Riemann surfaces: Lectures given at the Institute of Mathematical Sciences, New York University 1957-1958

RIEMANN SURFACES

Lipman Bers 1957-58

Notes by E. Rodlitz and R. Pollack

Courant Institute of Mathematical Sciences New York University

The Courant Institute publishes a number of sets of Lecture Notes.

A list of' titles currently available

may be found on the last pages of this volume.

ii

FO1 EWORD

These lecture notes represent the content of a course taught at the institute of Mathematical Sciences,

New York University, during the academic year 1957-1958. A The notes reflect the lectures as actually given. proper course on Riemann surfaces would need about twice the time that I had at my disposal. For that reason, many important topics were not covered; in particular, Abel's theorem, the Jacobi inversion problem, classification of open Riemann surfaces, qussiconformal mappings, the problem of moduli.

I owe a debt of gratitude to Professor Robert Osserman, who replaced me on several occasions; Lectures 14, 23, 24 and 25 were given by him. I am also very grateful to Mr. Richard Pollack and Miss Esther Rodlitz, who prepared these notes.

L. B.

iii

TABLE OF CONTENTS 1

Lecture 1

Defintion of conformal structure, Riemann surface. Meromorphic functions on a Riemann surface. Sketch of the course. 9

Lecture 2

Euclidean polyhedral surfaces. Surfaces and n-manifolds Differentiable structures on manifolds. 15

Lecture 3

Isothermal coordinates. Formal complex differentiation. Beltrami equation.

Lecture 1.

29

Existence of solutions to the Beltrami equation. Lecture 5

36

Continuation of existence theorem. Properties of solution of the Beltrami equation. Tensors on an n-manifold.

Metric tensors.

Introducing a conformal structure on a 2-manifold with with a metric tensor. 46

Lecture 6

Defining a metric tensor on an n-manifold. Regular embeddings. Partition of unity. Embedding theorem. Embedding problems. Lecture 7

57

Regular, unramified function elements. Analytic continuation of function elements. Analytic functions in the sense of Weieratrass. Function defined by a polynomial equation P(z,w) 0 0. Algebraic and algebroid functions. Lecture 8 Branch points of a function. Analytic configuration. Function elements. Riemann surface of an analytic function. Compactness of the niemann surface of an algebraic function. iv

67

Lecture 9

76

Function field of a Riemann surface. Harmonic functions. Maximum principal for harmonic functions. Maximum modulos principal for analytic functions. Critical points. Primitive pair. Every compact Riemann surface is the Riemann surface for an algebraic function. Algebraic curves. Birational equivalence.

Lecture 10 Puiseaux series. Valuation rings. Valuation ideals.

90

Places.

Lecture 11 Topology of compact Riemann surfaces. Triangulation of the surface. Hyperelliptic surfaces. Topological polygon.

105

Lecture 12 Homotopy. Fundamental group. Covering space.

117

Lecture 13 Universal covering space. Defining subgroup. Covering groups.

129

Lecture 14 Harmonic functions. Sub-harmonic functions. Maximum principal. Perronts Principal. Dirichlet problem. Barriers.

136

Lecture 15 Uniformization for simply connected surfaces; compact case.

143

Lecture 16 Continuation of meromorphio functions. Gap values. Weierstrass points.

151

v

Lecture 17

157

Uniformization theorem; parabolic case. 163

Lecture 18

Uniformization theorem; hyperbolic case. Rado's theorem. Natural metric on a Riemann surface. Automorphio functions. Lecture 19

174

Fuchsian groups. Fractional linear transformations. Representation of a hyperbolic surface

as

tr.

Lecture 20

182

Conformal mappings of a surface into itself. Lecture 21

186

Fundamental region. Lecture 22

191

Inte-ration on Riemann surfaces. Differentials (real). Exterior differentiation. Closed and exact differentials. Conjugate differentials. Stoke's theorem. Harmonic differentials. Lecture 23

202

Dirichlet principle. Lecture 24

208

Proof of Dirichlet principle. Lecture 25

215

General uniformization theorem and applications. Interior functions. Lecture 26

222

Differentials (complex). Abelian differentials and their periods. Canonical polygon. Riemann's bilinear relations.

Lecture 27

232

Riemann period matrix. Properties of Abelian differentials. Lecture 28

247

Divisors and the Riemann-Rooh theorem. vi

Lecture j What is a Riemann Surface? Roughly put, it is a surface on which one can define analytic functions. We could now give many examples of hiemann Surfaces and by abstracting what is essential to all of them, arrive at a definition of Aiemann Rather than doing this we shall present a definition Surface. of Riemant Surface first, and then give many examples, checking that they satisfy the definition. First we need a few preliminary definitions. Definition 1. A hausdorff space S has conformal structure if, in the class of functions mapping open sets C C S Into E (the complex plane) a certain subset is distinguished. This distinguished subset is called the set of "Analytic functions". Furthermore, the not of Analytic functions must satisfy the following axioms. 1. To every point p c 3, there is an open set G, p E G C S,

and an analytic function f which maps G topologically onto an open set in E; every such a function f is called a local parameter. 2. Given an analytic function f defined on an open sot G, and given an open set Go C G, and a local parameter Z defined on Go; then the function fIG 0r.-1 (fJG denotes the func0

0

tion f restricted to the set Go) is analytic in the usual an analytic function f on G C S (In other words: sense. is an analytic function, in the usual sense, of every local parameter defined on any set Go C G). 3. The set of analytic functions cannot be enlarged without violating axioms 1 and 2. Note:

Axiom 3 means that any function f:G -- E (with G open in S) is analytic if it has the property expressed by Axiom 2.

1.2

2

Intuitively then, Axiom 1 tells us that the space 3 is locally like the complex plane, while Axiom 2 restricts the class of analytic functiuns to functions locally analytic in the usual sense, and Axiom 3 appears as a sort of uniqueness axiom. That is, the topological structure of S together with a set of local parameters dictate the conformal structure.

Now we can offer Definition 2.

A Riemann Surface is a connected Hausdorff space

S, which has conformal structure.

The next definition tells us when two Riemann Surfaces are essentially the same.

Definition.

Two Riemann Surfaces S and T are called confor-

mally egn1valent if there exists a homeomorphism $, which preserves conformal structure.

That is, if s and t are Analytic functions in S and T shall be analytic in 3 and T and respectively. respectively;

Definition 4.

A meromorphic function m on a Riemann Surface is

a mapping which takes on as values not only complex numbers but also oo, and can be represented in every domain of a local parameter as a quotient g/h where g,h are analytic functions (h ii 0).

Notice now that the Euclidian plane E is a Riemann Surface,

where E has the usual topology and a function is analytic if it is analytic in the usual sense. Theorem 1.

Every domain D, on a Riemann Surface 3, is a Riemann

Surface with the topology can D the relative topology, and the class of analytic functions, the restrictions of the analytic functions on S.

Proof.

Obvious.

Note.

The theorems stated in this lecture will not be proven now. Their proofs in fact will occupy many of the succeeding lectures. They are stated now, so that the structure of the subject will not be lost in the forest of details. Hence every domain in E is a Riemann Surface:

for example z-i

the unit disc (or the eonformally equivalent under d(z) - z+i upper half plane) is a Riemann Surface. Next, the full function theoretic plane E is a Riemann Surface. The basis for the topology of the space E will consist of neighborhoods of all points of E plus the neighborhoods of the point co; a "neighborhood of the point oo" is defined as the complement of any compact set in E, plus the point oo. A function f(z) is analytic at a point z (z # co) of E if f(z) is analytic in E; and f(z) is analytic at z = oo if f(1/z) is analytic at z - 0. With this topology and conformal structure, the full function theoretic plane E is a Riemann Surface, and, as is well known,homeomorphic (under stereographic projection) to the sphere. (E is also called the Riamann Sphere.) Now we might ask whether or not these three Riemann SurfaNotice first that E,U are homeomorces are really different.. phic but neither is homeomorphic to E (which unlike E,U is compact). Nert, the classes of analytic functions on the spaces E,U,E differ: the analytic functions on E arc the entire func-

tions, those on U may be the restrictions of functions meromorphic on E, and finally the only functions analytic on 9 are constants. Although E,U,E differ as Riemann Surfaces all three are simply connected. In fact we have: Theorem.

The Uniformization Theorem for Simply Connected Rie-

mann Surfaces (also known as the Riemann Mapping Theorem for Riemann Surfaces). Every simply connected Riemann Surface is conformally equivalent to one of E, U or E.

lie have seen that every domain of a Riemann Surface is a Riemann Surface.

We will now outline a method for obtaining new Riemann Surfaces from a given Riemann Surface. Given a topological apace X, and some equivalence relation between points of X, we can divide all points of X into equivalence classes: (p) will denote the equivalance class contain-

ing p, i.a. (p) = {xIx a X and x - p} . Denote by X the set of equivalence classes:

we have the natural mapping

A

A set will be called open in X if it is the image, u_zder 4, of an open set in X. Then X is a topological i.e. 3(p) = (p).

apace.

Consider two complex numbers w,ki' a E which are linearly j 0. We call z independent over the reals, i.e. Am z' if and only if z' = z + nw + mw' for some pair of integers n and m.

We can assume

3

u'/'

> 0.

For, if not, w N -w and S n -k/w' > 0.

Another way of expressing this equivalence relation is to consider the group G of transformations of the plane into itself generated by the translations,

z' =z+w

z' =z+w'

then z', z if there exists a T e G and z' = T(z).

Denote the the topology will be given by the discussion of the preceding paragraph. A point in E/G is a net of points,. (see diagram) set of equivalence classes by E/G:

0

A Now consider an open set G C E/G and a function

f: Go -- :s. We call a function f analytic if there is an analytic function f, defined, for all z such that [z] C Go, with the properties f(z) = f(z') for z .., z' and f(z) = f((z]). We could verify that in this way E/G is a Riemann Surface. The analytic functions f in correspond to analytic functions f doubly pericdic to E with periods w and w'. We can see explicitly what the Surface E/G is in the following day. Let zo C E; then each point (z) E E/G where z is not of the form z = zo + mw + nw' (mn = 0, m,n real) has a unique representative in the interior of the parallelogram B,

shown b:ilow.

,-O'z o +4) +c) t

,,. 11V

z

0

+(vt

a E

zo

A point (z') where z' is of the form z' = zo + nw + mw' (mn = 0, m,n real) has two representatives in B, but only one if we identify opposite points z' and z", and identify the four vertices. An open set not containing points (z'] will have a unique representative G in B. while an open set about (z'] will be re-

presented by G' U G" (see diagram) where 01, G" are connected by the prescribed identification. (The case where z' is a vertex is evident). We recognize B (with its identifications) as a torus.

1.6

6

and Wi.

We could have chosen two other complex numbers W In general we would get a different Riemann Surface.

Hence we

should really write E/G(tv,w' ) Assignment. E/G(w,..i') is conformally equivalent to E/G(+v11wl) if and only if there exist Inte'ers a,p,Y,b and at - PY = 1 with

tot=Yw+ow'

w1 = awf + Awl

Instead of haviii& the group 0 generated by two translations, we could larva a group 0 generated by one translation

z' = a + to In the same manner as before we obtain the Riemann Surface E/G(w).

E/U(w) can be represented by an infinite strip with opposite points identified. The surface is obviously a cylinder and the analytic

z'/

-yzo+ LJ

functions are the usual analytic functions which are simply periodic, with period w. Assignment.

All E/G(a,) are

conformally equivalent.

We cannot obtain any additional Riemann Surfaces from the finite plane E by defining equivalence classes under any Remark.

other group and operating as above.

The two groups considered

are the only groups of confcormal transformations mapping E into

£ which have the properties required, viz. they are 1. discrete. That is, there exists no element other than I

which maps any point arbitrarily close to Itself, i.e. I is an isolated point. 2. fixed point free. fixed point.

That is, no element other than I has a

1.7

7

Are there groups of transformations of c with this property? The only possitility is ill, which gives us ao new Riemann Surfaces. But U considered as the upper half plane has the group of non-E:uclidian motions

Z.

az+

YZ+

a,P,Y,8 real and ab - PY = 1, which has many sub-groups with properties 1 and 2.

::e

later prove the following

Uniformization Theorem for Riemann Surfaces. Any Riemann Surface is conforma].ly equivalent to an d/G, U/G, or E.

Theorem.

A uniformization theorem more general than the first we 'ave, and less genergl than the second, will relate to Riemann Surfaces of planar character (schictortige). Any Riemann Surface which is separated by every closed Jordan curve is said to be of planar character. Defin'_tion 5.

This definition is motivated by the observation that any sub-domain of 2 is separated by a closed Jordan curve. Every Riemanr. Surface of Planar Character is conforTheorem. mally equivalent to a sub-domain of E.

notice that this theorem is more general than the first we gave, since E,U are sub-domains of E. Let S be a connected topological space, with every point p E S contlined in an open set G c: S which is mapped homeomorphically onto an open set in E by a function b. Furthermore if two such functions dl, d2 are 'iven in overlapping regions G1, b21 411 and 42 be analytic G2 we require that in G1 fl G2, 1 functions in S. Under these conditions we call the functions d, defining local parameters. This last requirement is called the consistency condition. If we now call analytic, all functions f, which map an open set G C. S into E, and which are such that

in any open set Go C G, f1, - 4-1 is analytic in E, then S will 0 be a Riemann Surface.

1.8

8

A theorem to be proved is Theorem.

Let function f be called analytic functions in 3 if: 1. f maps an open set G C S into E,

2. in any open set Go C G, where the defining local parameter 4-1 is analytic in E.

4:00 , -> E then fIG 0

Then 3 will be the only Riemann Surface with the given class of defining local parameters.

2.1

9

Lecture 2 §1.

Introduction.

In this lecture we shall consider, through illustrative examples, the topological restrictions to be placed on a space so that it can be made into a Riemann surface.

Two observations which are evident from the definitions of the terms are: a) Every domain, in a space with conformal structure, is a Riemann surface. b) If a function is analytic. in a domain D, its restriction is analytic in any sub-domain Do C D. (See Lecture 1, definition of analytic function, Axiom 3.)

Euclidean polyhedral surfaces and Riemann surfaces. Consider a set of disjoint triangles (2-simplices) in E, with the natural orientation defined for each triangle. Denote by Y the set of points belonging to these triangles. We shall define a topological space X by an identification of elements in Y; we shall begin by making the following assumptions. 1) Given two distinct triangles Ol, A2 of Y, there are three possibilities: i) no point of 01 is identified with a point of A2 ii) an edge of Al is identified with an edge of 02 §2.

(then: the two edges must have the same length, and the identification is performed by a proper motion),

iii) one vertex of 'Ll is identified with one vertex of 02. 2) Every vertex is identified with at most a finite number of other vertices. 3a) Each edge of any one is identified with exactly one edge of another triangle. 3b) Given any two triangles Al,

many triangles "between them".

of Y, there are finitely

This means:

there exists a fi-

nite collection of triangles 'Ll, A21...,n such that each pair of successive triangles has a common edge.

2.2

10

Remark: from (3a), if we start with one triangle we may obtain a succession of triangles abutting each on each with one fixed common vortex through identification of*odies; In addition, such a succession must end according to (2). Notice that (1) makes Y an orientable complex, while (2) makes Y a star-finite complex.

The identifications (1)-(3b) yield a Hausdorff apace X, called a Euclidean polyhedral surface. X has very much more structure than is provided by the axioms defining a topological space; and condition (3b) ensures the connectadneas of X.

Such a space can be naturally (i.e. canonically, with no free choices) made into a Riemann surface. In order to so define a Riemann surface, we shall have to say what complex-valued funotions defined on a domain in X are analytic. Let us distinguish three kinds of points in X: 1. Inner points of triangles (not identified with any other points). 2. Inner points of edges (each of which is identified with one other point). 3. Vertices (each of which is identified with at least two other vertices). Definition.

A function will be called "analytic on a domain in analytic in the ordinary sens3 in the neighborhood of each point of the first kind; and if it is continuous at every point of the second or third kind. X" if it is:

Theorem.

The Euclidean polyhedral surface X, with the above definition of analytic function on a domain in X, is a Riemann

surface. Sketch of proof.

We have to show that there is a local. parameter in the neighborhood of every point of X. This is clear for points of the first kind. For a point of the second kind, we bring together the two edges determined by the point; the conti-

nuity prescribed by our definition permits us to define a local parameter (via analytic continuation) in some neighborhood of the point.

Next let zl be a point of the third kind, and let the equivalence class of zl consist of the three vertices zl,z2,z3. (The argument which follows is virtually unchanged for a larger equivalence class.) Denote by al,a2,a3 the vertex angles oorrespondinp to zl,z2,z3; also, form the quantities v - 21t/ t a3.

a3 and

Then define the mapping ei0

.1 = (z-z1)v

where z is a point that the sector of the sector 0 < 0 < the origin) in the

such that I7-z11 < r, and O1 an angle such radius r in the first triangle is mapped into val of the circle of radius rv (and center at c-plane. The similarly defined mappings

v i6

43 = (z-z3) e 3 (j = 2,3) will fill out the remainder of this circle: the remaining sectors are va1 < A < v(al+a2), v(al+a2) < 0 < v(al+a2+a3) = 2n. Then the functions r.

(j = 1,2,3) each to be taken in the appropriate region, together give a local parameter and any analytic function f in a neigh-

borhood of zl may be expressed as an analytic function of this local parameter. The definition of analytic function on a domain in X therefore coincides with that which we gave in Lecture 1 and our theorem is proved. can one realize every compact hisAn open question is: mann surface as a Euclidean polyhedral surface which can be embedded (without self-intersection) in E3. Orientable surfaces and Riemann surfaces. A Hausdorff space X will be called locally Euclidean if every point x E X has a neighborhood homeomorphic to some En (the same n for all x); n is called the dimension of X. X is further called an n-manifold if it also is connected. A 2-mani23.

fold is called a surface if it satisfies the second countability axiom (i.e., has a countable basis of open sets).

2.14

12

Let x be a point of a manifold. We have by definition a neighborhood G 3 x, topethor with a homoomorphism f: G - En which maps G topologically into a domain in E. Every such domain G will be called a coordinate patch and the corresponding function f on G to En will be called a coordinate mapping. Now let y (it x) be a second point of the manifold. Consider a coordinate patch G1 3 y and a coordinate mapping fl: G1 -y En. Suppose 0 fl G1 = H / 0. Then consider the resff1-1; we shall call f, fl simitriction of f, f1 on H, and form (The correslarly oriented if ffil is orientation-preserving. ponding patches are understood to intersect.) A manifold will be called orientable if it can be covered

by coordinate patches Gi, with corresponding coordinate mappings fi such that any two of the fi are similarly oriented if the If a covering of this description corresponding Gi's intersect. has been constructed, the manifold will be called oriented. We

state without proof: Theorem.

(T. Rado)

Every orientable surface is homeomorphic to (The homeomorphism is called a

a Euclidean polyhedral surface. triangulation.)

A result belonging to the same order of ideas as Rado's theorem is: Theorem.

Every orientable surface can be made into a Riemann

surface.

Apply Rado's theorem, and recall that we have shown how to define a conformal structure for a Euclidean polyhedral surface. Proof.

A second theorem by Rado will later be proved: Theorem.

/. Riemann surface is a surface.

Remarks.

A Riemann surface is a 2-manifold (it is connected,

and every point has a neighborhood homeomorphic to an open set in E). The proof of Rado's second theorem will therefore amount to demonstrating (as we shall later do) that every hiemann surface has a countable basis of open sets.

2.5

all..

13

CO structure.

Let f be a real-valued function defined on an open set 3 in Euclidean spice. The classes Cp are defined as follows: C° is the class of continuous.functions on S; f E Cn (n = 1,2,3,...) f with all its partial derivatives up to order n is means: continuous on S. Flirther, f E Ca (0 < a < 1) means: on every compact subset of S, f satisfies a H81der condition of exponent a, i.e. if x,y E any compact subset of S then If(x)-f(y)I < f E Cr*' (n = 1,2,3,... and 0 < a < 1) means: f has continuous partial derivatives up to order n, and those of order If f has derivatives n are 1181der dontinuous with exponent u. of all orders, we write f E C-00; and if f is a real analytic

function, we write f E C. Now let p have as possible values n, n+a, co, 'u as above. We shall say that the n-manifold Xn has Cp structure it certain real-valued functions defined on open sets of Xn are distinguished as functions of C. The class of functions of Cp must satisfy the following axioms: Axiom 1.

Given any point P t Xn, there is a coordinate

patch G 30 P with the corresponding function f on 0 to an open set of En (U is called an allowable coordinate natch, and f an allowable coordinate map.) such that

f(F) = (x1(P),...,xn(P)) where the functions xi(P) E Co. Axiom 2.

Given a Cp function g defined on the open set

G1 and kive.n an Allowable coordiiiate woo f defined on the allowable patch 0 C G1, then the mapping gf-1: fG -> gCl is a Cp function in the ordinary sense.

If a function is a CO function Axiom 3. (Maximality) (in the ordinary sense) in terms of allowable coordinates, then it is a Cp function.

2.6

14

Let p > 1; we now define an orientable space with Cp Let f,g be allowable coordinates in the intersecting 0); then gf-1 is a homeomorallowable patches G. Go (G 1) Go phism of one domain in QT into another, viz.: n structure.

gf-1(x) _ (yl(xl,...,xn),...,yn(xl,...,xn))

.

Form the Jacobian Set of the mapping:

(3)

for any f,a as defined the Jacobian will be

non-vanishing (since det(ayi/3x')dot (axi/ayj) = 1) and contiIn the case of a negative nuous, therefore always of one sign. (positive) Jacobian, the maopinps f,g will be called oppositely (similarly) oriented. An orientable space with Cp structure is one in which there are these two classes of mappings; upon the choice of one of these, we call the space oriented. fotice that this is in accordance with the previous definition of orientable space, since a C1 homeomorphism of a domain in En with positive Jacobian is orientation preserving. We have now proved the Theorem.

Every Riemann surface is (is:

not "can be made into")

an oriented CLO space.

yn

Then a homeomorphiam f: Xn _ Yn is called a Cp homeomorphism if the functions giving the allowable coordinates of one space in terms of those of the other is a Cp function. If in particular p = oo and Xn, Yn satisfy the second countability axiom and are homeomorphic, Now let Xn,

both have Cp structure.

then there arises a question: does there exist a Coo homeomorIf n = 2, the answer is yes (as we shall

phtsm between Xn, yn?

see later); if n = 7, the answer (given by Niilnor) is no.

Lecture

Now we shall consider a problem in the plane which Is related to the problem of making a 2-manifold with a Riemann metric into a Riemann Surface. Isothermal Coordinates.

Suppose that in a neighborhood of the origin In the x,yplans there are defined four real-valued continuous functions which are elements of a symmetric, positive definite matrix. gik Such a matrix defines a Riemannian metric ds2 = g11dx2 + g22dy2 + 2g12dxdy (details will be iven later). Vie shall say, that a transformation of coordinates given by

u = u(x,y)

,

v = v(x,y)

>0 with ra(1190 -)kx,Y)

and

(du2+dv2) = A(r,y)(1'11dx2 +

g22dy2) = ads2

where X is conttrruous, introduce isothermal coordinates coordi. Hates to this neighborhood (and the functions u,v are called isothermal parameters). The problem that we want to solve is:

given a neighborhood G of the origin and given continuous func-

tions gik (i,k = 1,2) which are elements of a symmetric positive definite matrix, can we introduce isothermal coordinates in a neighborhood of the origin contained in G? A special case first observed and studied by gauss is the following. Consider a surface in 3 space (coordinates x,y,4) given by 4 = 4(x,y). Then we have ds2 = dx2 + dy2 + d42

_

(1+42)d X2 + 24.4. dxdy + (1+42) dy2

The problem of finding locally on this surface a conformal mapping to the iuclidean plane is our problem of introducing isothermal coordinates, where

all

.

1+4

,

P..22' 1 + 42 ,

C12 = 4x4y

3.2

16

Formal Complex Derivatives.

The introduction of formal complex derivatives will simplify the writing of our theorems. If we have two differentiable functions u(x,y) and v(x,y) we write u + iv = w(z,z) where z = x+iy,

(Although w(z) occasiorally is meant to connote that

z = x-iy.

w is an analytic function of z, we will simply write u+iv = w.) Now we define wz and w_ by

z wz =

(wx + iwy)

(wx - iwy)

That this definition is a natural one, is seen by blindly following the rules of differentiation, viewing w as a function of z and z, and z,i as functions of x and y. We would then obtain,

wx = wz + w and solving for w

z

wy =

(wz - w

a This procedure

gives us our definition.

w

z would be justified if u and v were real analytic.

Notice that w(z0) = 0 is equivalent to ux= vy and Uya -vx z

which are the Cauchy Riemann equations. Theorem 1.

Hence we have

If w is monogenic at zoo that is, if the finite

limit

w(z 4h) - w(z o

h->0

exists then wx and wy exist at z° and w (zo) = 0 and w'(zo)=wz(zo).

z Proof.

see any text on function theory.

Now we ask what is the condition on w = u+iv for u and v to be isothermal parameters? Write

dz=dx+idy then we have, dx = dx and dy in

dz=dx - idy

(dz+ddz), dy = -

(dz-dz).

Substituting for

3.3

17

ds2 = g11dx2 + 2g12dxdy f g22dy2 we get dal = dz2(1(g11-922)

-

2 912) + d1 2(t(911-922) + 2 912)

+ dzdz(.(g11+922))

- lA dz + B dz12 where 1A12

Z

Idz + 8 di12

=

1-+

911922 - 912]

and =

(911-g22) + ig12

2

B

(911+g22) +

write

11

g11g22-g12

= µ, then

A 1u12 =

2 (gll+g22)

911922-912

2 (911+922) } 118-g hence ds2 = 1(x,y) ldz + µ dz12

where

1µl < 1

.

Note that du2+dv2 = dwdw = Idw12 and dw = wzdz + w d% z = wz(dz + (w /wz)dz). So it appears that we require

z 1.

= µ wz

W

and

wz Yr 0

z This equation is called the Beltrami equation. Theorem 2.

911,822 >

Let 91119221912 0. Let

C1 be such that 811922-912 > C and

µ(z) ?(211-922)

+ 1912

2 (911+922) + if w(z) E C' is a solution of

gllg22-g

W_ = µwz near the origin, and wz(0) ' 0, then in a neighborhood

a

of the origin, writing w(z) = u+iv, u and v are isothermal parameters, that is

v) 1. a(u x,y > 0 2. dug+dv2 = X(x,y)(glldx2 + 2g12dxdy + g22dy2). Proof. d(x,y)

d(z,z) - wzwa

and since w is a solution of w

=

1wz12 -

wzwz =

1wz12 -

= uwz

Iu121wz12 = 1wz12(l - Iu12)

and since wz(0) / 0 and w r. C' and Iu12 < 1, a u.v 2.

1wz12

> 0.

We have seen that when wz 1 0 ww1

d.)12

dug+dv2 = Iwz12 1(dz + z

but

I

W_

wz

(g11-922)

= U =

- 1912 -911922-g12

2 (911+g22) + so that Iw z 12

du2+dv2 =

2(811+922)+

(dz2

-911922-912

(911-922-2i ,12) + dz2 1+(`'11-822+2'812) + dzdz 2(911+922)

and replacing dz, dz by dx+idy and dx-idy du2 + dv2 = X(x,y)(dx2g11 + dy2g22 + 2P12dxdy)

3.5

19

Theorem 1 shows that the problem of finding isothermal parameters to introduce isotherrircl coordinates is reduced to

finding a solution to the Beltrnmi equation. Gauss proved the existence of a solution to the Beltrami equation when the and therefore also µ are real analytic gSk functions, that is, when µ can be expanded in a power series in z,z.

In general a classical, that is continuously differentiable solution need not necessarily exist if µ is only continuous. However a generalized solution can be obtain3d if p is measurable. We will demonstrate here the existence of a classical solution when µ is HBlder continuous. Now we present a succession of lemmas which will enable us to prove it. Lemma 1.

Let w(z) b C1 and D a simply connected region bounded by the sufficiently smooth curve r, then

w d z = 21 J

w_ dxdy z

I

That this is true can best be sean by writing everything in real notation. Then the only change in the hypothesis is that u,v E C1 and the conclusion becomes r Proof.

I

1r

wdz=21Sjw_dxdy D

z

(u+iv)(dx+idy) = 21

I

11r

SI

2(wx+iwy)dxdy =

C

(-wy+iwx)dxdy

D

(r

u dx - v dy + i v dx + u dy = J(_u_v)dxdY + i D

equating real and imaginary part v dy - u dx =j (uy + vx) dxdy

r

D

uX vy)du3y D

3.6

20

(ux - vy) dxdy

v dx + u dy = iJJJJJJ

each of which is only Green's Theorem. note that if w was an gnalytic function w

would be 0 and

z we would have only stated the Cauchy Integral Theorem. Hence we can look at this as a generalization of the Cauchy Integral Theorem. The next is in the same way a generalization of the Cauchy Integral Formula. Lemma 2.

Same hypothesis as Lemma 1 with z a fixed point in D.

WW

w(

-Z

2111

F

d4 - Ic 1

JJ

w^(2:)d4d9 r. 2; - Z

D

1 w To prove this we consider the function W(Z) = 27ET Z-- Z in the region D-D'

Proof.

r

Then Lemma 1 tells us (taking care of the doubly- connectednesa of D-D'` by appropriate cross cuts) 1

w(4)df

r

z+

f

1

2: -z r

-

-z -o

1

w

it D-D'

and continuing precisely as in the proof of the Cauchy Integral Theorem (by estimating the difference between the first Integral and w(z) and taking the limit as r -> 0) we obtain the desired result.

21

3.7

Lemma

If (z I

R and

(31i) implies (35)

11W2 111+¢ w

2

Cn+2+a where w

J

dgdj

(integration here and hereafter over 1410 for 0 H defined by

(X,y)

if p

1.

F, (p) = 0

2.

if p E D, with allowable coordinates

,

X2 + y2 < R ,

.D

P1(p) a (li)IF(2 J),...)

To

6.8 9(J) = 0

and

R

E(J)

3J+2 ° Q)

3J+3

if ,( is not one of 33+1, 33+2, 33+3

R2r (X2+-+7-) x

E3 3+1

F.

53

R2r(x2+ 2)7 R2

= R2'(x2+' 2

The functions F3 have the following properties.

1. E J) = EkJ)(p) are C a functions (in the e structure of M) 2. F.( J)

vanishes identically on the boundary of the disc D3

3 P1 # P2; Pl,P2 E D3 -) F3(P1) I F3(P2) is a homeomorphism

4. F31D 3

5. The rank of the Jacobian of F3 is maximal in D3 (i.e., - 2).

The proofs of 1-4 are trivial. (E33+l)x =

For 5, we get the following

R2f + 2x2f'

(E3J+2)x = 2xyf' (E3J+3)x = 2xf' (E3J+1)y = 2xyf'

(E3J+2)y = R2f + 2y2ft (E3J+3)y = 2yf'

hence R2f+2x2f'

2xyf'

2xf'

2xyf'

R2f+2y2f'

2yf'

J =

If p has coordinates (0,0) then

6.9

54

2xyf'

R2r+2x2r'

=

12xyf'

R4r2

o

R2r+2y2re

If p has coordinates (x,y) x j 0 then, 2xyf'

2xf'

R2f+2y2f'

2yf'

= 2xR2ff' 1 0

I

And finally if p has coordinates (x,y), y R2r+2x2f

2xr'

2xyf'

2yf'

0 then

= 2yR2ff' / 0 00

Now we define the mapping F: M --- ti by F(p) =

f

P'(p),

which is really only a finite sum since IDt is a locally finite covering of M. F clearly gives a regular embedding of M Q.E.D.

into N.

Now we shall give a second proof of Theorem 1. We shall define the metric tensor gik without first constructing an embedding.

Definition 6.

/

partition of unity of _class Cr (o' < p) subordi-

nate to R locally finite covering fDis a sequence of real valued functions fdii defined on M,satisfying

rCr

1. 2.

di > 0

for all i for all i

OD

Claim.

3.

¢i = 1

1..

The support of

A partition of unity exists.

is contained in Di.

6.10

Proof. Define 3(p) _

Define

(p)

55

0

if

pAD

f(a-2 -2

if

pfD

CO

Q.E.D.

i(p)

di(p)

Now back to Theorem 1. Proof (2).

For each positive integer j we define a tensor

gg r by gik) -

0

if p t Di

bik4j(p)

if

peD

It is obviously symmetric and positive-semi-definite.

Now we

gik)(p) and making gik a covariant define gik by gik(p) (p) tensor it is clearly symmetric and positive definite since must be positive for some J. Remark.

Where in these proofs have we used 6 < W?

We used it

since we only have f E C. Query? Do we get every Riemann Surface in this way? That is, given a Riemann Surface S can we define a metric tensor on it

which gives us back the same Riemann Surface? Yes! For suppose S is a given Riemann Surface. We assume Rado's theorem. We repeat the reasoning given above, using as local coordinates the real and imaginary parts of local parameters. We notice then that the gik) satisfy the relations 911 - g22

912 = 921 - 0

For one coordinate system, and hence for any other (since the transformation from one system to another is governed by the Cauchy-Riemann equations). Hence the real and imaginary parts of local parameters are irothermal coordinates with respect to (Note that the gik constructed are Coo.) We formulate this gik' result as

6.11

56 Theorem 2.

Any Riemann Surface is conformally equivalent to an

orientable surface with a C°0 metric tensor gtk. The embedding problem for Riemann Surfaces consists in asking whether every Riemann Surface is conforwally equivalent to a surface of class Co embedded in EN (with a conformal structure defined by the metric of the embedding space). On the basis of the result just stated one can conclude that,

1. The answer to the embedding problem is yes for a'= oo and a sufficiently large N.

This follows from a theorem of Nash

which asserts that a Cco surface can be embedded isometrically in EM (N sufficiently large). Indeed, an isometric embedding is a fortiori a conformal embedding.

This follows from 2. The answer is yea for fl = 1 and N = 3. the theorem of Kuiper which asserts the existence of an isometric C1 embedding of a given surface into E3. (Wa shall show 0

metric leads to a conformal structure.) We shall show later that on every Riemann Surface one can define a real analytic metric tensor gik which respects the conformal structure. (And has constant positive curvature.) The embedding problems for Cr > 0, N = 3 is still open later how a C

7.1

57

Lecture 3 In this and the subsequent lecture we shall arrive at another way of obtaining Riemann Surfaces. A way which is closely motivated by the problem in function theory which motivated Riemann to first conceive the idea. But first we prove the following theorem. Theorem 1.

A single non-constant meromorphic function w defined

on the Riemann Surface S, determines the conformal structure of S. Proof:

"e shall accomplish the proof by using w to determine a local parameter at every point p e S. Suppose at po r. 3, w(po) = a < co, suppose also that we have a local parameter z defined at p such that z(p0) = 0. Then we can write

w=a+

am 1 0

akzk m

akz k = amz (l + m

w-a=

a

m+l

as+2 z2 +

z +

. .

.)

m

in

We can pick b such that bi° = am, and an m-th root of +1 (1 + z + ...) can also be taken, hence s

w-a = bmzm(1 + b z + b2z2 + ...)m = (bz + bblz2 + ...)m = 4(z)m I and since b / 0, 4 is a local parameter for sufficiently small values of z. Hence an m-th root of w(p)-w(p0) is a local parameter. Suppose now that w(p0) = oD. Then we have

zm a-M

+ zrn-i

...

am

then taking 1

(1 + b1z + b2z2 + ...)°1 =

aam+1 $ + ...

1 +

m

0

58

7.2

-

and b so that bm = 1A-m, we have

w=

bmzm(l + b1z + b2z id + ...)m

=

1 (bz + bblz

=

+ ...)m

1 Vz)'a

Since b ( 0 for sufficiently small values of z, Z is a local parameter. Hence if w(p0) = cc, F )1/i° is a local How do we determine m? It is simply the number of times values close to a are taken on in a neighborhood of p0. Q. Since a non-constant meromorphic function w on S determines the conformal structure of S, we can view w as presenting 3 as a covering of the extended plane G. Namely If w(p) = a we say

parameter.

that "p lies above a". In terms of a local parameter r., w = a + gy. If m = 1 then w is locally (near p) a homeomorphism.

If m > 1, then every point near a corresponds to m distinct In this case we call p a branch point of order

points near p. m-1.

(That is we view p as being mapped m times onto a.)

If p

Is a pole, that is w(p) = oo, we have the same situation over the north pole. The inverse of w is not properly defined, it ib multi-valued. The study of analytic functions in the plane leads unavoidably to the consideration of multi-valued functions. This problem was resolved, both by Wierstrass and by Riemann. "e shall make some of these elementary ideas precise, and see first what *eirstrass and Riemann were led to. A revular unramified function element over a finite point, or simply a function element (we shall see the need for all the adjectives when we confront singular, ramified Definition 1.

Co

in(z-a)n

function elements) is a non-constant power series

which we denote by 9, which converges for some zo / a. We call a the center of 0 and we write Z(Q) = a. Note that A is the

power series, not the function determined by the value to which it converges.

However, we associate with 0 the real number R(Q) If 1p-ai < R(Q) then

which is the radius of convergence of 0.

CO

there is a uniquely determined power series 00 =

iz_ n=

bn(z-p)n.

fl

59

7.3 00

00

We can determine Q'by writing

> n-

an(z-a)n =

an((e-p)-(a-p)) n=

and expanding by the binomial theorem. know from elementary function theory that R(Q') > R(Q) - JA-aj > 0. 0' is called an It follows that if fp-aj < H(0)/2, im`nediate continuation of 0. then 0 is also an immediate continuation of 0'. Also if 0 is an immediate continuation of 0' and Z(Q) = Vol), then 0 = 0'. Definition 2. !;'e shall call the function element 0', Z(0') = p, an analo*.ic continuation of the function element 0, Z(Q) o ii, if there is a curve f (f, a continuous function: I -> E) which has a as Initial point (f(0) = a) and $3 as end point (f(l) = p) and if associated with each t: 0 < t < 1 there is a function element Qt, Z(Qt) = f(t), such that conditions 1 and 2 are satisfied.

1. 00= 0, 01= of 2. Given a t0 c I, there exists e > 0 such that if t f.

and ft-t0

I

< e then 0t is an immediate continuation of Ot 0

Theorem 2. If analytic continuation along the curve f is possible, it is unique. That is, if 0', Z(Q') = p, is an analytic

contin-ation of b, Z(Q) = a along f, and if A is another analytic continuation of the same kind (obtained by a different associa-

tion t Qt) then 3 = 01. Let 0t be the continuation which yields 0', and at be

Proof:

the continuation which yields G.

n Qt = Qt

. 1. T is not empty, since 0 = 0o = a0, i.e. 0 E T. 2. T is open. = Qtop then by condition 2,

If t0 C T, that is 0t

Proof:

Let T be the set {tlt a I.

0

< min(e,t) definition 2, there exists c, a such that for It-t 0 0t, et are respectively immediate continuations of 0t and Ct e 0 But 0t = at , hence 0t = gt. I

0

0

3. T is closed. Proof: Suppose to -a t00 and to e T. By condition 2, definition 2 there exists c.1, such that if t-tCOI < min(e,t) = e', then

7.4

60

is an immediate continuation respectively of Ot,

and Qt

0t

co

0o

at' Ot

Hence for n sufficiently large Itn -t00 < e'/2 hence is an immediate continuation of Qt = Gt . Since and Ct I

n

n

00

T is a non-empty subset which is both open and closed in I which is connected, the only thing that avoids a contradiction a.E.O. Hence O1 = 31. is T - I. Unfortunately however it is possible to have 0' /.V both analytic coztlruations of 0. By Theorem 2, the only way this can happen is if the cortinuati.on took place Along different curves. This unpleasant situation is why we are forced to consider multi-valued functions. single multi-valued function?

Just what all we mean by a This we will see after we prove

the following theorem, which tells us when continuation along two curves give the same result. Lemma 1.

If continuation of the function element along the

curve f to the function element 91 is possible, then there exists e > 0 such that if if is any curve satisfying

If(t) - ?(t) l a = f(0) = f(0)

<e

for 0 < t < 1

P = f(1) = f(1)

then continuation of Q is possible along ? and yields Q'. Proof:

First we note that R(Qt) is a continuous function of t,

since

IR(Qt) - R(Qt,)I < If(t) - f(t')I

.

Hence there exists 6 > 0 such that R(Qt) > 6 for 0 < t < 1. If ?ts any curve such that Choose a such that 0 < c < 6/44.

(f(t) f(t)1ae we associate with each t a function element 0 by choosing $t to be the immediate continuation of Qt with center f(t,).

Since R(Qt) > L4e, and HC9t) > R(Qt) - If(t)=f(t)I > 14e - e = 3e, we have R(at) > 3e.

Ve will be through when we show that the family of function elements Vt actually gives an analytic continuation along i. That is, given a to we must find a a such

61

7.5

that if It-t0I < w than Ot is an immediate continuation of Ot 0

Take w = w(c) where w(t) is the minimum of the modulus of conti-

nuity of f, f Then, if It-tol < w we have

if(t0) -S(t) I < I$(to) - f(t0) I + If(t) - f(t0) I < 2e hence 0 is an immediate continuation of Ct . 0 immediate continuation of It . Q.E.D.

Hence Vt is an

0

Theorem 3. If 0' and A are analytic continuation of 0 along the curve f and I' respectively. Then V = 0' if f can be continuously deformed into ?i through curves with end points a,p along which continuation is possible. That is, if there exists a function F(x,t), F: I x I -? E which satisfies the conditions, 1. F(t,O) = f(t)

,

F(t,l) = ?(t)

2. F is continuous in s and in t. 3. F(0,s) = a , F(l,s) = p for all a 6 I II.. If s r. I then continuation Is possible along F(t,e)

then 0' =1. = 0'1. Let S be the set {s Is £ 1, 0l S is not empty s since 01 0 = 0'. That is, 0 6 S. By Lemma 1, S is open. Also

Proof:

as in Lemma 2 we see that R(0t s) is a continuous function of S. Hence G = Of Q.E.D. Hence S is closed. Hence S = of all function elements 0 obtained Now consider the set S0 I.

0

by analytic continuation from 00. If we consider any particular point a £ 1 then a may be the center of no, one, or many function elements of S0 . This is the approach of ''ierstrass to multi-valued functi8ns, and we call a set S0 a complete analytic

function in the sense of "eirstrass.

o

At this point riemann enters the picture (not historically, but logically) And we can imagine him saying (as we might), "I don't like multi-valued functions. I would rather think of single-valued functions. I shall construct (find) a surface on which I can view these multi-valued functions in the plane as single-valued functions on the surface."

62

7.6

If we are to construct a surface we shall need points. for points we will have function elements. late will denote by W the set of all function elements. Furthermore we need a topo-

in fact a topology which makes ?'' a hausdorfi space.

logy on

"7e will do this by defining a basis of open sets in W.

If 00 F W, Z(Q) = a we shall mean by the disc of radius e > 0 with center 00 (denoted by Qe(00)) the set of all

Definition 3.

0E

Z(0) = p., such that Ip-al < e and 0 is an immediate con-

tinuation of 00.

From Definition 3 follows our definition of open sets. A set M c b' is called open if every point in M is the center of a disc contained in M.

Definition 1I.

Now we must see if this definition makes '' a hausdorff space.

We check the axioms for a Hausdorff apace. (1) If jOjjJEJ is a family of open sets then any point 0

U 0j is contained in a 0k.

Hence 0 is the center

jEJ

of a disc LC 0k C U 0.

j

Hence U O

is open.

j

(ii) If 01 and 02 are open sets, and if 0 E 01 n 02, then

0 e Del(0) C O10 O F A2(0) C 02.

If we take

e = min(a1,e2) we have 0 E ae(0) C 011% 02. (iii) (Separation Axiom) If 0 7( 3 we have two possibilities. 1. a = Z(Q) # Z(9) = p. Then if we take e < (a-p)/2

then A,(0) I1 De(3) = 0. 2. a = p.

Then if we take e < min(2 0

ACM r) ACM = 0 since if *6

.

,

H(

De(Q) and

),

E

then both 0 and

would be immediate continuations of A, and hence would be equal. Impossible! We would next like to give W a conformal structure.

That Clearly the function Z (center of 0) defined in a OE(0) by Z(0) - a is a local parameter. is we must give a system of defining local parameters.

7.7

63

Let us consider the set Sg G W consisting of all points in 0

W which are and points of curves with initial point 00. Clearly 30 is connected. It is also open. For If 0 E SO then there 0 0 exists a curve f: I -> W, f(o) = 0°, r(I) = 0. If 3 E Qe(0) then we can join II to 0 by a curve g whose support is in A.M. Then the curve t' defined by f(2t)

for

0 < t < 1/2

g(2t-l)

for

1/2 1 t < 1

is a curve joining 0' to 00.

Similarly S0 is o is a connected component of W. and hence is a Hence 0' E S0 .

0

closed.

Hence S0 0

Riemann Surface.

Furthermore our ambiguous use of S0 is reasonable since as sets they are exactly identical. ° Consider a SO . If 0 6 S0 and Z(0) = a, we can write 0

00

0 =

0

We can define on S0

an(z-a)n.

n-f tions Z and V: S0 - E.

two single-valued func-

Namely Z(Q) = a, V(0) = a0

0

If Z 4 F(30 ) then the multi-valued function V(Z-1(z)) = w(z) is 0 also some way connected with the Rieman.i Surface S0 . In fact 0 30 the complote analytic function in the sense of Wierstrass is 0

The Riemann Surface S0

what we mean by w.

is called the Rie0

mann Surface of the regular anramified function elements of SO 0

U'e now prove a theorem which characterizes some of the Riemann Surfaces S0 in an algebraic way. And more important, 0

will indicate how we should extend the Riemann Surfaces S0 0

We shall mean, by a polynomial in w of degree n, with maromorphic coefficients a polynomial, P(w) = wn + An-1(z)wn-1 + ... + A0(s)

where the coefficients Ai(z) are meromorphic functions defined on E. The discriminant of such a polynomial is a meromorphio

7.8

64

Hence unless it is identically zero, it has discrete Hence if the polynomial is irreducible, (for if the then discriminant were identically 0 we would have P = PM-Q) the discriminant has discrete zeroes. Ve shall say that a function element satisfies the polynomial, if replacing w by the power series of the function element, the resulting expression is identically zero, and if also its center is not one of the critical points of the polynomial. By the critical points of the polynomial we mean the zeroes of the discriminant, and the singularities of the coefficients. function. zeroes.

The totality of solutions of an irreducible polynomial P of degree n is a complete analytic function in the sense of Wierstrass. And furthermore except at the critical points, over every point there are exactly n function elements of this complete analytic function. And furthermore, in the neighborhood Theorem

of the finite excluded points, the value of the function elements become infinite of at most a finite order. Conversely, if 3p is a complete analytic function which over every point except co and a certain discrete set has exactly n function elements, such that if z is a finite exceptional point, and z1 - zc (all zi 0

non-exceptional) and over each zi we pick one of the n function 00

elements 01 = S; av(zi)(z-zi)v the sequence av(zi) becomes V=

infinite of at most a finite order, then S0 is the totality of solutions of an irreducible polynomial of degree n with meromorphic coefficients. First we prove part of the converse. Over each non-exceptional point z0 we have n function elements Proof:

OD

3(zo) = z av(zo)(z-z0)v, j = 1,2,...,n.

Since in a neighbor-

v=

hood of z we can make this numbering consistent, the functions 0 av(z) are analytic. Hence the n functions

65

7.9

Bn_1(z) _ -T so (z) J=

Bn-2(z)

a0 (z)a02(z)

B0(z) = (-1)n ao(z)ao(z)...ao(z) the elementary symmetric functions of as are analytic functions. If we let w(z) be the sum of any one of our n function elements

A3(zo), or what is the same, any ao(z) we find

0 = 1 (w(z) - ao(z)) = wn + Bn-1wn'l + ... + Bo 3=1

Furthermore the Bi are meromorphic since at non-exceptional points they are analytic, and since in the neighborhood of finite exceptional points the roots of the polynomial become infinite of finite order, these points at worst are poles of the Bi. That is, the Bi are meromorphic. After we prove the remainder of the theorem we shall easily see that the polynomial is irreducible. At every non-critical point z 0 the polynomial P has n disBy the impltoit function theorem (see tinct roots wl,w2,...,wn.

the appendix to lecture 7) to each root w, there corresponds a av(z-zo)v

uniquely determined function element 0

= w3 + 7

-

V= l

which is a solution of P. Furthermore by the theorem on the permanence of functional values, any analytic continuation of a solution is a solution. Hence the totality of solutions consists of k complete analytic functions S. . But by the part of the SGT there corresponds a polynoconverse that we proved, to each mial with meromorphic coefficients P4 of which Se

Is the tota-

lity of solutions. Hance P = P1P2...Pk. But we assumed P irreducible. Hence k = 1. Now we can finish the converse, for it the polynomial B = wn + Bn_lwn-l + ... + B 0 could be factored

66

7.10

into irreducible factors ffl,$2,...,IF

then by the rest of the

theorem to each Bi there would correspond a complete analytic function So But a complete analytic function SO is the totalI*

ity of solutions.

And since a complete analytic function is Q.E.D.

connected, H is irreducible.

4!e see that at all finite points the coefficients hove at most poles and the roots become infinite of at most a finite order. However at oo if the coefficients have only poles, that is they are rational functions, then the roots can become If the coefficients are infinite of at most a finite order,

rational we say that the complete analytic function which is the solution of the polynomial equation is an algebraic function, Otherwise we call it an alpebroid function. The Riemann Surfaces S9 have holes at the excluded 0

points.

That is the surface is not compact.

In the next

lecture we shall oompactify some of the surfaces. Then we shall be able to talk of "the Riemann Surface of an analytic function",

8.1

67

Lecture 8 Now we want to consider what happens at the excluded points. First we consider a situation which is both more general and

more particular than the one we wish to consider. 00

Suppose the function element 0 =

an(z-a)n with la-b, 1 we say that the function element (z,w) is ramified. If the representation is of type 2 we say that the function element (z,w) has center at oo. In either case, depending upon whether or not the series for w has any non-zero coefficients corresponding to negative powers, the function element (z,w) is called regular or singular. Although in the strictest sense It Is not true that the unramified, regular function elements with finite center are not elements of 'd, it is clear In what sense they are elements of W. It is clear that the elements of W which are not elements of % are isolated elements of P. The following lemma shows us that 'P is also a Hausdorff space. Lemma 1.

Given a function element (z',w') in a sufficiently

small neighborhood of the function element (z,w) with center a, we can determine the function element (z,w). Proof.

shall prove the lemma for "a" finite.) Take a neighborhcod of (z,w) so small that apart pcssibly from (z,w) all

function elements in the neighborhood are regular and unramifted. Let (z,w) be given as

8.7

73

ao

E = a + tm

w

bntn

Then (z',w') would be given by

z' _

00

a"tn

OD

w,

n

where T is an analytic function of t, namely

t=t - tI and tl sufficiently small.

By proper parameter changes

A0

AnIfn we obtain

m Z, = al +

Been

w' =

.

In the a' plane we have the function element ao

w' _

Bn(zo-al) n

which can be continuod along every curve in the disc punctured at a. Then we obtain m function elements over each point. Using the mapping at the beginning of the lecture we can throw them back to the t plane where we got our function element (z,w), CO

z = a + tm

bntn

w =

Q.E.D.

Furthermore in a sufficiently Small neighborhood of (z,w) OD

Z

ant"

w

2L

bntn

the function F(z,w) = t is a local parameter.

fence W has

conformal structure. A is called an "Analytic ConfiguA connected comvonent of ration". The elements of W in an analytic configuration are a

8.8

74

complete analytic function in the sense of I-'eieretrass.

That is

if two elements of W can be connected by a curve in 1'.!, then they

can be connected by a curve in W. This is clear, for since the support of a curve is compact, it can have only a finite number of singular ramified elements (elemunta not in W) and a suffi-

ciently small neighborhood of these points will contain only points of W. Hence by changing the curve slightly we can con-

nect the two elements in W by a curve in 1'1. (zt,wi) e W

points not in W

Hence a function element 90 i W uniquely determines an which contains the complete analytic

analytic configuration 'SQ

0

function Se. The Riemann Surface "S'2 face of the function Co.

is called the Riemaru Sur-

Notice that our "normalization" of the function elements of S distinguishes z from w. This is a distinction that we have made in order to visualize what is happening. No such distinction The analytic configurations have more structure exists in l". than a Riemann Surface, namely it distinguishes two meromorphic functions on the surface: the functions z and w. Hence two distinct analytic configurations might be (as in fact they frequently are) conformally equivalent as Riemann Surfaces. Theorem 1. compact.

The niemann Surface of an algebraic function is

8.9

7.5

There were only a finite number of points excluded as In SA each of these points is the center of elements of Sc,. center of only a finite number of function elements. Hence if Proof.

we have any covering of O by open sets then we can find a finite subcovering which covers the centers (since 9 is compact) and a finite number of these subcoverings will cover L. Hence 0 Q.E.D. is compact. Theorem 1 has two converses. One a weak converse and the The week converse is trivial, the other a strong converse. strong converse we shall prove in the next lecture. (weak converse) If an analytic configuration Is Theorem 2. compact, then it is the Riemann Surface of an algebraic function.

Let us exclude the points of E over which there are ramified or singular elements of the analytic configuration 3, end the point co. Since the surface Is compact this is a finite The number, which Is finite, of regular unramified function set. elements over a non-excluded point, a, is the same as the number over any other non-excluded point. Furthermore, since 3 is Proof.

compact, In the neighborhood of an excluded point the value of the function elements can become infinity of at most a finite order. Hence S Is the Riemann Surface of an algebraic function. Q.E.D.

Theorem 3.

If an abstract tiiemann Surface is (strong converse) compact then it is conformally equivalent to the Riemann Surface

of an algebraic function.

9.1

76

Lecture 9

In this lecture we shall prove the following theorem, which is the strong converse mentioned at the and of Lecture This theorem states that every compact Riemann Surface S, is conformally equivilant to the analytic configuration of an algebraic function. 8.

In order to prove our theorem we shall need the following lemma which we shall later prove. Lemma 1.

For every compact Riemann Surface S, there exists

an integer K0, such that for any point p E S and k > Ko there exists a meromorphic function on S with a pole of order k at p, and no other poles on S.

We shall now begin a systematic study of the meromorphic functions on a Riemann Surface S, (later we will discuss only compact Riemann Surfaces), which will eventually yield a demonstration of our theorem. Definition 1.

On a Riemann Surface S, we shall denote by {s}, the set of meromorphic functions defined on S.

Remark. {S) contains C, the set of constant functions Lemma 2.

{S) is a field with the natural operations of ad-

dition and multiplication. Proof.

Obvious.

Definition 2.

If at a point p E S, the function z a LS - C,

has the value co,then in terms of a local parameter t, which vanishes at p,, a can be represented in some neighborhood of p $s z = . We shall say that p is a pole of order m of z. tm We shall also say that z has m poles pt p. Definition 3.

If at a point p s S, the function z E &sj - C,

has the value 0, then in terms of a local parameter t, which vanishes at. p, z can be represented in some neighborhood of p as z - tm. We shall say that p is a zero of order m of z, or alternatively a takes on the value 0 m times at p.

9.2

77

Similarly, it at p e S, 5 {S} -C and z -z e (S) - C has a zero of order m, we say that z takes on the values Z m times at p.

Lemma 3.

If on the domain DG S, the function s 4E (D) - C has

at least k poles (counted with the proper multiplication) then, there exists an A, such that if 1i > A, then z takes on the value r at least k times in D. Proof.

There are at least n point pl,...,pn,such that at

pi, z has a pole of order mi, andmi >`- k.

Since the poles

i=l

of a meromorphio function are isolated: there exist numbers ei > 0 such that the ei discs about pi don't intersect and in terms of the local parameters ti, which vanish at pi, z can be represented in the discs about pi as z = l/tii

Itil < ei

Take e < min (1, el,...,en) and consider the discs Ltil < to The mapping -mi maps this disc (in the plane) fit! < e, mi times

onto the domain

mi

< It!,

e

mi
A = l

e

n then z takes on the value t at least

mi > k times in the

i=1 union of these discs (on D). least k times in D.

r

Hence z takes on the value t at

Q.F.D.

Lemma . If on the domain DC-- Z, the function z F- t.61 - C. takes on the value r, at least k times, then there exists an e > 0 such that; if it' e, then z takes on the value 4' at least k times.

E {D} - C has at least k poles in D. Proof. The function z The lemma follows immediately from Lemma 3. A real valued function u defined on an open set GCS, is called a harmonic function, if every point p E 0 Definition la..

is contained in an open set 0pC G. and in Gp, u on be represented

78

9.3

as u = Re(z), where z is an analytic function. Lemma $.

(The maximum principle for harmonic functions).

If u is a harmonic on the domain DC S, and if D has compact closure 1, and if u is continuous on the boundary of D, then the maximum M, of u, in 13 is taken on by u at an interior point, if and only if a is constant in D.

If u is constant in D then by the continuity of u in

Proof.

$, and the compactness of 5, u = M at all points of D. Conversely, suppose for p e D, u(p) = M. Then in the neighborhood of p we have u = Re(z) where z is analytic in this neighborhood In terms of a local pargmetar t which vanishes at p, z of p. can be represented as

z - a+tm, It d<e. Hence

but since

U = Re(a + tm)

- Re(a) + Itl' cos in arg t t(p) - 0 u=M+ ItIm coamargt

But for some t, for which ItI < e, cos m arg t is positive.. Hence in the disc corresponding to Iti < e we must have,

u=M. Note.

Since cos m arg t must also take on negative values

we are also proving the minimum principle. Hence the set of points value M, is open.

0M C :D at which u takes on the

By the continuity of u it is also closed.

Hence since D is connected, either GM = or 0M = D. assumed that u(p) = M, hence 0M = D, and u : M on D.

But we Q.E.D.

(The maximum modulus principle for Analytic function) If z is a function, analytic on the domain D; and If D is comLemma 6.

pact, and if IzI is continuous on the boundary of D, then

9.4

79

denoting by M the maximum of I,z( in 5,

(z( = M at a point of

D if and only if z is a constant on D. Proof. Clearly, if z is a constant on D, then by the continuity of Iz( in B, (z(=M at every point of D. Conversely,

if (z ( - !1 at p G D, then in terms of a local parameter t which vanishes at p, z can be represented in a neighborhood of p as

z=a+tm hence

(t(<e,

(z(2 = Re2(z) + Im2(z)

= (Re(a) + It (mcoo m argt)2+ (Im(a) + jtjm sin m argt)2 = Re2(a) + Im2(a) + It (2m+ 2(Re(a) It (m cos m arg t + Im(a)itjm sin in arg t)

since

t(p) = 0

and

(z(p)( = K

(z12=M2+ It(2m+u where u is a harmonic function which vanishes at p. By lemma 5, u, unless it is identically 0 must take on positive values in any neighborhood of p. Hence z is a constant in this neighborhood of p and (z( = M. Furthermore, if an analytic function Is constant in an open set, then it is identically constant in the domain in which it is defined. z is a constant in D. Lemma 7.

Hence

If D is a domain in S. and if 5 is compact, and if

E {D} - C is continuous on the boundary of D, and if a has at most k poles in D then there exists an A, such that if iz( > A the z takes on the value z at most k times. z

Proof.

Let the points pl,...,pn be the poles of z, with the

orders ml,...,mn.

Then n mi < k.

There exist local

i-1

parameters ti which vanish at pi, such that for ei so small

80

9.5

that the Ei discs about pi don't intersect, and don't intersect the boundary of D. We shall also require that ei < 1. In the neighborhood of pi, z is then represented as /timi

z =

It11 < Ei

Let M be the maximum of IzI in 1. Choose e' so small that 1/elk >M. Choose e < min (e',el,e2,...Icr} ). Denote by ni the

conformal disc which is the pre-image of Itil < e, and r the boundary of Di.

DI , we have

Then by Lemma 6 in 1) -

i= 1

n Hence if 141 > A, z can take on the value Z only in U D. 1=1 But in Di if IZJ > A, Z is taken on at most mi times and n

F mi

< k.

Hence If 141 > A, then a takes on the value r. at

most k times in D. Lemma 8. z e 1,D}

If D C S is a domain with compact closure, and if - C takes on the value Z at most k times in D, then e then z takes on

there exists an e > 0, such that if the value Z' at most k times. Proof.

Consider f = 1/z-Z a JD} - C, and the lemma follows

immediately from lemma 7. Lemma 9. ";'_van z 4 1D} - C, 6 compact, denote the set of values Z E 2 which z takes on at least n times by L5(n). Then either

L(n) = d or L5(n) = B. By lemma ii, with k = n, L5(n) is open. Since 1`i is compact, an application of lemma 6, with k = n-1, shows that the Froof.

complement of L5 (n) is open.

Since L5(n) C

,

and t is connected,

the conclusion follows. Lemma 10. Given a F {D} - C, a compact, denote the set of values Then either r. C 2 which z takes on at most n times by he(n).

Ma(n) _ 4 or Ma(n) _*2.

81

9.6

Froof.

By lemma 8 with k = n, MX (:a) is open. By lemma Since Ms (n) C

k = ntl, the complement of hz(ii) is open.

with and

is connected, the conclusion follows. Lemma 11.

Given a z 6 D - C, b compact, denote the set of

values r. E E which z takes on exactly n times by Ez(n).

Then

Ez(n) = 4 or Ez(n) ;- t Proof.

Ez(n) = Lz(n) n Mz(n).

Hence the conclusion follows

from lemmas 9 and 10.

Definition 5. Given a compact Riemann Surface S, (henceforth we shall refer only to compact niemann Surfaces), and z E {S} - C, z can have only a finite number of poles; the sum of these poles (counted with the proper multiplicity) is called the order of z. Lemma 12. Proof.

If z E {S} - C has order n then En(z) = E.

n

co e En(z), hence En(z) = E.

is If z e fS} - C has order n, and if Definition 6. Is taken on by z at n distinct points pl,...,pn a S, then called a non-critical value of z; and the points pl,...,pn are called non-critical points of z. All other numbers 4 6 1 and

points p r. s are called respectively critical values and

critical points of z. Lemma 13. Given a z E CS} - C of order n, and a rational symmetric function R(rl,...,9n) of the n indeterminates f l' " ''n then given any w r. {S} - C, the function L(4) = R(w(pl(r.)), w(P2(r')), ..., w(Pn(Z)))

is a rational function defined on 9. Proof.

First of all we must state what we mean by the points

given a value, the points pi(2). What we mean is: pn(2:) are the n points (not necessarily distinct) where z takes

on the value Z. Although we write.pi(Z), we cannot really call this a function of 4, at least not globally. However Y, we

917

82

Furthermore since R is symmetric, L(Z) is a well defined function depending only on Is it a rational function? Well, at a non-critical value r. of can associate these n points with it.

z, we get n distinct points pl(Z),...,pn(t), with some particular ordering. At pi(Y.), z can be represented in terms of a local parameter ti which vanishes at pi as, z = 4 + ti

Iti1 < ei

Hence for values 4', IZ'-41 < e < min

we can call pi(e),

i

the point which corresponds to ti = 41-4.

Hence near these

non-critical points the functions (which locally they are) w(pi(r.)), are rational functions of Y,. Hence except for the critical values of z, L(r,) is a rational function of a rational function. Since at any point p G 3, both z,w can become infinite of at most a finite order, we see that L(4) is a rational function on E. Theorem 1.

If z F {S} - C has order n, and if w is any element of {S} - C,'then there exists a polynomial P(E1,E2) over g whose degree in is n, such that z,w satisfy the algebraic equation 2 P(z,w) = 0 on S. Consider the elementary symmetric functions n) on n indeterminates E1,...99n). If we replace gi BI(Eit.*by w(pi(r.)), (which was described in lemma 13), then we see Proof.

that

n

Til

(w(p) - w(pi(z(p)))) = 0 a wn + an-1(z)wn-1 +...+ ao(z)

for all p c S.

Furthermore, by lemma 13, the ai(z) are rational functions of z. Definition 7.

Given a z e {S} - C and a p e S, then in terms of

a local parameter t, z can be represented in a neighborhood of p as,

This series is called the series induced by z at p relative to t.

9.8

83

Definition 8.

Given z,w s fS} - C, z,w is called a primitive pair, if at every point p E S, the pair of series induced by z,w relative to any local parameter t which vanishes at p, is a function element. That is, a representative of an element of

And if furthermore, at distinct points we obtain represen-

tatives of distinct elements of. Clearly a primitive pair z,w induce a natural 1-1, topological, conformal map of S into 'P. The following theorem tells us when two functions z,w form a primitive pair. Theorem 2.

The two functions z,w E {S} - C are a primitive pair if and only if the polynomial in w of theorem 1 is irreducible over the field of rational functions in z. Proof.

Suppose the polynomial an-1(z)wn-1

P(w) = wn + is reduo'ible.

+...+ ao(z)

Then P1a

P =

P22 ...Pa k

Remove from S the critical points of z and the poles of z and Call 3 so mutilated So. At a point p e So, the pair of series induced by z,w relative to some parameter t which vanishes at p is

w.

=cacti 00

+t

w

Since different values of t give different values of z, the pair of series is a function element. In fact, a regular, unramified function element over a finite point. That is, an element of W, which we will write 00 (1)

w =

ai(z-Z)i

Hence we have a mapping Z: So - W. Clearly z is a continuous map. Also since S is locally Euclidean and connected, the removal of a finite number of points from 8 leaves So connected.

84 Hence x($0) is connected.

949 Hence x(S0) is contained in a com-

plete analytic function in the sense of '-'eierstrass, which is a

Since the function element (1) satisfies P it must satisfy one of the irreducible factors, say Pl. The degree of PI is less than n. by Theorem 4, lecture 7, P1 determines a complete analytic function Sp. since %(So) C So, But then at the n PI is satisfied by w at every point of So. connected component of W.

distinct points in S0 where z takes on a value Z, w can only have degree of P1 number of different induced series. Hence z,w are not a primitive pair on S0, hence certainly not on S. Conversely, suppose F irreducible. Then P defines a complete analytic function So. Remove from SG the points where the diaoriminnnt of P is 0. Call the mutilated So, 30. Remove Call from So, the points whose image under x are in SQ - 130. the mutilated So, So. Since SG to an algebraic function it is

compact, and hence we have removed for SQ and S0 only a finite number of point a. Hence Wig, 50 are connected. x(30) is open and contained in *S'0, since neighboring points in t0 are given by small values of the local parameter t, and the same t gives the induced function elements in S0. Also Y(`S0) is closed, for if the sequence of images x(pi) - 900 14- ~30, then for a less than half the radius of convergence of 8co we have a pn, Such that x(pn) induce 90D. But that means that near pn is a point POO at which the induced aeries is

GOD 6 L. Hence _4g0) = S9.

9co.

Also p00 a

0 since

Since P has degree n, ~G has n

function elements over each point in z(/a0).

Hence at the n points where z has the same value, w has different values. Hence z,w are a primitive pair on'tQ. A point of S .So is either 1. a critical point of z 2. a pole of z 3. a pole of w 4. a non-critical point of z, corresponding to a noncritical value Z, at which w takes on less than n distinct values at the n distinct points pi(z). The argument, that shows that the series induced by sew at

any of these points is a function element, is the Same for points of all four types. To illustrate, we shall present the argument

for a point of type 1. Let the critical value be Z. of a local parameter t which vanishes at p we have

z

In terms

w=sit t OD

+tm

This pair of series could fail to be a function element only if for arbitrarily small, distinct tl,t2; z(tl) = z(tj) and w(t ) = w(t2). But the points corresponding to tl,t2 are points of io, since the excluded points a=a isolated. But we have seen that on' o w takes on distinct values at points where z has the same value. The same argument shows that at distinct excluded points, (z,w) induce distinct function elements. Hones (z,w) are a primitive pair on S.

Theorem 3.

Given a z E {S} - C there exists a w r.

{S}

- C such

that z,w are a primitive pair. Proof.

Take a non-critical value Z of z. Let n = order of z. Then corresponding to r. we have n distinct points pie S at which the value of z is Z. Pick n distinct integers ki all greater than the Ko of lemma 1. By lemma 1 there exists a wi c {S} - C which has a pole at p1 of order k1 and no other poles in S.

Take w =

wi E {S} - C.

In terms of local para-

meters ti which vanish at p1 we have z = Z + ti

w = 1/tiki

Take a to, such that (toy < e < min (ci).

It iI < ei

Corresponding to

i

ti = to, we have n distinct points qi at which z(qi) = Z+to and the values of w at these n points are all distinct. Hence the irreducible polynomial with coefficients, rational functions of z, which w satisfies must be of degree at least n. Hence, by theorems 1 and 2, z,w are a primitive pair. Theorem 4. Every compact Riemann Surface S is confonmally equivalent to the Riemann Surface of an algebraic function. More

precisely, given any primitive pair (z,w) on the Riemann Surface

9.11

86

S, there exists a canonical conformal homeomorphiam of S onto the analytic configuration of the algebraic function determined by the algebraic equation satisfied by (z,w). Proof.

By theorem 3 there exists a primitive pair z,w on S. By theorems 1 and 2, w satisfies an irreducible polynomial P of degree n = order of z whose coefficients are rational functions of z. By theorem 4, lecture 7, P determines an algebraic function So, which determines a So, the analytic configuration of an algebraic function. We have seen, that the pair z,w induce Clearly X is continuous, 1-1, and a canonical map X. S "ao. fS) is open. But since S is compact, X(S) is also closed. Hence z(S) = 30. Clearly, % gives a conformal equivalence between S and to. Theorem

z,w E {S} - C are a primitive pair it and only if every function f E 131 - C is a rational function of z,w. Proof.

Suppose z,w are a primitive pair.

Consider a non-criti-

cal value Z of z, such that w has n = order of z distinct values Given f a {S} - C at the n distinct points we have the following n equations n-1

av(V w(pi(z))v

f(pi(Z)) _

The determinant of this system of equations is, 1

w(plW)

1

w(P2(z))

1

w(pn(r.))

w(pl(Z))n-1

...

w(P2()n-1

...

w(pn

which is the Vandermonde determinant, and hence is non-zero when the second column has distinct entries. Hence at all but a finite number of points on H, the av(z) are determined as rational functions of z. Since at the excluded points the av can become infinite of at most a finite order, the av are rational functions of s.

9.12

87

Conversely, sup'-ose every function f f {S} - C is a rational function of z,w. By theorem 3, there exists a primitive pair l,w. By assumption we have two rational functions

R1(g1, 2), R2(E11 2) such that r. = R1(z,w)

,

W= R2(z,W)

Suppose that at a point p e S, z,w do not induce a` function element. That is, at p, z,w induce the series, z =

antn

,

w =

bntn

and for arbitrarily small t1,t2 we get the same values for the pair z,w. But inserting the power series in R1,R2 we would have Similarly if the same situation with r,u which is impossible. at two distinct points z,w induce the same function element, then at these same two points we would have ?,w inducing the same function elements which is impossible. Hence z,w are a primitive pair.

Q.E.D.

Definition 9. An algebraic curve is the set of points (z,w) E E x E which satisfy a polynomial equation P(E1,92) = 0.

Definition 10. Two curves 1. z Anm znwm = 0 and

2.

B3k4 J k = 0

are called bi-rationally equivalent if there exist rational functions R1, R2, P1, P2 such that writing

z = R1(:,w)

r. = P1(z,w)

w = R2(1:,u%)

(= P2(x,w)

we have

r. = PI(R1(Z,w), R2(r,w))

w= P2(RR2(r,i, ) for all (1:,w) which satisfy T BJ

3 k = 0 and

9.13

88

z = R1(P1(z,w), P2(z,w))

w = R2(P1(z,w), P2(z,w)) for all z,w which satisfy Z Aan znwm - 0. Theorem 6.

If (z,w) and (r,,u,) are primitive pairs on the sur-

face 3 and satisfy the irreducible algebraic equations

1.

Anm z

2.

B,kl: jwk

rm = 0 = 0

then the algebraic curves corresponding to equations 1 and 2 are bi-rationally equivalent. Froof.

Obvious using theorems 4 and 6.

Note that theorem b is another way of stating that given the primitive pairs (z,w), (g,w) on S, there is a canonical conformal equivalence between the analytic configurations into which they each map S.

Theorem 7.

If there exists in ;S} a function of order 1 then 3

is conformally equivalent to the Riemann Sphere. There is only 1 pole of order 1, hence S is conformally equivalent to the Riemann Sphere. Proof.

Theorem S.

If there is a g c ZS} such that every function in {S} is a rational function of g, then S is the Riemann Sphere.

Suppose the order of g = n. Then the order of every function f E {S} is a multiple of n. But by lemma 1 there are functions of all orders larger than K0. Hence n = 1, and the theorem follows from theorem 7. Proof.

10.1

89

Lecture 10 In Theorem 5 of Lecture 9 we showed that we could define `primitive pair" in purely algebraic terms. As a consequence we recognize that the field l.Si for S compact is something quite familiar.

For we have seen that if a,w are any primitive

pair then the Riemann Surface S is conformally equivalent to the analytic configuration determined by the irreduciblq

algebraic equation

wn

an-1(z)wn-1

+

+,.,ao(z) = 0 - P(w)

Furthermore we saw that every

which is satisfied by z,w . f E {S} is of the form 1

f => a,(z)wi i=0 Hence the field {S) is obtained by first constructing a transcendental extension C[z] , and then constructing the quotient field C(z) of the integral domain C[z] . Forming a simple algebraic extension of C(z) by adjoining a root of the polynomial P(w) , we obtain a field isomorphic to 1S } .

If we had started with an arbitrary ground field 0 , rather than C , this construction yields a entity called an "algebraic function field of one variable over G " . The following theorem is an almost immediate consequence

of the foregoing remarks. Theorem 1. If F = {S } and Ft = {St} , S and S' compact, are isomorphic under an isomorphism 4 , which leaves the constants fixed, then S is conformally equivalent to St .

Let z,w be a primitive pair in F . Then 3 is conformally equivalent to the analytic configuration determined by the algebraic equation Proof.

10.2

9o

Since every element t c: Ft can be written

n-1 (z )wk

k=0 every element ft - 4(f) E Ft can be written, n-1

ft =Z__ ak(zt )wtk k=0 since $ preserves constants where zt = 4(z) and wt = 4(w). Hence zt and wt are a primitive pair on Ft and satisfy the same irreducible equation as a,w . Hence St is conformally

equivalent to the same analytic configuration as S , and hence S and St are conformally equivalent. Note that this isomophism is canonical. Theorem 1 tells us that if we are given a field F , which

we know to be the field of meromorphic functions on a compact Riemann surface S , then the Riemann surface St is uniquely determined by the algebraic structure of F up to conformal equivalence. However given such a field F, where we are told which elements are constants (remember that we had to know that the isomorphism 4 preserves constants). A Digression on Formal Laurent Series and Formal Puiseaux

Series Consider the set or formal power series in an indetermiBy formal we mean that there is no question of conThat is, the elements are strictly speaking sequences of complex numbers. Clearly with the natural definition of addition and multiplication this set forms a ring. Also the set of formal Laurent series form a field. Furtherseries, two typical elements more the set of formal Puiseaux nate t

.

vergence.

of which 00

aa

T-antn/q ,

-M

00

b - 2"bntn/r -M

10.3

91

form a field where a + b and a - b are of the form co

L

0 nto/(q,r]

-L

where [q,r] is the ]east common multiple of q and r. The following is a true theorem which we shall not prove. Theorem.

The field of formal Puiseaux series, with coeffic-

ients in any algebraically closed field, is algebraically closed. However as a simple consequence of some earlier work, we have. Theorem 2.

Consider an irreducible algebraic equation, p(w) = wn + an-lwn-1 +,...,+ so _ 0

where ai E C(s) .

If the element m

w=T_ aJ/r M in the field of formal Puiseaux series is a root of P = 0, then the series 00

w

=j-a3sj/r -M

converges for small Iz$

.

In the field of formal Puiseaux series this equation P(w) - 0 can have only n roots. But we have seen that in the analytic configuration determined by this equation over 0 there were the function elements Proof.

10.4

92

w =

a = 1,...,r

and

ql + q2 +,...,+ qr a n .

Since corresponding to qs we can multiply z by qs, qs roots of unity, we have exhibited all n roots. By the definition of a function element the corresponding series converge. A ring l c F the field of meromorphio function on a compact Riemann surface, is called a valuation ring if, Definition 1.

1. CCSl# F

2. fl Ur-1 -P w EE .

Lemma, 1.

1 4==?,, w-1

The set of non-units P of a valuation rl,ngSZ.is an

ideal. Proof. Claim 1.

If x E

and u e P , then xu F P .

Suppose

(xu)'l a v E 52.., then 1 a (xu)v = x(uv) and since uv e SZ, x if P . Contradiction. Claim 2.

If x,y E P then x - y e P .

In F we have the identities

x(l-x)=x-y=y(T Since.. is a valuation ring either 1 - x or

- 1 is in fl. ,

hence by Claim 1, x - y E P . The set of non-units P of a valuation ring is called a valuation ideal.

Definition 2.

Note.

Given a point p E S , where P a 43} , p determines a

valuation ring C?.. Namely-nis the not of functions regular at p .

Then the valuation ideal P of

tiona which have a zero at p .

is the not of funa-

10.5 Lemma 2.

93

A valuation ideal P uniquely determines the valuation,

rings._ of sh ich P is the set of non-units.

write &_1 =SI

P

Hence we can

.

We have only tp show that given an x c F we can decide whether or not x E 91. Clearly if x e P then x red. Also Proof.

Panda EPthen x e !a.

if x

(Obviously also

Definition 3.

A valuation ideal P is called a place if given, any f E.S?P , then there exists a Z E C such that f E P . Another true theorem which we shall not prove, (we shall not need it), is: Theorem.

Every valuation ideal is a place.

From now on we shall refer only to valuation ideals which are places. Note.

If

Lemma

P

and

Z =4' . Proof.

If f- C E P and f- Z'E P then

(f - z') - (f - 4) = Z - Z' E P , hence Z - r' = 0 and Z = Z' Hence we can define a mapping

'X, p:

i C by ')(f)

if and only if

f-ZeP. Lemma 5.

The mapping ")C is a honomorphism.

Proof.

'(, (f) +')(, (g)

Claim 1.

Proof.

(f + g)

If

f-ZCP and g-Z'EP then

(f - 0 + (g - 41 1 - (f + g) - (r + z') e P

10.6

94

X. Cr) % (g) ' % (fg)

Claim 2. Proof .

If

f -ZeP and g-r.'e P then

4)(g - z') - 2441 + z'r + Zg

fg -

(f - Z)(g - 41) - M' - g) + 41(f - r) E P Definition 4.

If f - Z e P we shall call r the value of f at

P and we shall write

Z : f (P) If A, x e P then we shall say x is equivalent Definition 5. to y . That is, we shall write Lemma 6.

is an equivalence relation.

Proof b Claim 1.

Proof. Claim 2. Proof. Claim 3. Proof.

CC P , hence 1 E P ,

If

x...y

then

hence x~x y. #x

Obvious

If x.vy and

then

y..., a

y,X E SZ. and

L,

hence

3.1 ya s hence

a

'

Z.L xy

Z e L-1 x

10.E

95

If Y E P we shall say that y precedes x .

Definition 6.

And

we shall write 7 1 x .

Lemma 1. < induces a linear ordering on the equivalence classes determined by r...

Let us denote the equivalence class containing x by

Proof.

If (x] # (y] the not both 9 and X are in .SZ . Hence either y or a E P , hence either x < y or y < x , and we have (x] .

If (x) < (y) and y c x , then

(x] < (y] or [y] < x .

yi,

E P C Sl for some xr a (x) and y'

[y] , hence

xl...yt x: , hence (x] _ [y] . Also if (x) < (y) and x' E (x) Y1 C (y) , then for X E P and x..,x' , y....y' hence

,

8' x.r'eP x' =Y. x X' y Lemma 8.

If tl,...,ta E P then

1, ti,

are linearly independent over C(x) where Proof.

Suppose 1,

over C(x) .

x = t1t2,...,ts are linearly dependent

Then there exists po, ...,ps-1 C- C(x) not all 0

such that po + Pitt +,.c.,+ ps-1 (tlt2,...,ts-1) = 0

We

can assume that the pi are polynomials in x , for if they weren't we could multiply the equation by the least common multuple of the demoninator. We can further assume that not all pi are divisible by x , for otherwise we could factor out a sufficiently high power of x . Let us write aj Pi

(0)

Then pj - aj is divisible by x .

Pj - aj

.

That is

x Q, j

10.8

96 or

P, a 6' +xQ' hence

ao + altl +,...,+ as-ltl,...,ts-1 =

X(

2

o3te,...,tit )

.

0

But since x e P ao + alt +,...,+

as-ltl, ...,ts-1

-xu E

hence

ac E P hence

ao = 0 Let ak be the smallest non-zero ai, then aktl,...,tk +,...,+ as-1t1,...,ts-1 . xu hence

ak +,...,+

as-1t1,...,ts-1 -

tk+l,...,ts-1 u E P

hence

ak e P hence

ak=0 hence all ai = 0.

Contradiction.

The a of Lemma 8 is less than a

Corollary 1.

K(x) which

depends only on x . Proof. There is a y C F , such that 7f y is a primitive pair. Then by the comments at the beginning of the lecture F can be viewed as a vector apace of dimension K(x) - order of x

over C(x) .

Hence there are at most K(x) linearly independent

elements of F

over C(X) .

10.9

97

Lemma 9.

There exists an element x E P of smallest order. That is, there exists an x E P , such that if y E P then

x m , then

If m'-m > 0 the right hand side is in P while the left hand Hence m, - m , and we have

side is not.

i=g or

If t

SZP then f-le S2 P , hence

98

or

but since g is a unit of I-1P , f = t1°g

Definition S. If for f r: F - l0} we have

at a place p ,

f = tag

then we call m the order of f at P , and we write

vp(f) a in and we define vp(0) = oo

Lemma 11.

.

f e S2.p if and only if vp(f) > 0 and f a P if and

only if vp (f) > 0 Proof.

Obvious.

Now, given an F which is the field of meromorphic functions of a compact Riemann surface S , we shall show that there is a 1-1 correspondence between the points of S and the places of F . Hence we can transplant the topology and conformal structure of S onto the places of F . It is even possible to give an a priori definition of the topology and conformal structure place on § , the set of places of F , which is the same as the one we obtain by the method previously However the definidescribed. However we shall not do that. tion that would be given is as follows: The functions f e F are complex valued functions on with the following definition. For f 6 AP Definition

defines f (P) .

If f

S')_ P

the we define f (P) = co . We give

S the weakest topology that makes all the functions t 6 F

continuous.

That is, a sub-basis of open sets is the family

of sets

={G I G=f-1(0) forf6F, and 0

open in

We give 'S a conformal structure by stating that a function f op F for which at some P e SS , f(P) = co is meromorphic on S . This does determine a conformal structure for l , since

we saw in Lecture 7 that a non-constant meromorphic function on a Riemann surface determines the conformal structure of the Riemann surface. We would have to show that 1T is a compact Riemann surface. That is, we must show that it is a compact, Nausdorff space, and that it satisfies the axioms for a Riemann surface. This is done directly in Chevalley. Now we proceed to establish the 1-1 correspondence between S and 3 . Definition 9. Given f e F, P e S and t a uniformiier at P ante is associated we say that the formal Laurent series

with f if

f

k -N

ISP 00

and we shall write

t

antn

.

and t a uniformizer at P Lemma 12. Given f E F, P e then there is a unique formal Laurent series '5 a_tn such that OD

7 anto Proof.

By Lemma 10, f can be uniquely expressed at

f . tag , g e SIP - P .

Fence we need only show that for

10.12

100

there is a uniquely determined series.

g E a p - P

In fact

the series will be a formal power series,

be ( 0

T bntn 0 If g re

SZ.p - P , then there is a uniquely determined b0E C

Suppose bo,...,bk were uniquely deter-

and that g - b0 E P .

mined so that

k g -

bits

0t

= gk e P . is uniquely expressible as

Then by Lemma 10 and 11, gk

m > 0 , gm E np ,

gk = tmgm ,

hence

there is a uniquely determined bm and that Hence

9k

t

= gm, and

gm - bm E P .

k g

0t

biti - bmt1°

=gm - bme P .

+m

are uniquely determined.

Hence bi , i = 1, 2,...

are all uniquely

Definition 10. If f "_"

CO

Hones

determined.

Tantn write

-N 00

Op(f) =Tantn -N Lemma 13.

The mapping

4 : F --3 L the field of formal Laurent

series, is a homomorphism.

10.13

101

Proof

Claim 1.

CID

antn

(f) =

Mp(g)

.

-M then

k

f - antn ak rs

=

-N tN+

bntn

-M tM+,

P and bk

hence if M > N

k

-/[

g

E P

k

f + g -

(an + bn)tn

-M

E P

hence 00

bn)tn

(f + g) _

$

(an +

11

-M Claim 2

k

kI

fg -

bntn)

antn)

N

-M

tN+M+ k+k'

f(Lbtn)

2(tantn)(tbntn) = akbk' -

-M

-M

N

+

t N+ M +k+k I

t i1+M+k+k

g(/ante) +

M

tn+M+ +kl

k

kv

ak

s akbkl + =ll+

+

t---

T- E

P

10.14

102

hence

p(f) p(g) = 0p(fg)

Q.E.D.

The uniformizar t is a meromorphic function on S .

Put

?'so that t, 'r form a primitive pair.

Hence we have A

rao

tn'e=0

Ot (t)=t ao

p (if) _ =%tn N

Since p is a homomorphism,

4p(t)

,

Op(-C) satisfy 0

Hence the pair (ap(t) , 1p(-Y)) is a function element on the analytic configuration Si,.y, determined by the alge-

braic equation

T

0.

Anm tnTm

Furthermore it is clear that a change in the choice of a local uniformizer t', instead of t corresponds to a permissable parameter change since t Now we can define a tt .

canonical may 0 : S --j S .

Where

F = {S} . The map of

will be the composition of two maps, ml

02

,

01:

S -ySt,.t

$2:

St,r

Where 0l is defined by

1(P) _ ($ (t) 110,

,

#p('Y))

S

10.15

103

where t is a local uniformizer and t,'t'are a primitive pair. And 2 is the conformal homomorphism which is determined by the primitive pair t, "C. Neither of the maps X1,¢2 are canonical since they depend upon the choice of 't. However 4) is canonical. For we need ohly note that at a point p e S there is a function z e {S) which vanishes at p and no other point of S . This is an immediate consequence of Lemma 1, Lecture 9 . This statement shows that a point p E S determines a unique place P = 4)'(p) namely the set of functions which vanish at p Clearly the map

:.S --- s into

is canonical.

We will show that 4) and 4)t are inverses.

This

will show that Qf is 1-1 , onto and canonical.

To do this we need only show that if z E P then z vanishes at $(z) . Clearly t , a local uniformizer at the place P , vanishes at l(P) and hence at 4,(P) . However if z e P then

m i-1

Hence z also vanishes at 4)(p) . Is the mapping 4) topological. Clearly there exists an a > 0 such that if 1Zll,1Z21 < a

then t takes on the values Z10t2 at distinct places and at distinct points. Clearly these points and places correspond to one another under 4 . Somewhat surprising is that we now obtain a concrete result from all this formal construction. Namely that the

local uniformizer 'if at a place P has a simple 0 at the point Since at p we have a local parameter p corresponding to P Z which vanishes at p . That is

4(p) = 0 and 'r a Zk 0

104

Suppose k 1 1 is in 4k .

10.16

.

Every P E F is a power series in 2', that

But by Lemma 1, Lecture 9 they ara functions with poles of all orders greater than KO . Hence k divides all integers greater than Ko, hence k = 1 ,

105

11.1

Lecture 11

The topology of compact Riemann surfaces In Lecture 2 we defined a Euclidian polyhedral surface.

Clearly, we can do the same on a sphere with spherical triangles rather than Euclidian triangles.

We saw in Lecture

9 that any compact Riemann surface S is equivalent to the analytic configuration determined by an irreducible algebraic equation.

P(z,w) = w11 +

an-1(z)wn-1

+,...,+ ao(z) = 0

satisfied by a primitive pair (z,w) .

We can tringulate

the sphere into a complex so that the excluded points are vertices of triangles, and so that no edge has excluded points for both of its vertices.

Now if we make n copies of the

sphere with this triangulation, pasting the triangles together according to which values the function z continues into upon crossing the edge.

Hence we have triangulated the Riemann

surface S and Rado's theorem has been proved. Note.

The same triangulation could be effected with the

knowledge of a single non-constant meromorphic function on S for compact Riemann surfaces. The Euler characteristic 'X,

is a topological invariant.

We can compute it using the triangulation we have just given. Let ao,al,a2 denote the number of 0,1,2, dimensional simplioes, (i.e. vertices, edges, and triangles) in the triangulation of the sphere.

Also, let ao,a1,a2 denote the number of

vertices, edges and triangles in the triangulation of the Riemann surface S .

Since the sphere is topologically equi-

valent to the surface of a tetrahedron

(sphere) = ao - al + a2 = 4 - 6 +4 = 2 clearly

a2=na2, al=nal. Suppose the vertex zo on the sphere has branch points of orders ml,m2,...,ms over it, then there are s vertices. on

S corresponding to the vertex zo for the sphere,

Since

(ml + 1) + (m2 + 1) +,...,+ (ma + 1) - n or

mi +e =n i=1 or

s=n-

MI.

1=1

we have

ao=nao - B. Where B , the branch number of S , is the mum of the orders of all branch points on S .

Hence

Y _(s) = s.2-al+ao = na2-nal+nao-B = 2n - B

,

Assuming we kne' that the Buler characteristic of a closed

orientable manifold is 2 - 2g , g non-negative, it follows that

2-2g=2n-B or

2(n + 1

B

that is, B is even.

-g) ,

Suppose we have the equation

w2 = (z-e1)(z-e2),...,(z-e2x)

then n = 2 and B = 2.

.

ai 1 e' if i # j

Since

2-2g=2n-B we have

1-g=2-.Q or

g is the genus of the surface, we sea

that in this case the topology of the surface can be read off from the equation. Definition 1.

A Riemann surface is hyperell.iptic if it can

be represented,

w2 = polynomial in z Theorem 1.

.

A closed Riersann surface S is hyperelliptio it

11.4

108

and only if there exists a meromorphio function on it with

oxactly 2 poles. Proof.

Let z be a runetion on S with 2 poles.

There exists

a w on St such that z,w are a primitive pair.

Hence they

satisfy an irreducible equation

w2 + 2p(z)w + q(z) - 0 or

(w - P(z))2 = P(z)2 - q(z) = 4(z)

denote w - p(z) by wl

Clearly wl,z are a primitive pair,

2

and w l = 4(z) , where 4(z) is a rational function of z Hence

wi = PA(Z)

P,S polynomial in z

2 _ a(z)2(z-el),...,(z-er)

wl s(t)2(z-er+1),...,(z-et) or

wl = Y(z)2(z-el),...,(z-ej) wlY

where y(z) is a rational function.

Denoting

j

by w2 .

We have

w2 = (z-el)(z-e2),...,(z-eX)

and z,w2 are a primitive pair.

Since the "only if" part of

the theorem is obvious, Thoorom 1 is proved.

Now with the preceeding information we proceed to give certain "topological" normal forms for compact Riemann surfaces.

First consider the following triangles.

Take a circle

and inscribe a triangle, we label this triangle

then construct triangles

with

4o ,

we

which have one edge in common and the vertex opposite this

or

edge bisects the corresponding are.

11

i

-Consider a set U which is a union of the following type.

U = (Qo) U (A,1 ) U (i 2i2 ) U, ..., U (Anin) 1

where Akihas an edge in common with Ok-li

k-1

1c

Clearly

this set is convex. Definition 2.

A topological polygon with identification is

11.6

110

a topological space homeomorphic to a convex polygon with edges identified so that each edge is identified with exactly one other edge. Lemma 1.

Every compact triangulation surface is a topological

polygon with identification. Proof.

We know that in the triangulation of the surface, each

edge of each triangle is identified with exactly one other edge of another triangle.

Hence we can enumerate the trian-

gulation so that Ti and Ti+l have an edge in common.

Then

we can map To onto Ao , T1 into the proper Q1, and so forth. In this way we get a topological polygon with identifications. Lemma 2.

Order the vertices of the polygon P1,1.z00.0,Pn so

that in this order we go once around the polygon.

Then if

the edge Pi Pi+l is identified with the edge Pk Pk+1 , they are identified so that Pk is identified with Pi+1 and Pk+1 is identified with Pi

4' Pk+l

Proof.

Since a surface is orientable, the triangles are

oriented coherently.

Hence if we look at the triangles which on the surface they must be as is

contain PkPk+l and PiPi+l required for the lemma. Definition 3.

Assign a letter to each edge, so that identified

edges have the same letter, but one we call the inverse of the other.

For instance if in Lemma 2 we called PkPk+1

we would call PIPi+1 "a-1"

.

an

Then going once around the

polygon we would obtain an expression of the following type abca-1 c-1

bd .

This expression is called the symbol of the

topological polygon.

Clearly the symbol uniquely determines

the topological polygon with identifications. We shall now describe a sequence of reduction steps to show that our polygon can be represented by a symbol having one of the following two normal forms.

a a-1

(1)

(ii)

n g

-1

-1

a2j-l a23 a2J-1 a2,

We shall let Greek letters denote sub-symbols.

a = a b o a-1 1.

a a a-1 P is homeomorphio to a P.

e.g

11.0

112

The following diagram demonstrates the proof of this stop

2.

We can arrange it so that all vertices are identified.

That is all vertices correspond to the same point on the

surface. Proof.

Suppose we have a vertex Q not identified with P.

We

shall reduce the number of Q Vertices by one while increasing

b/'

the P vertices by one R

If we make a out joining R to P and paste back the two pieces along b we obtain

R

r

If in our symbol

a

a-l b b-1 occur in the following

order,

as 3 b ya-l 6b-171 we say that 3.

si, a-l, b,

b-1 separate

Given a, there exists a b such that

a

a-1

b b-1 separate.

For if not we would have a

7 0-1

/

where a had no edge identified with an edge of P .

But this

is impossible., for then the two vertices of C could not he

11.10

114

identified, which they were ire step 2 . I}.

We change

a a p b y a-1 0 b-l ? to e f e-1 f-1

F

Hence it will be clear that we obtain one of our normal forms.

Proof .

we out along C and attach at b , our symbol then becomes

aC

a-10Y C-1 07a

a C.

a-1 Cr C-1 V

or

that is

c

115

we cut along d and attach a , and we have

C-ld-led -C'c or

o f a-1 f-1

Q.E.D.

F.

Hence every compact surface is a topological polygon with symbol

a a-1

(1)

or g

M a2j-1 a23 a2J-1 a23 3=1 -1

-1

g is called the genus of the polygon, and the polygon with symbol

a

a-1

has genus 0 .

We shall now show that different normal polygons are not homeomorphic.

We shall do this by computing the Euler characa2,_1 a23

teristic of the polygon

a2;

We can

3=1

triangulate it as in the diagram and count 0,1,2 simplices, -l

a2i-1

11.12

116

we get a2 = 36g

al = 46g + 6g = 54g

ao12g+!{g+2-16g+2 hence

'}(,

= a2 - Cl + ao

2g

Hence since % is a topological invariant, polygons with different normal forms gre not homeomorphio. It is easy to see that the polygon of genus g is

homeomorphio to the sphere with g handles.

12.2

118

If a,p are two curves such that, the and point of a coincides with the initial point of p, we define the

Definition 7. product

asp

of the two curves to be the curve

a'(2t) for 0 0 or if and only if f(x) < f(x+a)

f(x-a) +

that is a convex function lies

below the linear function with the same endpoints.

137

14.2

Two dimensions: linear harmonic

subharmonto

+ uyy = 0

uxx

uxx + uyy > 0

.

However we want to allow functions to be stibharmonic which do not necessarily have derivatives. Following the analogy we prove the following theorem. For a continuous function v(p) defined in a region G Theorem 1. (an open connected set on a Hiemann Surface) the following are equivalent:

1.

For every region D C G, such that 15 compact, D C G, if u

is harmonic in D, continuous in D and if v < u on 15-D then

v 2g-1 is a gap value then

16.6

156 either

s-1

or

1

is a gap value

either

s-2

or

2

Is a rap value

either

s-g

or

g

is a gap value

and we already have g gap values which are as many as there can be.

On a surface of genus g the numbers 1,2,...,g are usually If at a point p a number k > g is a gap value

the eap values.

then p is called a

point.

Theorem . On every compact Riemann Surface there is a finite number of Weierstrass points.

Now we have more than enough material to finish Theorem 1 connected surface For, since g = 0 for a S, the proof of Theorem 1 of this lecture shows us that the function which we constructed in Lecture 15 is the real part of a single valued meromorphic function f on S. Furthermoro the order of f is 0 or 1. But it is not 0 since f is Lion-constant. From Lecture 9 we know that a single vslued meromorphic function on a Riomann Surface takes on all values the same number of times. Since f, by the maximum principle, takes on infinity only once on S, it follows that f is a 1-1 conformal mapping of S onto the sphere. Hence the theorem Is proved. of Lecture 15.

Lecture 17 An open Riemann Surface S is called parabolic if there does not exist on S a non-constant negative subharmonlo function. Definition 1.

It follows than that on a parabolic Riemann Surface there does not exist a non-constant subharmonic function which is bounded from above. Parabolic Surfaces are also called surfaces which have a null ideal boundary. The maximum principle when stttad in the following way is true for bounded subharmonic functions on a parabolic Riemann Surface. Lemma 1. Lot z be a local parameter which vanishes at p F S, a parabolic Riemann Surface. Let !1 correspond to IzI < r and Y correspond to Izi - r. If 1.

u is subharnonic in S-0

2.

u