1Conformal transformations

III Schramm--Loewner Evolutions



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.