III Schramm--Loewner Evolutions

4Phases of SLE

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.