obtain the bounds
|a
2
|,
a
2
+
1
z
0
2.
By the triangle inequality, we must have |z
1
0
| 4, hence |z
0
|
1
4
.
The 1/4 theorem bounds the distortion in terms of the value of
f
0
(0). Con-
versely, if we know how much the distortion is, we can use the theorem to derive
bounds on f
0
(0).
Corollary.
Let
D,
˜
D
be domains and
z D
,
˜z
˜
D
. If
f
:
D
˜
D
is a conformal
transformation with f(z) = ˜z, then
˜
d
4d
|f
0
(z)|
4
˜
d
d
,
where d = dist(z, D) and
˜
d = dist(˜z,
˜
D).
Proof.
By translation, scaling and rotation, we may assume that
z
=
˜z
= 0,
d = 1 and f
0
(0) = 1. Then we have
˜
D = f (D) f(D) B(0, 1/4).
So
˜
d
1
4
, 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
X
n=1
n|a
n
|
2
.
Proof.
In the ideal world, we will have something that helps us directly compute
area
(
f
(
D
)). However, for the derivation to work, we need to talk about what
f
does to the boundary of
D
, 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
(0
,
1), and define the curve
γ
(
θ
) =
f
(
re
) for
θ
[0
,
2
π
]. Then we
can compute
1
2i
Z
γ
¯z dz =
1
2i
Z
γ
(x iy)(dx + i dy)
=
1
2i
Z
γ
(x iy) dx + (ix + y) dy
=
1
2i
ZZ
f(rD)
2i dx dy
= area(f(rD)),
using Green’s theorem. We can also compute the left-hand integral directly as
1
2i
Z
γ
¯z dz =
1
2i
Z
2π
0
f(re
)f
0
(re
)ire
dθ
=
1
2i
Z
2π
0
X
n=1
¯a
n
r
n
e
iθn
!
X
n=1
a
n
nr
n1
e
(n1)
!
ire
dθ
= π
X
n=1
r
2n
|a
n
|
2
n.
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
K
is a compact hull, then there exists a unique conformal
transformation
F
:
C \
¯
D
to
C \ K
that fixes infinity and has positive derivative
at
, i.e.
lim
z→∞
F
(
z
)
/z >
0. Let
H
be the set of compact hulls containing 0 with
lim
z→∞
F
(
z
)
/z
= 1. If
K H
, then the Laurent expansion of the corresponding
F
is
F (z) = z + b
0
+
X
n=1
b
n
z
n
.
Observe there is a correspondence between the space of such
F
and
U
by sending
F to 1/F (1/z), and vice versa.
Proposition. If K H, then
area(K) = π
1
X
n=1
n|b
n
|
2
!
.
Observe that the area is always non-negative. So in particular, we obtain the
bound
X
n=1
n|b
n
|
2
1.
In particular, |b
1
| 1. This will be how we ultimately bound |a
2
|.
Proof.
The proof is essentially the same as last time. Let
r >
1, and let
K
r
=
F
(
r
¯
D
) (or, if you wish,
C \ F
(
C \ r
¯
D
)), and
γ
(
θ
) =
F
(
re
). As in the
previous proposition, we have
area(K
r
) =
1
2i
Z
γ
¯z dz
=
1
2i
Z
2π
0
F (re
)F
0
(re
)ire
dθ
= π
r
2
X
n=1
n|b
n
|
2
r
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
2
).
Lemma.
Let
f U
. Then there exists an odd function
h U
with
h
(
z
)
2
=
f(z
2
).
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
z 6= 0
f
0
(0) z = 0
.
Then
˜
f
is non-zero and conformal in
D
, and so there is a function
g
with
g(z)
2
=
˜
f(z). We then set
h(z) = zg(z
2
).
Then
h
is odd and
h
2
=
z
2
g
(
z
2
)
2
=
f
(
z
2
). It is also immediate that
h
(0) = 0 and
h
0
(0) = 1. We need to show
h
is injective on
D
. If
z
1
, z
2
D
with
h
(
z
1
) =
h
(
z
2
),
then
z
1
g(z
2
1
) = z
2
g(z
2
2
). ()
By squaring, we know
z
2
1
˜
f(z
2
1
) = z
2
2
˜
f(z
2
)
2
.
Thus,
f
(
z
2
1
) =
f
(
z
2
2
) and so
z
2
1
=
z
2
2
. But then (
) implies
z
1
=
z
2
. So
h
is
injective, and hence h U.
We can now prove the desired theorem.
Proof of theorem. We can Taylor expand
h(z) = z + c
3
z
3
+ c
5
z
5
+ ··· .
Then comparing h(z)
2
= f(z
2
) implies
c
3
=
a
2
2
.
Setting
g
(
z
) =
1
h(1/z)
, we find that the
z
1
coefficient of
g
is
a
2
2
, and as we
previously noted, this must be 1.
1.4 Half-plane capacity
Definition
(Compact
H
-hull)
.
A set
A H
is called a compact
H
-hull if
A
is compact,
A
=
H
¯
A
and
H \ A
is simply connected. We write
Q
for the
collection of compact H-hulls.
A large part of the course will be about studying families of compact
H
-hulls,
and in particular random families of compact
H
-hulls. In this chapter, we will
establish some basic results about compact
H
-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
g
A
: H \ A H with |g
A
(z) z| 0 as z .
This conformal transformation
g
A
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
φ
:
D H
be a conformal transformation which is bounded on
bounded sets and sends
R D
to
R
. Then
φ
extends by reflection to a conformal
transformation on
D
= D {¯z : z D} = D
¯
D
by setting φ(¯z) = φ(z).
Proof of proposition.
The Riemann mapping theorem implies that there exists
a conformal transformation
g
:
H \ A H
. Then
g
(
z
)
as
z
. By the
Schwarz reflection principle, extend
g
to a conformal transformation defined on
C \ (A
¯
A).
By Laurent expanding g at , we can write
g(z) =
N
X
n=−∞
b
N
z
N
.
Since
g
maps the real line to the real line, all
b
i
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
0
+
X
n=1
b
n
z
n
,
with b
1
6= 0. We can then define
g
A
(z) =
g(z) b
0
b
1
.
Since
b
0
and
b
1
are both real, this is still a conformal transformation, and
|g
A
(z) z| 0 as z .
To show uniqueness, suppose
g
A
, g
0
A
are two such functions. Then
g
0
A
g
1
A
:
H H
is such a function for
A
=
. Thus, it suffices to show that if
g
:
H H
is a conformal mapping such that
g
(
z
)
z
0 as
z
, then in fact
g
=
z
.
But we can expand g(z) z as
g(z) z =
X
n=1
c
n
z
n
,
and this has to be holomorphic at 0. So c
n
= 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
A
(z) z).
Thus, we have
g
A
(z) = z +
hcap(A)
z
+
X
n=2
b
n
z
n
.
To justify the name “capacity”,
hcap
(
A
) had better be non-negative and
increasing in
A
. These are both true, as we will soon prove. However, let us first
look at some examples.
Example. z 7→
z
2
+ 4t
is the unique conformal transformation
H\
[0
,
ti
]
H
with |
z
2
+ 4z z| 0 as z . We can expand
p
z
2
+ 4t = z +
2t
z
+ ··· .
Thus, we know that
hcap
([0
,
2
ti
]) = 2
t
. This justifies our previous funny
parametrization of [0,
ti].
Example.
The map
z 7→ z
+
1
z
maps
H\
¯
D H
. Again, we have
z +
1
z
z
0
as z , and so hcap(H
¯
D) = 1.
Proposition.
(i) Scaling: If r > 0 and A Q, then hcap(rA) = r
2
hcap(A).
(ii)
Translation invariance: If
x R
and
a Q
, then
hcap
(
A
+
x
) =
hcap
(
A
).
(iii)
Monotonicity: If
A,
˜
A Q
are such that
A
˜
A
. Then
hcap
(
A
)
hcap
(
˜
A
).
Proof.
(i) We have g
rA
(z) = rg
A
(z/r).
(ii) Observe g
A+x
(z) = g
A
(z x) + x.
(iii) We can write
g
˜
A
= g
g
A
(
˜
A\A)
g
A
.
Thus, expanding out tells us
hcap(
˜
A) = hcap(A) + hcap(g
A
(
˜
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
2
hcap(
¯
D H) = r
2
.
So we know that hcap(A) diam(A)
2
.
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
B
t
be complex Brownian motion. Define the
stopping time
τ = inf{t 0 : B
t
6∈ H \A}.
Then
(i) For all z H \ A, we have
im(z g
A
(z)) = E
z
[im(B
τ
)]
(ii)
hcap(A) = lim
y→∞
y E
y
[im(B
τ
)].
In particular, hcap(A) 0.
(iii) If A
¯
D H, then
hcap(A) =
2
π
Z
π
0
E
e
[im(B
τ
)] sin θ dθ.
Proof.
(i)
Let
φ
(
z
) =
im
(
z g
A
(
z
)). Since
z g
A
(
z
) is holomorphic, we know
φ
is
harmonic. Moreover,
ϕ
is continuous and bounded, as it
0 at infinity.
These are exactly the conditions needed to solve the Dirichlet problem
using Brownian motion.
Since
im
(
g
A
(
z
)) = 0 when
z
(
H \ A
), we know
im
(
B
τ
) =
im
(
B
t
g
A
(B
τ
)). So the result follows.
(ii) We have
hcap(A) = lim
z→∞
z(g
A
(z) z)
= lim
y→∞
(iy)(g
A
(iy) iy)
= lim
y→∞
y im(iy g
A
(iy))
= lim
y→∞
y E
iy
[im(B
τ
)]
where we use the fact that
hcap
(
A
) 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
H
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,
˜
D C
be domains, and
f
:
D
˜
D
a conformal transforma-
tion. Let
B,
˜
B
be Brownian motions starting from
z D, ˜z
˜
D
respectively,
with f (z) = ˜z. Let
τ = inf{t 0 : B
t
6∈ D}
˜τ = inf{t 0 :
˜
B
t
6∈
˜
D}
Set
τ
0
=
Z
τ
0
|f
0
(B
s
)|
2
ds
σ(t) = inf
s 0 :
Z
s
0
|f
0
(B
r
)|
2
dr = t
B
0
t
= f(B
σ(t)
).
Then (B
0
t
: t < ˜τ
0
) has the same distribution as (
˜
B
t
: 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
D
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
) =
1
2π
1 |z|
2
|e
z|
.
Similarly, on H, starting from z = x + iy, the exit distribution is
f(u) =
1
π
y
(x u)
2
+ y
2
.
Note that if x = 0, y = 1, then this is just
f(u) =
1
π
1
u
2
+ 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
t
6∈ H \A}.
Then
If x > Rad(A), then
g
A
(x) = lim
y→∞
πy
1
2
P
iy
[B
τ
[x, )]
.
If x < Rad(A), then
g
A
(x) = lim
y→∞
πy
P
iy
[B
τ
(−∞, x]]
1
2
.
Proof. First consider the case A = and, by symmetry, x > 0. Then
lim
y→∞
πy
1
2
P
iy
[B
τ
[x, )]
= lim
y→∞
πy P
iy
[B
τ
[0, x)]
= lim
y→∞
πy
Z
x
0
y
π(s
2
+ y
2
)
ds
= x,
where the first equality follows from the fact that Brownian motion exits through
the positive reals with probability
1
2
; 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
A
= u
A
+ iv
A
. We let
σ = inf{t > 0 : B
t
6∈ H}.
Then we know
P
iy
[B
τ
[x, )] = P
g
A
(iy)
[B
σ
[g
A
(x), )]
= P
iv
A
(iy)
[B
σ
[g
A
(x) u
A
(iy), )].
Since
g
A
(
z
)
z
0 as
z
, it follows that
v
A
(iy)
y
1 and
u
A
(
iy
)
0 as
y . So we have
P
iv
A
(iy)
B
σ
[g
A
(x) u
A
(iy), )
P
iy
B
σ
[g
A
(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
A
(x) x +
1
x
if x > 1
x +
1
x
g
A
(x) x if x < 1.
Moreover, for all A Q, we have
|g
A
(z) z| 3 Rad(A).
Proof.
Exercise on the first example sheet. Note that
z 7→ z
+
1
z
sends
H \
¯
D
to
H.
This corollary gives us a “zeroth order” bound on
g
A
(
z
)
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
g
a
(z)
z +
hcap(A)
z
c
Rad(A) · hcap(A)
|z|
2
.
Proof. Performing a scaling if necessary, we can assume that Rad(A) = 1. Let
h(z) = z +
hcap(A)
z
g
A
(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
A
(z)) =
im(z)
|z|
2
hcap(A).
Let B be a complex Brownian motion, and let
σ = inf{t 0 : B
t
6∈ H \
¯
D}
τ = inf{t 0 : B
t
6∈ H}.
Let
p
(
z, e
) be the density with respect to the Lebesgue measure at
e
for
B
σ
.
Then by the strong Markov property at the time σ, we have
im(z g
A
(z)) =
Z
π
0
E
e
[im(B
τ
)]p(z, e
) dθ.
In the first example sheet Q3, we show that
p(z, e
) =
2
π
im(z)
|z|
2
sin θ
1 + O
1
|z|

. ()
We also have
hcap(A) =
2
π
Z
π
0
E
e
[im(B
τ
)] sin θ dθ.
So
|v(z)| =
im(z g
A
(z))
im(z)
|z|
2
hcap(A)
=
Z
π
0
E
e
[im(B
τ
)]p(z, e
) dθ
im(z)
|z|
2
·
2
π
Z
π
0
E
e
[im(B
τ
)] sin θ dθ
.
By applying (), we get
|v(z)| c
c hcap(A) im(z)
|z|
3
.
where c is a constant.
Recall that
v
is harmonic as it is the imaginary part of a holomorphic function.
By example sheet 1 Q9, we have
|
x
v(z)|
c hcap(A)
|z|
3
, |
y
v(z)|
c hcap(A)
|z|
3
.
By the Cauchy–Riemann equations,
re
(
h
(
z
)) satsifies the same bounds. So we
know that
|h
0
(z)|
c hcap(A)
|z|
3
.
Then
h(iy) =
Z
y
h
0
(is) ds,
since h(iy) 0 as y . Taking absolute values, we have
|h(iy) =
Z
y
|h
0
(is)| ds
c hcap(A)
Z
y
s
3
ds
c
0
hcap(A)y
z
.
To get the desired bound for a general
z
, integrate
h
0
along the boundary of the
circle of radius |z| to get
|h(z)| = |h(re
)|
c hcap(A)
|z|
2
+ 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
B
be a complex Brownian motion, and
A
¯
D
be connected,
0 A, and A
¯
D 6= . Then for z D, we have
P
z
[B[0, τ] A = ] c|z|
1/2
,
where τ = inf{t 0 : B
t
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 \ 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
|g
A
(z) z| 3Rad(A),
and so
diam(g
A
(
˜
A \ A)) diam(A) + 6Rad(A) 8 diam(
˜
A).
If
d <
1, fix
z H
so that
U
=
B
(
z, d
)
˜
A \ A
. It then suffices to bound
the size of g
A
(U) (or, to be precise, g
A
(U \ A)).
Let B be a complex Brownian motion starting from iy with y 2. Let
τ = inf{t 0 : B
t
6∈ H \A}.
For B[0, τ ] to reach U , it must
(i)
Reach
B
(
z,
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
U
before leaving
H \ A
. By the Beurling estimate,
this occurs with probability cd
1/2
.
Combining the two, we see that
lim sup
y→∞
P
iy
[B[0, τ] U 6= ] cd
1/2
.
By the conformal invariance of Brownian motion, if
σ
=
inf{t
0 :
B
t
6∈
H}, this implies
lim sup
y→∞
y P
y
[B[0, σ] g
A
(
˜
A \ A) 6= ] cd
1/2
.
Since g
A
(
˜
A \ A) is connected, by Q10 of example sheet 1, we have
diam(g
A
(
˜
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 (
A
t
)
t0
is a family of compact
H
-hulls. We say that (
A
t
)
is
(i) non-decreasing if s t implies A
s
A
t
.
(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
s
(A
t
\ A
s
)) ε.
This is a continuity condition.
(iii) parametrized by half-plane capacity if hcap(A
t
) = 2t for all t 0.
We write
A
be the set of families of compact
H
-hulls which satisfy (i) to (iii).
We write A
T
for the set of such families defined on [0, T ].
Example.
If
γ
is is a simple curve in
H
starting from 0 and
A
t
=
γ
[0
, 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
γ
so that
hcap
(
A
t
) = 2
t
for all t 0. Upon doing so, we have A
t
A.
Theorem.
Suppose that (
A
t
)
A
. Let
g
t
=
g
A
t
. Then there exists a continuous
function U : [0, ) R so that
t
g
t
(z) =
2
g
t
(z) U
t
, g
0
(z) = z.
This is known as the chordal Loewner ODE , and
U
is called the driving
function.
Proof.
First note that since the hulls are locally growing, the intersection
T
st
g
t
(
A
s
) consists of exactly one point. Call this point
U
t
. Again by the
locally growing property, U
t
is in fact a continuous function in t.
Recall that if A Q, then
g
A
(z) = z +
hcap(A)
z
+ O
hcap(A)Rad(A)
|z|
2
.
If x R, then as g
A+x
(z) x = g
A
(z x), we have
g
A
(z) = g
A
(z + x) x = z +
hcap(z)
z + x
+ O
hcap(A)Rad(A + x)
|z + x|
2
. ()
Fix ε > 0. For 0 s t, let
g
s,t
= g
t
g
1
s
.
Note that
hcap(g
T
(A
t+ε
\ A
t
)) = 2ε.
Apply (
) with
A
=
g
t
(
A
t+ε
\ A
t
),
x
=
U
t
, and use that
Rad
(
A
+
x
) =
Rad(A U
t
) diam(A) to see that
g
A
(z) = g
t,t+ε
(z) = z +
2ε
z U
t
+ 2ε diam(g
t
(A
t+ε
\ A
t
))O
1
|z U
t
|
2
.
So
g
t+ε
(z) g
t
(z) = (g
t,t+ε
g
t,t
) g
t
(z)
=
2ε
g
t
(z) U
t
+ 2ε diam(g
t
(A
t+ε
\ A
t
))O
1
|g
t
(z) U
t
|
2
.
Dividing both sides by
ε
and taking the limit as
ε
0, the desired result follows
since diam(g
t
(A
t+ε
\ A
t
)) 0.
We can ask about the reverse procedure given a continuous function
U
t
,
can we find a collection
A
t
A
whose driving function is
U
t
? We can simply let
g
t
be the solution to this differential equation. We can then let
A
t
= H \ domain(g
t
).
On the example sheet, we show that this is indeed a family in
A
, and
g
t
are
indeed conformal transformations.
2.2 Schramm–Loewner evolution
We now take
U
t
to be a random process adapted to
F
t
=
σ
(
U
s
:
s t
). We can
then define
g
t
(
z
) as the solution to the Loewner equation for each
z
, and define
A
t
accordingly. We then obtain a random family A
t
in A.
Definition
(Conformal Markov property)
.
We say that (
A
t
) satisfy the confor-
mal Markov property if
(i) Given F
t
, (g
t
(A
t+s
) U
t
)
s0
d
= (A
s
)
s0
.
(ii) Scale-invariance: (rA
t/r
2
)
t0
d
= (A
t
).
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 (
A
t
) satisfy the conformal Markov property, then there
exists κ 0 so that U
t
=
κB
t
, where B is a standard Brownian motion.
Proof.
The first property is exactly the same thing as saying that given
F
t
, we
have
(U
t+s
U
t
)
s0
d
= (U
s
).
So
U
t
is a continuous process with stationary, independent increments. This
implies there exists κ 0 and a R such that
U
t
=
κB
t
+ at,
where B is a standard Brownian motion. Then the second part says
(rU
t/r
2
)
t0
d
= (U
t
)
t0
.
So U satisfies Brownian scaling. Plugging this in, we know
r
κB
t/r
2
+ at/r
d
= r
κB
t
+ at.
Since r
κB
t/r
2
d
= r
κB
t
, we must have a = 0.
This finally gives us the definition of SLE.
Definition
(Schramm–Loewner evolution)
.
For
κ >
0,
SLE
κ
is the random
family of hulls encoded by
U
t
=
κB
t
, where
B
is a standard Brownian motion.
When κ = 0, we have U
t
= 0. This corresponds to the curve γ(t) = 2
ti.
Usually, when talking about SLE
κ
, we are instead thinking about a curve.
Theorem
(Rhode–Schramm, 2005)
.
If (
A
t
) is an
SLE
κ
with flow
g
t
and driving
function
U
t
, then
g
1
t
:
H A
t
extends to a map on
¯
H
for all
t
0 almost surely.
Moreover, if we set
γ
(
t
) =
g
1
t
(
U
t
), then
H \ A
t
is the unbounded component of
H \ γ([0, t]).
We will not prove this. From now on,
SLE
κ
will often refer to this curve
instead.
In the remainder of the course, we will first discuss some properties of
SLE
κ
,
and then “prove” that the scaling limit of certain objects are
SLE
κ
’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
κ
,
the curve
SLE
κ
has that property. We can then deduce that the scaling limit is
SLE
κ
for this κ.
What we will do is to prove properly that the
SLE
κ
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
X
t
= M
t
+ A
t
,
where
M
is a continuous local martingale and
A
is continuous with bounded
variation.
The important concepts we will review are
(i) Stochastic integrals
(ii) Quadratic variation
(iii) Itˆo’s formula
(iv) 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
F
t
, which satisfies the “usual conditions”, namely
F
0
contains all P-null sets, and F
t
is right continuous, i.e.
F
t
=
\
s>t
F
s
.
Stochastic integral
If
X
t
=
M
t
+
A
t
is a continuous semi-martingale, and
H
t
is a previsible process,
we set
Z
t
0
H
s
dX
s
=
Z
t
0
H
s
dM
s
+
Z
t
0
H
s
dA
s
,
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]
t
= lim
n→∞
d2
n
te−1
X
k=0
(M
(k+1)2
n
M
k2
n
)
2
,
and is the unique continuous non-decreasing process of bounded variation such
that
M
2
t
[
M
]
t
is a continuous local martingale. Applying the same definition
to a bounded variation process always gives zero, so if
X
is a semi-martingale, it
makes sense to define
[X]
t
= [M + A]
t
= [M]
t
.
Also, we have
"
Z
( · )
0
H
s
dM
s
#
t
=
Z
t
0
H
2
s
d[M]
s
,
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
2
, then we have
f(t) = f(0) +
n
X
k=1
(f(t
k
) f(t
k1
))
for any partition 0 =
t
0
< ··· < t
n
=
t
. If we were to do a regular fundamental
theorem of calculus, then we can perform a Taylor approximation
f(t) = f(0) +
n
X
k=1
f
0
(t
k1
)
(t
k
t
k1
) + o(t
k
t
k1
)
f (0) +
Z
t
0
f
0
(s) ds
as max |t
k
t
k1
| 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
t
) = f(B
0
) +
n
X
k=1
f(B
t
k
) f(B
t
k1
)
.
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
t
is non-zero. We write this as
f(0)+
n
X
k=1
h
f
0
(B
t
k1
)(B
t
k
B
t
k1
)+
1
2
f
00
(B
t
k1
)(B
t
k
B
t
k1
)
2
+o((B
t
k
+B
t
k1
)
2
)
i
.
Taking the limit, and using that E[(B
t
k
B
t
k1
)
2
] = t
k
t
k1
, we get
f(B
t
) = f(0) +
Z
t
0
f
0
(B
s
) dB
s
+
1
2
Z
t
0
f
00
(B
s
) ds.
More generally, if
X
t
=
M
t
+
A
t
is a continuous semi-martingale, and
f
C
1,2
(R
t
× R), so that it is C
1
in the first variable and C
2
in the second, then
f(t, X
t
) = f(0, X
0
) +
Z
t
0
s
f(s, X
s
) ds +
Z
t
0
x
f(s, X
s
) dA
s
+
Z
t
0
x
f(s, X
s
) dM
s
+
1
2
Z
t
0
2
x
f(s, X
s
) d[M]
s
.
evy characterization of Brownian motion
Theorem
(L´evy characterization)
.
Let
M
t
is a continuous local martingale with
[M]
t
= t for t 0, then M
t
is a standard Brownian motion.
Proof sketch.
Use Itˆo’s formula with the exponential moment generating function
e
iθM
t
+θ
2
/2[M]
t
.
Stochastic differential equations
Suppose (Ω
, F, P
) and (
F
t
) are as before, and (
B
t
) is a Brownian motion adapted
to (
F
t
). We say a process
X
t
satisfies satisfies the stochastic differential equation
dX
t
= b(X
t
) dt + σ(X
t
) dB
t
for functions b, σ iff
X
t
=
Z
t
0
b(X
s
) ds +
Z
t
0
σ(X
s
) dB
s
+ X
0
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
t
g
t
(z) =
2
g
t
(z) U
t
, g
0
(z) = z
for U
t
=
κB
t
, where B
t
is a standard Brownian motion.
4 Phases of SLE
We now study some basic properties of
SLE
κ
. Our first goal will be to prove the
following theorem:
Theorem. SLE
κ
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
X
= (
B
1
, . . . , B
d
) be a
d
-dimensional
standard Brownian motion. Then
Z
t
= kX
t
k
2
= (B
1
t
)
2
+ (B
2
t
)
2
+ ··· + (B
d
t
)
2
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
dZ
t
= 2B
1
t
dB
1
t
+ ··· + 2B
d
t
dB
d
t
+ d · dt.
For reasons that will become clear soon, we define
Y
t
=
Z
t
0
B
1
s
dB
1
s
+ ··· + B
d
s
dB
d
s
Z
1/2
s
.
Then tautologically, we have
dZ
t
= 2Z
1/2
t
dY
t
+ d · dt.
Observe that
Y
t
is a continuous local martingale, and the quadratic variation is
equal to
[Y ]
t
=
Z
t
0
(B
1
s
)
2
+ ··· + (B
d
s
)
2
Z
s
ds = t.
By the evy characterization,
Y
T
is a standard Brownian motion. So we can
write
Lemma.
dZ
t
= 2Z
1/2
t
d
˜
B
t
+ d · dt.
where
˜
B is a standard Brownian motion.
This is called the square Bessel stochastic differential equation of dimension
d.
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
d
, written BES
d
,
is
U
t
= Z
1/2
t
.
Applying Itˆo’s formula, we have
dU
t
=
1
2
Z
1/2
t
dZ
t
1
8
Z
3/2
t
d[Z]
t
The square Bessel SDE lets us understand the first term, and we can calculate
[Z]
t
= 4Z
t
dt
Simplifying, this becomes
Lemma.
dU
t
=
d 1
2
U
1
t
dt + d
˜
B
t
.
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
U
t
6
= 0, and
U
1
t
fails
to be Lipschitz as
U
t
0. To avoid this problem, we say this is defined up to
when it hits 0.
Proposition. Let d R, and U
t
a BES
d
.
(i) If d < 2, then U
t
hits 0 almost surely.
(ii) If d 2, then U
t
doesn’t hit 0 almost surely.
This agrees with what we expect from the recurrence of Brownian motion,
for integral
d
. In particular, for
d
= 2, a Brownian motion gets arbitrarily close
to 0 infinitely often, so we expect that if we are just a bit lower than
d
= 2, we
would hit 0 almost surely.
Proof. The proof is similar to the proof of recurrence of Brownian motion. For
a R
>0
, we define
τ
a
= inf{t 0 : U
t
= a}.
We then consider
P
[
τ
b
< τ
a
], and take the limit
a
0 and
b
. To do so, we
claim
Claim. U
2d
t
is a continuous local martingale.
To see this, we simply compute using Itˆo’s formula to get
dU
2d
t
= (2 d)U
1d
t
dU
t
+
1
2
(2 d)(1 d)U
d
t
d[U]
t
= (2 d)U
1d
t
d
˜
B
t
+
(2 d)(d 1)
2U
t
U
1d
t
dt +
1
2
(2 d)(1 d)U
d
t
dt
= (2 d)U
1d
t
d
˜
B
t
.
Therefore
U
t
is a continuous local martingale. Since
U
tτ
a
τ
b
is bounded, it is a
true martingale, and so optional stopping tells us
U
2d
0
= E[U
2d
τ
a
τ
b
] = a
2d
P[τ
a
< τ
b
] + b
2d
P[τ
b
< τ
a
].
If d < 2, we set a = 0, and then
U
2d
0
= b
2d
P[τ
b
< τ
0
].
Dividing both sides as b, we find that
U
0
b
2d
= P[τ
b
< τ
0
].
Taking the limit b , we see that U
t
hits 0 almost surely.
If d > 2, then we have
P[τ
a
< τ
b
] =
U
0
a
2d
b
a
2d
P[τ
b
< τ
a
] 0
as a 0 for any b and U
0
> 0. So we are done in this case.
If
d
= 2, then our martingale is just constant, and this analysis is useless.
In this case, we consider
log U
t
and perform the same analysis to obtain
the desired conclusion.
Now take an SLE
κ
curve. For x R, consider the process
V
x
t
= g
t
(x) U
t
= g
t
(x) g
t
(γ(t)).
Essentially by definition,
x
starts being in
A
t
when
V
x
t
= 0 (formally, we need
to take limits, since x is not in H). We thus define
τ
x
= inf{t 0 : V
x
t
= 0}.
This is then the time
γ
cuts
x
off from
. We thus want to understand
P
[
τ
x
<
].
We can calculate
dV
x
t
= d(g
t
(x) U
t
) =
2
g
t
(x) U
t
dt
κ dB
t
=
2
V
x
t
dt
κ dB
t
.
This looks almost like the Bessel SDE, but we need to do some rescaling to write
this as
d
V
x
t
κ
=
2
V
×
t
/
κ
dt + d
˜
B
t
,
˜
B
t
= B
t
.
So we get that
V
×
t
/
κ BES
d
, with
d
= 1 + 4
. Thus, our previous result
implies
P[τ
x
< ] =
(
1 κ > 4
0 κ 4
.
Theorem. SLE
κ
is a simple curve if κ 4, and is self-intersecting if κ > 4.
Proof.
If
κ
4, consider the probability that
γ
(
t
+
n
) hits
γ
([0
, n
]). This is
equivalently the probability that
γ
(
t
+
n
) hits
A
n
. This is bounded above by
the probably that
g
n
(
γ
(
t
+
n
))
U
n
hits
H
. But by the conformal Markov
property, g
n
(γ(t + n)) U
n
is an SLE
κ
. So this probability is 0.
If
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
m
(A
m
) U
m
[n, n]] = 1.
The probability that SLE
κ
self-intersects is
P[g
m
(A
m
) U
m
[n, n] and g
m
(γ(t + m)) U
m
[n, n] for some t].
By the conformal Markov property,
g
m
(
γ
(
t
+
m
))
U
m
is another
SLE
κ
given
F
m
. So this factors as
P[g
m
(A
m
) U
m
[n, n]]P[γ(t) [n, n] for some t].
Since this is true for all
m
and
n
, 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
κ
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
κ
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
x
= τ
y
].
where we assume 0
< x < y
. If
τ
x
=
τ
y
, this means the curve
γ
cuts
x
and
y
from
ft the same time, and hence doesn’t hit [
x, y
]. Conversely, if
P
[
τ
x
=
τ
y
] = 0,
then with probability one,
γ
hits (
x, y
), and this is true for all
x
and
y
. 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
g
(
x, y
) =
g
(1
, y/x
) as
SLE
κ
is scale-
invariant. So we may assume x = 1. Moreover, we know
g(1, r) 0 as r ,
since P[τ
1
< t] 1 as t , but P[τ
r
< 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
r
= τ
1
} is equivalent to the event
sup
t<τ
1
V
r
t
V
1
t
V
1
t
< . ()
Proof.
If (
) happens, then we cannot have that
τ
1
< τ
r
, or else the supremum
is infinite. So (
)
{τ
1
=
τ
r
}
. The prove the proposition, we have to show that
P
τ
1
= τ
r
, sup
t<τ
1
V
r
t
V
1
t
V
1
t
=
= 0.
For M > 0, we define
σ
M
= inf
t 0 :
V
r
t
V
1
t
V
1
t
M
.
It then suffices to show that
P
M
P
τ
1
= τ
r
sup
t<τ
1
V
r
t
V
1
t
V
1
t
M
= P[τ
1
= τ
r
| σ
M
< τ
1
] 0 as M .
But at time
σ
M
, we have
V
r
t
= (
M
+ 1)
V
1
t
, and
τ
1
=
τ
r
if these are cut off at
the same time. Thus, the conformal Markov property implies
P
M
= g(1, M + 1).
So we are done.
So we need to show that
P
sup
t<τ
1
V
r
t
V
1
t
V
1
t
<
(
> 0 κ (4, 8)
= 0 k 8
.
We will prove this using stochastic calculus techniques. Since ratios are hard, we
instead consider
Z
t
= log
V
r
t
V
1
t
V
1
t
.
We can then compute
dZ
t
=

3
2
d
1
(V
1
t
)
2
+
d 1
2
V
r
t
V
1
t
(V
1
t
)
2
V
r
t
dt
1
V
1
t
dB
t
, Z
0
= log(r 1).
Here we can already see why κ = 8 is special, since it corresponds to d =
3
2
.
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)
(V
1
σ(t)
)
2
.
So the map
t 7→
˜
B
t
=
Z
σ(t)
0
1
V
1
s
dB
s
is a continuous local martingale, with
[
˜
B]
t
=
"
Z
σ(t)
0
1
V
1
s
dB
s
#
t
=
Z
σ(t)
0
1
(V
1
s
)
2
ds = t.
So by the evy characterization, we know that
˜
B
t
is a Brownian motion.
Now let
˜
Z
t
= Z
σ( t)
.
Then we have
d
˜
Z
t
=
"
3
2
d
+
d 1
2
V
r
σ(t)
V
1
σ(t)
V
r
σ(t)
#
dt + d
˜
B
t
.
In integral form, we get
˜
Z
t
=
˜
Z
0
+
˜
B
t
+
3
2
d
t +
d 1
2
Z
t
0
V
r
σ(s)
V
1
σ(s)
V
r
σ(s)
ds
˜
Z
0
+
˜
B
t
+
3
2
d
t.
Now if κ 8, then d = 1 +
4
κ
3
2
. So we have
˜
Z
t
˜
Z
0
+
˜
B
t
.
So we find that
sup
t
˜
Z
t
˜
Z
0
+ sup
t
˜
B
t
= +.
So we find that
sup
t<τ
1
V
r
t
V
1
t
V
1
t
= .
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 +
ε
2
.
Then we have
˜
Z
0
=
log
(
r
1) =
log
(
ε/
2). We will show that for small
ε
, we have
g
(1
, r
)
>
0. In fact, it is always positive, which is to be shown on the example
sheet.
We let
τ = inf{t > 0 :
˜
Z
t
= log ε}.
Then
˜
Z
tτ
=
˜
Z
0
+
˜
B
tτ
+
3
2
d
t τ +
d 1
2
Z
tτ
0
V
r
σ(s)
V
1
σ(s)
V
r
σ(s)
ds
˜
Z
0
+
˜
B
tτ
+
3
2
d
t τ +
d 1
2
Z
tτ
0
e
˜
Z
s
ds
˜
Z
0
+
˜
B
tτ
+
3
2
d
t τ
d 1
2
(t τ)ε
=
˜
Z
0
t +
˜
B
tτ
+

3
2
d
+
d 1
2
ε
(t τ).
We let
Z
t
=
˜
Z
0
t +
˜
B
t
+
3
2
d +
d 1
2
ε
t.
We then have shown that
Z
tτ
˜
Z
tτ
.
We assume that
ε >
0 is such that
3
2
d +
d1
2
ε
<
0. Then
Z
t
is a Brownian
motion with a negative drift, starting from
log
ε
2
, so (by the second example
sheet), we have
P
sup
t0
Z
t
< log ε
> 0.
So we know that
P
sup
t0
˜
Z
t
< log ε
> 0.
So we have
g
1, 1 +
ε
2
> 0.
This concludes the proof of the theorem.
SLE on other domains
So far, we only defined
SLE
κ
in the upper half plane
H
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.
An
SLE
κ
in
D
from
x
to
y
is defined by setting it to be
γ
=
ϕ
(
˜γ
), where
˜γ
is an
SLE
κ
in
H
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
6
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
D
with hexagons of size
ε
such that
x
and
y
are lattice points. We
perform critical percolation on the hexagonal lattice, i.e. for each hexagon, we
colour it black or white with probability
p
=
1
2
. We enforce the condition that
the hexagons that intersect the clockwise arc of
D
from
x
to
y
to all be black,
and those along the counterclockwise arc must be white.
x
y
Then there exists a unique interface
γ
ε
that connects
x
to
y
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
γ
ε
exists in distribution
and is conformally invariant.
This means that if
˜
D
is another simply connected domain, and
˜x, ˜y
˜
D
are
distinct, then
ϕ
:
D
˜
D
is a conformal transformation with
ϕ
(
x
) =
˜x
,
ϕ
(
y
) =
˜y
,
then
ϕ
(
γ
) is equal in distribution of the scaling limit of percolation in
˜
D
from
˜x
to ˜y.
Also, percolation also satisfies a natural Markov property: if we condition on
γ
ε
up to a given time
t
, then the rest of
γ
ε
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
γ
satisfies the
conformal Markov property, and so must be an
SLE
κ
. So the question is, what
is κ?
To figure out this
κ
, we observe that the scaling limit of percolation has a
locality property, and we will later see that
SLE
6
is the only SLE that satisfies
this locality property.
To explain the locality property, fix a simply-connected domain
D
in
H
(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
H
(with black boundary on
R
<0
and
white boundary on
R
>0
), up to the point we hit
D \ H
. In other words,
γ
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
κ
should have an analogous
property. To be precise, we want to find a value of
κ
so that the following is
true: If
γ
is an
SLE
κ
in
H
from 0 to
, run up until it first hits
D \ H
, then
ψ
(
γ
) is a (stopped)
SLE
κ
in
H
from 0 to
where
ψ
:
D H
is a conformal
transformation with
ψ
(0) = 0,
ψ
(
y
) =
. This is the locality property. We will
show that locality holds precisely for κ = 6.
Suppose that (A
t
) A has Loewner driving function U
t
. We define
˜
A
t
= ψ(A
t
).
Then
˜
A
t
is a family of compact
H
-hulls that are non-decreasing, locally growing,
and
A
0
=
. 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(
˜
A
t
) =
Z
t
0
γ(ψ
0
t
(U
s
))
2
ds,
which should be somewhat believable, given that hcap(rA) = r
2
hcap(A).
˜
A
t
has a “driving function”
˜
U
t
given by
˜
U
t
= ψ
t
(U
t
), ψ
t
= ˜g
t
ψ g
1
t
, g
t
= g
A
t
.
We then have
t
˜g
t
(z) =
t
˜a
t
˜g
t
(z)
˜
U
t
, ˜g
0
(z),
To see this, simply recall that if A
t
is γ([0, t]) for a curve γ, then U
t
= g
t
(γ(t)).
To understand
˜
U
t
, it is convenient to know something about ψ
t
:
Proposition. The maps (ψ
t
) satisfy
t
ψ
t
(z) = 2
(ψ
0
t
(U
t
))
2
ψ
t
(z) ψ
t
(U
t
)
ψ
0
t
(U
t
)
z U
t
.
In particular, at z = U
t
, we have
t
ψ
t
(U
t
) = lim
zU
t
t
ψ
t
(z) = 3ψ
00
t
(U
t
).
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 :
Z
u
0
(ψ
0
s
(U
s
))
2
ds = t
.
Then we have
t
˜g
σ( t)
(z) =
2
˜g
σ(t)
˜
U
σ(t)
, ˜g
0
(z) = z.
It then remains to try to understand d
˜
U
σ(t)
. Note that so far what we said
works for general
U
t
. If we put
U
t
=
κB
t
, where
B
is a standard Brownian
motion, then Itˆo’s formula tells us before the time change, we have
d
˜
U
t
= dψ
t
(U
t
)
=
t
ψ
t
(U
t
) +
κ
2
ψ
00
t
(U
t
)
dt +
κψ
0
t
(U
t
) dB
t
=
κ 6
2
ψ
00
t
(U
t
) dt +
κψ
0
t
(U
t
) dB
t
.
After the time change, we get
d
˜
U
σ(t)
=
κ 6
2
ψ
00
σ(t)
(U
σ(t)
)
ψ
0
σ(t)
(U
σ(t)
)
2
dt +
κ d
˜
B
t
,
where
˜
B
t
=
Z
σ( t)
0
ψ
0
s
(U
s
) dB
s
is a standard Brownian motion by the evy characterization and the definition
of σ(t).
The point is that when κ = 6, the drift term goes away, and so
˜
U
σ(t)
=
6
˜
B
t
.
So (
˜
A
σ(t)
) is an SLE
6
. Thus, we have proved that
Theorem.
If
γ
is an
SLE
κ
, then
ψ
(
γ
) is an
SLE
κ
up until hitting
ψ
(
D \ H
)
if and only if κ = 6.
So κ = 6 is the only possible SLE
κ
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
G
= (
V, E
) be a graph with uniformly
bounded degree, and pick
x V
and
n N
. The self-avoiding in
G
starting
from
x
of length
n
is the uniform measure on simple paths in
G
starting from
x
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
G
=
Z
d
, it has been shown that
for d 5, the scaling limit is just Brownian motion (Hara and Slade).
In
d
= 4, the same is conjectured to be true. In
d
= 3, there is no conjecture
for the continuous object which should describe its scaling limit. In this course,
we are only interested in
d
= 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
6
is the locality property. In self-avoiding walks, the special property
is restriction. This is a very simple concept if
G
0
= (
V
0
, E
0
) is a subgraph
of
G
with
x V
0
, then the self-avoiding walk on
G
conditioned to stay in
G
0
is a self-avoiding walk on
G
0
. 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
x
conditioned to leave
D
at
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
H
, we start with a complex Brownian motion
B
=
B
1
+
iB
2
,
with
B
1
0
= 0 and
B
2
0
=
ε >
0. We then condition
B
on the event that
B
2
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
ˆ
B = (
ˆ
B
1
,
ˆ
B
2
)
where
ˆ
B
1
is a standard Brownian motion and
ˆ
B
2
BES
3
, and the two are
independent.
The key property about Brownian excursion we will need is the following:
Proposition.
Suppose
A
be a compact
H
-hull and
g
A
is as usual. If
x R \ A
,
then
P
x
[
ˆ
B[0, ) A = ] = g
0
A
(x).
Proof.
This is a straightforward computation. Take
z
=
x
+
with
ε >
0, and
let B
t
be a Brownian motion. Define
σ
R
= inf{t 0 : im(B
t
) = R}.
Then the desired probability is
lim
ε0
lim
R→∞
P
z
[B[0, σ
R
] A = | B[0, σ
R
] R = ].
By Bayes’ theorem, this is equal to
lim
ε0
lim
R→∞
P
z
[B[0, σ
R
] (A R) = ]
P[B[0, σ
R
] 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
|g
A
(z) z| 3Rad(A).
Thus, using conformal invariance, we can bound
P
g
A
(z)
[B[0, σ
R+3Rad(A)
] R = ] P
z
[B[0, σ
R
] (A R) = ]
P
g
A
(z)
[B[0, σ
R3Rad(A)
] R = ].
So we get
im(g
A
(z))
R + 3Rad(A)
numerator
im(g
A
(z))
R 3Rad(Z)
.
Combining, we find that the desired probability is
lim
ε0
im(g
A
(x + ))
ε
= g
0
A
(x).
The restriction property
We now return to understanding the restriction property. We assume that
κ
4,
since this is the range of κ-values so that SLE
κ
is simple.
Recall that Q is the set of all compact H-hulls. We define
Q
+
= {A Q :
¯
A (−∞, 0] = ∅}
Q
= {A Q :
¯
A [0, ) = ∅}
For A Q
±
= Q
+
Q
, we define ψ
A
: H A H by
ψ
A
(z) = g
A
(z) g
A
(0).
This is the version of
g
A
that fixes 0, which is what we need. This is the unique
conformal transformation with
ψ
A
(0) = 0, lim
z→∞
ψ
A
(z)
z
= 1.
We will need the following fact about SLE:
Fact. SLE
κ
is transient, i.e. if γ is an SLE
κ
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
κ
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
V
A
= {γ[0, ) A = ∅}.
Definition
(Restriction property)
.
We say an
SLE
κ
satisfies the restriction
property if whenever
γ
is an
SLE
κ
, for any
A Q
±
, the law of
ψ
A
(
γ
) conditional
on V
A
is that of an SLE
κ
curve (for the same κ).
Observe that the law of
γ
is determined by the probabilities
A 7→ P
[
V
A
] for
all A Q
±
.
Lemma. Suppose there exists α > 0 so that
P[V
A
] = (ψ
0
A
(0))
α
for all A Q
±
, then SLE
κ
satisfies restriction.
Proof.
Suppose the assertion in the lemma is true. Suppose that
A, B Q
±
.
Then we have that
P[ψ
A
(γ[0, )) B = | V
A
] =
P[γ[0, ) (ψ
1
A
(B) A) = ]
P[γ[0, ) A = ]
=
(ψ
0
(ψ
1
A
(B)A)
(0))
α
(ψ
0
A
(0))
α
=
(ψ
0
B
(0))
α
(ψ
0
A
(0))
α
(ψ
0
A
(0))
α
= (ψ
0
B
(0))
α
= P[V
B
],
where we used that
ψ
ψ
1
A
(B)A
= ψ
B
ψ
A
.
So the law of ψ
A
(γ) given V
A
is the law of γ.
We now have to show that
SLE
8/3
satisfies the condition in the lemma. Let
F
t
be the filtration of U
t
=
κB
t
. Then
˜
M
t
= P[V
A
| F
t
]
is a bounded martingale with
˜
M
0
= P[V
A
]. Also,
˜
M
t
1
V
A
by the martingale convergence theorem. Also, if we define the stopping time
τ = inf{t 0 : γ(t) A},
then we get
˜
M
T
= P[V
A
| F
t
] = P[V
A
| F
t
]1
{t<τ}
= P[V
g
t
(A)g
t
(0)
]1
{t<τ}
by the conformal Markov property.
Observe that if
M
t
is another bounded
F
t
-martingale with the property that
M
t
1
V
A
as t , then M
t
=
˜
M
t
for all t 0, since
M
t
= E[1
V
A
| F
t
] =
˜
M
t
.
Given what we were aiming for, we consider
M
t
= (ψ
0
g
t
(A)g
t
(0)
(0))
α
1
{t<τ}
Lemma. M
tτ
is a continuous martingale if
κ =
8
3
, α =
5
8
.
These numbers are just what comes out when we do the computations.
Proof.
Recall that we showed that
g
0
A
is a probability involving Brownian ex-
cursion, and in particular is bounded in [0
,
1]. So the same is true for
ψ
0
A
, and
hence
M
tτ
. So it suffices to show that
M
tτ
is a continuous local martingale.
Observe that
M
tτ
= (ψ
0
g
tτ
(A)g
tτ
(0)
(0))
α
So if we define
N
t
= (ψ
0
g
t
(A)g
t
(0)
(0))
α
,
then it suffices to show that
N
t
is a continuous local martingale by optional
stopping. We write
ψ
t
= ˜g
t
ψ
A
g
1
t
,
where ˜g
t
= g
ψ
A
(γ(0,t])
. We then have
N
t
= (ψ
0
t
(U
t
))
α
.
In the example sheet, we show that
t
ψ
0
t
(U
t
) =
ψ
00
t
(U
t
)
2
2ψ
0
t
(U
t
)
4
3
ψ
000
t
(U
t
).
By Itˆo’s formula, we get
dN
t
= αN
t
(α 1)κ + 1
2
ψ
00
t
(U
t
)
2
ψ
0
t
(U
t
)
2
+
κ
2
4
3
ψ
000
t
(U
t
)
ψ
0
t
(U
t
)
dt
+ αN
t
ψ
00
t
(U
t
)
ψ
0
t
(U
t
)
·
κ dB
t
.
Picking
κ
=
8
3
ensures the second d
t
term vanishes, and then setting
α
=
5
8
kills
the first dt term as well, and we are done.
We next want to establish that M
t
1
V
A
as t .
Recall that
M
t
is the probability that a Brownian excursion on
H \ γ
[0
, t
]
from
γ
(
t
) to infinity does not hit
A
. So we would expect this to be true, since
on
V
A
, as
t
increases, the tip of the SLE gets further and further away from
A
,
and so it is difficult for a Brownian excursion to hit
A
; conversely on
V
C
A
, 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
r
< almost surely for all r > 0 since SLE
8/3
is transient.
Lemma. M
tτ
1 on V
A
as t .
Proof.
Let
ˆ
B
be a Brownian excursion in
H \ γ
[0
, σ
r
] from
γ
(
σ
r
) to
. Let
B
be a complex Brownian motion, and
τ
R
= inf{t 0 : im(B
t
) = R}, z = γ(σ
r
) + iε.
Then the probability that
ˆ
B hits A is
1 ψ
0
σ
r
(U
σ
r
) = lim
ε0
lim
R→∞
P
z
[B[0, τ
R
] H \γ[0, σ
r
], B[0, τ
R
] A 6= ]
P
z
[B[0, τ
R
] H \γ[0, σ
r
]]
, ()
We will show that this expression is
Cr
1/2
for some constant
C >
0. Then we
know that
M
σ
r
τ
1 as
r
on
V
A
. This is convergence along a subsequence,
but since we already know that M
tτ
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
im
(
z
) =
r
+ 1 without hitting
R γ
[0
, σ
r
]. Afterwards, the gambler’s ruin
estimate tells us the probability of reaching
im
(
z
) =
R
without going below the
im(z) = r line is
1
Rr
.
In fact, we shall consider the box
S
= [
1
,
1]
2
+
γ
(
σ
r
) of side length 2 centered
at
γ
(
σ
r
). Let
η
be the first time
B
leaves
S
, and we want this to leave via the
top edge
`
. By symmetry, if we started right at
γ
(
σ
r
), then the probability of
leaving at
`
is exactly
1
4
. Thus, if we are at
z
=
γ
(
σ
r
) +
, then the probability
of leaving via ` is >
1
4
.
What we would want to show is that
P
z
[B(η) ` | B[0, η] γ[0, σ
r
] = ] >
1
4
. ()
We then have the estimate
denominator
1
4
· P
z
[B[0, η] γ[0, σ
r
] = ] ·
1
R r
.
Intuitively, (
) must be true, because
γ
[0
, σ
r
] lies below
im
(
z
) =
r
, and so if
B
[0
, η
] doesn’t hit
γ
[0
, σ
r
], then it is more likely to go upwards. To make this
rigorous, we write
1
4
< P
z
[B(η) `]
= P
z
[B(η) ` | B[0, η] γ[0, σ
r
] = ] P[B[0, η] γ[0, σ
r
] = ]
+ P
z
[B(η) ` | B[0, η] γ[0, σ
r
] 6= ] P[B[0, η] γ[0, σ
r
] 6= ]
To prove (
), it suffices to observe that the first factor of the second term is is
1
4
, which follows from the strong Markov property, since
P
w
[
B
(
η
)
`
]
1
4
whenever im(w) r, which in particular is the case when w γ[0, σ
r
].
To bound the numerator, we use the strong Markov property and the Beurling
estimate to get
P
z
[B hits A without hitting R γ[0, σ
r
]] P
z
[B[0, η] γ[0, σ
r
]] · Cr
1/2
.
Combining, we know the numerator in () is
1
R
C · r
1/2
· P[B[0, η] γ[0, σ
r
] = ].
These together give the result.
Lemma. M
tτ
0 as t on V
c
A
.
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
A
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
tτ
ψ
0
t
(U
t
) = 0.
For m N, let
t
m
= inf
t 0 : |γ(t) β(s)| =
1
m
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
L
t
= g
t
(`) U
t
.
Note that a Brownian motion starting from a point on
`
has a uniformly positive
chance of exiting
H \ γ
[0
, t
m
] on the left side of
γ
[0
, t
m
] and on the right side as
well.
On the second example sheet, we see that this implies that
L
t
m
{w : im(w) a|Re(w)|}
for some
a >
0, using the conformal invariance of Brownian motion. Intuitively,
this is because after applying
g
t
U
t
, 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
H
from 0 to
hits L
t
m
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
A
(0))
5/8
.
There is a rather surprising consequence of this computation. Take
γ
1
, . . . , γ
8
to be independent SLE
8/3
’s. Then we have
P[γ
j
[0, ) A = for all j] = (ψ
0
A
(0))
5
.
Note that this is the same as the probability that the hull of
γ
1
, . . . , γ
8
des not
intersect
A
, where the hull is the union of the
γ
j
’s together with the bounded
components of H \
S
j
γ
j
.
In the same manner, if
ˆ
B
1
, . . . ,
ˆ
B
5
are independent Brownian excursions,
then
P[
ˆ
B
j
[0, ) A = for all j] = (ψ
0
A
(0))
5
.
Thus, the hull of γ
1
, . . . , γ
8
has the same distribution as the hull of
ˆ
B
1
, . . . ,
ˆ
B
5
.
Moreover, if we take a boundary point of the hull of, say,
γ
1
, . . . , γ
8
, then
we would expect it to belong to just one of the
SLE
8/3
’s. So the boundary of
the hull of
γ
1
, . . . , γ
8
looks locally like an
SLE
8/3
, and the same can be said for
the hull fo
ˆ
B
1
, . . . ,
ˆ
B
5
. 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
4
s.
To define the Gaussian free field, we have to do some analysis.
Notation.
C
is the space of infinitely differentiable functions on C
C
0
is the space of functions in C
with compact support.
If D is a domain, C
0
(D) is the functions in C
0
supported in D.
Definition
(Dirichlet inner product)
.
Let
f, g C
0
. The Dirichlet inner
product of f, g is
(f, g)
=
1
2π
Z
f(x) · g(x) dx.
This defines an inner product function on
C
0
. If
D C
is a non-trivial
simply-connected domain, i.e. not or C, we can define
Definition
(
H
1
0
(
D
))
.
We write
H
1
0
(
D
) for the Hilbert space completion of
C
0
(D) with respect to ( ·, ·)
.
Elements of
H
1
0
(
D
) 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
1
0
(D):
Proposition.
(i)
Conformal invariance: Suppose
ϕ
:
D
˜
D
is a conformal transformation,
and f, g C
0
(D). Then
(f, g)
= (f ϕ
1
, g ϕ
1
)
In other words, the Dirichlet inner product is conformally invariant.
In other words,
ϕ
:
H
1
0
(
D
)
H
1
0
(
˜
D
) given by
f 7→ f ϕ
1
is an
isomorphism of Hilbert spaces.
(ii)
Inclusion: Suppose
U D
is open. If
f C
0
(
U
), then
f C
0
(
D
).
Therefore the inclusion map
i
:
H
1
0
(
U
)
H
1
0
(
D
) is well-defined and
associates
H
1
0
(
U
) with a subspace of
H
1
0
(
D
). We write the image as
H
supp
(U).
(iii) Orthogonal decomposition: If U D, let
H
harm
(U) = {f H
1
0
(D) : f is harmonic on U}.
Then
H
1
0
(D) = H
supp
(U) H
harm
(U)
is an orthogonal decomposition of
H
1
0
(
D
). 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
(
U
) and
g H
harm
(
U
).
Then
(f, g) =
1
2π
Z
f(x) · g(x) dx =
1
2π
Z
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
1
0
(
D
), and
f
0
the orthogonal projec-
tion of
f
onto
H
supp
(
U
). Let
g
0
=
f f
0
. We want to show that
g
0
is harmonic.
It would be a straightforward manipulation if we can take ∆, but there is no
guarantee that f
0
is smooth.
We shall show that
g
0
is weakly harmonic, and then it is a standard analysis
result (which is also on the example sheet) that
g
0
is in fact genuinely harmonic.
Suppose ϕ C
0
(U). Then since g
0
H
supp
(U), we have
0 = (g
0
, ϕ) =
1
2π
Z
g
0
(x) · ϕ(x) dx =
1
2π
Z
g
0
(x)∆ϕ(x) dx.
This implies g
0
is C
on U and harmonic.
We will define the Gaussian free field to be a Gaussian taking values in
H
1
0
.
To make this precise, we first understand how Gaussian random variables on
R
n
work.
Observe that if
α
1
, . . . , α
n
are iid
N
(0
,
1) variables, and
e
1
, . . . , e
n
is a stan-
dard basis of R
n
, then
h = α
1
e
1
+ ··· + α
n
e
n
is a standard n-dimensional Gaussian random variable.
If x =
P
x
j
e
j
R
n
, then
(h, x) =
n
X
j=1
α
j
x
j
N (0, kxk).
Moreover, if
x, y R
n
, then (
h, x
) and (
h, y
) are jointly Gaussian with covariance
(x, y).
Thus, we can associate with
h
a family of Gaussian random variables (
h, x
)
indexed by
x R
n
with mean zero and covariance given by the inner product
on R
n
. This is an example of a Gaussian Hilbert space.
We now just do the same for the infinite-dimensional vector space
H
1
0
(
D
).
One can show that this is separable, and so we can pick an orthonormal basis
(f
n
). Then the Gaussian free field h on D is defined by
h =
X
j=1
α
j
f
j
,
where the α
j
’s are iid N(0, 1) random variables. Thus, if f H
1
0
(D), then
(h, f)
N (0, kfk
2
).
More generally, if
f, g H
1
0
(
D
), 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
1
0
(
D
) with mean zero and covariance (
·, ·
)
.
We can’t actually quite make this definition, because the sum
h
=
P
j=1
α
j
f
j
does not converge. So
h
is not really a function, but a distribution. However,
this difference usually does not matter.
We can translate the properties of
H
1
0
into analogous properties of the
Gaussian free field.
Proposition.
(i)
If
ϕ
:
D
˜
D
is a conformal transformation and
h
is a Gaussian free field
on D, then h ϕ
1
is a Gaussian free field on
˜
D.
(ii)
Markov property: If
U D
is open, then we can write
h
=
h
1
+
h
2
with
h
1
and
h
2
independent where
h
1
is a Gaussian free field on
U
1
and
h
2
is
harmonic on U.
Proof.
(i) Clear.
(ii)
Take
h
1
to be the projection onto
H
supp
(
U
). This works since we can take
the orthonormal basis (
f
n
) to be the union of an orthonormal basis of
H
supp
(U) plus an orthonormal basis of H
harm
(U).
Often, we would rather think about the
L
2
inner product of
h
with something
else. Observe that integration by parts tells us
(h, f)
=
1
2π
(h, f)
L
2
.
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
x
, and so
log |x y|
is a Green’s function for ∆. Given a
domain
D
, we wish to obtain a version of the Green’s function that vanishes on
the boundary. To do so, we solve
˜
G
x
= 0 on
D
with the boundary conditions
˜
G
x
(y) = log |x y| if y D.
We can then set
G(x, y) = log |x y|
˜
G
x
(y).
With this definition, we can define
1
ϕ(x) =
1
2π
Z
G(x, y)ϕ(y) dy.
Then ∆∆
1
ϕ(x) = ϕ(x), and so
(h, ϕ) (h, ϕ)
L
2
= 2π(h,
1
ϕ)
.
Then (h, ϕ) is a mean-zero Gaussian with variance
(2π)
2
k
1
ϕk
2
= (2π)
2
(∆
1
ϕ,
1
ϕ)
= 2π(∆
1
ϕ, ∆∆
1
ϕ)
= (2π
1
ϕ, ϕ)
=
ZZ
ϕ(x)G(x, y)ϕ(y) dx dy.
More generally, if ϕ, ψ C
0
(D), then
cov((h, ϕ), (h, ψ)) =
ZZ
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
H
(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
D,
˜
D
be domains in
C
and
ϕ
is a conformal transformation
D
˜
D. Then G
D
(x, y) = G
˜
D
(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
h
, i.e.
{x
:
h
(
x
) = 0
}
are
SLE
4
’s. It takes some work to make this precise, since
h
is not a genuine
function, but the formal statement is as follows:
Theorem
(Schramm–Sheffield)
.
Let
λ
=
π
2
. Let
γ SLE
4
in
H
from 0 to
.
Let
g
t
its Loewner evolution with driving function
U
t
=
κB
t
= 2
B
t
, and set
f
t
= g
t
U
t
. Fix W H open and let
τ = inf{t 0 : γ(t) W }.
Let
h
be a Gaussian free field on
H
,
λ >
0, and
𝒽
be the unique harmonic
function on
H
with boundary values
λ
on
R
>0
and
λ
on
R
<0
. Explicitly, it is
given by
𝒽 = λ
2λ
π
arg( ·).
Then
h + 𝒽
d
= (h + 𝒽) f
tτ
,
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
h
are
SLE
4
’s. First, observe that since
𝒽
is harmonic, adding
𝒽
to
h
doesn’t change how
h
pairs with other functions in
H
1
0
(
H
). All it does is to
change the boundary conditions of
h
. So we should think of
h
+
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
h
is well-defined
as an element of L
2
.
Similarly, (
h
+
𝒽
)
f
tτ
is a Gaussian free field with the boundary condition
that it takes
λ
to the right of
γ SLE
4
and
λ
to the left of it. Thus, we can
think of this as taking value 0 along
γ
. The theorem says this has the same
distribution just
h
+
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
4
.
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
0
(W ),
((h + 𝒽) f
tτ
, ϕ)
d
= (h + 𝒽, ϕ).
In other words, writing
m
t
(ϕ) = (𝒽 f
t
, ϕ), σ
2
0
(ϕ) =
ZZ
ϕ(x)G
H
(f
t
(x), f
t
(y))ϕ(y) dx dy,
we want to show that
((h + 𝒽) f
tτ
, ϕ) N(m
0
(ϕ), σ
2
0
(ϕ)).
This is the same as proving that
E
h
e
((h+𝒽)f
tτ
)
i
= exp
m
0
(ϕ)
θ
2
2
σ
2
0
(ϕ)
.
Let F
t
= σ(U
s
: s t) be the filtration of U
t
. Then
E
h
e
((h+𝒽)f
tτ
)
F
tτ
i
= E
h
e
(hf
tτ
)
| F
tτ
i
e
iθm
tτ
(ϕ)
= exp
m
tτ
(ϕ)
θ
2
2
σ
2
tτ
(ϕ)
,
If we knew that
exp
m
t
(ϕ)
θ
2
2
σ
2
t
(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
m
t
(
ϕ
) is a martingale
with
[m
·
(ϕ)]
t
= σ
2
0
(ϕ) σ
2
t
(ϕ).
To check that m
t
(ϕ) is a martingale, we expand it as
𝒽 f
t
(z) = λ
2λ
π
arg(f
t
(z)) = λ
2λ
π
im(log(g
t
(z) U
t
)).
So it suffices to check that
log
(
g
t
(
z
)
U
t
) is a martingale. We apply Itˆo’s formula
to get
d log(g
t
(z)U
t
) =
1
g
t
(z) U
t
·
2
g
t
(z) U
t
dt
1
g
t
(z) U
t
dU
t
κ/2
(g
t
(z) U
t
)
2
dt,
and so this is a continuous local martingale when
κ
= 4. Since
m
t
(
ϕ
) is bounded,
it is a genuine martingale.
We then compute the derivative of the quadratic variation
d[m
·
(ϕ)]
t
=
Z
ϕ(x) im
2
g
t
(x) U
t
im
2
g
t
(y) U
t
ϕ(y) dx dy dt.
To finish the proof, we need to show that d
σ
2
t
(
ϕ
) takes the same form. Recall
that the Green’s function can be written as
G
H
(x, y) = log |x y| + log |x ¯y| = Re(log(x y) log(x ¯y)).
Since we have
log(f
t
(x) f
t
(y)) = log(g
t
(x) g
t
(y)),
we can compute
d log(g
t
(x) g
t
(y)) =
1
g
t
(x) g
t
(y)
2
g
t
(x) U
t
2
g
t
(y) U
t
dt
=
2
(g
t
(x) U
t
)(g
t
(y) U
t
)
dt.
Similarly, we have
d log(g
t
(x) g
t
(y)) =
2
(g
t
(x) U
t
)(g
t
(y) U
t
)
dt.
So we have
dG
t
(x, y) = im
2
g
t
(x) U
t
im
2
g
t
(y) U
t
dt.
This is exactly what we wanted it to be.