These are my live-TeXed notes (reorganized according to my tastes) for the second part of the course Introduction to Riemann surfaces and Teichmuller theory by Professor Feng Luo, June 28 — July 14, at Math Science Center of Tsinghua University.
Classification of Riemann Surfaces
Surfaces (assumed to be connected and orientable) of finite type (i.e., with finitely generated fundamental group) have a complete topological classification due to Mobius ([1, Theorem 1.1]):
(Topological classification of surfaces)
- Any closed surface is homeomorphic to the connected sum of a sphere with
tori, where
is called the genus.
- Any compact surface is obtained from a closed surface by removing
disjoint disks, where
is the number of boundary components.
- Any surface is obtained from a compact surface by removing
points from the interior, where
is the number of punctures.
Among these,
,
,
,
are the only cases with abelian fundamental group
. However, each single topological class may contain various analytic structures.
Let

be a topological surface. Denote by

the set of biholomorphism classes of Riemann surfaces that are homeomorphic to

. This is called
Riemann's moduli space.
We can restate the uniformization theorem as follows.
(Uniformization Theorem)
Let
be a Riemann surface with nontrivial fundamental group. From general algebraic topology property, we know that there exists a simply connected surface
and a covering map
such that
acts freely and discontinuously on
: for any
,
is a homeomorphism and
if and only if
for some
; for any
there exists a neighborhood
of
such that
for any nontrivial
. In particular, for any nontrivial
, we have
for any
.
is a Riemann surface such that
is analytic. The Uniformization Theorem tells us that
is either
,
or
.
The main goal of this section is to prove the following classification of Riemann surfaces with simple topology.
To prove Theorem 3, we need the following two lemmas.
- For any
, we have
.
- Any continuous map
can be lifted to
such that
, where
and
are the universal covering maps. Furthermore, if
is a homeomorphism then
is so. Obviously, if
is analytic then
is so, hence if
then
.
(a) is direct. For (b), since

is simply connected and

is a covering map, we choose

to be the lifting of

. Furthermore, suppose

is a homeomorphism with

. Taking

,

and a lift

,

, we get a lifting

similarly. Then

so

and

are both the liftings of

sending

to

. The uniqueness of the lifting ensures that

are inverses.
¡õ
Let

, then

for any

if and only if all eigenvalues of

are real.
Since

, the eigenvalues are of the form

. The condition that the eigenvalues are real implies that

. The fixed points condition implies that

, namely

, too.
¡õ
Now we are in a position to prove Theorem 3.
(Proof of Theorem 3)
(a) Since every Mobius transformation

has a fixed point in

,

cannot act on

freely.
(b) Suppose
. The automorphism
has no fixed points, hence
and
, namely
. If
, then
by
. If
with
, then topologically
is
. We claim that
. If
is a rational multiple of
, then
for
, therefore
, which contradicts the fact that
and
are generators of
. If
is an irrational multiple of
, then
is dense in the line
, so that
has limit points which contradicts the discontinuity. So
where
is a lattice. A similar argument shows that
cannot have more than two generators.
(c) Suppose
.
- If
, with
, then every fixed point of
is inside
by freeness. After conjugating by some
, we may assume that
has the form
, where
(by Lemma 2) or
.
If
, then
. So
by
If
, then
, where
. Then
is biholomorphic to
where
, by composing the logarithm map, the strip-scaling map and the exponential map 
- If
, then
since
. Up to conjugation, we may assume that
or
(
). If
, then
and
, so
cannot act discontinuously; if
, then
and
with
. If
, then
; if
, then
cannot act discontinuously.
We claim that no two
's are not biholomorphically equivalent. If
then
by Riemann's theorem on removable singularities. If
, suppose
is a biholomorphism, then
. By Lemma 1,
lifts to
and
. Comparing the eigenvalues, we know
, therefore
. The claim follows.
¡õ

.
![$\Mod(S^1\times\mathbb{R})\cong
(1,\infty]$](./latex/latex2png-Teichmuller_96290683_-5.gif)
where

consists of two points

and

.
The second part of the corollary follows directly from Theorem
3 and the first part follows Theorem
3 along with the following lemma.
¡õ
If

then

if and only if

are equivalent under the action of

.

: by change of lattice basis and Remark
1.

: composing

by an automorphism of

, we may assume
![$h[0]=[0]$](./latex/latex2png-Teichmuller_243782259_-5.gif)
. Let

be the lifting of

such that

using Lemma
1. Hence

, which implies that

and

are equivalent under the action of

. Moreover,

ensures that they are actually equivalent under

.
¡õ
By now, we have completely described Riemann's moduli spaces
for all topological surfaces
with abelian fundamental groups.
Quasi-conformal maps
We fix the following conventions in this section:
are open in
.
is a
- smooth function. Denote
,
,
,
and so on. The following proposition can be thought of the chain rule with respect the variables
and
.
Let

and

be

-smooth,

. Then
.
.
The derivative

of

is given by

.
By definition and Proposition
1, we have

The lemma follows.
¡õ
It is easy to see that every
-linear map
is given by
, where
satisfies
and
. Moreover,
. If
, then
. It follows that the Jacobian of
is equal to
. So
preserves orientation if and only if
.
Geometrically,
sends circles to ellipses. The following easy lemma describes which of the radii of a circle are sent to the axes of the corresponding ellipse.
For

(assume that

), we have

The second equality holds if and only if

, if and only if

. The first equality holds if and only if

, if and only if

.
The
dilatation of

, denoted by

, is the ratio of axes of the ellipse

, which is equal to

.
Note that
will switch the major axis and the minor axis. Also, it is easy to see that
and
.
The
Beltrami coefficient is defined to be

. We have

and

.
Assume

preserves orientation. Define

to be the dilatation of

and

to be the Beltrami coefficient of

. Then

.
Suppose

is

-smooth, then

where

.
By definition,

The lemma follows.
¡õ
If

, then

is conformal and

. If

, then

is conformal and

.
Say

is a
quasi-conformal map if

is an orientation-preserving homeomorphism,

-smooth outside a discrete set

and there exists

such that

for any

. Denote

, then

.
The concept of quasi-conformal maps can be generalized to Riemann surfaces using local coordinates. A natural question is to find the minimum dilatation quasi-conformal maps (in given a homotopy class) between Riemann surfaces with different analytic structures. This is the content of Teichmuller theory and the main theorem is the following.
(Teichmuller Theorem)
Suppose

are compact Riemann surfaces of genus

and

is a given homeomorphism. Then there exists a unique quasi-conformal map

such that

is homotopy equivalent to

and

has the smallest dilatation among all such

. Moreover,

is a constant independent of

.
This theory is motivated by the classical Grotzsch problem.
(Grotzsch problem)
Let

be rectangles with vertices

and size

,

. Consider all

-smooth diffeomorphism

such that

. We have:

where

is the affine map.
For any
![$y\in[0, b_1]$](./latex/latex2png-Teichmuller_221554913_-5.gif)
, let

be the horizontal line segment

, then

So using the triangle inequality, we obtain

Now the Jacobian

, so

by Cauchy's inequality. The equality holds if and only if

is affine:

is a constant and

is a constant, therefore

and

are constants. The theorem now follows.
¡õ
Let

be an annulus of module

(

). Then

by

, where

. So if
![$f:R_1=[0, r_1]\times [0, 2\pi]\rightarrow R_2=f(R_1)$](./latex/latex2png-Teichmuller_12340067_-5.gif)
is a

-smooth diffeomorphism, using the same proof as in Theorem
6, we know that

and the equality holds if and only if the lift

is affine given by

.
How can we generalize the solution to the Grotzsch problem to arbitrary Riemann surfaces?
Teichmuller used quadratic differentials to solve this problem.
Quadratic differentials
A holomorphic 1-form

on

is an analytic function

such that

for any

. Similarly, a quadratic differential

on

is an analytic function

such that

for any

. More generally, for a Riemann surface

with analytic charts

, a
holomorphic 1-form on

is a collection of analytic functions

such that

, i.e.,

. Similarly, a
quadratic differential on

is a collection of analytic functions such that

.
There are no holomorphic 1-forms on

(

). All holomorphic 1-forms and quadratic differentials on

are of the forms

and

where

(

).
A holomorphic 1-form at
gives an
-linear map from the tangent space
to
. Similarly, a quadratic differential
at
gives a quadratic form
from
to
. A quadratic differential provides the following geometric data.
- A horizontal direction in
, namely
such that
. So
.
- A vertical direction in
, namely
such that
. So
.
- A flat Riemannian metric
near
with
, given by
. This metric is conformal to the standard Euclidean metric, i.e., angles in
are the same as angles on the Riemann surface
.
is locally isometric to the Euclidean space
. Indeed, given
, define
, then
and
and
.
(Teichmuller)
Suppose

is a nonzero quadratic differential on a Riemann surface

, then for any

, there exists an analytic coordinate

at

with

such that

in

is given by

where

. We call this analytic coordinate the
natural coordinate for

.
We may work locally and assume that

,

and

where

is analytic. If

, we have just proved that

where

. Now suppose that

. Let

where

is analytic and

. Our goal is to write

as

where

is analytic,

and

. Namely, we would like to solve the equation

Formally, it is equivalent to

or

We need to justify that this is well-defined. Suppose

where

. Then

Hence

is single-valued and satisfies Equation
1.
¡õ
For a nonzero quadratic differential

, we denote

.
The metric

near

is a cone with cone angle

. The cone minus the vertex has a flat metric. Suppose the cone angle is

, then the pullback metric of

via

is

. For example, the metric

is obtained by gluing three half planes along the origin.
Teichmuller Uniqueness Theorem
Let
be a Riemann surface,
a quadratic differential on
and
, where
. Then there exists a new Riemann surface
such that the underlying set of
and
are the same, and there exists a quadratic differential
on
such that
is
-quasi-conformal. Let us construct the analytic charts for
as follows. Pick any point
.
Type I
. By Theorem 8, there exists a natural coordinate
of
at
with
and
. Define a new analytic chart
for
by
, in other words,
,
.
Note that if another
is analytic with
and
such that
, then
, which implies that
. So if
and
are two natural coordinates of Type I for
, the the transition function
between them is analytic: we know that
where
is a constant, therefore
, so
. Hence
,
is a well-defined quadratic differential on
.
Type II
. Choose a natural coordinate
at
such that
by Theorem 8. Define a chart for
by
This is well-defined since the right hand side is
and
, so we can just choose the principal branch. Moreover,
is a homeomorphism
. Indeed,
, so 
We claim that the transition function
for a Type I chart
and a Type II chart
is analytic. Suppose
at
. We may assume that
. Then we find that
in
, so
. Hence
is analytic and
.
The following uniqueness theorem can be regarded as the general answer to the Grotzsch problem.
(Teichmuller Uniqueness Theorem)
Let

be a compact Riemann surface with a nonzero quadratic differential

. Let

be the new Riemann surface. If

is a quasi-conformal map homotopic to the identity map, then

and the equality holds if and only if

.
Denote
for simplicity.
(Key Lemma)
Suppose
![$\alpha:[0,1]\rightarrow Y$](./latex/latex2png-Teichmuller_116178443_-5.gif)
is an embedded horizontal arc in

and
![$\tilde\alpha:[0,1]\rightarrow Y$](./latex/latex2png-Teichmuller_156900614_-5.gif)
such that

,

are homotopic relative to

. Then the length of

is no greater than the length of

in

.
The problem occurs due to the presence of singularities (negative curvatures). To prove this Key Lemma, we need the following version of Gauss-Bonnet Theorem.
(Gauss-Bonnet)
Let

be a polyhedral surface obtained by isometric gluing of Euclidean triangles. For a vertex

of

, define its curvature

Then

.
(Proof of Key Lemma 7)
Let us minimize the length of

. We may assume that

is a piece-wise geodesic path homotopic to

such that the vertices of

are in

. Moreover, any vertex other than

and

has cone angle at least

(otherwise we can decrease the length), so

for these vertices. Now by the Gauss-Bonnet Theorem, we get

, impossible unless

.
¡õ
Let

be a

-quasi-conformal map homotopic to the identity map. Then there exists a constant

such that for any horizontal arc

in

, we have
Let

be a

-smooth homotopy between

and

. Let

. Using Key Lemma
7, since

is also horizontal in

and

, we know that

This completes the proof.
¡õ
Let

be the partial derivative with respect to the natural coordinate

in

,

. Let

be measured in

. Then
Let

be

minus all horizontal lines through

(several branches at one singular point), then

has full measure in

. For any

and

, define

be the horizontal line centered at

of length

. Then

is simple since the horizontal direction is unique outside

. We choose the arc length parametrization, namely

, so

, where

under the natural coordinate. Using Key Lemma
7, we know that

where

is a constant. But by the chain rule

we know that

So

The lemma then follows by letting

.
¡õ
(Proof of Theorem 9)
By Lemma
9 and Cauchy's inequality, we have

since

when

is compact. The theorem follows.
¡õ
Teichmuller spaces
Let us return to the study of Riemann's moduli space
consisting of all Riemann surfaces
homeomorphic to a fixed topological surface
. We observe that it is the equivalent to study all complex structures on
.
Let

Let

be a homeomorphism, then we can pull

back to another complex structure

on

. Define

Then the Riemann moduli space

.
We say two complex structures

and

are Teichmuller equivalent if there is a biholomorphism

such that

is homotopic to the identity map. The set of all Teichmuller classes is called the
Teichmuller space, denoted by

. Define

Then

is the same as

.
The group

is called the
mapping class group of

. Then

.
There is a subtle problem concerning "markings" on the Teichmuller space since we fix the underlying set
.
What is the space of all congruence classes of triangles in the plane? Let

be a fixed triangle. Then

which is isomorphic to

. But the Riemann moduli space

is isomorphic to

.
Instead of fixing the underlying set
, we may consider the equivalence classes of marked Riemann surfaces.
Fix a topological surface

. A
marked Riemann surface 
by

is a pair

where

is a homeomorphism.

and

is Teichmuller equivalent if there exists a biholomorphism

such that

are homotopic. So the Teichmuller space

is the set of all Teichmuller equivalent classes of

.
Our goal is to show that
is isomorphic to the Fricke space
Fricke showed that
can be viewed as a subset in a
-dimensional manifold
, and then one can prove that
is actually homeomorphic to
(cf. Uniformization Theorem 14).
Hyperbolic geometry
This subsection is a crash course on the basics of hyperbolic geometry. Define a Riemannian metric
on
.
The distance

Since the cross ratio is invariant under a Mobius transformation, we have

where

are the intersection points of the semicircle through

with the

-axis.
Let
be a compact Riemann surface of genus
. The uniformization theorem tells us that
. The metric
is invariant under
, so it induces a Riemannian metric
on
(thus such an
is called a hyperbolic surface) and now it makes sense to talk about lengths on
.
Suppose

is a closed hyperbolic surface, then for

, we have

, or equivalently,

has two distinct real eigenvalues (which are called
hyperbolic).
By the proof of Lemma
2, we know that

. If

, after conjugation, we may assume that

. Since

is compact, we can find finitely many small balls

covering

(so

's are contractible). By Lebesgue's Lemma, there exists

such that any loop of length less than

in

is in some

(hence homotopic to a point). But for

, consider the image of

, whose length is

. It is homotopically nontrivial, a contradiction.
¡õ
Uniformization theorem
We are now going to talk about the remaining two main theorems, namely the uniformization theorem and the measurable Riemann mapping theorem. Teichmuller asserted that the Teichmuller space
is isomorphic to
, but this result was not rigorously proved until the work of Ahlfors and Bers in 1950s. We will utilize some corollaries of the measurable Riemann mapping theorem (MRM) to prove the uniformization theorem in this section and discuss the MRM in the last section.
Let
be a closed surface of genus
. Let
be the universal covering of
, then there exists a biholomorphism
such that
acts freely and discontinuously on
. Moreover,
induces a biholomorphism on the quotient
. In summary, we get a map
The image
is called the Frick space of
. By Remark 10 , the map
is injective. So we can identify
with
.
(Fricke)

is a subset of the

dimensional smooth manifold

.
Let
![$\pi_1(\Sigma_g)=\langle a_1, b_1,\ldots, a_g, b_g: [a_1, b_1]\cdots[a_g,b_g]=1\rangle$](./latex/latex2png-Teichmuller_246024894_-5.gif)
. Let

be a discrete faithful representation and

,

. We get
![\begin{equation*}
A_1B_1A_1^{-1}B_1^{-1}=[A_g, B_g]^{-1}\cdots [A_2, B_2]^{-1}. \tag{2}
\end{equation*}](./latex/latex2png-Teichmuller_47218397_.gif)
After conjugation, we may assume that

, where

by Lemma
10. Suppose

. By the proof of Theorem
3, we know

in

, since

is nonabelian. But

, so

(consider

) and

(consider 0).
Now
commutes with
, so after conjugation we get 
We may assume
for
. If we can solve uniquely
from Equation 2, then
. Indeed, we compute that
and one can show that
,
,
,
can be solved successively using the fact that
,
.
¡õ
By the Riemann-Roch Theorem 7,
is a
dimensional
-vector space and has a natural
-norm. Let
. Denote
, where
. Define the Teichmuller map ![$$\Phi: \QD_1(X)\rightarrow \Teich(\Sigma_g),\quad \eta\mapsto [X_\eta]=[(X_\eta,\Id)].$$](./latex/latex2png-Teichmuller_85191659_.gif)
Now the Teichmuller Uniqueness Theorem 9 can be restated as follows.
(Teichmuller Uniqueness Theorem)

is injective.
If

and
![$[(X_{\eta_1},\Id_1)]=[(X_{\eta_2},\Id_2)]$](./latex/latex2png-Teichmuller_118188258_-5.gif)
, then there exists a biholomorphism

such that

and

are homotopic. By the Teichmuller Uniqueness Theorem
9, we know that

therefore

. It follows from the construction of

that

.
¡õ
Combining Theorem 11 and Theorem 12, we obtain the following picture: ![$$\xymatrix{\QD_1(X) \ar@{^{(}->}[r]^-\Phi & \Teich (\Sigma_g) \ar[r]^-\Psi_-{\sim} & \mathcal{T}_g\subseteq M^{6g-6}.}$$](./latex/latex2png-Teichmuller_111308510_.gif)
The next two lemmas will be proved using MRM in the next section.

is continuous and proper.

is connected.
Now the uniformization theorem will follow easily from the above two lemmas and Brouwer's famous invariance of domain theorem.
(Brouwer's Invariance of Domain)
Let

be an open set in

and

be a one-to-one continuous map into a manifold of the same dimension, then

is open in

and

is a homeomorphism.
(Uniformization Theorem)

is a homeomorphism. In particular, the Teichmuller space

and the Fricke space

are homeomorphic to

.
By the continuity in Lemma
11 and Invariance of Domain Theorem
13, we know that the image of

is open in

. By the properness in Lemma
11and the fact that the target

is a locally compact Hausdorff space, we find that

is a closed map and thus the image of

is also closed in

. But

is connected by Lemma
12, hence the image of

is the whole of

, which completes the proof.
¡õ
Measurable Riemann Mapping Theorem
As promised in the last section, we shall discuss the Measurable Riemann Mapping Theorem and utilize it to finish the proof of the uniformization theorem.
Let
be a quasi-conformal map. By definition, the supremum norm of the Beltrami coefficients
. The following inverse question asks about the existence of such quasi-conformal maps.
(Beltrami Equation)
For

with

, is there any quasi-conformal map

such that

?
The Measurable Riemann Mapping Theorem gives affirmative answers to this question.
(Measurable Riemann Mapping Theorem)
Let

,

, then there exists a unique quasi-conformal map

satisfying the normalization

,

and

such that

.
This deep theorem was proved by Lavrentev for
-functions
and by Morrey for measurable functions
.
MRM can be also generalized to Riemann surfaces. From Sullivan's geometric point of view, the MRM simply says the following: place an ellipse on each tangent space on a Riemann surface, then there exists a unique quasi-conformal map sending this ellipse field to a circle field. In other words, for a pair
, we need to construct a new Riemann surface
satisfying the above geometric condition.
Suppose

are Riemann surfaces and

is a quasi-conformal map. Then the
Beltrami differential (of

)

is well-defined on

.
Suppose

, where

are analytic. Let

. By the chain rule, we have

Then

So

is well-defined.
¡õ
Now let
be the vector space of all Beltrami differentials on
. It has a natural
-norm given by
. We have a pairing
So
can be viewed as the dual of
in some sense.
Looking at the universal covering and using the normal MRM Theorem 15, one can prove the following MRM on Riemann Surfaces.
(MRM on Riemann Surfaces)
Let

be a Riemann surface and

be a Beltrami differential on

with

. Then there exists a Riemann surface

and a quasi-conformal map

such that

and

is unique up to biholomorphism.
Finally, here comes the proof of Lemma 11 and Lemma 12, which in turn completes the proof of Theorem 14.
(Proof of Lemma 11)
By Remark
15,

is the composition of the continuous MRM given by Theorem
16 and the continuous map

, hence continuous. If it is not proper, suppose

satisfies

(

) and

converges to

. Then we get quasi-conformal maps

such that

. But

for some constant

as

converges to

, which contradicts the assumption that

.
¡õ
(Proof of Lemma 12)
We use MRM to prove that

is connected. Namely, we need to show that there is a continuous process from one

to another

such that every matrix representation of

is discrete and faithful. This can be done by using

.
¡õ
References
[1]Benson Farb and Dan Margalit, A Primer on Mapping Class Groups, http://www.math.utah.edu/~margalit/primer/.
[2]Lars V. Ahlfors, Lectures on Quasiconformal Mappings (University Lecture Series), American Mathematical Society, 2006.