1Conformal transformations
III Schramm--Loewner Evolutions
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
iθ
[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
iθ
) =
1
2π
1 − |z|
2
|e
iθ
− 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!