The Weil-Petersson Metric
on Teichmüller
and Moduli Space
Ken Bromberg
Notes written by
Yael Algom-Kfir,
William Malone,
Erika Meucci
University of Utah Fall 2008
Contents
1 The Hyperbolic Plane, Models and Isometries
1.1 Geometric Viewpoint . . . . . . . . . . . . . . .
1.1.1 Inversions in Circles . . . . . . . . . . .
1.1.2 Poincaré Model . . . . . . . . . . . . . .
1.1.3 Deriving the Riemannian metric on ∆ .
1.1.4 The Upper Half Space Model . . . . . .
1.2 Analytic Viewpoint . . . . . . . . . . . . . . . .
1.2.1 Models of Hyperbolic Geometry . . . . .
1.2.2 Length and Area . . . . . . . . . . . . .
1.2.3 Isometries and Uniqueness of Geodesics .
1.3 Classification of Isometries . . . . . . . . . . . .
1.4 Gauss-Bonnet Formula . . . . . . . . . . . . . .
1.5 Problems: . . . . . . . . . . . . . . . . . . . . .
1.6 Solutions: . . . . . . . . . . . . . . . . . . . . .
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5
5
6
8
10
12
14
14
14
15
18
19
21
22
2 Quasi-Conformal Maps
2.1 Dilatation of a linear map . . . . . . . . .
2.2 Regular Quasi-Conformal Mappings . . . .
2.3 An optimization problem . . . . . . . . . .
2.3.1 The Module of a Quadrilateral . . .
2.3.2 The Module of an Annulus . . . . .
2.4 Extremal Length . . . . . . . . . . . . . .
2.4.1 Proof of the Compactness Theorem
2.5 Grötzsch’s Theorem in the general form . .
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25
25
25
27
32
33
35
39
43
3 Teichmüller Space and Moduli Space
3.1 Riemann Surfaces . . . . . . . . . . .
3.2 Special Case: the Torus . . . . . . .
3.3 Quasi-Conformal Map on the Torus .
3.4 Hyperbolic and Conformal Structures
3.4.1 Hyperbolic Structures . . . .
3.4.2 Conformal Structures . . . . .
3.5 T (S) is homeomorphic to R6g−6 . . .
3.6 Length functions . . . . . . . . . . .
3.7 Fenchel-Nielsen Coordinates . . . . .
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45
45
48
50
51
52
54
55
55
63
3
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4
CONTENTS
3.7.1 Dehn Twist . . . . . . . .
3.7.2 Thick-Thin Decomposition
3.7.3 Bers Constant . . . . . . .
3.8 T (S) is complete . . . . . . . . .
3.9 Constructing Riemann Surfaces .
3.10 1-forms . . . . . . . . . . . . . . .
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69
69
72
73
76
78
4 Quadratic Differentials
4.1 Holomorphic quadratic differentials . . . . . . . . . . . . . . . . . . .
4.2 Branched Double Covers . . . . . . . . . . . . . . . . . . . . . . . . .
4.2.1 Strebel Differentials . . . . . . . . . . . . . . . . . . . . . . . .
81
81
85
87
5 The Measurable Riemann Mapping Theorem
5.1 Bers’ Simultaneous Uniformization . . . . . . . . . .
5.1.1 Conformal Maps . . . . . . . . . . . . . . . .
5.1.2 Bers’ Embedding . . . . . . . . . . . . . . . .
5.1.3 An addendum to the infinitesimal Measurable
ping Theorem . . . . . . . . . . . . . . . . . .
93
94
96
97
6 The
6.1
6.2
6.3
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. . . . . . . . .
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Riemann Map. . . . . . . . . 101
Weil-Petersson metric
103
Norms on Q(X) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103
Incompleteness of the Weil-Petersson metric . . . . . . . . . . . . . . 104
Other properties of the Weil-Petersson metric . . . . . . . . . . . . . 106
Chapter 1
The Hyperbolic Plane, Models and
Isometries
There are many ways to define Hyperbolic Geometry. We will present two viewpoints.
In the Geometric Viewpoint, we start from the definition of isometries of the Hyperbolic Plane and then we find a metric for such space. Conversely, in the Analytic
Viewpoint, first we define the metric in the Hyperbolic Plane and then we find the
isometries of this metric space.
1.1
Geometric Viewpoint
In this section we will use a construction from Euclidean geometry to introduce
“straight” lines for hyperbolic geometry as well as a group of diffeomorphisms that act
nicely on these lines. Finally, we will derive the Riemannian metric invariant under
this set of diffeomorphisms. This take on Hyperbolic Geometry is from Thurston’s
Book.
Fact 1.1.1. (From Euclidean Geometry) P~A· P~B (Euclidean dot product) is indepen2
dent of the choice of the line. In the following picture this says that P~A · P~B = P~ t .
t
O
p
M
A
B
2
Figure 1.1: P~A · P~B = P~ t .
5
6
CHAPTER 1. THE HYPERBOLIC PLANE, MODELS AND ISOMETRIES
Proof.
P~A =
P~B =
=
~
~
PA · PB =
=
=
=
~
P~M + MA
~
P~M + MB
~
~
P M − MA
|P M|2 − |MA|2
(|P O|2 − |OM|2 ) − (|OB|2 − |OM|2 ) (Pythagorean theorem)
|P O|2 − |OB|2
|P O|2 − radius2
which is independent of A and B.
1.1.1
Inversions in Circles
Let c be a circle such that c ⊂ R2 ∪ ∞ =Riemann Sphere= Ĉ. Define ic : R2 ∪ ∞ →
R2 ∪ ∞ via the following:
c
P
R
P′
O
Figure 1.2: Definition of ic .
~ · OP
~ ′ = R2 . Define ic (P ) = P ′ , ic (0) = ∞, ic (∞) = 0.
such that OP
Properties:
0) ic is a diffeomorphism.
1) ic |c = id.
2) ic interchanges the inside and outside of the circle.
3) Lines through the origin (∪∞) are invariant under ic .
4) Circles orthogonal to c are ic -invariant.
5) If h : R2 ∪ ∞ → R2 ∪ ∞ is a homothety (x → λx) (λ > 0), then ic h = h−1 ic .
The only property that is not obvious is number 4.
7
1.1. GEOMETRIC VIEWPOINT
Proof. (of 4.) Since the two circles c, c′ are orthogonal, there exists a point t ∈ c ∩ c′
~ · OP
~ ′ =
and the tangent line at t to c′ goes through O. The fact then implies OP
2
2
′
~
|OT | = R . Thus ic (P ) = P .
c
t
P
P′
c′
O
Figure 1.3: Circles orthogonal to c are ic -invariant.
More Properties:
1 ic is conformal (preserves angles).
2 Circles not containing 0 go to other such circles.
3 Circles through 0 are interchanged with lines(∪∞) not through 0.
Proof. Of 1. Find circles cu and cv that are ⊥ c and tangent to u and v respectively.
u
c
P
P
u′
O
v
′
v′
Figure 1.4: ic is conformal.
Since ic preserves cv and cu , ic (P ) = P ′ = (other intersection point). ic∗ (u) = u′
and ic∗ (v) = v ′ where u′ , v ′ are tangent to iu , iv at P ′ . Thus \(u, v) = \(u′, v ′ ).
8
CHAPTER 1. THE HYPERBOLIC PLANE, MODELS AND ISOMETRIES
Ft
p
c
Fp
Figure 1.5: Circles not containing 0 go to other such circles.
Of 2. Start by showing that {circles, lines} = clines tangent to c map to clines of
the same form. Consider the family Ft of clines tangent to c at a point p. Also
let Fp be the family of clines perpendicular to c at p.
When a curve from Ft intersects Fp , they meet perpendicularly. ic preserves
Ft , so it preserves the line field perpendicular to Ft . So ic preserves the corresponding integral manifolds, and these are Ft .
Now let c′ be any circle not through 0. There is a homothety h such that h(c′ ) is
tangent to c. Then ic (c′ ) = ic (h−1 [h(c′ )]) = hic (h(c′ )) where ic (h(c′ )) is a circle
tangent to c.
Of 3. Mimic the proof of 2.
1.1.2
Poincaré Model
Let ∆ be the open unit disc in R2 . A geodesic is a diameter or an arc of a circle
perpendicular to ∂∆.
A reflection rγ : ∆ → ∆ in a geodesic γ is the restriction to ∆ of the inversion ic
where γ ⊂ c or the euclidean reflection when γ is a straight line.
An “isometry” of ∆ is a composition of reflections. These naturally form a group
to which we associate the following notation:
Isom(∆) = group of “isometries” of ∆
Isom+ (∆) = index 2 subgroup consisting of orientation preserving “isometries”.
9
1.1. GEOMETRIC VIEWPOINT
0
∆
Figure 1.6: The Poincaré Disc Model.
P
0
Figure 1.7: Isom+ (∆) acts transitively on ∆.
Properties.
1. Isom+ (∆) acts transitively on ∆.
2. Any isometry takes geodesics to geodesics and Isom+ (∆) acts transitively on
the set of all geodesics.
3. The stabilizer of 0 ∈ ∆ consists of the restrictions to ∆ of Euclidean isometries
fixing 0.
4. Any two points in ∆ are on a unique geodesic. (Postpone until we know the
metric)
Proof. Of 1. Use the intermediate value theorem on geodesic segments that meet
the geodesic from 0 to P perpendicularly and then compose with a reflection
through 0 and P .
Of 2. The first part follows from the properties of inversions. For the second part we
can assume without loss of generality that one of the points is 0 by conjugation
by the isometry from the first property. Now since we can find an isometry
that takes P to 0 by property one and the first part says that this is a geodesic
10
CHAPTER 1. THE HYPERBOLIC PLANE, MODELS AND ISOMETRIES
P
0
Figure 1.8: Any isometry takes geodesics to geodesics.
through 0 (ie a diameter). Then rotate (composition of two reflections) to land
on the desired geodesic.
Of 3. O(2) ⊂ Isom(∆) and O(2) = {group of isometries fixing origin}. O(2) is
topologically the wedge of two circles.
Let ϕ ∈ Isom(∆) fix 0. Compose ϕ (if necessary) with an element of O(2) so
that ϕ is orientation-preserving and leaves a ray from 0 invariant.
Claim 1.1.2. ϕ = id
Proof. (Of Claim) Since ϕ preserves angles, it preserves all rays through 0. This
implies that all geodesics are preserved. Now given any point represent it as
the intersection of two geodesics so that every point is fixed.
The claim implies that Stab(0) = O(2).
1.1.3
Deriving the Riemannian metric on ∆
Definition 1.1.3. Let ρ1 and ρ2 be two Riemannian metrics on X. We say that ρ1 is
conformally equivalent to ρ2 if there exists a continuous map φ : X → R+ such that
for all x ∈ X:
< vx , wx >ρ1 = φ(x) < vx , wx >ρ2 .
This is equivalent to the condition that the angles measured with respect to the two
metrics are equal.
Corollary 1.1.4. Up to scale, there is a unique Riemannian metric on ∆ which
is invariant under the isometry group. This metric is conformally equivalent to the
Euclidean metric. Geodesics in the Riemannian metric are Euclidean lines and circles
perpendicular to ∂∆ intersected with ∆. Circles in the Riemannian metric (i.e., the
locus of points equidistant from the center) are Euclidean circles.
11
1.1. GEOMETRIC VIEWPOINT
x
y
Figure 1.9: Because in a Riemannian metric space local geodesics are unique, a
geodesic must be contained in the fixed point set of an isometry that fixes x and y.
Proof. Let ρ be a Riemannian metric invariant under the isometry group. Notice that
ρ is uniquely determined by its value at a base point (say 0 ∈ ∆) since the group
acts transitively on X. After rescaling we may assume that ρ is equal the Euclidean
Riemannian metric at that base point. Now at 0, Euclidean angles and ρ angles
coincide. Since inversions preserve Euclidean angles, ρ-angles and Euclidean angles
agree at every point. Indeed, if v, w ∈ Tx ∆ let g be an isometry taking x to 0, then
\ρ (v, w) = \ρ (dx g(v), dxg(w)) where the later two vectors lie in T0 ∆. Hence,
\ρ (dx g(v), dxg(w)) = \Euc (dx g(v), dx g(w)) =
\Euc (d0 g −1(dx g(v)), d0g −1(dx g(w))) =
\Euc (v, w) .
Thus, ρ and Euc are conformally equivalent.
Since rotations about the origin are isometries for both the Euclidean and the
hyperbolic metrics on ∆, Euclidean circles about the origin are hyperbolic circles.
Since Euclidean circles are preserved by hyperbolic isometries, they are hyperbolic
circles even when they are no longer centered at the origin (however, the Euclidean
and hyperbolic centers of the circle do not coincide).
Let γ be a hyperbolic geodesic from x to y. Then x and y either lie on a circle
perpendicular to ∂∆ or on a diameter of ∆. Let C be the diameter or arc of the circle
intersected with ∆. Then g = iC is an isometry fixing x and y. Therefore, it takes
γ to another hyperbolic geodesic from x to y. In a Riemannian metric, geodesics
are locally unique. Therefore, if x and y are sufficiently close, g(γ) = γ so γ must
be the subarc of C between x and y (see Figure 1.9). Now even if x and y aren’t
close, γ must be a local geodesic, so at every point it must coincide with a subarc of
a diameter or a circle perpendicular to ∂∆. But two such distinct curves cannot be
12
CHAPTER 1. THE HYPERBOLIC PLANE, MODELS AND ISOMETRIES
Figure 1.10: Geodesics in the upper half space model are semicircles centered on the
real line, and vertical lines.
tangent at a point in ∆ and so a hyperbolic geodesic must be a subarc of a diameter
of ∆ or a circle perpendicular to ∂∆.
Question 1.1.5. Write down an explicit expression for the metric.
1.1.4
The Upper Half Space Model
We define
H2 := {z ∈ C|Imz > 0}.
).
One can apply an inversion to ∆ to obtain H2 (for example: φ(z) = 1−z
z−i
The isometry group of H2 consists of reflections about these geodesics.
Theorem 1.1.6. The following map is an isomorphism:
+
2
Φ: P SL2 (R)
−→ Isom (H ).
a b
az+b
7−→ z 7→ cz+d
c d
Proof. First consider the above map with SL2 (R) as its domain. We must show
(among other things) that it lands where it’s supposed to - orientation preserving
isometries. Consider the images of the following matrices:
1 r
−→ (z → z + r) this is a composition of two reflections in geodesic lines
•
0 1
perpendicular to R. Thus it is an orientation-preserving isometry.
0 1
1
•
−→ (z → −z
) this is a composition of reflection in the unit circle: z1
−1 0
and in the y-axis: −z. Thus it is an orientation-preserving isometry.
1 0
0 1
1 −r
0 −1
Notice that
=
therefore lower triangular mar 1
−1 0
0 1
1 0
λ 0
is sent
trices also land in orientation-preserving isometries. In addition,
0 λ−1
to the map z → λ2 z which is a composition of two inversions centered at the origin
13
1.1. GEOMETRIC VIEWPOINT
φ(i)
i
0
Figure 1.11: If φ doesn’t stabilize i, compose with maps in the image of Φ to get one
which does.
(much like the composition of two reflections in perpendicular lines is a translation).
Finally,
by
every matrix in SL2 (R) can be written as a product
elimination,
Gaussian
1 r
λ 0
1 0
hence the map on SL2 (R) is well defined. Next, a
and
,
of
0 1
0 λ−1
r 1
matrix is in the kernel of this map if b = c = 0 and a = d. Thus this map descends
to a monomorphism from PSL2 (R).
a b
∈ Stab(i),
We wish to compute which matrices land in Stab(i). Suppose Φ
c d
ai+b
then ci+d
= i which implies a = d, b = −c. Since the determinant equals 1, we get
a b
2
2
∈ P SO(2) = SO(2). Therefore, Φ maps SO(2) onto
a +b = 1 and therefore
c d
the stabilizer of i in Isom+ (H). Now let φ : H2 → H2 be any orientation-preserving
isometry. After composing φ with a translation and a dilation (which are both in the
image of Φ) we can assume that φ fixes i, and we’ve already shown that such a map
is in the image.
Now we are in a better position to figure out what the invariant Riemannian metric
should be. At i, choose the metric to coincide with the standard inner product. Since
horizontal translations are isometries, the metric should only be a function of y. Since
dilation by λ is an isometry the vector v at i and the vector λv at λi should have
the same norm. Thus we obtain the metric ds2 = y12 dt2 where dt2 is the Euclidean
metric.
Exercise 1.1.7. In ∆, the hyperbolic metric is: ds2 =
4
dt2 .
(1−r 2 )2
Remark. By the formula:
4 < v, w >= kv + wk2 − kv − wk2,
the norm determines the inner product when it exists.
14
CHAPTER 1. THE HYPERBOLIC PLANE, MODELS AND ISOMETRIES
Figure 1.12: Geodesics in the Upper Half Plane Model.
1.2
1.2.1
Analytic Viewpoint
Models of Hyperbolic Geometry
In the last section we started from the isometries of the Hyperbolic Plane and we
deduced its metric. In this section, we start with the metric of the Hyperbolic Plane
and we find its isometries.
There are two main models of the hyperbolic geometry: the Upper Half Plane Model
and the Poincaré Disc Model.
In the Upper Half Plane Model the metric is
gH2 (z) =
1
gE2 ,
(Im(z))2
where gE2 denotes the Euclidean metric, and the geodesics are the intersections of H2
with the Euclidean lines in C perpendicular to the real axis and the intersections of
H2 with Euclidean circles centered on the real axis (see Figure 1.12).
In the Poincaré Disc Model the metric is
gD (z) =
4
g 2
(1 − |z|2 )2 E |disc
and the geodesics are the intersections of D with the Euclidean lines in C through
the origin and the intersections of D with Euclidean circles perpendicular to ∂D (see
Figure 1.13).
1.2.2
Length and Area
We will identify z ∈ C with (x, y) ∈ R2 .
Definition 1.2.1. If γ : [a, b] → H2 is a C 1 -path, we define the length of γ
lH2 (γ) :=
Z
γ
1
|dz| =
Im(z)
Z
a
b
|γ ′ (t)|
dt.
Im(γ(t))
15
1.2. ANALYTIC VIEWPOINT
0
∆
Figure 1.13: The Poincaré Disc Model.
For example, if γ : [a, b] → H2 , γ(t) = it, then
lH2 (γ) =
Z
γ
1
|dz| =
Im(z)
Z
b
a
|γ ′(t)|
dt =
Im(γ(t))
Z
b
a
since |γ ′ (t)| = 1 and Im(γ(t)) = t.
1
b
dt = ln
t
a
Definition 1.2.2. If X ⊂ H2 , we define the area of X
Z
Z
1
1
area(X) :=
dx dy =
dx dy.
2
2
X Im(z)
X y
For example, if X ⊂ H2 is the region bounded by the three Euclidean lines
{z ∈ H2 | Re(z) = 1, Im(z) > 1}, {z ∈ H2 | Re(z) = −1, Im(z) > 1} and {z ∈
H2 | Im(z) = 1} (see Figure 1.14), then
area(X) =
Z
X
1.2.3
1
dx dy =
y2
Z
1
−1
Z
∞
1
1
dy dx = 2.
y2
Isometries and Uniqueness of Geodesics
We know the metric of the Hyperbolic Plane and so we can consider the group
Isom+ (H2 ) of the orientation preserving isometries of H2 .
Theorem 1.2.3. Isom+ (H2 ) ∼
= P SL2 (R).
Proof. (Sketch) We define
+
2
ϕ: P SL2 (R)
−→ Isom (H ).
a b
az+b
7−→ z 7→ cz+d
c d
16
CHAPTER 1. THE HYPERBOLIC PLANE, MODELS AND ISOMETRIES
X
i
−1
0
1
Figure 1.14: The set X.
We have to check that ϕ is a well-define bijective homomorphism.
az+b
We start proving that ϕ is well-define, i.e. γ(z) = cz+d
∈ Isom+ (H2 ). In other words,
we want to show that
|γ ′ (z)|
1
=
.
Im(γ(z))
Im(z)
Since
ad − bc
a(cz + d) − c(az + b)
=
γ ′ (z) =
2
(cz + d)
(cz + d)2
and
az + b
(az + b)(cz + d)
aczz + adz + bcz + bd
=
=
, (a, b, c, d ∈ R)
2
cz + d
|cz + d|
|cz + d|2
we have
|γ ′ (z)| =
Hence,
Now, since
ad − bc
|cz + d|2
and Im(γ(z)) =
ady − bcy
(ad − bc)y
=
.
|cz + d|2
|cz + d|2
|γ ′ (z)|
(ad − bc) |cz + d|2
1
=
=
.
2
Im(γ(z))
|cz + d| (ad − bc)y
Im(z)
n
az + b
o
Ker(ϕ) =
∈ P SL2 (R) =z =
cz + d
n
o
a b
=
∈ P SL2 (R) b = c = 0, a = d = Id ∈ P SL2 (R),
c d
ϕ is injective.
We leave, as an exercise, to prove that ϕ is a surjective homomorphism.
a b
c d
17
1.2. ANALYTIC VIEWPOINT
Using this theorem we can prove the following
Proposition 1.2.4. There exists a unique hyperbolic geodesic between two points
in H2 .
Proof. Without loss of generality, we can suppose that the two points are ia and ib.
We want to prove that γ : [a, b] → H2 , γ(t) = it is the unique geodesic through these
two points.
Obviously, γ is a geodesic. We need to prove the uniqueness.
By contradiction, we suppose that there exists a geodesic f : [a, b] → H2 such that
f (a) = ia, f (b) = ib.
ib
γ
f
ia
Figure 1.15: The path γ and f of the Proposition 1.2.4.
p
We consider the map ϕ(reiϑ ) = ri, i.e., f (x, y) = (0, x2 + y 2). Since ϕ is an
orientation preserving isometry, if we prove that ϕ is a contraction, then l(ϕ(f )) <
l(f ) and so f is not a geodesic.
From this argument it follows that a geodesic between ia and ib is contained in the
imaginary axis. Now, we observe that γ is the smallest path between ia and ib that
is contained in the imaginary axis.
Hence, we need to show that ϕ is a contraction and we are done.
We consider a tangent vector v = (x, y). Since the Jacobian of ϕ is
!
0
0
D(ϕ) =
√ x2 2 √ 2y 2
x +y
and so
D(ϕ)(v) =
we have
kvk(x,y) =
p
0
√
x2 + y 2
=
y
x
x2 +y 2
r
1+
0
√
y
x2 +y 2
x 2
y
!
x +y
x
y
=
p 0
x2 + y 2
,
p
x2 + y 2
= kDϕ(v)k(0,√x2 +y2 ) .
>1= p
x2 + y 2
18
CHAPTER 1. THE HYPERBOLIC PLANE, MODELS AND ISOMETRIES
Since the tangent vectors shrinks, ϕ is a contraction and this concludes the proof of
the proposition.
Since the Hyperbolic Plane is uniquely geodesic, we can define the distance between two points p and q as
dH2 (p, q) := minγ {l(γ) | γ : [0, 1] → H2 , γ(0) = a, γ(1) = b}.
1.3
Classification of Isometries
We can ask the following question: How many fixed points an orientation preserving
isometry has in H2 ?
We can compute the fixed points solving
az + b
=z
cz + d
i.e. cz 2 + (d − a)z − b = 0.
We have two cases. If c = 0, then we have only one fixed point at ∞ if d = a and
b
if d 6= a.
two fixed points ∞ and (d−a)
Now, if c 6= 0, then
p
−(d − a) ± (d − a)2 + 4bc
.
z=
2c
Let ∆ := (d − a)2 + 4bc. Since ad − bc = 1,
∆ = (d − a)2 + 4bc = d2 + a2 − 2ad + 4bc = d2 + a2 + 2ad − 4 =
2
= (d + a) − 4 = trace
a b
c d
− 4.
We have three cases:
a b 1. ∆ < 0 ⇔ trace
< 2, then we get two conjugate complex roots and
c d
hence, we have only one fixed point in H2 . In this case, the isometry is called
elliptic;
a b 2. ∆ = 0 ⇔ trace
= 2, then we get one real root and hence, we have
c d
only one fixed point in the real axis of H2 . In this case, the isometry is called
parabolic;
a b 3. ∆ > 0 ⇔ trace
> 2, then we get two real roots and hence, we
c d
have two fixed point in the real axis of H2 . In this case, the isometry is called
hyperbolic or loxodromic.
19
1.4. GAUSS-BONNET FORMULA
Since two Möbius transformations M and N that are conjugate (i.e., there exists
a Möbius transformation P so that M = P ◦ N ◦ P −1 ) have the same number of fixed
points, modulo conjugation, the standard forms are
cos(ϑ) sin(ϑ)
(rotations) in the elliptic case;
1.
− sin(ϑ) cos(ϑ)
1 b
(translations) in the parabolic case;
2.
0 1
a 0
(dilatations) in the hyperbolic (or loxodromic) case.
3.
0 a−1
1.4
Gauss-Bonnet Formula
The first remarkable result in hyperbolic geometry is the Gauss-Bonnet Theorem.
Theorem 1.4.1 (Gauss-Bonnet). Let P be a hyperbolic triangle (i.e., the sides are
geodesics) with interior angles α, β and γ. Then
areaH2 (P ) = π − α − β − γ.
Proof. We divide the proof on two steps:
Step 1): We consider the triangle P as in Figure 1.16 (in this case γ = 0).
b
a
β
α
−1
1
0
Figure 1.16: Particular geodesic triangle P .
We have
areaH2 (P ) =
Z
P
1
dx dy =
y2
Z
cos(β)
cos(π−α)
Z
∞
√
x2 +y 2
1
dy dx =
y2
20
CHAPTER 1. THE HYPERBOLIC PLANE, MODELS AND ISOMETRIES
=
Z
cos(β)
dx
cos(π−α)
p
=
x2 + y 2
Z
β
π−α
−
sin(t)
dt = π − α − β
sin(t)
as we wanted to prove.
Step 2): Since Möbius transformations preserve the hyperbolic area, we can suppose
that the triangle P = ∆(abc) has the geodesic ab included in the circle of center 0 and
radius 1 and the geodesic ac included in the vertical geodesic a∞, as in Figure 1.17.
c
γ
β′
α
a
β
b
Figure 1.17: Triangle P of Step 2).
We denote with β ′ the angle cb̂∞.
Using step 1), we have
areaH2 (P ) = areaH2 (∆(ab∞)) − areaH2 (∆(cb∞)) =
= π − α − (β + β ′ ) − [π − (π − γ) − β ′ ] = π − α − β − γ
and this concludes the proof of the theorem.
This theorem can be generalized to polygons.
Theorem 1.4.2. Let P be a hyperbolic polygon (i.e., the sides are geodesics) with
interior angles α1 , . . . , αm . Then
areaH2 (P ) = (n − 2)π −
n
X
αk .
k=1
Now, we know that a surface of genus g is made by identifying edges of a 4g-gon
so that all vertices are identified.
Consider a regular geodesic 4g-gon in the Poincaré disc model ∆ concentric to the
disc (fundamental polygon). The angle sum of a small 4g-gon is ∼ (4g − 2)π which
is grater than 2π if the genus g > 1. For a large 4g-gon the angle sum is ∼ 0. Since
21
1.5. PROBLEMS:
the angle sum changes continuously, we can find a 4g-gon with angle sum 2π.
As the sides have the same length, we can glue sides by orientation-preserving isometries. In this way, we get a so called hyperbolic atlas for the surface. We use the
unique isometries joining the edges to make charts for the edges. As the angle sum
is 2π, we get a chart for the vertices.
Moreover, applying the isometries to the fundamental 4g-gon gives a tiling of ∆.
11111
00000
00000
11111
0000
1111
00000
11111
0000
1111
0000
1111
1111
0000
0000
1111
0000
1111
0000
1111
0000
1111
0000
1111
Figure 1.18: Charts of edges and vertices in a genus 2 surface.
1.5
Problems:
1. For r > 0 consider f : [0, 2π] → C given by f (t) = reit , which parameterizes the
Euclidean circle with Euclidean center 0 and Euclidean radius r. By definition,
if γ : [a, b] → C is a path,
Z
|dz|
.
lD (γ) :=
2
γ 1 − |z|
Calculate lD (f ).
2. Given s > 0, let Ds be the open hyperbolic disc in D with hyperbolic center 0
and hyperbolic radius s. Show that the hyperbolic area areaD (Ds ) of Ds is
Z
4
2 1
areaD (Ds ) :=
dx
dy
=
4πsinh
s .
2 2
2
Ds (1 − |z| )
Hint: First prove that for 0 < r < 1
1 + r dD (0, r) = ln
1−r
and, hence, that r = tanh[ 12 dD (0, r)].
From this observation it follows that the hyperbolic radius s of Ds is related to
22
CHAPTER 1. THE HYPERBOLIC PLANE, MODELS AND ISOMETRIES
the Euclidean radius R by R = tanh2 ( 12 s).
Then, using the polar coordinates, calculate
Z
4r
dr dϑ.
areaD (Ds ) =
2 2
Ds (1 − r )
3. Consider
+
2
ϕ: P SL2 (R)
−→ Isom (H )
a b
az+b
7−→ z 7→ cz+d
c d
and prove that ϕ is a surjective homomorphism.
Hint: It is easy to prove that ϕ is a homomorphism, but it is more difficult
to prove that it is surjective. To prove surjectivity, convince yourself that an
isometry φ : M → N between 2-dimensional Riemann surfaces M and N is
determined by the image of a point p and the image of a vector in Tp M.
So if we prove that we can take the pair (i, (0, 1)), where (0, 1) is the vertical
unit vector in Ti H2 , in any other pair (p, v) through an element in P SL2 (R),
then we are done.
If p = x + iy, you can go first from i to iy and then from iy to p.
4. Two elements M1 and M2 in P SL2 (R) are conjugate if there exists some P ∈
P SL2 (R) so that M2 = P ◦ M1 ◦ P −1 .
Suppose that M and N are elements in P SL2 (R) conjugate by P , so that
M = P ◦ N ◦ P −1 . Prove that N and M have the same number of fixed points
in H2 .
5. Prove that the hyperbolic area in H2 is invariant under the action of orientation
preserving isometries. That is, if X be the set in H2 whose hyperbolic area
areaH2 (X) is defined and if A ∈ P SL2 (R), then
areaH2 (X) = areaH2 (A(X)).
1.6
Solutions:
1. We have
lD (γ) =
Z
γ
|dz|
=
1 − |z|2
Z
0
2π
2πr
|f ′ (t)|
dt =
.
2
1 − |f (t)|
1 − r2
2. For 0 < r < 1, we consider f : [0, r] → D given by f (t) = t. As the image of
f is the hyperbolic segment in D joining 0 and r, we have that, by definition,
dD (0, r) = lD (f ). Hence,
Z r
2
dD (0, r) = lD (f ) =
dt =
2
0 1−t
23
1.6. SOLUTIONS:
Z rh
1 + r
1 i
1
dt = ln
.
=
+
1+t 1−t
1−r
0
Solving for r as a function of dD (0, r), we have r = tanh[ 12 dD (0, r)].
Since the hyperbolic radius s of Ds is related to the Euclidean radius R by
R = tanh( 12 s), then, using the polar coordinates, we have
areaD (Ds ) =
Z
Ds
4r
dr dϑ =
(1 − r 2 )2
Z
Z
R
0
Z
0
2π
4r
dr dϑ =
(1 − r 2 )2
R
4r
4πR2
= 2π
dr =
=
2 2
1 − R2
0 (1 − r )
4πtanh2 ( 12 s)
2 1
=
4πsinh
=
s .
2
1 − tanh2 ( 12 s)
3. First we prove that
+
2
ϕ: P SL2 (R)
−→ Isom (H )
a b
7−→ z 7→ az+b
cz+d
c d
is a homomorphism.
Let
Then
ϕ
a b
c d
a b
c d
′ ′ a b
∈ P SL2 (R).
,
c′ d ′
′ ′ ′
aa + bc′ ab′ + bd′
a b
=
=ϕ
·
ca′ + dc′ cb′ + dd′
c′ d ′
(aa′ + bc′ )z + (ab′ + bd′ )
=
=
(ca′ + dc′)z + (cb′ + dd′ )
a b
=ϕ
◦ϕ
c d
a
c
a′
c′
a′ z+b′
c′ z+d′
a′ z+b′
c′ z+d′
+b
+d
b′
.
d′
=
Now we want to prove surjectivity. We observe that an isometry φ : M → N
between 2-dimensional Riemann surfaces M and N is determined by the image
of a point p and the image of a vector in Tp M.
So if we prove that we can take the pair (i, (0, 1)), where (0, 1) is the vertical
unit vector in Ti H2 , in any other pair (p, v) through an element in P SL2 (R),
then we are done.
Let p = x + iy = x + ieλ . First we go from i to x + iy through
!
λ
e2
0
1 x
A=
.
−λ
0 1
0 e2
24
CHAPTER 1. THE HYPERBOLIC PLANE, MODELS AND ISOMETRIES
Notice that A∗ (0, 1) is a vertical unit vector. So, if we want to get the vector v,
we need to do a rotation sending (p, A∗ (0, 1)) to (p, v). Hence, the matrix that
we are looking for is
!
λ
0
e2
1 x
cos(ϑ) sin(ϑ)
∈ P SL2 (R).
−λ
0 1
− sin(ϑ) cos(ϑ)
0 e2
4. If x is a fixed point for N, then
M(P (x)) = P ◦ N ◦ P −1 (P (x)) = P (N(x)) = P (x)
and so P (x)) is a fixed point for M. Moreover, since N = P −1 ◦ M ◦ P , if x is
a fixed point for M, then
N(P −1 (x)) = P −1 ◦ M ◦ P (P −1(x)) = P −1 (M(x)) = P −1 (x)
and so P −1 (x)) is a fixed point for N.
Hence, M and N have the same number of fixed points in H2 .
5. The proof is an application of the change of variables.
Since a, b, c and d are real numbers and ad − bc = 1,
A(z) =
(az + b)(cz̄ + d)
az + b
=
=
cz + d
(cz + d)(cz̄ + d)
acx2 + acy 2 + bd + bcx + adx
y
+i
2
2
2
(cx + d) + c y
(cx + d)2 + c2 y 2
=
so we can write
A(x, y) =
The Jacobian is
acx2 + acy 2 + bd + bcx + adx
(cx + d)2 + c2 y 2
D(A)(x, y) =
(cx+d)2 −c2 y 2
((cx+d)2 +c2 y 2 )2
−2cy(cx+d)
((cx+d)2 +c2 y 2 )2
,
y
(cx + d)2 + c2 y 2
2cy(cx+d)
((cx+d)2 +c2 y 2 )2
(cx+d)2 −c2 y 2
((cx+d)2 +c2 y 2 )2
!
and the determinant of the Jacobian is
det(D(A)(x, y)) =
1
.
((cx + d)2 + c2 y 2 )2
Then, if we define m(x, y) = y12 , we have
Z
Z
1
areaH2 (A(X)) =
dx dy =
m ◦ A(x, y) |det(D(A))|dx dy =
2
A(X) y
X
Z
((cx + d)2 + c2 y 2 )2
1
=
dx dy = areaH2 (X).
2
y
((cx + d)2 + c2 y 2 )2
X
Chapter 2
Quasi-Conformal Maps
2.1
Dilatation of a linear map
Suppose D is a diagonal 2-dimensional real matrix. Then: D(x, y) = (ax, by) and if
(x, y) satisfy x2 + y 2 = 1 then (ax, by) satisfies a12 x2 + b12 y 2 = 1. Hence, a diagonal
matrix takes a circle to an ellipse.
Now, let B be an arbitrary matrix. By the singular value decomposition, B can
be written as B = SDR−1 , where D is a diagonal matrix and R, S are rotations i.e., orthonormal matrices. Therefore, B takes circles to ellipses. Let t1 , t2 be the
eigenvalues of D.
There are two cases:
• t1 = t2 . Then D is a scalar matrix and B takes circles to circles.
• t1 6= t2 . In this case, R(e1 ) and R(e2 ) are taken by B to vectors tangent to the
axes of the ellipse. Both R(e1 ) ⊥ R(e2 ) and B(e1 ) ⊥ B(e2 ).
Let
λmin := min kB(v)k kvk = 1
λmax := max kB(v)k kvk = 1 .
Then, by the definitions above, {λmin, λmax } = {t1 , t2 }. Suppose λmin = t1 , then
vmin = R(e1 ), vmax = R(e2 ). We will later need:
|B| = |SDR−1| = |S||D||R−1| = |D| = λminλmax ,
where |A| := det(A).
Definition 2.1.1. If B is a linear map the dilatation of B is D(B) =
2.2
λmax
.
λmin
Regular Quasi-Conformal Mappings
Here we follow the conventions from [8].
Definition 2.2.1. D ⊆ C is called a Jordan domain if π1 (D) = 1 and ∂D is a Jordan
curve (i.e., ∂D is a simple closed curve).
25
26
CHAPTER 2. QUASI-CONFORMAL MAPS
Definition 2.2.2. Let w : D → D ′ a homeomorphism of two Jordan domains D, D ′
which extends to a homeomorphism on the boundary. Suppose that wx , wy exist and
are continuous. Denote by T w(z) the matrix of partial derivatives. The directional
derivative at angle α will be denoted ∂α w(z). Let λmax (z) := max {∂α w(z) |α ∈ [0, 2π)},
λmin (z) := min {∂α w(z) |α ∈ [0, 2π)} The dilatation of w at z is:
Dw (z) := D(T w(z)) =
λmax (z)
.
λmin (z)
The homeomorphism w is a regular K-quasi-conformal map if
sup{Dw (z)} ≤ K.
z∈D
We drop the adjective regular if we allow w to have isolated singularities.
R−1
R(e2 )
R(e1 )
D
e2
λe2
e1
λe1
R
Figure 2.1: Every linear transformation can be decomoposed as a rotation R−1 followed by a dilatation D followed by another rotation S. Therefore a circle centered
at the origin will map to an elipse.
Example 2.2.3.
0) f is conformal on D ⇐⇒ T f doesn’t distort angles ⇐⇒ T f
takes circles to circles ⇐⇒ Df (z) = 1 for all z ⇐⇒ f is 1-quasi-conformal.
1) Consider two rectangles: R1 is the rectangle with vertices at 0, a, a + i, i and
R2 is the rectangle with vertices at 0, a′ , a′ + i, i. We assume a′ > a. Define
27
2.3. AN OPTIMIZATION PROBLEM
′
f : R1 → R2 by f (x + iy) = aa x + iy we call f the affine map from R1 to R2 .
We compute
a′ ux vx
0
= a
.
uy vy
0 1
a′
a
Therefore, λmax =
conformal map.
2.3
and λmin = 1 for all z. So f (z) is a regular
a′
a
quasi-
An optimization problem
Theorem 2.3.1 (Grötzsch). Let w : R1 → R2 be any homeomorphism with continuous partial derivatives which extends to a homeomorphism on the boundary and
′
which takes (0, a, a + i, i) to (0, a′ , a′ + i, i), then w is K-quasi-conformal and K ≥ aa .
′
Moreover, if Kf = aa then f is the affine mapping of R1 onto R2 .
Proof. The fact that w is K-quasi-conformal follows from the compactness of R1 . We
(z)
a′2
now assume that for all z ∈ R1 , λλmax
≤ K and prove that a′ ≥ Ka
. We will need
min (z)
the following calculations:
1. Jw (z) = |T w(z)| = λmax (z)λmin (z) ≥
1
λ (z)
K max
≥
2. by the Chauchy-Schwartz inequality,
Z a
Z a
21 Z
2
1 · wx (x + iy)dx ≤
1 dx
0
Thus
0
0
1
|wx (z)|2
K
a
12
|wx (x + iy)| dx .
2
Z
2
1 a
|wx (x + iy)| dx ≥ · wx (x + iy)dx
a
0
0
With equality if and only if wx (x + iy) is a scalar.
R a
3. 0 wx (x + iy)dx = |w(a + iy) − w(iy)| ≥ a′ the last inequality follows since
w(a + iy) is on the line x = a and w(iy) is on x = 0 (here we use the fact that
vertices are mapped to vertices).
Z
a
We now compute:
2
Z
Z
1
|wx (x + iy)|2 dxdy =
K
R1
R1
2
Z 1 Z a
Z 1 Z a
1
1
2
dy
dy wx (x + iy)dx ≥
|wx (x + iy)| dx ≥
=
K 0
aK 0
0
0
Z 1
′2
1
a
≥
dy|a′|2 =
aK 0
Ka
′
a = Area(R2 ) =
Jw (z)dxdy ≥
The inequality is strict except for when: wx (x + iy) = c a scalar, λmax (z) = |wx (z)|
(z)
for all z and λλmax
= K for all z. So vmax = x hence vmin = y, and wy (x + iy) = Kc .
min (z)
′
Therefore, w(x + iy) = cx + i Kc y. But since vertices are mapped to vertices, c = aa
and w is the affine map from R1 to R2 .
28
CHAPTER 2. QUASI-CONFORMAL MAPS
Remark. There is another formula for the dilatation of w. First we define the
differential operators ∂z, ∂ z̄:
z̄ = x − iy
y = 2i1 (z − z̄).
z = x + iy
x = 12 (z + z̄)
∂z
∂z
+ fy ∂y
= 12 fx + 2i1 fy = 12 (fx − ify ). Similarly,
Using the chain rule: fz = fx ∂x
fz̄ = 21 (fx + ify ). We have
∂
∂
∂
∂
1
∂
1
∂
.
=
−
i
=
+
i
∂z
2 ∂x
∂y
∂ z̄
2 ∂x
∂y
The Beltrami coefficient of f (z) is
µf (z) =
fz̄
(z).
fz
Notice that if f is holomorphic and fz 6= 0 everywhere, then µf (z) = 0.
If f is orientation preserving, then Jw (z) > 0 for all z. Any easy computation shows
that Jw (z) = fz2 − fz̄2 . Therefore, fz (z) 6= 0 and µf (z) < 1.
Fact 2.3.2. Df (z) =
1+|µf (z)|
1−|µf (z)|
=
|fz |+|fz̄ |
.
|fz |−|fz̄ |
1+|µ (z)|
|fz |+|fz̄ |
Proof. It is enough to check that Df (z) = 1−|µff (z)| = |f
for f (x + iy) = x + iKy
z |−|fz̄ |
since f is locally of this form. We know that in this case Df = K. We have:
fz̄ =
fz =
Hence,
µf =
and so
1
(1
2
1
(1
2
+ i(iK))
− i(iK)).
fz̄
1−K
=
fz
1+K
1+
1 + 1−K
1+K
1−K =
1 − 1+K 1−
K−1
1+K
K−1
1+K
= K = Df .
Notice that kDf k∞ ≤ K if and only if kµf k∞ ≤ k < 1. Hence, f is K-quasi.
conformal if kµf k∞ ≤ K−1
1+K
A proof of Grötzsch’s Theorem using this definition can be found in [5].
Corollary 2.3.3. If a′ > a there is no conformal mapping from R1 to R2 which takes
vertices to vertices.
Proof. If there were a conformal map f : R1 → R2 , Caratheodory’s theorem garantees
that it extends to a homeomorphism of ∂R1 to ∂R2 . Up to this point there is no
contradiction, in fact, the Riemann mapping theorem implies that there is such a
map. However, if we further require that vertices are mapped to vertices then by
′
Grötzsch’s theorem Kf ≥ aa > 1 which is a contradiction.
29
2.3. AN OPTIMIZATION PROBLEM
We give the definition of absolute continuous function:
Definition 2.3.4. A function f : [a, b] → R is absolute continuous if E ⊂ [a, b] with
µ(E) = 0 implies that µ(f (E)) = 0.
Theorem 2.3.5. If f is absolute continuous, then f ′ exists almost everywhere.
Definition 2.3.6. A function f : Ω → C = R2 , f (x, y) = (f1 (x, y), f2(x, y)) is
absolute continuous on lines (ACL) if for all rectangles R = [a, b] × [c, d] ⊂ Ω the
following holds:
1. For almost everywhere y ∈ [c, d] the functions x 7→ f1 (x, y) and x 7→ f2 (x, y)
are absolute continuous;
2. For almost everywhere x ∈ [a, b] the functions y 7→ f1 (x, y) and y 7→ f2 (x, y)
are absolute continuous.
Remark. Notice that (1) implies that there exist partial derivatives of f respect to
x in almost all the lines y = c and (2) implies that there exist partial derivatives of
f respect to y in almost all the lines x = c. If f is absolute continuous on lines, by
Fubini, fx and fy exist almost everywhere.
Definition 2.3.7. A function f : Ω → C = R2 is K-quasi-conformal if it is a
homeomorphism and
1. f is absolute continuous on lines;
2. the essential supremum kµf k∞ ≤
K−1
.
1+K
Notation 2.3.8. We will write quasi-conformal instead of K-quasi-conformal when
it is not important the value of the constant.
A conformal map is 1-quasi-conformal, moreover
Theorem 2.3.9. A 1-quasi-conformal map is conformal.
Example 2.3.10. f : C → C, f (x + iy) = x + iKy is obviously K-quasi-conformal.
√
Example 2.3.11. We consider f : C → C, f (x+iy) = √1K x+i Ky and the following
diagram:
C
f˜
C
/
ϕ
ϕ
›
C
f
/
›
C
where ϕ(z) = z 2 . Notice that we can lift f to fe. What does fe look like?
Looking at the diagram we see that the image of the blue curves in Figure 2.2 go
to vertical lines through ϕ, then they are expanded via f and, after the lifting, they
come back to the original curves but with the expansion given by f . In a similar way,
the image of the red curves in Figure 2.2 go to horizontal lines through ϕ, then they
30
CHAPTER 2. QUASI-CONFORMAL MAPS
e
Figure 2.2: Description of f.
are contracted via f and, after the lifting, they come back to the original curves but
with the contraction given by f .
fe is not differentiable in 0, but it is differentiable at any point in C \ {0}. It
K−1
is easy to see that fe is absolute continuous on lines and kµf k∞ ≤ 1+K
, so fe is
K-quasi-conformal.
Remark. We can replace ϕ(z) = z 2 by ϕn (z) = z n for n ∈ Z, n ≥ 2.
r
eiϑ is not K-quasi-conformal
Example 2.3.12. The funcion f : ∆ → C, f (reiϑ ) = 1−r
for any K. In fact, there are not quasi-conformal maps from ∆ to C.
Notice that we already knew that there are not conformal maps from ∆ to C by
Liouville’s Theorem.
Example 2.3.13. Let C 1 be the middle 3rds Cantor’s set and let C 1 be the Cantor
3
4
set described in the Figure 2.3.
0
1
5
8
0
3
8
1
Figure 2.3: Definition of C 1 .
4
We denote with m the Lebesgue measure. We have
m(C 1 ) = 0,
3
m(C 1 ) = 1 −
4
∞ X
1 n
n=1
4
2
1
1
= .
=1−
1
4 1− 4
3
31
2.3. AN OPTIMIZATION PROBLEM
There is a map φ : I → I such that φ is a homeomorphism and φ(C 1 ) = C 1 . By the
3
4
previous observation, φ is not absolute continuous on lines.
We can define a measure ν by ν(E) = m(φ(E) ∩ C 1 ). So ν(C 1 ) 6= 0 and ν is a
4
3
measure on R. Now, we define ψ(x) = ν([0, x]). The graph of ψ is as in Figure 2.4.
2
3
0
1
Figure 2.4: The graph of ψ.
Let Ψ(x, y) = (x, ψ(x) + y) = x + i(ψ(x) + y). In particular, Ψz̄ = 0 almost
everywhere since Ψx = 1 + iψ ′ (x) (when ψ ′ (x) is defined) and Ψy = i, but Ψ is not
absolute continuous on lines and so it is not quasi-conformal.
Facts about quasi-conformal maps:
We will explain why the quasi-conformal maps are important.
The proof of the Theorems 2.3.15 and 2.3.16 can be found in [5].
Definition 2.3.14. The total derivative of f : Ω ⊂ R2 → R2 at x ∈ Ω is a linear
function Dfx : R2 → R2 such that for every ε > 0 there exists δ > 0 so that if |h| < δ,
then
f (x + h) − f (x) − Df (h) x
< ε.
|h|
Theorem 2.3.15. If f is quasi-conformal, then f has a total derivative almost everywhere.
Theorem 2.3.16. If f : Ω → C is quasi-conformal, then for all φ ∈ C0∞ (Ω), where
C0∞ (Ω) denotes the set of C ∞ (Ω)-functions with compact support in Ω,
Z Z
Z Z
fz · φ = −
f · φz ,
(2.1)
Ω
Z Z
and fz and fz̄ are locally in L2 .
Ω
Ω
fz̄ · φ = −
Z Z
Ω
f · φz̄
(2.2)
32
CHAPTER 2. QUASI-CONFORMAL MAPS
Definition 2.3.17. If f : Ω → C has fz and fz̄ such that 2.1 and 2.2 hold, then fz
and fz̄ are called distributional derivatives.
We can give another definition of quasi-conformal maps:
Definition 2.3.18. The map f : Ω → C is quasi-conformal if it is a homeomorphism
and
1. f has distributional derivatives fz and fz̄ locally in L2 ;
2. k ffz̄z k∞ ≤
K−1
.
1+K
b be open and connected, a1 ,
Theorem 2.3.19 (Compactness Theorem). Let Ω ⊂ C
b be disjoint compact sets. Define
a2 , a3 ∈ Ω and let A1 , A2 , A3 ⊂ C
F(K; A1 , A2 , A3 ) := {f : Ω → C | f is K-quasi-conformal and f (ai ) ∈ Ai , i = 1, 2, 3}.
Then F is compact in the compact-open topology (i.e., if {fj } ∈ F, then there exists
a subsequence {fnj } such that for every compact B ⊂ Ω, {fnj } converges uniformly
on B).
We need the Arzelà-Ascoli Theorem, but in order to use the Arzelà-Ascoli Theorem
we need another definition of quasi-conformality. Before we give the definition of
quasi-conformal maps, we must introduce the module of a quadrilateral.
2.3.1
The Module of a Quadrilateral
Definition 2.3.20. A quadrilateral (Q, q1 , q2 , q3 , q4 ) is a closed Jordan domain q with
four distinguished points.
Proposition 2.3.21. For every quadrilateral (Q, q1 , q2 , q3 , q4 ) there is a rectangle R =
[0, a] × [0, b] and a homeomorphism h : Q → R which extends to a homeomorphism
on the boundaries such that h is conformal in int(Q) and h(q1 ) = 0, h(q2 ) = a,
h(q3 ) = a + ib, h(q4 ) = b.
Moreover, ab is independent of h. It is denoted M(Q, q1 , q2 , q3 , q4 ) or m(Q) and called
the modulus of the quadrilateral (Q, q1 , q2 , q3 , q4 ).
Example 2.3.22. Let Q = H2 the Upper Half Plane and the four points are −x, −1, 1, x
where 1 < x ∈ R. Let Ra = [0, a] × [0, 1] be a rectangle. One way to see that there
exists a conformal map h : Q → Ra for some a is the following. Map Q to the unit disc
U and −x, −1, 1 to 1, i, −1 using the Riemann mapping theorem. x will be mapped
to some point p on the arc (−1, 1) determined by the cross-ratio1 [−x : −1 : 1 : x].
Similarly, map Ra onto U and 0, a, i + a to 1, i, −1. i will be mapped to q some point
on the arc (−1, 1). As we vary a, the cross-ratio [0 : a : a + i : i] varies continuously,
and q moves continuously along the arc. There exists a rectangle Ra with the same
cross ratio as (H2 , −x, −1, 1, x) and the construction described above will yield a map
1
Here we must believe that the cross-ratio of four points is invariant under conformal maps.
2.3. AN OPTIMIZATION PROBLEM
33
with the right properties.
Surprisingly, in this case we can actually write down the map:
Z z
dw
p
f (z) =
(w + x)(w + 1)(w − 1)(w − x)
0
is a conformal map from H2 onto some rectangle [−K, K] × [0, K ′ ]. For details see
[9].
Proof of Proposition 2.3.21. By the Riemann mapping theorem there’s a conformal
map which takes (Q, q1 , q2 , q3 , q4 ) onto (H2 , r1 , r2 , r3 , r4 ). We claim that there is a
Möbius transformation φ : H2 → H2 taking r1 , r2 , r3 , r4 to −x, −1, 1, x. The geodesics
[r1 , r3 ] and [r2 , r4 ] intersect at z0 with angle α. Let z1 (x) be the intersection point
of [−x, 1], [−1, x]. If x is close to 1 the angle at the intersection is close to π, if x is
close to ∞ then the angle at the intersection point is close to 0. Thus there is one x
for which the angle at the intersection point is α. Let φ be the element of P SL2 (R)
determined by taking z0 to z1 (x) and the tangent vector to [r1 , r3 ] at z0 to the tangent
vector to [−x, 1] at z1 (x). φ takes r1 , r2 , r3 , r4 to −x, −1, 1, x.
Now, we use the map in Example 2.3.22 to map it onto a rectangle. Since all of the
maps above are conformal, so is their composition, and by Caratheodory, it extends
to a homeomorphism on the boundary.
As for the uniqueness, it follows from Grötzsch’s theorem (or by the reflection principle).
Definition 2.3.23. We say that f : Ω → C is K-quasi-conformal if it is a homeomorphism onto its image and for all quadrilaterals Q
1
m(Q) ≤ m(f (Q)) ≤ Km((Q)).
K
This definition is equivalent to the previus one.
2.3.2
The Module of an Annulus
An annulus is just a topological annulus A, that is, it is a surface homeomorphic to
S 1 × (0, 1).
Proposition 2.3.24. There exsists a conformal map f : A → A(r, R), where A(r, R)
is the standard annulus (see Figure 2.5).
Notice that we can rescale the radii of the annulus A(r, R) through a conformal
map so that the radius r is equal to 1 and we call this new annulus A0 . Up to scaling,
the standard annulus A0 associated to A is unique.
e of A and A
e0 of A0 and we use the UniProof. We consider the universal cover A
e
e0 (see Figure
formization Theorem to go from A to a disc and from the disc to A
2.6).
34
CHAPTER 2. QUASI-CONFORMAL MAPS
R
r
Figure 2.5: The standard annulus A(r, R).
1 0000
0
11111
0
0
1
0
1
0
1
0
1
000000
111111
0
1
0
1
000000
111111
Figure 2.6: Proof of the Proposition 2.3.24.
Definition 2.3.25. We define the modulus of the annulus A as
m(A) :=
1
R
log( ),
2π
r
where r and R are the radii of the standard annulus A(r, R) associated to A.
Since the covering map π : R → A is the exponential map, m(R) = m(A).
The modulus of an annulus is useful because if f : X → Y is a K-quasi-conformal
map between two surfaces, then there exists a relationship between the length of a
curve γ in X and the associate annulus AX in the covering space:
lγ (X) =
π
.
m(AX )
If we can control how m(AX ) changes, then we can control the length of the curve
(see Figure 2.7).
35
2.4. EXTREMAL LENGTH
AX
γ
Y
X
Figure 2.7: Universal covering associate to γ.
Definition 2.3.26. We say that f : Ω → C is K-quasi-conformal if it is a homeomorphism onto its image and for all annulus A
1
m(A) ≤ m(f (A)) ≤ Km((A)).
K
2.4
Extremal Length
Definition 2.4.1. Let Ω ⊂ C be a region and ρ : Ω → [0, ∞) be a function (not
necessarily continuous or smooth). For every v ∈ Tz Ω, then |v|ρ = ρ(z)|v| is a
conformal metric, where |v| is the Euclidean length.
If γ(t) with a ≤ t ≤ b is a piecewise smooth path, then
Z b
Z b
′
|γ|ρ =
|γ (t)|ρdt =
ρ(γ(t))|γ ′ (t)|dt.
a
a
Similarly,
area(ρ) =
Z
ρ2 dxdy.
Ω
Let Q(z1 , z2 , z3 , z4 ) be a quadrilateral and Γ be the set of piecewise smooth paths in
Q from the side z1 z4 to z2 z3 .
Let ρ be a conformal metric on Q, then
LΓ (ρ) := infγ∈Γ |γ|ρ.
In order to get extremal length we need to normalize.
Definition 2.4.2. The extremal length E(Q, Γ) is given by
supρ
LΓ (ρ)2
.
area(ρ)
36
CHAPTER 2. QUASI-CONFORMAL MAPS
z4
z3
Γ1
z2
z1
Figure 2.8: A path Γ1 in Γ.
Lemma 2.4.3. If λ ∈ (0, ∞), then
LΓ (λρ)2
LΓ (ρ)2
=
area(ρ)
area(λρ)
The proof follows trivially from the definition. We would like to see that E(Q, Γ) =
m(Q).
Lemma 2.4.4. Let Q, Q′ be quadrilaterals and f : Q → Q′ be a conformal map
taking marked points to marked points. Then E(Q, Γ) = E(Q′ , Γ′ ).
Proof. Let ρ′ be a conformal metric on Q′ . Define ρ on Q to be
ρ(x) := |f ′ (x)|ρ′ (f (x)).
Then since Γ′ = f (Γ) we see that
LΓ′ (ρ′ )2
LΓ (ρ)2
=
area(ρ)
area(ρ′ )
2
′ 2
LΓ (ρ)
LΓ (ρ )
′
′
and hence E(Q, Γ) = supρ area(ρ)
= supρ′ area(ρ
′ ) = E(Q , Γ ).
Lemma 2.4.5. If R is the rectangle in C with vertices 0, a, a+ib, ib, then E(R, Γ) = ab .
Proof. Let ρ be a conformal metric on R. Then by Cauchy-Schwartz we have
!2
!2
ZZ
ZZ
Z Z
Z
b
Area(ρ)·ab =
R
ρ2 ·
Thus
R
12 ≥
a
b
ρ·1dxdy
0
0
≥
LΓ (ρ)dy
0
LΓ (ρ)2
a
≤ .
area(ρ)
b
This implies that E(R, Γ) ≤ ab . If ρ is the constant function 1, then
LΓ (ρ)2
a
=
area(ρ)
b
and hence E(R, Γ) = ab .
= (b·LΓ (ρ))2 .
37
2.4. EXTREMAL LENGTH
Corollary 2.4.6. m(Q) = E(Q, Γ).
Proof. Extremal length is invariant under confomal maps. By Grötzsch’s Theorem,
we can end Q conformally to the rectangle in C with vertices 0, a, a + ib, ib. Then
applying Lemma 2.4.5, m(Q) = ab = E(R, Γ).
Let Qt (z1 , z2 , z3 , z4 ) = Q(z2 , z3 , z4 , z1 ). Let Γt denote the paths for Qt . Then
1
notice that m(Qt ) = m(Q)
.
m(Qt ) = E(Qt , Γt ) = supρ
LΓt (ρ)2
1
=
,
area(ρ)
m(Q)
so m(Q) = infρ Larea(ρ)
2 . If ρ is a conformal metric on Q, then
t (ρ)
Γ
LΓ (ρ)2
area(ρ)
≤ m(Q) ≤ 2
.
area(ρ)
LΓt (ρ)
If ρ = 1, the above inequality is called Rengel’s inequality.
Example 2.4.7. Let Q be the whole quadrilateral and Q0 the subquadrilateral as in
Figure 2.9.
Q
Q0
Figure 2.9: The subquadrilateral Q0 .
Let ρ0 be a confomal metric on Q0 and ρ the conformal metric given by
ρ0 (x)
x ∈ Q0
ρ(x) =
0
x ∈ Q − Q0 .
Then
m(Q0 ) =
area(ρ0 )
area(ρ)
= 2
≥ m(Q).
2
LΓt (ρ0 )
LΓt (ρ)
Definition 2.4.8. (Extremal length for annuli) In this case, Γ will be the set of paths
from one boundary component to the other boundary component. Then
E(A, Γ) = sup
ρ
LΓ (ρ)2
.
area(ρ)
Let A(r, R) := {(t, θ) ∈ R2 | r ≤ t ≤ R, θ ∈ [0, 2π)}.
38
CHAPTER 2. QUASI-CONFORMAL MAPS
Figure 2.10: A fundamental domain of the universal covering of the annulus A.
Figure 2.11: The annulus A in the Example 2.4.11.
Lemma 2.4.9. Let A = A(r, R). Then E(A, Γ) =
1
log( Rr ).
2π
Proof. Apply the same argument to some fundamental domain of the universal covering of the annulus which is a strip.
1
. This metric arises by taking the
The nice metric on the annulus is ρ(z) = |z|
natural map from the bi-infinite Euclidean cylinder and mapping it to C. Under this
1
conformal map the standard Euclidean metric pushes forward to |z|
.
t
In this case, Γ := {simple closed paths seperating the boundary components}.
LΓt (ρ)2
.
Let E(A, Γt ) = supρ area(ρ)
Lemma 2.4.10. If A = A(r, R), then
E(A, Γt ) =
2π
1
=
.
R
m(A)
log( r )
Proof. Apply Cauchy-Schwartz to ρ and
1
.
|z|
2
LΓ (ρ)
Define m(A) = supρ area(ρ)
= infρ Larea(ρ)
2.
t (ρ)
Γ
Example 2.4.11. Let A be an annulus with diam(inner boundary) ≥ ǫ and diam(A) ≤
R.
If ρ = 1, then Area(ρ) ≤ π( R2 )2 and LΓt (ρ) ≥ 2ǫ. This implies that
m(A) ≤
πR2
.
(2ǫ)2
Theorem 2.4.12. Let ∆ = {z ∈ C | 0 < |z| ≤ 1}. If f : ∆ → C is quasi-conformal
and bounded then f extends continuously to 0.
2.4. EXTREMAL LENGTH
39
Proof. Let A = {z ∈ C | 0 < |z| ≤ 1}, then area(A) = ∞. Let Cr = {z ∈ C | |z| = r}.
f extends continuously if and only if for all ǫ there exists a δ such that if r < δ then
diam(f (Cr )) < ǫ. For the standard annulus A(r, R) we see that
1 1
1
m A r,
=
→ ∞ as r → ∞.
log
2
2π
2r
But if f does not extend then m(f (A(r, 12 ))) is uniformly bounded above. Since
m(f (A(r, 12 ))) ≥ K1 m(A(r, 12 )), m(A(r, 21 )) is uniformly bounded and this gives a contradiction.
Corollary 2.4.13. There does not exist a quasi-conformal map from C to ∆.
Proof. Assume that f : C → ∆ is such a map. Let g(z) = f ( z1 ) then g is defined
on ∆∗ and is bounded. This implies that g extends to zero and hence f extends to
b and ∆ are not homeomorphic.
infinity. This is a contradiction since C
2.4.1
Proof of the Compactness Theorem
We gave three definitions of quasi-conformal map: the analytic definition, the quadrilateral definition and the annulus definition.
By Grötzsch’s Theorem, the analytic definition implies the quadrilateral definition.
Moreover, if we cut an annulus A along two arcs, then we get two quadrilaterals Q1
and Q2 . Comparing m(A) with m(Q1 ) + m(Q2 ) it is easy to see that we get the
equality if A = A(r, R). So, using the standard annuli, we see that the quadrilateral
definition implies the annulus definition.
We will give a sketch of the proof of the Compactness Theorem 2.3.19 using the
quadrilateral definition and the fact that the quadrilateral definition implies the annulus definition.
Proof. We will divide the proof in many steps.
Step (0): We reduce to the case where Ai = {ai }.
If f ∈ F, then there exists a unique Möbius transformation φ such that φ(f (ai )) = ai
and
{φ ∈ Möb = P SL2 (C) | φ−1(ai ) ∈ Ai }
is compact.
Let assume that we proved Step (0).
b which is the metric conformal to the standard
We will use the spherical metric on C
metric on the sphere.
Theorem 2.4.14 (Arzela-Ascoli). Let X and Y be two metric spaces and fn : X → Y
be continuous maps such that
1. fn are equicontinuous i.e., ∀ ε > 0 ∃ δ > 0 such that if ∀ x0 , x1 ∈ X so that
d(x0 , x1 ) < δ, then d(fn (x0 ), fn (x1 )) < ε for all n;
2. {fn (x)} is compact for all x ∈ X.
40
CHAPTER 2. QUASI-CONFORMAL MAPS
Then fn has a uniformly convergent subsequence.
Step (1): Let fn ∈ F. Then fn has a subsequence that converges uniformly on
compact sets.
S
We know that there exist countable compact sets C1 ⊂ C2 ⊂ · · · with Ω = Ci .
Let assume that we have Step (1) for a compact set. Then inductively we construct
subsequences {fi,j } such that {fi,j } converges uniformly on Ci and {fi+1,j } is a subsequence of {fi,j }. By the diagonal argument, {fi,i } converges uniformly on all Cj .
Since any compact set is contained in some Cj , {fi,i } converges uniformly on all compact sets.
It remains to prove:
Step (1’): Let fn ∈ F and C ⊂ Ω be a compact set. Then fn has a subsequence that
converges uniformly on C.
We want to apply the Arzela-Ascoli Theorem.
• Equicontinuity: Fix ε > 0. Choose r > 0 such that ε < r and
1. d(ai , aj ) > 2r for i, j ∈ {1, 2, 3};
2. If z ∈ C, B(z, r) ⊂ Ω.
Then for every z ∈ C, at most one of a1 , a2 , a3 is in B(z, r). We can assume that
a2 , a3 ∈
/ B(z, r).We assume that the ball B(z, r) is so small that the euclidean
metric is equivalent to the spherical metric.
Let A := A(z, δ, r) denote the annulus with center z, inner radius δ and outer
radius r.
2
Choose δ > 0 such that m(A(z, δ, r)) > Kπr
.
ε2
′
′
Let z ∈ C such that d(z, z ) < δ.
Let f ∈ F. Since f is K-quasi-conformal, we have
m(f (A(z, δ, r))) ≥
1
πr 2
m(A(z, δ, r)) > 2 .
K
ε
Let η := min{d(f (z), f (z ′ )), d(a2 = f (a2 ), a3 = f (a3 ))}. Notice that d(a2 =
f (a2 ), a3 = f (a3 )) > 2r.
We want an upper bound of m(f (A(z, δ, r))) in terms of η:
m(f (A(z, δ, r))) = infρ
and so
4πr 2
πr 2
area(ρ)
≤
=
LΓt (ρ)2
(2η)2
η2
πr 2
πr 2
>
.
η2
ε2
Hence, η < ε < r.
If η < 2r, then η = d(f (z), f (z ′ )) and so d(f (z), f (z ′ )) < ε.
b is compact.
• {fn (x)} is compact for every x since the sphere C
41
2.4. EXTREMAL LENGTH
fn(A)
A
z2
11
00
z1
1
0
z3
1
0
11
00
fn (z2 )
11
00
fn (z3 )
01
fn (z1 )
Figure 2.12: The spherical arcs in A and their images in fn (A).
By the Arzela-Ascoli Theorem, there exists a subsequence {fnj } which converges
uniformly to f and f is continuous.
Step (2): Let fn ∈ F and fn → f uniformly on compact sets. Then, at each z ∈ Ω,
f is either locally injective or locally constant.
By contradiction, we fix B(z, r) ⊂ Ω and we assume that there exist z1 , z2 , z3 ∈ B(z, r)
such that f (z1 ) 6= f (z2 ) = f (z3 ). We connect these points with spherical arcs as in
Figure 2.12.
Since f (z1 ) 6= f (z2 ), there exists r such that d(fn (z1 ), fn (z2 )) > r for every n.
Moreover, εn := d(fn (z2 ), fn (z3 )) → 0, because f (z2 ) = f (z3 ).
Assume 2εn < r. Let An := A(fn (z2 ), 2εn , r).
Both the boundary components of An intersect both the boundary components of
fn (A). Hence, each boundary components intersect the annulus fn (A) twice. Define
1
, if z ∈ fn (A) ∩ An
|z−f2 (z)|
ρn (z) :=
0,
otherwise.
Since LΓ (ρn ) ≥ 2 log( 2εrn ), we have
2π log( 2εrn )
area(ρn )
m(fn (A)) ≤
→ 0,
≤
LΓ (ρn )2
(2 log( 2εrn ))2
where LΓ (ρn ) is the set of closed curves that separate the boundary components.
We proved that any sequence fn ∈ F has a subsequence that converges uniformly on
compact sets.
We need to show: if fn ∈ F, fn → f , then f ∈ F.
Step (3): f is injective.
We know that at every point f is either locally constant or locally injective. Obviously,
locally constant is an open condition. If zi → z and f is locally constant at zi , then
f is not locally injective at z because locally injective is an open condition and so f
is locally constant at z. Therefore, locally constant is a closed condition. Since Ω is
connected, if there is one point where f is locally constant, then f is constant.
Since f is not constant because f (ai ) = ai for i = 1, 2, 3, f is locally injective.
Now, let z ∈ Ω and B = B(z, r) ⊂ Ω such that f is injective on B.
Since f is locally injective we have the following fact:
42
CHAPTER 2. QUASI-CONFORMAL MAPS
R
ω
Q
Figure 2.13: The rectangle R and the quadrilateral Q.
1
0
0
1
00
11
00
11
11
00
00Q
11
00
11
00
11
00
11
Qn
0
1
11
00
11
00
1
0
0
1
Figure 2.14: The quadrilateral Qn .
Fact 2.4.15. There exists ε > 0 such that fn (B) ⊃ B(f (z), ε) for every n sufficiently
large.
Since fn is globally injective, if z ′ ∈ Ω \ B, then fn (z ′ ) ∈
/ B(f (z), ε). Therefore,
′
f (z ) 6= f (z). Hence f is injective and f (ai ) = ai , for i = 1, 2, 3, since fn (ai ) = ai , for
i = 1, 2, 3 for every n.
The last thing to show is the K-quasi-conformality of f .
Step (4): f is K-quasi-conformal.
We will show that if Q is a quadrilateral, then m(fn (Q)) → m(f (Q)) as n → ∞.
Let R a rectangle and Q a quadrilateral with vertices in the squares as in Figure 2.13.
We put the standard euclidean metric on R.
Fact 2.4.16. Given ε > 0 there exists δ > 0 such that if ω < δ, then |m(Q)−m(R)| <
ε.
Let Q ⊂ Ω.
Fact 2.4.17. For large n there exists Qn ⊂ Q such that fn (Qn ) ⊂ f (Q), m(Qn ) →
m(Q) as n → ∞ and fn (Qn ) is in a δ-neighborhood of f (Q).
So, taking the limit as n → ∞,
1
m(Qn )
K
↓
1
m(Q)
K
≤ m(fn (Qn )) ≤ Km(Qn )
↓
↓
≤ m(f (Q)) ≤ Km(Q).
This concludes the proof of the theorem.
43
2.5. GRÖTZSCH’S THEOREM IN THE GENERAL FORM
2.5
Grötzsch’s Theorem in the general form
We proved Grötzsch’s Theorem in the case of differentiable functions. The aim of
this section is to give the generalization of Grötzsch’s Theorem in the case of almost
everywhere differentiable functions and one application of this theorem.
Lemma 2.5.1. Let Q = Q(z1 , z2 , z3 , z4 ), Q1 = Q(z1 , z5 , z6 , z4 ) and Q2 = Q(z5 , z2 , z3 , z6 )
such that Q = Q1 ∪ Q2 , Q◦1 ∩ Q◦2 = ∅. Then m(Q) ≥ m(Q1 ) + m(Q2 ). We get the
equality if and only if Q1 and Q2 are rectangles.
Proof. Let ρ = 1 on Q and ρ1 = ρ|Q1 , ρ2 = ρ|Q2 . By the definitions of ρ, ρ1 and ρ2 ,
LΓt1 (ρ1 ) ≥ LΓt (ρ),
LΓt2 (ρ2 ) ≥ LΓt (ρ).
Therefore,
m(Q1 ) + m(Q2 ) ≤
≤
area(ρ1 ) area(ρ2 )
+
≤
LΓt1 (ρ1 )2
LΓt2 (ρ2 )2
area(ρ1 ) + area(ρ2 )
area(ρ)
=
≤ m(Q).
2
LΓt (ρ)
LΓt (ρ)2
Hence, m(Q) ≥ m(Q1 ) + m(Q2 ).
The last part of the statement is trivial.
We now can prove Grötzsch’s Theorem in the general form:
Theorem 2.5.2 (Grötzsch). Let R1 and R2 be rectangles in C with vertices (0, a, a +
′
i, i) and (0, a′ , a′ + i, i), a′ > a. If f is K-quasi-conformal, then K ≥ aa . Moreover,
′
K = aa if and only if f is the affine mapping of R1 onto R2 .
Proof. We consider two rectangles Q1 and Q2 in R′ as in Figure 2.15.
R′
Q1
R
f −1
Q2
x
f −1(Q1)
f −1(Q2)
x̃
Figure 2.15: The rectangles Q1 and Q2 in R′ .
By Lemma 2.5.1 and the fact that f is K-quasi-conformal,
1 ′
1
a = (m(Q1 ) + m(Q2 )) ≤ m(f −1 (Q1 )) + m(f −1 (Q2 )) ≤ m(R) = a
K
K
′
and so K ≥ aa .
We have the equality K =
Q1 and Q2 scaling by K.
a′
a
if and only if f −1 (Q1 ) and f −1 (Q2 ) are obtained from
44
CHAPTER 2. QUASI-CONFORMAL MAPS
So f −1 (Q1 ) and f −1 (Q2 ) are rectangles if and only if there exists x̃ such that f −1 (x +
iy) = x̃ + iỹ. We have
m(f −1 (Q1 )) = x̃,
and
Then
x
≤ x̃ ≤ Kx,
K
m(f −1 (Q2 )) = a − x̃
(a′ − x)
≤ (a − x̃) ≤ K(a′ − x).
K
1
x̃
≤ ≤ K,
K
x
1
a − x̃
≤ ′
≤ K.
K
a −x
x̃
1
x
If xx̃ > K1 , then K1 > aa−x̃
′ −x , which is a contradiction. So x = K , that is x̃ = K .
Similarly, for the y-coordinate we use f and we draw a horizontal line to define the
rectangles Q1 and Q2 in R. The calculation is the same as for the x-coordiante.
Using the geometric definition of the modulus and the Theorem 2.5.2 it is easy to
prove the following
Theorem 2.5.3. If f is 1-quasi-conformal, then f is conformal.
It is possible to prove the theorem using the analytic definition. The problem is
that we don’t know that the composition g ◦ f of a quasi-conformal map f and a
conformal map g is quasi-conformal. The difficult part to show is the ACL condition.
So the proof uses the geometric definition.
Proof. We consider a square Q. So f (Q) is a quadrilateral. Since f is 1-quasiconformal, there exists a map g which sends the quadrilateral f (Q) to a square Q′ .
Since g ◦ f sends the square Q to the square Q′ , by Grötzsch’s Theorem in the general
form, g ◦ f is the affine map. Hence, f is conformal.
Chapter 3
Teichmüller Space and Moduli
Space
3.1
Riemann Surfaces
Let S be a topological surface, i.e., a Hausdorff paracompact 2-manifold. A chart on
S is a pair (U, φ) where U ⊆ S is open and φ : U → C is a homeomorphism onto its
image. An atlas A is a collection of charts such that ∀x ∈ S there exists (U, φ) ∈ A
with x ∈ U.
Let (U0 , φ0 ), (U1 , φ1 ) be charts, where U0 ∩ U1 is non-empty. The map φ1 ◦ φ−1
0 :
φ0 (U1 ∩ U2 ) → φ1 (U1 ∩ U2 ) is called a transition map. Note that transition maps are
homeomorphisms of subsets of C.
Definition 3.1.1. A Riemann surface is an atlas where all the transition maps are
holomorphic1 .
Example 3.1.2.
0) C with an atlas containing only the identity id : C → C.
1) Ω ⊂ C with an atlas containing only the inclusion.
b = the one point compactification of C. The atlas consists of two charts id :
2) C
U = C → C and id : V = C → C with the transition map φ : C \ {0} → C \ {0}
defined by φ(z) = z1 .
3) Let f : C2 → C be holomorphic. One can endow f −1 (0) with a complex
structure2 .
Notation 3.1.3. We will always denote by S a topological surface and by X a surface
with a complex structure (a Riemann surface).
1
2
Note that since they are also homeomorphisms they are actually biholomorphic.
“inverse function thm”?
45
46
CHAPTER 3. TEICHMÜLLER SPACE AND MODULI SPACE
Definition 3.1.4. Let X be a Riemann surface. A map f : X → C is holomorphic
if for every chart (U, φ) the map f ◦ φ−1 is holomorphic.
The morphism f : X → Y is holomorphic if for ever chart (U, φ) of X and every chart
(V, ψ) of Y the map ψ ◦ f ◦ φ−1 is holomorphic (where defined).
b is holomorphic, we say that f is meromorphic.
In the special case that f : X → C
b is the set
Fact 3.1.5. From the course on complex analysis we know that Mero(C)
of rational functions.
Definition 3.1.6. Given a Riemann surface X, Aut(X) is the set of holomorphic
maps f : X → X which are homeomorphisms. This is in fact a group since a
composition of homlomorphic maps is holomorphic, and as for the inverse: if f is
holomorphic and injective then f ′ (z) 6= 0, we can find a holomorphic inverse for f in
a neighborhood of z. Collecting the charts we define f −1 .
Definition 3.1.7. SL2 (C) is the set of 2 × 2 matrices with complex entries and
determinant equal to 1. PSL2 (C) = SL2 (C)/ ∼ where A ∼ B if they differ by a
non-zero (complex) scalar.
b = PSL2 (C)
Theorem 3.1.8. Aut(C)
a
c
identified via the isomorphism
az + b
b
→
d
cz + d
Theorem 3.1.9 (The Uniformization Theorem). If X is a simply connected Riemann
b or the unit disc ∆ = {z||z| < 1}.
Surface, then X is biholomorphic to C, C
b is not homeomorphic to C or ∆, therefore it is different as
Remark. Notice that C
a Riemann surface. Liouville’s theorem implies that any holomorphic map from C
b ∆ are
to ∆ must be constant, thus it cannot be a homeomorphism. Therefore C, C,
distinct Riemann surfaces.
We can compute Aut(∆) and Aut(C) relatively painlessly from Theorem 3.1.8,
and
b is simply connected,
Theorem 3.1.10 (The Riemann Mapping Theorem). If D ⊆ C
and its complement contains more, than one point then there is a biholomorphic
homeomorpism h : D → ∆.
b with an isolated singularity
If φ ∈ Aut(C), then it is a holomorphic function on C,
only (possibly) at ∞. If the singularity was essential, or a pole of order greater
than 1 then in a neighborhood of ∞, φ would not be injective. Thus φ must be a
b which is in fact linear. Therefore,
meromorphic function on C
Aut(C) = {a + bz | a, b ∈ C}.
As for ∆, first recall by that by the Riemann Mapping Theorem ∆ is biholomorphic
to H2 = {z|Im(z) > 0} the upper half plane. If φ ∈ Aut(H2 ), then by the Schwartz
47
3.1. RIEMANN SURFACES
b which preserves the
Reflection Principle we can extend it to an automorphism of C,
upper half plane. This is exactly the subgroup PSL2 (R).
Aut(∆) ∼
= Aut(H2 ) = PSL2 (R)
Now we wish to classify all closed Riemann Surfaces X. Any such surface will be
orientable so X is homeomorphic to S2 , T2 , or Sg for g > 1. Since X is a Riemann
b C, ∆.
e By Theorem 3.1.9 there are three cases: X
e = C,
surface - so is X.
b then X is topologically a sphere so X = C.
b
e=C
If X
3
e = C then X is homeomorphic to a torus (if we allow open manifolds then
If X
X could also be an annulus). Here there are many different Riemann structures
we can endow T2 with. Given X and thinking of π1 (X) as automorphisms of C, i.e.,
z → z + bx + iby , we have a homomorphism π1 (X) ֒→ C. If we identify π1 (X) with Z2 ,
this homomorphism is determined by (1, 0) → b1 and (0, 1) → b2 where b1 , b2 ∈ C. So
we get a parallelogram with vertices at 0, b1 , b2 , b1 + b2 . It degenerates only if bb21 ∈ R
/ R
(but that is impossible if X is a torus). Conversely, given b1 , b2 ∈ C with bb21 ∈
we get a torus with a complex structure Xb1 ,b2 by identifying opposite sides of the
parallelogram. When are Xb1 ,b2 and Xc1 ,c2 the same Riemann Surface?
For example, if there exists a λ ∈ C such that ci = λbi then the automorphism
f :C → C
z → λz
descends to an automorphism f : Xb1 ,b2 → Xc1 ,c2 . Thus Xb1 ,b2 is biholomorphic to
/ R. Also note that X1,λ is biholomorphic to
X1,λ where λ = bb21 . Notice that λ ∈
X1,−λ after switching orientation. In conclusion, after fixing the orientation one can
parameterize Xb1 ,b2 by H2 , the upper half plane.
If X0 , X1 are bi-holomorphic with parameters λ1 , λ2 respectively, does it follow
that λ1 = λ2 ? No. The problem lies when we identified π1 (X) with Z2 . If we had
made a different choice we’d have gotten a different parameter. We fix this by making
our choice of generators explicit.
Let T2 be an oriented torus. Fix generators for π1 (T2 ) so that their order matches
with the orientation. Consider the set of pairs (X, f ) so that X is a Riemann surface
on the torus and f : T2 → X is an orientation preserving homeomorphism. There
is a map from the set {(X, f )} to the upper half plane: Given (X, f ), lift the paths
α, β that represent the generators in π1 (X) and post-compose with an automorphism
which takes the endpoint of α̃ to 1. Then λ is the endpoint of the lift β̃.
We’ve seen that the map (X, f ) → λ is onto, since given λ we can glue opposite
sides of the parallelogram with vertices at 0, 1, λ, λ + 1 to get a Riemann surface on
the torus Xλ .
e = C. The proof involves understanding the
In fact if X is homeomorphic to the torus then X
discrete abelian subgroups of PSL2 (R).
3
48
CHAPTER 3. TEICHMÜLLER SPACE AND MODULI SPACE
We define (X0 , f0 ) ∼ (X1 , f1 ) if there exists a bi-holomorphic map φ : X0 → X1
such that φ ◦ f0 is homotopic to f1 .
3.2
Special Case: the Torus
We consider
T 2 = S 1 × S 1 = R2 /h(x, y) 7→ (x + 1, y), (x, y) 7→ (x, y + 1)i.
Let (X, f ) be a marked Riemann surface on T 2 . We have
f˜
R2
›
f
T2
/
e
X
›
/
X
e is the universal cover of X and it is not canonically isomorphic to C. Indeed,
where X
we can choose f˜ so that f˜(0, 0) = 0 and f˜(1, 0) = 1. Then f˜(0, 1) = τ and if we put
an orientation on X, we can take Im(τ ) > 0.
We have the following
Lemma 3.2.1. Let (X0 , f0 ) and (X1 , f1 ) be two marked Riemann surfaces with parameters τ0 and τ1 respectively.
(X0 , f0 ) ∼ (X1 , f1 ) if and only if τ0 = τ1 .
Proof. First, we suppose (X0 , f0 ) ∼ (X1 , f1 ) and so there exists a conformal map
φ : X0 → X1 such that φ ◦ f0 ∼ f1 . Since φ ◦ f0 ∼ f1 , (φ ◦ f0 )∗ = (f1 )∗ as maps
π1 (T 2 , x0 ) → π1 (X1 , f1 (x0 )) (we may suppose that φ(f0 (x0 )) = f1 (x0 )). We consider
the following diagram
R2
›
T2
f˜
f
/
/
C
φ̃
›
φ
X0
/
C
›
/
X1
e = 0, that is, φ(z) = az for some
and without loss of generality, we suppose that φ(0)
complex number a.
e = a = 1 and so
Since f˜0 (0, 0) = 0 = f˜1 (0, 0) and f˜0 (1, 0) = 1 = f˜1 (1, 0), we have φ(1)
e 0 ) = τ1 .
φ(τ
On the other hand, id ◦ f˜0 ∼ f˜1 in an equivariant way. Let φe be the linear homotopy
e
and define φ to be the map induced by φ.
Hence, there is a bijection from equivalence classes of marked Riemann surfaces
on T 2 to the Upper Half Plane.
For any τ = a + ib, we can define (Xτ , fτ ) by defining first f˜τ : R2 → C,
x
1 a
˜
fτ (x, y) =
y
0 b
49
3.2. SPECIAL CASE: THE TORUS
and then considering the map fτ induced by f˜τ so that the diagram
R2
›
T2
f˜τ
C
/
›
fτ
/
Xτ
commutes.
s q
∈ P SL2 (Z) = SL2 (Z) and define f˜ : R2 → C
Now, let
r p
x
s q
1 a
˜
.
f (x, y) =
y
r p
0 b
Then, we consider the map f induced by f˜ so that the diagram
R2
›
T2
f˜
f
/
C
›
/
Xτ
commutes. If z0 := f˜(1, 0) and z1 := f˜(0, 1), then
z1
pτ + q
=
,
z0
rτ + s
where τ is the parameter associate to Xτ .
So we defined an action of SL2 (Z) on {(X, f )}.
SL2 (Z) is the so called Mapping Class Group of T 2 .
We denote by Dif f + (T 2 ) the group of orientation-preserving diffeomorphisms of T 2
and by Dif f0+ (T 2 ) the subgroup of diffeomorphisms of T 2 homotopic to the identity.
Definition 3.2.2. The Mapping Class Group of the Torus is denoted by MCG(T 2 )
and it is defined as
Dif f + (T 2 )
MCG(T 2 ) :=
.
Dif f0+ (T 2 )
We have the following
Lemma 3.2.3. Let f0 , f1 : T 2 → T 2 ∈ Dif f + (T 2 ), then f0 ∼ f1 ⇔ (f0 )∗ = (f1 )∗ .
Proof. If f0 ∼ f1 , then obviously (f0 )∗ = (f1 )∗ .
If (f0 )∗ = (f1 )∗ , first we lift the maps to the universal cover and we take the linear
homotopy. Then we project down the homotopy and we get f0 ∼ f1 .
Hence, [f0 ] = [f1 ] in MCG(T 2 ) if and only if (by definition) f0 ∼ f1 if and only if
(by the lemma) (f0 )∗ = (f1 )∗ .
Fact 3.2.4. Every automorphism of π1 (T 2 ) = Z2 is the induced map of some diffeomorphism of T 2 .
50
CHAPTER 3. TEICHMÜLLER SPACE AND MODULI SPACE
Indeed, we have Aut(Z2 ) = GL2 (Z) and Aut+ (Z2 ) = SL2 (Z) and for every M ∈
GL2 (Z) we have the following commutative diagram
R2
M
›
/
T2
R2
›
/
T 2.
Hence, MCG(T 2 ) = SL2 (Z).
MCG(T 2 ) acts on marked Riemann surfaces of the torus in the following way: if
ϕ ∈ MCG(T 2 ) and (X, f ) is a marked Riemann surface of the torus, then (X, f ) 7→
(X, f ◦ ϕ−1 ). (X, f ◦ ϕ−1 ) is a marked Riemann surface since (X, f ) ∼ (X1 , f1 ) ⇔
(X, f ◦ ϕ−1 ) ∼ (X1 , f1 ◦ ϕ−1 ).
We can sum up what we have proved in the case of the torus.
We have a bijection between H2 := {z ∈ C | Im(z) > 0} and the Teichmüller space
T(T 2 ) for T 2 .
2
2
2
The Mapping
Class
Group MCG(T ) acts on T(T ) inthe following
way: if τ ∈ T(T )
p q
s −q
and ϕ =
∈ MCG(T 2 ) = SL2 (Z) so ϕ−1 =
and
r s
−r p
ϕ·τ =
pτ − q
.
−rτ + s
Let (X0 , f0 ) and (X1 , f1 ) marked Riemann surfaces and ϕ : X0 → X1 . If (ϕ ◦ f0 ) fi
f1 ⇔ (X0 , f0 ) fi (X1 , f1 ), then there exists ψ ∈ MCG(T 2 ) such that ϕ ◦ f0 ∼ f1 ◦ ψ −1 .
Definition 3.2.5. The quotient of T(T 2 ) under the action of MCG(T 2 ) is called the
Moduli Space M1 of the torus and is the set of unmarked conformal structure of T 2 .
Since
T(T 2 )
H2
M1 =
=
,
MCG(T 2 )
P SL2 (Z)
the Moduli Space of the torus is in bijection with the region in Figure 3.1.
3.3
Quasi-Conformal Map on the Torus
Let f : X0 → X1 a diffeomorphism between manifolds.
Definition 3.3.1. f is a K-quasi-conformal map if for all charts (U, φ) on X0 and
(V, ψ) on X1 , the composition ψ ◦ φ ◦ ψ −1 is K-quasi-conformal where it is defined.
We want to define the Teichmüller distance in T(T 2).
Definition 3.3.2. Let Kf be the quasi-conformal constant for f . The Teichmüller
distance between the marked Riemann surfaces (X0 , f0 ) and (X1 , f1 ) is
1
dT eich ((X0 , f0 ), (X1 , f1 )) := log(inf (Kφ )),
2
where the infimum is taken over the diffeomorphisms φ : X0 → X1 with φ ◦ f0 ∼ f1 .
3.4. HYPERBOLIC AND CONFORMAL STRUCTURES
1
0
1
0
0
1
1
0
0
1
0
1
1
0
0
1
0
1
0
1
0
1
0
1
0
1
0
1
1
0
0
1
0
1
0
1
0
1
−1
51
1
0
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
0
1
1
Figure 3.1: Moduli Space of the torus.
This distance determines a metric on T(T 2) which is equal to the hyperbolic metric
on H2 and this is the reason of why we have 12 in the definition of the Teichmüller
distance.
The infimum is realized in the case of the torus, but it is not realized in general.
Moreover, we will prove that when the infimum is realized, the diffeomorphism that
realizes the infimum is unique and “behaves nicely”, in the sense that it is a stretch
in one direction and a shrink in the other direction.
In the case of the torus we have this nice result which is not true in the general case.
Theorem 3.3.3. Let X be a conformal structure in T 2 . Given any two points z0 and
z1 in X, there exists a conformal map φ : X → X isotopic to the identity such that
φ(z0 ) = z1 .
e∼
Proof. We lift X to the universal cover X
= C and we define
φet (z) = z + (e
z1 − ze0 )t,
where ze0 and ze1 are two lifts of z0 and z1 . Then we project φe1 down. By construction,
this map satisfies the thesis of the theorem.
3.4
Hyperbolic and Conformal Structures
We will define the hyperbolic and conformal structures on a manifold and we will prove
that conformal structures, Riemann surface structures, and hyperbolic structures all
coincide when the dimension of the manifold equals 2 and genus ≥ 2.
52
3.4.1
CHAPTER 3. TEICHMÜLLER SPACE AND MODULI SPACE
Hyperbolic Structures
Definition 3.4.1. (First Definition) A hyperbolic structure on a manifold M is a
Riemannian metric with constant sectional curvature equal to −1.
Theorem 3.4.2. There is a unique simply connected n-manifold (Hn ) with a complete hyperbolic structure.
An important fact is that the isometry group of Hn acts transitively. In other
words, if x, y ∈ Hn , then there exists an isometry φ : Hn → Hn with φ(x) = y. This
shows that Hn is a homogeneous space (every point looks like every other point).
Corollary 3.4.3. Let M be a complete hyperbolic n-manifold. Then every point
x ∈ M has a neighborhood isometric to a neighborhood of Hn .
Definition 3.4.4. (Second Definition) A hyperbolic structure on a manifold M is an
atlas with chart maps mapping into Hn and transition maps given by restrictions of
isometries of Hn .
At this point the focus will shift to the case where n = 2. In this case H2 ⊂ C
and isometries are holomorphic. This immediately implies that a hyperbolic structure
determines a Riemann surface structure.
Lemma 3.4.5. Let X h → X r be the natural map from the space of hyperbolic
structures to the space of Riemann surface structures. Then it is the case that X0r =
X1r if and only if X0h = X1h .
Proof. (⇐) This direction is obvious.
(⇒) Since X0r and X1r also carry a hyperbolic structure, we see by Theorem 3.4.2
that the universal cover of X0r and X1r is H2 (specifically we will use the disc model
∆).
f0r
∆=X
X0r
φe
f1r = ∆
X
X1r
Now there is a map conformal automorphism φe : ∆ → ∆. We saw earlier that
conformal automorphisms of ∆ are hyperbolic isometries. Thus the map φe descends
to a hyperbolic isometry φ : X0r → X1r .
3.4. HYPERBOLIC AND CONFORMAL STRUCTURES
53
At this point we would like to see which Riemann surfaces carry a hyperbolic structure explicitly. To do this, we first recall the Uniformization Theorem which states
b or ∆ = H2 . Knowing the
that every simply connected Riemann surface is either C, C,
classification of surfaces and the fact that all Riemann surfaces are orientable (holomorphic maps are orientation-preserving) we can classify closed Riemann surfaces by
their universal covers.
Theorem 3.4.6. Let X be a closed Riemann surface.
e=C
b if and only if X = C.
b
1. X
e = C if and only if X ∼
2. X
= T 2.
e = ∆ if and only if X has genus≥ 2.
3. X
Proof. The only orientable surface covered by the sphere is the sphere. When comb has a fixed point (implies
bined with the fact that every conformal automorphism of C
group of deck transformations is trivial) gives the first statement.
The deck transformation group (z 7→ az + b such that there are no fixed points)
e
of X = C is abelian. The only surfaces with abelian fundamental group are T 2 and
b but C
b is not covered by the plane which proves the forward implication of 2).
C,
The backward implication of 3) holds since these surfaces have non-abelian fundamental group. This only leaves the backward implication of 2) and the forward
implication of 3) which are equivalent.
To do this we need to examine abelian subgroups of P SL2 (R) = Isom+ (H2 ).
Exercise 3.4.7. Let γ, β ∈ Isom+ (H2 ). [γ, β] = id if and only if f ix(γ) = f ix(β)
where f ix(γ) = {z ∈ ∆ | γ(z) = z}.
By the classification of isometries we get three cases.
1. f ix(γ) = z ∈ ∆ if and only if γ is elliptic. But deck transformations have no
fixed points.
2. f ix(γ) = z ∈ ∂∆ if and only if γ is parabolic. Another exercise is if g0 and g1
are parabolic then there exists an h such that h ◦ g0 ◦ h−1 = g1 .
In the upper half space model H2 we can assume by the exercise that γ(z) =
z + 1. Thus if [γ, β] = id then β(z) = z + λ with λ ∈ R.
Exercise 3.4.8. Any injective map of Z2 in R has dense image.
e = H then π1 (X) ֒→ P SL2 (R). The image
If X is a Riemann surface and X
must also be discrete or else it is not a deck transformation of H2 . In fact the
same problem comes about with elliptics.
3. f ix(γ) = f ix(β) = {z0 , z1 } ⊂ ∂∆. There is a unique geodesic Λ ⊂ ∆ with
endpoints z0 and z1 . The set of orientation-preserving isometries preserving Λ
looks like R with the additive structure.
54
CHAPTER 3. TEICHMÜLLER SPACE AND MODULI SPACE
These arguments show that there are no discrete subgroups of P SL2 (R) isomorphic to Z2 . Thus the torus T 2 cannot be covered by ∆ which finishes the 2 missing
implications.
e = ∆ then X r has a hyperbolic
Lemma 3.4.9. If X r is a Riemann surface and X
h
h
r
structure X with X = X .
e = ∆, around any point p ∈ X r we can find a neighborhood that lifts
Proof. Since X
to the universal cover. This neighborhood and the lift together form a chart around
the point p mapping into ∆. The only thing left to check is that the transition
maps are hyperbolic isometries. Given two charts of the above form p ∈ (U, ϕ) and
q ∈ (V, ψ) it might be the case that we choose lifts that do not intersect. But there
is a deck transformation taking one lift of U to the lift that intersects the chosen lift
of V . This transformation is, by definition, a hyperbolic isometry.
This discussion gives an extension of the Uniformization Theorem.
Theorem 3.4.10. If X is a closed Riemann surface with genus(X) ≥ 2, then X
has a hyperbolic structure. If genus(X) = 1, then X has a Euclidean structure. If
e = ∆ or X
e = C. If π1 (X) is non-abelian, then
π1 (X) = Z, {1} is abelian, then X
e
X = ∆.
3.4.2
Conformal Structures
We would like to define a notion of angle.
In Rn with the usual dot product structure the regular notion of angle θ = \(v, w)
is given by < v, w >= |v||w|cosθ. If V is a vector space with an inner product < , >,
then define θ = \(v, w) to be
< v, w >:= |v||w|cosθ.
Two inner products < , >0 and < , >1 are conformally equivalent if \0 (v, w) =
\1 (v, w) for all v, w ∈ V .
Exercise 3.4.11. < , >0 and < , >1 are conformally equivalent if and only if there
exists λ > 0 such that < , >0 = λ < , >1
Definition 3.4.12. A conformal structure on V is an equivalence class of inner products.
Definition 3.4.13. Let g0 , g1 be Riemannian metrics on M. Then g0 , g1 are conformally equivalent if there exists λ : M → R+ such that g0 = λg1 and λ is smooth.
Example 3.4.14. Let H = {z ∈ C|Im(z) > 0} and λ : H → R is given by λ(z) =
1
. If gE is the Euclidean metric on H ⊂ C then the hyperbolic metric on H
Im(z)
denoted gH is conformally equivalent to gE since gH = λgE .
3.5. T (S) IS HOMEOMORPHIC TO R6G−6
55
Definition 3.4.15. A conformal structure on a manifold X is an equivalence class
of Riemannian metric.
Theorem 3.4.16. (Isothermal Coordinates due to Gauss) Let g be a Riemannian
metric on Ω ⊂ C = R2 then there exists a diffeomorphism φ : Ω → Ω′ such that
φ∗ g = λgE where λ : Ω → R+ .
Now assume that X has a conformal structure and let g be a metric in the conformal class. Let (U, φ) be a chart and let φ∗ g be the metric on φ(U). Using isothermal
coordinates, we can assume that
φ∗ g = λgE with λ : φ(U) → R+
defines an atlas. Check that transition maps are holomorphic which will give a Riemann surface structure on X, and check that the Riemann surface structure does not
depend on the choice of g. This shows that conformal structures, Riemann surfaces,
and hyperbolic structures all coincide when the dimension of the manifold equals 2
and genus ≥ 2.
3.5
T (S) is homeomorphic to R6g−6
Today, we think of T (S) as the space of hyperbolic structures on S. We aim to prove:
Theorem 3.5.1. T (S) is homeomorphic to R6g−6 where g is the genus of the surface
S.
3.6
Length functions
Definition 3.6.1. A map γ : S1 → X is an essential curve if γ is not homotopic to
the constant map.
Lemma 3.6.2. Let γ be an essential curve in X, either:
1. γ is homotopic to γ ′ where Imγ ′ is a geodesic in X, or
2. ∀ ε > 0 there is a γε homotopic to γ such that the length of γε in X is smaller
than ε.
Proof. We will see that if X is closed (compact with no boundary) case 2 is impossible.
Now γ : S1 → X is not homotopically trivial, therefore γ∗ (π1 (S1 )) is isomorphic to
Z. Let Xγ be the corresponding covering space. Xγ is topologically an annulus.
Metrically, there is an isometry φ of H2 such that Xγ = H2 / < φ >. φ is either
hyperbolic or parabolic:
1. If φ is a hyperbolic isometry, let γe′ be the (unique) axis for φ. Then γe′ descends
to a closed loop γ ′ in Xγ which is a local geodesic. In Xγ : γ∗ = γ∗′ so there is a
homotopy from γ to γ ′ which descends to X.
56
CHAPTER 3. TEICHMÜLLER SPACE AND MODULI SPACE
H2
y=
Xγ
1
ε
ε
p
0
1
Figure 3.2: γ is homotopic to a loop of arbitrarily small length.
2. If φ is a parabolic isometry Xγ is as in Figure 3.2 and γ is the core curve which
is homotopic to curves of arbitrarily small loops.
Now we are in a position to define length functions:
Definition 3.6.3. Let γ be an essential closed loop on S, the length function corresponding to γ is a map lγ : T (S) → [0, ∞) defined by lγ (X, f ) is the length of the
geodesic homotopic to f ◦γ in the hyperbolic structure X or 0 if no such representative
exists.
Exercise 3.6.4. Show that this is a well defined function on T (S), i.e., if (X, f ) ∼
(Y, g) in T (S), then lγ (X, f ) = lγ (Y, g).
Lemma 3.6.5. If S is closed, then lγ (X, f ) > 0 for every (X, f ) ∈ T (S), and every
essential curve γ.
Proof. We define the injectivity radius of a hyperbolic surface at a point: v
Definition 3.6.6. Let X be a hyperbolic structure on S, and p ∈ X. If D =
diam(X) then B(p, D) is X, on the other hand, since the exponential function is
injective in a small enough neighborhood of p, there is some ε > 0 such that B(p, ε)
is homeomophic to a disc. The injectivity radius of X at p, denoted inj(X, p) is
sup{r | B(p, r) is homeomorphic to a disc}.
We have defined a continuous map inj : X → (0, ∞). Since X is compact, it takes
on a maximum and a minimum. The minimum of this function is denoted inj(X).
Notice that for any closed loop in X, lγ (X) ≥ 2inj(X) and we get the lemma.
Definition 3.6.7. A pair of pants is a surface homeomorphic to S2 \ 3 discs.
Facts about a Pair of Pants P:
3.6. LENGTH FUNCTIONS
57
Figure 3.3: A pair of pants is homeomorphic to a thrice punctured sphere.
1. χ(P) = −1
2. Every essential simple closed curve (ssc in the sequal) is isotopic to a component
of the boundary (One way to see this is to think of the curve as a Jordan curve
in S2 \ {x1 , x2 , x2 } the curve separates this space into two components and
partitions {x1 , x2 , x3 }. The partition tells us which boundary component the
curve is homotopic to).
3. If φ : P → P is an orientation preserving homeomorphism which doesn’t permute the boundary componants then φ is homotopic to the identity. See Figure
3.4.
Figure 3.4: The fundamental group of P is F2 =< α, β > where α is the red curve and
β is the blue curve. Since φ doesn’t permute boundary components, α, φ(α) are freely
homotopic to the same boundary curve so they are homotopic. Similarly, β, φ(β) are
homotopic. Thus φ is homotopically trivial.
Fact: If φ : S → S is homotopic to the identity, then it is isotopic to the identity.
If γ1 , γ2 are s.c.c which are homotopic, then they are isotopic. (see [6]).
Proposition 3.6.8. Given a, b, c ≥ 0, there exists a unique X hyperbolic structure
on P where the boundary components are geodesics with hyperbolic lengths a, b, c
respectively.
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CHAPTER 3. TEICHMÜLLER SPACE AND MODULI SPACE
Proof A of proposition 3.6.8. We will prove:
Lemma 3.6.9. Given a′ , b′ c′ > 0 there is a unique hyperbolic right angled hexagon
H with alternating side lengths a′ , b′ , c′ .
Given a, b, c, take two copies of H( a2 , 2b , 2c ): H1 , H2 . Glue the sides with the unspecified edge lengths, to get a space which is topologically a pair of pants X. X is
in fact the two sheeted (metric) covering of H1 and thus is itself a hyperbolic surface. Since lifts of boundary components are geodesics in the upper half plane - the
boundary is geodesic, and the components have the specified lengths.
Uniqueness: Let X be a hyperbolic pair of pants with boundary componants
A, B, C with lengths a, b, c. Let δAB be a geodesic arc connecting A, B which is
perpendicular to both componants (to find such an arc - lift A, B to H2 and consider
the shortest arc between them. By a surjery argument one can see that it will be
perpendicular to both geodesics). Let δBC and δAC be geodesic arc perpendicular to
B, C and A, C. Cutting along δAB , δBC and δAC we get two right angled hexagons.
Moreover, they will have three alternating side lengths which are equal - namely the
ones that we don’t specify. Thus, by the uniqueness of H(a′ , b′ , c′ ) those hexagons are
isometric. So X can be constructed by gluing two isometric right-angled hexagons of
appropriate alternating side lengths. Uniqueness now follows from the uniqueness of
the hexagons.
Proof of Lemma 3.6.9. Figure 3.5 shows how to construct a pentagon with specified
alternating side lengths a, b.
b
M
L1
x
a
L2
z
L3
y
Figure 3.5: The green geodesics are the ones whose lenght is specified.
Start with the geodesic line L1 and draw the (non-geodesic) line M such that
every point on M is a hyperbolic distance b away from L1 . Pick any point x and
59
3.6. LENGTH FUNCTIONS
draw a geodesic segment of length a starting at x and perpendicular to L1 . Let L2
be a common pependicular to this segment and the real line which intersects the real
line at point y. Let L3 be a geodesic with endpoint at y and tangent to M at point
z. Complete the pentagon by drawing the shortest geodesic from z to L1 (its length
is b). We get a geodesic pentagon with alternating side lengths a, b.
We alter this construction slightly to get a hexagon: move L3 away from L2 so
that the distance on the real line between them is r. I.e., L3 (r) is a Euclidean circle
perpendicular to the boundary at distance r from L2 which is tangent to M (see
Figure 3.6). Denote the tangency point z. The last side of the hexagon still has
length b. Let t(r) be the hyperbolic distance between L2 and L3 (r). t(0) = 0 and
limr→∞ t(r) = ∞. It is also not hard to see that t is a strictly increasing function.
Therefore, by the intermediate value theorem, there is an r such that t(r) = c. Thus
we get a hexagon with side lengths a, b, c.
b
L1
L3
t
a
L2
r
Figure 3.6: The green geodesics are the ones whose lenght is specified.
Uniqueness of the hexagon. Suppose we have another such hexagon. By composing with a Mobius transformation we can assume that the left endpoint of the edge
of length a is at x and the edge to the right of it is a segment of L1 - the y axis.
Uniqueness now follows form the fact that t(r) is strictly monotone increasing.
We have constructed a polygon as needed in the case where a, b, c > 0 and where
a, b > 0 and c = 0. The case where a > 0 and b = c = 0 is left to the reader.
Proof B of proposition 3.6.8. Consider a hyperbolic isometry z → eλ z which takes
the vertical line x = 1 to the line x = eλ . The quotient is topologically an annulous.
Figures 3.7 and 3.8 show how we can get different metric structures. Notice that in
Figure 3.8 we get an incomplete metric structure. Indeed, the length of the green
60
CHAPTER 3. TEICHMÜLLER SPACE AND MODULI SPACE
ri + eλ
i+1
1
eλ
Figure 3.7: When we glue without sheering we get a complete metric space.
eλ i + eλ
hi + eλ
i+1
0
1
eλ
Figure 3.8: When we glue with sheering we get an incomplete metric space. The
length of the boundary curve depends on the amount of sheering. The bigger r is the smaller the boundary curve becomes.
3.6. LENGTH FUNCTIONS
61
curve in Figure 3.9 is (eλ − 1)(1 + λ1 + λ12 + · · · = λ < ∞. The length of the boundary
curve changes with λ.
y = eλ
len = l2 = l1 /eλ
y = e2λ
len = l1 = l0 /eλ
y=1
len = l0 = eλ − 1
0
1
eλ
Figure 3.9: The length of the green curve is < ∞. the length of the boundary curve
is λ.
Now consider two ideal triangles. An ideal triangle is one with all three vertices
on the boundary. All ideal triangles are isometric. For each triangle, the midpoint of
a side is the intersection point of the side with the circle inscribed inside the triangle.
We glue the triangles together along their boundaries with an orientation reversing
isometry which takes the midpoints d, e, f to the points at a distance x, y, z from
d′ , e′ , f ′ respectively. We get a pair of pants. By changing x, y, z we can control the
side lengths of the boundary curves. In fact, the lengths of the boundary curves will
be |x + y|, |y + z|, |x + z|. See Figure 3.11.
Given a, b, c we can solve |x + y| = a, |y + z| = b, |z + x| = c and perform the
construction described above. Actually we get eight different solutions for x, y, z.
Does that contradict the uniquness of the hyperbolic structure? The answer is no by
Lemma 3.6.10.
Lemma 3.6.10. Every hyperbolic pair of pants with positive boundary lengths has
eight decompositions into two ideal triangles.
Proof. Pick two points from each boundary circle and connect them with non-intersecting paths. For each boundary circle, pick a direction and start twisting the circle in
that direction. In the universal cover this amounts to sliding the points (equivariantly
along the geodesic). In the limit (in the universal cover) we get an ideal triangle which
descends to an ideal trangle downstairs (see Figure 3.12). The complement will also
be an ideal triangle4 . So any hyperbolic structure on a pair of pants with positive
boundary lengths can be gotten by gluing two ideal triangles. Notice that we had 8
4
why? Notice that the comlement of the region we spun is also a hexagon.
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CHAPTER 3. TEICHMÜLLER SPACE AND MODULI SPACE
d
f
d′
f′
e′
e
y
x
z
f′
d′
e′
Figure 3.10: We glue the triangles together along their boundaries with an orientation
reversing isometry which takes the midpoints d, e, f to the points at a distance x, y, z
from d′ , e′ , f ′ respectively.
ex+y
ey
1
ey
1
e−x
Figure 3.11: The length of the boundary curve is |x + y|.
3.7. FENCHEL-NIELSEN COORDINATES
63
choices when twisting the boundary componants which is exactly the descrepency in
the existence proof. Thus given boundary lengths - the hyperbolic structure on the
pair of pants is unique.
Figure 3.12: We spin the endpoints of the hexagon until we get a decomposition into
two ideal triangles.
3.7
Fenchel-Nielsen Coordinates
Let S = Sg be a closed surface of genus g ≥ 2. We recall that
T(S) := {(X, f ) | X is an hyperbolic surface f : S → X, f ∈ Homeo+ }/ ∼,
where (X0 , f0 ) ∼ (X1 , f1 ) if there exists an isometry φ : X0 → X1 such that φ◦f0 ∼ f1 .
Here there is an outline of what we are going to prove in this section.
As before, we consider cutting along disjoint simple closed curves (actually we consider
geodesics with respect to the hyperbolic metric on X). When there are no more
simple closed curves in the remaining open sets, then we will prove that we are left
with pairs of pants. We proved that the hyperbolic structure of a pair of pant is
complitely determined by the lenghts of the curves in the boundary of the pair of
pant.
Since S is reconstruited by gluing all the pairs of pants together suitably, we can
consider a system of coordiantes for the Teichmüller space T(S) given by the set of
lengths used in the above decomposition of S in pairs of pants and the set of the twist
coefficients used to glue the pieces of the decomposition together.
This system of coordinates gives the homeomorphism
T(S) ∼
= R6g−6 .
We have the following definitions:
Definition 3.7.1. If X is a surface with boundary, we say that γ is peripheral if it
is homotopic to a boundary component of X.
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CHAPTER 3. TEICHMÜLLER SPACE AND MODULI SPACE
Definition 3.7.2. The geometric intersection number of two homotopy classes of
curves α and γ is the minimum number of intersections between curves representing
the two homotopy classes.
Let ℘ be a maximal collection of disjoint simple closed curves such that
1. If α, β ∈ ℘, α fi β;
2. If γ is an essential simple closed curve and γ ∈
/ ℘, then there exists α ∈ ℘ such
that i(α, γ) > 0.
Lemma 3.7.3. If χ(S) < 0, then S \ ℘ is a collection of pants.
Proof. Let X be a component of S \ ℘.
Since (2) holds, ℘ doesn’t contain essential curves and so X 6= D 2 .
Moreover, X 6= S 1 × [0, 1] because otherwise there are two elements α, β ∈ ℘, α ∼ β
and this contradicts (1).
By the classification of surfaces, if X is not a pair of pants, then X has an essential,
non-peripheral simple closed curve.
By what we have just said, X is a pair of paints. Since this is true for every component
of S \ ℘, S \ ℘ is a collection of pants.
Hence, we get a so-called pants decomposition ℘. Examples of pants decomposition
are shown in Figure 3.13 and Figure 3.14.
Figure 3.13: Pant decompositions of a genus 2 surface.
Figure 3.14: Pant decomposition of a genus 4 surface.
Recall that χ(P ) = −1, where P is a pair of pants.
If S \ ℘ = P1 ⊔ . . . ⊔ Pk , then χ(S) = −k = 2(1 − g) and so k = 2(g − 1) Since the
65
3.7. FENCHEL-NIELSEN COORDINATES
number of pairs of pants is 2(g − 1), we have 3g − 3 curves in ℘.
Let l℘ : T(S) → (R+ )3g−3 be the length function which sends an element (X, f ) ∈
T(S) to the length of the curves in ℘.
If (X0 , f0 ), (X1 , f1 ) ∈ T(S) have l℘ (X0 , f0 ) = l℘ (X1 , f1 ), then there exists φ : X0 → X1
such that φ ◦ f0 ∼ f1 . The map φ takes curves in ℘ to curves in ℘ and the length of
curves in ℘ is equal on X0 and X1 .
We have the following fact:
Fact 3.7.4. Let P be a pair of pants. If φ : P → P is a orientation-preserving
homeomorphism that preserves the boundary components, then φ is homotopic to
the identity.
Hence, we have an isometry between a pair of pants in X0 and the correspondent
pair of pants in X1 . The problem is that maybe we cannot extend this isometry to
agree on the boundary of the pair of pants because there can be twists in the gluing
of the boundary. We want to study what happens in more details.
Theorem 3.7.5. Let γ0 and γ1 be a collection of disjoint essential simple closed
curves on S such that the curves in γ0 are homotopic to the curves in γ1 . Then there
exists a homeomorphism f : S → S such that f ∼ id and f (γ0 ) = γ1 .
This theorem is true even if the collection γ0 contains homotopic curves. We refer
to [6] for the proof of this Theorem.
Let P be a pair of pants in S \ ℘. Then there exists a hyperbolic structure Y on P
and isometries i0 : Y → X0 and i1 : Y → X1 that map Y to P .
By the Theorem 3.7.5, we can assume that φ takes geodesic representatives of ℘ on
X0 to geodesic representatives of ℘ on X1 . Hence, we have
Y CC
i0
›
X0
CC i1
CC
CC
φ !
/X
1
and the composition i−1
1 ◦ φ ◦ i0 is almost well-defined since if x ∈ P , then φ ◦ i0 (x)
is in the image of i1 . We want that φ ◦ i0 (x) is the image of only one point of i1 but
we can have a problem to find this point if x is on the boundary of P . It is easy to
see that we can make i−1
1 ◦ φ ◦ i0 well-defined everywhere (see Figure 3.15).
−1
We assume that i1 ◦ φ ◦ i0 is well-defined. Since i−1
1 ◦ φ ◦ i0 ∼ id, let ψt be the
homotopy such that ψ1 = id, ψ0 = i−1
◦
φ
◦
i
.
We
define
0
1
φ(x), if x 6= P
φt (x) =
i−1
1 ◦ ψt ◦ i0 (x), if x ∈ P.
Notice that φt is not continuous.
What do we need to do to get continuity?
Lemma 3.7.6. Every orientation-preserving homeomorphism of S 1 is isotopic to the
identity map.
66
CHAPTER 3. TEICHMÜLLER SPACE AND MODULI SPACE
P
x
i0
i−1
1 ◦ φ ◦ i0 (x)
i1
φ
X0
X1
Figure 3.15: The map i−1
1 ◦ φ ◦ i0 is well-defined since given a point x in P , we can
follow its image under this map and we know in which boundary component of P it
is going to land.
Since there exists ε > 0 such that the ε-neighborhood of each geodesic in ℘ is an
embedded annulus, we consider a collar ε-neighborhood for each curve in the boundary
of the pairs of pants and we make φ an isometry outside the collar neighborhood.
Proposition 3.7.7. Let φ0 , φ1 : X0 → X1 be isometries outside the ε-collars. Then
there exists a homotopy φt such that φt = φ0 = φ1 outside the collars.
Proof. The first part of the proof involves topology. We pick a base point x0 in the
boundary of one collar. We consider the universal covering π : H2 = Se → S and we
choose x
e0 ∈ π −1 (x0 ). We lift φe0 , φe1 : H2 = Se → Se = H2 so that φe0 (e
x0 ) = φe1 (e
x0 ).
e
e
e
e
e
Since φ0 and φ1 are such that φ0 (e
x0 ) = φ1 (e
x0 ), if Y is a component in Se of the
preimage of collars, then φe0 (Ye ) = φe1 (Ye ).
Now we use geometry: since φe0 and φe1 are isometries, φe0 |Ye = φe1 |Ye .
Take the linear homotopy from φe0 to φe1 . This homotopy descendes to a homotopy
φ t : X0 → X1 .
The situation (described in Figure 3.16) is the following: we can foliate a collar
with geodesic orthogonal to the boundary of the annulus. An arc can go around many
times in the image of the collar. In the universal covering we pick a point and we
have a orizontal segment. The image of this segment gives us a number a − b called
twist coefficient, which doesn’t depend on the point that I picked, on φ and on the
marking.
67
3.7. FENCHEL-NIELSEN COORDINATES
a
b
Figure 3.16: The red segment on universal covering of the annulus on the left side
denote a preimage of the dashed segment, while the red segment in the right correspond to the image of the red segment. The number (a − b) is the twist coefficient of
the blue curve in the annulus.
68
CHAPTER 3. TEICHMÜLLER SPACE AND MODULI SPACE
If we consider (X0 , f0 ), (X1 , f1 ) ∈ T(S), the pants decomposition ℘ = {γ1 , . . . , γ3g−3 }
and we assume l℘ (X0 ) = l℘ (X1 ), then we have the twist coefficients
{τi (X0 , X1 )}i=1,...,3g−3 .
Lemma 3.7.8. Under the above hypothesis we have:
1. If τi (X0 , X1 ) for every i, then X0 = X1 ∈ T(S);
2. If X0 , X1 , X2 ∈ T(S), then τi (X0 , X1 ) + τi (X1 , X2 ) = τi (X0 , X2 ), for every i.
Proof. First we prove (1). Since τi (X0 , X1 ) for every i, we can homotope φe to be an
isometry also in the collars. The homotopy descends to φt : X0 → X1 .
The proof of (2) follows by the observation that
(a − b) + (b − c) = a − c,
where a, b, c are the numbers in the Figure 3.17.
a
b
b
c
Figure 3.17: Proof of (2) in the Lemma 3.7.8.
Theorem 3.7.9. l℘−1 (a1 , . . . , a3g−3 ) ∼
= R3g−3 with the affine structure (i.e., it is a
vector space without 0).
Notice that all the twist coefficients are realized. We can construct an element
in T(S) with a certain twist coefficient τi just cutting S along the i-th curve in the
pants decomposition ℘ and gluing the curves that we got with a twist τi .
Putting together all that we know till now we get
3.7. FENCHEL-NIELSEN COORDINATES
69
Theorem 3.7.10 (Fenchel-Nielsen Coordinates). T(S) is an affine (trivial) bundle
over (R+ )3g−3 with the fiber of dimension 3g − 3:
R3g−3 ֒→
T(S)
↓
+ 3g−3
(R )
.
Corollary 3.7.11. dimR T(S) = 6g − 6.
3.7.1
Dehn Twist
Definition 3.7.12. Let γ be an essential simple closed curve on S and A := S 1 ×
[0, 2π] a collar neighborhood of γ, where γ = S 1 × {π}. We define the Dehn Twist
Dγ as
x, if x 6= A
Dγ =
(ϑ + t, t), if x = (ϑ, t) ∈ A.
Figure 3.18: The Dehn Twist.
Let γ ∈ ℘ and consider (X, f ) ∈ T(S) and (X, f ◦ Dγ ). Notice that l℘ (X, f ) =
l℘ (X, f ◦ Dγ ), but the twist, called the Dehn twist of γ, is
τγ ((X, f ), (X, f ◦ Dγ )) = lγ ((X, f )) = lγ ((X, f ◦ Dγ ))
Hence, Dγ is an example of a non-trivial twist.
Remark. Sometimes the twist coefficient is defined as our twist coefficient divided
by the length of the curve lγ . With this notation the Dehn twist is 1.
3.7.2
Thick-Thin Decomposition
Theorem 3.7.13. Let X be a complete hyperbolic surface (without boundary unless
explicitly stated). Then there exists an ǫ > 0 such that every component of X <ǫ =
{set of points where injectivity radius is less than ǫ} is either:
1. an annular neighborhood of a simple closed geodesic of length < 2ǫ;
2. neighborhood of a puncture.
70
CHAPTER 3. TEICHMÜLLER SPACE AND MODULI SPACE
Equidistant curves
Figure 3.19: The equidistant curves from the core curve of the annulus.
Definition 3.7.14. The injectivity radius at a point x ∈ X is
inj(x) := supr {r | D(x, r) is an embedded disk}.
Definition 3.7.15. (Second Definition of injectivity radius)
1
inj(x) := inf {length of all essential closed curves that contain x}.
2
The equivalence of the two above definitions can be easily seen when considering
disks upstairs that project to non-disks downstairs (ie where a simple closed curve
through x exists).
Let γ be a hyperbolic isometry, then H2 /γ ∼
= annulus. Moreover, the injectivity
radius is a constant function on equidistant curves from the core curve of the annulus.
Another fact is that the injectivity radius increases under covering space projections.
Fact 3.7.16. Let π̂ : X̂ → X be a covering space with π̂(x̂) = x, then injX̂ (x̂) ≥
injX (x).
X
Proof.
γ
Figure 3.20: Annular neighborhood of γ will embedd under covering map.
If γ is a parabolic hyperbolic isometry then H2 /γ ∼
= annulus. Since every
parabolic is conjugate to the map z 7→ z + 1 in the Upper Half Space model we
see that horocycles are the simple closed curves on the cusp.
3.7. FENCHEL-NIELSEN COORDINATES
71
111
000
000
111
000
111
000
111
000
111
000
111
000
111
0
1
z 7→ z + 1
Figure 3.21: The horocycles are the simple closed curves on the cusp.
Recall that horocycles in H2 are circles tangent to ∂H2 (include horizontal lines
in Upper Half Space model). In Euclidean space the curvature of a circle of radius r
is 1r .
In hyperbolic space the curvature of a circle of radius r is coth(r), and the curvature
of a curve lying at distance d from a geodesic is tanh(d).
In order to prove the theorem it suffices to study the geometry of a pair of pants since
there is a pants decomposition of the surface.
Lemma 3.7.17. (Collar Lemma) If γ is a simple closed geodesic on X then γ has a
collar neighborhood with
1 + cosh λ 2 ,
radius = log sinh λ2
where λ = length(γ).
Proof. Extend γ to a pants decomposition. Let P be a pant with γ ∈ ∂P . We need
to show that γ has sufficiently large collar neighborhood on P . We are going to
decompose P into two ideal triangles.
λ
λ
λ
The radius of the circle centered at 1+e
is e 2−1 and thus sin(θ) = eeλ −1
= tanh( λ2 ).
2
+1
π
Consider the path ei t for θ ≤ t ≤ 2 , then the length of this path is
Z π
1 + cosh λ 1 + cos(θ) 2
dt
= log 2 .
= log λ
sin(t)
sin(θ)
sinh
θ
2
Question 3.7.18. What is the length of the collar boundary?
Answer 3.7.19. Consider the retraction reiθ 7→ ir. Under this map vectors of length
cosh(d) gets mapped to vectors of length 1. This implies that the length of the collar
boundary is
"
!#
1 + cosh( λ2 )
λcosh log
.
sinh( λ2 )
72
CHAPTER 3. TEICHMÜLLER SPACE AND MODULI SPACE
z 7→ eλ z
θ
0
eλ
1
geodesic being stabilized
Figure 3.22: Decomposition of P into two ideal triangles.
z 7→ eλ z
what is
this
distance
θ
0
1
1+eλ
2
annulus
eλ
geodesic being stabilized
Figure 3.23: circle centered at
1+eλ
2
with radius
eλ −1
.
2
Should also get the same answer by calculating the length of the path et eiθ for 0 ≤
t ≤ λ.
Moreover, as λ → 0 we see that
"
!#
)
)
sinh( λ
1+cosh( λ
2
2
λ
+
1 + cosh( 2 )
)
)
sinh( λ
1+cosh( λ
2
2
λcosh log
=λ
→ 2,
λ
2
sinh( 2 )
when λ → 0.
If we assume that X has a cusp, since the length of λ is 2 and geodesics are shorter
than horocycles, the injectivity radius is less than 2.
3.7.3
Bers Constant
Let X be a finite area complete hyperbolic surface. Then there exists a constant B
and a pants decomposition P depending only on χ(X) such that the simple closed
3.8. T (S) IS COMPLETE
73
geodesic
horocycle
Figure 3.24: A horocycle and the geodesic through the points.
curves in P have length less than or equal to B.
Proof. We know that each pair of pants can be decomposed into 2 ideal triangles and
thus Area(X) = 2π · |χ(X)|. Also
Area(D(x, R)) =
Z
R
0
So given x ∈ X,
2πsinh(r)dr = 2π(cosh(R) − 1).
2π(cosh(inj(x)) − 1) ≤ 2πχ(X)
which implies that there exists B0 such that every x ∈ X has an essential closed curve
of length ≤ B0 through x. Cut along this curve and get a surface with boundary. Now
we proceed on the surface with boundary. Since the area of the collar neighborhood
is
Z R
X
λi sinh(R) < 2π|χ(X)|.
λcosh(r)dr = λsinh(R) ⇒
0
bdry components
The disk that realizes the injectivity radius of a given point might or might not now
intersect the boundary. If it does not then we can proceed as before. If the disk
intersects the boundary we get three cases. The first is when the “disk” intersects
the boundary in a point. Here we do not get a simple closed curve and hence choose
another point. The second is when the “disk” intersects a boundary component in
two points. Join them by a non-trivial arc. The regular neigborhood of the boundary
union the arc is a pair of pants. The third case is when the “disk” intersects the
boundary in one point in two different components. Here do same thing as in case 2:
join the boundary components by a non-trivial arc. The regular neigborhood of the
boundary components union the arc is a pair of pants.
3.8
T (S) is complete
We want to prove that T (S) is complete. First we need the following proposition:
74
CHAPTER 3. TEICHMÜLLER SPACE AND MODULI SPACE
Proposition 3.8.1. Let (X, f ), (Y, g) be points in T (S) and φ : X → Y a K-quasiconformal map such that φ ◦ f is homotopic to g. Then for every essential simple
closed curve γ in S:
1
lg(γ) (Y ) ≤ lf (γ) (X) ≤ Klg(γ) (Y )
K
where lα (X) denotes the hyperbolic length of the curve α in the hyperbolic surface
X.
e and Ye respectively and φ : X →
Proof. We lift X and Y to the universal coverings X
2
2
e = H → Ye = H . Let X̂ be the cover of X with π1 (X̂) =< f (γ) > and
Y to φe : X
Ŷ be the cover of Y with π1 (Ŷ ) =< g(γ) >.
The deck translation t corresponding to γ has the form z → eiλ z where λ = lγ (X).
If A is the annulus in Figure 3.25 and R is the associated rectangle, then m(A) =
m(R) = πλ .
H2
λ
R
A
log
0
1
π
λ
e
Figure 3.25: The annulus A in the Proposition 3.8.1.
Hence,
π
π
π
1
≤
≤K
K lg(γ) (Y )
lf (γ) (X)
lg(γ) (Y )
and so
1
lg(γ) (Y ) ≤ lf (γ) (X) ≤ Klg(γ) (Y ).
K
Using the geometric definition of a quasi-conformal map it is easy to prove the
following
Fact 3.8.2. If φ is K-quasi-conformal and ψ is K ′ -quasi-conformal then φ ◦ ψ is
K · K ′ -quasi-conformal.
Recall that
∃φ : X → Y a K-quasi-conformal map
dT ((X, f ), (Y, g)) = log inf K such that φ ◦ f is homotopic to g
1
2
Theorem 3.8.3. T (S) is a complete metric space.
3.8. T (S) IS COMPLETE
75
Proof. Let (Xi , fi ) be a Cauchy sequence. For all ǫ > 0 there exists N such that if
i, j > N then there exists a K-quasi-conformal map φi,j : Xi → Xj (homotopic to
fj ◦ fi−1 ) such that 12 log K < ǫ.
1
Step 1: Let {(Xi , fi )}∞
i=1 be a sequence of points in T (S) so that dT (X1 , Xi ) < 2 log(J)
(here we’re suppressing the markings). Let φ1,i : X1 → Xi be a J-quasi-conformal
map in the right homotopy class. We show that there exists a Riemann surface X1,∞
and a subsequence φ1,ik that converges to a map φ1,∞ : X1 → X∞ . Moreover, there
are maps φi,∞ : Xi → X∞ and φj,∞ ◦ φi,j is homotopic to φi,∞ .
We want to apply the compactness theorem, but we can’t do so directly. We must
lift the φ1,i to maps of the complex plane and must set things up so that three points
fi is isometric to H2 . Fix a loop γ in S
are taken into compact sets. For each i, X
and let γi be the geodesic representative of fi (γ) in Xi . Pick a point p on γ1 and let
f1 with H2 so that i projects to p1 and the y-axis projects to
pi = φ1,i (p1 ). Identify X
g
fi with H2 so that the y-axis projects to γi . Define φ
γ1 . Identify X
1,j to be the lift of
2
φ1,j which takes i to the lift of pj which is closest to i ∈ H .
The deck translation tj corresponding to γj has the form z → eiλj z where λj =
lγj (Xj ). By Proposition 3.8.1, J1 λ1 ≤ λj < Jλ1 . Thus 0 < λmin
√ ≤ λj ≤ λmax . Define
1
iθ
√
M(s, t) = {re | s ≤ r ≤ t, 0 < θ < π}. Then M = M( λ1 , λ1 ) is a fundamental
√
1
g
g
λmax ). In particular, φ
domain for t1 containing i. φ
1,j takes i
1,j (M) ⊆ M( √λmax ,
into a compact subset of C. In addition, since lifts commute with the action of the deck
1
n
n
g
g
g n
g
translations, φ
1,j ◦ tγ1 = tγj ◦ φ1,j . Thus, φ1,j (tγ1 (i)) ∈ tγj ◦ φ1,j (i). So λmin √λmax ≤
√
√
1
n
n
n
n
g
g n
|φ
1,j (tγ1 (i))| ≤ λmax λmax . Thus φ1,j (tγ1 (i)) ∈ M(λmin √λmax , λmax λmax ). For n
√
√
1
λmax < λnmin √λ1 making the three sets
large enough λ−n
max λmax < λmin and
min
√
√
√
1
−n
n
n √ 1
√ 1
√
,
λ
,
λ
)),
M(
λ
),
M(λ
M(λ−n
max
max
max
min λmax , λmax λmax ) disjoint (see
min λmax
λmax
Figure 3.26). Now, we may apply the compactness theorem to get a subsequence jk
g
g
and a map φg
1,∞ so that φ1,jk → φ1,∞ and the limit map is a J-quasi-conformal homeomorphism from the upper half plane to itself.
g
φ
1,j
i
i
g
Figure 3.26: We constructs the lifts φ
1,j so that three points are taken to compact
sets.
76
CHAPTER 3. TEICHMÜLLER SPACE AND MODULI SPACE
Now, for each j we have an injective homomorphism ρj : π1 (S) → Aut(H2 ). We
have the following commutative diagrams:
g
φ1,j
fj ∼
f1 ∼
X
= H2
= H2 −−−→ X


ρ (g)
ρ (g)
y1
yj
X1
φ1,j
−−−→
Xj .
g −1
Therefore, define ρ∞ : π1 (S) → Aut(H2 ) by ρ∞ (g) = φg
1,∞ ◦ ρ1 ◦ φ1,∞ . Since
ρ∞ is discrete and faithful, we can define X1,∞ := H2 /ρ∞ (π1 (S)). We get a point
(X1,∞ , φ1,∞ ◦ f1 ) ∈ T (S).
Step 2: Notice that in our previous construction the ρ1 action was topologically
conjugate to both ρ2 and ρ∞ . Therefore, ρ2 and ρ∞ are conjugate through the map
g
−1
g
φ
1,2 ◦ φ1,∞ . Thus X2,∞ = X1,∞ and we denote this limit by X∞ .
In conclusion (by passing to a subsequence), we get a sequence of maps φi,∞ : Xi →
X∞ such that φi,∞ ◦ fi ∼ φj,∞ ◦ fj and where φi,∞ is (1 + ǫ)-quasi-conformal. This
concludes the proof that T (S) is complete.
3.9
Constructing Riemann Surfaces
There are many ways to construct Riemann Surfaces:
• via Hyperbolic Geometry: glueing pair of pants;
• we can construct a Riemann surface X using the universal cover H2 : H2 /Γ = X;
• using polynomials in C2 (algebraic geometry);
• sufaces as quotient spaces.
We will think of surfaces as quotient spaces.
Let P be a polygon in C and identify pairs of sides with orientation-preserving Euclidean isometries. Then this determines a Riemann surface structure since the link
of a vertex is a compact 1-manifold and hence S 1 .
Definition 3.9.1. A traslation surface X is a Riemann Surface such that for any two
charts (U, φ) and (V, ψ) the transition map
ψ ◦ φ−1 : φ(U ∩ V ) → ψ(U ∩ V )
is a traslation.
It is clear that both on the inside of the polygon P and on the edges that the
transition maps are Euclidean isometries. Around the vertices define the chart by
2π
mapping a neighborhood of the vertex by the map z 7→ z α where α is the sum of
77
3.9. CONSTRUCTING RIEMANN SURFACES
Glue opposite
sides
Figure 3.27: Gluing of a polygon.
||
||||
|||
|
||||
|
||
|||
Figure 3.28: Gluing by traslations.
the angles around the vertex. We also want to require that the Euclidean isometries
that we are using are pure translations. One way to visualize such maps is by using
subdivided rectangles.
Let T (C) be the group of translations of C and X = (P, G) where P is our polygon
and G ⊂ T (C) are the gluings. Let φK (x+iy) = Kx+iy. Then XK = (φK (P ), φK (G))
defines a new polygon with gluings still in T (C) (the map φK stretches in the xdirection which does not affect the gluing map). This map φK descends to a map
between the quotient Reimann surfaces.
Theorem 3.9.2 (Teichmüller’s Theorem). Let X and XK as above. Then
1
dT (X, XK ) = logK
2
and if φ : X → XK is K-quasi-conformal then φ = φK .
In other words, the Teichmüller Theorem says that if ψ : X → XK , ψ ∼ φK and
ψ is K ′ -quasi-conformal, then K ′ ≥ K and if K ′ = K, then ψ = φK .
Definition 3.9.3. φK is called the Teichmüller map.
The proof of this theorem is at its heart the same as the proof of Grötzsch’s theorem. In order to get a bijection between the space of rectangles and the Teichmüller
space of a surface, we allow our gluings to live in T + (C) which is all translations and
rotations by π.
78
3.10
CHAPTER 3. TEICHMÜLLER SPACE AND MODULI SPACE
1-forms
Let ω be a complex 1-form on S and γ : [a, b] → S a path. Then we can integrate the
1-form over this path via
Z
Z
b
ωγ(x) γ ′ (t)dt
ω=
γ
a
at each point x ∈ S where ωx : Tx S → C.
In R2 the two standard 1-forms are dx(u, v) = u and dy = (u, v) = v. Every 1-form
on Ω ⊂ R2 is of the form f dx + gdy for f, g : Ω → C. In C we have the following
1-forms:
dz = dx + idy
dz(u, v) = u + iv = dx(u, v) + idy(u, v)
dz̄ = dx − idy
dz̄(u, v) = u − iv = dx(u, v) − idy(u, v)
1
dx = 2 (dz + dz̄)
dy = 2i1 (dz − dz̄)
1
1
f dx + gdy = (f − ig)dz + (f + ig)dz̄.
2
2
Hence, every complex value 1-form on Ω is of the form
1
1
(f − ig)dz + (f + ig)dz̄.
2
2
Now, for f : Ω → R(or C) we have
1
1
df = fx dx + fy dy = (fx − ify )dz + (fx + ify )dz̄.
2
2
Definition 3.10.1. Let f : Ω → C be a map. Then
1
1
fz = (fx − ify ) and fz̄ = (fx + ify ).
2
2
The Cauchy-Riemann equations say that f is holomorphic if and only if fz̄ = 0.
If we are going to change coordinates, then for φ(z) = w and α = α(w)dw we have
φ∗ αz (u, v) = αφ(z) φ∗ (u, v).
If φ = (φ1 , φ2), then the Jacobian is
u
(φ1 )x (φ2 )x
.
v
(φ2 )x (φ2 )y
If φ is holomorphic, then (φ∗ )z (u, v) = φ′ (z)(u + iv) and φ∗ α = φ′ (z)α(φ(z))dz.
Analogously, if β = β(w)dw̄, then φ∗ β = φ′ (z)β(φ(z))dz̄.
Therefore, if ω is a C-valued 1-form on a Riemann surface, there is a canonical
decomposition into holomorphic and anti-holomorphic parts. We have seen that the
tangent bundle of X tensored with C as a decomposition in holomorphic and antiholomorphic tangent bundle.
79
3.10. 1-FORMS
Let α be a section of the holomorphic tangent bundle then it “only has dz part in
local coordinates”. In local coordinates, α = α(z)dz and dα = 0 ⇐⇒ dα ∧ dz ⇐⇒
αz dz ∧ dz + αz̄ dz̄ ∧ dz ⇐⇒ αz̄ = 0. Thus, α is holomorphic
R if and only if it is closed.
Now, let α be a holomorphic 1-form. Then the integral γ α with γ(0) = z0 , γ(1) = z
is well defined since α is closed. If we are not at a zero of the 1-form, then this defines
a chart.
Let α be a holomorphic 1-form on X. In local coordinate, α = αU,φ dz with αU,φ a
holomorphic function. Let p ∈ X and (U, φ) be a chart with p ∈ U. If αU,φ (φ(p)) = 0,
then for all charts (V, ψ) with p ∈ V αV,ψ (ψ(p)) = 0 since (ψ ◦ φ−1 )′ (z) 6= 0 and
αU,φ (z) = αV,ψ (ψ ◦ φ−1 (z))(ψ ◦ φ−1 )′ (z).
Hence, α(p) = 0 is well-defined, the oder of a zero is well-defined and moreover, the
zero’s are isolated.
Let αU,φ dz represent α on a chart where α 6= 0. For any p ∈ U there exists β defined
on a neighborhood at φ(p) such that β ′ = αU,φ .
Since β ′ (φ(p)) 6= 0, β is a local homeomorphism. Shrinking the chart (if necessary),
(U, β ◦ φ) is a new chart such that α = αU,β◦φ dz. What is αU,β◦φ ?
Since
αU,φ (z)dz = αU,β◦φ (β(z)) · β ′ (z)dz = αU,β◦φ (β(z))αU,φ (z)dz,
αU,β◦φ = 1.
Hence, if p ∈ X and α(p) 6= 0, then there exists a chart (U, φ) with p ∈ U such that
α = dz on U. Let Aα be the set of charts with α = dz and let Bα be the set of charts
(U, φ) such that (V, ψ) ∈ Aα and φ = φK ◦ ψ.
Since αU,φ (z) = αU,β◦φ (β(z)) · β ′ (z), β ′ (z) = 1 and so β(z) = z + const.
Hence, Bα defines a new Riemann traslation surface on the surface S.
In conclusion, Aα determines a Riemann surface X, Bα determines a Riemann surface
XK and φK : X → XK . We are in the same situation as in the case of the polygons.
What does it happen in the zero’s? If α(p) = 0, then there exists a chart such that
α(z) = z n dz, n ≥ 1. Indeed, take the map z 7→ z n and pullback the 1-form dz. We
get nz n−1 dz. α is the pull-back of dz using the map z 7→ z n .
How do we get the rotation by π? To find an answer to this question we need to talk
about the quadratic differentials.
80
CHAPTER 3. TEICHMÜLLER SPACE AND MODULI SPACE
Chapter 4
Quadratic Differentials
We will define quadratic differentials and in particular, holomorphic quadratic differentials. Moreover, we will prove that the set of holomorphic quadratic differentials
with area less than 1 is homeomorphic to T (S).
4.1
Holomorphic quadratic differentials
Definition 4.1.1. A quadratic differential Φ is a collection of functions {φU } on local
charts such that
φU (z) = φV (f (z))f ′ (z),
where f is the transition map.
A holomorphic quadratic differential is a quadratic differential Φ such that the functions {φU } are holomorphic.
If φU (z) in U, φV (z) in V , U ∩V 6= ∅ and f (z) = w is the (holomorphic) transition
map, then
φU (z)dz 2 = φV (f (z))f ′ (z)2 dz 2 .
Definition 4.1.2. Let p, q ∈ Z. A (p, q)-differential Φ is a collection of functions
{φU } on local charts such that if φU (z)dz p dz̄ q in U and φV (z)dw p dw̄ q in V , then
q
φU (z)dz p dz̄ q = φV (f (z))f ′ (z)p f ′ (z) dw pdw̄ q ,
where f is the (holomorphic) transition map.
If φU ’s are holomorphic, then Φ is called holomorphic (p, q)-differential.
The cases that we are interested in are (1, 0), (2, 0), (−1, 1) (Beltrami differentials),
(1, 1) (2-forms), (0, 0) (functions).
Remark. The set of (p, q)-differentials form a C-vector space.
Let Q(X) denote the set of holomorphic quadratic differentials in X.
In the holomorphic setting we can apply the Riemann-Rock Theorem and we have:
Theorem 4.1.3. dimQ(X) = dim(T (S)).
81
82
CHAPTER 4. QUADRATIC DIFFERENTIALS
Quadratic differentials behave very nicely. Indeed, they satisfy the following properties:
1. If Φ is a (p, q)-differential, then Φ is a (q, p)-differential.
2. If Φ is a (p, q)-differential and Ψ is a (p′ , q ′ )-differential, then Φ · Ψ is a (p +
p′ , q + q ′ )-differential.
√
3. If Φ is aR (2, 0)-differential, then |Φ| = Φ · Φ is a (1, 1)-differential. Hence,
Area = X |Φ| is defined since dzdz̄ = (dx + idy)(dx − idy) = −2idxdy,
φU (z)dzdz̄ = φV (f (z))f ′ (z)f ′ (z)dzdz̄ = φV (f (z))|f ′ (z)|2 dzdz̄,
and so
Z
φU dzdz̄ =
4. If α is a 1-form, then Area =
R
Z
X
−2iΦdxdy =
Z
φV dwdw̄.
|α|2 .
R
If α is a holomorphic 1-form, then we defined a chart at p ∈ X with α(p) 6= 0 by γ α,
where γ is a path from p to q. R
Since α is closed, the integral γ α is well defined.
If Φ is a quadratic differential where φU represent Φ in a chart U about p and p ∈ X
2
with Φ(p)
R √6= 0, then there exists αU such that αU = φU and we can define a chart
taking γ Φ, where γ is a path from p to q. This gives a family of charts. Indeed,
there exists a (transition) map β such that β ′ = αU . If (U, ψ) was the original chart
containing p, then (U, β ◦ ψ) is the new chart (possibly after shrinking U since we
2
c
want a simply-connected domain) and on this chart Φ is φc
U dz . What is φU ?
Since
′
2
2
2
2
2
c
c
φU (z)dz 2 = φc
U (β(z))β (z) dz = φU (β(z))αU (z)dz = φU (β(z))φU (z)dz ,
φc
U = 1.
Let AΦ be the set of charts with transition maps given by translations and rotations
by π (we ignore the zero’s for the moment). In charts, Φ = dz 2 . Let BΦ be the set of
charts (U, φ) such that (V, ψ) ∈ AΦ and φ = φK ◦ ψ.
If (U, ψ0 ) and (V, ψ1 ) are charts such that Φ = φU dz 2 on (U, ψ0 ) and Φ = φV dz 2 on
(V, ψ1 ) with φU = φV = 1, then the derivative of the transition map (ψ1 ◦ ψ0−1 )′ = ±1.
So ψ1 ◦ ψ0−1 (z) = ±z + const, that is the transition maps are given by translations
and rotations by π.
A consequence of the Teichmüller Theorem is the following.
Let Q1 (X) := {Φ ∈ Q(X) | area(Φ) < 1}. We define TX : Q1 (X) → T (S) in the
1+k
following way: if Φ ∈ Q1 (X), let k := area(Φ) and K := 1−k
. The quadratic
differential Φ determines AΦ and so BΦ . This gives a Riemann surface Y and a map
fΦ : X → Y . Let TX (Φ) = (Y, fΦ ). TX is obviously continuous.
By the Teichmüller Theorem, TX is an injective and proper map since dT (X, TX (Φ)) =
1
log(K).
2
4.1. HOLOMORPHIC QUADRATIC DIFFERENTIALS
83
Hence, TX is a proper, injective, continuous map between two copies of Rn of the
same dimension (since Q1 (X) is open in Q(X) and dimQ(X) = dim(T (S))). By the
Invariance of Domain, TX is a homeomorphism.
Remark. If Φ is fixed, then λ 7→ TX (s(λ)Φ) is a geodesic, where s(λ) is a function
in λ. Moreover, there exists a unique geodesic between any two points in T (S).
Example 4.1.4. Let λ ∈ ∆ = {z ∈ C | |z| < 1} define a map from the complex plain
λ)
z
= λ.
to itself by φλ (z) = z + λz. The Beltrami differential in this case is µ(φλ) = (φ
(φλ )z
λ
λ
Let P be the polygon in Figure ?? and denote P = φ (P ). The gluing maps are
gj (z) = z + cj , where c1 = 2, c2 = 1, c3 = 2i, c4 = i. We define the gluing on P λ so
that the following diagram commutes:
φλ
P −−−→

gi
y
φλ
Pλ

g λ
yi
P −−−→ P λ .
Let X λ be the Riemann surface defined by quotienting P λ by the gluing giλ. The
λ
map φλ : P → P λ descends
to a map f : X0 → Xλ . Using Teichmüller’s Theorem,
1+|λ|
we get dT (X 0 , X λ) = log 1−|λ|
= d∆ (0, λ) where d∆ (·, ·) denotes the distance from
0 to λ in the disc model of the hyperbolic plane. Similarly, for each λ, λ′ ∈ ∆
′
′
′
define f λ,λ = f λ ◦ (f λ)−1 : X λ → X λ . Again by Teichmüller’s Theorem, we get that
′
dT (X λ , X λ ) = d∆ (λ, λ′). Therefore, the map λ → X λ defines an isometric embedding
from ∆ to T (S).
Now we prove the Teichmüller’s Theorem.
Proof of Teichmüller’s Theorem. Let f : X → X λ be the Teichmüller map as previously defined and let K be its quasi-conformal constant. Let g : X → X λ be any
other L-quasi-conformal map from X to X λ . We wish to prove that L ≥ K with
equality iff g = f .
Let α(p) be the direction of maximal stretch for f at p. Define λ(g, p) = |∂α(p) g| where
∂α(p) g is the directional derivative of g in the α direction1 . For notational simplicity,
assume that f is defined in a way where in every chart α(φ(p)) is the x-direction2 .
Following the method of proof of Grötzsch’s theorem, in local coordinates we have:
|g|2 ≤ (|gz | + |gz̄ |)2 =
|gz | + |gz̄ |
(|gz |2 − |gz̄ |2 ) ≤ LJp (g),
|gz | − |gz̄ |
(4.1)
where Jp (g) is the Jacobian of g at p. Therefore, we have:
λ(g, p)2 ≤ LJp (g).
And by invoking Schwartz’s inequality we get:
1
2
exists almost everywhere?
Now f is not given in terms of the map φλ .
(4.2)
84
CHAPTER 4. QUADRATIC DIFFERENTIALS
R
λ(g, p)
X
2
R
R
2
λ(g,
p)
1≤
X
X
R
≤ X LJp (g)dx · Area(X) = LArea(Y )Area(X).
≤
Note that the area of X and Y are calculated with respect to the volume form which
corresponds to the vectors of maximal/minimal stretch of f 3 . Now ,if we also had
R
Area(Y)
= K. We need the following lemma.
Area(Y) ≤ X λ(g, p) we would get L ≤ Area(X)
Lemma 4.1.5. Let (X, ρ) be a compact metric space, and f : X → X a homeomorphism homotopic to the identity. There exists a constant M such that for every
geodesic segment γ, |f (γ)|ρ > |γ|ρ − 2M, where the constant M only depends on f .
This Lemma says that the image of a geodesic will not get too short.
e f˜ : X
e → X,
e where X = X/π
e 1 (X).
Proof. We lift f to the universal cover X:
We can choose f˜ such that f˜ ◦ α = α ◦ f˜ for all α ∈ π1 (X).
˜
e Then,
By compactness, there exists M such that dρ̃ (f(x̃),
x̃) ≤ M, for all x̃ ∈ X.
since α ∈ π1 (X) is a Deck transformation, we have
˜
dρ̃ (f˜(x̃), x̃) = dρ̃ (α ◦ f˜(x̃), α(x̃)) = dρ̃ (f(α(x̃)),
α(x̃)),
for all α ∈ π1 (X).
˜
If x ∈ X, define ψ(x) = dρ̃ (f(x̃),
x̃), where x̃ is the preimage of x. The function ψ is
continuous on X, so there exists M such that ψ(x) ≤ M for all x ∈ X. Hence, if γ is
a geodesic from x0 to x1 , and respectively, x̃0 and x̃1 are the preimages, then
˜ 0 )) + dρ̃ (f(x̃
˜ 0 ), f˜(x̃1 )) + dρ̃ (f˜(x̃1 ), x̃1 ) ≤ 2M + dρ̃ (f(x̃
˜ 0 ), f˜(x̃1 ))
dρ̃ (x̃0 , x̃1 ) ≤ dρ̃ (x̃0 , f(x̃
and projecting this inequality to X, we get
|γ|ρ ≤ 2M + |f (γ)|ρ .
Consider the map h = g ◦ f −1 : Y → Y . Define
λ(h, q) := |gx (q)|.
Since f −1 shrinks by K in the x-direction, we get λ(h, f (p)) = K1 λ(g, p) which implies
Z
Z
Z
Z
λ(g, p) =
Kλ(h, f (p))dX =
Kλ(h, q)Jp (f )dY =
λ(h, q).
(4.3)
X
We must show that
X
R
Y
Y
λ(h, q) ≥ Area(Y ).
Proposition 4.1.6. If h : Y → Y , h ∼ id, then
3
Huh?
Y
R
Y
λ(h, q)dAY ≥ Area(Y ).
85
4.2. BRANCHED DOUBLE COVERS
Proof. First, we need to do the setup.
We consider the geodesic flow on Y in the x-direction. This flow is well defined
since we have a well defined x-direction and y-direction. This flow defines a map
ϕt : Y → Y in the following way: if (U, φ) is a chart, p ∈ U, for small |t| < ε define
ϕt (p) := φ−1 (φ(p) + t).
We are working with the singular Euclidean metric and so the flow is not defined in
the singularities. If the flow is defined, then ϕt0 ◦ ϕt1 =Tϕt0 +t1 .
Let Ωt := {p ∈ Y |ϕt (p) is defined} ⊂ Y and let Ω := t∈R Ωt . We have area(Ωt ) =
area(Y ) and in general Y \ Ω will be dense. For example, think about an x-direction
with irrational slope in the torus.
With this setup we can prove the theorem. First, since ϕt is an isometry we have
Z
L
−L
Z
λ(h, ϕt (p))dAY dt =
Ωt
Z
L
−L
Z
Ωt
Z
λ(h, p)dAY dt = 2L
λ(h, p)dAY
(4.4)
Y
and also, by Lemma 4.1.5,
Z
L
−L
λ(h, ϕt (p))dt ≥ 2L − 2M.
The equality 4.4 and the disinequality 4.1.5 imply:
2L
Z
λ(h, p)dAY =
Y
=
Z Z
Ωt
Hence,
L
−L
λ(h, ϕt (p))dtdAY ≥
Z
Y
λ(h, p)dAY ≥
Z
Z
L
−L
Ωt
Z
λ(h, ϕt (p))dAY dt =
Ωt
(2L − 2M)dAy = 2(L − M)area(Y ).
2(L − M)
area(Y ) → area(Y )
2L
as L → ∞.
This concludes the proof of the Teichmüller’s Theorem.
4.2
Branched Double Covers
Let X = (P, G). We want to build a double cover of X.
Let P ′ = r(P ), where r(z) = −z + c is a rotation.
Define G ′ in the following way:
1. If g ∈ G and g(z) = z + c, then g ∈ G ′ and r ◦ g ◦ r −1 ∈ G ′ ;
2. If g ∈ G and g(z) = −z + c, then r ◦ g ∈ G ′ and g ◦ r −1 ∈ G ′ .
(4.5)
86
CHAPTER 4. QUADRATIC DIFFERENTIALS
So we have X ′ = (P ⊔ P ′ , G ′ ) and a map π : X ′ → X such that
X ′ CC
π
›
X
CCh◦π
CC
CC
!
h /
X.
We can apply the same argument in the proof of the Teichmüller’s Theorem with
h ◦ π instead of h.
Theorem 4.2.1 (Teichmüller Existence ad Unique Theorem). Let X, Y ∈ T (S).
There exists a unique map f : X → Y such that f is K-quasi-conformal and
dT (X, Y ) = 12 log(K).
Proof. Define TX : Q1 (X) → T (S) in the following way. Given Φ ∈ Q1 (X), let
fΦ : X → YΦ be the Teichmüller map such that
(fΦ )z̄
= area(Φ).
(fΦ )z
We have the following
1. TX is continuous;
2. dimR (Q1 (X) = 6g − 6 (Riemann-Roch Theorem);
3. dimR (T (S)) = 6g − 6 (Fenchel-Nielsen coordinates);
4. Tx is injective (the Teichmüller map is unique);
5. TX is proper (if Φi is such that area(Φi ) → 1, then dT (X, TX (Φi )) → ∞).
By the Invariance of Domain, TX is a homeomorphism.
Let Φ ∈ Q(X) with area(Φ) = 1. Define
TΦ : H2 ≡ ∆ → T (S),
TΦ (λ) := TX (λΦ).
We have shown that TΦ is an isometry between the hyperbolic disc and T (S) (i.e.,
∆ is a totally geodesic subspace of T (S)) and so if X, Y ∈ T (S), there exists an
isometric embedding H2 → T (S) such that X and Y are in the image, then there is
a geodesic connecting them, because there is a geodesic connecting their preimage.
Remark. Here we think H2 as a complex manifold.
Our next goal is to show that the Teichmüller space is not negatively curved and
is not CAT (0).
First, we need to introduce the Strebel differentials.
4.2. BRANCHED DOUBLE COVERS
87
Figure 4.1: Glue the opposite sides of the polygons to get a quadradic differential on
S2 .
4.2.1
Strebel Differentials
Glue the opposite sides of the polygons in Figure 4.1 to get a Quadradic Differential
on S2 . Notice that horizontal lines close up to simple closed curves. Up to homotopy,
there are respectively two and one homotopy classes of horizontal trajectories going
from left to right.
Definition 4.2.2. Let Φ ∈ Q(X), Φ is a Strebel differential if all but finitely many
(nonsingular) horizontal trajectories are simple closed curves.
An ǫ-neighborhood of a closed trajetory is an annulus foliated by closed trajectories. The number of pssible annuli is 3g − 3. A Strebel differential decomposes S into
collection of cylinders representing homotopy classes γ1 , . . . , γk . These cylinders are
standard Euclidean cylinders. An example of a non-Strebel differential can be found
by rotating any Strebel differential by some complex number that is not a multiple of
π
. In addition, there are quadradic differentials that when multiplied by any complex
2
number are not Strebel.
Theorem 4.2.3 (Strebel). Given X, a collection of disjoint simple closed curves
γ1 , . . . , γk and heights h1 , . . . , hk , there exists a unique Strebel differential Φ on X
with cylinders γ1 , . . . , γk and heights h1 , . . . , hk .
Recall that given any conformal annulus A with 0 < m(A) < ∞, there exists a
Euclidean metric on A with height h.
Proof. Let A = {A1 , . . . , Ak } be a collection of disjoint open annuli on X with core
curves γ1 , . . . , γk . Define
X h2
i
.
m(A) :=
m(Ai )
Step 1: There exists an A that realizes inf {m(Aj )}, where the infimum is over all the
possible collection of disjoint open annuli.
88
CHAPTER 4. QUADRATIC DIFFERENTIALS
Step 2: Let v be a smooth vector field on X and ψt the flow. This determines a family
At of annuli. We would like to show that
dm(At )
| t=0 = 0.
dt
To do this we will show that
dm(At )
| t=0 =
dt
Z
Φvz̄ = 0,
X
where Φ is a quadradic differential (not necessarily holomorphic) and vz̄ is a
(−1, 1)-form.
R
Step 3: If for all smooth vector fields v, X Φvz̄ = 0, then φ is actually holomorphic.
The key step is to apply the following lemma for forms instead of functions.
Lemma 4.2.4. (Weyl’s Lemma) Let f be a C-valued function on Ω ⊂ C (need
some regularity assumptions on f ). If for all φ ∈ Cc∞ (Ω), then
Z
f φz̄ = 0 ⇐⇒ f is holomorphic up to a set of measure 0.
Ω
Proof. (of Step 1) Let An be a sequence of families that limits to the inf {m(Aj )}.
The inf {m(Aj )} is not ∞ since m(Ai ) 6→ 0, thus m(Ai,j ) 6→ 0 as j → ∞. Let gi,j be
the conformal map from a standard annulus to Ai,j . The maps gi,j → gi (after maybe
passing to a subsequence) determine a family A that realizes the inf .
Proof. (of Step 2) Let A = (A1 , . . . , Ak ) and define a quadradic differential Φ on X
be letting Φ|X be the quadradic differential given by a Euclidean structure on Ai
with height hi and Φ|X−{A1 ,...,Ak } ≡ 0. Now, Φ is
R a (2,0) differential. If µ is a (-1,1)differential, then Φµ is a (1,1)-differential and X Φµ is defined. Let v be a smooth
∂
in local coordinates. Note vz̄ is
(−1, 1)-form (AKA a vector field). Then v = v(z) ∂z
a (−1, 1)-form. Let ψt be the flow of v. We want to calculate
Z
dm(At )
|t=0 =
Φvz̄ .
dt
X
Let At be the collection of annuli given by the images of the standard annulus through
ψt ◦ gi for every i (see Figure 4.2).
The modulus of the annulus At is
m(At ) =
h
h
⇒ c(t) =
.
c(t)
m(At )
◦ ψt ◦ g0 =
Then gt−1 ◦ψt ◦g0 is the flow on the annulus. Lift this map to the map gt ^
e
e
ψt . The t = 0 derivative of ψt is a vector field e
v and vez̄ = vz̄ since all maps were
conformal. In particluar,
Z
Z h Z c(0)
Φvz̄ =
1 · vez̄ .
−1
A
0
0
89
4.2. BRANCHED DOUBLE COVERS
ψt ◦ g
ih
get
x − axis
Figure 4.2: The annulus At .
X
g0
gt−1^
◦ ψt ◦
g0
gt−1 ◦ ψt ◦ g0
gt
−1
Figure 4.3: The map gt ^
◦ ψt ◦ g0 = ψet .
90
CHAPTER 4. QUADRATIC DIFFERENTIALS
e + iy) + c(t), Im(ψet (x)) = 0, Im(ψet (x + ih)) = h, and
Now, ψet (x + iy + c(0)) = ψ(x
1
vx + ie
vy ). Thus,
vz̄ = 2 (e
e
Z h Z c(0)
Z h
Z h
vx dxdy =
e
(e
v(c(0) + iy) − ve(iy))dy =
c(t)dy = −hc′ (0),
0
0
0
since
Moreover,
ve =
Z
dψe
d
e + iy) + c(t))
|t=0 ⇒ (ψet (x + c(0) + iy) − ψ(x
dt
dt
= vet (x + c(0) + iy) − e
v(x + iy) − c′ (0).
c(0)
0
and you get
0
Z
h
0
ZZ
Now,
m(At ) =
vy dydx =
e
Z
c(0)
0
v(x + hi) − ve(x)dx = 0
e
h2 m′ (A)
1
.
1·e
vz̄ = − h · c′ (0) =
2
2m(A)2
h
−m′ (A0 ) · h
h
⇒ c(t) =
⇒ c′ (0) =
.
c(t)
m(At )
m(A0 )2
Thus,
d
dt
X
h2i
m(Ai,t )
=2
Z
Φvz̄ .
X
Now, our goal is to show that T (S) is not non-positively curved.
Theorem 4.2.5 (Masur). Let X ∈ T (S). Then there exist geodesic rays rt , qt so
that r0 = q0 and dT (rt , qt ) ≤ B for all t ∈ R.
Corollary 4.2.6. T (S) is not non-positively curved.
Proof. Pick disjoint simple closed curves γ1 , γ2 and choose Strable differentials Φ1 , Φ2
so that their heights are 1, 1 in Φ1 and 1, 2 in Φ2 (see Figure 4.4).
Let Nǫ1 be an ǫ-neighborhood of γ1 , γ2 in the singular Euclidean metric defined by
Φ1 , and let Cǫ1 = Φ1 r Nǫ1 . Similarly, let Nǫ2 be an ǫ-neighborhood of γ1 , γ2 in the
singular Euclidean metric defined by Φ2 , and let Cǫ2 = Φ2 r Nǫ2 . Then Cǫ1 is a disjoint
union of two annuli each of which has height 1 − 2ǫ and Cǫ2 is a union of two disjoint
annuli with heights 1 − 2ǫ and 2 − 2ǫ. Define a map f0 : Φ1 → Φ2 homotopic to the
identity so that f0 takes Nǫ1 to Nǫ2 , and f0 is an affine map when restricted to each
annulus of Cǫ1 taking them to the annuli of Cǫ2 . Now, let rt be the geodesic in T (S)
defined by stretching Φ1 in the y-direction by t. Let gt be the Teichmüller geodesic
defined by stretching Φ2 in the y-direction by t. Let ft : rt → gt be equal to f0 when
restricted to Nǫ1 and the affine map on the complementary annuli. Now, notice that
the quasi-conformal constant of ft is bounded above by the quasi-conformal constant
of f0 (on Nǫ1 they are the same and on the rest ft stretches less than f0 , therefore
d(rt , gt ) ≤ log(K(ft )) ≤ log(K(f0 )).
91
4.2. BRANCHED DOUBLE COVERS
Φ1
1
Φ2
1
1
2
Figure 4.4: Masur’s example of rays in T (S) emanating from a single source and
staying a bounded distance from each other.
92
CHAPTER 4. QUADRATIC DIFFERENTIALS
Chapter 5
The Measurable Riemann Mapping
Theorem
Definition 5.0.7. Let B(Ĉ) denote the set of complex valued measurable functions
µ on Ĉ with kµk∞ < ∞ and denote B1 (Ĉ) = {µ ∈ B(Ĉ) | kµk∞ < 1}. Let B(H2 )
denote the set of complex valued measurable functions µ on H2 with kµk∞ < ∞ and
denote B1 (H2 ) = {µ ∈ B(H2 ) | kµk∞ < 1}.
We have seen that if f : C → C is a K quasi-conformal map then
fz̄ K−1
fz < K+1 . Miraculously, the converse is also true.
fz̄
fz
∈ B1 (C) and
The Measurble Reimann Mapping Theorem (Morrey, Ahlfors-Bers, Glutysk).
Given µ ∈ B1 (Ĉ), there exists a unique quasi-conformal map f : Ĉ → Ĉ such that
1.
fz̄
fz
=µ
2. f (0) = 0, f (1) = 1 and f (∞) = ∞.
We call f the standard solution for µ given by the Measurble Reimann Mapping
Theorem.
Corollary 5.0.8. Given a µ ∈ B1 (H2 ) where H2 denotes the hyperbolic plane in the
half plane model, there exists unique quasi-conformal map f : H2 → H2 such that
1.
fz̄
fz
=µ
2. f extends continuously to ∂H2 and f (0) = 0, f (1) = 1 and f (∞) = ∞.
Proof. We note that any quasi-conformal map f : H2 → H2 extends continuously to
the boundary. We wish to invoke the Measurble Reimann Mapping Theorem. Define

µ(z), if Im(z) > 0
ν(z) =
µ(z̄), if Im(z) < 0
93
94
CHAPTER 5. THE MEASURABLE RIEMANN MAPPING THEOREM
Since kνk = |µ| < 1, we get a standard map f : C → C such that ν =
claim that f : H2 → H2 . Define g(z) = f (z̄). It is easy to verify that
gz̄
(z)
gz
=
fz̄
. We
fz
fz̄
(z̄) =
fz
ν(z̄) = ν(z). Moreover, g(0) = f (0̄) = 0, g(1) = f (1̄) = 1 and g(∞) = f (∞)
¯ = ∞.
Therefore, g is another normalized solution for ν. By uniqueness, it follows that
f (z) = g(z) = f (z̄). Thus for all r ∈ R, f (r) = f (r) hence f takes the reals to the
reals, and since it is quasi-conformal it is orientation preserving, so f takes the upper
half plane to itself.
Let f : X → Y be a quasi-conformal map between Riemann surfaces. Let (U, φ)
be a chart in X, and (V, ψ) be a chart in Y . Let fU,V be the map f in local coordinates.
(fU,V )z̄
We wish to understand how µU,V = (fU,V
changes as we vary the charts in U and V .
)z
Observation 5.0.9.
1. Let (V ′ , ψ ′ ) be another chart with V ∩ V ′ 6= ∅. One can
verify that µU,V ′ = µU,V . Morally, this is because the quasi-confomal dilitation and directions of maximal stretch don’t change and these are exactly the
absolute value and argument of µU,V . We therefore denote µU,V simply by µU .
2. Let (U ′ , φ′ ) be another chart on X with U ∩ U ′ 6= ∅. Let χ = φ′ ◦ φ−1 : φ(U) →
′ (z)
. Therefore, µ(z) is a
φ′ (U ′ ). Since fU = fU ′ ◦ χ we get µU (z) = µU ′ (χ(z)) χχ(z)
(−1, 1) form! In addition, |µ| = |µU | is independent of the chart. Thus |µ| is a
function on X and we denote its essential sup by kµk∞ .
Definition 5.0.10. B(X) is the set of (−1, 1) forms on X with kµk < ∞ and B1 (X)
is th set of (−1, 1) forms on X with kµk < 1.
5.1
Bers’ Simultaneous Uniformization
Theorem 5.1.1. Let X be a Riemann surface and µ ∈ B1 (X). Then there exists a
Riemann surface Y and a quasi-conformal map f : X → Y such that µf = µ.
e∼
Proof. X
= H2 let τ : π1 (S) → PSL2 (R) denote the deck transformation action. Lift
µ to µ
e ∈ B1 (H2 ) (we can do this since X has a single chart so µ can be thought of
as a function). Let fe : H2 → H2 be the normalized solution of µ
e. By the change of
coordinates formula, for each γ we have:
τ (γ)′ (z)
=µ
e(z)
τ (γ)′ (z)
e. So geγ is
Let e
gγ (z) = f ◦ τ (γ)(z). By a direct calculation we get µgfγ = µ
another solution for µ
e. For each γ let ρ(γ −1 ) be the Mobius transformation taking
g (0), e
e
g (1), e
g (∞) to 0, 1, ∞. Then µρ(γ −1 )◦eg = µ
e since post composing with a conformal
map doesn’t change the Beltrami differential. ρ(γ −1 ) ◦ e
g = ρ(γ −1 ) ◦ fe ◦ τ (γ) is a
normalized solution for µ
e. By uniqueness we get fe = ρ(γ −1 ) ◦ fe ◦ τ (γ) or ρ(γ −1 ) =
fe ◦ τ (γ) ◦ fe−1 . We get that ρ : π1 (S) → PSL2 (R) is a representation and that it is
topologically conjugate to τ . Let Y = H2 /ρ then fe descends to a map f : X → Y
with µf = µ, so f is a quasi-conformal map.
µ
e(τ (γ)(z))
5.1. BERS’ SIMULTANEOUS UNIFORMIZATION
95
Let µ ∈ B1 (Ĉ) define a path ∆ → B1 (Ĉ) by mapping λ → λµ. Let fλ : Ĉ → Ĉ
be the normalized solution. Now, fix z and let φz (λ) = fλ (z).
Theorem 5.1.2. φz (λ) is a holomorphic function and
Z
z(z − 1)
1
′
µ(w)
dx dy
φz (0) = −
π C
w(w − 1)(w − z)
where w is the variable of integration and w = x + iy.
Let X = H2 /Γ where Γ ⊂ P SL2 R ⊂ P SL2 (C). To recall the proof we took µ and
e = H2 together with an equivarience property
lifted it to µ
e on X
µ
e(γ(z))
γ(z)
=µ
e(z) (∗)
γ ′ (z)
that holds for all z ∈ H2 . We also saw that for H2 ⊂ Ĉ we can extend µ such that
(∗) holds for all z ∈ Ĉ. Some possible extensions are
µ(z), if Im(z) > 0
1. ν(z) =
µ̄(z), if Im(z) < 0
µ(z), if Im(z) > 0
2. ν(z) =
0,
if Im(z) < 0
We also defined fe : H2 → H2 and there exists ρ : Γ → P SL2 R such that
fe ◦ γ = ρ(γ) ◦ fe ∀γ ∈ Γ
where ρ is a representation of Γ = π1 (S).
Let fe : Ĉ → Ĉ be the normalized solution for ν, then there exists ρ : Γ → P SL2 C
e The proof of this follows exactly the proof given before.
such that fe ◦ γ = ρ(γ) ◦ f.
Let U = {upper half plane} and L = {lower half plane}, then if U/Γ = X and
L/Γ = X̄ the actions are conjugate by the map z 7→ z̄. X̄ is really X with the
opposite orientation. In fact fe desceds to a homeomorphism f + : X → fe(U)/ρ(Γ)
e
and f − : X̄ → f(L)/ρ(Γ).
ν desceds to Beltrami differentials ν + , ν − on X, X̄ such
that µf + = ν + and µf − = ν − .
Definition 5.1.3. A discrete subgroup of P SL2 (R) is a Fuchsian group and a discrete
subgroup of P SL2 (C) is a Kleinian group.
Let ν + ∈ B1 (X) and ν − ∈ B1 (X̄) and lift them to ν ∈ B(Ĉ). Consider the
normalized solution fe : Ĉ → Ĉ with Y + = fe(U)/ρ(Γ) and Y − = fe(L)/ρ(Γ). Then
we get f + : X → Y + and f − : X̄ → Y − with µf + = ν + and µf − = ν − .
Definition 5.1.4. A Kleinian group Γ is quasi-Fuchsian if there exists a Fuchsian
group Γ and a quasi-conformal map fe : Ĉ → Ĉ such that Γ = fe ◦ Γ0 ◦ fe−1 .
Theorem 5.1.5. (Bers’ Simultaneous Uniformization Theorem) For any X ∈ T (S)
and Y ∈ T (S̄) there is a unique quasi-fuchsian group uniformizing X and Y .
This theorem implies that QF (S) = T (S) × T (S̄). Philosophically, this will give
us a complex structure on T (S).
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CHAPTER 5. THE MEASURABLE RIEMANN MAPPING THEOREM
5.1.1
Conformal Maps
Let Ω ⊂ Ĉ and f : Ω → Ĉ be a holomorphic map with f ′ (z) 6= 0 for all z ∈ Ω (this
implies it is a local homeomorphism). We want to estimate how close f is to a Möbius
transformation (think that this is similar to curvature). The key is to approximate f
at a point by Möbius transformations and then to see how they differ.
Fact 5.1.6. Given w ∈ Ω there exists a unique φ ∈ P SL2 (C) such that f (w) = φ(w),
f ′ (w) = φ′ (w), and f ′′ (w) = φ′′ (w). The fancy way to say that the derivatives match
up to the second derivative is that the 2-jet of f and φ agree at w.
Define Mf : Ω → P SL2 (C) where Mf (w) is the Möbius transformation that best
approximates f at w. If f is Möbius, then Mf is a constant function. Mf will turn
out to be holomorphic. Now, lets think about the Lie Algebra of P SL2 (C). Suppose
φt ∈ P SL2 (C) is a path with φ0 = id. Then,
φt (z) =
a(t)z + b(t)
˙ = (ȧz + ḃ) − z(ċz + d)
˙ at t = 0.
and φ(z)
c(t)z + d(t)
This last formula gives a vectorfield. φ−1
0 ◦ φt is a path at the identity and it will
∂
be a
give the best approximation at time t since translating back to id. Let v(z) ∂z
holomorphic vectorfield on Ω.
′′
2
∂
Now, Mv (w) = (v(w) + v ′ (w)(z − w) + v (w)(z−w)
) ∂z
and Mv : Ω → sl2 (C) (sl2 (C) is
2
the Lie Algebra of SL2 (C) and so of P SL2 (C)). Now, by differentiating with respect
to w, we get
Mv′ (w) = v ′ (w) + v ′′ (w)(z − w) + v ′ (w)(−1) +
′′′
= v 2(w) (z − w)2.
v′′′ (w)
(z
2
− w)2 − v ′′ (w)(z − w)
′′′
This will have a fixed point at w and v 2(w) (z−w)2 is holomorphic. Now, let f : Ω → C
be holomorphic. If we think f : Ω ⊂ R2 → R2 , then the condition Dv f = (a +
ib)D ∂ f is the condition for a function to be holomorphic with v = (a, b). Hence, the
∂x
directional derivative is completely determined by the derivative in the x-direction.
Let f : Ω → C. Then
∂
Mf′ (w) = Sf (w)(z − w)2
∂z
and Sf (w) is called the Schwarzian derivative. If you want you could calculate Sf (w)
explicitly. If φ ∈ P SL2 (C), then Mφ◦f = φ ◦ Mf and Mf ◦φ = Mf ◦ φ. The point is
′
that Mf′ = Mφ◦f
, but this is not true for Mf ◦φ . Indeed, if φ1 αt ∈ P SL2 C, then
φ−1 ◦ α0−1 ◦ αt ◦ φ = (Mf ◦ φ)−1
0 ◦ (Mf ◦ φ)t
and this is some kind of adjoint action of the Lie group on the Lie Algebra.
Now, shift our attention back to Beltrami Differentials. Let µ ∈ B1 (X̄) and lift it
to Ĉ. Extend µ to zero on U to get µ
e. Let fe be the normalized solution for µ
e. fe is
conformal on U. For γ ∈ Γ,
fe ◦ γ = ρ(γ) ◦ fe ⇐⇒ fe = ρ(γ) ◦ fe ◦ γ −1 .
5.1. BERS’ SIMULTANEOUS UNIFORMIZATION
97
Let S fe be the Schwarzian derivative of fe. Then
e
S f(γ(z))
= S(ρ(γ) ◦ fe ◦ γ −1 )(γ(z)).
Now, by our earlier computation
e
S fe(z)γ ′ (z)2 = S f(γ(z))
and S fe descends to a holomorphic quadratic differential on X. This gives a map
T (S) → Q(X), where the space of quadratic differentials is a complex vector space,
and puts a complex structure on T (S) (since it is injective of the right dimension).
5.1.2
Bers’ Embedding
We have a map T (S) → Q(X) called Bers’ embedding and dimT (S) = dimQ(X).
Since Q(X) is a complex vector space, the Bers’ embedding gives T (S) and complex
structure (and so a differentiable structure). Is the structure independent on X? This
question is the same to ask if given the embedding T (S) → Q(X) and T (S) → Q(Y )
there is a holomorphic map Q(X) → Q(Y ). We will see that the answer to this
question is yes.
Let µ ∈ B1 (X). We lift µ to µ̃ on the lower plane and we extend to 0 on the upper
half plane. Let f˜ be the normalized solution to the Bers’ equation and consider the
Shwarzian derivative S f˜ ∈ Q(X) in X. We defined a map BX : B1 (X) → Q(X).
Notice that B1 (X) is infinite dimensional, while Q(X) is finite dimensional.
Theorem 5.1.7. Let µ0 , µ1 ∈ B1 (X) with solutions f0 : X → Y and f1 : X → Y
suth that f0 ∼ f1 . Then BX (µ0 ) = BX (µ1 ).
b→C
b and f˜1 : C
b →C
b respectively. Since
Proof. We lift the maps f0 and f1 to f˜0 : C
b
b
f0 ∼ f1 , there exists a map g : C → C such that
b=
C
=
f0
›
b
C
== f1
==
==
g ffi
/ b
C
and g is conformal on f0 (upper half plane) and f0 (lower half plane).
It is not difficult to prove that g is quasi-conformal.
Since g is quasi-conformal and and g is conformal on f0 (upper half plane) and f0 (upper half plane),
g is conformal. If g is a Möbius map, then f˜0 = f˜1 on the upper half plane.
We have BX : B1 (X) → Q(X) and T0 (B1 (X)) = B(X). If µ ∈ B(X), then, for
small t ∈ R, tµ ∈ B1 (X) and BX (tµ) is a path in Q(X). Let ḂX (µ) be the tangent
vector to the path BX (tµ) at t = 0, hence ḂX (µ) ∈ T0 (Q(X)) = Q(X). We lift µ to
µ̃.
b Let f˙µ (z)
let f˜t be the normalized solution at tµ̃. For a fixed z, f˜t (z) is a path in C.
98
CHAPTER 5. THE MEASURABLE RIEMANN MAPPING THEOREM
be the tangent vector at t = 0.
By the Measurable Riemann Mapping Theorem,
Z
1
z(z − 1)
˙
fµ (z) = −
dx dy,
µ̃(w)
π Cb
w(w − 1)(w − z)
(5.1)
where w = x + iy.
∂
is a vector field which conformal on the upper plane. By the equality 5.1, we
f˙µ (z) ∂z
have
6Z
1
(3)
2
˙
ḂX (µ) = fµ dz = −
µ̃(w)
dx
dy
dz 2 .
π Cb
(w − z)4
What is ker(BX )?
Notice that if µ is a (−1, 1)-differential and Φ is a (2, 0)-differential, then µΦ is a
(1, 1)-differential and we can integrate. Hence, it is well defined the pairing
B(X) × Q(X) → R C.
(µ, Φ)
7→ X µΦ
Let
N(X) := {µ ∈ B(X) |
Z
X
µΦ = 0, ∀ Φ ∈ Q(X)} ⊂ B(X).
The pairing we have defined induces a pairing on the quotient:
B(X)/N(X) × Q(X) → C.
Since we killed all the elements in B(X) that go to 0,
dim(B(X)/N(X)) ≤ dim(Q(X)).
Moreover, we have the following
Lemma 5.1.8.
dim(B(X)/N(X)) = dim(Q(X)).
Proof. If Φ ∈ Q(X), the
Since if Φ 6= 0, then
Φ
|Φ|
∈ B(X).
Z
X
Φ
Φ=
|Φ|
Z
X
|Φ| > 0
and therefore, dim(B(X)/N(X)) ≥ dim(Q(X)).
We want to show that ḂX : B(X)/N(X) → Q(X) is an isomorphism, so that we
can identify B(X)/N(X) with the tangent space TX (T (S)).
In order to do that, we need to write N(X) in another way.
Let µ ∈ B(X) and lift µ to µ̃ ∈ B(H2 ). By the infinitesimal version of the Riemann
∂
Mapping Theorem, there exists a vector field ṽ(z) ∂z
in H2 such that ṽz̄ = µ̃. We have
2
X = H /Γ and
γ ′ (z)
µ̃(γ(z)) ′
= µ̃(z).
γ (z)
5.1. BERS’ SIMULTANEOUS UNIFORMIZATION
99
Hence, we have
∂ ′
(γ (z)ṽ(z) − ṽ(γ(z))) = γ ′ (z)ṽz̄ (z) − ṽz̄ (γ(z))γ ′ (z) =
∂ z̄
γ ′ (z)µ̃(z) − µ̃(γ(z))γ ′ (z) = 0
and so the vector field γ ′ (z)ṽ(z) − ṽ(γ(z)) is holomorphic.
Moreove, notice that γ ′ (z)ṽ(z) − ṽ(γ(z)) ∈ sl2 (R).
Since the vector field is holomorphic, it descends to a vector field on X. When we
flow the vectro field in X we have a map homotopic to the identity, so we have a
trivial deformation in the Teichmüller space.
Now, we consider X to be the annulus in Figure 5.1.
Figure 5.1: The annulus X.
Consider < γ >= Γ = Z. where γ(z) = λz, λ ∈ R. Then
µ̃(γ n (z))
λ
λ
= µ̃(λn z) = µ̃(z).
λ
λ
Fact 5.1.9. If Γ ⊂ P SL2 (C), µ ∈ B1 (C) such that
µ(γ(z))
γ ′ (z)
= µ(z),
γ ′ (z)
and f is the normalized solution, then f ◦ Γ ◦ f −1 ⊂ P SL2 (C).
Let γt (z) = λtz and f˜t be the normalized solution in H2 . By the Fact 5.1.9,
f˜t ◦ γ f˜t−1 = γt ∈ P SL2 (R).
Let ṽ(z) = f˙µ (z). Consider the flow given by γt , take the derivative and denote it by
γ̇(z) ∈ sl2 (R). If we differentiate f˜t ◦ γ f˜t−1 = γt , we get
γ ′ (z)ṽ(z) − ṽ(γ(z)) = γ̇(z).
If it is equivariant, i.e. equal 0, then we have a trivial deformation in the Teichmüller
space. If it is not equivariant, i.e. not equal 0, then we have a non-trivial deformation
in the Teichmüller space.
We have the following
100
CHAPTER 5. THE MEASURABLE RIEMANN MAPPING THEOREM
Theorem 5.1.10. µ ∈ N(X) if and only if there exists a vector field v on X such
that vz̄ = µ.
Proof. ⇐: Let Φ ∈ Q(X). By the assumption and the intebration by parts, we have
Z
Z
Z
µΦ =
vz̄ Φ = −
vΦz̄ = 0,
X
X
X
since Φ is holomorphic (i.e. Φz̄ = 0).
⇒: Let µ ∈ B(X) a Beltrami differential which does not arise as the z̄ derivative of
any vector field of X. We wish to prove that µ ∈
/ N(X). Lift µ to µ
e ∈ B(H2 ). Let
v be the unique (up to normalization) vector field such that e
e
vz̄ = µ
e. Let X = H2 /Γ
since we’re assuming that e
v does not descend to a vector field on X, there exists an
element γ ∈ Γ such that
γ ′ (z)e
v−e
v(γ(z)) 6= 0
(5.2)
Recall the above expression is an element of sl2 (R). We can normalize the group Γ
so that γ(z) = λz for some λ > 0.
b by letting µ
Extend µ
e to C
e = 0 on the lower half plane. Note that for all α ∈ Γ we
still have:
α′ (z)
µ
e(α(z)) ′
=µ
e(z)
α (z)
Indeed it is true for all α ∈ Γ in the upper half plane since µ
e is the lift of a Beltrami
differential on X. And it is true for z in the lower half plane since α ∈ Gamma
preserves the lower half plane and µ
e is identically 0 there.
b r {0, ∞}/ < γ > is a torus. µ
Now, Tγ = C
e descends to µγ in B(Tγ ). On the torus
there is a unique quadratic differential up to scale which is the one descending from
e = 12 . Indeed,
φ(z)
z
e n (z))(γ ′ (z))2 = φ(z)
e
φ(γ
Let φγ denote the quadratic differential
of Tγ that φe descends to.
R
Now, comes a calculation that Tγ µγ φγ 6= 0. To make this calculation, consider
b This
A(1, λ) denote the annulus whose inner radius is 1 and outer radius is λ in C.
b
is a fundamental domain for the < γ > action on C r {0, ∞}. The map z → ez
1
takes the rectangle with side lengths 2π, log λ to A(1, λ). Pulling back
R z 2 we1 get
the quadratic differential 1 on the rectangle. One can compute 0 6= rec 1µrec ez =
R
R
e=
µ
e
φ
µ φ . The fact that ve satisfies equation 5.2 translates to the fact that
A(1,λ)
Tγ γ γ
the terms don’t
R cancel in the first integral to show that it is non-zero.
We have that Tγ µγ φγ 6= 0 but we must show that there is a quadratic differential ψ
R
on X so that X µψ 6= 0.
Poincaré Series. Let Γ be any discrete subgroup of PSL2 (C) (in this proof Γ =
π(X)or < γ >). Let f : H2 → C be holomorphic. The poincare series of f is:
Θf (z) =
X
α∈Γ
f (α(z))(α′ (z))2
101
5.1. BERS’ SIMULTANEOUS UNIFORMIZATION
This
R series might not converge, however there are many cases that it does, for example,
if H |f |dxdy then Θf (z) is finite. In addition, notice that this is a way of producing
quadratic differentials:
X
Θf (β(z))β ′(z)2 =
f (α ◦ β(z))α′ (β(z))2 β ′ (z)2 = Θf (z)
α∈Γ
Let F be a fundamental domain for Γ. Let µ
e be a Γ-equivariant Beltrami differential,
and let ψe = Θf . By its construction ψe is Γ equivariant and descends to a quadratic
differential ψ on C \ {0, 1}/Γ or H/Γ. Note that:
Z
Z
Z
Z
e
µψ =
µ
eψ =
µ
eΘf =
µf
F
F
C\{0,1} or H
Going back to our situation, notice that we can actually write
λ−1
priately chosen f . Indeed, let f = z(z−1)(z−λ)
then f (z) = − 1z
Θγ f (z) =
X
n
f (λ z)λ
2n
n∈Z
Therefore,
0 6=
Z
1X
=
z n∈Z
µ γ φγ =
Z
Tγ
1
1
−
−n
z−λ
z − λ1−n
C\{0,1}
H
for an approHence
=−
1
z2
µ
ef (z)
But since µ
e = 0 on the lower half plane
Z
Z
Z
µ
ef (z) =
µ
ef =
µΘΓ f
C\{0,1}
1
as Θf
z2
1
1
z−1 z−λ
(5.3)
X
Thus there exists a quadratic differential on X namely ΘΓ f so that it’s pairing with
µ is non-zero. Thus µ ∈
/ N(X). We remark that we need to worry a little about the
convergence of ΘΓ f , but comes out from the equality in 5.3.
5.1.3
An addendum to the infinitesimal Measurable Riemann
Mapping Theorem
We give a sketch of why the theorem is true. First let us compute the z̄ derivative of
1
. One can show that for all holomorphic φ:
z
Z
1
1
−
φz̄ (z) dxdy = φ(0)
π
z
Therefore, by integration by parts, ∂∂z̄ 1z is the atomic infinite measure at 0, the Dirac
delta function.
Now, given a Beltrami differential µ we wish to find v so that vz̄ = µ.
Z
1
µ(w)
v(z) = −
dxdy
π C w−z
102
CHAPTER 5. THE MEASURABLE RIEMANN MAPPING THEOREM
1
is a
If µ is smooth with compact support we get vz̄ (z) = µ(z)1 . In PDE jargon − πz
fundamental solution for fz̄ = µ.
The normalized solution of the infinitesimal measurable Riemann mapping theorem
is
Z
z(z − 1)
v(z) =
µ(w)
dxdy
w(w − 1)(w − λ)
C
z(z−1)
1
z
= w−z
− w−1
+ z−1
. The z̄ derivative of
We can simplify this expression: w(w−1)(w−λ)
w
the last two terms is zero, so with the new formula we still get vz̄ (z) = µ(z). We can
also easily see that v(0) = v(1) = 0. It also works out that v(∞) = 0
1
Why are these the conditions?
Chapter 6
The Weil-Petersson metric
We can represent infinitesimal deformations of T (S) as Beltrami differentials. We
have proven the pseudo-theorem: B(X)/N(X) = TX T (S) this is only a pseudo theorem since we haven’t defined a differential structure on T (S) yet. However, we may
ask when does the infinitesimal deformation represented by µ ∈ B(X) trivial? It is
iff µ ∈ N(X).
6.1
Norms on Q(X)
The cotangent bundle of T (S) is Q(X) = TX∗ (T (S)). We can define several norms on
it.
φ ∈ Q(X) is a (2, 0) form. |φ| is a (1, 1) form. If ρ2 is the area form of a metric ρ,
then it is a (1, 1) form. Let ρX denote the hyperbolic metric on X. Then ρ|φ|
is a
2
X
function on X. More explicitly, let φe denote the lift of φ to Q(H2 ) (the upper half
e
e is a function on H2 which we denote by |φ|H2 .
plane model). Then |φ| = Im(z)2 |φ|
ρH2
Since φ is γ equivariant, so is |φ|H2 which descends to |φ|ρX .
Definition 6.1.1. The Lp norm of φ ∈ Q(X) is the Lp norm of the function |φ|ρX
with respect to the hyperbolic area.
R
R
2
Example 6.1.2.
• p = 1: kφk1 = X ρ|φ|
|φ| which is the Euclidean area.
2 ρX =
X
X
This turns out to be the Teichmüller co-metric.
• p = 2: The norm comes from an inner product:
Z
φ ψ 2
ρ .
< φ, ψ >2 =
2
2 X
X ρX ρX
qR
|φ|2
. This is the Weil-Petersson co-metric.
Hence, kφk2 =
X ρ2
X
Let V be a vector space, and V ∗ its dual. Let (·, ·) : V × V ∗ → C be the pairing
of V with V ∗ . Given a norm k · k∗ on V ∗ , one can pull it back to define a norm on V
as follows:
(v, w)
kvk = inf ∗
.
w∈V kwk∗
103
104
CHAPTER 6. THE WEIL-PETERSSON METRIC
In fact, all we are using in this definition is the existence of the pairing (and not the
fact that it is non-degenerate). Thus we can use this definition whenever there exists
such a pairing. For example, for V = B(X) and V ∗ = Q(X).
This construction yields norms on TX (T (S)).
6.2
Incompleteness of the Weil-Petersson metric
The goal of this section is to prove the following theorem due to Wolpert and Chu.
Theorem 6.2.1. The Weil-Petersson metric is not a complete metric.
Proof. Let X be a Riemann surface obtained from the square R with vertices 0, 1,
i and i + 1. We assume that the two vertical sides are identified by a horizontal
translation. The top an bottom are identified by some more complicated gluing
pattern. Let Xt be the path in Teichmüller space obtained by squeezing the rectangle
to one of width e−t . We call this rectangle Rt . Hence, we have a map f˜t : R → Rt ,
f˜t (x + iy) = e−t x + iy, which descends to a map ft : X → Xt . This is actually a
Teichmüller geodesic but this will not be important. We will see that this path has
finite Weil-Petersson length.
First, we note that a fairly straightforward calculation shows that the derivative of
the path at time t is the Beltrami differential µt = −1/2, where we are writing µt in
the coordinates given by theR rectangle we are using to build the surface.
|
µt φ|
t
where the sup varies over all non-zero quadratic
Recall that kµt kW P = sup Xkφk
2
differentials φ ∈ Q(Xt ). Note that µt φ is a form of type (1, 1) (essentially a 2-form)
so the integral makes sense. For the L2 -norm kφk2 we need to use the hyperbolic
metric. Also note that the ratio is scale invariant.
We can assume that φ is some holomorphic function on Rt and we can assume that
Z
φ = −2e−t ,
Rt
where the integral is with respect to the standard Euclidean area form. Let φ0 = φ+2.
So
Z
Z
Z
φ+
φ0 =
and
Rt
Rt
Rt
Z
µt φ =
Xt
Z
Rt
2 = −2e−t + 2e−t = 0
(−1/2)(φ0 − 2) = e−t .
We now need to estimate kφk2 . Let ρ be the conformal factor in Rt for the hyperbolic
metric on Xt . Note that this is a finite area metric on the rectangle and is necessarily
incomplete. We can extend ρ to an incomplete hyperbolic metric on the entire strip
between the R-axis and horizontal line at height i.
Let ρS be the complete hyperbolic metric on the this same strip. The Schwarz Lemma
6.2. INCOMPLETENESS OF THE WEIL-PETERSSON METRIC
105
(of course!) implies that ρS > ρ. We also note that by symmetry ρS is constant along
horizontal lines. Finally, one can check to see that
Z
φ0 g = 0
Rt
for any function g that is constant along horizontal lines. To see this we let C be a
rectangular contour with vertical sides at 0 and e−t and horizontal sides at heights
ia and ib with 0 ≤ a < b ≤ 1 (see Figure 6.1). Since φ is holomorphic, by Cauchy
Theorem,
Z
φ0 dz = 0.
C
Rt
Hb
ib
C
ia
Ha
Figure 6.1: The contour C with vertical sides at 0 and e−t and horizontal sides at
heights ia and ib.
The function φ0 will extend to a periodic function, with period e−t , on the entire
strip between the R-axis and the horizontal line at height i so the two vertical sides
of the contour will cancel each other out and therefore, if Ha is the horizontal line
segment of length e−t starting at ia, we have
Z
Z
φ0 dz =
φ0 dz = A,
Hb
Ha
for all 0 ≤ a < b ≤ 1 (A is a constant). We then see that
Z 1Z
Z
φ0 dz ds = A
φ0 =
0=
0
Rt
so we must have
Z
Hs
φ0 dz = 0.
Hs
Notice that if
Z
Hs
φ0 dz = 0 ⇒
Z
φ̄0 dz = 0.
Hs
If g is constant along horizontal lines then
Z
Z 1Z
Z
φ0 g dz ds = 0 and also
φ0 g =
Rt
0
Hs
Rt
φ̄0 g.
(6.1)
106
CHAPTER 6. THE WEIL-PETERSSON METRIC
We use this to bound kφk2 . We calculate:
Z
|φ|2
2
kφk2 =
by definition
2
Rt ρ
Z
|φ|2
(since ρS > ρ)
≥
2
Rt ρS
Z
|φ0 |2 − 2(φ0 + φ̄0 ) + 4
=
(remember φ = φ0 − 2)
ρ2S
Rt
Z
|φ0 |2 + 4
(by equalities 6.1)
=
ρ2S
Rt
Z
4
|φ0|2
2 −t
≥
=
c
e
(since
is positive).
2
ρ2S
Rt ρS
So we have
and
kφk2 ≥ ce−t/2
|
R
Xt
µt φ|
kφk2
≤
e−t
e−t/2
=
ce−t/2
c
for all non-zero φ ∈ Q(Xt ). Note that we could explicitly calculate c by explicitly
writing down ρS . This isn’t especially difficult but also isn’t necessary so we will skip
it.
Finally, we have
e−t/2
kµt kW P ≤
c
and
Z ∞
Z ∞ −t/2
e
2
kµt kW P dt ≤
dt = < ∞.
c
c
0
0
In particular, the path Xt , 0 ≤ t ≤ ∞, has finite Weil-Petersson length.
6.3
Other properties of the Weil-Petersson metric
The Weil-Petersson metric is a Riemannian metric on T (S), then there are unit speed
parameterization geodesics g : (a, b) → T (S).
We have the following property.
Theorem 6.3.1. The map lγ ◦ g : (a, b) → (0, ∞) is strictly convex, where lγ is the
length function associated to an essential simple closed curve γ.
This is a consequence of the fact that the Weil-Petersson metric is a Kähler metric
(geometrically means that the metric on the tangent space approximates to order 2).
Proof. Let µ ∈ HB(X). Consider the path η : (a, b) → T (S), η(t) = tµ for t small.
We can compose η with lγ to get a function lγ ◦ η : (a, b) → (0, ∞). Since the WeilPetersson metric is Kähler, the sign of the second derivative of lγ ◦ η is equal to the
107
6.3. OTHER PROPERTIES OF THE WEIL-PETERSSON METRIC
sign of the second derivative of lγ ◦ g if g(0) = 0 and g ′ (0) = µ.
We start doing a general setup. Let µ ∈ B(X). Then tµ is a path in T (S). For fixed
t, tµ ∈ T (S). Consider l(t) := lγ (tµ).
We want to find a formula for ¨l(0).
Let Σ := {z | 0 < Im(z) < π}. By the Riemann Mapping Theorem, we can consider
the conformal map Σ → H2 . The pull back of the hyperbolic metric ρH2 is the
1
hyperbolic metric ρ = sin(y)
on Σ.
Notice that the traslation of length |x| z 7→ z + x, x ∈ R is an isometry and
• if Im(z) = π2 , the metric is Euclidean;
• if |Im(z) − π2 | > 0, the metric is > Euclidean.
The map f t : Σ → Σ has Beltrami differential tµ. We want to normalize the setting
in the following way. We consider a lift γ̃ of γ and we suppose that γ̃ is as in Figure
6.2, in which there is the disc model of the hyperbolic space (that is the covering of
X) and its conformal annulus. The curve γ̃ is the red curve.
∞
−∞
γ̃
∞
0
Figure 6.2: The curve γ̃ is the red curve in both the disc model of the hyperbolic
space and its conformal annulus.
Let f (z, t) = f t (z) and f t (z + l(0)) = f t (z) + l(t).
We define ν so that
1. ν(z) = µ(z) for z ∈ Σ;
2. ν(z̄) = ν(z);
3. ν(z + 2πi) = ν(z).
We are extending µ(z) reflecting along R and by periodicity in each strip (2nπi, 2(n +
1)πi). We didn’t defined the value of ν(x), x ∈ R, but it is not important since we
are working with a Beltrami differential.
Let F t be the normalized solution to tµ, F t (0) = 0, F t (1) = 1, F t (∞) = ∞. Notice
that F t fixes R and it maps the line of height kπ, k ∈ Z to a line parallel to R.
If we fix z and we move t, then F t (z) is holomorphic. We want to change the
normalization so that we get smoothness.
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CHAPTER 6. THE WEIL-PETERSSON METRIC
Consider h(t) := Im(F t (x + iπ)) = Im(F t (iπ)).
By the Measurable Riemann Mapping Theorem, h(t) is a smooth function. By the
uniqueness of the Measurable Riemann Mapping Theorem, we get the following
t
(z)
and t 7→ f (z, t) is smooth for each z ∈ C.
Lemma 6.3.2. f t (z) = π Fh(t)
Let z = x + iy. We fix y such that 0 < Im(z) < π and we calculate
Z l(0)
fxt (z) dx = f t (l(0) + iy) − f t (iy) = l(t).
0
So we have
Z
π
0
Z
l(0)
t
f (z) dx dy =
0
Z
π
l(t) = πl(t).
0
Hence, if we call F the fundamental domain of the action at time t = 0 i.e., F =
(0, l(0)) × (0, iπ), we have
Z
F
fxt (z) dx dy = πl(t).
Now, fix x. We calculate
Z π
Im
fyt (z) dy = Im(f t (x + iy) − f t (x)) = Im(f t (x + iy)) = π.
0
Then
Z
F
We know that
We have
fyt (z) dx dy = πl(0).
1
fz̄t = (fxt + ifyt ).
2
Z
π
Re
fz̄t (x + iy) dx dy = (l(t) − l(0)).
2
F
(6.2)
We want ¨l(0) and the idea is to use this integral representation to find it. Recall that
the notations are
∂f
f t (z) = f (z, t) and f˙t (z) =
(z, t).
∂t
Tanking the derivative in 6.2, we have
Z
π˙
l(t) = Re
f˙z̄t (x + iy) dx dy .
2
F
Since fz̄t = tµfzt , f˙z̄t = µfzt + tµf˙zt . At t = 0, f˙z̄t = µ and so at t = 0,
Z
π˙
l(0) = Re
µ dx dy .
2
F
˙ =
Theorem 6.3.3 (Gardiner). For γ ∈ π1 (S) there exists Φγ ∈ Q(X) such that l(0)
Re < η, Φγ >.
6.3. OTHER PROPERTIES OF THE WEIL-PETERSSON METRIC
109
Since we want ¨l(0), we need to take another derivative.
Observe that f¨z̄t = µf˙zt + µf˙zt + tµf¨zt .
At t = 0, f¨z̄ = 2µf˙z , then
Z
π¨
l(0) = Re
µf˙z dx dy .
4
F
We are going to apply this formula for µ ∈ HB(X).
Assume that µ ∈ HB(X). This means that
1
µ(z) = 2 φ(z) = sin2 (y)φ(z),
ρ
where φ is holomorphic with φ(z + l(0)) = φ(z). So
X
φ(z) =
an ecnz (Fourier series) .
n∈Z
˙
Let f(z)
=
∂f
(z, 0).
∂t
We have
1. (By the Measurable Riemann Mapping Theorem) f˙(z) is a continuous function
on Σ;
˙ + iy)| < K|x|;
2. (growth bound) Since f t (z + nl(0)) = f t (z) + nl(t), n ∈ Z, |f(x
3. f˙z̄ = µ = sin2 (y)φ(z);
4. f˙(x) and f˙(x + iπ) are real numbers.
Let solve the equation f˙z̄ = sin2 (y)φ(z). Since
1
1 1
sin2 (y) = − ez−z̄ − e−z+z̄ ,
2 4
4
we get
1
1
1
µ(z) = φ(z) − ez φ(z)e−z − e−z φ(z)ez
2
4
4
z
−z
Notice that e and e are holomorphic functions and φ(z), φ(z)e−z and φ(z)ez are
anti-holomorphic functions.
Integratig µ(z) in z̄, we have the solution
Z
Z
Z
1 z +∞
1 −z z
1 z
−τ
φ(τ ) dτ + e
φ(τ )e dτ − e
φ(τ )eτ dτ .
w(z) =
2 z0
4
4
z
−∞
Since we can add to w(z) all the holomorphic functions that we want, we change w(z)
to
Z z
1
1
Z z
Z +∞
z
−z
−τ
f˙ = Re
φ(τ ) dτ + e Re
φ(τ )e dτ − e
φ(τ )eτ dτ .
2
2
z0
z
−∞
If w1 and w2 are two solutions of the equation, since (3) holds for both the solutions,
they are holomorphic functions. From (2) and (4), we have the uniqueness of the
solution.
P
Now, we put φ(z) = n∈Z an ecnz in f˙ and we substitute in µ(z) = sin2 (y)φ(z). Doing
an estimate of ¨l(0), one can conclude the proof of the Theorem.
110
CHAPTER 6. THE WEIL-PETERSSON METRIC
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[5] Imayoshi, Y. and Taniguchi, M., An Introduction to Teichmüller Spaces,
Springer-Verlag;
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