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]):

Theorem 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.

Definition 1 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.

Theorem 2 (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.

Theorem 3 Suppose that is a Riemann surface such that its fundamental group is abelian and nontrivial. Then
1. .
2. If , then or , where is a lattice in .
3. If , then , the annulus of module , where . Moreover, if and only if .

To prove Theorem 3, we need the following two lemmas.

Lemma 1
1. For any , we have .
2. 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 .
Proof (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
Remark 1 Let be a lattice in and where , then . By lemma 1, we know .
Lemma 2 Let , then for any if and only if all eigenvalues of are real.
Proof 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 (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 .

1. 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
2. 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

Remark 2 in the third case can be viewed as the length of the hyperbolic geodesic (cf. Example 7).
Remark 3 Here is another proof of the claim in the proof using the following beautiful theorem due to Caratheodory.
Theorem 4 (Caratheodory) The Riemann mapping extends to a homeomorphism .

Now suppose is a biholomorphism. Then by Caratheodory's Theorem 4, extends to a homeomorphism . Applying the Schwarz reflection theorem repeatedly, this map extends to a biholomorphism . Thus it is a scalar, which has to be 1 since maps to . So .

Corollary 1 . where consists of two points and .
Proof The second part of the corollary follows directly from Theorem 3 and the first part follows Theorem 3 along with the following lemma. ¡õ
Lemma 3 If then if and only if are equivalent under the action of .
Proof : by change of lattice basis and Remark 1. : composing by an automorphism of , we may assume . 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.

Remark 4 Suppose that and is a genus closed Riemann surface. Riemann computed that . To see that intuitively, let , then , where the fundamental group is generated by () subject to . The total number of degrees of freedom is . Since for , the total dimension is no more than . More generally, let be a topological surface of finite type (homeomorphic to a compact space) such that is generated by elements, then .

## 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 .

Proposition 1 Let and be -smooth, . Then
1. .
2. .
Lemma 4 The derivative of is given by .
Proof 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.

Lemma 5 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 .
Definition 2 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 .

Definition 3 The Beltrami coefficient is defined to be . We have and .
Definition 4 Assume preserves orientation. Define to be the dilatation of and to be the Beltrami coefficient of . Then .
Lemma 6 Suppose is -smooth, then where .
Proof By definition, The lemma follows. ¡õ
Corollary 2 If , then is conformal and . If , then is conformal and .
Definition 5 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 .
Remark 5 if and only if is analytic. Hence the ratio of the two axes is conformally invariant.

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.

Theorem 5 (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.

Example 1 (Grotzsch problem) Let be rectangles with vertices and size , . Consider all -smooth diffeomorphism such that . We have:
Theorem 6 where is the affine map.
Proof For any , 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
Example 2 Let be an annulus of module (). Then by , where . So if 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 .
Question How can we generalize the solution to the Grotzsch problem to arbitrary Riemann surfaces?

Teichmuller used quadratic differentials to solve this problem.

Definition 6 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 .
Example 3 There are no holomorphic 1-forms on (). All holomorphic 1-forms and quadratic differentials on are of the forms and where ().
Remark 6 The following theorem is classical.
Theorem 7 (Riemann-Roch) Let be a compact Riemann surface of genus , then the vector space of all quadratic differentials on is of complex dimension (and real dimension ).

Roughly speaking, we will see a correspondence between and , which explains the significance of quadratic differentials in Teichmuller's construction (cf. Remark 4and Theorem 14).

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.

1. A horizontal direction in , namely such that . So .
2. A vertical direction in , namely such that . So .
3. 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 .
4. is locally isometric to the Euclidean space . Indeed, given , define , then and and .
Theorem 8 (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 .
Proof 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
Definition 7 For a nonzero quadratic differential , we denote .
Remark 7 Let be a quadratic differential on a compact Riemann surface , then we obtain a flat Riemannian metric on , so the area of is well-defined: We also denote it by , then is a -normed vector space.
Example 4 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 .

Remark 8 Teichmuller proved that all analytic structures on a topological surface can be obtained via this construction.

The following uniqueness theorem can be regarded as the general answer to the Grotzsch problem.

Theorem 9 (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.

Lemma 7 (Key Lemma) Suppose is an embedded horizontal arc in and 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.

Theorem 10 (Gauss-Bonnet) Let be a polyhedral surface obtained by isometric gluing of Euclidean triangles. For a vertex of , define its curvature Then .
Proof (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
Lemma 8 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
Proof 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
Lemma 9 Let be the partial derivative with respect to the natural coordinate in , . Let be measured in . Then
Proof 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 (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 .

Definition 8 Let Let be a homeomorphism, then we can pull back to another complex structure on . Define Then the Riemann moduli space .
Definition 9 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 .
Definition 10 The group is called the mapping class group of . Then .
Example 5 For , then and .

There is a subtle problem concerning "markings" on the Teichmuller space since we fix the underlying set .

Example 6 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.

Definition 11 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 .
Remark 9 If are homotopic, then and are Teichmuller equivalent, so we may replace by its homotopy class.
Remark 10 If , then if and only if , where . This can be proved by looking at the universal covering . So a marking on the Riemann surface can be regarded as a choice of generators of the fundamental group under the equivalence of conjugation.

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 .

Example 7 The -axis is a geodesic in . Let be a path in with staring point . Then and equality holds if and only and , if and only if is along the -axis.

Note that is an isometry of . Thus we know that all vertical lines are geodesics. Note that and are also isometries. Since all Mobius transformations are composition of these three types of isometries, we know that . Actually one can show that . Using the transformation and the translations, it follows that all geodesics are either vertical lines or semicircles perpendicular to the -axis.

Remark 11 has constant Gaussian curvature since acts on transitively. Indeed, for a Riemannian metric , the Gaussian curvature is given by . A computation then shows that .
Example 8 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 .

Lemma 10 Suppose is a closed hyperbolic surface, then for , we have , or equivalently, has two distinct real eigenvalues (which are called hyperbolic).
Proof 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 .

Theorem 11 (Fricke) is a subset of the dimensional smooth manifold .
Remark 12 has a natural topology induced from , hence we can use to define a topology on such that is a homeomorphism.
Proof Let . Let be a discrete faithful representation and , . We get 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

Now the Teichmuller Uniqueness Theorem 9 can be restated as follows.

Theorem 12 (Teichmuller Uniqueness Theorem) is injective.
Proof If and , 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:

The next two lemmas will be proved using MRM in the next section.

Lemma 11 is continuous and proper.
Lemma 12 is connected.

Now the uniformization theorem will follow easily from the above two lemmas and Brouwer's famous invariance of domain theorem.

Theorem 13 (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.
Theorem 14 (Uniformization Theorem) is a homeomorphism. In particular, the Teichmuller space and the Fricke space are homeomorphic to .
Proof 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.

Question (Beltrami Equation) For with , is there any quasi-conformal map such that ?

The Measurable Riemann Mapping Theorem gives affirmative answers to this question.

Theorem 15 (Measurable Riemann Mapping Theorem) Let , , then there exists a unique quasi-conformal map satisfying the normalization , and such that .
Remark 13 The in question is absolutely continuous, and is smooth whenever is so.

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.

Lemma 13 Suppose are Riemann surfaces and is a quasi-conformal map. Then the Beltrami differential (of ) is well-defined on .
Proof 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.

Theorem 16 (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.
Remark 14 Moreover, Ahlfors-Bers showed that the solution depends analytically on . This will imply that is a complex manifold (see [2]).
Remark 15 We can interpret the Teichmuller map in terms of Beltrami differentials. For a quadratic differential , the associated at is constructed by stretching by a factor of along the horizontal direction . On the other hand, we know that transforms as and thus it is a Beltrami differential. The quasi-conformal map associated to the Beltrami differential in Theorem 16satisfies , hence it locally behaves as stretching by a factor of along the horizontal direction too. In summary, the Teichmuller map can be obtained by taking of the Beltrami differential .

Finally, here comes the proof of Lemma 11 and Lemma 12, which in turn completes the proof of Theorem 14.

Proof (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 (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.