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
z--1
drdYL
SS
z
IzI<
Proof.
Take w(z) = z and D the disk of radius R about the origin. Then by Lemma 2 i =
1
rt-z
dg
1
dd
I2I-R
Izl
but 2
dL
- 2iT
2a`i
t-zT 3 0
J
ICI=R
Iii=R
since the sum of the residues = 0. Lemma
I'
Iz12=zz=-1it
Id
rdr, -z
!L
1
Then
As in Lemma 3, take w = zz. dE:
Izl2 = ZZ = R2
Ii -R
r,-z
R2
_
- dr-di
I
Izl R
z-z
but
t l=R since the sum of the residues is 1. Lemma 5.
If 0 < El,E2 < 1 and E1+E2 d 2 and zl / z2 and
Iz1I < R and 1z21 < R, then
_
d_d_y SS
ICI
114-z21.
I4-z1I
1
2 (zl-z2I
1
2
What is essential in this result 1s that c(El,E2) depends only on e1 and E2, not on R.
3.8
22
We see that the integral exists by changing to polar For then the extra r in the numerator takes care of one r in the denominator and the exponent of r in the denominator is less than 1. Let 28 = 1z1-z21, D = r. I141 < R11 and consider Proof.
coordinates.
01 = ziz E D
and
I4-z11 < 6}
A2 - 1y14eD
and
1z_z21 < 8}
p3=D-(Al
L2)
then
then changing to polar coordinates, and since the integrand in Ll and A2 is independent of 0, we obtain +t 2-2
< 8
Q -
f 27(
Sf
(('a
+e 2-2
rE-1
2x
r t1-2 r dr = 8
I
0
and similarly
t.}
a
dr = 6 1
+e 2-2
27E'
el
0
271
e2
42 both of which are of the required form. Now Since the function (r,-zl)/(4-z2) is regular in IL3 its maximum and minimum modulus Hence
are achieved on the boundary.
g<
__Z2 zl
<2
for
A3
i
e
Hence, 1
63
<
2'2 t2
d1d
L3
14-zl 1
1
2
3.9
23
Now since the integrand is positive, if instead of inte-
grating over 3 we integrate over
the whole exterior of Ol we only obtain something still larger. Namely first change to polar coordinates, then 2-e2
Q
el+e2-3
r
2a
)
(
2
dr =
)
(
2-e2 2A
6
c1 -e 1
E1+e2-2 O
2
3
Q.E.D. Note.
Lemma 2 decomposes any function in class C1 into a sum of two functions. One of which is analytic. The remaining lemmas will show that we have a great deal of information about the other summand.
Consider the function
w(z) = -
21211
dtdj
I?I
Id < R
1.
Ip(z)I < M
2.
for all z1,z2 such that JzI < R
for
0 < a < 1
!p(zl)-p(z2)I < KIzl-z21a
Now form
6(z) = -1
J
(r)- (z
dFdj
z)
which is defined since p satisfies a H81der condition. now prove three lemmas about w and
Lemma 6.1.
Iw I
,
in
< 2RM
Proof. 1w(z)
dEd it
z_T
;! < R.
We will
3.10
24
since the inteprand is positive, the right hand side becomes still larger if we integrate in the disc of radius R about z, i.e.
lw(z)) <
M
d$
dd < -,fir 141
Ir.-zl
i z1
and changing to polar coordinates
sR
M2n1 dr=2MR 0 2K Ra lol <
Lemma 6.1.1.
a
Proof.
a Id
as in 6.1 <
j
dEdl
1
a
and changing to polar coordinates
= n 27E
((R
ra-1 dr
=
2Q
Ra
.
0 Lemma 6.2.
and wz exist, and in fact
w z
w
Proof.
z
= p and wz = o'
At a fixed point z, let w(z) = w(z)-p(z0)z.
Then by the
definition of w and Lemma is.
PW-P(zo) w(z) =
Jd
Z-z
dJ;d
j
lr.l
We show that w(z) has a complex derivative at zo.
We have
3.11 w(z +h) - w(z )
°
25
(p(r,)-P(zo)) -;:I
° I?I
P(O-P(zo) (Zzo-h
" Z y 0 d4d'
dFdj
r.-zo
It I
0 is
(F (2
o
) _ _In
I pw-P(z )
d dh
( -zo )
kR
I
hence we look at 1
n
P(4)-P(zo)dd
"za
-z0-h
Izl CR
1
P(_)-P(zo)dP.dq
A
o (2-z)
1J P) pp
-zo
o
Z yo) dx*(
(--z0-h
114 J
_ iit i
I
JJ IdI
P(O-Ozo) 2
IP(z)-P(z
jh[ R
141
dgdq
(--zo) (2:-zo-h)
Iz-z°i
a dr
Ir.-zo-hl
and since p satisfies a H81der condition < _IhIK
12--- °-I a ICI
71
ICI
dd
14-z°I 1z-zo-hl
I4-zoI
Iz-zo-hl
(and now by Lemma 5, with el = a, z1 = zo, z2 = zo+h, we have)
3.12
26
c(a,l)Ihla+l-2
(hl
which goes to 0 as h -- 0.
=
c(a,l)Ihla
Hence we have
wz(z0) _ T(z0) _ _1C SS P(Z
(z
(,-z0) 2
z0.
But at zo, wz = wz = 6r(z0). Hence by Theorem 1 at zo
0
dEd7
Furthermore w is monogenic at
0=w =w z - P(z0) z
hence
w (z0) = p(z0) 2
Also the point z0 was an arbitrary point in the disc, hence throughout the disc
w, _
w
cy-
z
=p
which is as was desired. Lemma 6.3.
w is H81der continuously differentiable with the same exponent a < 1 as p, that is 1.
(wZ(zl) - wZ(z2)I < c(a)Klzl-z2ja
and
2.
Iwz(z1) - wz(z2)I < c(a)Klzl-z2(a
N.B.
c(a) depends only on a, not on R.
Proof. Since w = p there is nothing to prove about w_.
z
wz = T we must show I6'(a1)
z - r(z2) j < c(a)Klsl-z2ja
For a fixed zl,z2 set ) ( P z2 - P zl ) 0=. z1 - z2
hence, by the H81der condition on p I31 =
and let
z l) - P(z2)1 < Klzl IP(
1-
-
z21
z2ja-1
Since
3.13 W(Z) = w(z) + pza
27
d
-
+ BR2
(4-z Izl
by Lemma 4 with p(z)
p(z) + Pz and by Lemma 6.2
wz = -1
(P )
nPaz)
dgdr
14 I
and write a(z) = wz(z) = wz(z) + OF, hence
(z 2 )
-
a_ It1
6(z 1)
14 I
_
-1
1f
--
P(g)-^(7 2) P(4)-P(z1) z )2 _ z 2) ( 1)
(
d d$
(^P(2:) -p(z2))(r'-Zl)2-(A(r.)-P(zl)Xr-z.2)2 d
I2I
noting that p(z2) = p(zl)
_
P(_) P(z1)(z2-11 )(24-z1-z2)
SS
-1
d dl
(r,-z1) (2:-z2)
ICI
{P(4)-P(zl))(z2-z1)((4-zl) + (?-z2))
I
d&d't
JJS
(4-zl) (r-z2)
ItI
noting the following P(z1) = P(4) - P(z2) and
P(Z) - P(zl) = P(r.) - P(zl) + P(r.-zl)
P(z2) = P(4) - P(z2) + P(4-z2) we obtain
I = I1 + 12 + 13
where
I1=-
(z2-zl)
it
j
p(r.)-P(zl)
Izi
(z-z2)
ddb
3.14
28
hence
,' l i<
d dfi
n l z2-zl I K 26
a
k-sll
1zI
-
It-z2l
and applying Lemma 5 Klzl_z21a
< const (a)
and in the same manner 1121 < constp(a) Klzl-z2Ia
13 = -n2 (z2-zl)P
j
!
ItI
_
-n (z2-z1)P
z21zl
z
dxd
-z2
1) 1
(r z2 - r zl)dFdr IzI
and applying Lemma 3
= -2c3(z2 - zl) hence 1131 = 21131
and since
I(3I <
Iz2-zl! Klzl-22"-1
<
2KIz1-z21a
hence I;;(z2)-Q(zl)I = IF(z2) - a(zl) + Az2 - OZ11
< III < II1 +1I21+1131 < const(a)Iz1-z2la hence
IQ(z1) - 6(z2)l- const(a)Izl-z2Ia - IGIIz2-zll
-< const(a) Iz1 -
z2 Ia
.
29
4.1
Lecture 01. Introduction. We showed last time that the problem of introducing isothermal coordinates in the neighborhood of the origin is equiva-
lent to that of demonstrating the existence of a solution, near the origin, of the Beltrami equation
w- = µ(z)w(z)
z
(1)
µ(z) > 0 ,
(µ(z)I < 1
for
1z1 < R
.
In this lecture and the next, we apply the succession of lemmas proved in Lecture 3 to give the existence proof. Consider the equation
M_ = µ(z)wz + Y(z)w + NO
(2)
z of which (1) is a special case. Theorem.
We shall prove:
If µ,Y,6 a Cn+a (0 < a < 1; n = 0,1,2,...) then the
system consisting of (2) with the conditions (3)
w(0) = a
(1})
wz(0) = b
possesses, in the neighborhood of the origin, a solution w(z) of class Cn+l+a. :Notice that the theorem asserts existence but not uniqueness.
The proof will be carried out in two stages:
we prove
the case n = 0 (i.e. µ,Y,6 * Ca and conclude w r. Cl+a) of the theorem today, and shall prove the p.eneral case by carrying out
an induction on n in Lecture 5.
We can in full generality assume µ(0) = 0. To see Then introduce a new independent variable 4 = z + µ0z near the origin. Since Z = z + µ0z, we find Remark.
this, let µ(0) = µ0.
wz = wZµo + w
wz=wr. + µ0w and applying (2), we have
(2')
(µ-µo)wI+ l - µ;o
=
w
+bw l -14µo
But (21) is of the same form as (2), and the coefficient of w4 vanishes at the origin. We may therefore take µ(0) = 0 in (2), in full generality. 62.
Some heuristic consideration. Before proceeding to a formal proof of our theorem, let
us investigate the consequences of assuming that a solution of Cl+a (2) exists. Let w(z) be a function in a disc jzI < R; then according to Lemma 2 of Lecture 3: w_(4)dgdrj
w(z) =
(5)
1
r, - z f T w(K)d J
1
-
'rt
IaI
zl=R
-z
'
where the first term on the right side of (5) is an analytic function. Next, let w(z) satisfy (2). Using (5), we have (where all surface integrals are over Izi < R and the integration
is with respect to g, as in (5)):
(61
w(z)
1 f n{
-z
_ 11
w(
(
-z
r.
z
i 9L +f (a) +4
(4) Y_ Z +x 1(-µLWW-+x 1
(
(r.)
)w
where f(z) Is an analytic function, with the obvious property f(0) = w(0) = a. To see that f(z) is analytic, differentiate (6) with respect to a:
(7)
Wz
a
from lemma 6.2 of Lecture 3, z
+yw+8+f
7
31
4.3
But w satisfies (2), so that f- =_ 0, and f(z) is analytic; also
z f(O) = w(O) - a, as above.
Applying lemma 6.2 again:
wz(0) = f'(0)
(8)
or
f'(0) = b
83.
Proof of the existence theorem (n = 0). Now consider the integro-differential equation
(9)
w(z) =-2Lzt
iJJ
-4---Z --µ(r')w (,-) z
- 1 J`y(r)w(Z) -
r-z
7 J
f,
1 is `
J,
( 1 1 +nJIY_5.1+z ((
(
b(
w
+ 1L( w(
+ z
Z-z
+. f(z)
b(
+
x e
where f(z) is a given analytic function, with f(0) = a, f'(0) =b. If we can show that a solution w(z) (of the proper class) of (9) exists, then w(z) satisfies (2), (3), (4). Notice that the three terms of (9) ir.volving 6 are analytic functions: combine these with f(z) and call the result g(z).
We shall now apply the method of successive approximations to prove that (9) has a unique solution w(z) for every g(z) with H81der continuous derivatives. Consider the class of complex-valued functions w(z) defined in the disc IzI < R (R > 0).
(10) (11)
IIwII(R) = o
Introduce the norms (where 0
max
Izl
I1wII(R) = 11wI1(R)+ Ra l.u.b.
Iw(z)I
1z121
z1z2
Iw(zl) - w(z2)1 IZ1 -
z21
4.4
32
(12)
IIw i11R) = Ilw IIo)+ Rllwz 11(R) + RIw_IbR)
(13)
IIw I1Ra = IIw IoR) + R jwz 11"') + Rllw_1"')
z
z
Assignment:
let t = 0,1,1+a; prove that
Iw(z) 11t(R) = liw(RzIM It
(14)
(Notice that (14) allows us to take R = 1 in our norms.)
Also
show that
if w10; 110It-0
(15)
IIw It>0
(16)
Ilxwlt = Pd 0wIt
(17)
Ilwl-W211t < Iwl It + 1lw211t
(A real) .
With properties (15), (16), (17) we have a metric space; our space is also linear and complete, and therefore a Banach
space. We can also prove monotonicity: IMItR) < conat. IIw ItR' )
(18)
if R < R', t < t'; and, finally, thst we have an "almost Banach"
algebra: (19)
IIwlw211tR)
< const. Iwl IIt uw2lt
(If the constant were 1, in (19), we would have a Banach algebra. Now let us assume there exists an a (0 < a < 1) such that
µ,y,b r- Ca and g E
Cl+a.
We shall work w.+.th the Banach space
defined by the norm (13), and will prove the existence of a solution w i C1+a of (9) in the disc Izl < R. To begin with, let w(z) he anv function, and form
(20)
W(z) = + 1
1
1
Sly )w
+ z
(Z)w(Z
1
--
iLWW (Z)
1
(5
+ 1
n
4.5
33
which we shall alternately write 6
W = Tw = 2_ W
(20')
pi
j
Comparing (20') and (9) (with g defined as f, plus the three integrals involving 6) we see that the equation to solve is
w=Tw+g .
(21)
Our task will be to find an estimate for the norm of T, i.e. for
11T1 = l.u.b. 0q." rl+a
(22)
wI(R) 1+a .
We notice that T is a linear operator, so that it will be suffi-
cient to compute 1wJ
(i = 1,...,6), then to apply the triangle
inequality.
We first prove
(23)
ttWlIl+a < const. Ra1wttl+a
The proof is as follows.
(24)
W1 (
z
By definition (see (20))
)_
-x-l
I
u
ryl
1
and from lemma 6.2 of Lecture 3, together with the bound KRa for 1µj, we have
(25)
tiWl1o < 2KRa9wt!l+a
We next want (see (11), (.l})) RI(W1)z1to; but we may compute and apply lemma 6.3 of Lecture 3 to obtain
(26)
R1I(W1)zoo < const. Rdwz is 1µ a
< const. t(w I1+a Ra where we have used ttuga < const. aR. We next need RI(Wl)Ho; z
4.6
34
Finally and we have our estimate for this term. (Rl+a). (1181der cont. of (see (11)) we need an estimate for
but Wi =
aWl/aa).
1`wz
Using 9Wl/aa = µ(z)wz, we have const. RIIwzIla IlAa as
the estimate, which leads to
(27)
aw
Ra
(H81der coast. of ,a z1) < cons t.IIwIIl+aRa
An application of lemma 6.3 leads to an estimate of the form of (27) for aW1/ai. Combining our results, we see ($n view of (11), (13)) that we have proved (23). Next consider
(28)
w3
=
w(o) Also
We see that (23) holds for W3.
(29)
IIW5(z) I10 < RIJaWl/az1o < cont. Ra9wl1+a
and the bound for RIIaw5/az Qo follows as easily. (30)
w2
x Jj IL41 -w z
Finally, consider
Z
From lemma 6.1 of Lecture 3:
(31)
Jw2 i!o < 2RIIw
80
< 2RIIw 111+a
Then 3W2/az = yw, so that (32)
Ell-AW2h z IIa < RIIyw Da < const. RIIw 9a < comet. R8w ll+a
by (18). For our estimate of (33)
aW2
ai
1
y(
4) - Y(Z)W(Z)
( - z)
we use our lemma on d (see Lecture 3), as before. results: (31+)
IIw2111+a _< conet. R1w Bl+a
Combining our
35
4.7
Notice that W4 = W2(C); and the w6 estimate follows from that for W2 as well. Remark.
We are interested in small R only; i.e. Rd > R and
(31i) implies (35)
11W2 111+¢ <_ cons t . Ilw IIl+a Rd
In sun, we have shown (35)
IITi < const. Rd < 0 < 1
for small R. We first apply (35) to show that the solution of (21) is unique. Suppose there is a w' such that
w' = Tw' + g .
(21')
Then
11w-w' 11+a = IIT(w-w') nl+a -<
11T 11
liw-w' 11+a
which is impossible unless 1iw-w'111+a = 0.
-<
9I1w-w' 11+a
It follows (see (15))
that w = w'. To prove existence, form the series
w = g + Tg + T2g +
(36)
...
and we have (in norm) (37)
11w11
so that w is well-defined and in our space. (38)
Finally, we have
Tw = Tg + T2g + ...
and from (36), (38): (39)
w - Tw = g
Q.E.D.
5.1
36
Locture I Given three funct.icns defined for small values of Iµ) < 1, lc= 0,1,2,..., and given IaI, µ(z), Y(z), b(z) E two constants P.,b 3 R > 0, and a furctinn w(z) E Ck+l+a, w dofined for all z such that Iz1 < R and satisfying the conditions w(0) = a, wz(0) = b and the equation w = µwz + Yw + b. Theorem 1.
Ck+a,
z In the last lecture the theorem was proved for k = 0. We complete the proof by induction on k. Assume that the theorem is true for k = n. In order to prove that this implies the truth of the theorem for k = n+l we consider two cases. Proof:
Case 1.
Y
0, µ, b e Cn+l+a
if we differentiate the differential equation = µwz + 6 we get
(Remark: w z
w zz
= w
zz
+ µzwz + 8z
= µwzz
That is, if w is a solution of the equation we are considering. then wz is a solution of the general eq'.iation where the smoothness condition on the coefficients is that of the case where we know there Is a solution by the induction hypotheses. Hence we can expect to push through the induction by constructing a function whose z derivative is a solution of the appropriate equation)
n+a µ, b e C n+l+a --> µ, µz, bz a C. Hence
by the Induction hypotheses
2 R > 0, and a function a satisfying e(0) = at, crz(G) = b' and
cr = µKz + µza' + bz z
for
IzI
and we will decide later what we want a', b' to be.
W(z) _ -, 1.
S
{
°
drdat
hence
P(z)
a
J Z¢ dgd i 14 JcR
Consider
5.2
37
then by Lemma 6.2 of Lecture 3
but
z
YI(W-) z
hence
Wz=T.
Assume for the moment that W 6 Cn+2+a (we shall prove it at the conclusion of the argument). Consider
(1_-µWz-6]z=W_ - (µWz)z - bz=tr zz
z
z
-
- µWz - S) is an analytic function of z.
Hence (14
Since the
z
partial derivatives are continuous and the Caixchy-Riemann equations are satisfied. Write W
z
- 0z - 6 = f(F)
where f is analytic. Let F(F) be any analytic function such that dF/dz = f(F), and form w(z) = W(z) - F(i)
,
then w
z
- µwz - b = W
- f(F) - µWz - 6 = f(i) - f(z)= 0
F
hence w(z) satisfies the equation w_ = µwz + b and w(O)=W(O)-F(O), z
wz(0) = (0), so we want a' = b and since F wis determined only up to a constant we can have w(O) = a. General case (y may or ma,., not _= 0).
Case 2.
We proceed
in an elementary fashion and look for a solution w of the form We require w = µwz 4 w + 6 or after dividing by e z
w = a%g.
g_ + a-e- = µEz + µ%zg + Y8 + eC or
z
z
g(%_ - µd`z - Y) + EZ = µ8z + e-A6
5.3
38
from case 1 we obtain solutions A,g a Cn+2+a of a_ = µaz + Y
a(0) = 0
Az(0) = 0
with g(O) = 0
gz(0) = b
with
Z
and
g_ = µGz + e-%0 z
% Hence w = a g is in Cn+2+a and w_ = µwz + yw + 0 and wz(0) = a,
wz(0) = b. z Now in order to complete the theorem we must show E C n+2+aW . This will be accomplished by proving the following
lemma. Lemma 1.
p E Cn+l+a
>w
2
Cn+2+a where w
J
dgdj
(integration here and hereafter over 141
The proof is by induction on n. ay Lemma 6.2, Lecture 3, w_ = p E C1+a hence 1. n = 0. C1+a. z we must show wz E 1
w
"11
-a
(P 1814-z12)dEd
+
PZ lg ft-zJ2dxdq
which can be written by the extension of the Cauchy integral theorem (Lemma 2, Lecture 3) as, P
w
lgk-zI2dZ + n 1 pZ l3Ir-zf2dEd*
and now it is legitimate to differentiate under the integral,
wz =
2ni
z
dZ +
P
dFdp
The first integral is in Coo, and the second is in Cl+a (by Lemmas 6.2 and 6.3, Lecture 3) hence w C C2+a. Ck+l+a As 2. Assume the lemma true for n = k > 0, p E above we obtain
5 !} w
z
8
39 P
zd +n
wz=-
P zd.-
by the induntion hypothesis the second integral is in Ck+l+a Q.E.D. since p ti Ck+1+¢ -> P2: E Hence w F Ck+a.
Ck+2+a,
Also, if the coefficients µ,Y,6 are in Cw then the CauchyKobolevsky theorem tells us that we can get a C' solution. In fact since the system is elliptic, a unique solution, if the complete Cauchy data are prescribed. The same methods would have ;riven us a solution to w = pw + vw + )-w + bw + e (1µj + jvj < 1). Which when z
z
z
written in real form is seen to be the most general elliptic linear partial differential equation of first degree. In any case we now know that we can always obtain a (in Cl+a solution to the Beltrami equation. the small) w_ = µwz, z
<1, p£Ca, a>0.
We have the following six properties of the solutions to the Beltrami equation. 1.
if w b Cn+a, defined for smail values of z, then
R > 0 and a solution w e Cn+l+a defined for jzj < R which is a homeomorphism. Take a solution w, such that wz(C) 1 0. Proof:
Then in a whole neighborhood wz # 0, we have seen that with w = u + iv a(R,Y
jwzj2
jw_j2
=
jwzj2(l
-
jp12) > 0
hence w is a homeomorphism. 2.
if w and w are solutions, and w a homeomorphism, then
w is an analytic function of w. w^ has at least continuous first partial derivatives, and Proof: since w a homeomorphism
5.5
40
a1
aw az
=
+
az aw
aw
aw dz az aw
&(ww
9z
z
)z az ww z z+ w zwz) z,z)
zz)
"d(w
µwzwz + µwzwz
d(z,z) 0
hence w = f(w), f analytic. if w is a solution and f is analytic, then 'Z' = f(w)
3.
is a solution. Proof: ^W- = f'(w)w_ = f'(w)µwz z
z
µwz = f'(w)µwz i.e.
if w,w are solutions then
4.
w / 0,
w/w are solutions.
w+w and w -w are solutions since the equation is linear.
Proof:
Write w = w w = µ(wzw + wzw) = z
z
z
Write
µwz
w
= --- - µ- -z -ww
ww
W
5.
z
(WWZ
WZw)
_µ(
w )z
If w is a solution and µ F Cn+a then w e
Proof:
µw z Cn+l+a.
By property 1 there exists a solution w E Cnfl+a which is a homeomorphism. Hence, by property 2, w = f(w) with f analytic.
Hence w a n+l C+a.
6.
Proof:
if w is a solution and p E COa then w e C. µ E Cn+a for all n hence by 5, w b Cn+1+a for all n,
hence w E C. 7.
Proof:
w w if w is a solution and µ E C then w e C same as 5, using the Cauchy-Kowalevsky theorem.
Tensors on an n-manifold. Suppose at a point p E M (an n-manifold with CP structure, p > 1) we have a function which associates with each allowable coordinate map defined at p,n real numbers. Let us denote by ai the n numbers associated with the allowable coordinates (xl,x2,...,xn) and by ai the number associated with the allowable coordinates (xl,x2,...,xn). (It is possible that xi = xi Since M has CP structure, p.> 1, the xi and xi are differentiable functions of one another. Noticing the two distinct ways in which partial derivatives and differentials respectively transform under a transformation of coordinates, we say that the at p.)
rule which associates n real numbers with each allowable coordinate map is a contra -variant vector, or a contra)-variant tensor of rank 1 if either ax ai = a -_-i covariant 3
axi
or
_ a
ai = s
ax axi
contravariant
i
Similarly, we define a contra -variant tensor of rank 2 defined at a point p E N as a function which associates n2 real numbers with each allowable coordinate map defined at p such denotes the n2 numbers associated with the allowable that if yis the numbers associated with coordinates (xl,x2,...,xn) and
r
Yij
the allowable coordinates (xl,x2,...,xn)
Y1
or
"xk x2 Ykl axi 'Zx j
covariant
42
5.7
axi o1X Yij =
yA
Xk
contravariant
In the same way we could define tensors of rank m = 1,2,... Note that we can define a tensor at a point p t M by prescribing n2 numbers yij in one allowable coordinate patch and letting the transformation rule define it in all other allowable coordinate patches.
We would also obtain contral )'variaut
tensors of rank 2 by saying let yij = alpj where a1, pj are both components of a contra -variant vector.
We say that a tensor is symmetric, anti-symmetric or Y11 positive definite according to whether or not the matrix (yij) is symmetric (yij = yji), anti-symmetric (yij = y 1) or positive definite ((tlt2...tn) # 0 > yijtitj > It is easily observed that if the matrix corresponding to one allowable coordinate map defined at p e M is symmetric, anti-symmetric or
positive definite, then the matrix corresponding to some other
allowable
map defined at p is so, too.
We shall say a tensor is defined over the whole manifold M if it is defined at every point of Mi. We could in this case talk of tensors of class C r. However it is clear that this makes sense only if a-< p-1. If a covariant, positive definite tensor gik is defined over the whole manifold we say that the manifold carries a Riemannian metric or a metric tensor gik. By means of the metric tensor Fik we can measure lengths of sufficiently smooth curves in the following way. First we define the length of a curve r whose support lies in a single allowable coordinate patch. Let r have its support lie entirely in the allowable coordinate patch G. In G, r is the set of points (x1(t),x2(t),...,xn(t))
We define an element of length do by ds2 =
gikdxidxk
0 < t < 1
5.8
43
and define the length of r , .P( r) as the integral of ds along
r.
That is
gik at- at- dt
P (r) _ 4
Now the support of any curve r consists of a finite number of pieces rl,(Z,...,rn with common endpoints, where each piece lies in a single allowable coordinate patch. We define 1( r) by
c(ri)
.c(r) _
It is easily seen that this definition is legitimate. In a similar way we define the angle a between the curves by
r and r' which intersect at t
0
dxi dx'k
cos U -
gikdt t dx
cTx' d k
dxk 1/2
gik dti dt )
(
gik WE- dt )
where everything is evaluated at to. It is apparent from this formula that if we have two metric tensors gik and gik such that '2gik' then the measurement of angles given by gik and gik gik = would be the same, although ds = Xds. When two metrics are
proportional in this sense, they are called conformallyenuivalent.
Theorem 2. An orientable 2 manifold M with Cn+l+a structure which carries a metric tensor of class Cn+a can be made Into a
Riemann Surface in a natural way. Proof:
With the allowable coordinates (x,y) we associate the And with the corresponding
point z = x+iy in the complex plane. gik we associate the function µ(z,7) = µ(z) =
911 - g22 +
g11 + 922 +
21g12
2/g,1922-9712
5.9
44
then g e Ck+a and we have seen that 10 < 1 and it is apparent from the definition of p that a conformally To make M into
equivalent metric would give rise to the same µ.
a tiemann Surface we must distinguish which functions mapping a domain of M into the complex plane are to be called analytic. Namely, a function w defined on a domain D of M is to be called analytic if in a neighborhood of every point p @ D, w_ = µws E
This definition of which functions are analytic seems to depend upon which allowable coordinates are being used. Fortunately we have the following lemma. Lemma. If in f. neighborhood of the point p with allowable coordinates (x,y) the function w satisfies w = µ(z)wa then if the L
same neighborhood is given by allowable coordinates (a,y) then writing = x + 17 we have
w = 1+(Z)w Proof:
dz = zr. d4 + a dZ
da = zr. dr. + ad Z
ds2 = Idz + µ(z)dzl2 = Idr.(z
+ µ(z)zr) + dZ(zZ + µ('z)zI?
and since p is the same for conformally equivalent metrics
z4 + µ(z)az zZ + IL(z)sZ
wZ = wzzZ t w zZ = wzZ + µ(z)zZ)
z wZ = wszZ + wyzZ = wZ(aZ + µ(s)E)
5.10
Z_
45
+ µ(z)a
µ(Z)wZ = w4(zZ +
µ(z)z
w
= µ(r)w
R.E.D.
Now we have to check the axioms 1,2,3 for conformal structure. But properties 1,2,3 respectively of the solutions of the Beltrami equation state precisely what is required. Where did we use orientability? In the very definition of It. In fact,.this definition di3tinguishes between the first coordinate and the second, (since it contains the term g11-g22) and such a distinction is possible only on an orientable manifold. (On a non-orientable manifold we have that if (x1,x2) is an allowable coordinate system then so is (x2,x1). Quer l
Can any orientable 2 manifold with Cp structure (p > 1)
Because of Theorem 2 this only be made into a Riemann Surface? asks whether a Riemann metric can be defined on any orientable If we are willing to 2 manifold with Cp structure (p > 1). believe the theorem of T. Rado that any Riemann Surface is a
surface, it is apparent that the answer must be no, since the manifold would have to have a countable basis of open sets In In the next lecture we shall prove that order to be a surface. it is in fact sufficient that the manifold be a surface.
6.1
46
Lecture 6
The object of this lecture is to prove the following theorem.
On an n-manifold M with Cd structure (1 < d- < w) that satisfies the second countability axiom we can define a metric tensor gik E CC'l Theorem 1.
We must first develop a number of concepts and tools which will enable us to prove theorem 1. In the sequel everything we say shall apply to non-orientable n-manifolds with C3 structure.
However, in order to draw pictures to illustrate what we are doing and since that is all we need to answer the query of 5, we shall talk about orientable 2-manifolds with Cd" structure.
Definition 1.
By a disc of class C or simply a smooth disc, we mean the inversa image (under an allowable coordinate map 4,
with the associated allowable coordinate patch 0) of the interior of a closed disc entirely contained in 4(0). See diagram 1.
6.2
47
Definition la. On a Riemann 3urface S, we shall mean by a conformal disc, the inverse image (under a local parameter 4) of
the interior of a closed disc entirely contained in the image of 4.
If a 2-manifold M with C4r structure satisfies the second countability axiom, then M can be covered by a coutable (finite or infinite) number of smooth discs.
Lemma 1.
If M is compact, then we can cover every point by a Proof. smooth disc and select from this a finite subcovering. Whether
M is compact or not, it certainly has a countable basis of open sets. That is there exists a sequence {Gi} of open sets, such that every open set in M is the union of a subsequence of the G1. An open set G C M is small if G is contained in
Definition 2.
a smooth disc. Definition 2a.
An open set G C 5 (a Riemann Surface) is small
if G is contained In a conformal disc. Now we continue the proof of Lemna 1. Suppose we have an open set r c M, then every point p E r is covered by a smooth (Since p c r implies that there Is an allowable disc D C r coordinate patch o c r). But the smooth disc is open (the topological image of an open set), hence D is the union of a subsequence of the sequence {G,J, hence p is contained in a small G1. We see then, that a subsequence of (Gij consisting only of small GI already form a basis of open sets. So without any loss of generality we assume that all GI In the sequence {G1} are small. Then we have
Co
G2,
n
n
n
D1
D2
Dk
...,
Go
hence M C U ai C U Di. 1=1
Gk
Gl
1=1
Q.E.D.
6.3
48 Lemma la.
A Riemann Surface S can be covered by a countable
number of conformal discs.
Using the theorem of T. Rado (yet to be proved), S has a countable basis of open sets. The remainder of the proof is Proof.
exactly the same as the proof of Lemma 1. If M is a manifold with 0*' structure that satisfies the second countability axiom, then we can construct a sequence
Lemma 2.
aD
of domains fGij such that:
1. 91 C G1+1' 2. M = U G1, 1=1
ik
3. Gk = U Di where the D1 are smooth discs.
i=1 If M is compact, then the sequence (Gif with G1 = Mj
Proof.
I = 1,2,3,... satisfies the conditions of the lemma. case, we have a sequence of smooth discs
In any
which cover M. k Furthermore we can rearrange the sequence so that DI) fl Dk+l D1
('Ul1
Since if for some k we could not find a Dk+1 we could have
{ 0.
k (a)
CO
( U D1) (1
i= +1
1=1
connected.
D1) = 0 which is impossible since M Is
(
k2
Lot G1 = D1, G2 = U D where k2 is the smallest 3=1 J k
integer k greater than one such that
U.
There
Dj
JD1'
certainly exists such a
k
covers M and
since
is com-
kt+l pact.
Similarly we define GR+l = U
DJ where kf+l is the
smallest Integer k greater than k, such that Ut C exists such a k since (}t
is compact.
.
There
Also condition (a) Insures
that the 01 are domains. Lemma 23.
If S is a Riemann Surface, then we can construct a
sequence of domains 1G1J such that,
1. G ' i C 0 1+1
2. M
1k
co G1
1=1
where the Di are conformal discs.
3. Gk = U DI i=1
6.4
49
Using the theorem of T. Redo and continuing as
Proof.
in
Lemma 2.
Defininition 3. If ZGi} is an open covering of the topological apace M such that every point p c M is contained in a neighborhood which intersects only a finite number of Gi, then the covering (G,} is called a locally finite oovariggi If M is a 2-manifold with C(r structure that aatiefitae the second countability axiom, then there is a locally finite
Lemma _j.
covering of M by a countable number of smooth discs.
If M is compact then M has a finite covering by discs,
Proof.
which is certainly locally finite. In any case, we have the sequence of domains L Gi provided by Lemma 2. Since G1 is compact and G1:= G2, we can cover Z*1 by a finite number of smooth discs D1,D2,...,D
"l,
all contained in G2.
Now G3 - G1 is open,
cover each point In G3 - G1 by smooth discs contained in G3 - G1. 11
G2 - L'
i=1
is a compact subset of G3 - G1, select from the cover-
i1
ing cf G3 - G1 a finite subcnvering of G2 smooth discs Di1+1' DJ1+2,.... DJ2 .
UJ 1=1
Di .
Call these
Note that none wf the smooth
discs Dj1+1,...,D,2 intersect U. D
D
2
G3 D8
D4
6.5
50
Suppose that we have obtained discs D1,D2,...,D,i such that, it
TIC U Dk
1.
k=1
Then 01+2
for l < I c i, j1+l < k
Ei 0 Dk = 0
2. -
< 31 1 by a
G'1 is open, cover each point in Gi+2 i1
smooth disk and since Ut+1 - U Dk is a compact subset of k=l
i
G1+2 - t,i we can select a finite subcovering of 71 1+1
Dk.
k=1
Number these finite number of discs D , +1, D , +2,...,D. . i i i+l we have
.li+l
1.
Dk
1+1 1
and
2.
Now
1
Gt n Dk = 0
for 1 < I < i+l, j1+1 < k -4 ji+l
Hence we have defined a sequence 1D1l of smooth discs. 1Dif covers M. since it covers all the domains in kGi} which cover M. Also every point p c M is contained in a Gk, and Gk intersects at most 3k D1, hence {Di} is a locally finite covering of N by smooth discs. R.E.D. Lemma 3a.
If S is a Riemann Surface, then S has a locally finite
covering by conformal discs. Proof.
Use the theorem of T. Rado and continue as in the proof
of Lemma 3.
Embedding.
Definition
.
By a reoulsr embedding of class
CIr
(6 < p) of a
2-manifold M with Ca-structure (p > 1) into EN, we mean a homeomorphism 4, 4: M --> EN such that, given any point p t M in an allowable coordinate patch with allowable coordinates (x1,x2) and 3(p) _ (y
C EN.
6.6
51
1.
yi = yi(xl,x2) are Cd functions for all i..
2.
The rank of the matrix axi is maximal.
(that is = 2)
Given a regular embedding of M into an EN we could measure lengths of curves, by defining the length of a curve as the length of its image in EN. Hence it is not at all surprising that the embedding induces a metric tensor gik. Namely,
axi 2xk
ik J,i
Hence we see that Theorem 1 would be proved if we proved that we could regularly embed any orientable 2-manifold into an This was done by Whitney. We shall accomplish our purpose EN. by giving a regular embedding in f?, the Hilbert space of square summable sequences _Z_
That is _ E H means _ _
gi < co and we define the norm of cc
by
gi
17=1
We denote by Hk the subspace of H which consists of all points, such that all coordinates beyond the k-th are 0. Definition r. By a regular embedding of class C a (6 < p) of 2-manifold M with Cp structure (p > 1) into H we mean a homeomorphism d, $: M - H such that, If C is a compact subset of M, then 4(C) C Hk for 1. 2.
some k. If G is an allowable coordinate patch with compact clo-
sure, then given any point p E G with allowable coordinates (xl,x2) and 4(p) = have a. Ci = Ci(xl,x2) C C'
,0,0,...) we
for I = 1,2,...,k_
b. The rank of the matrix ex' is maximal (that is
6.7
52
Once a;ain we see that such an embedding induces a metric in M, and a metric tensor gik' Big °
2Xi axk
We now give a proof of Theorem 1 by constructing an embedding of the 2-manifold M Into H. There exist functions f(t) defined for 0 < t < 1
Proof (1).
satisfying
1.
f4
coo
2. f>0 for 0
f(O) - 1, f(1) = 0
4.
f'(t) < 0
5.
f'(0) = 0 ,
for
0 < t < 1
fk(1) = 0
for
k = 1,2,...
In fact one such function is
l r(t) _
°
e x e( x1) dx - t
-1e
e
e (
dx
o e (7t-1)2 dx
0
Let tDi} be a locally finite covering of M by smooth discs. each disc we associate the mapping F! F,: M -> 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,
can be continued along every curve in the domain Iz-bl < R shell then say that the function element 9 (R might be oo). has a branch point at z = b. Consider the set PQ of all function elements which are obtained from 9; by continuation in this domain. Then the number of function elements in Fp with center at any point c G Iz-bj < R is the same as the number of function elements in PA which have center a. This is so because there is a curve in the domain joining a and c, and distinct function elements with center at a must continue to distinct function elements at c along this curve. Otherwise we would continue back along the curve from c to a and the same function element at c would have different continuations at a along the same If the number of function elements curve, which is impossible. over each point is one, then in the domain Iz-bl < R there is defined a single valued function g(z) which is the value of a unique function element with center z. In this case we say that If over each point the order of the branch point b, is zero. there are infinitely many function elements we say that the order of the branch point b, is infinite. At thts point it might seem that there would be an uncountable number of function elements over each point. Later we shall see that there are at most a countable number, and we call a branch point of infinite order a If over each point there are m logarithmic branch point. (1 < m < oo) function elements, then we any that b is a branch It is obvious that the set PQ is an open, point of order m-l. connected subset of Se, and hence a domain on a Riemann Surface. Hence Pp is a Riemann Surface. we shall now show that all of these Riemann Surfaces are conformally equivalent. We exhibit a mapping which shows PO conformally equivalent to the punctured disk and depends only on in. We shall call Pp an unrainified,
8.2
68
First we give a topounbounded covering of the punctured disc. Consider in the logical mapping of the punctured disc onto Pp. t-plane, the disc Itl < Ri/m. Consider a function element 9 a PC with center a on the real axis. For simplicity we assume
b = 0.
Izl < R Rl/m is written uniquely as Any t c Itl <
t
=
It le i(2mk+ A)
= ltle id
for some A satisfying 0 < A < 2n/m and for some k = To find *(t) we continue 9 elong the curve -'which is the image of the curve r under the mapping z = tm. So that for every t we get an element of P9. The mapping 4, is onto, for at any z 6 0 < Izl < R we have m function elements with this center, 1 i (2Ttk + arg z) they are the images of Izlm e m m , k = 0,l,2,...,m-1. Are these ri images distinct? Yes, for if two values of k gave the same element of F` there would ue less than m elements of PO with center z, since any other curve with endpoint z is homotopic to one of the m we nave used. r'urthermore, for t' close to t, (t,t' not real) the k is the same and +y(t') would be an immediate continuation of *(t). For t,t' both real and close, *(t') is
clearly an immediate continuation of +i(t). For t real and arg (t') suffi^iently small or sufficiently close to 2R, and tat sufficiently close to (t'), (t' is correspondingly just moving away
69
8.3
from *(t) or just coming back close to it. Hence the mapping is topological. What does the function v(Q) (value of Q, i.e. if Q =
an(z-a)n then v(Q) = ao) look like if transplanted to
nthe t-plane?
It is a function regular in 0 < Itl < Rl/m, hence
o,
v(*(t)) _ = a0 3 and converges for 0 < Iti < Rl/m.
But since
we go from the t-plane to the z-plane by z = tm or a = b+tm, we
m
can write this as v(Z(Q)) = = ai(
(with Z(Q) - a),
m
-co
(Z(Q) = the center of 9, i.e., if Q =
an (z-a)" then Z(Q)=a).
And corresponding to each of the m values of rzwe get each of the m values of v(Z(Q)). Similarly, for a branch point at infinity we consider Izi > R, and we obtain 00
v(Z(Q)) _
If in either case there are only a
a (
-ca
J m 1z-b )J.
finite number of negative exponents we say that the complete analytic function SQ (P0 c SQ) has an algebraic singularity
over b. Now we return to consider the excluded points of the polynomial equation An-1(z)wn-1
P(w) = wn +
+ ... + A0(z) = 0
with meromorphic coefficients Ai. At a finite excluded point a, we have a punctured disc 0 < Iz_a( < R whicn contains no excluded points. Consider a function element Q with center in 0 < Iz-al < R, which satisfies P(w) = 0. If we consider PQ we will have 1
m1 function elements which satisfy P(w)=0.
If m1
another function element Q2 with center in 0<Jz-aj
w could be represented by
5_m
%P
8.4
70
But since w(s) can become infinite of at most finite order at z = a, we really have V /M
00
Furthermore if the complete analytic function 8G corresponding to the equation P(w) is algebraic, at co we will also have, j)(E)y/m
w =
The series that we have just been writing, power series, in a fractional power of the argument with a finite number of negative powers is called a "Puiseaux Series". What appears then to be a natural thing to do, would be to adjoin these new "function elements" to our space W. ("Analytiaches Gebilde")
Analytic Configuration.
We consider a new space classes of pairs of Laurent tions.
Its elements are equivalence series, satisfying certain condi-
Namely (z,w) c W means
and, there exists r > 0 such that for 0 < Itl < r both series converge 1. 2. if t1, t2 are in 0 < (tl < r then at least one of the equalities (a), (b) fail to hold.
a.
anti =
nt2 `-N
b.
,
tn
ao
n1 '; on2 tn
6.5
71
we shall consider two pairs eq'lvalsnt if they are obtained
by replacing T for t where Go
r =
5:
al 1 0
antn
n=
that is, if they are equivalent under a proper parameter change.
We shall now call these elements of 1'1 function elements. (Recall that what we have been used to calling function elements were really "regular unramified function elements over a finite point".) Remgrk.
This procedure can be carried through for n-tuples. Then the elements of the space are called "curve elements". This will be found in H. "eyl: Naeromorphic Curves.
A
Now we want to make W into a 'lausdorff space.
As before we obtain a basis of omen sets by first defining the neighborhoods.
Definition 1.
(z',w') is in an c neighborhood of (z,w) ant e
z =
w =
n Plat
if for a Itl1 c e we replace t by tl + (t-tl) = tl + t and
expanding in T we obtain (z',w'). 0' Is called open if every point in M is Definition 2. A set h C W contained in an E neighborhood contained in M.
Definition 1 depends upon which representation we use for the function element (z,w). However, this does not affect Definition 2, since a neighborhood determined using a particular representation contains a neighborhood determined by using any other representation. Now it is easy to see that Definition 2 makes W a topological space. (i)
If JOIJ is a family of open sets, then any function
element In ( Oi is contained in some Ok and hence in a neighborhood contained in 0k which in turn is contained in U Oi.
U 01 is open. i
I
Hence
8.6
72
(ii) if O1 and 02 are open sets, and the function element (z,w) a O1 (0 02 then (z,w) is contained in an el neighborhood contained in 01 and an e2 neighborhood contained in 02, hence an e = min (e1,e2) neighborhood of (z,w) is contained in 01 Pi 02.
Hence 01 (1 02 is open. Following the same route as the discussion at the beginning of lecture 7 it is clear that by a proper parameter change any function element can be represented in one of the following two ways. (1)
t = a + tm
w =
(2)
z =
t
w =
2 ntn (o
Pntn
If the representation turns out to be of type 1, we call "a" tie center of the function element (z,w). If m = 1 we say that the function element (z,w) is "unramified". If m > 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 <
e
.
Hence if I?1 > 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
Pick a non-zero x in P .
If x is not of lowest order,
then there exists a t1 E P such that t1 < x . = t2 E P or x = t1t2 .
That is
Similarly if t1 is not of lowest
1
order, we obtain x = t1t2t3 . But by the corollary to Lemma 8 we cannot continue this procedure indefinitely. we obtain a tk of lowest order.
Hence
Definition 7. An element t of least order at a place P is called a local uniformizer at P . Lemma 10. At a place P , every f e F - {0) can be uniquely written, f = t=og where t is a local uniformizer at P , and
g is a unit of ..C4 . Proof. Suppose f e 1-4p If f y P then f= z°f It f a. P , then by the corollary to Lemma 8 there is a m for which mt E P . That is tm = g e lI - P , or f = tig . Suppose f = t1°' g' , and ml > 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
Definition 8.
The Inverse, a-1, of a curve a is
a-1 W = a(l-t) 4 Consider the set of cloned curves beginning and ending at p e X . The relation, a homotopic to p, denoted by a....p is an equivalence relation. Lemma 1.
Proof 1.
1.
a"-a: Choose F(s,t) = a(t)
2.
a--p > p-a:
3.
for all a.
Choose FI(s,t) = F(1-s,t).
and p,-
a..,, r:
establish the homotopic
If F and G, respectively, a..-p, p.,,Y, choose
1-
`F(2a,t)
F
j
lG(2s-l,t)
0 < a < 2
2 < a < 1
Definition 9. Denoting the equivalence (homotopy) class containing a by [a]p , where p is the initial and and point of
a, we define the product of two homotopy classes with the same base point p by the equation [a]p (PIP = [ap)p
.
Lemma 2. Definition 9 is proper, that is, it is independent of the choice of representatives of the homotopy classes. In
and p...3
other words we must show that if Proof.
then
Obvious.
Definition 10. Defining curve [e] p by
[a]pl
e(t)=p
as
[a
p
and the identity
O
the set of homotopy classes of closed curves with and point p form a group, called the fundamental group of X with base point p .
It is denoted by
n(X,p) .
Since X is arcwise connected it Is clear that the fundamental group with base point p is isomorphic to the fundamental group with base point q, for any point q in X. However this isomorphism is not in general canonical. For if a is any curve going from p to q we have the isomorphism W given by (Jp([a)q) _ (Pap-1]p .
Clearly the isomorphism W
Such an isomorphis is called an allowable Clearly any inner automorphism is an allowable
depends only on [R]. isomorphism,.
isomorphism.
If f : X --),Y is continuous, then f induces a
Lemma 3.
homomorphism fp of it(X,p) into it(Y,f(P)) defined by * fp ((a]p)
[f^a]f(p) (f e a means f(a(t)) )
Proof.
We must show two things.
1.
If a...a' then faa-,.fall
.
12.1}
120
2 f aQ.fO3 1.
foa.p
If F(s,t) establishes the homotopy between a and at, the
f* F
establishes the homotopy between
fea and foal* 2.
Lemma L.
Obvious.
If pl,p2 a X, f : X -! Y is continuous, and if p
is a curve from pl to p2, then
t f - pfpl = fp2 p This is expressed by saying that the following diagram is It is indicated by shading the diagram.
f* pl
f*
P2
Proof.
Lemma 5.
Trivial since If
X - Y
fpl,f *
and
are homomorphisms.
f : X --3 Y
is the identity then
fp = 1 . Proof.
Lemma 6.
Obvious. If
X 14 Y -+ Z , f,g continuous then
(gf)* = g f *
12.5
Proof.
121
(Sf )p[a]p = [?fa]gf
p
fp[a]p = [fall g; [fa]f = [gfa]gf p Lemma 7.
If
f : X -y Y
p
p
is a homeomorphism then f* is an
isomorphism. Proof.
By Lemma 5 (ff-1)* is the identity, and by Lemma 6
(ff-1)*
=
f*f-1
hence f* is an isomorphism.
Theorem 1. X is simply connected if and only if trivial; i.e. consists solely of the identity. Proof.
it(X,p)
trivial.
1.
X simply connected > i(X,p)
2.
7c(X,p) trivial > X simply connected.
Suppose
is
a(0) = p(0) = p
a(l) = a(p) = q Since a(X,p) is trivial,
aR ti ep .
That is, there exists
a continuous F(s,t) with F(0,t) = ap-l
Consider F(a,2) = y
p
F(l,t) - ep .
12.6
122
II.
Covering Spaces
Definition 1.
fl
is called a covering it
1.
fl is onto.
2.
II is continuous. is open.
3.
Example 1.
X " X
Let X be a Riemann surface S, and X be the Riemann
Then a non-constant meromorphic function on S is a
sphere.
covering.
Definition 2.
A covering TI : X -4 X
is called unramified
or unbranced if it is locally a homeomorphism.
That is, every
p e R has a neighborhood VP such that 71V
point
r
p means it restricted to Vp), is a homeomorphism.
(IT iV P
Note that Example 1 is not an unramified covering. Renoeforth when we say covering we mean an unramified covering.
ode shall call this condition I. Suppose p a X
TI(p) = p e X .
and
that "p lies over p," or
Then we shall say
Furthermore a curve a beginning at p has for an image a = T'(a), a curve beginning at p. How about going backwards? That is, suppose a is a curve beginning at p e X , if p lies over p is there a curve a beginning at p and that T(a = a ? If the answer is yes we say that the curve a has been "lifted" to 'a. Theorem 2.
covers p".
If UP are curves beginning at p e X
TFa=TTp then 5-.p. Proof
a N
I--* X
I-XN
and
Let T denote the set of
t e I
for which a(t) = p(t) .
0 e T ; hence T is not empty. By Condition G. T is open, and by the continuity of a and b, T is close. Hence T = I. Q.E.D.
(Condition 5) A covering is called unbounded Definition 3. if every curve a In X can be lifted to every point covering its initial point.
Example 2.
X is shown by the solid line
, X by the
broken line - - - - .
p:. 1\
r TT is the projection along a radius. This covering is unbounded, however it is not unramified, hence p* does. not have a neighborhood in which Ti is a homeomorphism. Example 3.
X is shown by
the solid line -, X'by the broken line - - -
p
-
\
.tea
1218
124
is the projection along a radius. This covering is unramified but not unbounded since a
cannot be lifted to p We now consider the following problem. Given X., we want to describe all unbounded, unramified coverings of X. Condition
I.
Every point
has a neighborhood VP and
p e X
that T-l(Vp) is a union of disjoint open sets in X, and TT is a homeomcrphism of each onto Vp.
That is, for every
there is a neighborhood Vp and that
p e X
T[
1.
-1 (V
P
) = U VP^, a
a
Tj('a) - P
2. Vp
3.
is a neighborhood of pa a
Vp
is a homeomorphism onto VP CL
The Vp
Note.
Theorem 3.
are called distinguished neighborhoods.
If conditions 1-4 are satisfied, then
Condition 5 (ua) Condition 51. Pr,,of .
a)
51=5
Take a curve a in X with initial point p. the curve a restricted to the interval
Denote by at
0 < ti < t
.
Let
T denote the set of t for which at can be lifted. T contains 0, hence T is not empty. Look at a distinguished neighborhood of a point a(t) for which t is either in T or an accumulation point of T and it is clear that T is both open and closed.
12.9
T=I.
Hence
5 > 51
b) Suppose at, at
125
[a]-[al]
in X.
, preserving end points,
Then if we lift a,a' to [a] = [at] in X.
Looking at
the diagram below which shows the deformation of a into a', we see that lifting the curves at establishes a homotopy between a
and _C0, since by Condition t}, If is locally a homeomorphism.
This means
that we can cover each curve at by a neighborhood and move at slightly as at moves in this neighborhood. Since X is locally simply connected, given a p E X there is a neighborhood VP of p which is simply connected. Consider all
aT
pae
and all curves 3 with end point q lying in Vp.
Then lifting
to pa we get a
qa which depends only on pa and
q e VP .
Denote by Vp
a
the set p
of all qa. Clearly all the Vp of
1
l
are identical or disjoint, and
each V is in a 1-1 corresponwith VP.
Hence
P
is a
_ff IV
Pa homeomorphism onto.
It is also clear that
U Vpa = TJ'1(Vp ) Q.E.D.
126
12.10
Henceforth we shall assume that all coverings satisfy Conditions 1,2,3,4,(5,51). Two coverings
Definition 4. lent,
(1T1,X1) ;(172,X2)
"T: X2 %1
(Tf1,X), (TT2,X2)
are equiva-
if there exists a homeomorphiam
such that the diagram below is valid. X1
N
r
x2
T
X
Note.
Suppose two homeomorphisms Z', a (fl2= TT1Z, TFZa -ffl T )
coincide at a single point, say p2 of X2, than Z- 01. Proof . -r' (P2) = 0-(p2)
Take any q2 in X2 .
Since X2 is arcwise connected there is
a curve a2 joining p2 and '92 .
T2a2 = a has initial point
peX
1r1 2: a2 = 172a
since (Tr1,X1) -(lT2,X2 )
TT1Qa`2 ='R2a
since (TT1,X1),.,(JT2,X2)
Hence 71 r a2 = 1r1 a2
= a,
Z'a2
oat,
by the uniqueneat
127
12.11
of lifting, and in particular, 'Cq2 Definition 5.
Let
(TT,X)
=
,Tq2 Q.E.D.
be a covering of X.
Consider all
homoomorphisma Z: X_ c such that IT-C_ IT. Clearly they form a group. This group is called a covering group of T, and it is denoted CTT
From the last comment, it is clear that the group CTT is fixed point free.
That is, no It E CTT has any fixed points
except for the identity. Definition 6.
Two points
p1,p2F X are called equivalent,
p1- 2 , under C1T if there is a
C E CTT such that
T(Pl) ° P2 Furthermore, Condition 51 implies that the group properly discontinuous.
is
That is, every point of X is contained
in a neighborhood in which no two points are equivalent. For if two points pl,p2 in the same distinguished neighborhood
were Vp equivalent, their image under 17 would be the same would not be a homeomorphism.
and fl J.V
In fact, we see
p
that the elements ' of
only permute the distinguished
neighborhoods. Since by Condition 2 the mapping TI : X --),X is continuous, we can consider the induced map
TTp
s 'a(X,ji) ---1 lt(X,p) , TT(p) - p
12.12
128
It is easily seen that this is an "isomorphism into";
that is, it is 1-1, since homotopic curves in X are lifted to homotopic curves in X.
Hence
Gp a subgroup of
p
t(X,P) .
Theorem
If
lT(p1) = TT(p2) = p , then G'
is a conjugate
1
of G9 2. 11
17p1[P"laP] _ where
(a] s GP2
[P]"lTa] [P] ` Gpl
As a runs through %, we Get G-l .
.
Hence G51 = [P)-1 Gp-2 [P]
If Q is a subgroup of t(X,p) conjugate to Gp, then there is a point 33 covering p,
Theorem 5 (Converse of Theorem 11).
such that
Q = GV .
Proof. Since Q is conjugate to Gp, Q = [(3] Gp (p]"1, for some (P] a 7t(X,p). Clearly if we lift P to (i, a curve beginning
at p and ending at p', then GpF
.
Lecture 1 Definition 1.
A covering (TI,X) of X is called a universal
covering if i[ is simply connectei, in which case X is called a universal covering surface. Definition 2.
A covering (TT,X) of X is called regular if each closed curve a on X is the image under if of only closed curves on lX or of only open curves on X.
The group Gp = (a(7C,p)) is called the projected or defining subgroup of a(X,p). Liter we will prove a theorem which justifies this terminology. Notice that curves homotopic on X are lifted to curves homotopic on X. It follows that any covering space of a simply connected space is simple conr.ected. That is, all covering spaces of a simply connected space are universal covering spaces. know by Theorem 5, Lecture 12 that a closed a on X Is lifted to a closed curve a on if and only if [a] a Gp where is any point on a. Similarly, let a and G be two closed curves which begin and end at p E X. Let a and be curves beginning
at P and let T(a) = a, 'gy'(p) = p and fl(p) = p. Then a and 5 have the same end point if and only if [a)[P]-1 E Gp, that is, if and only if [a] and [p] lie in the same coset of n(X,p) modulo G.,.
Theorem 1.
The number of sheets in a covering (fl,X) of X is equal to the index of any projected subgroup GP in x(X,p). Froof. Let SpvJ be the set of points on X covering p e X. Pick Draw a curve av, beginning at pv and a fixed point pv ending at IS., for each pv, e It follows from the remarks preceding the statement of Theorem 1 that to each coset of it(X,p) modulo G9 there corresponds one and only one curve av, Cpv}.
V
and therefore one and only one point pv,.
Moreover, since each
curve av, is projected onto a curve in some one of these cosets, it follows that the number of points covering p is the index of Note that the index of G4 is therefore independent:of the G5 V pv Py choice py.
13.2
130
Connect q to p by a curve y. Let Qµ be the set of points covering q. Let y be the curve y L lifted to q,. The and point of Yµ is a point pv(µ) and v(µ) ' v(J.' if / µ', since lifting is possible only uniquely. Now let q a X, q # p.
Therefore number of number of points in
the number of points in qµ} is not greater than the points in 1pvf. Similarly lifting y to each pv the points in Jpv is not greater than the number of {qµ}. n.E.D.
Note that if X is simply connected, a(X,p) is trivial and the index of the projected subgroup is one. Hence the covering
(T,X) is one sheeted and TI is a homeomorphism. If X is simply connected, n(X,p) is trivial and the proHence the index of jected subgroup contains only the identity. Gp is the order of tt(X,p).
Then a closed curve at p is in
Supnose 17 is regular.
if and only if it is in all 0p
GP
v fore 0
= ('for all pv, e
pv
vr
for all pv, E
and by Theorem k, Lecture 12,
PV1
v
must be a normal subgroup of n(X,p).
G
There-
By the same theorem
v
it follows that if G is a normal subgroup for any
v; than
v
IT must be regular. Theorem 2.
(
11X1)
02,X2) <-> G
Trl(pl) = 1T2(p2) = p Proof.
1.
01,X1)
(IT2,X2)
homeomorphism t such that
X2
Xl
71
= G- where l
p2
= GG p2 91
There exists a
and 't(pl) = D2.
than there is a closed curve al
If [a] E Gp 1
on Xl such that a = TTl(al ) _ Tt2T(al) . Hence (a] a G G. C G-P2. Similarly GC G_ . Therefore G.__P2 a G-Pl pl P1 P2
2. G
1
and 2
8"2,X2).
p =) (171,X1) 2
171
All
Connect pl to ql by al. Let a = T,(61) and lift a to a2 at If q2 is the endpoint of For a2 we will show that q2 is independent of the choice of ai. Take a point ql on }C1.
T52.
suppose pl connects to ql. closed curve at p and since
Then if "1'( '1) ='1P0 ap-1 is a
is closed on X1,
ap-1
E Gp . 1
But
G
Therefore if p2 is p lifted to X2 at p2, then
52 .
x2321 is a closed curve on X2, and q2 is independent of the choice of al. Let 2: X1 " X2 so that 42 = C(ql) as described by the process above. It is clear that 't is one to one since 'Cl is described by switching the indices 1 end 2 in the previous paragraph. But z is also locally a homeomorphism. For we can take a neighborhood RI of '41 so small that 7l maps N1 homeomo:-phically onto a neighborhood N of q which is in turn so small that ri is lifted to a neighborhood N2 of 'q2 yields a homeomorphiem of N2 But then T is a homeomorphism of X1 -> X2. Since by and N.
construction 72Z =l the theorem. is proved.
132
13.4
Definition 3.
Let (T11,-X1) and (TT ,X2) be coverings of X and X2 such that (TT,xl) suppose there exists a mapping T(: 1
is an unbounded, unramified covering of X2 and such that X1. Then we say that the covering T12(TT(pl)) for pl 02,X2) is suborA?ihate to the cover g (TT1, 1 ).
An argument si nilar to that in part one of Theorem 2 shows that if (772,X.2) is subordinate to (711,x then G pl '2 Furthermore arguing as in the second part of Theorem 2 we see that G C G_ implies that (712,X2) is subordinate to (171'X1)' "1
p2
It follows that every covering is subordinate to the universal covering, since the universal covering surface has a trivial fundamental group. Theorem 3.
(Main Theorem) Let X Le a locally simply connected, arcwiso connected space. Then, given p E X and a subgroup
Gc: n(X,p), there exists a covering (TT,k) of X with G as its defining subgroup. proof.
Let a and a be curves from p to q oil X.
If [ap-1] a G
then we say a is equivalent to ,3, that is, a - p. Hence all curves beginning at p are divided into equivalence classes [(a,q)]. Let X he the space whose points the equivalence classes [(a,q)]. Ce.fine the mapping T7: X -- X by TT([(a,q)]) = q
.
Since X is arcwise connected, TT is onto. We wish to define a tupologv on X.
Call a neighborhood of a distinguished neighborhood of q if it is simply connected. Let a' be a curve from q which lies wi.thi.z a fixed distinguished neighborhood of q and has endpoint q'. We call the set of points q
+i
((aa',q') ] a distinguished neighborhood of the point ((a, q) ] on X. It is clear that using these distinguished neighborhoods on X as a subbase one gets a topoloa} on X. With this topology X is a Hpusiorff space. For if q and q' are distinct, these are
disjoint distinguished neighborhoods of q and q', and therefore
disjoint distinguished neighborhoods of any two points ((a,q) ]
13.5
133
and ((a, q')] on X.
The distinct points ((a,q)) and ((p,q)] have disjoint distinguished neighborhoods consisting of the points [(aa',q')) and ((pa',q')) respectively. For if E '1 which means that [(na',q')I = [(PP',q')I then G. But is a closed curve lying in a simply connected neighborhood of q. Therefore (ap-1) s G and our two points on X could not be distinct. It is evident that X is locally simply connected. To show that X is arci:ise connected, we will show that every point ((a,q)] on can be joined to the point [(ep,p)] by a curve. Since a: I - X, let at = al and let at be the point [(at,at(t)]. Now consider the curve a(t) = at for 0 < t < 1. Clearly a(t) joins [(ep,p)] and [(a,q)]. It follows that (,X) is an unbounded, unramified covering of X. 'io see that u is its defining group, note that a from ((e ,p)] to ((a,p)) Is closed on ?C if and only if curve that is, if and only if (aep1] = (a] E G. (ep,p)J = aa'(Pp')-1
(ca'p'-1p-1]e.
a'p'-1
U.E.D.
Theorems 2 and 3 justify our definition of G as the defining croup of a covering. We recall that the covering group C of a covering (fl,X)
consists of all home omorphiams r:}( --> X which commute with T F. Since an element of C is determined by what it does to a single point, the group C is fixed point free. Let l; be the defining
suboronp of (T1,X) and let :v = N(',) be the normalizer of G, that is, N is the set of all (a) a n(X,p) such that
Clearly, v is a normal subgroup of
[a]G(a]-1
= G.
and 'e can therefore
form the factor group N(G)/(}. Theorem
There exists a canonical isomorphism between N(G)/G
and C.
Proof. We shall give a canonical homo:norpntsm of i':(G) onto G with kernel G. Consider [a) C. i1(G). want to construct an element of C.
Let q t X and let curve p Join D to q, where
covers the point p. Let p = 1T(p) and let pl be the endpoint of a lifted to )i; let ql be the endpoint of p lifted to pl. Then
TT(o)
= f(nl) - p and Tf(i 1) = q.
13.6
134
want to show that ql depsnds only on q and not on the !Je wish to show that That is we wish to .., E G since
to q. join choice of the curve p. Let the and point of -Fl(") lifted Lop 1 is 41 (apY-a-) G. But = show that (ag(cy)-
pY_
pY-
a-1)
Fence since (a) a N(G), [apY-1 = [a](pY-11a]-1 3ut then lifted to p is closed, which proves the
is closed.
apY-1a-1
E G.
point.
Now let o: X -> X be defined by 4(q) = q1.
Clearly
t T? _ r, and r,' is a homeomorphis-n, so that o E G. We now define T : X -- X by 'c = T-1 and we map [a] onto T. Note that if a1a2 = a3 where al, a2, a3 r (G) then '.2crl = "3 or 631 = 61021 acid
l`"2 = 73 so that the mapping Is a homomorphism.
Since the kernel of this mapping Is clearly G, we have C is onto in order to only to check that the mapping of .,(C) complete the proof of Theorem 4. Let T be any element of C and (p) = p1. Then for any p E G, P is let a join p to p1 where 1 closed, so thRt = (aiiaand [a) E N(G). But then the image of [a) is T. Q.E.D. BLra_l
A Croup of mappings on a set is called transitive if any
two elements of the set can he mapped onto each other by an element of the group. Given any equivalence class of points on X, C can be viewed as a group of transformations on that class. It is evident that C will he transitive if and only if the defining group G is a normal subgroup of n(X,p). If (fl,R) Is a universal covering therefore, C is transitive and isomorphic to it(X,p). As an example, consider
a torus as the following rectangle with the indicated identifications. The complex
+ i
plane covers this torus in an obvious manner and gives a universal covering of the torus.
0
1
13.7
135
But now we are in a position to represent any apprupriste space by its covering space with identifications under the covering group. In the subsequent lectures therefore we shall inveattpate only simply connected Riemann Surfaces. Then we shall complete the classification of niemann Surfaces by deterndning, for a given Riemann Surface, the simply connected Riemann Surface which is its universal covering space.
14.1
136
Lecture 14 This lecture will he spent in developing some of the machinery that we will need in subsequent lectures. ye shall assume the following as known about harmonic functions u(x,y), that is, functions for which Du = uxx + uyy - 0. 1. 1.
Poisson's formula 2A
u(z) =
`
Y.(z,r.)u(r.)dQ
.. 0
where K(z,4) is the Poisson kernel. 2. If f is a continuous function defined on the boundary of a circle, then 2n
u(z) _1i J0
is harmonic inside the circle and has boundary values u(r,) = f(Z).
If un(z) is a non-decreasing sequence of harmonic functions, that is uhf, and u = lim un then either u = co, or u is harmonic. 4. Harmonic functions are invariant under conformal mappings. 3.
Hernsck's Principle:
A consequence of 4 is that it is meaningful to speak of harmonic functions on a niemann surface.
II. Subharraonic functions on a Riemann surface. The definition of subharmonic functions can be motivated by the following analogy with convex and linear. functions. One dimension:
linear function
f is linear if and only if
f° = 0
convex function f !s convex if and only if f" > 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 a = 0 2.
12%
v(0) <
v(reie)dQ 0
for all r < e. In 2, replace every neighborhood N(p) by some neighborhood 2a. 3.
If D r G, u harmonic in D, v < u in D, then either
v
v _= u
in D
1
} trivial
2 > 2a1
2 =) 3. v < u or v - u < 0 in D. Let E be the set of points p such that (v-u)(p) = C. tinuous.
£ is closed since v-u is con-
E is open since p E E inplies w(p) = 0 where w = v-u,
hence by 2a,
2'a w(p) = 0 <
w(retQ)dp
21-2
J0
14.3
138
and since by assumption w < 0 everywhere in D, we have
0 3 ==
Hence w < 0 on the boundary of D.
Let w = v-u.
1.
.
Let m = max w(p). There exist po £ D such that w(po) pi D If po D, then m < 0. If po C D, then v < u+m in D. v(po) = u(po) + m.
M.
But
Hence
for all p C D
v(p) a u(p) + in W(p) _-
in
M<0 A continuous function v defined In a region G is subharmonic if v satisfies any and hence all of the above conditions. Definition 1.
Mvximum Principle for S ibharmonic Functions. Weak form.
1.
If v is subharmonic in 0 and v < N. then either
v < M or v = M. Proof. Obvious from condition 3. Stron¢ form. If G is compact, v subharmonic in G, and v(p) < M, then v < i when r = U-G, p e G.
2.
lim
p-9Er roof.
Same as 3 ) I.
If vi(n), I =
sre subharmonic functions then the
following are subh-irmonic functions. n 1.
7-
Xiv1(p) = w(p),
X1 > 0
2.
w(p) =
3.
for po F G and id(p0) corresponding to Izl < z
max
i=1,2, ...,n
(vi(p))
v pv
for
p
tv(po
r2a t(z,2:)v(r,)d9
0
for p t N(p0)
14.4
139
1) follows from condition 2, 2) follows from condition 3, and 3) follows from 2a. Ferron's Principle. If G is any region on a Riemann surface and f(q) any real valued function on r = Zr`-G. Let 9 be a family
of all continuous functions v defined in G such that 1. v is subharmonic
2. pTEr
Assume
v(P) < NO
a)
3' non-empty
b)
sup V63
v(p) < co
all p c G
h(p) = sup vEI
v(P)
Then
is harmonic in G.
Remark.
If
< h and we assume
Ifi
3.
v(P) < N,
then a) and b) hold automatically, since -I9 E 3. Proof.
Take p
a G.
0
Choose v(p0) <-> (zj < F.
to shall show
Choose {vtJ , vn 3 such that that h(p) is harmonic in 14(p0). f Let V (p) = v (p ) --> h(p ). max (v (p)). Then n o n o i=1,2,...,n i
1. Vn
3'
2.
vn < V n
3.
VnT
n = 1,2,...
Then wn is harmonic in N(po) and wnI, and n vn < wn, and wn E. 1. By Harnack's principle
Let wn(p) = PV
.
w = lim wn(p)
is harmonic in N(p.)
Since vn(p) < wn(p) < h(p), h(po) = lim vn(Po) < w(Po) < h(po)
14.5
140
Hence h(po) = w(po). Let p' be any other point in N(p0). Let max Let Vn(p) = (vn(p)vi(P)). Let vn(p') -. h(p').
i'=1,2 .. ,n
Vn(p), w < wn. Hence wn = PV,. Fence wnT, and Vn(p) n But at po, w(po) = h(po) w' = lint wn is harmonic in ir(po). = w'(po), while w(p) < w1(p) in N(po), hence by the maximum In particular, w(p') = w'(p') principle, w(p) = w'(p) in N(po). = h(p') and hence w(p) = h(p) in N(p0). Q.E.D.
If 67 is compact and f continuous on r = Fr-G, and
Theorem 2.
the Dirichlet problem has a solution u, then h = u where h is the function obtained from Parron's Principle. f is bounded, hence h exists. u E 1, hence u < h. If Mm v-u < 0. Hence v < u, hence h(p) < u(p), v e j, then Froof.
hence h = u.
p--qcr-
A boundary point q E r is regular if for every N(q), there exists a function P(p) subharmonic in G such that 1 . 1 3 ( p ) < (, Definition 2.
2. 3.
3(p) s -1 lint p(p) = 0
for p j r(q) .
p--?q
Definition 3. Theorem 3.
r is regular if every point q of r is regular.
If G is an arbitrary region on a aliemann Surface,
and f(q) defined on r, the boundary of G, and f is bounded, and qo is a regular point of r at which f is continuous, then
0 Proof.
I.
lint
h(?) = f(q0)
h(p) > f(qo).
There exists an N(qo) such that
p-g0 f(q) > f(qo.) - e for q E 2,'(q0) n r Let P(p) correspond to
N(qo).
Let
w(p) _ (f(q0) + M)P(p) + f(q0) - e where lf(q)J < M.
qE r
Then w(p) E 7, hence w(p) < h(p),
hence
lim
h(p)
f(qo) - t
h(p)
f(q0)
,
p-'q0 hence
li.m
p-qo I7.
lfm h(p) < f(qo).
Since f(qo) < M, choose a so
p-'q0
that f(qo) + t < M. Choose i:(q0) so that. f(q) < f(qo + t Take any v 6
in i1(go) n r
then
v(p) + [M - (f(qo) * WP(P) < f(qo) + in N = N(q0) !) 0, by the maximum principle applied to i4, since
the left hand side is subharmonic in N.
Hence In N
v(p) < -[N - (f(q0) + c))p(p) + f(q0) + e
,
hence b(p) < -(n! - (f(q0) + e))R(p) + f(qo) + e
,
hence
Ii;a h(p) < f(q0) +
.p-'go and
Mm h(p) < f(qo)
Q.F.D.
p--qo
If r is compact, regular and if f is continuous on then h solves the Dirichlet problem.
Corollary. r,
Proof.
Theorem
Obvious.
If q f: y C r where y is an analytic are, then q is
regular.
Let q be the endpotnt of en analytic are. local parameter at p. Then Proof.
part of y near q i.e oD aiti, al j 0, 0
Y
Let z be a q
14 2
14.7
then on a sufficiently small circle in the z plane, z maps the circle 1-1 onto the Riemanu Surface. Then we have for the image of this neighborhood on the surface,
U
Now we map the circle by z' =,fa onto
z'(q)
and then by th3 linear fractional. transfor.^iution which map
0 -> 0, -1 -> -1-1, and 1 -a -1+i.
That is, we have in the
r, plane
then 0 = P,e (r) is the required function.
N.E.D.
Lecturo 15 In the course of the next four lectures we shall prove the Uniformization Theorem for Simply Con_zectei Riemann Surfaces. ','e begin by stating the following theorem which shall be proved in Lectures 15 and 16. Theorem 1. Every compact simply connected Riemann Surface is eonformally equivalent to the sphere. In order to prove Theorem 1, we shall first need the following. Theorem 2.
Let q be any point on a compact Riemann Surface S. Then there exists a function u, harmonic on S-q, which in terms
of a local parameter t that vanishes at q, can be expressed in a neighborhood of q as
u = Re
(
1.) + v
where a is an arbitrary constant, n a positive integer, and v a function harmonic in th9.t neighborhood. Consider p such that 0 < p < 1 < R, where 4 is a local Let Cp, C1, CR be the images on S of 141 = p, jr.E = 1, 141 = R. Outside Cp we can solve the Lirichlet Problem for zn -arbitrary continuous function on Cp. On Cp we take u(O(p,O) = a cos(p+p) where a,p are real numbers chosen so Proof.
parameter for I2:1 < R.
that 1.
max u ( p ) (1,9) = Cl
2.
min u(p)(1,9) _ -1
1
Ci
Ve shall prove the theurem fox a = 1, n = 1; that is, u = Re (1/c) + v. To get u = Re (i/Z) we would have to take u(p) = a sin (A+p); to obtain u = Re (1/e) + v, we would have to take u(p) = a cos n9 + The function f(P)
) = u(P) - 1uy0 to an analytic funcHence
tion of S for p < d < R.
144
15.2
5-an(P)n -00 The function F(P)(2:)
f(P)(z) dz
= f
za
is analytic in the annulus n.id Re (F(P)) = U(P). llenoe
hence writing 4 = to
(P)
-
u(P)(teiP) = a°
+
, we have 8(p)
00
In t + k-1
(a(P)tk + k
a(p)t-k)
-k
(P)tk + b(kp)t-k) and
27 a(P) + a(a) in t =
(1)
0
rt
(2)
a(P)tk k
+
a(p)t-k = 1 f -k
n
+_(P)(teig) dA
j
21t
u(P)(teie) cos k9 d6
1
JO 271
{3)
b(P)tk + b(k)t-k = i
u(P)(te1
)
sin kQ dQ
0
By the maximum principle, In(P)l < 1 outside C1. Taking t = 1, (1) implies
1.(01 < 2 Taking t = R, (1) Implies (5)
.
cos kA
sin kA
X5.3
14 5
Taking t = 1, (2) implies akF)
(6)
+ a(k' = Al ,
1111 _ 2
for k = 1,2,...
,
For t = p, (2) implies akp)Pk +
(7)
° 0
R(k)P-k
for
k = 2,3,...
For t = R, (2) implies ak(P)R k
(8)
(P) -k = A2 + a-k R
1x21 _ 2
for k
1,2,...
From (6) and (8) we obtain IA1
k
a(P) k
_
)2 IRk
R1
1
AR 1
k -
2
A-
k al - A R 2
R7 R-k)
hence jakP)j < CR'k, k = 1,2,... where C does not depend on p. Similarly, Ib(P)J < CR"k, k = 1,2,... From (7) and (8) we have k Ip
(P)
s -k
Rk
-
I
x 0
x p2kRk
(
2 pir P-cam2k Rk
2
,
R-kl
hence Ia(k)I < Cp2kR-k, k = 2,3,... Similarly lb(k)I < Cp2kR-k k = 1,2,... And la-11 < C. 't'hroughout the previous discussion C has been a constant independent of p. .
With the bounds that we have obtainod for the coefficients it follows that for p < r < 1 < R the functions u(p) are uniformly bounded, independent of p. For 1uPI < 1 + In r + C
In r+4C 11
2(()k CD
+
(ir)k+ E
r .
Q.E.D.
15.ti
146
Hence by the maximum principle the functions u(p) are uniformly bounded outside Cr. Lemma 1.
If U. is a family of uniformly bounded harmonic functions on a domain D of a compact Riemann Surface S, then (Lis a normal family. Proof.
If u is a function harmonic in a coordinate patch 141 < R, 4 = x + iy then au/ax is a harmonic function of Z. the mean value theorem
By
u(rei8) de
u(0) = 2nr
J
hence
2n
u(reic) r do
ru(C) _ 0 hence R
Jo ru(C)dr = 11
R .2n
0°00
u(reic ) r do dr
1I
1
u(0) =
u(reic) dxdy
nR
relation for the harmonic function 2)u/ax, we obtain,
!u
ax 0
n-1R2
_
1
j
u ax dxd y
uI dy < -1
(21(RM) =
2M s
where M is a uniform bound for the elements of Q. In order to complete the proof of Lemma 1, we need to derive
Harnack's Inequality-.
If u(z) is harmonic for )z) < R and continuous for )z) < R, then
u(O) R-r c
min
r<)zI
u(z) < max u(z) < u(O) R'-+rr - r
Froof.
To simplify the calculations, we assume that R = 1. Using the Poisson integral formula, we get 2n u(reic) =
2
1
1 - r
0
2rcos(0-iy)+r
1
u(ei*) dip
27[
1r2 2 -It
{l - r)
0
u(ei'') dip
21E
1+r I = __F 2 WE
u(ertr)dT so
max
u(z) <
il-+t_rr
min
u(z)
-r u(O)
r<)z)
u(O)
Similarly r
Q,.F.D.
Hence we see that a family of harmonic functions which are uniformly bounded at a point are uniformly bounded in a diso abbut that point. Suppose D is a compact subset of D. by a finite number of conformal discs
Then we can cover b in each of which we have a local parameter defined for )z) < 8 where 8 is some positive number. Then from the previous comments, in each Li the family of partial derivatives are uniformly bounded (these are not functions on the surface, but functions of the chosen local parameters). By Arzela's Theorem there is a subsequence un I which converges uniformly on L3 . Similarly there
is a subsequence un
of {ul which converges uniformly on A2. \' Continuing in this manner we obtain a subsequence Iun j which
converges uniformly on A1,D2,...,dk and therefore on all of 0.
15.6
148
Hence, given a compact subdomain D1 C D we have a subsequence full which converges uniformly on D1. Consider a sequence of compact subdomains Di, such that Di Di+l and ao
J Di = D.
i=1
Suppose we have a subsequence 1un of unj which
converges uniformly on D
.
i
Then by what was said previously,
we can find a subsequence fun+ll of lull which converges uniformly on Di+1.
The sequence {u n; cclearly converges uniformly on
every compact subset of D.
Q.E.D.
The functions u(p) satisfy the conditions of Lemma 1. Hence taking a sequence pi - 0 we have a sequence of functions u
(pi)
u
(pi)
which converges to a harmonic function u on S-q. The series which represents u near q is the series whose coefficients are the limits of the coefficients of the approximating function
u=
a_
Hence near q,
cos 0
l t
a_
Re (
a
+ a In t + 2 +
Co
(ak cos kA + bk sin kQ) tk
) +alnt+v
where v is harmonic. From the representation of u near q it follows that a_1 / 0. For if a_1 = 0, u > 0, u could not have a minimum outside of a small circle, and if u < 0, u could not have a maximum Furthermore u is not a constant since outside a small circle.
it has max u = 1, and min u = -1. C1
C1
Ve need the following result to show that a = 0. Lemma 2.
On a compact Riemann Surface S, if a function u is
regular harmonic outside a disc A which Is contained in a conformal disc A, and if u is continuously differentiable on the boundary of A, then the integral around the boundary of 0 of the normal derivative of u is 0. In order to prove Lemma 2 we need the following definition.
15.7
Definition 1.
149
A density is a rule which assigns to every point
and local parameter a number 3(z) such that 4(z)dxdy is invariant. That is, if we use the local parameter r, = E, + iq
instead of z = x + iy, we should have 4(z )dxdy = 3(z(r.)) If f is a function on S then of depends on the point
Example.
aLid local parameter.
However,
Qr =
2
t} 32az
and
f
=
a
Pf se's + -L i vz
az ,;z
az az
hence
AS =
;2f = 4 "2f
2
dr.:z 1jZl
or
-Af (x,y)
= Lf
i(x ti'
(t,r) `
F. )a
We see that while it does not make sense to integrate a function over a domain on a Riemaan Surface, it does make sense to integrate a density. Nov we shall prove Lemma 2. Ve consider a finite covering of S by conformal discs Gi for which we may assume that A r- G1. Then we consider a partition of u.u ty subordinate to this cover-
ing. do this in such a way that the functions ini and their vanish Ti athe complement of G Furnormal derivatives thermore we require that :e 1 is identically 1 in a disc
contains Q. Then since 7u; i G=
!
S-Q
uL
r
1,
UAWi=
S-Q
i
X
which
'L&
Gl
uQWI
+ 1*1
Gi
.
150 Denoting the boundary of Gi by yi and the boundary of by b, we apply Green's identity to obtain
(u ni - wi au) ds = 0
uAidi = Gi
Yi
For i = 1, U&I'l = Y'b
(u and - u '1 n)ds = - Q+ an ds
Hence
fd8=O
Q.E.D.
Now if we apply Lemma 2 to the function
u = Re (.)+bint+v and a conformal disc S about q we have
0duds=b 2Intds=b rds 8 hence b = 0.
,JJ
Q.E.D.
J
Lecture 16 Continuation of Moromorphic Functions Definition 1. By a function element on a Riemann Surface 3, we mean a domain D C S and a meromorphic function f defined on D. We denote the function element by (f,D). Definition 2. If two function elements (f,D), (f',D') are such that 1) D fl D' f'DnD' the two 0 , 2) = f!IDnD' , then function elements are called immediate continuations of one another. Proceeding as in Lecture 7 we define continuation along a curve and prove that if continuation is possible then it is possible in only one way. Suppose we have a set of function elements an 8 the following properties, 1.
p e S > p is contained in the domain of a function element.
continuation of a function element is possible along every curve in S. 3, any two function elements in this set are continuations of one another. Consider a point p E 3 and a conformal disc about p given by a local parameter r. with 4(p) = 0. We shall call a series 2.
co
anZ
a Laurent series on S, if the series converges for some 0. From now on, what we shall mean by a "function element" is an equivalence class of Laurent series, where two Laurent series are called equivalent if one is obtained from the other by an allowable parameter change; that is, a parameter change of the form co
where
ane
ane converges for some 4 jf 0.
al { 0
152
16.2
As in Lecture 7, we may doftne immedinto analytic continuation, and placo the scime topology on the set consisting of all function elements. Let us call this space of all function elements S.
It Is a Hausdorff space and locally simply connected. "lift" the conformal structure of S to S to give a conformal structure to S. We have a natural mapping
t'e
T(: S --> So. The image of a function element under Ti is its center.
The connected components of S are Riemann Surfaces.
On
each of these connected components a meromorphic function is defined, namely, the function whose value at a point (function element), is the value of the function element at its center.
Consider a connected component S of S.
The i:vvpping U: S --> S
is a covering of S. If the function elomeiit which defines S can be continued along every curve on S then this covering is unbounded.
From Lecture 14 we know that every unbounded covering That is, TI is a homeo-
of a simply connected space is trivial. morphism.
So. Suppose we have a tpl'p2' "''pn1 = harmonic function h defined on So, and in a neighborhood of pi, ni). At every point there is a function element h = Re (a1/2:
Consider S -
(f,D) such that h = Re (f) in D. Since. h is defined on all of S this function element can be continued alung every curve. Hence we obtain an unbounded covering of S. When does continua-
tion along curves y, y', with the same end point, yield the same
result? Clearly v-hen y, y' are homotopic.
(In Lecture 14, we
showed that homotopic curves in the base space, when lifted are homotopic.) Suppose a is a closed curve on S beCinning and ending at the center of a function element f. Continuation of f
along a, will, in general, add some imaginary constant to f. (Re f is a single valued harmonic function on S.) That is, after continuation we obtain f + ic, ubere c deeds upon the homotopy class of a. Hence if p Is the center of the function element f we can consider the mapping 4f defined on xr(S,p) by f(ta)) 3 c
.
16.3
153
Clearly 4f([a)[P)) = 4f([aIl) = 4.,([a)) + 4f([P))
.
Fence we have a homomorphism of :.(S,p) onto a subgroup of the additive group of real numuers. Clearly the kernel of 4 Is the
commutator subgroup of x(S,p), that is 4 is an isomorphism of the one dimensional homology group of S onto this subgroup. In a previous lecture we Saw that a non-constant single valued meromorphic function on a compact d+emann Surface permits The space the surface to be triangulated in a particular way. of this complex is homeomorphic to a certain topological polygon (with the usual identifications) of 1g sides, where g is the genus of the surface. Now we shall construct a single valued non-constant meroFrom Lecmorphic function on any compact fiiemann 3u.--face S. ture 15, we know that there exists a harmonic function $ with a first order pole at p E S. Similarly there is a harmonic function 4' with first order pole at q e S. Let s be any point on S iy a local parameter at a. Then jX - i4 and and z = x I
*X - idly are analytic functions of Z.
Fowever
X-4Q)/(*X-i*y
is a meromorphic function independent of the local parameter z. That is, it is a function on 3. For if r, is another local parameter at a, then
z
3
3z dz
Consider now the corresponding topological polygon of
lea sides vi.th the usual identifications. Then any closed curve beginning and ending at
p is homotopic to a product of the ai and bi. That is, the fundamental Croup has these 2g curves as gonerations.
16.4
154
Consider 2g
Let a be a harmonic function defined on S. generators al,a2,...,a2g of These 2g
the group 7t(S,p).
a2g
curves aro called a homology Locally, at p, there is an anatytlc runction r such that d = Re (f). If we basis.
3
continue f around ai we will,
in general add an imaginary
constant to f. CR11 this constant (¢)c .
Then
will be the real pazt of a stngle-valued
i
C, I = 1,2,...,28.
analytic function if of
ai
+ (V)
ai
Clearly
and if A is a constant, (X4) ai = x(d) ai .
Cn a closed Riemann Surface S of genus g there exists at every point p 6 3 a non-constant meromorphic function with a pole of order at most g+l at p, and at no other poles. Theorem 1.
vote that this is the theorem we used without proof in Lecture 9. We know from Lecture 15 that there exist harmonic functions 41 and i1 regular everywhere in S-p, and at p Froof.
Re () + v,
J = 1,2,...,g+l
z
dpi=Re (i) +uj
j=1,2,...,g+1
Form +1
_
(a1
4.
+ ,,. J*J )
where R3, µJ are real numbers.
f is harmonic with a pole at p Consider the equation
of at most order g+l. (
= 0
)Q v
for
v - 1,2,...,28
16.5
155
That is, (1)av =
f i;y
0
(a1(41)av +
If these equations are satisfied then is the real part of a single valued analytic function. But we have 2g equations and the 2g+2 unknowns X v µi, 1 = 1,2,...,g+). fIenee there is a non-trivial solution and I+ is the real part of a non-constant
single valued analytic function whose only pole is at p and there the pole is of order at most g+l. Q.E.D. Actually somewhat more is true. Given any g+l distinct oositive anteaers 0 < kl < k2 <...< kg+1 and a point p on a compact itiemann Surface of genus g, there exists a single valued non-constant analytic function whose only pole is at p and the drder of the pole there is one of the numbers ki. Theorem 2.
Proof.
Exactly as for the previous theorem.
Definition 3. Consider p 1 3 a compact
Surface.
Then
n is called a pap value at p, if there does not exist a single valued meromorphic function with pole of order exactly n at p
and no other poles. Theorem 3.
(Weiorstrass)
At every point p
S of a compact Riemann Surface of Bonus g there are exactly g ;dap values. i_-
We shall not prove this theorem. However it is obvious from the last theorem we proved that there are at each point, at most g gap values. Theorem I. On a compact Riemann Surface of genus g all gap values are less than 2g-1.
Proof. 4!e first note that if k is a gap value and if fa+n = k then either m or n is a gap value.
Par otherwise the product of
functions fm and fn w'iich have poles of order m and n respectively will have a pole of order a+b = k. Hence if a > 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
3.
max u < m, Y
then u < m in S-L. Let
Proof.
v =
f max
(u,m) - M
m-M
in S-Q
i n/,
Then v is a napative subharronic function in S.
Hence v = m-M.
Hence u < M in S-o. In the same way we would have a mininium principle for sup-
erharmonic functions bounded from below. "e also obtain the following maximum modulus principle for bounded analytic functions. Let z be a local parameter which vanishes at p E S, a parabolic Riemann Surface. Let A correspond to IzI < r and Y correspond to (z) = r. If Lemma 2.
17.2
158 1.
2.
3.
f is analytic in S-4 IfI < M in S-G max If) < m Y
then
if1
Proof.
< m in 3-U. Let
max (lg If 1, lg m) - lg M in S-Q
lgm - lgiinA
then v is a negative subharmonic function in S.
v = lg m - 1g N.
lg
Hence
If1
Hence
< ig a in S-,L. Hence if I
<m
in S-L. y.
if
Let f be an arbitrary continuous real valued function on !-'e shall call u a proper soluti.on to the Dirichlet problem 2.
u is harmonic in S - (L+y) u is continuous in S-A
3.
min f' < u < max f in S-L
1.
Y
Y
Let z be a local parameter which vanishes at p c S, a parabolic Riemann Surface. Let Q, correspond to 1zI < r and Lemma 3.
If f is an arbitrary real valued conticorrespond to 1z1 = r. nuous function on y, then there exists a unt.yue proper solution to the Dirichlet problem. Proof.
Let R > r.
Let Q '
correspond to Iz1 = R. 1.
correspond to r < 1z1 < R.
Let v be a function such that
v harmonic in /'
2.
v= f ony
3.
v = min f on S -
Let w be a function such that
w Is harmonic in L' 2. w= f onY 3. w = ,ax f on S 1.
Let Y'
17.3
159
Then v is subharmonic in 3-L`j and w is superharmonic in S-Q. Let f be the family of all functions u such that 1.
u is subharmonic in S--
2.
v
.
Then by Perron's Principle, f = sup u is a proper solution to ulf the Dirichlet problem. Suppose f' is another proper solution to the Dirichlet problem. Then by the unaximum principle for subharmonic functions f - f' C, and by the minimum principle for superharmonic functions f - f' > 0. fence f = f'. Q.E.D. Now using precisely the same mattod as we used for compact surfaces we can obtain at each point p t S functions up. up harmonic in S - ip}, bounded outside any neighborhood of p and near P.
u' = Be (1) a a ig Jz + v' p z and
u'p'=Re (Z) +blg Hence
up - u'p' = Be (2) ) + v
up =
a
Now, if S is simply connected we have up = Re (fp) where fp is analytic in S, has a simple pole at p, and has bounded real part outside any compact sot containing p. Hence we can take a sequence of real numbers An so large that fp = 1/(? - A satisfies fp has a single pole at pn 1. n is bounded outside ovary compact set 2. Ifp n containing pn I
3.
pn--* P
Let z be a local parameter which vanishes at p.
Let p
n the functions fp A are non-constant in the region 8 corresponding to the annulus
correspond to zn, and suppose IzaI < r < R.
17.4
16o r < IzI < R. 1.
2.
Hance we ca' normalize the functions fp so that n = 2 max If 8 pn I
min I f pn 8
I
=1
are uniformly bounded in 8. Hence there n to which converges uniformly in 8. exists a subsequence fp n can be written terms of the local parameter z, fp n
The functions fp
'
'
s
fpno
= hT(z) n = z nZnz)
where sn,(z) are regular functions.
Then
sn, (z) = (z-zn, )hl,, (z ) therefore
Isn-smI _< Ihn-hmI Iz -znI + Ihml Izn - zmI Hence sn converges uniformly to a in b and therefore in (Q+y+b) corresponding to IzI < R. Also 1,/(z-zn) converges to 1/z.
Therefore hn converges uniformly to s/z = h.
Since fp
n is bounded outside (A+y+B), Lemma 2 guarantees that the sequence fp converges on S-p to fp. Since fp is non-constant n
and S is parabolic, fp has a pole at p.
Hence we have proved
At each point p
3, a sLeply connected parabolic Riemann Surface, there exists a function fp meromorphic in S and satisfying Theorem 1.
1.
2.
fp has a simple pole at p, If pl is bounded outside every compact set containing p.
"e shall call a function satisfying conditions 1 and 2 an admissible function at p.
Suppose we have two functions fp and fp with these properties at the point p. Then, in terms of a local parameter z which vanishes at p,
fpa+j azn n oD
z
f'p
"n=
=2+rzn z
t _=U
whence n fp - fp is bounded in a neighborhood of p and hence throughout S. Since S' is parabolic, (a/a)fp - fp = c, a constant. That is, fp = a'fp + P'. If q is close to p, then tho function 1/(fp - fp(q)) is sn admissible function at q. Hence we see that for a given fp the set F consisting of points q £ S at which the fq are linear fractional transformations of
fp is closod, open and non-empty. Hence F = S. Lemma 3. An admissible function is univalent. Proof.
Suppose fq (p) = fq (p'). p
= a(f) p where X
We know f q
p
o
is a linear fractional transformation.
Hence
X(fp(p)) _ X(fp(P'
hence fp(p) = fp(P')
hence
P=P'
Q.H.D.
it follows than that in admissible function f is a conformal homoomorphism of S into the sphere. However f can only omit one point.
For if f omits two points zo, z1, consider the map-
ping of f into the z sphere followed by
z-zl /z
io
This is a single-vnlued function on S once we pick its value at one point. Hence if the value 41 1s taken on, the value -Z1 is not taken on. Furthermore, since the
17.6
162
map is open, a whole neighborhood of -Zl is not taken on. letting Z1 be the image of some point on 3 the function
h = Re (-
77-1
Hience
)
is harmonic on S and bounded from above, which is impossible on a parabolic Riemann Surface. In conclusion then, we have proved the following theorem. Theorem 2.
Every simply connected parabolic Riemann Surface is
conformally equivalent to the plane.
In the next lecture we shall complete the uniformization theorem by showing that all other simply connected Riemann Surfaces are conformally equivalent to the open disc.
Lecture 18 Definition 1. A Riemann surface S is called hyperbolic if it is not parabolic, that is, if there does exist a non-constant
negative subharmonio function on S.
Hyperbolic Riemann surfaces are also said to havo a "positive ideal boundary".
Let 0 be a conformal disc on the hyperbolic Riemann surface S. Then there exists a function win S - Q satisLemma 1.
fying 1. w is harmonic in S - 0 and is continuous on
o-L1=r
2.
Q = 1 on
3.
O<W<1
Proof.
I
in S - 0
Let * be a non-constant positive superharmonic func-
i has a positive minimum on f' . We can assume
tion on S..
Then t < 1 at some point in S - L. Let F be the family of all functions defined in S - 0 satisfying
min i = 1.
1.
v subharmonic
2.
O
We must show F is not empty. Let correspond to jzj = r, and y to jz, = R > r; and let 8 be the region corresponding
to r <
IzI
< R. in 8
V = 0 Clearly v e F.
in 3 - (AU8)
Hence, by Perron's Principle w = sup v is vtF harmonic and has the properties required. Q.E.D.
18.2
164
Definition 2.
A function G(p,q) is a Green's function on S
if, 1.
G(p,q) is a harmonic function of q in S
2.
G(p,q) > 0. If z is a local parameter which vanishes at p then
3.
in a neighborhood of p,G(p,q) + log !zI is harmonic. 4.
If H has properties 1, 2, 3, then H > G.
Theorem 1. At every point f on a hyperbolic Riemann surface S, there exists a Green's function G(p,q). Proof.
Let F be the family of functions such that u e F if
and only if
1.. 2..
u>0 u subharmonic on S - lp'}
3.
u has compact support
4.
If z is a local parameter which vanishes at p then u + log jzi is subharmonic near p.
Let Q correspond to )zj < R where z is a local parameter which vanishes at p. Then the function R
log
in
Izl
U =
inS -
0
/j
is in F, so that F is non-empty. We must now show that F is uniformly bounded in the exterior of any open set about p,
and therefore bounded at every point of S - -spy .
Let r
correspond to Izj = R, y to Izi = r < R and b to r < IzI < R.
Now let w be the function described in Lemma 1 with Q corresponding to Izd < r. If we set X = max C.), we get 0 < X < 1.
r-
ForueF, (max u) 0 - U Y
165
18.3
is superharmonic outside ,&, and (max u)(J - u > 0 Y Hence
max u < max [(max u) w) r - r Y
l(max u) Y
But
max u < log 1 + max v Y r Y
where v is subharmonic in A. Hence + A(max u)
max u < log r + log l + Max u < log -
hence
(1 - A) max u < log 1 Y
rB
or
max u < 1 Y Since u < 1
log r8
1 1 log rR
everywhere outside 1zle r.
Applying
Perron's Principle to the family F we obtain g = sup u ueF harmonic in S - p. At. p, g + log Iz{ is harmonic. Hence g has properties 1, 2, 3. Suppose H has properties 1, 2, 3. Consider u - H, u e F. This function is subhariuonic everyOutside a compact set u - H < 0. Hence u - H < 0 where. everywhere. Thus for u e F. u < H. and g = sup u < H. Q.E.D. Note that since H - G > 0 and harmonic we either have
H > G or H3 G. Definition 3.
We shall call a function Pp admissable if
18.4
166
1.
fp(P) = 0
2. IfpI < 1 3.
fp analytic on S
4.
If hp has properties 1, 2, 3 then (hpI < IfpI
Lemma 2. If p is any point on a simply connected hyperbolic Riemann surface than there exists an admissable function fp. Proof.
Consider the Green function G(p,q).
Near p,
G(p,q) = -log Iz_f + u where u is harmonic.
In some neighborhood of p, u has a conjugate harmonic function v. Form the function fp = ze-(u
+ iv)
Then log IfpI = log jzJ - u = -G.
Since G has a conjugate H at every point of S - {p} , we may continue fp. At any point q e S - fpf ,
fP = e-(G + M)
It is clear that fp(q) = 0 if and only if p = q. fp is analytic on S, and ifpI < 1 since G > 0.
Furthermore Now suppose
that hp is a function satisfying properties 1, 2, 3. -log fhpj.
Consider
Since hp has a zero at p, -log Ihpl > G = -log Ifp{
from which it follows that Ihp{ < Ifp1.
The first inequality
results from the fact that -log Ihp, satisfies the first three conditions of a Green's function. Consider
4(P) =
f (p) - f q (g) 0 q° 1 - f lglfqo (P) qo
It is easily seen that j4(p)J < 1, 4(q) = 0.
Hence
167
19.5
I4(p)I
Ifq(P)I
If we let p = go we obtain
.
Ifq (q)I .< Ifq(go)I 0
Hence
Ifq (q)I < Ifq(g0)I 0
and
I4(go)I = Ifq(go)I
That is,
IfI
= 1, so
.
ti-.at
Ifg1
4 = e-i9fq Therefore fq (p) = fqo (q) implies that 0
p = q ,
and fp maps S 1-1 conformally into the unit disc. All that is left to show is that fp maps S onto the unit disc.
he could now complete the uniformization theorem stated in Lecture 15 by utilizing the Riemann mapping theorem. However this is not necessary, for the following simple argument shows that fp maps S onto the unit disc. Let a be a boundary point of fp(S) in the w plane.
We
can select a single valued branch of log(w - a) = 4. This maps Let the image of S into A domain contained in red < log 2. a be 2; o.
Now we map the half plane Re4 < log 2 conformally
into the unit disc taking 1: o into 0.
Call this composed
mapping of S into ICI < 1, g. Since IgI < 1 and g(p) = 0 it follows that IgI < Ifp1. Let pn correspond to an a sequence
168
18.6
of points in fp(S) which converge to a. Ig(pn)! -4 1, hence Ifp(pn)I -+ 1.
It follows that
Hence every boundary
point a has modulus 1.
Q.E.D.
As a consequence of the uniformization theorem we can now prove Radots Theorem. Rado's Theorem. of open sets.
Every Riemann suface S has a countable basis
Proof.
Consider S the universal covering surface of S. S is conformally equivalant to the sphere, plane, or disc all of which have a countable basis. Fence S has a countable basis, hence S has a countable basis.
Consider an arbitrary Riemann surface S and its universal covering surface S. We recall that the covering group G is a group of conformal homeomorph.isms of S onto itself which is transitive and fixed point free. We now ask: What Riemann surfaces have the sphere as
their universal covering surface?
The only conformal homeomorphisms of the sphere onto itself are of the form
z+
az+b
cz + d
ad-bc#0.
The fixed points of this mapping are given by the roots of cz2 + (d - a)x - b, so that every such map has at least one fixed point. Thus G contains only the identity mapping.
Since G is isomorphic to the fundamental group of the original surface S, it follows that S itself was the sphere. We might note that in this case, (S} , the field of meromorphic functions on the sphere is the field of rational functions. We know that on any Riemann surface S we can introduce a conformally invariant metric ds2 = x2(dx2 + dy2)
169
a6.7
The Gaussian curvature then is
Tt a -
log a xc
-
Can we define on S a conformally invariant metric such that K is a constant? If so, is the choice unique: can, then the Gauss-Bonnet formula give us
J1
If we
KdA = 2n)C = 2%(2 - 2s)
S
where g is the genus of the compact orientable surface S. We see that g = 0 implies K > 0. g = 1 implies K = 0, and g > 1
that K < 0. We know that the sphere has a metric with constant Gaussian curvature. Suppose the universal covering surface of the Riemann surface S in the plane. Then the only elements of G are mappings of the form z+ = z + b. The metric ds2 = %2(dx2 + dy2), X a constant is invariant under transla-
tion and hence under the covering group.
Hence on any surface whose universal covering surface is the plane we can uniquely
define a conformal metric for which K = 0. Suppose the universal covering surface of S Then G consists of mappings of the form
=az + b
fz +a
Ia12 > Ib12
The Poincare metric Idz12
ds2 a
(1 - Iz12)2
is invariant under G, since
.
is the disc.
18.8
170
Jdz!(JaL2
Is?
Jb12
Ibz + a12
Idz'1
1-
-
12
1
az+b2 ($z + a
Idz,(jai2 -
Eb12)
'bz + al2 - faZ + b12
IdzI
1 - (z12 Hence
X =
1
and
log (1 - Iz12)
0 log a
azcaz
hence
K = -4 Hence the surfaces S for which the disc is the universal covering surface carry a confor:al metric with constant negative curvature. Hence the universal covering surfaces of closed surfaces of genus 0, 1, and g > 1 are the sphere, the plane, and the disc respectively. The problem of classifying all closed fiemann surfaces is reduced to the problem of classifying all properly discontinuous fixed point free groups of conformal self homeomorphisms
of the plane and the disc.
The remainder of this lecture is
devoted to the case of the plane. Since the fixed points are given by the roots of cz2 + (d - a)z - b, we must have c = 0, if is to be the But unless a = 1, we only fixed point. Then z' = az + b. b as a fixed point too. Thus G consists only 1 - a of mappings of the form z' = z + b; b is called the period of the mapping. G will be discontinuous if the set of periods
have z =
Since G is a group, the set of periods must be a module (vector space) over the ring of integers. Thus each group G corresponds to a set of ccmplex numbers M, M being the above-mentioned module over the integers. is discrete.
Let us consider one such M. that lies closest to the origin.
Let tJ be an element of M
( w need not be unique).
It is possible that M contains only elements of the form k c
,
k = 0, +1, +2,..9 .
z' =z+kw. Definition 14..
Then G is the set of elements
A function f is called automorphic under a
group G, if for all g e G, f(z) = f(g(z)). . Then G contains the translaThe functions sin z, cos z, etc. are automorphic functions under G.
Example. tion
Let M = .(z + 21tk)
z' = z + 2nk.
If f e {S} then there is a naturally defined function e { S.which is automorphic under the covering group.
Con-
versely if f e iS } is automorphic under the covering group then there is a corresponding function f e {S} . Hence if S is the plane and the covering group G consists of the transformation at = z + kci
k = 0, +1, _2;...
18.10
172
then .[S) , the field of meromorphic functions, consists of
entire functions with period W . Setting r, _ 3, and letting S be the Z-plance, G consists of the translations 2:1 = Y + k k = 0, +1, +2,...
takes the strip
The mapping w = 21i log
0 < Red < 1 onto the once punctured w sphere.
There the field
of meromorphic functions are functions of the form cc
anwn
n=-co 00
n=-0o where w = e2ni4.
bnwn
Hence (S} consists of functions of the form
I:cc a 500
b n=_00 n
a2nn
e2z[inZ
Now suppose M contains w, which is not of the form k W, That is, w does not lie on the same line through the origin
as c does.
Let (.it be an element of Pi which is not among the
element ke.w, but is closest to the origin.
more than one such wt.
see that
1
There might be
Choose one closest to w .
Then we
, w generato all elements of M.
For if there were another s)" there would have to be one
in the parallelogram 0, W, c + w , i,it which contradicts the choice of Cit.
Hence the group G is generated by two trans-
lations with periods
w
W, C.1t .
Clearly Im ( 3, ) > 0. Call
The group can also be generated by (nl, Wt, where
G)1= acJ+RWt
18.11
G:i=YW+6wt
,
173
a6-PY=1
Hence the group is determined by 'e up to,
Lt=
az+ d . YC+ 6
{S} will then correspond to the field of elliptic functions
with periods ',.r, wt .
19.1
174
Lecture 19 Theorem 1.
The fundamental group it
of a Riemann Surface
S
is countable . Proof:
Since the plane and the disc are countable unions of compact sets it follows that any Riemann Surface S is the countable union of compact sets Di. Since the set of points equivalent to a point p c S is discrete, it follows that there can only be a finite number of such points in each D.i. Hence there can only be a countable number of points equivalent to p.
However, the number of points equivalent to p is the number of elements in the covering group which is isomorphic to the fundamental group. Q.E.D. We shall call a discontinuous fixed point free group of conformal mappings of the disc onto itself a Fuchsian Group. Frequently in the literature this is called a fixed point free Fuchsian Group since it not always assumed that the group is fixed point free. 1e shall call two Fuchsian Groups G1 ,G2 similar if there exists a conformal mapping A of the disc onto itself such that G1 = AG2A-l. We shall consider the disc on the Poincare model of the NonEuclidean (hyperbolic) plane. We could just as well use the upper half plane. shall denote the Non-Euclidean plane by U and use both the model of the upper half plane and the disc. If G Is a Fuchsian Group then we shall denote by U/G the Riemann Surface obtained from U by identifying equivalent points under G. Theorem 2.
U/G1 is conformally equivalent to U/G2 if and only if Gland G2 are similar.
Proof:
to U/G2.
(a) G1 similar to G2 = U/G1 is conformally equivalent Consider the projection maps.
175 19.2 111
: U --- U/Gl
T2 : U --- U/G2 and suppose G, = A G2 A-1.
`<<e shall construct a map f:
Take pl a U/G1, and z e U such that T z= pl. Let f(pi) = TT2A-lz. Suppose Tll zl = p1, then there exist U/Gl > U/G2.
41 a Gl, g2 a G2 such that glz = Ag2A-lz = zl.
Hence
g2A-lzl and therefore T72A-lzl = Tj2A-1z. It follows then that f is a properly defined mapping. Clearly f is a compound map; hence we need only verify that f is 1 - 1. Suppose f(pl) = f(p2) and
1 z1 = P1 , ll z2 = P2.
f(pl) = 12 A-lzl = T2A-lz2 = f(p2).
Then
Hence there is a g2 a G2
such that g2A-lzl = A-lz2 , or Ag2A-1z1= Z2-
But since
Ag2A-1 a G1 it follows that q z1 = T2 z2 which shows that
p1 = P2 (b)
U/Gl conformally equivalent to U/G` = G1 is similar to G2.
There exist maps T('1
:
II
72 : U
f
U/Gl
-)
U/Gl
) U/G2
1 -i
Conformal
U/G2
It is clear that locally there exists a :nap such that the diagram is commutative.
U --- i)
176 19.3
Since U is simply connected we can continue f# throughout
U.
Consider a g1 a G1. Then TT2fef*-1
2f9'
X,
g1f
f
#-1
TT2 1
-1
Hence f glf
hence G1 =
= fTTl Y
a G2.
Similarly g2 a G2 implies f g2f
-1 f#G2f,
e G1;
Q.E.D.
A group G of transformation on a topological space S is called discontinuous if for any s e S the set KGs` = sl) there exists a g e G such that sl = g Its discrete. A group G of transformation on a topological space S is called discrete if there does not exist a sequence gni, gn e G such that for all s e S gn(s) converges to s. In order to prove, among other things, that a group of non-Euclidean motions is discontinuous if and only if it is discrete we shall discuss the general linear fractional transformation and in particular the non-Euclidean motions. The general linear transformation A(z) = az+b , which is cz
normalized so that ad - be = 1 a > 0, when looked upon as a mapping of the finite plane,has as fixed points, the roots of cz2 + (d - a) z - b = 0. Since co is taken into a/c, it is not a fixed point unless c = 0. Thus we see that a linear transformation (which is not the identity) mapping of the extended plane, has either one or two fixed points. we shall classify these transformations according to 1) the number of fixed points and 2) the behavior of the mapping near the fixed points. First we consider the case where the mapping has two fixed points. Let us assume that they are 0, co. Then the mapping is, A(z) = Az, X / 0, 1 is called the multiplier or invarient of the transformation. We divide the mappings into three classes: 1) Hyperbolic when A > 0. 2) Elliptic when JX - l and 3) Loxodromic otherwise.
177
19.4 If the mapping is hyperbolic then all lines through the fixed points remain fixed, and the family of circles orthogonal to these lines is left fixed as a family. Also the points move away from the origin and towards infinity if we specify 1 > 1.
Hence the origin is called the repelling fixed point and inIf two finite ppinta v-, finity the attracting fixed point. are the fixed points then the mapping is
z-
A(z) A( Z)
- C
z -"L
Here all circles through , are left fixed and the family of orthogonal circles remains fixed as a family, and i- is the reIf a hypelling fixed point, r"the attracting fixed point. perbolic transformation A is to be a non-Euclidean matrix, In this case A is called and ' must lie on the unit circle. a non-Euclidean translation. It is easily seen that A carries
the non-Euclidean straight line joining 7 and T into itself, and has no fixed points inside the disc. If the mapping is elliptic then, assuming 0 and
to be the fixed points, we see that the family of straight lines through these points remains fixed, as a family, and each cir-
cle orthogonal to these lines is carried into itself. As a non-Euclidean motion an elliptic transformation has one fixed point, no fixed non-Euclidean line. Such a transformation is called a non-Euclidean rotation about that fixed point.
If the mapping is Loxodromic, that is A = calea a > OR e / 2ni then the mapping is the composition of an elliptic transformation and a hyperbolic transformation. Hence it is clear that there will be no fixed line unless A is an odd multiple of a in which case any line circle through the fixed point is left fixed, however in this situation the interior and the exterior of the fixed circles are enlarged. Hence we see that a non-Euclidean motion cannot be given by a LoxDdronio motion.
178 19.5
There remains the transformation with one fixed point. It is clear that this condition requires a = 0 if we let co be the one fixed point. Hence the transformation is of the form A(z) = az + b, However unless a = 1 we have the two fixed points lba , co . Hence the Parabolic transformation with oD as its only fixed point is A(z) = z + b. This transformation is translation of the plane. The lines parrallel to the line
through 0, b are fixed and the family of porpindicular lines is fixed. To see what a parabolic transformation is as a nonEuclidean motion, it is easiest to use the upper half plane as the model of the non-Euclidean plane. Then we require that b is real and there are no fixed non-Euclidean lines or points. As a non-Euclidean motion the parabolic transformation is called a limit rotation (is a rotation around infinity). Since the most general non-Euclidean motion is of the form A(z) =
e'.+b
it follows that a non-Euclidean motion is either Sz+a a translation, a rotation or a limit rotation. A possible element of a r'uchsian group, the rotations must be excluded since they have fixed points inside the disc.
When do two non-Euclidean motions A,B commute? Suppose B is a limit rotation with fixed point cn . Then B(z) = z + µ µ real and ? 0.
Let A(z) = az+b c z+d
Then if AB = BA we have,
az+aµ+b=aa+b+ cz+cµ+d cz+d Let z = m, then=+L a ' c
= ac
-c=0 Hence A is of the form A(z) = as + b, and
179
19.6 AS = BA 4 a z+ aµ + b
az+b + µ
4 aye 1
=:b a
=:A(z)
z+U
Hence the only transformation other than the identity which commutes with limit rotation is a limit rotation with the same fixed point.
Suppose B(z) = Xz , A > 1,
then A3 = BA 7 cxz+d let. z = 0;
-
cz+d%b
then 4b = A a
b
0or ca
b =0ord=0 suppose b = 0, then a%z _ a%z caz+d - cz+d then
let z = co,
c - A c
c = 0 since ad-bc)k 0 hence A(z) =
Z.
d d = 0, then
Suppose
=
aXz+b
c=z or
aX.+Xb
cz
acXz2 + bez = ac12z2 + bca2z
s+ b c cAz
=
a c
=
a
-
r,7.z+Xb 0z
c
A
0
a c
180 19.; b
_
cxz
ab cz
=, be = 0
impossible
A(z) = a1z
Hence
.
Suppose B(z) is a rotation, we use the disc as the nonEuclidean plane and assume the origin fixed, hence B(z) = eiez, and
A(z) =
az+b
1bJ2
la12 -
= 1
bZ+a aoiez+'+
AB = BA
be
=
z-a
let z = 0, then
ae1oz+be1e bz+a
b
= eie b
h
a
b=0 or a=0 if a = 0 it is impossible that Jal2 - lb!2 = 1
hence b = 0,
hence 1;(z) = s z = ataz
a
Thus
have proved that if two non-Euclidean motions
commute, they are of the same type and have the same fixed points.
Theorem 3.
A group of non-Euclidean motions is discontinuous
if and only if it is discrete. Proof:
a) not discrete
not discontinuous:
If the group is
not discrete there exists a sequence of motions Tn(z) such that T n(z) --?
of any Tn.
for all z.
Pick a z which is not a fixed point
Assume then z is 0.
Then the set Tn(0); are all
181
19.8
different from 0 and 0 in a limit point Hence the group is not discontinuous. not discrete.
of the
set.
b) not. discontinuous
For since the group is not discontinuous
there exists a z such that the set of points equivalent to za an accumulation point, that is an accumulation point in the Let 0 be the accumulation point.
Then we have a sequence
of distinct elements zn such that zn--4
0, and all zn are equi-
disc.
valent.
Denote by Tn the element of the group 0 which takes
7,n+1
zn
Denote by An the mapping
An(z) _
z-zn
1-znz
Then the mapping Cn
-1 = ATn n+1 n
leaves the origin fixed and is therefore a rotation around tho origin. That is Cn(z) = eien z. A subsequence of a16n cone1n is that subsequence and let the limit verge assume that ata If = 1, then Ari 4 1, Cn be eta . and therefore eia Tn --a 1, hence G could not be discrete. If r 1,
c(z) = ataz, then Cn ----a C,
Tn
----3PC-1Tn+1Tn
and T-n+l Tn # I-since the za are distinct.
__4 1
182
20.1
Locturo 20
Lot S be a Riemann Surface and let G be the group of conforraal homeomorphisms of S --> S.
Our object in this li.;oture
shall be to show that with the exception of the following seven cases G is discrete. The seven exceptional cases are: 1)
The sphere, in which case G consists of trappings A(t) = cz+a with ad-be / 0 and we normalize the group so that a = 1. Thus we see that the group depends continuously on six roal parameters. The group is a six-dimensional real Lie Group
2)
The plane, in which case the Group consists of mappings !(z) = az+b which is a four-dimensional real Lie Group.
3)
The cylinder, the group of conformed self-transformations consist of the helical motions. For consider the cylinder as the plane identified by a group generated by the translation w = z+l. Then an element of the group of conformal self-transformations must satisfy the functional equation.
w(z+l) = w(z)+l. I,.)
Clearly the mappings A(z) = z+b do this.
The Torus, that is any compact surface of genus 1. vering group is a group generated by z! = z+l
The co-
In t>0
zs =z+z
.
:Ionce a conformal self-transformation A must satisfy, A(z+=,
= AA(z)+l and A(z+'.") = A(z)+ 'z; .
Once again this is satis-
fled by the mapping A(z) = z+b. $)
The disc.
The group of conformal self-mapping consist of A(z) = az+b bbz+a
which is a three-dimensional real Lie Group.
Iat2-JbI2 = 1
20.2
6)
The Punctured Disc.
The group of
salt-mappi_.;;s is
the group A(t) = eio2 which is a one-dimensional rear. Lie Group. 7)
The annuli form a whole class of conforr._.:.ly
The Anrulas.
distinct surfaces, as do the toruses in I}). However we can represent them as a one-parameter family in which the radius
r < 1 is the real parameter determining a conformal equivalence class.
The group of conformal self-mappings consist of
A(z) =
and
B(z) = ei4z
z
We shall now proceed to show that
these are the only possible e_:ceptions. Suppose we have two Riemann Surfaces S, St with universal covering surfaces 5, St. If f is a continuous map of S into St then t,-are ti
is a map w which maps S continuously into St so that the following diagram. holds.
S J
S
w J t
f
-> S t
Clearly w exists locally, for locally we may define w =
_ t-1f 7i . -V
by Since S is simply connected w can be contir.ued to all of Furthermore we see that w is determined by ;.:e :aodromy theorem.
its value at a single point. ups of S, S' respectively, Let G, Gt be the covering then all possible w are given by the various htw for At E Gt.
say that f induces w.
184
20.3
Suppose now that we have another surface Se, with the universal covering surface S", and a continuous iaap g of St into Site
Then we obtain the map p which is induced by g.
That is, we
have the following diagram
S
V1
s
>
St
> Sit
>
SI
>s»
That is,
871
= Tt114
fiT
= 1T' w
and
therefore gflT=gTTtw
= IT"ir,.,
hence pw is induced by gf.
Thus we see that if f is a homeomorphism, then the induced map w is a homeomorphism and if f is conformal then w is conformal. Consider A E G; then wA is a mapping of w(S) = SQ into itself. Call tnis map At. Clearly AI E.G'. That is we have a mapping
X: G -->G given by wk = x(A)w.
)( is a homeomorphism for,
%((AB)w = 47AB =)( (A)wB = X(A)/I (B)w
We say that the map w induces the homeomorphism )(.
note the following, AtwA = A'X(A)w = A'X(A)At-'Atw or
(A'w)A = AIX(A)A'-IA'w
We may also
185
20.4 In other words if w induces )( then Alw induces Furthermore if f is a homeomorphism then )( is an isomorphism.
Let us now consider the following situation, S = S+, S = St = U the non-Euclidean plane, and f is a conformal self-
mapping of S. Then for A E G, wA = X(A)w = wAw-1 = )((A) E G. Now suppose there exists a sequence fn of such mappings converging to the identity. Then it is possible to choose a sequence wn of. induced maps which converge to the identity. Choose an element A E G different from the identity. Then wnAwn1 A-1 E G converges to the identity. Since G is discrete for all sufficiently large n, wnAwn1 A-1 = 1. That is wnA = Awn for large enough n. Suppose G contains a limit notation, then for all sufficiently large n, wn is a limit rotation with the same fixed point. Hence all the elements of G are limit rotations with the same fixed point. Hence if we let U be the upper half plane G
The mapping 4 = o 7tiz/a maps U onto the punctured disc in the 4 plane in such a way that each set of equivalent points are mao ed onto the same point. Hence S is the punctured disc and this gives us case 6) desis generated by A(z) = z+a, a real.
cribed above. Suppose G contains a non-Euclidean translation.
Then by
the same reasoning we see that G consists solely of non-EuclJdean translations with the sane fixed point. If we take the fixed points to be 0, oc and U to be the upper half plane then G is generated by A(z) = eo'z. The mapping 4 = log z naps the upper half plane conformally onto the strip -oo < Re r < oo, 0 < Im g < ,[. The non-Euclidean translations now appear as
Euclidean translations of this strip parallel to the real axis. e-2 %
map the strip onto the annulus Finally the ma ping w = 1 < (w) < e2 'A /a so that equivalent points are taken into equivalent points. Thus S is an annulus which is just case 7). The only alternative left for G is that it contain neither limit rotation nor non-Euclidean translation, that is, 0 contains only the identity and S is the disc. Thus we have proved that the seven exceptions listed are the only possible ones.
186
Lecture 21 Definition 1. Given a topological space D and a group G of mappings of D into D, then a fundamental region of the group 0 Is a set f c- D which satisfies the following properties:
1. Every point of D is equivalent to some point of
under G.
2. No two interior points of
are equivalent under G.
Given a Riemann Surface S with hyperbolic covering surface U and covering group G, there exists a fundamental region of the group G. Theorem 1.
Proof:
Pick a point z
0
a U.
Enumerate the elements Al a G, We divide U into classes
and denote A1(zo) = zi and AO = 1.
jas follows, '4e shall say that z is in L! j if z is not further (in the Poincare metric) from zj than from any other zi. The set..j is called the J M Foincare set with center zj. Since the set zo,z1,...1, is discrete, every z Is in some_.; j. Since i is the intersection of the closed half planes consisting of points satisfying
d(z,z) < d(z,zk) It follows that
k = 0,1,...
, is closed and convex.
Since zi is obviously an interior point of .'i It follows has a non-empty interior. Clearly z is a boundary point of J if and only if there exists a zk 7A zj such that
d(z,zk) = d(z,zj)
Furthermore, if d(z,zk) = d(z,zi) and d(z,zk) > d(z,zi) for all I # j,k
,
it i.s clear that z is the center of a non-Euclidean
line segment which enjoys the same property.
If z has the same distance from tnree or more points zi then we say that z is a vertex of any of the corresponding I. Clearly z can be equidistant from only a finite number of zi since the zi are discrete. For the same reason we see that the set of vertices is discrete. Furthermore we can say that Lfi is a polygon in the sense that its boundary consists of vertices
187
21.2 and non-Euclidean straight lin.:s.
"e now assert that _'.Q (in
fact any L%I) is a fundamental region and is called alternatively the normal, Poincare, or metric polygon. Since any z e U is in some !-s, say
k,
it follows that A-1 (z) a 'o
since the Let z', z° be
elements of G preserve the
two points of
tY
are equivalent under G. There exists an
Ai such that Ai(z') = z". z
a
is
Hence z" a
but z" a _p, hence
`o and therefore z" is a. boundary point of - o
In
the same manner we see that z' is also a boundary point, and the theorem is proved. Since the elements of C preserve distance, we see that they interchange the.'i and hence a boundary point which is not a vertex is equivalent to e:;actly one other boundary point. Furthermore the same reasoning shows that one edge is equival-
ent as a whole to another edge or is equivalent to itself:
But
an edge could be equivalent to itself only under a non-Euclidean translation which could take an interior point into another interior point, which is impossible. Hence if the number of edges is finite, we see that the number of edges is even. Notice however that the number of edges depends upon which z we call z0. Nevertheless, certain AI e G, call them Bi, identify by the edges of ;o. Since we go from . o to an adjacent i
such a B., and since we can get to a.:i adfaceut to L11 by a
BIBiB., we see that the Bi generates G. Ve also see that .., is compact if and only if the Riemann Furthermore, if S is compact, it then
Surface 3 is compact.
follows that
has only a finite number of sides since the
vertices can have no. limit point. Suppose the fundamental region has 2n edges and 2n vertbxz.
Corresponding to the vertex Yi is an angle ai. The vertices are separated into equivalence class. The sum of the angles corresponding to a particular equivalence class must be 21t,
since after identification it must correspond to a single
188 21.3
Since the sum of the neighborhood on the Niemann Surface S. angles in a non-Euclidean triangle is less than i, a glance at the following figure
a1{3
demonstrates the inequality 2n
ai < 2nn - 21t = 2(n-1) it
1=1 However, we also know 2 - 2g = x = K + 1 - 3n + 2n = K - n + 1
where K denotes the number of equivalence classes of vertices. 2n
Clearly
ai = 2nK < 2(n - 1)
at
i=1
hence K
has the disc as its universal covering surface, than g > 1. Theorem 2.
If S is a closed hyperbolic fliemaiin Surface and G
the Fuchsian group which Is isomorphic to the covering group of S. then G contains only translations.
189
21.4 Let A be any element of G other than the identity. Ve That is, we are looking for will seek to minimize d(z,A(z). Proof:
a z
0
such that d(zo, A(z0)) < d(z,A(z))
unit disc.
for all z # zo in the For any z there is a T e G such that T(z) = z e a
the fundamental region. Then d(z,Az) = d(Tz,TAz) = d(z,TAT-1s). Hence the problem reduces to minimizing the expression
d(z,TAT-lz) over z e L,, T e G. minimum problem has a solution.
shall now show that this Consider d = inf D(z,TAT-lz) zec
TeG
Then we have sequences Tn a G, zn e:1 such that d(zn,TmATm1 2-4 d. Since
is compact, we can select a convergent subsequence of the znt which converge to i. Thus d(z,TnATnlz) d. Call
TnATnl = Bn. Thus d(z,Bnz) d. Since G is discrete, only a finite nu_nber of Fn move z a short distance. Hence for a particular B, d(z,Bti) = D. And therefore for the corresponding T we have d(z,TAT-lz) = d > 0. Since B is a limit rotation on a translation B2 is not the identity. Consider the points z,B(z), and B2(z): The points r. and 4' are the midpoints of [B(i),32 (i)] respectively. the segments [z,B(z)] ,
B(4) = r' since B preserves distance.
Letting T-1(4) = z and
we have d(Y.,BZ) = d(R,TAT"lr.) = d(T(4),TAz)
= d(z,Az) > d
190
21.5
But by the triangle inequality d(?,B(i)) + d(Bz,E4)
-
2_
d + d
d
It follows that B leaves the line fixed, implying that B is a translation from which it follows that A is a translation since Hence d(r,,Br.) = d andz,B(i), and B2(2) are colinear.
TAT-1 has the same number of fixed points as A. Q.G.D.
191
Lecture 22.
In the theory of functions of a complex variable, it makes sense to talk about the integral of a function f(z) over a curve ro, or over a plain domain -7. That is, the expressions
{
4f(z) dxdy
f(z) dz I J
are well defined. It will not, however, make sense to talk about intorrating functions over a domain on a Riemann surface, as the following simple example will show: Let b be a domain lying within a single parameter disc, with local parameter z = x + iy, and let f be a continuous function defined at least in this paruineter disc. Suppose we decided
to define the integral of this function over jJ
f(p) = ,
as
fl(x,y) dxdy
z(.U)
where z(f)) is the plane image of Z undar the mapping of the local parameter, and fl(x,y) = f(p(x,y)), where p is the inverse In order for this integral to of the local parameter mapping.
be well defined, we must have f1(x,y) dxdy =
I
z(Z)
f2(u,v) dudv
w(19
where w = u + iv is any other local parameter. But It Is quite clear that in general, these integrals are not equal. In fact, since w is an analytic function of z, and since
)"(x.y
u u,v
=
1dz12 dw
we obtain fl(x,y) dxdy = Sl
z (4)
1
f2(u,v)
dude
w(i)
so that we must have tdz/dwl2 % 1 for the two integrals to be equal.
192 22.2 We are able to solve this problem by introducing differentials on the surface. They will be characterized by the property of being invariang undez local parameter change, which will qualify them for the role of "quantities to be integrated". In this soction we shall briefly outline some of the important properties of differentials. For a more complete discussion of this, the reader may look at almost any standard text on Riemann Surfaces.
Definition 1.
Let .a be a domain on a Riemann surface S.
A
first-order differential, or a form of de,rree 1, on . will be an express ion
a = u(x,y) dx + v(x,y) dy
which satisfies the following transformation rule: local parameters, then
If we change
a = u(z,V) d! + v(x,v) dy where
u(3:,-Y) = u[x(
,y), y(ix,y)] `X + v[x(x,':'), y(X,y)] .Z ?X
aX
X + v(x(X,Y), y(X,Y)] °y
u[x(x,Y), Y(x,y)]
(This is just the transforn:ation rule tort holds for an analytic change of variables z = z(w) in an ordi.-iary line integral, 1
u(x,y)dx + v(x,y)dy =
j
[u Ax- + v dt] dt
Jo
in the complex plane.)
For a complex valued function f(p) and a differential a defined in g, the expression
fa = fu dx + fv dy A simple calculation shows that fa satisfies the required transformation rule. is a first order differential.
193
22.3
Definition 2. A form of de,;rce 2 is an expression f(x,y)dx A dy which satisfies the transformatiou rule
f(x,y)dx /\ dy = fy(5,Y)1
o(x v
dx A dy
Requiring anti-commutativity, i.e., dx A dy = -dy A dx, we get dx A dx = dy /\ dy = 0, so that we may define the product of two forms of degree 1, a = u dx + v dy, and js = p dx + q dy to be
(udx+vdy) A(pdx+qdy) = (uq - pv) dxAdy where (uq - pv)dx /\ dy is a form of degree 2, as is easily checked. In general, on any surface of real dimension n, we may define forms of degree m to be expressions u il...im dxil /\ ...
A
dxim
which remain invariant under a coordinate change. We arain assume anti-comnutativi.ty, so that multiplication may be defined in the natural way. It is easily seen that for m > n, a form
of degree m must be identically 0. For convenience, a function defined on a domain of the ;even surface will. be called. a 0-form.
`-'e shall now define the exterior deri native d of a differential form, which will be a homomorpnism of m.-forms into (m+l)forns. (Two forms of degree m are ad3ed in the obvious way.)
The reader might note the close resemblance of d to the cobound0*, which is a homomurphism of m-cochaina into
ary operator
(m+l)-cochains. nefinttion 3..
a) If 4 is a C-form n
dx d4 = 17J=T Cxi i where dd is a 1-form.
(This should be checked)
194
22 .4
b) Let F u
'l" 'gym
dx
A ... A dx
i1
d( u Jr.. im dx31A .../\dx im ) _r-nn
im
be an m-form.
it .. Jm
r
dx .Adx
Then
1 ...Adx im
is an (m+l)-form. In particular, the exterior derivative of a form of degree 1, a = u dx + v dy, on a Riemann surface is the form of degree 2
da= ( VX
-
ay
) dx/\dy
This is obtained from Definition 3b as follows
A 4y/\dx + ?x dx/\dy + 22 dy/\dy
d(p dx + q dy) = ax (
CIX
- ,y) dx A dy
y
Me have used the fact that dx A dy = -dy A dx,
dxAdx=dyAdy= 0. A straightforward calculation shows that d2 = 0, i.e. for any m-form v;,
d2uw
= 0.
'"e shall now define integration of forms of degree 1 and 2 on a Riemann surface. As a matter of convenience, we shall, unless otherwise stated, assimie all forms to be defined over the entire surface. Integration of a 1-form u. i) Let 4. be a differentiable curve lying entirely in the domain of a single local parameter, i.e. °, is the image of h(t) E C1, 0 < t < 1, on S. (,!e may define the integral of a
over C. to be
fa=
pdx+gdy
where z = x + iy is a local parameter of a donwaiil containing C.
By a conformal parameter change z = z(i), we introduce a new local parameter. Since a is a 1-form, it remains invariant under this change of varisblos, proving that the integral is independent of the local parameter.
195
22.5 ii) Now assume that 6 extends beyond a single parametric We subdivide the unit t-interval into subintervals (ti_l,ti), I = 1,...,n, to = 0, to = 1 so that each (ti-1'ti) n
disc.
lies in a parametric disc. 'Ihen C=
Y
vi, where ti Is the
ere define
image of (ti_l,ti) under h(t).
a=a n
S r,
i
To show that the integral is independent of the particular subdivision, we make use of a refinement of the given subdivision. Since each 1 lies entirely in a parametric disc, we obtain m
f
a=
Jr
a
the i;1 resulting from the refinement. There,j=1 fore the integral remains the same for a refinement of the given If we are given two suitable subdivisions, we may subdivision. apply this result to a common refinement, thus proving that the integral is well-defined. where
Integration of a 2-form P.
i) As in the case of the 1-form, if P lies entirely within a coordinate patch, we may define the integral of the 2-form over the domain 13 as follows. li
(3
-b
p(x,y) dxdy
'J
z(.f)
The integral, so defined, Is independent of the local parameter. ii) If the domain .Z'-extends beyond a single parametric disc, and if F is compact in S, we use a partition of unity to define the integral. Let us denote by h(p) that part of the parametric disc around p which contains all points for which
Izi < 1/2. Since 1 is compact, a finite n'ui,ber Nl,...,Nk of such Let be a partition of unity with neighborhoods covers it. respect to this covering. Than we define
J-l
196
22.6
((
k
k
a
JJ
if xa
N flr
must still 'U P are defined in 1). h 0.0 1 show that the integral does not depend on the particular cover-
where the integrals JJ
ing and partition of unity that we choose. Let NJ, J, be any other suitable covering and partition of unity respectively. N
0 except on
1, and since
Then since j
J i) N. we get
k
£ 1
a
1 xi JI
1
`3-1
j-
1=1
N
f J PIJ i N£,
Nl, t
talc b
iii) If 7,1s not compact, we assume that 3 is real, nonnegative and define
It
a = l.u.b. J . 'L,7- Fi x., where A has compact closure.
p = ¢f - 3-
I
P J
0
For arbitrary real 3, we write
for complex ¢, wo write 3 = fle p + i Im (3.
We
assume that the reader is familiar with these arglaments and so A more comprehensive account of this omit all further details.
and almost everything also we discuss in this lecture can be found in Springer, Introduction to Riemann Surfaces, Chapter 6. Definition 4. differential
Let f 6 C1 on S.
"e shall call the first order
dx + of dy
df = 3x
(where z = x + iy is a local parameter) the total differential of f.
197
22.7 Definition 1.
An exact differential is the total differential
of a function t E C2 or, S. Thus if a is exact, then locally a = u dx + v dy and If, however, ou/'3y = .v/ax at every point of S, ^ou/ay = av/rx. it does not follow that a is exact. For this condition only says that a = df locally - and continuation of f to the rest of S may not result in a single-valued function on S. A first order differential a is said to be of class Cn if locally, a = u dx + v dy where u(x,y) v(x,y) E Cn in the parameter disc in which z = x + iy is a local parameter.
Pefini.tion 6.
Definition 7. A differential a E Cl which is locally exact (a = df locally) is called a closed differential.
Definition 8. The conjugation operator * maps a 1-form into a 1-form as follows: If a = u dx + v dy, then *a = -v dx + u dy (The reader should verify that
.
is a 1-form.)
Pronerties of the conjugation operator *. 1) If a,P are 1-forms, then
*(a + P) _ .ja + ii) If f is a function, then
*(fa) = f'( a)
iii) a = *(era) = -a
.
iv) If a = u dx + v dy, P == p dx 4 q d
a
v) a
locally, then
P = P * a = (up + vq) dx /\ dy a = (u2 + v2)dx A dy
The reader should verify that
. and a*P are forms of degree 2.
198
22.8
Definitions-.
formally, we define
t'riting d = (3X)dx + ,Ud =
(:ax)dy
For f 2 C1, x(df) = (id)f, so that we may write 'df without dan^er of confusion. Definition 10.
If f e C2, and if a = =xlf, then a is called
coexact.
Definition 11. co-closed.
If a C C1 is locally coexact, we say that a is
We shall state without proof two equivalent definitions of closed and co-closed differentials. Definition 7'.
A differential a E C1 is closed if da = 0 on S.
Definition 11'.
A differential a t C1 is co-closed if d*a = 0.
Definition. 12.
For f a C2, we define the Laplace Operator,
0= d * d: *df= (
2 +-) dx/\dy 3y
2 Jx
A straightforward calculation (which we omit) gives us Stokes's Theorem on S: Theorem.
boundary
be a region on S with compact closure, whose
Let
consists of finitely many piecewise analytic arcs.
Then for a E Cl in d+, r
Taking a = fR, we obtain, from Stoke's Theorem rr
(1)
f dp =
and taking p = *'dg, g f' "
(2)
JJ
{
1j
df +1
f¢
C2 in . , we get (
f Ag = 1
f
f* dg -
df * dg
199
22.9
Interchanging f end g in (2) and subtracting, we get the familiar Green's formula
(3)
(fig - ggf) _
(t
(f dg - Q-udf)
(See Springer for details.) We shall now state, without proof, the following properties of intef:rals of closed and exact differentials. Property 1.
If a is exact and w a curve given by h(t),
0
where a = df on S. Property 2.
If a is exact and
. a closed curve,
a=0 Property 3.
If u is a closed differential, and if 4 is homoto-
pic to t'' then a= .
Property
4
If a is closed an3 if
S
PropertP 5.
' a 1Y
is ho-notopic to a point then
a=0
If C. is homologous to C' and a a closed differential
then
a=J
a
200
22.10 Definitio> 11.
If a is a closed different!
'
and
a closed
curve, we d.:.fine the period of a on & to be I a . (I From Property 5 we conclude that the period of a differential on a closed curve depends only on the homology class of the
curve. Again we state without proof the A closed differential a is exact if and only if all its periods are zero.
Theorem.
Now we come to the most important part of this discussion: the introduction of harmonic differentials. These differentials will play a vital part in the discussion of the Koebe Unifo rmization Theorem. A differential a E C1 is called harmonic if it. is both closed and co-closed, i.e. if da = d*a = 0.
Definition 14.
Definition 15. A function f E C2 is called harmonic if 'Lf = 0 in terms of local parameters. Theorem.
A differential a e C1 is harmonic if and only if in every parama$er disc there is a harmonic function f such that a = df in that disc.
a is harmonic <_> da = d*a = 0 ) a is locally exact, i.e. a = df locally, and d*(df) = 0. Proof.
i) Necessity:
ii) Sufficiency: a = df locally, for f harmonic locally a is closed (locally exact). Since f is harmonic, d*a = dTdf = 0, so that a is co-closed.
Definition 16. Let a,(i be two Cl forms of degree 1 on S. we define their scalar product to be (a, a) =
ff a *'3 3
.
Then
201
22.11
Now it is clear that the first order differentials of class C' form an infinite dimensional complex vector space. becomes a normed vector space when we define fall= VT-a,-a7.
space is not complete.
It
This
Its completion is the Hilbert Space of u:;hu
all measurable differentials w on S for which Ilu:II2 =
JJS
exists in the sense of Lebesgue. By a measurable differential we mean one which can be represented locally by u(x,y)dx + v(x,y)dy, where u and v are measurable functions of the local parameter.
202
Lecture 23
It started by consid"That, after all, is uniformization? ering algebraic curves F(z,w) = 0 and one sought functions f,g rind a domain D so that as t ranges over D z = f(t),w = g(t) gave all points of the curve. Also, t is called the uniformizing variable. When we start with a function element w = ` an zn, we obtain a r.iemann Surface and lonally we have w =
bn to
The uniformization theorem that we have already proved enables us to do what is asked in the first paragraph. For D we hr±ve the universal covering surface and the projection map p gives is the functions f and g. However, +,,hnt we would really like is a mapping from a That plane domain which is not only conformal but also 1-1. is, we would like each point of the Riemann Surface to correspond to one and only one value of the uniformizing variable. z =
an tn.
Hence we see that we are asking when, is a Riemann Surface con-' formally equivalent to a plane domain? Clearly, a necessary
condition for a Riemr.nn Surface S to be conformally equivalent
to a plane domain is that S be topologically equivalent to a plane domain. The Koebe uniformization theorem states that this condition is also The property of plane domains which we use to characterize plane domains is, the Jordan curve theorem.
Definition 1: tar j
A Riemann Surface 3 is called planar or schlich-
if every Jordan curve on S separates S.
The subsequent three lectures shall be devoted to proving Koebe's general uniformization theorem which we state as follows: Theorem Every schlichtartig Rie:nann Surface is conformally equivalent to a plane domain. To accomplish this, we noed a function f analytic on S with at most one pole and 1-1. Suppose we have such a function f with pole at p0. Then we obtain a function u = V+:a.f which satisfies 1.
u harmonic in S - p
2.
a
3.
-Z is harmonic near po. u has a single valued conjugate v.
!}.
f = u + iv is 1-1.
203
23.2 Conversely, if we con find a u with properti@s 1-4, we obtain the required function f. !.e shall use the Dirichlet Principle to accomplish this. The Dirichlet Principle was first used to solve the first boundary value problem or Dirichlet problem. That is, we are given a domain G with boundary r, and a function f continuous on r. Among the function u defined on G + r and equal to f
on r1wo seek one such that-..u = 0 in G.
The auler-Lagrange
equations show that the extremal function for the functional (fx2 + fy2)dxdy must satisfy,.%u = C. So the Dirichlet D(f) = '
principlTat first stated that we choose an extremal function for the Dirichlet integral D(f) and we have solved the Dirichlet problem. However, it was rapidly pointed out that there need not be an extremal function among the functions taking on the given boundary values. In fact, Hadamard gave the following First we note that in the unit circle example. D(rn cos n8) = D(rn sin n8)=nn, and if
m u = r (an cos n8 + bn sin ne)rn, then n=1
D(u) = x
CD
a==y
n(an 2 + b n2 ). m
Hence if we take as boundary values f(8) _ m=1 the solution to the Dirichlet problem is, U
m
VT- sin m4 e m=l m ca
however
sin m48 -, then
D(u) _
m4
= m.
m-1 m Hence »e see a Dirichlet problem which has a solution and yet the harmonic function with the given boundary values has a divvaraent Dirichlot integral. We shall use the Dirichlet Principle in the following form.
Dirichlet Principle:
Let S be any Riemann Surface and r-.-) any
closed differential on S. Assume that there exists a closed differential'? such that JJW - *1 < co, then there exists a function f0 on S such that
204
23.3 a)
D(fn) < o
b)
W.J+ dfo = coo is harmonic. Lot 401 = (Jo - *QT, then 1101111 1101 + df 11 for all f with
`.
D(f) < co . cf)
We s::all consider a condition
and show that c) is
equivalent to c').
et) if 61 = cJo
D(f) < ao Proof:
,
then (t 1,df) = 0 for all f such that
cc')
1I,Jl+ adf112 = IG,;1112 + 2A((Jl,df) + x21(df112
which is > 11W, 112 for all A if, and only if (Wl,df)
0.
As an immediate corollary of Stokes' theorem we have
(wl,;:df) = 0 for f e C1 W co-closed (WI,df) = 0 for f e C0 ')harmonic (co,df) = ((,,1,-*df) = 0 for f e Ca (c) closed =
The way to remember this information is through the following picture.
df
f E Co
*df
205
23.4 We should now ask, when does this picture tell the whblo That is, when can we uniquely decompose any differential into the sum of one which is exact, one which is co-exact, and one which is harmonic. story?
Let us consider the unit circle Uj
lzi < 1)
zI
Let H = the set of harmonic differentials on U Let = the set of differentials 4 a C° in U, that is, the differentials which are continuous in ff.
Leto = the set of differentials 4 e Co in U.. Let F = the set of functions f which are in 0° in TY,
f e C2 in U,,
f = 0 on lzl = 1 and D(f) < a). Then, Lemma 1:
Given 4 e 1, there exists ai e
Proof:
_To
such that
a0.
Iii - 411
Obvious.
Lemma 2:
Given f e F, there exists f1 a Co such that
lldfi - df 11 -4 0. Proof:
for 0 < e < consider a function c4e(R') defined for -1< t <1 which satisfies
(i for-l+e < t < 1 - e
Ofort>
-
C, t <-1+2
2. e a C2. Ve now consider the function de(p,e) _ (p). quence ei - 0, the sequence of function fi = Consider
lldf - dfill =
I<1 (f(1
(f(1 - de
1-e
For a sea
2
iy
4 ei'z + [f(1 -
4ei))y
i
f(
But
S5
1-a
(f2 + f2)dxdy =
1J
y
1-e
x
-'
(f2 + 1
r
Co.
f2)rdrdB e
23.5
206
Hence if
(fr+f2)drd9 -a0, then so does
Jr'
1-a<Jzl<1 But
(f2 +f72 )dxdy.
1-ei
SS
If (1 -1 ei 112 + If (1 -,b ei ) ] g r
1-ei
<2
fr(1 -je 2+f2(1 - ¢E )r +f9(1 - $e B i 1-ei
2
The only term which might not go to 0 is if'2 1-eizl
Certainly we can arrange that (1 i
f2(1-4e )2 i
1-ei<'zl
<
C
We proceed to show this.
ff
Then
f2
Sf fr. f2 _ < e2 S i-eiIzI<1 1-eiz(<1
By Schwartz's inequality
f1 f9)dj')2 < 1-r
f2(r,9)
.
1-ei
ei
We shall be finished if we show that
-- 79 ei
i
fy(2,9)dfi
Hence 1
r
1
Sa1 f2(r,£)dr < ` (1 -r) 1
f2(P,9)d?dr 1
1
1
'pi
pl
f2(P,9)I (1 -f1)- (1-,P)2] P1 1
(l -pl)2 JA1 f2(.P,Q)dp
pl
207
23.6
Thus
f
2 < e2
1-e< z1<1
If w closed,
Lemma
JJ
Q.E.D.
f2
1-a
, f e F then (W ,*df)
0.
Choose fi a Co such that
Proof:
{l dfi - dffl> 0 . Then,
(w,*df) ° (w,*dfi) ° (w,df-dfi) Lemma
that Proof:
0.
Given 4 e ro,, then there
:
sts w e H; f1,f2 e F such
= (.i) + df1 +*df2 Let d*4 _ )31 dxdy and d4 = p2 dxdy, then 01,2 a Ca.
Then there exists a function t,) in
I z I
< 1 such that
,,
z
= 1p1
and there exists a function 4 in Iz < 1 such that
(see
_
Lecture 3).
Af i = !}f i
zz
Let fi =
='
=
01.
,
then fZ =
z =
z
(<_) _ z
l u),
hence
Also, we know that there exists a function g1 such that
'Lg1 = 0 in U and g1 = f1 on lzI = 1.
f1 e F and &f1
Let f1 = fi - g1, thcn
Similarly, we can find f2 such that
Qf2 =J)2 and f2 e F. Let co =4 - df1 - *df2, then d w = 4 - d*df 2 '*J'2 dxdy - P--2 dxdy = 0 . Similarly, d*) = 0, hence w is harmonic.
Q.E.D,
208
Lecture 24.
Given a sequence of differentials 4i continuous
Theorem 1.
in 7), necessary and sufficient conditions that there exists a
harmonic differential
e C such that
>o
0 are
1) I14i - 41 0 -* 0 2) There exists a sequence ei =i0 such that
1(4i , df)I is I(41
< eillldf II
i
for all f e Co in U.
Proof: a) Necessity.
1) obvious, 2) If there exists such an
then
(:,,o,df) = (,:o,adf) = 0 Let ei = 11$i -i,,011, then
I($I,df)l = 1(4 - a,df)l < !4 and similarly for I(41,;df)I. Comment.
weakly (x
eiudfll
In Hilbert space we say a sequence x11 converges to x if for all linear functionals L tien
we have L(xn) -- L(x). Strong convergence is the usual convergence in norm. Strong convergence clearly implies weak
However, the converse Is not true, for consider the following example in .2. convergence.
Let In =
Since for any vector
A = (al,a2,...) we have (A,In) = an, it is clear that Inweak 0, however it is just as clear that In does not converge to 0 strongly. However it is true that weak convergence of a Cauchy sequence does imply strong convergence. b) Sufficiency. We would like to show that there exists an
G) 0 a
FI such that (di,3) --b
have this we are through, .for
for all d e
Once we
209
24.2 Ddi
cdi -;..!0,41 -WO) = c4i - ds,di - o) + (4, - o.d)
410-U4i Define L(4) = lim
i-*m
(41,4)
--01 + 1(4 - 0,4)I
j0-
for 4 e
Then L is a continuous linear functional. For if the sequence 41
is a Caichy sequence, then there is an M such that Q4i II < M. Hence I (4i.4) I < 114i ll Hence L(4) < M 11411.
1141
_
Hence if
M 114-u.
(Idi
- 4 n --,4 0
L(41) - L(d) = L(4i - 4) < M11di - 411-+ 0.
Since
o
is dense in cI a,emnma 1, Lecture 23) it is enough
to get our result for 4o e §0. For if (;40 - b 11 - 0, then L(4o) -4 L(c(), and (1.b,411) -+ (0,4)If 4 e 1o, then 4 = .,+ dfl + vdf2.
Hence
L(dfl) + L(3*d 2) = L(,,) (by condition
L(4) =
2 of the hypothesis). And (-.0,4) = Hence we need only show that there exists L(...:)
that for all .,c H
Let:Jn = d lie(zn) = d(rn cos n8) Letc.'n = d Im(zn) = d(rn sin n8)
Let An = L('..,n) and B. = L(Wn). In lecture 23 we saw that ;Iwn11 w
Hence
n ,
=
11i'
}
is an orthonormal sequence.
e H such
210
24.3
By Bessel's inequality we have for all 4i that nI(1,
In_ )I2' 1141112<M2
Hence if we let 1 -4 m we see that A2 71p <
M2
n
and we have the same for Bn. CO
$ n (An + B2
Hence
< CD
n-
m
71n=1-
A
B
cos n6 + n sin ne)rn
Let u = 1
then D(u) = -
ri
o,
,
n (An+Bn) <do
t1=1
Let cdo = du.
Then
and
Bn = L(LJn).
in H, we have for all
An = L(Wn)
Since the set of 'n and(j is dense
e H, (W01W) = L(W). Q.E.D.
Diri.chlet Princiale. Let S be any Riemann Surface, (.) an arbit-
rary closed differential on S.
ferential (T such that that a)
If there exist a closed dif-
m, then there exists fo such
D(f0) < m
i f C.`o =0 + df, then o is harmonic, c) ¶fo 1 =(,,'o - :: tj , then 11011(1 l'='1 + df q for all f such that D(f) < co.
b)
Lot d = inf I!c - G+ df Ii2 over all f such that D(f) < co. choose fi with D(fi) < co such that ift.0, = W+ dfi then
Proof:
Q`1i - 1112 = di --3 d.
211
21.1}
Xdf 112 > d for all f with O (f) < co and all X.
1'.Ji
Then
That is
df) + di - d > 0 for all X.
X2 jdf 12 +
df)2 < (di - d) 0d fp2
Hence
=i
Let
' ),df) I < Ei qdf 11. In particular we may set
Hence
f = fi --f' and't)i - , = d(fi - f i ) hence
1!.
.
D(fi - f')
i - se,:ali - W j) I _ Ei Il(ji - oj II -10j)I <_ ejG(.h -(sj11
hence Ilk .1 - ;J'j V2 = (f!i
L)j, ;,)i -' J ) wi) - (Wi - *(, )1 -i.)J)
=
< (e hence
II(A)i - (,)j Il
+a
)
IIGJI - (
II
< Ei + c, andc.)i is a Cauchy sequence.
Let U be a conformal disc on S and let (fU be the norm taken over U. Then clearly 1-,, - i,)' IIU < 04:1 - W3 U < Ei + E 6
If f E Co in U, then (*(,df) = 0 Hence 1(")i,df)I = 1(!.)i - r(,df)l < eI0dfIand I(wi,df)I = 0. a harmonic differential in U Hence by theorem 1 there exists 0. Suppose V is a conformal disc consuch that II-)i --.,TI IIU
tained in U. We then have an W V and IIf.}V
hence
It.)V
-WTj1v
V-' iCV + IL)U
i8v
0,
-WUI = 0 and-)v -(JU'
Suppose U r'l V # 0. For each p a U V, choose W such that p c t} C U r V, then at p,WU =W%V. Hence we see that we have
24.5
212
obtained a harmonic differential(.)o on the whole surface S.
Then
Let D be a region on S with comiact closure. IL ) i - c_. ?(
"A
WO
4
.
Now
1
1-3o '(A
i
-'j "31
11D
'-OD +
(r
;J
i - i IID IIc'i - i.;o DD + If") -
<
_)
< (ei + c
)
o)D
1
Ili, i - o 1ID +
: o DD
II-A - ')o 16
L.'o'ID Nc.;i -(-".I'D
Letting f - m ; we have or
-1.)016 < ci I{:,1i -,.)016 D rl
hence
i - 00 0D 1 Ei
-`poll < Ei
hence
0c..'1 -I-)o0
0
Suppose we can show that there exists an fop with D(f0)< co,
such that c.-,o = (,) + dfo. I{
+i.,i -
- xr, Il =
Hence II i,)0
Then we would be through, for
02
< d and hence
o
*'-III
II L,J0 -
I(2
= d.
Hence if(.) i = r.lo - * 6 then
*o'+d(f+fo)11
II:)14= n:o- *5II=iaU0 - kit+(df)I =
Let
want to show that
= 4j o - c,j:
r= dfo, and D (f0)
=
0'41+ dfo for all f with D(f)<®.
is exact, that is,
0 Ir8< Go .
I r9 = I{: 4o-(-)II < IIc,)o
11"i -w II < oo.
Hence we need only show that t' is exact.
2' is exact if
213
24.6 and only if
-r`''=
0 for all closed curves C.
Recall that
dfi =:J1 -w. hence jd f
i - ell =
l{
i
0.
hence r is the limit of exact differentials.
that the linear functional
If we can show
is a. continuous linear
functional, we will be finished. For then LC(dti) --. LC(:) but LC(dfi) = 0, hence C is exact. We shall accomplish this by showing that LC can be represented as a scalar product for all
closed differentials 4). Since any closed curve is homologous to a sum of simple closed analytic curves, we need only do it for such curves. Let C be a simple closed analytic curve. That is, C is the analytic image of Ifs = 1. Since the map is analytic, we can extend it into the interior corresponding to 0 < p < IfI
1. gcCOD
2. g s 0 0< r< p 3. g(l) = 1 4. g monotone increasing 5. g(n)(p) = g(n)(1) = 0 for all n.
g(r)
1
214
21.7
let h(r,9) = g(r) < 1 let a = e dh in A corresponding to 0 < z
`,0 in S - A
(c), *4) = ti ((-N A **4
then
S
=-c.fL.nb S A dh.
A A
A
but d(hw) = dh i W+ hdt..w ; if 1:' is closed, then d (hla) = dh A w = - < : A dh. Hence d(Wa) _
(w, *4) A
aA
h:.)= S tJ C
hence if we call 4C = *4, then L(o)) = )u?= (4C,,:)) for any closed C
differential Corollary. Proof:
Q.E.D.
If C divides the surface S, then 4C = *4 Is co-exact.
extend h to be 1 on one side and 0 on the other.
#4C =-4 =-4h.
Then
215
Lecture 25 u with the following proper ties: 1.
u is harmonic in S - pb
2.
u-
3.
u has a single valued conjugate v.
!..
f=u+ivisl-1.
i) is harmonic near p°
Let z be a local parameter which vanishes at p Let g
(al +
:
+...+ zn) in 0 < (zi < r1 define g suitably
for rl < Izi < r2 and 0 elsewhere.
g e C2.
0
By "suitably" we mean that
Let 6) = dg. a a (Z ++... n) in0< Izi
Let h = )suitably in r1< Iz1 < r2 elsewhere. Let Cr = dh.
Clearly t.!- * u a C. Applying the Diriehlet
Principle, we obtain a function fo such that cs)+ dfo
is
So = S - po. Hence '-:o = dg + df = d(g + fo).
harmonic in
Let u = g + fo.
Then u is harmonic in So.
Near pop
77 bnzn) + c log Iz F, but since D(f0) < co, actually
u =vat _(
07
+...an
u = fps a + &
If &
z
+
bn zn).
= du is exact, then u has a single valued conjugate.
That is, we would like to show that for any closed curve C , C C
+
0.
Once again we see that it is sufficient to show
°
this for any simple closed analytic curve C.
From the previous
lecture, we see that we can associate with C the differential
216
25.2
'dho. Hence
But
-(*4 ,GO)= -(dho,u)o).
by condition c' l of the Dirichlet Principle (4,10- *i5,df) = 0 for all f with D(f) < co. where h'
If, instead of J we use cT' = dh',
is different from h only in that we have it go to 0
in a different annulus rl' < jzj < r2.
In fact, we choose the
annulus so that the carrier of h' does not intersect the carrier
of o.
Then,
(dho,!JQ) = (dho,,;.o - *(f)+(dho,*(f- *0')+(dho, i) = 0. Hence *(Jis exact and *u)= dv. Hence we have f = :i1 + iv.
Ve
now want to show only that f is 1 - 1. Tde first show that if p1,p2 are any two points of S at
which df 4 0, then f(pl) 4 f(p2). f(pl) = f(p2) = ul + vl.
Assume the contrary, that
The level curves u = ul,v = v1 di-
'vide S into regions in each of which one of the following four sets of inequalities hold:
1. 2.
u>ul
,
u>u1
,
3. 4.
u
,
v>v1 vv1
,
v
Since df ==0 at pl(or p2), locally f-1 yields a 1 - 1 conformal mapping of a neighborhood of (ui,v1) onto a neighborhood of pl,p2.
Thus the neighborhood of pl,p2 is divided into four
regions by u = ul,v = v1.
a neighborhood of p
0
The same is true of p0, for f maps
1 - 1 conforn:.sily onto a neighborhood of co.
And this neighborhood of infinity is divided into four parts by u = ul,v = vi.
217 25.3
We prove now that S is divided into only four regions by these curves. That is, we prove that f-l(u < ul, v < v1) has only one connected component. Suppose one of these regions has more than one connected component. Then there is a region, call it D, such that 1.
2. 3.
U > ul, v > vl in D; v1 on the boundary of D; u = ul, or v po is not a boundary point of D.
Consider functions 4(C), 'ft L') a Coo for - ao <
< cop
such that 1. 0 = 3(ul)= 4' (ul)= 4"(u1) = Vs(v1) = V' (v1) _ O(v)
lu'I,
2. I4I,
1* 1,
1*1I <M
3. 4'i' > 0 except at (u1,v1). f4(u(p)
Define F(p)
tV
(v(p)) for p e D
in S
D. Then F = 0 in a neighborhood of p0. LO
Furthermore, Fx = 4',yux + 4j'vx = 4' *ux - Wuy Similarly, Fy = 6tuy + Jt'ux Thus F2 + Fy = (4'2;2 + 42*'2)(U2 + u2) < Klux + u9). Hence DS(F) = DD(F) < KDD(u) < Co.
Hence by c
c' , ((:jo - *Lf ,dF) = 0. ' 'e c.n choose a U'
so that its carrier does not intersect the carrier of F.
0= =
;%a, dF)
t
o, dF) = (du,dF)
vxFx + UY Fy
ux = uy = 0
contradiction.
Hence S is divided into four regions by u = ul, v = vl. these regions
Hence,
Label
218
25.4
I: u>u1 , v>v1 II: u < ul , v > vl III: u ul , v < vl And each of these regions has po,p,,p2 on its boundary. lying entirely in I and We can join pl to p2 by an are al pl to pl by an are allying entirely in IT. The curve al a2 = a is a simple closed curve and hence must separate S into two parts, D1, D2, since S is achlictartig. One of these parts, say D1, contains po. There are points of the regions II and IV in the neighborhood p0, so II and IV are both, completely contained in D1. Otherwise if II (or IV) meets D1 and D2 we would have II = (D1(II)L)(D2`)II) which is impossible, since II is connected. However, in a neighborhood. of pl, we
see that points of II lie on one side of a and points of IV lie on the other side. Hence II and IV cannot both be contained In D1. Contradiction. Hence f(pl) We have shown, moreover, that it cannot vanish at any point of S. for if df = 0 at some point pl, in the local parameter z about pl, f(z) = anzn +...,an / O,n > 1. But in the neighborhood of z = 0, f takes on each value n times at n distinct points of S, which contradicts what we have just proved. Hence we conclude that f maps S 1-1, conformally into the sphere. 74f(p2).
Q.E.D.
219
25.5
will denote An Application of the Uniformization Theorem: domains in the complex plane (For the more general case of don mains on a Riemann Surface, see S. Stoilow: Lecons sur les Principes Topologigue de l.s Thoorie des Fonctions Analytiques.). Definition 1:
A continuous, non-constant, complex valued
function f(z), defined on.i), will be called an interior function, or said to be tonolcgi.cally equivalent to an analytic function,
if there is a homoomorphtsm Z:
such that the function
f(z) = g(X(z)) = g(w)
is an analytic function on L. X is called the homeomorphic part of f, g the holomorphic part of f. We shall give two sets of necessary and sufficient conditions for a function f(z) to be an interior function. The sufficiency of the first set of conditions, due to Stoilow, will
not be demonstrated. A continuous, non-constant, complex valued function f;z), defined on, ;'2, is an interior function if, and only if, (Stoilow)i)f takes open sets into open sets; ii) f does I not transform a continuum into a single point; or equivalently: II In a neighborhood of all points of a discrete set r, f(z) is a ho_aeomorphism. The necessity of conditions I and II follows irranediatoly from the properties of analytic functions. The sufficiency proof for II is divided into three parts: Assume that f is a local homeomorphism at all points (1) As a domain in the complex plane, &)is a Tiiemann surface '-7e with the natural conformal structure of the plane itsalf. shall consider a new Riemann surface. f, whose points and topFor its open sets) are identical with those of/9. This zo e/%f we define local parameters by r,o f(z) - f(zo). is clearly a homeomorphism of a small enough disc (zi < p, and a neighborhood of zo. Also, any analytic function g(r.0) serves as a local parameter. Certainly two local parameters at a
220
25.6
point depend analyticnfly on each other; similarly for overlapping parameter patches, using the fact that an analytic function of a linear function is analytic. Therefore is a Riemann surface, on which f(z) is analytic. Furthermore, 4f is of planar character. It follows from the Koebe Uniformization Theorem that there is a conformal map,`.(z) of-Of onto a plane region A(with the usual conformal structure), and that f(z) = g(X(z)) = g(w) is analytic on6. (2) Now assume that d9 is a disc, and that f is a local homeomorphism at all points of,&except possibly at z = 0. Letting- be the punctured disc, we consider,Sr, and as in (1),
get a homeomorphism X:S --n
where (fit is a doubly connected
plane domain. Let be a domain whose closure lies in the union of and z = 0. Assume that z = 0 actually is in as is shown in the diagram below:
X(z) or
Y(z)
The image X(,&') contains ,((o).
;
X()y,)
Now 'k(o) is either a single
point or a continuum. If XC(o) were a continuum, then x(ti' ) would be confor:nally equivhlent to an annulus in the c-plane,
giving us the situation shown in the following diagram:
221
2S.7
conformally
/
\ inconformall
xar) ro
Let L
/
a sequence of points in the annulus which approach on the inner boundary. Corresponding to the rn are 0 From the continuity of f(z) at zn in b which approach a = 0. z = 0, we have the value f(o) corresponding to h(so), where h(r,) is analytic in the annulus. Since 4o can be any point on the inner boundary, we conclude that h() a constant, which is a contradiction. Therefore X.(o) is not a continuum, but a single point, Consequently, we may remove the singularity of g(w) at w = X(o). With no loss of generality, we may assume 7(o) = 0, so that near w = 0, ' be
a point
g(w) = awn(1 + a1w + a2w2 +...) Therefore, in some neighborhood of z = 0,
Z0 =M,/f(z) - f o) is a local parameter. (3) Now suppose that f(z) is a local homeoinorphism on k' = j- _, where is a discrete set. Again we consideri''f with local parameter ?o = f(z) - f(z0) around each z a ,Cq'. 0 Putting discs around the zi in we use (2) to remove the singularities and take i
4i =
m
i f z) - M.7
as local parameters there.
26.1
222
We return to the discussion of differentials on S, with the purpose of defining Abelian Integrals, and their periods; and ultimately of proving the Riemann-Roch Theorem. Definition 1:
A differential of type (m,n), m,n integers, on a
Riemann Surface S is an assignment of a measurable function 4)(z)
to every local parameter defined in a domain Z) such that if S is any other local parameter in then (2:)dedrn
(z)dzmdzn = where dives
belonCs to the parameter 4. = 4(,)(a;,)m(=)n.
d.,
A formal calculation
d
Definition ?: A differential of type (m,0) 4(z)dzm is called is holomorphic, i.e. regular holomorphic (meror:orphic) if
analytic (meromorphic) in the domain of the local parameter z. Definition 3: A meromorphic differential 4)(z)dz of type (1,0) is called an Abelian differential. If 4)(z) is holomorphic, then 4)dz
is said to be an Abelian differential of the first kind.
If 4(z)dz is an Abelian differential of the first kind in some parameter domain and 4) = u +iv, we may write
¢dz = (u+iv)(dx+idy) = (udx -vdy)+i(vdx+udy) where it is easily verified that both (udx -vdy) and (vdx + uciy) are differen-
tials; and if we write a = udx -vdy, we see that *a = vdx +udy, The Cauchy-Riemann equations ux = vy, uy = -vx imply that a and *a are real harmonic differentials. so that 4)dz = a + i-a.
That is
da = uydy /adx - vxdx A dy = - (uy +vx )dx at dy = 0 proving that a is closed.
Similarly dea = 0. Conversely, if a Is a real harmonic differer.tial, then *a is also a real harmonic
differential, since as = dl** = 0 =* d*a = d**a = -da = 0. The differential *a is the contuaate harmonic differential to a, and and the differential (,. = a + i*a is an Abelian differential of the
first hind, as can be checked by means of the Cauchy-Riemann It is also easily seen that 41W = -i W for u) equations. holomorphic.
26,2
223
In a previous chapter we defined
,
a for closed
differentials a. By the linearity of the integral j W is automatically defined for Li holomorphic. Definition 4:
An Abelian integral of the first kind 16 0 on S
is the integral of an Abelian differential of the first kind. As an example of an Abelian integral of the first kind, consider
where (w(z))2 = (z -a1)(z -a2)...(z -a2g+2).
The Riemann
surface of w(z) is the hyper-elliptic surface of genus g, with branch points at al,a2,..,,a2g+2. To prove that this is an Abelian integral of the first kind, we must show that the integrand, expressed as a differential on the Riemann surface, has no singularities. At a branch point, say al, that t =
z - al as a local parameter.
dz
-z- 4
... 1z -a2g+2) r
Then
22=
al a2)... (e + al
proving the differential is regular there.
2z+2)
224
26.3
Let Al and A2 be differentials of type (m,n) and (k,,Z) reepeoti-vely; then we define
i) the product of Al and /'2, /x 142, to be differential of type (m+k,n +M, i.e. if locally Al - 4idzmdzn, then 4142dzm+kdZn"
A1A2 =
y
ii) the quotient nl//\2 is to be the differential 2dzm-kdin-R.
Two differentials of the same type may be added and
subtracted in the obvious way: $dzmdzn +- +ydzmdzn = (11 *)dzmdzn
.
The following properties are easily checked of two meomorphic differentials of the i) The quotient
w w
same kind is a meromorphic function on S. ii) If t is a meromorphie function and CO a meromorphic differential, then 4 c, is a meromorphic differential. iii) If a meromorphic differential cti) can be represented in a neighborhood of a point by a local pr'ameter z, so that 00
00
aizi)dz or cj =I E aizi)dz, i.e. W has either
either w =1
M
-n
a pole of or\der n or/ a zero of order m; then for any other local parameter
= 2:(z), the numbers n or m remain the same.
Theorem 1:
Let co be an Abelian differential on S.
Then if
N(W ) is the number of zeros counted with their proper multiplioities, and P(cJ) the number of poles with their multiplicities, we have
N(w) - P(W) - 2g -2 Proof: We must show two things:
First, that for any two
Abelian differentials 4,)l, W21N(W 1) - P(W1) o N(° P-) - P(cJ2);
second, that for a particular 4U, N (L4) ..P(c) - 2g -2.
225
26.4 1, 11) 2, i02
0 are Abelian differentials on
1/u?2 = f is a meromorphic function on S.
S
If N(f) and P(f) are the number of zeros and poles of f counted with their multiplicities, then we know that N - P = 0. It follows that
0 = N - P = N(uJ 1)- P(C.J1)-(N(v.)2)- P(u02)) ii) Let = f(z) be a meromorphic function on S. Then d? fzdz is an Abelian differential. We assume that the n poles of ? are distinct. (If not, for A large enough, = A at n distinct points, so that we may replace r, by-A l .) Z defines an n-sheeted covering of the plane. At a branch point of order (k-1)
4(z) = zk + 0(zk+l); dZ _ (kzk-1 +...)dz so that the branch number b z N.
akzk and d4
+
z
k=1
-
z
At a pole of dZ, we must have k ak
+
zk-l.
Thus, p = 2n,
1
so that
N - P = II - 2n = 2 (n F P-l) - 2n = 2g - 2. For g = 0, N - P = -2, implying that there are no Abelian differentials of the first kind on the sphere, which we already know from Liowville's Theorem. In Lecture 11 we found that every compact hiemann surface S of genus g is homeomorphtc to a regular 49 sided plane polygon with certain identifications. We would like to get a similar Clearly, it conformal representation of the given surface. will not be a plane polygon for g > 1. For in this case, the sum of the interior angles of the 49 aided polygon is greater than 2a, and since all vertices are identified, is. they all represent the same point on S. the mapping would not be conform-
al there. It is possible to represent S as a 49 sided non-Euclidean polygon with identifications and the usual defining relation: 1
1=1
ai bi ail bit= 1
226
26.5
This is celled the canonical polygon. However, this involves a rather detailed procedure. We shall represent S as a "polygon-like" domain of the non-:,uclidean plane. This will be a domain bounded by 49 twice continuously differentiable arcs with the usual identifications and relation. In fact, it will differ from the canonical polygon only in that its sides will not be non-Euclidean straight lines. It has already been shown that S may be represented by
the so-called normal or metric polygon in the non-Euclidean plane. This polygon has, in general, more than 4g sides. We map this polygon by means of a twice continuously differentiable mappinS, onto a regular plane polygon with the same number of sides and identifications. Repeating the arguments in Lec-
ture 11, this may be put in the normal topological form, by means of mappings that are in C2 or better.
The aides of this
polygon correspond to the curves al, bl, all, bl3... in the non Euclidean plane with the usual identifications and the relation
I ai bi ail bil
,
which gives us the desired representation of S.
Thaorem 2:
As a vector space over the real numbers, the space of real harmonic differentials on surface of genus g has dimension 2g. Proof:
Let (al,bl,...,ag,bg)be a canonical dissection.
numbers Al =
a , Bi =
I
a
The
are the periods of the differen-
bi ai
Since every tial a around the curves of the homology basis. closed curve C' is homologous to some linear combination, of the
ai and b1, the period of the differential a around a is a linear combination of the periods AI and Bi. First we show that the dimension is not greater than 2g.
Let a and p be two real harmonic differentials such that
a=
Ai= ai
j ai
P
,
a=
Bi=
i
{3
i
Then a - j3 has all its periods equal to zero.
227
26.6
We form the function
p
F(p)
S (a -P) po
F(p) is a single-valued harmonic function with no singularities, and is therefore a constant, implying that a - P = 0.
To show that the dimension is actually 2&, we must produce 2g linearly indeppAdent real. rmonio functions. These differentials (il,...,R will have the following properties:
ai=6S o
f
J
b'
a1
0
bJ
ai
We have already seen that if o- is a closed differential with period Ai and Bit then there exists the following decomposition
a'= a+df where a is harmonic with periods Ai.. Bi and df is exact.
Therefore it will only be necessary to construct closed differentials with the desired periods, and then appeal to the decomposition theorem to get the harmonic differentials. We shall construct a
closed, differential 0' whose periods agree with those we desire for al. The others are constructed similarly. We define first the function (0
1
h(z)
In C2
it 0
_1
a1
z e I or IV
z e II z e III z e p2 q z e p1 ql, p3 q3 a1
where the polygon is partitioned into the regions I, II, III and IV, as, shown i:. the accompanying diagram.
26.7
228
Now we define the f-
L0
z
p2
z
CpLq
Clearly cre Cl and dc' = 0 so that cc is closed. Since cY= 0 outside III, q2
ql
al
0-=0, iTl;
c.h _ l; ai
S
b
(T= 0,
i _ l,. .,g.
i
The vector space of holomorphic differentials has real dimension 2g.
Theorem 3:
Proof:
For every real harmonic differential a,() = a + i*a
is a holomorphic differential. Conversely, every holomorphic differentials.! determines a harmonic differential a such that
w=a+Ia..
This establishes a,i iso,norphinm between the space of holomorphie
differentials and the space of harmonic differentials, implying that they have the same dimension. Theorem h: The vector space of holomorphic differentials has complex dimension g. Proof: obvious We shall investitate certain relations between the periods of Abelian differentials, and give necessL:ry conditions for a collection of complex numbers to be the periods of a certain normalized basis for these differentials. First, we observe tort, given the real part of a. period of
a holomorphic differential
rJ ,
the imaginary part is uniquely
determined by the decomposition r.)= a + i:.a, where a is a uniquely
determined real harmonic differential. Furthermore, we shall see that the periods over the ai uniquely determine the periods over the bi.
To see this, we
need to derive the Bilinear Relations of Rtemann. Theorem $:
Letr.,,r,,'be two closed C1 differentials with periods
Ai, Bi, and Ai, Bi respectively.
Then
229
P6.8 .
i
(a)
(A D '
- A' B') = 0
if W 4 0 is holomorphic, then
i E (A3
(b)
B3A ) > 0
Since the polygon is 41 is a closed Cl differential. simply connected, there is a function f(z) such that W = df Proof:
In general f(p) 4 f(q) where p and q are identified
in Tj .
points on the boundary r of Tj , so that f is not single valued on the Riemann surface. We obtain
f(p) - =(q)
(1)
I
=
1w = B1
p
s
since
=
J+
0
q rl Similarly for the other bi and ai.
It follows from the simple connectivity of TT , that r' And since f.v' is a closed differential
is homotopic to a point.
in TT , we have
r fwS = 0
26.9
230
(2)
l fw I
+
ai+bi+ail+bil
fool +
+
ai
f0 -
ft,' + Bi
; ai
wt + A 1
f0
)(f+B1t - i (f-AS)1 ai
i =
c
bil
'ai
bi
bi
0 = -BiAi + AiBi bi
where we used (1) to get the second equality.
Summing, we obtain
part (a) of the theorem
(3)
0
AiBi - BiAi
which is called tie first bilinear relation of Riemann. If 'w,c i1
are holoiaorn`;ic, then Stokes' Theorem applied to where the
are the trirnCles shown in the accompany-
ing
r proving that a holor:,orphic differential and the complex conju-
-ate of a holor.orphic differential are orthoZonal with respect.
to the scalar product we have already introduced in Chapter 22.
26.10
231
Part (b) is similarly obtained: R
(
II* w+ = i
)
Tf
=i
I,
n
ft.i
=
(A,'a J
-
BiKj, )
I-
so that for 4j= J, 4 0, we get (5)
II'II2 = i 7
(Ai
Bi - B, Ai) > 0
which is called for second bilinear relation of Riemann. Theorem 6: If "and w+ are ti-o holomorphic differentials for which Ai = Ai, I. = 1,...g, then Bi = B That is, the ai-periods
of a holomorphic differential uniquely determine the bi-periods.
Proof: Let 0 = 4)-,v+
.
Then A; = jo.
IYrom ( 5)
aj
i 0 II? = i = A' 9; - B9 A; = 0, implying that 9 = 0 and that &.J =6J .
:From formula (5) we obtain Theorem 7: If c) is a holomorphic differential all of whose periods are real, thenWJ = 0. Theorem: If 4- is a holomorphic differential all of whose
ai-periods (or bi-periods) vanish, then do = 0.
232
27.1
Lecture 27 Definition 1:
wl, ..., wg
A basis
for the Abelian differentials
of the first kind is called a normalized basis if Theorem ].:
Sa iwj
= ail
There exists a normalized basis for the Abolian
differentials of the first kind. Proof:
Let
w,,
..., wg
w3 =
be any basis with Is
We must find complex numbers
Ail
such that
form a normalized basis.
g ,
g
c,i
i
wi = 1 ajiwj This means we must solve
systems of equations n lcit A.k =
k = 1, ik
This is possible if and only if
..., g
i=1, ..., g det 'Ajkl A 0.
If the determinant
were equal to zero, then there would be a non-trivial solution for A
k
c3 AJk = 0
implying that
w =
clw1
which in turn implies n the w1 Definition 2: (T1j)
where
Let
for the surface
(albla2b2... )
.
S
has all its a-periods equal to zero,
w = 0 , contradicting the independence of
wl, ..., wg S. w1
T4
1, ..., g
be a normalized basis.
The matrix
is called the Riemann Period matrix
with respect to the particular homology basis
?33
27.2
Necessary and sufficient conditions for a matrix to be a Riemann period matrix are not known.
We can, however, derive two
Any matrix satisfying these two conditions
necessary conditions.
will be called a Riemann Matrix. Theorem 2:
(a)
(b)
If
is a Riemann Period Matrix, then
('tij)
't11 _''i (.im'ri3) is positive definite. i.e. . '7i 'Yi, . 0 for all real vectors ..., xg) +
(xi,
Proof:
(a)
(0,
..., 0)
.
Near the end of the last lecture, we found that
0 = (w, We) = - i i A1Bi - B1Ai i=1
Setting
w = wi
,
wI
= 0)k ,
0 = (co, jwk) since
Aui = 63i
and
we get
it AJ1 Bki
[ tki- Tfk]
- BJi Aki
Bki
= zk1 Here we use the second bilinear relation, '
(b)
11 w112
= 1
kl Ai3
- i > 0
for
sr
0
27.3
234
Let
w = xiwl + x2w2 + ... +
all zero.
Since the
wi
x909 , where
xi
are real and not
are a normalized basis, we have
AJ=Jaw=xi and
JL BJ
kJ
k=r1
Thus
Iw112
Definition 3.
L x( k=l xkTkj j=1
=i
J
)
-
A meromorphic differential is said to be of the
second kind if all. its poles are of order at least 2.
That is,
locally it looks like a-7
R
w =
where
n > 2
(
zn
+ ... +
z2
+ regular function)dz
.
Hence, locally,
w = dF , where
F
is a single valued mero-
morphic function.
We shall show that for any point p e S and for a meromorphic differential of the second kind which is regular on 3 - p and at
p
looks like
235
27.4
n +
regular function
L
It has been shown that for on
n > 2
there is a real function. h(z)
satisfying
S
(i)
(ii)
is harmonic on
h(z)
S -p
dx (zri1) + harmonic function near p
h(z) =
Then
dh +. i *dh = (hx - ihy)d(x+iy) is such a meromorphic differential, as is easily checked. Indeed, h.- ihy
is locally analytic on
S -p, i.e. the Cauchy-
Riemann equations are satisfied, and it has the proper singularity.
By taking linear combinations of such differentials, we see that for arbitrary points 2,,,,,n,, ,
pl,...,pk, and numbers al
1
,,,,,ak
jk
there exists a meromorphic differential with
principal part
a/n Q z
at p1 ,
1 = 1,...,k.
+ ... +
+
(
z3
n
) dz
z
Two differentials with the same singulcrty
differ only by a holomorphic differential, so that a differential,
of the second kind is uniquely determined by its singularities
and its a j-j,eriods.
27.5
236
An Abelian differential of the second kind is said
Definition 4:
to be normalized if all its
a,- periods are zero.
Any meromorphic differential is said to be of the
Definition 5: third kind.
A differential of the third kind is called normalized
Definition 6: if (i)
(ii)
it has no singularities of order greater than 1. at two distinct points
singularities (iii)
and
zz
p
and
q
on
S
it has
,
respectively.
- zz
a,- periods are zero.
all its
p and
We shall show that for any two points
q
there exists
such a normalized differential of the third kind. AFirst, assume that
p
and
q
lie in the same parameter disc
of generality we may assume that and
z = + b
respectively.
valued function
(i)
(ii)
h
and
q
With no loss
correspond to
z = - 6
Consider the f:netion
h (z) = log 0
defined in Vol .
p
yv .
1-6 1
I
z+b
We have already shown that there exists a single h
on
3
such that
is harmonic outside
h - ho
is harmonic in
Then, as in the previous construction,
(hx - ihy)d(x + iy)
27.6
237
is an Abelian differential on S with the proper singularities. Subtracting the appropriate holomorphic differential, we can make the a3-periods zero.
The uniqueness is demonstrated in the usual
way by taking the difference of two such differentials, which turns out to be a holomorphic dif-
p = pl
ferential whose aJ-periods vanish.
For the case where p and q do not lie in the same parameter disc, we join the two points by a curve,
cover the curve by means of a chain of parameter discs so that
each one contains the center of
Q = Pk
its predecessor, and finally,
choose from this chain or cover-
ing, a finite chain or covering (Heine-Borel Theorem).
Let p = p1, and let p2 be in the fiat two parameter discs, p3 in the second and third, and so on. ferentials of the third kind
Form the nonaalized dif-
which are such that Wl
has singularities at p1 and p2, W2 has singularities at p2 and p3
etc. Then O= Z;3i is the desired differential. Lemma 1:
The sum of the residues of an Abelian differential on a
compact Riemann surface vanishes. a Proof:
Suppose
hood of a point p.
zl + regular function)dz in some neighborThen if a is a closed curve around p which
does not surround or pass through any other singularities, we have 1
)n a-1
27.7
238
so that a-1 is independent of the local parameter.
Triangulate S by means of regions df so that (i)
each Z\,j is contained in a parameter disc
(ii)
each singularity lies in the interior of some
(iii)
each Lj contains only one singularity.
Then the sum of the residues is equal to
j
where Yf = apj.
YJ
But since the surface is compact the integrals
over the various boundaries cancel each other.
That is, for each
side of yj we get contributions
C) and -
+
CO from the two ad-
jacent triangles. Lemma 2:
Let(J3 be a differen-
tial of the third kind satisfying conditions (i) and (iii) of Definition 6.
Then
w%=LcGf3 where therJ3 are normalized differentials of the third kind. Suppose that CJ is to have poles at p1,...,pk with residue cl,...,c?.
Let q be a point on S distinct from the pi and let
have poles at pi and q with residues + ci and - of respectively. Then I cicoi has poles of order one at the pi with residues oi.
Furthermore it has no other poles. it's re-ular at q.
For, since I of - 0 (Lemma 1)
239
27.8
Uniqueneas follows in the usual way.
Lemma
Let
w be an Abelian differential (third kind).
Then
w may be written uniquely as
w= wl+w2+w3 where
wl
is of the first kind,
w2
of the second kind, and
w3
Proof:
has residues
Suppose that
co
is a normalized differential
is of the type described in Lemma 2.
ol, ..., ck
at
pl, ..., pk .
Form k
as in Lemma 2.
with
Then
a, - periods
w - w3 Al,
is a differential of the second kinc
..., Ag .
Let
w2 ,
...,
wg2
normalized basis for differentials of the first kind.
wl
l
AJ w1
the differential
w-w3-wl=w2 is a normalized differential of the second kind.
be a Letting
24O
27.9
Theorem 3: (Bilinear relation for differentials of the first and third kind) let
Let
be a differential of the first kind, and
wl
be of the third kind with residues
w3
Pl' ..., Pk
respectively.
poles.
(a1bl...)
the
Let
pi
aV b3
wl
I
i
Suppose further that
J
i
has no other
wl
bi
ari
w3
Bi
w3
Ai
bi
Jai
from
w3
at
and call
Bl
J
Choosing a point
..., ck
be a homology basis chosen so that none of
lie on any of the
Al
cl,
q
on
S ,
and letting
be a fixed path
(bi
to p i , we have
q
(A1 B1 - B1 Ai) = 2:ci
k
cj
J wl
.
01 Proof:
We shall assume that
region of a Fuchsian group,
S
is given as the fundamental
In fact we use the canonical repre-
Since the canonical
sentation introduced in the previous chapter.
polygon
n
is simply connected, we may conclude that there is a
regular function f(q) = 0
.
f
such that
wl = df
in
n
and such that
Using formula (2) of Theorem 5 of the previous chapter,
we get
i=1
Al Bi - Bi Ai
k
fw3 = 2%1
rj
27.10
where But
rj
is the residue of
241
at
fc,33
p'
,
r3 = f(pJ)Cj
so that
f(pJ) = I tl , which proves the theorem.
Theorem
(BJi'inear relation for differentials of the first and
second kind).
w2
Let
a single pole at
p
be a differential of the aeoond kind with
and principal part
d
, n > 2
.
Let
wl
z
be any differential of the first kind.
In a neighborhood of p
wl = (a0 + a z + ...)dz I
.
Then setting
Al =
Bl =
L(J1
we have
cal
an-2
(Al B2 J J
n-Proof:
B2 A2) 3
2,1
Again we use formula (2), Theorem 5 of the previous
chapter to get r
,E(A; B2 - B1 A2) =
where
wl
= df inn ,
which is equal to
a n-z n-1
and
rq
J fw2 - 2ai rq
is the residue of
fw2
at
q
24 2
27.11 An immodiate result of. Theorem 4 is
Theorem w2
suppose that
In addition to the hypotheses of Theorem
is normalized
ai
w2 = Ai = 0)
and
wl
= (0
(Al
= bk J
Then if
akz + ... )dz
w l k = (ao +
Near
p, k 2
J b
For n - 2
,
an
k
n-1 w1
so that
_2 = ao _ (dz)z=0
wl f w2 = tai (dz)z=0
bk It is natural to ask whether there exist meromorphic functions or Abelian differentials with prescribed singularities or This question is in part answered by
zercs, and if so "how many".
the Riemann-Roch theorem, which we shall eventually prove in its most general form. Let nk
nl,
on
S
But first we consider some special cases.
pl, , pk be
k
distinct points on
be positive integers.
S
,
and let
Consider the sot of functions
(it may be empty of course) which are regular everywhere
except possibly at the
pi , where they are permitted to have poles
of at most order
If
where
a
ni
.
f
is a complex number.
is such a function, If
f
and
g
then
Xf
is too,
are such functions
27.12 f + g
then
213
is one, proving that these functions form a vector
space over the complex numbers.
Similarly the set of Abelian
differentials of the first kind which have zeros of at least order ni
pi
at
form a complex vector space.
of the first space by w1
a_id the second by B , we shall prove
A
k A - B = 1 + 7lnj -g .
that
Denoting the diversion
nj =1 .
First we assume that
Let
be the normalized differential of the second kind which is
regular everywhere except at
part a2 .
where it has as its principal
pj
Consider the normalized differentials of the second
Z
kind
where the
ti
are complex numbers. w2 = 0
such that
We would like to find t3
This means that we rmist solve the following
.
system of liAear equations k
1
ti
w =0
i = 1, ... ,
g
.
bi If the system has a non trivial solution, then for a particular solution,
w2
is an exact differential (its periods are all zero),
so that there exists a function at the
Clearly
pi f
except possibly
S
where it may have simple poles and such that is unique up to an additive constant.
we have b from the arbitrar$ constant. let
f , regular on
p = rank () w2)
A = k + 1 - p ,
wZ = df
Thus, if we the
1
resulting
244
27.13
B , we form the expression
To find a similar expression for
siWi
where the wi are a normalized basis for the holomorphic differenWe would like to choose the
tials. p, .
so that
si
wl = 0
at the
If we assume that in some neighbcrhood of p, each
Wi = (aI
+ alfz + a2fz2 + ... )dz
we are left with the following homogeneous system to solve
(,)I
siaof = 0
(p;) =
,( = 1, ..., k
3=1 Thurefore if we let p = rank (aof) , we get at the case and
W2)
( f
n = 2
B
g - p .
Looking
in Theorem 5, we see that the two matrices
are transposes.
Thus
(apj)
p =p and
bi
A - B = 1 + k= g Now let us assume that
ni + 1
.
necessarily.
of the proof is essentially the same as for
Since the Idea
nj = 1 , we shall omit
certain details and explanations which we think that the reader can quite easily supply for himself.
For
i = 1,
..., k , we form the
following normalized differentials of the sooond kind which have the prescribed singularities at
pi
and no others.
2#5
27.1k
as
2
z2
k
Let
= dz
2
wi2
,
wil
w2 =
...
2
z
1
2 wij
f 1tip
as ni+
wins =
,
z
We want
.
i.=1 Therefore we want to find
w2
to have zero
b, - periods:
such that ti,)
tt'b
z
2
wi3=0
1- 1, ..., g
Therefore the dimension of the space of functions on regular except possibly at the
have poles of at most order ni
k A
=1
where
p = rank (
bI
Set
wl =
siwi
.
which are
where they are permitted to
is
ni+1 - p
)
cc
We recall that at
wi = ((x o of
pi
S
p1
+ ai jz + all z2 + ... ?.e want to choose the
) dz
si
so that
wl(p,)
i=1 has a zero of order n1 equations
.
This means we must solve the system of
24 6
27.15
f = 1, ..., k
.
If rank (ant = p , then the dimension of the space of holoiaorphic differentials with zeros of at least order B = g - p
.
is at p1 nI Again Theorem 5 allows us to conclude that'we have
transposed matrices, so that
A-B
p = p
k 1=1
and
n i+1-g
In a previous lecture we found that an Abelian differential of the first kind has exactly k
1
ns > 2g - 2, .re have B = 0
2g-2 zeros.
Therefore, if
28.1
247
The relation A-B = kthat we derived in the-previous chapter is a special case of the Riemann-Roeh theorem for ni > 0. In this chapter we shall complete the proof of the general theorem. In order to do this, it is convenient to introduce the notion of a divisor. Definition 1: Let pl,...,pk denote k distinct points on S and let all.... ak be integars. Then the formal symbol
will be called a divisor. Definition 2: teger
ai is called the order of a at pi.
The degree of a divisor a = pli...pkk is the in-
k
deg a =
5 i=1
ai
may define multiplication and division of divisors in the obvious way: If a = pal...pkk, b - gR...gJ then akk
ab =
1
a li...q/
a
a
; b =
p
a
- 01...q/- aj
11 ...p kkQ1
Requiring multiplication to be commutative, and letting 1 denote the divisor which is of order zero at all points, we see that the divisors form a multiplicative Abelian group. Frc.. the definitions of multiplication and division we have deg ab = deg a + deg b ; deg b = deg a - deg b Definition 3:
A divisor a is called integral if the ai > 0.
We shall say that b divides a if a/b is an integral divisor. The symbol b/a is to be read " b divides all or "a is a multiple of b." Definition 4:
248
28.2
DAfini:ion 5: Let f -- 0 ob a meimorphic function on S, and let 0 be an AbeliF_n differential on S. We define th3 divisors of f and CL' to be vpI' (f) vp (f) {f} =P1 " ....pk v
v
glgl
(CL,j
....q1 A
where the pi are the zeros and poles of f of order vo (f).
(see gaga 98). Siiiilarly for (Z.
1
It is easy to see th^t the following relations hold for meromorphic functions f and g, and an n differential 63: (f)(g) _ (fg)
(f):W) _ (fw) (c) = 1 ,
a is constant.
-le know that on a compact Riemann surface S;
deg(f) = 0 and deg(cc) = I vqi (6)) = 2; -.
p e3
vP (f) = 0 is
t:ze divisor of a me_,omornhic function, If a Definiti^n 6: then a is said to be a div?ser.
Ahe p.incipal divisors fom a subprou? P of the group of divisors. We form the quotient group of divisors modalo principal divisions. An element of the quotient group is an equivalence class of divisors, Denoted by a whore a,b a a} if and only if there exists a meroraorphic function f such that
a/b = (f). a - b is to be read "a is equivalent to b," and will be used to express the fact that a,b e "al . The class {al = A will be called a divisor class. Clearly, a b ==:>deg a deg b.
28.3
21.9
Definition 7: The class of divisors that are divisors of Abeldifferentiala is called the Canonical class and is denoted by 1.11
Let a be a divisor.
Consider the space of me.orcor-hie
functions Fa such that f err^a if and only if 1/aj (f) . That is, a(f) is an integral diviaor, cr (f) = c/a where o 's integral. Clearly Fa is a vector spece over the complex numbers. Definition 8:
The dirnene4on of a divisor a is the dimension of
the vector space Pa th^.t is dim a = diia FA. Theorem 1:
If a,-- b then dim a = dim b.
Proof:
b
a -
a/b = (g) where g is some meromorphie func-
tion on S.
h e Fb => -1h ar'(h) = c/b where c is integral. Since h/g is a meromorphic function, we have (h/g) = (h)(1/g) _ o/az Thus h/g a Fa giving a 1-1 correspondence h <-> h/g between elements of Fb and Fa. Therefore
dim b=dirnFb=dim Fa=dirra . 3ccause of the previous theo: em we icy talk about the dimension of a divisor class, A = ka }. That is we nay define din A = dim a where a is any in A.
Definition 9:
The index of F. divisor a, written ind a is the
dimension of the complex vector space Da, w7hare Da consists of Abelian differentials !' such test al((o), or (c) = ac, where c is integral. As before we can write ind A = ind a.
We nm1t the proof
as it is similar ;o the proof of Theorem 1.
If 1 and cJ2 are Abelian dif£erencials on S. then Cal/c'`2 is a raeromo phic function, so thot ,J1 ti 43 2 and deg deg(4)2).
26.4
250
The following theorem was proved in Chapter 26. We 'hall reprove it using the language of -divisors and the special case
of the T?ieman-Roc's theorem proved to the previous chapter. Theorem 2 :
If (t`is an Abelian differential on a compact sur-
face S of genus g then deg («)) = 2g-2 .
Let G) be a holomorphic differential. Then [J= plk...P k ai > 1. Using the notation of the previous chapter we have Proof:
A = dim (w) , B = ind (W) and k A-B =
ai+l-g
Let Wi, (a`2..,c;,91 be a Saris for Abelian differentials of the
first kind.
Then
dim(W1)-ind(W1)
=
Ind, d, if (O is an Abelian is integral so that differential in P(O1) then a holomorphic function on S and i) therefore a constant. It is also easy to see that di:n(t,) ) = g.. For W 1 .2 First we show that ird (r.)1) 1 = 1.
'
l Li
are 2 linearly independent functions in
To see that. the: a are at most g independent;, functions, suppose that f e F(IJI) . I'hien f c is r. holomorphie differential and
ar
28.5
251-
1 Therefore g-1 = deg(cJ)+1-g or
deg(iCl) = 2g-2
But if CJ is any Abelian differential deg(! ) = deg(C Theorem 3:
= 2g-2
ind a = diia((6.`)/a), or as we shall usually write it,
ind A = dim W/A, where W = Proof:
)
(63)1.
We must show that
dim Da = dim F
a a
F((J)/a consists of meromorphic functions which are multiples of
a/(O), and Da is composed of Abelian differentials which are multiples of a. Thus for every integral divisor o, we have a function f e F((,,)/a and a differential cJ e Da such that (f) = c(a/('J)) and (L.`) = ca. This establishes a 1-1 correspondence between the two vector spaces, ,:uving that their dimensions are the same.
Definition 10: Let A be a divisor class and let W = Then W/A is called the complementary class of A. The oren !.:
(F'iemann-Roeh).
nus g _u d 1st a be a i.:visor.
Let S be a compact surface of ge-
Than
dim .?. = ini_ A + deg A -g+l where A is the divisor class of which a is a representative. There is a more symmetric but probably less useful form of the theorem contained in the following:
28.6
25 2
The expressions
Lemma 1; (i)
dim A = ind A + deg A -g+l
(ii)
dim A - 2 deg A = dim
A
-
z
deg
a
are equivalent. Proof:
ind A + deg A-g+l = dim
+ 2 deg A +
deg W
2
A since deg W = 2g-2.
deg A -
But this is equal to
dim A11 +
1 deg Adeg A 2 2
which proves the equivalence.
Lemma 2:
If A = a } and B
then ab
= W.
Proof:
=
b j are complementary classes,
From definition 10, B =
Hence
deg ab = deg a + deg b = deg A + degA = deg W In the previous chapter we proved the Riemann-Roch theorem for an integral divisor a, and hence for any A which contains an integral divisor. From Lemma 1 we see that Theorem 4 holds for A if and only if it holds for W/A.
Hence the theorem is true for A if either A or 1./A contains an integral divisor. Now assume that neither A nor W/A contains an integral divisor. For this case dim A = 0. If dim A # 0, it must be that for any a e A there is a meromorphic function f for which
(f) _ (l,'a)Q where c is integral. But since f is meromorphic, c
a, and hence e e A which contradicts the assumption that A
contains no integral divisors.
By the same argument dim A = 0.
253
28.7 It follows from Theorem 3 that tzzd A = 0 %!so.
sins we have
already proved the Riemann-F'oc'_i theorem for a divisor class A.
which has the property that eitier A itself or W/A contains an integral divisor, it cuffices to show that %.hen neithor A nor %;/A contains integral divisors, the expression
Jag A = g-1 holds. It is clear that any diviscr a a A may be written uniquely as a = b/c where b = 41 ...qkk, c = qi / r, ,
qi # qj, r; / rj, for any pair of i, ', and (ii, -I,j > 0. This will be called the reduced form of a. Finally we shall let k degb='1=1 Ri=r, dego
j=1
Yj=s
Now let us consider the space ;a for an arbitrary divisor a = b/c, written in reduced form. f e Fa <:-a c/bl(f) or (f) = (c/b)d where d is integral. Hence f is a meromorphic
function that is required to have zeros at the points ri of order at least yi, and is permitted polca at qi of order no greater than (ii. We shall show that such functions exist, i.e.,
dim A / 0, it and only if r-s > g-l. deg a = deg A = r-s
But
,
so that fox- .ri A = 0, (which is the case we are now conside. ing), we shall have proved that deg A < g-l. in order to construct f we use the normalized differentials C.?ii of the second kind which were introduced in the previous chapter. At qi, i = 1,...,k,
Oil - a2 , <.?i2 - z j , ..., W 1pi
=
p-,+1 Z
The differential
28.8
254
k
Pi
j=
i
i3
We restrict the so that Sb n= or ai3 n = 1,...,g. If r > g this is possible. Then there exists a meromorphic function f on S, unique up to an additiv3 constant has zero ai - periods.
such that ti)= df.
These functions form a complex vector space
of dimension r+l-g. Requiring f to have a zero at each ri of order yi, imposes s more conditiuns. Thus for P. to be non-
empty it is necessary and sufficient that r+l-s-g > 0. In a similar manner we get deg W/A < g-l. But -then, 2(g-1) = dog W = deg A + deg W/A < 2(g-1) so that deg A = g-1, which completes the proof of the Riemann-Roch theorem. To see how the Riemanii-Roch theorem answers some questions about the existence of meromorphic functions with prescribed poles and zeros, we shall include a few simple applications. Application 1:
Since ind a = dim Da > 0 for an arbitrary divi-
sor a, it follows t:. at dim A > deg A-g-l, from which we may immediately conclude that: (a)
dim A > 0 if deg A > g-1
(b)
A > 1 (i.e., Fa contains non-constant functions) if deg A > g.
0 there is a non-constant f e Fa with (f) = b/a where b is integral. But then deg -a = dog b > 0, which implies that dim A = dim a = 0 if dog a < 0. Application 2:
If dim A
Application: If deg It > 2g-2 then deg
deg W - deg A =
= 2g-2 - deg A < 0. From hpplication 2, ind A = dim W/A = 0 . FpplicetionJ: Let a = p. Then ind a = g if and only if all differentials of the first kind vanish at p. We shall show
28.9 that this can not happen.
255
For, if ind a = g, the Riemann-Roch
theorem gives u3 that
dt:ma =dega+l=2. This means that there ¶s a non-constant meroi.:c:phic function on S with a single simple .ole, which implies that S i.r
confor many
Thus, for a = p. ind a < g.
equivalent to the sphere.
Application:
If a is a non-trivial integral divisor, then there is a divisor b = p (i.e., there is a point p e S) such
that b l a.
Then Da C Db and ird a < ind b, so that dim a
< deg a+l
if a is intogral, with equality if and only if g = 0. Application b:
For an integral divisor a, dim a > 1, so that
deg a + ind a-g > 0. If a = pl,p2 "'pn' p1 # pJ for I # J, then deg a = n and ind a > g-n. For al = Pi this becomes ind a1 > g-l. From Application 11, we have ind al < g-l so that ind al = g-l for al
= pl. Dal and ind a2 < ind al. Let a2 = plp2 Then al Ia2, Da However, if ind a2 = ind al = g-1, ?here exist g-1 linearly .IZ dependent differentials in Da which vanish at p2. If g > 2
there is a non-trivial differential (x:e Dal.
Just take p2 to
Then ind a2 < g-2, which together with the previous result gives ind a2 = g-2. In the same way, we can easily show that for n < g, there are n points pl,...,pn such that if an = pl...PnP ind as = g-n. be, a point at which 'A.) # 0.
For n = g, ind ag = 0 and therefore dim ag = 1
25 6
28.10
proving that there exist g points on a surface of genus g such that there are no non-constant meromorphio functions having at worst poles of order 1 at these points. In Chapter 13 we stated the Weierstrass Gap Theorem and proved part of it. We showed that at any point p on a compact Riemann surface S, there are no more than g positive integers ni such that there does not exist a meromorphio function having as its only singularity a pole of order ni at p. Now we shall show that for each p e S there are exactly g such ni, and that 0 < nl... < ng < 2g, Let p be any point on S. Then dim pk = ind pp + dog pk-g+l For k = 0, ind 1 = g, so that
dim1=1 tc:iich just says that the only everywhere regular analytic func-
tions on a compact surface are constants.
For k.b 1
dim p = ind p-g+2 .
But this means there is an analytic function on S with a simple pole at p and regular everyelse. The existence of such a function means that S is co:rormally equivalent to the finite plane. Thus ind p - g-1 ind p = g, then dim p = 2.
a7d dim p = 1.
Now consider pk, k > 1. dim pk = ind pk+k-g+l
dim pk+1 = ind pk+l+k-g+2
.
If ind pk = ind pk+1 then d1la
pk+1 = dim pk+1
so that there exists a function with a pole of order k+1 at p
28.11
25;
and regular everywhere else. If, however, ind pk+1 = ind pk-1, we have dim pk+l = dim pk, so that there is no function on S, regular everywhere except at p, and having a pole of order k+l there. What this all means is that, when going from k to k+l, ind pk = ind pk+1 implies that there does exist a function with pk+1+l a pole of order k+l at p, and ind pk Q ind implies that there does not exist such a function.
Since deg W= 2g-2 for any Abelian differential CJ, we have ind p2g-1 = 0. This means that in going from k to k+l, it must happen that ind pk = ind pk+1+1 exactly g times, for k = = kl,...,k9 where ki < 2g-l. Therefore there are g integers
(k1+1) < (k2+1) < ... < (kg+l) < 2g such that there exists no meromorphio function on S having a single pole at p of order ki+l. The numbers k1+1 = mi are sometimes referred to as gap values or defective values. As we have seen, for g > 1, 1 is always a gap value. In fact we shall prove Except for a finite number of points, the numbers nJ = J, j = 1,...,g are the gap values. (These exceptional points are called the Weierstrass points). Theorem_:
If p is a Weierstrass point then ind pg > 0. There are Then, since infinitely many points pJ such that ind pg > 0. S is compact, there is a point p on S which is a limit point of the set [pjj . Let z be a local parameter, Proof:
around p and assume that the pJ all lie in the domain of this
parameter (if they do not, just eliminate those that lie outside it - there are only finitely many - and let the pi be those that are left). Now consider a basis wl,...,(Jg for AbeIn terms of the paramelian differentials of the first kind.
Clearly, the j are g linearly independent functions in the parameter patch. Since we have assumed that ind pn > 0, there are g complex numbersn),...,ag) such that ter z, WJ = 41 dz.
28.12
258
Cv = Aln)W1+A2n)W2+...+Agn)cJg = Ain)4l(z)dz+...+A(n)4g(z)dz
vanishes at z - zn where zn corresponds to p
Thus we have
n i.
Aln)41(zn)+...+X(n)4g(zn) = 0 A1(n)j2(zn)+...+A9(n)49!(zn)
0
0
.
0
where k(zn) denotes the k derivative of 4 at zn. A,n) have been assumed to exist
Since such
41(z)...4g(z) 41(z)...41(z)
W(z) = 0
dot
-1(z)
at z = zn.
W(z), the Wronskian of 41,...,4g is an analytic function in the parameter patch which vanishes at infinitely many points in every neighborhood of p. Thus W(z) __ 0 in some neighborhood of p, implying the linear dependence of 41,...,cg, which contradicts the assumption of their independence. It is easy to see that for g = 1 there are no Weierstrass points.
Here 2g = 2, so that the existence of a Weierstrass point would imply that the forms was oonformally equivalent to the sphere. That is, if p were a Weierstrass point, there would be a function, regular everywhere except at p, having there a simple pole. Theorem;
We state without proof a theorem by Hurwitz
For g > 2 there do exist Weierstrass points.
the number of such points,
If n is
219
28.13 2g+2 < n _ (g-1)g(g+1)
with n = 2S+2 only when S is hype elliptic, i.e., whenever S May be represented by a two sheeted branched covering of the plane with at least 4 branch points. the Weierstrass points.
if fact the branch points are
LECTURE NOTES PUBLISHED BY THE COURANT INSTITUTE OF MATHEMATICAL SCIENCES Price
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ARTIN, E. - Elements of Algebraic Geometry, 1955, 142 pp. ARTIN, E. - Modern Higher Algebra (Galois Theory), 1947, 173 pp. BAUMSLAG, G. - A Universal Approach to Groups and Rings, 1963, 100 pp. BERKOWITZ, J. - An Outline of the Theory of Functions of a Real Variable, 1956-57, 114 pp. BERS, L. - Introduction to Several Complex Variables, 1964, 218 pp. BERS, L. - Riemann Surfaces, 1957-58, 259 pp. BERS, L. - Topology, 1957, 256 pp. COURANT, R. - Calculus of Variations, rev, by J. Moser 1962, 280 pp. COURANT, R. - The Theory of Functions of a Complex Variable, 1948, 233 pp. FRIEDRICHS,K. 0. - Spectral Theory of Operators in Hilbert Space, 1959-60, 205 pp. FRIEDRICHS, K. 0.,. SHAPIRO, H. N., et al. - Integration of Functionals, Seminar, 1957, 220 pp. GLIMM, J. - Lectures on Harmonic Analysis (non-Abelian), 1965, 110 pp. JOHN, F. - Partial Differential Equations, 1952-53, 211 pp. JOHN, F. - Ordinary Differential Equations, 1964-65, 233 pp. LAX, P. - Partial Differential Equations (with Appendix by A. Douglis), 1950-51, 258 pp. LAX, P. - Theory of Functions of a Real Variable, 1959, 154 pp. NEUMANN, H. B. - Special Topics in Algebra: Universal Algebra, 1961-62, 78 pp. NIRENBERG, L. - Functional Analysis, 1961, 133 pp. RICHTMYER, B. D. - Introduction to Electromagnetic Theory, 1964, 109 pp. SCHWARTZ, J. T. - Algebraic Number Theory, 1961, 114 pp. SCHWARTZ, J. T. - Non-Linear Functional Analysis, 1963-64, 246 pp. SCHWARTZ, J. T. - Differential Geometry and Topology, 1965-66, 274 pp.
SCHWARTZ, L. - Functional Analysis, 1964, 212 pp. STOKER, J. J. - Differential Geometry, 1948, 158 PP. STOKER, J. J. - Topics in Non-Linear Elasticity, 1964, 96 pp.
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