Real Elliptic Curves (North-Holland Mathematics Studies 54)

REAL ELLIPTIC CURVES

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NORTH-HOLLAND MATHEMATICS STUDIES

54

Notas de Matematica (81) Editor: Leopoldo Nachbin Universidade Federal do Rio de Janeiro and University of Rochester

Real Elliptic Curves

N O R M A N L. ALLING University of Rochester Rochester, New York 14627, U.S.A.

NORTH-HOLLAND PUBLISHING COMPANY - AMSTERDAM

NEW YORK

OXFORD

North-Holland Publishing Company, I981 All rights reserved. No part of this publication may be reproduced, stored in a retrievalsystem, or transmitted, in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without the prior permission of the copyright owner.

ISBN: 0 444 86233 1

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Library of Congress Cataloging in Publication Data

A l l i n g , Norman L., Real e l l i p t i c curves. (Not as d e matemgtica * 81) (North-Holland mathematics s t u d i e s ~. : 54$ I n c l u d e s index. 1. Curves, E l l i p t i c . I. Title. 11. S e r i e s : Notas d e matemgtica. (North-Holland P u b l i s h i n g Company) j 81. 111. S e r i e s : North-Holland mathematics s t u d i e s ; 54. QALN86 no. 81 [QA567] 510s [516.3'521 81-9655 I S B N 0-444-86233-1 AACR2

PRINTED IN THE NETHERLANDS

FOR KATHARINE

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PREFACE

I n P a r t I of t h i s monograph w e w i l l s k e t c h t h e 1 8 t h C e n t u r y s t u d y , by E u l e r and o t h e r s , of w h a t l a t e r became known a s e l l i p t i c I n P a r t I1 w e w i l l c o n s i d e r t h e work o n i n v e r t i n g

integrals.

c e r t a i n o f t h e s e i n t e g r a l s , by G a u s s , A b e l , and J a c o b i , t o form W e w i l l c o n s i d e r t h e work of W e i e r s t r a s s on

e l l i p t i c functions.

h i s e l l i p t i c f u n c t i o n s , t h e work o f Riemann on Riemann s u r f a c e s , and t h e work of K l e i n and o t h e r s on t h e e l l i p t i c modular f u n c t i o n .

Until. a p p r o x i m a t e l y 1840 t h e p a r a m e t e r s i n t h e s e i n t e g r a l s were real.

I n t h e l a t t e r h a l f o f t h e 1 9 t h c e n t u r y and d u r i n g m o s t o f

t h e 2 0 t h , t h e r e a l case was l a r g e l y n e g l e c t e d .

The p u r p o s e o f

t h i s monograph i s t o g i v e a v e r y t h o r o u g h t r e a t m e n t o f t h e r e a l We w i l l p r e s e n t t h e t h e o r y o f r e a l e l l i p t i c

e l l i p t i c case.

Many of t h e s e t h e o r e m s a r e new.

c u r v e s i n P a r t 111. Let

be an a l g e b r a i c f u n c t i o n f i e l d ( i n one v a r i a b l e )

E

o f g e n u s 1, o v e r t h e r e a l f i e l d face; then

Y

IR.

Let

Y

be i t s Klein sur-

i s e i t h e r a t o r u s , o r i t i s a n a n n u l u s , a Msbius

s t r i p , o r a Klein b o t t l e .

Y

w i l l be c a l l e d a

i.e.,

c u r v e w h e t h e r o r n o t i t h a s any r e a l p o i n t s : n o t t h e boundary Y

is not a torus.

aY

of

Let

Y

s

i s nonempty o r empty.

whether o r Assume t h a t

b e t h e number of components of

i t w i l l b e c a l l e d t h e s p e c i e s of

modulus t ,

elliptic

a p o s i t i v e r e a l number. vii

Y

Y.

s

aY;

a l s o has a geometric and

t

characterize

viii Y

Norman L. Ailing

up t o d i a n a l y t i c e q u i v a l e n c e .

D e f i n i n g e q u a t i o n s and t h e i r

modulus w i l l be s t u d i e d a n d compared t o a n a l y t i c d i f f e r e n t i a l s of

Y

t.

Automorphisms and

w i l l also be c o n s i d e r e d .

TABLE OF CONTENTS

Page vii

Preface Chapter 0.

Research Historical and bibliographic notes Prerequisites and exposition Indexing Acknowledgments

0.10 0.20 0.30 0.40 0.50

PART I.

ELLIPTIC INTEGRALS

Chapter 1. 1.10 1.20 1.30

Chapteh 2. 2.10 2.20 2.34 2.40 2.50

Chapter 3. 3.10 3.20 3.30 3.40

PART 11. Chapter 4. 4.10 4.20 4.30 4.40

1

Introduction

Examples of elliptic integrals

11

Some integrals associated with an ellipse The simple pendulum The lemniscate integral 21

Some addition theorems Examples of addition theorems The arcsine integral Fagnano's theorem Euler's addition theorem Other addition theorems Development of some discoveries made prior to 1827

33

Linear fractional substitutions Generalized Legendre form Some of Legendre's work Gauss's arithmetic - geometric mean ELLIPTIC FUNCTIONS 59

Inverting the integral

...

Abel's Recherches Jacobi's Fundamenta Nova Gauss's work on elliptic functions The question of priority ix

...

Norman L. Alling

X

Chapter 5. 5.10 5.20 5.30

Chapter 6. 6.10 6.20 6.30 6.40 6.50

Chapter 7. 7.10 7.20 7.30

7.40 7.50 7.60

Chapter 8. 8.10 8.20 8.30 8.40 8.50 8.60

Chapter 9. 9.10 9.20 9.30 9.40 9.50

Origins Definitions Properties of theta functions The introduction of analytic function theory

10.40 10.50

85

Early history Lattices in Q: Fields of elliptic functions Some applications of Cauchy's and Liouville's work Theta functions treated with analytic function theory Weierstrass's work on elliptic func$ions

103

+...

Introduction Weierstrass's Vorlesun en Weierstrass's t eory reordered Representation of doubly periodic functions An addition theory for 'p A relation between Weierstrass's 0 function and Riemann surfaces

135

Introduction Definitions Some properties of the Riemann sphere Some properties of @/L Surfaces of genus one The divisor class group The eiliptic modular function

167

Introduction Definition and elementary properties Reflection of J across acl D c Modular functions An inversion problem

Chapter 10. Algebraic function fields 10.10 10.20 10.30

75

Theta functions

Definitions and introduction Extensions The Riemann surface of complex algebraic function field A theorem of coequivalence The Riemann-Roch theorem

189

Table of Contents

PART 111.

REAL ELLIPTIC CURVES

Chapter 11. Real algebraic function fields and compact Klein surfaces

Chapter 12. The species and geometric moduli of a real elliptic curve

Chapter 13. Automorphisms of real elliptic curves

237

The automorphism group of Yslt The orbit subspaces of Yslt Orthogonal trajectories

Chapter 14. From species and geometric moduli to defining equations

251

Introduction Species 2 and 1 Species 0 Other quartic defining equations

Chapter 15. The divisor class group of 15.10 15.20 15.30

217

The extended modular group Species Geometric moduli Real lattices

12.10 12.20 12.30 12.40

14.10 14.20 14.30 14.40

207

Real algebraic function fields Klein surfaces Symmetric Riemann surfaces A theorem of coequivalence

11.10 11.20 11.30 11.40

13.10 13.20 13.30

xi

Introduction Calculations on Applications

YsIt

XT,

Chapter 16. Analytic differentials 16.10 16.20

273

283

Introduction Computations

Chapter 17. From defining equations to species and moduli 289 17.10 17.20 17.30 17.40 17.50

Introduction Determining the species Transformations of defining equations a. and X The invariant B

Bibliography

335

Index

341

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CHAPTER 0 INTRODUCTION

0.10

RESEARCH

In 1971 the Foundaticns of the theory

of

Klein surfaces [6], by Newcomb Greenleaf and the author, appeared.

In it a short description was given of compact Klein

surfaces of genus one, here called pp. 60-661.

elliptic curves 16,

These curves are of the following topological types:

a torus, an annulus, a Mobius strip, and a Klein bottle.

Ana-

lytic tori have, of course, been extensively studied in the classical literature.

These will be called classical elliptic

curves; the others will be called =-classical. The study of non-classical real elliptic curves is the research problem that motivated this study.

These curves are,

of course, characterized topologically by the number of boundary components they posess.

This number will be their species.

non-classical real elliptic curve

Y

of species

s

terized up to dianalytic equivale'nce [6, 951.2-1.41 real number and

t > 0

t,

its geometric modulus.

Given

s

A

is characby a positive

E

{2,1,0)

the non-classical real elliptic curve of species

and geometric modulus

t

will be denoted by

Yslt.

s

Further ,

are dianalytically equivalent if and only if Y s r t and Y S l I t l s = s' and t = t'. These results can be found in Chapter 12. The field "functions" on

E(Y

) (or E for short) of all meromorphic s,t Ys, (defined in [6, 11.31) is an algebraic

1

2

Norman L. Alling

function field (in one variable) over the reals of genus 1, whose field of constants is IR. x

and

y

such that

over IR(x).

As such it contains elements

.

E =IR(x,y

Necessarily

An algebraic equat on for

called a defining equation for

E.

is algebraic

over IR(x)

y

If

y

s =

2 or 1

Weierstrass p -function and its derivative are in E = IR(9,n') :

If

s = 0

will be

then the E.

Further,

and, of course, the following holds:

then this is not the case.

An elliptic function Q

can be defined using, for example, the Weierstrass zeta function (14.32:2), such that

E(Y

Ort

) =

IR(Q,Q') and, for which the

following holds: (2)

(Q')2

=

-(Q 2 + a 2) (Q2 + b 2 ) ,

To calculate

a

and

b

with

in terms of

a > 0

and

b > 0.

t, Jacobie's theory of

elliptic functions proves very useful.

These results can be

found in Chapter 14. Let (3)

P(x) E Ax4 + 4Bx3 + 6Cx2

be of degree E :: IR(x,y)r

n

=

where

field over IR exist unique

+

4Dx + E

E

IR[x]

3 or 4, and have distinct complex roots.

y2

=

P(x) : then

E

is an algebraic function

of genus 1, whose constant field is IR. s

E

{2,1,0) and

E(Yslt) are IR-isomorphic.

t > 0

Let

s

such that and

t

tively the species and geometriE modulus of

E

and

t

from the coefficients of

P(x).

There

and

be called respecE.

The problem

addressed and solved in Chapter 17 is the following: s

Let

compute

On developing ideas

that go back at least to Euler and Legendre one can see (53.2)

Introduction

that

y2

3

can be transformed, using linear fractional

= P(x)

F2

transformations with real coefficients, to (4)

F(2)

=

where

=

Lu I V I W (2,k) u,v and w

5 (I)

=

where

(-l)u(l-(-l)vit2)( l - ( -wlk2ii 2 ) ,

are in

and

{0,1}

k

is

(0,1]

E

Legendre's modulus. Computing the species n

let

= 3

from ( 3 ) or ( 4 ) is very easy.

s

be regarded as a real root of ( 3 ) .

m

the number of real roots of (3); then r > 0

then it turns out that

s

=

r

r/2.

Let

2 , or 0.

= 4,

Assume that

If r

be

If r

=

0.

is then necessarily either positive definite or negative

P(x)

definite.

If positive definite then s = 0 (17.20).

definite then

s = 2,

and if negative

Going back to ideas of Cayley

and Boole (c. 1 8 4 5 1 , (5) let

=

G2(P)

let

-

AE

G 3 ( P ) E ACE

and let

A(P)

4BD + 3C2 ,

+

2BCD

G32 ( P )

-

-

AD2

-

B2E

-

C3 ;

2 7 G23 ( P ) .

Then these are invariants, in the sense of classical invariant theory, of weight 4 , 6 , and 12, respectively. (6) Let

then

J(P)

=

2 G3(P)/A(P):

is

-

of course

J(P)

the action of sense.

In 5 1 7 . 5 0 we define A ( P ) > 0,

let

B

if

A ( P ) < 0,

let

B

-

1.

2

B (P) = J ( P )

an invariant of weight

0

under

i.e., an invariant, in the contemporary

GL2("):

(7) If

Clearly

-

B (P)

Let

as follows:

Po(xo)

E

lR[xo]

be of degree

4

Norman L. Alling

3 or 4 , with distinct complex roots, and let y o = P (x ) . 0 0

where

Let

so

Eo :W(xo,yo),

be the species of

Eo.

The main

theorem of Chapter 17, and perhaps of the entire monograph,is the following: Theorem 2 (§17.55). s = s

only if

and

0

F to

F o :Eo(i)

and

Eo

are 1R-isomorphic if and

@(PI = @ ( P o ) .

It has been known and

E

for perhaps a century

are C-isomorphic if and only if

always contains several subfields K E(Yslt)i for different values of

[F:K] = 2

and

that

F = K(i).

2 , 3 , or 4 , depending on

F

3

E(i)

J(P) = J ( P o ) .

that are lR-isomorphic s

and

t,

such that

(The number of such subfields is E.)

Hence, J(P)

is too course an

invariant to characterize non-classical real elliptic curves. The automorphism group length in Chapter 1 3 . Go

of

G

G

of

YsIt

is considered at

The connected component of the identity

is isomorphic to the circle group.

The orbits of

Some of these ys,t constitute a foliation of Ys,t. orbits are distinguished: for example the boundary components

Go

in

for s = 2 and 1. However, also has distinYs,t guished orbits, in spite of the fact that as a topological space

Of

it is homogeneous. the orbits of

Go

The family of orthogonal trajecteries to in

Ys,t

also gives a foliation of this

space.

0.20.

Historical a n d bibliographic notes

The history of the theory of real elliptic integrals, functions, and curves has two intervals of substantial activity: from about 1750 to the latter part of the 19'th century: and

Introduction from about 1970 to the present.

5

The results presented in Part

I11 depend heavily on the work of Euler, Legendre, Gauss, Abel, Jacobi, Cauchy, Liouville, Weierstrass, Klein and others, which apply to the real case.

For example, Abel's defining equation

was (+'I2 = ( 1

(1)

-

c 24 2) (1 + e 24 2) ,

with

c

which can be used to define a Mobius strip

and

YlIt;

e > 0, where as

Jacobi's was (2)

(sn'I2

= (1

-

2 2 2 sn ) ( 1 - k sn 1 ,

with

which can be used to define an annulus

0 < k < 1,

YzIt.

(Hence, although

equations of type (1) and type (2) are equivalent over rc are not equivalent over IR.)

they

The initial inversion of real

elliptic integrals of the first kind to form elliptic functions, by Gauss, Abel, and Jacobi, was later superceded by using theta functions, having general complex parameters, to define elliptic functions; thus the case of complex defining equations become dominant, leaving the real case to languish, even though it is in the real case in which many of the applications to physics and engineering lie.

Thus the research in this monograph is

highly dependent on the results obtained on the real case between 1750 and late in the 19'th century.

Since there seems to be no

unified treatment of this subject, which prepares the way for the real case whenever possible, the author has written one in Part I and I1 of this monograph, as an introduction to Part 111. In addition,the author has written many historical and bibliographic notes throughout this volume.

These are included

in an effort to show what the founders of the subject actually wrote.

Further, those notes could serve as a guide to possible

Norman L. Alling

6

further reading.

It should be noted however that these notes

have not been prepared on the basis of the depth of historical research required to produce a full history of this subject. For example, as the references make clear, the author depended heavily on collected works, and with few exceptions did not go back to the original published versions, much less to manuscripts, letters,

... .

He also used some secondary sources extensively:

e.g., Klein's Vorlesungen uber die Entwicklunq der Mathematick

im 19.

Jahrhundert [401.

0.30

Prerequisites

and

exposition

This volume was written to be accessible to readers whose knowledge of mathematics encompasses at least the following: 1.

Analysis:

...

calculus, elementary real variable

theory, and the contents of a standard half year course on analytic function theory (see for example the contents of Ahlfors' Complex Analysis [2], Chapters I 2.

Algebra:

-

IV).

Elementary facts about groups, rings, fields,

and vector spaces as presented for example in van der Waerden's Modern Algebra vol. 1 [61]. 3.

Topology:

Elementary facts about compact spaces, con-

nected spaces, continuous maps, and surfaces. The reader only so equipped should be able to read this monograph, but it will require effort and not a little time.

A

more highly educated reader may want to skip, or at least only scan, large sections of Parts I and 11, and concentrate on Part 111; whereas an expert in this subject may want to go directly

to Chapter 12.

With this in mind a great many interval refer-

ences have been given:

thus it should be possible to pick up

Introduction

7

this text at almost any point and be able to find out the exact meaning of the notation and terminology easily.

Although largely

expository, Parts I and I1 are sprinkled with novel approaches and may even have a few new results.

0.40

Indexing

The decimal system of indexing is used to index the sections of this volume.

(According to Whittaker and Watson, who use this

system, it goes back to Peano 169, p.11 . I

The chapters are

numbered from 0 to 17, this being Chapter 0.

The first digit

to the right of the decimal point is the index of the major divisions of the chapter in question.

The second digit to the right

of the decimal point is the index of the division of the major divisions of the chapter into minor subdivisions: thus 53.12 refers to Chapter 3, main division 1, subdivision 2.

Should we

want to refer to all of the first main division of Chapter 3 we would refer to it as 53.1. For later reference, increased clarity, and occasionally for emphasis, many expressions have been displayed. how they are numbered consider an example.

TO describe

Inside 53.12 the

first displayed expression would be referred to merely as (1). Outside 53.12 it would be referred to as (3.12:l).

0.50

Acknowledgements

Thanks are due my colleagues Professors William Eberlein, Richard Mosak, Saul Lubkin, Arnold Pizer, and others who actively participated in a seminar held during the 1977-78 academic year at the University of Rochester on elliptic functions, in which a preliminary version of part of this monograph was presented.

The

Norman L . A l l i n g

8

a u t h o r i s a l s o g r a t e f u l t o t h e s t a f f s o f t h e Rush Rhees and t h e Chester Carlson L i b r a r i e s , a t Rochester, f o r t h e i r a s s i s t a n c e ; and f o r t h e e n l i g h t e n m e n t o f t h o s e i n c h a r g e o f book a c q u i s i t i o n a t t h e U n i v e r s i t y d u r i n g i t s f i r s t c e n t u r y , f o r most o f t h e b i b l i o g r a p h i c r e s e a r c h f o r t h i s volume was done u s i n g l o c a l l y h e l d books.

Deep t h a n k s a r e d u e Mrs. S a n d i A g o s t i n e l l i , M r s .

R o b e r t a Colon and M r s . Marion L i n d who so a b l y t y p e d t h i s manuscript.

The a u t h o r w i s h e s t o t h a n k D r . Mika S e p p a l a , o f

H e l s i n k i and P i s a , whose comments on t h e e a r l y c h a p t e r s o f t h i s work were v e r y h e l p f u l .

The a u t h o r i s a l s o i n d e b t e d t o t h e s t a f f

o f t h e North-Holland P u b l i s h i n g Company who have a i d e d i n t h e p r e p a r a t i o n and p r e s e n t a t i o n o f t h i s m a n u s c r i p t .

Finally,

s p e c i a l t h a n k s a r e d u e P r o f e s s o r Leopoldo Nachbin f o r e n c o u r a g i n g t h e author t o p u b l i s h t h i s unusual mixture of r e s e a r c h , scholars h i p , and e x p o s i t i o n i n t h e N o t a s d e Mathem6tica s e r i e s .

PART I ELLIPTIC INTEGRALS

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CHAPTER 1 EXAMPLES OF ELLIPTIC INTEGRALS

Some

1.10

Let

a

i n t e g r a l s a s s o c i a t e d w i t h a n ellipse.

and

b

be positive real numbers and let

then this set is the graph of an ellipse in standard position. a 1. b

Assume that (2)

( 1 - b 2/a 2)

k

and let

'.,

then

k

is the eccentricity of (1). Clearly

k=O

if and only if

a=b:

0<.k<1. Further,

i.e., if and only if (1) is a

a > b.

(a2 - b2)';

c

Let

then

a >c.

circle.

Assume now that

Ellipse

(a,c) is called the complementary ellipse to Ellipse

(a,b). (3)

Let

(4)

then

Clearly

be the eccentricity of Ellipse (a,c);

k'

k

2

+ (k'I2 = 1, and

0 < k' < 1.

Ellipse (a,b) is the complementary ellipse to

Ellipse

(a,c). Ellipse (a,b) is self complementary if and only if k = k' = 2 ' .

For any positive numbers

then

a'

a

and

is the arithmetic mean of 11

a

b

let

and

b,

and

b'

is the

12

Norman L. Alling

geometric mean of

a

and

b.

a* - 2ab + b 2

Since

=

(a - b) > 0,

a' > b'. (6) Let

Let

kl

y(El1ipse (arb)) t Ellipse (a',bl). be the eccentricity of

kl = (a - b)/(a

1.11

Let

maps

-

= (1 kl)/(l + k') ,

P(0)

Clearly

€IE I R .

of

+ b)

y(El1ipse (arb));then

be the vector

0 EIR + P( 8 )

maps IR

[Or2T) injectively onto

and

0 < kl < 1.

(acose, bsin8) , onto

for all

Ellipse (arb), and

Ellipse (arb). The arc length I

Ellipse (arb) is

On letting

Cp E 0 + ~ / 2 then (1) becomes

(2) j2'(a2cos2

Cp

+ b2sin2C p ) %dCp,

which equals

0

is an elliptic integral. Let us also write out (1) in Cartesian coordinates.

On letting

x

E

at

(6) 4 a \01 f1- k2 ti 7 d t

=

It is

(5) becomes

2 2 (1-k t )dt = 4aj0 [(1-t2) (1-k2t2 1 1 4 ~

1 dt 4al . 2 1 14 0 [(l-t2) (1-k7 t

-

4ak2

1

t2dt 2 4 I 0 [(1-t2)(1-k t2)1

Examples of Elliptic Integrals

13

Following Legendre [49] X

dt j0 [ (1-t2)(1-k2t2)1%

(7)

for

X E

[-1,1],

is called an elliptic integral of the first kind, and

IX

(8)

t2dt 2 2

0 [(1-t2)(1-k t 1 1

4'

for

X E

[-1,1],

is called an elliptic integral of the second kind. Legendre's time,

k

After

became known as Legendre's modulus and

kl

(1.10:3) became known as its complementary modulus.

if

x

then (7) and (8) are improper; however they do

1

= ?

converge.)

(Note:

Elliptic integrals of the first kind play a cardinal

role in the whole theory.

Let

a

f

arc sinx;

then (7) equals

In this form it is also called an elliptic integral of the first kind. --

(10) K then

K(k)

3

(11) K'

d8 jr2(l-k 2sin2f3)'

'

is known as a complete elliptic integral

K

kind). -

Let

Let Z

K(k').

Further , 2n

2n

dB

a 0 (1-k2sin20)'

d6

=

4K -.

a

(of the

first

Norman L. Alling

14 1.12

The history of the problem of computing the arc

length of an ellipse seems to go back at least 1655, when

J. Wallis is reported, by Fricke [25, p. 1821 , to have begun considering it. worked on it.

Later various members of the Bernoulli family Most of the results and notation found in S1.11

can be found in the three volume opus by Legendre [ 4 9 ] c. 1825, on the subject.

1.20

The

s i m p l e pendulum

Consider a point mass less rigid rod of length

m

attached to one end of a weightthe other end being attached to a

R,

frictionless fulcrum that restricts the rod in such a way that it always moves in a plane

P

through the fulcrum. Assume

further that the gravitational force vector is constant, of length

g,

and that

- if

also lies in the plane

P.

it is drawn from the fulcrum

- it

Assume further that the whole appa-

ratus is in a perfect vacuum, and that Newton's Laws hold. a

be the amplitude of the pendulum and assume that

Thus we have the following diagram, where

8

E

Let

0 < a
[-&,a].

Examples of E l l i p t i c I n t e g r a l s

(mass)

Since (2)

(acceleration)

x

2 2 m(Rd e / d t )

+

2

d 9/dt2

+

( r e s t o r i n g f o r c e ) = 0,

m g s i n 9 = 0.

Choosing u n i t s s o t h a t

(3)

+

15

g/ll = 1 g i v e s

s i n 0 = 0.

The " l i n e a r i z e d " e q u a t i o n a s s o c i a t e d w i t h ( 3 ) i s (4)

d29/dt2

+

9 =

Or

whose g e n e r a l s o l u t i o n i s

(5) c1

e ( t ) = clcos t + c 2 s i n t , and

c2

27~. Since

of

being c o n s t a n t s .

Note t h a t t h e p e r i o d of

limg+o(sin9)/9 = 1

(5) is

one can a r g u e t h a t t h e s o l u t i o n

( 4 ) should approximate t h e s o l u t i o n of

( 3 ) f o r "small"

9.

R e t u r n i n g t o ( 3 ) r w e c a n r e d u c e t h e o r d e r of i t by l e t t i n g

v :d9/dt (6)

be t h e a n g u l a r v e l o c i t y of

vdv/d0

m;

t h e n ( 3 ) becomes

+ s i n 9 = 0.

On i n t e g r a t i n g ( 6 ) w e o b t a i n

(7) v 2 = 2cos Note t h a t when

+A,

a constant.

A

9 = a r v = 0,

thus

see t h a t (7) g i v e s (8)

v = + 2 4(COS

9

- cos

a)

4

=+ - 2 ( s i n2 ( a / 2 ) - s i n 2 ( 9 / 2 ) ) ' 5 . (9)

Let

k :s i n ( a / 2 ) .

A = -2cos a .

From t h i s w e

Norman L. Alling

16 0 < k < 1,

Note that

since

0 < a < TI.

k

is sometimes called

the modular angle of the pendulum.

+

(10) Let

+

and let

z arcsin(sin(e/2) ),

be called the quasi-angle of the pendulum.

and note that as

0

runs through

[-a,a],

I$

Then

runs through

[-i~/2,~/2]. Note also that (8) is (12) de/dt = +2kcos Since

+

@. [ - I T / ~ , I Tcos+ / ~ ] , is always non-negative.

is always in

Since the pendulum swings back and forth, d8/dt changes sign, when

periodically

Let us impose the following initial condition:

t = O , d8/dt> 0

and

8=0;

thus when

t = O ,

$=O;

and

(12) becomes

T~

being the period of the pendulum,

Differentiating (11)

gives (14)

(1/2)cos(0/2)de = kcos + d$.

From (14) and (11) we obtain (15) de

=

-. (1-k sin 4 )

Solving (13) for (16) dt

=

dt

and using (15) gives

2d+ 2 4 , for (1-k sin I$)

-.r0/42t2.r0/4.

Examples of Elliptic Integrals

17

Hence we have

(18)

T

~

T~ = 4K(k).

,

the period of the pendulum, is a function of the modular

angle

k.

Clearly

Limk+O+K(k)

the period of ( 5 ) . Lim

-i0(k) = a ,

Further, which

-

= n/2,

thus

Limk-tl-K(k)= a ,

of course

-

Lim + ~ ~ ( k= )2 7 , k-t0 thus

makes sense physically.

k-tl-

We conclude this section with a very scanty table of function of

as a

a. K

a ___

~-

0

1.57079633

1"

1.57082623

2"

1.57091596

10"

1.57379213

20"

1.58284218

45"

1.63358631

90"

1.85407468

135"

2.4000945

179'

6.12778

1.21

K

...=n / 2

The history of the problem of finding the period of

the pendulum during the 17'th and 18'th century is apparently roughly as follows.

In 1 6 3 2 Galileo concluded that

nearly constant for "small" a. given

'I

0

as

is

In 1 6 7 3 Huygens seems to have

2 7 r ( I / c ~ ) ~ , for "small"

a.

in 91.20 seems to be due to Euler c. 1 7 3 6 . [ 6 8 , p. 721 for details.)

T~

The solution given (See e.g., Whittaker

18

Norman L. Alling Returning to the pendulum problem

1.22

given an amplitude

with

a,

a < TI, assume that gravity were

0

to exert its force upward on mass

m

with the same magnitude

g.

The pendulum would then have a new period

T :

could be computed by letting

tion given in 51.20. sin(a’/2) =cos(a/2)

T;

=

=

0’

Clearly

and using the solu-

thus

and hence

4K’(k).

It is natural to call

T

0 (See Whittaker

pendulum.

a ’ =- 7 - a

T I

Note that (1 -sin2 (ci/2))$ = (1- k2) $.,

(1) k ’ = sin(a’/2),

(2)

for a moment,

and

T;

complementary periods of the

68, pp.73-741, for more details. due to Appell, first seems to have

Also note that this result appeared in 1878.)

The

1.30

lemniscate integral

Consider the lemniscate given by (1)

(x2+y2I2

Let

xErcos 8

-

2xy and

=

0.

yFrsin8

,

for

r,O;

then (1) becomes

(2) r2 = sin 28. Forming the differential of (2), we obtain (3)

rdr

=

cos28d8,

for

8

E

[0,~1/21u [nl3~/2I: U.

Thus the arc length integral of the lemniscate is

Restricting our attention to tain

8 E U

allows us to use (3) to ob-

Examples of Elliptic Integrals

19

Thus (4) becomes

which we will call the lemniscate integral.

Hence the arc

length of the lemniscate (1) is

This integral is improper.

r

=

(8)

sin @

;

then (7) becomes

(1+sin @ ' 2@ ) %

41"' 0

To see that it converges let

which is not improper.

* Thus (7) converges.

To relate (8) to

complete elliptic integrals of the first kind note that 2 2 sin @ = 1 -cos @; thus (8) is

Letting

e

E

@+.rr/2

(10) 23/2j"/2 o

give us

d@ (1 - (%)sin213) %

Note that Legendre's modulus to its complementary modulus

1.31

k k'

associated with (10) is equal

.

Fricke [25, p.1821 asserts that Jacob Bernoulli con-

sidered the lemniscate integral as early as 1691.

Struik

[59, p.3761 also credits Jacob Bernoulli as having discussed 2 2 the curve given by x + y2 = a(x - y2)* in 1694 I and described

H. Attouch. D. Aze' and R. Wets

20

In the case of Convex Programming the quadratic augmented Lagrangian is given by the following formula

where

{

T - t s

+ r ( s , t ) = G:

t2

ifs2$, if

s

5 $,

The following proposition guarantees that the saddle points and saddle values are preserved when replacing L by L,, in the general (metric) setting.

Proposition 4.1. L and L, have same saddle value, and every saddle point of L is a saddle point of L,. Proof. Take (Z,y*) a saddle point of L; it is characterized by the following

inequality: sup L(?t,y*) 5 inf L(z,jj*).

Y'EY'

ZEX

By Proposition 3.1 sup L,(Z,Y*) = sup L(Z,Y*). Y'EY'

Y'EY'

Noticing that L, is greater than or equal to L , inf L(s,jj*) 1. inf Lr(z,jj*).

zEX

SEX

Combining the preceding inequalities induces

that is (Z,%*) is also a saddle point of

and

L,,

CHAPTER 2 SOME ADDITION THEOREMS

of

Examples

2.10

addition theorems

abound in the theory of the elementary functions as it was known in the 18’th century. (1) ex+Y = ex

ey,

For example, for all

x,y

E

IR.

This, of course, yields (2) log xy

log x

=

+

l o g y,

for all positive real

x

and

y.

Of more relevance to our study is the addition theorem for the sine, namely:

(3)

sin(a+B)

=

sinacosB

+

cos a sinB

for all

I

a,B

E

As written this involves both the sine and the cosine. ting (4)

c1

and

to

f3

sin(a + 6 )

=

[-71/2,71/2]

IR. Restric-

allows us to rewrite ( 3 ) as

(sin a) (1 -sin2B)’

+

(sin 0 ) (1 -sin2a)% ,

an expression involving only the sine.

From ( 3 ) and ( 4 ) we can

derive two versions of the double angle formula for the sine, namely : (5) sin2a

=

2sina cos a, for all

(6) sin2a = 2(sina) (1-sin2a) 4 I

21

a EIR,

and

for all

a

E

[-71/2,71/2].

Norman L. Alling

22

The

2.20

arcsine integral

Since this integral served as a prototype for all elliptic integrals of the first kind in the 18'th century, let us consider it with some care.

It is clear that

exists and defines a differentiable monotone increasing function on (-l,l), that is continuous on [-1,1].

-

course 0

E

(3)

the arcsinex.

Since

This function is

arcsin(sin0) = € I r

-

for all

[-1~/2,n/2] , we have sin0

e

dt

,

=

Restrict

for all

0

[-1~/2,n/21.

E

(1-t2?

0

and

a.

6

so that

a+ 6

a,$, and

all lie in

[-1~/2,~/21 ; then using (3) we obtain sincl dt (l-t2)%

0 Let

+

isin' 0

a :sin a., b :sin 8 , dt

(6)

where

dt = a+@ (1-t2)4 and

a(l - b2)'

+

(1-t2)JS

c :sin(a + 6);

dt

b(l -a2)'

C

'

=

then

dt

=

c,

by (2.10:3).

Clearly, equations ( 5 ) and ( 6 ) are a version of the addition theorem for the sine, in which the sine does not explicitly appear.

Now let a

dt

b=a;

then

C

dt 0 (1-t2 ) % ,

where

of

Some Addition Theorems

(8)

23

= c.

2a(l -a2)'

This holds for all

a

E

[-2-15,2-4], and is a version of the

double angle formula for the sine. L

f(u) : 2u/(l+u 1 .

Returning to (21, let seen that of (9)

is an order-preserving differentiable bijection

f

[-1,11

It is very easily

onto itself. 2

t :2u/(l+u

)

Let

, for all u

E

[-1,1];

then

Thus (9) rationalizes the integrand of the arcsine integral.

Since the arcsine integral seems to have served a s

2.21

a guide to Gauss, Abel, and Jacobi in their work on inverting elliptic integrals of the first kind, let us assume, for a moment, that all of standard calculus is available to us, except that the circular functions have yet to be defined or developed. For each

X E

(-l,l), define X

(1) arcsinx

to be

i,

dr (1-r2)+

We would like to show that the improper integral (2)

~+l 0

Consider

dr 2 4 (1-r ) t

E

converges.

[-1,1]

+

2t/(l + t

2

)

E

[-1,1],

and note that this

map is a differentiable, strictly increasing surjection. (3)

2t r=l+t2

for all I

Then (2) equals

t

E

[-1,11.

Let

Norman L . A l l i n g

24

where

i s a p o s i t i v e c o n s t a n t ( w h i c h w e know t o be

P

thus ( 2 ) converges.

ous on

i s d e f i n e d and i s continu-

Hence a r c s i n e x

and i s d i f f e r e n t i a b l e on

[-1,1]

n/2);

(-1,l).

Since t h e

(1) i s e v e n , a n d a l w a y s p o s i t i v e , a r c s i n x

integrand of

s t r i c t l y i n c r e a s i n g map o f

[-1,1] o n t o

[-PIP],

is a

such t h a t

(-l,l),

on

d arcsinx dx

(5)

=

(1 - x 2 ) -4

. (a theorem

Using t h e i m p l i c i t f u n c t i o n theorem, i n t h i s c o n t e x t ,

many 1 8 ' t h c e n t u r y m a t h e m a t i c i a n s f e l t f r e e t o u s e ) , w e see t h a t x

E

[-1,1]

+

arcsinx

0

E

[-PIP]

+

sin0

E

E

h a s an i n v e r s e f u n c t i o n

[-PIP]

[-1,1], w h i c h i s c o n t i n u o u s , s t r i c t l y i n c r e a s -

i n g , and such t h a t (6)

d

s i n e = ( 1 - s i n 2 8)'

(7) Thus,

(1- s i n 2 0 ) 4 ,

8

E

[-PIP]

i n c r e a s i n g on

(-PIP).

t o be

cos0

Now d e f i n e

on

+

for all

cos 0

[-P,Ol,

E

0

E

[-PIP].

[0,1]

is

continuous, and s t r i c t l y

and s t r i c t l y d e c r e a s i n g o n

[O,Pl,

such t h a t

(8)

d dB

cos0= -sine,

on

(-PIP).

C l e a r l y t h e s i n e i s a n e v e n f u n c t i o n a n d t h e c o s i n e i s a n odd function.

Clearly

Now l e t

s i n ( + P ) = fl, 2

cos(+P) = 0,

D 5 { ( a , @ )E W : a , @ ,

a n open connected s u b s e t o f

7RL.

and

Let

and

cos O = 1.

a + @E ( - P , P ) ] .

D

is

Some Addition Theorems (9)

f(a,B) EsinacosB + cosasinB,

for all

( a , @ )E D .

One immediately sees that

thus the gradiant of vector

(lrl).

f,

is zero, at each point of

consequence, f ( a , P ) c+0

Since

= g(c) =

c,

,

af/aB;

=

is always parallel to the

is constant on the line

f(a,@)

(-PrP).

grad f,

af/aa

Hence, the directional derivative of

direction ( - 1 r I - l then

25

D.

a+ @=c

Let in

f

in the

C E

(-P,P);

D.

As a

for some smooth function g

(c,0 )

D.

E

g (c)

=

on

sin c cos 0 + cos c sin 0 =

sinc; hence we have proved in this context

the addition theorem

for the sine, namely (10) sin(a +

@)

=sin acos @

+

cos a sin

@,

for all

(a,@) E D.

Since all the functions appearing in (10) are continuous on [-P,P], then by (101,

(11

=cOSar

For

let

for all

sin(cl +P)

let

then, by (lo),

(12

=

,

For

a

E

[-Pro].

be defined by using (11). Now

-cos a. let

sin(a-P)

be defined using (12).

so extended, the sine is continuous on [-2P,2P]. sees that the sine is differentiable at tal tangent at each such point.

fP,

Clearly

One easily

having a horizon-

One sees that the left hand

derivative of the sine at

2P

is

-1, and the right hand deriv-

ative of the sine at

is

-1.

Now let the sine be extended

to all of IR extended

-

-2P

by letting it be periodic of period

of course

-

4P.

So

this is the familiar sine function.

Thus all of the circular functions can be built up by inverting the arcsine integral. Bibliographic note.

The elegant proof of (10) is an adap-

tation of a proof given by Abel [l, vol.1, pp. 268-2691 in 1827.

26

Norman L. Alling

2.30

Fagnano's

theorem

Counte Giulio Carlo de'Toschi di Fagnano (1682-1766) was an amateur mathematician and was, (according to Struik [59, pp. 374-3831) the Spanish consel in his home town of Sinigaglia (Italy).

He published a series of papers during the

years 1714-1718 on the summation of the length of various curves, a problem introduced by Johann Bernoulli in 1698.

Much later

he republished this work in his Produzioni mathematiche, Pesaro (1750). It is this later work that appears in volume 2 of his collected works [23]. The result known as Fagnano's Theorem concerns the lernniscate integral.

2.31

Fagnano asserts the following Theorem 5

is the solution of

Fagnano breaks his demonstration into two stages.

Theorem1

[23, p. 3041 asserts that is the solution of

(3)

Fagnano demonstrates this by differentiating (3) to obtain

4 (5)

dx

=

(l-(l-z4)') dz 2 4 4 z (1-2 )

Some Addition Theorems

27

Then he asserts that ( 3 ) gives

Then combining (5) and (6) he obtains ( 4 ) , completing his argument in support of Theorem

23, p. 3051 asserts that

Theorem 3 of Fagnano

(7) x

= T'

.U.,L4

1.

is the solution of

(1-u 1

His demonstration is very much like that of Theorem 1.

Combin-

ing Theorem 1 and 3, he obtains Theorem 5, known as "Fagnano's Theorem", since the 18'th century.

2.32

Professor Carl L. Siege1 has given a more modern

version of Fagnano's Theorem in volume 1 of his excellent Topics

in Complex Function Theory [57, pp. 3-61, which we will now give in a slightly modified form. dz

for

c

E

Consider

[0,1].

On referring to (1.30:6), one sees that (1) represents the arc length of the lemniscate (1.3O:l) between the point given by z r = O , and the point given by r = c and 8 = (arcsinc ) / 2 . Let "

(2)

z2 =-

2X'

- g(x),

for all

X E

[0,11.

l+x Then

x

+

g(x)$

is an order-preserving injection of [ 0 , 1 ]

[0,1], that is differentiable on (0,l).

one can easily see [57, pp. 3-41 that

Cf. (2.31:3).

Onto

Then

28

Norman L. Alling

C

dz 0 (l+x4) 4

I

where

2 c

=

g(b).

Now consider (4) x2 -

2u2 1-u'

7 -

h(u),

for all

U E

[0,1).

(This, of course, is Fagnano's substitution (2.31:7).) h'(u) > 0, then

h(0)

h(uo) =l.

=

0,

Let

and a

E

Limu+l -h(u) = + m . Let [O,uol so that h(a)'=b;

Clearly

u = (2'-1)'; 0then

[571 P * 411 dx4 (5) J b 0 (l+x 1

=' "2

4 du 4 . 0 (1-u

Hence, we conclude that

This then is another version of Fagnano's Theorem in which (6) deals with doubling the arc length of a lemniscate and (7) is an algebraic relation between

a

and

c.

Note the striking

resemblance between the double angle formula for the sine, as given by (2.20:7) and (2.20:8); and the formula for doubling the arc length of a lemniscate, as given by (6) and (7). Clearly something very interesting is occurring here.

2.40

Euler's a d d i t i o n t h e o r e m s

In 1741 Leonhard Euler (1707-1783) responded to a call from Frederick the Great and took up residence in Berlin, where he remained until 1766.

(In 1766 he returned to Russia, at the

Some Addition Theorems

request of Catherine the Great.)

29

Fagnano sent his Produzioni

mathematiche to the Berlin Academy.

On December 23, 1751 these

results were put into Euler's expert hands [22, vol.XX, p. VII]. (Much later, in the 19'th century, Jacobi described this day as the "birthday" of the theory of elliptic functions [25, p. 1831.)

In the 1756/7 number of Novi commentarii academiae

scientiarum Petropolitanae [22, vol.XX, pp. 58-791, Euler considered the differential equation

asserting that its solution is (2) x2 + y2

+ (cxy)2

=

c2 + 2xy(l-c 4) 4

.

Euler gives essentially the following demonstration [22, vol.XX, p. 631: (3)

Differentiating (2) gives

-

xdx + ydy + c2 xy(xdy +ydx) = (xdy + ydx) (1 c4,) %

.

Collecting like terms gives (4)

( X + C2XY2 - Y ( ~ - 4c) 4)dx+ ( Y + C2x 2Y - x ( ~ - 4 c ) 4)dy = 0.

Solving (2) for

where the

*

y

and

x,

by the quadratic formula, gives

signs are independent of one another.

In order to

try to remove the ambiguity of sign in (5) note that if x = O l 2 then (2) reduces to y = c2; thus y = *c. Choose y to be c in this case.

Note that

Then (5) becomes

2 2 x + c xy -y(l-c4)'

using ( 5 ) , equals

-c (1- y4) 4

=

.

2 2 4 which, x(l+c y ) - y ( l - c ,') Similarly y + c2x2y - x(l - c4) 4 =

Norman L. A l l i n g

30

4s

c(l-x )

(7)

.

Thus ( 4 ) becomes

- c ( l - y 4 ) 5d x + c ( l - x 4 ) %d y = 0 ;

and hence

I n d e e d , E u l e r showed t h a t ( 2 ) i s a s o l u t i o n

p r o v i n g t h e theorem.

of (1).

2.41

P r o f e s s o r C. L. S i e g e 1 h a s g i v e n a n

r e c o n s t r u c t E u l e r ' s t r a i n of t h o u g h t . "

u,v,

and

(1) r

r

be i n [-1,1],

[57,

pp.

7-91.

l+u v

is j u s t Euler's (2.40:6).)

Assume t h a t

r

(Note t h a t t h i s

is held constant.

be c o n s i d e r e d t o be t h e i n d e p e n d e n t v a r i a b l e ;

v = v(u).

Let

2 2

=

u

attempt t o

and c o n s i d e r

a s a n e q u a t i o n r e l a t i n g t h e s e t h r e e numbers.

Let

'I...

Note t h a t

v ( 0 ) = r.

then

On d i f f e r e n t i a t i n g (1) and s i m -

p l i f y i n g we find t h a t

On i n t e g r a t i n g ( 2 1 , w i t h r e s p e c t t o

u,

from 0

obtain

Thus

J0

du

V

(1-u4 )5 + 1 0 * du

+

V

dv 4 % '

r (1-v )

to

u, w e

Sorne Addition Theorems r which by ( 3 ) is

dv

.

31

Hence we have shown that

This is a version for Euler's addition theorem for the lemniscate integral.

(Note: we have gone back to the 18'th and early

19'th century notation in which the variable in the integrand and the upper limit of integration are denoted by the same letter.

(Struik [59, p. 3 7 6 1 notes that the convention of indi-

cating the limits of integration, as we do now, is due to Fourier On letting

c. 1 8 2 2 . )

u=v,

to Fagnano's Theorem ( 2 . 3 2 : 6 )

Euler's addition theorem reduces and ( 2 . 3 2 : 7 ) .

Other a d d i t i o n theorems

2.50

Euler's addition theorems were evidently much admired. There are closely related addition theorems for 'elliptic integrals of the first kind due to Legendre, Abel, Jacobi, Gauss, and Weierstrass, which will be given in due course.

It will be

convenient to state Legendre's addition theorem.

(4) sinp

=

Let

0

k < l and let

sin+cos$A(k,$) + cos+sin$A(k,+) 2 2 2 1 - k sin +sin $

This Page Intentionally Left Blank

CHAPTER 3

DEVELOPMENT OF SOME DISCOVERIES MADE PRIOR TO 1827

Linear fractional substitutions

3.10

(1) Let

P(x) - A x4 +4Bx3 +6Cx 2 + 4 D x + E ~ K [ x l ,

where

is either the field IR

K

of

P(x).

if

P(x)

be called

P(x)

or

will be called admissible if

has no multiple roots in C.

&.

will regard

Let

C.

If

n

be the degree

n = 3 or 4 ,

P(x) EIR[X]

and

it will

Assume that

P(x)

is real and admissible.

as a root of

P(x)

if and only if

always has 4 roots in C u

{a},

we will usually take

to be

p4

p1,p2,p3, m.

and

p4.

n = 3. If

We

P(x) n=3

By hypothesis these 4 roots

are distinct. In about 1766 Euler [22, vol.XX, pp. 302-3071 considered (2) dx/P(x)’ in the case

(3)

=

dy/P(y)’,

n=4.

In order to simplify (2) Euler let

x : (az+b)/(cz+d).

The idea of using such a substitution is of course elementary, natural, and

-

in Euler’s hands

-

powerful.

We will break

off our historical account at this point, and introduce linear fractional substitutions in the language of 19’th and 20’th century mathematics.

3.11

Let

there exists

b

A E

A

be an integral domain. such that

Let

ab=1}; then 33

U(A) E {a

U(A)

E

A:

is the group

Norman L. Alling

34

of units of --

A.

Let

m,nc N(Z{l,2,...})

denote the set of all

m

by

and let Mm,n (A) matrices with entries in A.

n

-

Under the usual operations it is

of course

-

an A-module.

Further, MnIn(A), in which matrix multiplication is always defined, is an A-algebra.

The set of invertable elements of

M

(A) is frequently denoted by GLn(A) n,n general linear group of rank n over A. let

det(B)

denote the determinant of

(1) det(BC) = (det(B)) (det(C)), (2)

GLn(A)

(3)

Let

=

{BE M

SLn(A)

nln

5

(A): det(B)

{ B E Mn

,n

and called the Given

B;

for all E

B EM

n,n

then B,C E Mn

,n

(A).

U(A)?.

(A): det(B) =I]:

it is usually called the special linear group of rank A.

n

Of concern to us in this monograph will be the special cases

in which

n=2

(3.10), which

and

-

A=

2,

of course

C

Let

3.12

IR, or @ .

-

is

In general let

denote the Riemann sphere:

i.e., let are

union the complements of compact subsets of cl:,

e.g., [ 3 5 , ~01.11or [lo] for details.)

Clearly

con1

Z

conC a1

B

Let

is a group under composition.

of

1.

E

conz: f (a) = a } ,

:(conZa) n (con1 ) ,

B

and

1).

It will be refered

Given distinct

let

(2) conCcl5 If

on which ana-

is a conformal autohomeomorphism of

to as the conformal group

(-1

(See Chapter 8 of this work or

lytic structure has been put.

(1) con1 : if: f

K* :K-{OI

U(K).

in which the open neighborhoods of

C 5 C u {mi

in

(A),

arBl

and

y

35

Development of Some Discoveries Made Prior to 1827 : (con1 u,

con1 CllDlY

(3) Let z

Given

Let

(4)

C+

z = a + bi,

C,

Re(z)

5

(con1 ) . Y

n

{f E conz: f @? u

CO%~ f

E

a)

for unique

Im(z) E b, 8

a,

Im(z) L O } ,

5 { z E C:

and

f

corms If

E

{m}}.

a,b E D .

{z

E

C:

Im(z) > 01,

$u

cU+:

4 is called the upper half plane. (5)

=IR u

{m})

{m}.

Let

f (9)) = 4 1 .

co%Z:

Clearly the sets defined in (2), (3), and (5) are subgroups of conC.

Further,

con&

is clearly a subgroup of index 2 in

CO%C.

3.13

Given

(2) f ( z ) :h(M) ( z ) Since

1. m

M

If

f

is non-singular, f uOeC

is a pole of

is not a pole of

is an element in (3)

h

(az+ b)/(cz

f;

conC.

+ d) E

is a rational function of degree

f,

let

K(z).

let

f(u 0 ) E m .

f(m) :a/c.

Thus

If f:o

cf0, E

C

+

then

f(o)

E

It is easy to see that

is a homomorphism of

GL2(c)

into

conC, whose kernel

is C*I. (4)

Let

Clearly

Aff2(K) Aff2(K)

the affine group.

IG g)

f

E

GL2(K)1

is a subgroup of

. GL2(K).

It will be called

It is easy to see that

(5) h(Aff2 (C)) = conCm. (6)

then

Let

Tbf

r

b,

C

-

,

h(T b ) Z t b will be called translation by

b.

Clearly

Norman L. A l l i n g

36 d e t Tb = 1.

(7)

Let

h(Da) 5 d

9,

= (i

D~

for

a

E

K*.

&

w i l l be c a l l e d d i l a t i o n

a

a.

Clearly

= a.

d e t (D,)

F u r t h e r , it i s c l e a r t h a t

ITb:

(8)

b

K) u {Da: a

E

E

K*j

generate

h(S) 5 s

w i l l be c a l l e d

inversion.

that

i s o f o r d e r 2.

If

Let

s c1

:- d / c ( E K ) ;

then

Aff2(K).

-

M E GL2(K)

is in

MTaS

d e t ( S ) =1, a n d

Note t h a t

Aff2(K),

Aff2(K).

c # 0.

then

Hence

( 1 0 ) G L 2 ( K ) = Aff2(K) u (Aff2(K))S. Clearly

s

permutes

and

0

s (conCm)s = conC

Further,

0’

m

and

and maps

onto i t s e l f .

K*

t (conC ) t-l = conCa.

Hence

ow.

c1

w e have shown t h a t (11) t h e s u b g r o u p s

(conCa) a E C

z E C i s a f i x e d p o i n t of (12) cz2

+

c=O

If

and

addition, c#O f

(d - a ) z

or

i f and o n l y i f

- b = 0.

a=d,

b=O

f

are a l l c o n j u g a t e .

then

d f O and

t h e n every p o i n t of

afd,

f(z) =z+b/d. C

then ( 1 2 ) has a r o o t i n

If, in

i s f i x e d by C.

f.

If

Thus i n a l l cases

has a f i x e d p o i n t ; hence

( 1 3 ) c o n 1 = uaEcconCa. C l e a r l y ( 1 2 ) i s t h e zero polynomial i f and only i f f f l

t h e n ( 1 2 ) h a s a t most t w o r o o t s .

then

c#O

f # l

implies

(14) If

f

and h e n c e f

m

f = l .

If

I f it has t w o r o o t s

i s n o t a f i x e d p o i n t of

h a s a t most two f i x e d p o i n t s .

h a s t h r e e o r more f i x e d p o i n t s t h e n

f;

thus

Thus w e h a v e :

f=l.

37

Development of Some Discoveries Made Prior to 1827

Clearly

(15) the subgroups

conIw

=

-1 ta (conCmlo ) ta

(i.e., given

element

a

10'

conCm

acts simply transitively on

10

a'

and

(dciIia)in

conCw

=

Further, it is clear that C*

thus

E @ ;

I

(conCwla)aEc are all conjugate.

Ida: a € @ * I

(16) Clearly

a

for all

,a

in

there existsa unique

@*,

that maps

conCm

10

a').

to

ci

Combining

all of these observations we have the following: h

Theorem.

maps

GL2 ( C )

having as kernel C X I .

Given two triples

of distinct points of

C,

that

f ( a ) = a',

exactly triply transitively Given Let

M

-

E

GL2 (c)

M ' 5 1-1 'M;

then

and

on

E

SL2 (a:)

Ca',B',y')

f ~ c o n C such

i.e.,

f(y) = y':

conC

acts

C.

there exists M'

and

(a,B,y)

there exists a unique

f(B) = B ' ,

con1 ,

homomorphically onto

p

E

and

C*

h(M')

such that =

h(M)

.

.

p 2 = det (M)

Hence

(17) h(SL2 ( c ) ) = conC. Clearly

(kerh) n SL2(@)

{+I}.

2

@.

(18) PSL2(C)

and

con1

are naturally isomorphic.

co%C,

it is clear that

Further, it is evident that

(:(co%C)

n

of

Clearly

Turning now to

3.14

co%C.

is denoted by

SL2(C)/{+I}

and is called the projective special linear group

PSL2(@), rank

=

h(Aff2 (IR) )

=

h(GL2m))

co%Cm

(conZm)); thus, arguing as we did in 53.13 one sees

that (1)

h (GL2aR) )

=

co%C.

Now assume that zc + d # 0,

c

M

is in

GL2(lR).

For

z EC,

for which

Norman L. Alling

38

aczz + adz + bcz + bd 2 Icz +dl

h(M) ( z ) =

(2)

I

and hence

h(SL2@?)) c conb.

We immediately see that

6,

h(M) ( 0 ) =

+

GL2 (El)

4)

Given

M

E

i

GL;

Let

M' :p-'M;

(5)

h(GL:OR))

{z:

:

where {M

E

then

(kerh) n SL2 OR)

E

a}.

det(M) < 0

then

Let

G L 2 m ) : det(M) > 0 1

(I? there )

=

z

If

exists

M'

E

=

h(M')

=

p2 =

h(M).

.

det (M) Hence

conq. SL2 (lR)/{-tI} 5 PSL2 OR)

Let

{?I).

such that

SL2(lR) and

h(SL20R))

=

1-1 EIR

and let it

be called the projective special linear group of rank

2

Clearly

R.

(6) PSL20R)

and

conQ

is of index

ates

co%(C)/confi

con8 in

2

co%C,

and the image of

if and only if

The action of

of the action of

are naturally isomorphic.

on I R u

co%C

conC

on

C.

h(M)

gener-

det(M) < 0. {m}

is a great deal like that

Proceeding much as we did in

53.13, we obtain

acts exactly triply transitively on IR u i m 3 .

(7) co%z

Let the usual orientation of IR Ru

{m}.

Clearly

h(M)

induce an orientation on

either preserves or reverses this orien-

tation. (8) h(M) '

for all

z

( 2 )

=

det(M)/(cz +d)2 ,

for which

orientation of IR u

{m)

c z + d f 0.

Thus,

if and only if

h(M)

preserves the

det (M) > 0.

From ( 5 ) ,

Development of Some Discoveries Made Prior to 1827

39

and the remarks above, we obtain the following: (9) Each

fE

C

O

preserves the orientation of Q,

~

hence of its boundary, I R u

Each

{m}.

f

E

and

co%C

-con8

reverses the orientation of I R u (-1. Given

c1

c u+

E

(3.12:4),

let

conQa

(con&) n (conCa)

From (2) we see that ( 1 0 ) h(M) (i) = (ac + bd) + (ad - bc)i

c2 + d2 From this one easily sees that C

- mu

r

E

&.

The subgroups

M E SL2 W) R

and that

{a}),

such that

con&

co%I

acts transitively on

acts transitively on

(conQT).cES) are all conjugate. h(M)

E

~01-4~;

then

(

)

cose

Let

may be chosen in

sin0

Me.

-sin8 cose

Further,

0 EIR

-+

h(Me)

is a homomorphism of IR

since Q

onto

conQ

it is isomorphic to the circle

whose kernel is 2vZ; thus corni 1 group S (%lR/ZvZ). That is - of course

-

not at all surprising,

is conformally equivalent to the open unit disc in C.

Clearly if

f

E

conQi

Clearly the orbits of

fixes one point in I R u corni

then f = 1. are homeomorphic to S 1 ; or it

{i}.

-1 ( 1 ) M-l = (ad-bc)

If

0

Let

such that

(11) M =

is

Q.

detM= 1

d

(-c

-b a) cGL2(K).

then, of course, detM-'=

1.

We will be concerned with the equation 2 (2) Y = P ( x )

E

K[x]

,

{m},

40

Norman L. Alling

where

%

(3) Let

Let

Z

then

x

.

Then

~=P(c%+d)-~.

2

P 2 (c%+d)-4 = y2 = P ( x )

=

=P(h(M) ( % ) I ;

(c%+d)4 (P(h(M)( % ) ) I = :(%I

y2=

.

h(M) ( 2 )

h(M-I) ( x ) ;

y-y(c%+d)

Clearly (4)

is admissible (3.1).

P(x)

Lemma.

6(%)

Proof.

Let

hence

EK[%].

is admissible. 6

6(%).

be the degree of

Clearly

514.

Let

zj

(5)

Thus

h(M-l) ( p j ) ,

p j = h(M)

("p),

for each

distinct so are the four

G4

of (6)

can c# 0

in

Is.

j

Let

of course a/c

-

be zero even if

is a root of

Assume that

is non-singular).

to the finite roots of

(7) if M E Aff2(K) Now assume that

.

a # 0. then

P(x),

McAff2(K);

The finite roots o f

q2

then

(For example, if

x= 0

c=O

P(x)

then

p ( % ) is admissible.

M=S

= A-4B%+6CS2 - 4 D %

(3.13:9); 3

then

% = - l / x and

+E% 4 ,

is a factor of

P(x),

which is absurd, since

sible; thus

is always 3 or 4.

' j

' s

a#O#d

$ ( % ) ; thus

If

all the

and hence

transform, according to ( 5 ) ,

5=4.

EfO

and hence

From (4) one immediately sees that

then

If

'j"

then

:(%);

-

and

are denote the coefficient

i

Aa 4 +4Ba 3c + 6 C a 2c 2 +4Dac3 +Ec4

(since M

(8)

5

Since the four

j.

=

<4.)

ii = n.

1 (j (4.

for

are in C * .

E=O,

then

D=O

implies that

Assume that

P(x)

is admis-

n=4=5;

Using (5) we see that all

x

j

then Is

are

Development of Some Discoveries Made Prior to 1 8 2 7 in

G(%)

distinct,

is admissible.

then

E = 0,

and hence

Then

El

and

=m,

of

6 ( % ;)

and

6=4

A=O=E.

P(%)

.

thus

52,?31 $(%I

5,

and

Since

are roots of

P(x) Thus

Pcz).

5-L

and

5,

Clearly

E4

is admissible. M=S,

and

p2

0

pl.

(The case in which

are in =

6 = 3;

and

are distinct root

Assume that

is admissible,

are

Is

call it

n=3=fi;

are both roots of

m(-p4)

j

n=4

P(x);

E @ *

is admissible.

and

O(Epl)

(9) if

is a root of

0

5

Since the

Assume that

is treated similarly.)

are distinct.

6(%)

6(%).

each being a root of

@*,

41

are in

p3

@*

P(x)

n=3 then

and and

@*,

are distinct and (2);

is a root of

thus

Hence

g ( % ) is adm ssible.

Using ( 7 1 , ( 9 ) , and (3.13:lO

we see that

6(%)

is always ad-

missible, proving the lemma. Returning to Euler's 1 7 6 6 paper [22, vol.XX, pp. 3 0 2 - 3 0 7 3 , after this long but relevant

exposition of some mathematics in

the style of later centuries, let us consider again ( 3 . 1 0 : 2 ) . dx/P(x)"

can now be written as

dx/y.

Letting

x = h(M) (2)

leads to

.

(10) dx = (ad -bc)d% , (c% + d)

(11) @ Y

=

d% (ad-bc)-

3

thus

(cf. [ 2 2 , vol.XX, p. 3031.)

Euler then asserts that chosen so that

G(%)

M

(presumably in

is quadratic in

-2

x

.

GL2QR))

may be

Since his argument

is rather formal, and since some of the quantities that appear may be undefined for some values of the parameters, we will substitute another proof of this important theorem.

Norman L. Alling

42

form.

Generalized Legendre

3.20

Let

sible polynomial of degree 4 in IR[x].

P(x)

be an admis-

(The following theorem

and proof is an expanded version of a result that may be found in (69, pp. 513-5141.) (1) P(x) = S1 ( x ) S 2 (x)I ak # 0,

with (2)

S k ( x ) = ak(x

Thus

where

for

k=l

- rk) (x - ri)

Further, if

k

P(x)

and

and

k

2.

S1(x)

and

and

c k = a r r'

k k k ' .

has 4 real roots the

so that the roots of

+ 2bkx + c k e I R [ x l ,

for some rk

I

b =-a (r +ri)/2 k

2

Sk(x) = akx

S2(x)

S

ri

in c .

for each

k.

(x)'s may be chosen

k

do not interlace each

other. There exists A1'A 2 ' B1' either there exists a1eIR such that

and

Theorem.

B2 eIR*

for which

and 2 ; or there exists

ci

1

and

ci2

EIR,

with

(x-a2).-. LI

* (4) Sk(x) =Ak(x-al)'+B

ci1>a2, such that

for

k=l

and

2.

k

Let

Proof.

(5) dkI b:-akckl criminant of Since

t

and 2 ;

k = l

Sk(x)

and

then

2 dk = ak(rk

4dk

is the dis-

- ri)2/4.

Sk(x) must have distinct roots

(6) dk

Let

for

for each

k.

be another indeterminant and consider

(7) J(t,x) 2 (bl

G

S1(x) - tS2(x).

-

Clearly J(t,x) = (al ta2)x2

- tb2)x + (cl- tc2) ~IR[t,xl.

+

Development of Some Discoveries Made Prior to 1 8 2 7

(8) Let

then

DJ(t) :(bl - tb2)2 - (al - ta2) (cl - tc2) cIR[t]; is the discriminant of

4DJ(t)

sidered to be a quadratic in

x.

J(t,x), where it is con-

Then

DJ(t) = (bi-a2c2)t2 +(a c +a2c1-2blb2)t+ (b:-a

(9)

43

c

1 1

1 2

2

d2t + m t + d l , where

m:a

c 1 2

=

)

a2C1 - 2 b1 b2 '

+

(10) m = a a [r r' + r r' - (rl +ri) (r2 +r')/2].

1 2

A

Let

1 1

2 2

2

DJ(t); then-of course- A

be the discriminant of

is

n

L

m -4dld2.

A routine calculation shows that

DJ(t)

Lemma

and

has two distinct non-zero real roots,

t2. Proof (of the Lemma).

is

Since the constant term of is not a root of

dll which is non-zero (6), 0

Since we have assumed that the equation

P(x) = O

has 4 real roots.

is

S1(x)

n=4,

DJ(t).

the number of real roots of

4,2, or 0. and

DJ(t)

S2(x)

Assume first that

P(x)

were chosen so that their

roots do not interlace; thus without l o s s of generality we may rl > r 1 > r > r' From (11) we see that A > 0, 1 2 2' establishing the lemma in this case. Assume now that P(x) assume that

has 2 real roots. that

S1(x)

are real.

Without loss of generality we may assume and that r and r; 2 2 2 2 2 A =ala2/r1-r21 Irl- ril > 0,

is irreducible in IR[x] Then

r' =rl; 1

hence

proving the lemma in this case.

Now assume that

P(x)

has no

-

real roots. Then r' k = rk for each k. Hence 2 2 2 A=a1a2/r1-r2I b 1 - r 2 I > 0, proving the lemma. Continuing with the proof of the Theorem, it is perfectly

44

Norman L. A l l i n g

al-t.a

= O f o r j equal t o 1 3 2 Note: since the t a r e d i f f e r e n t and a 2 # 0 , j happen f o r b o t h j. Assume f i r s t t h a t possible t h a t

( 1 2 ) al - t 1 a 2 = 0 :

(13) alb2=a2bl,

cl-t

and hence

and hence

c = O

1 2

- a2bl) 2

S1(x) = t l S 2 ( x )

t h i s cannot

showing t h a t

b 1 - t1b 2 = O .

J(tl,x) = O

then,

2.

t 1= a1/ a 2 '

that

2 0 = a D (a /a ) = (alb2 2 J 1 2

Then,

Were

i.e.,

or

x

(for a l l

(for a l l

x

in

in thus

@);

@)

(7)

P(x)

would n o t be a d m i s s i b l e , which is a b s u r d , p r o v i n g t h a t

( 1 4 ) S1(x)

- tlS2

( x ) = (c l

- tlc2)

cIR*.

t l # t 2 , by t h e Lemma, w e may u s e ( 1 2 ) t o c o n c l u d e t h a t

Since

(15) a l - t

a f 0 ; 2 2

thus

J(t2,x)

J(t2,x),

Since t h e discriminant of that there exists ( 1 6 ) S1(x)

alcIR

i s of degree

2

in

x.

i s zero w e see

4DJ(t2),

such t h a t

- t 2 S 2 ( x ) = ( a l - t 2a 2 ) ( x -

S o l v i n g ( 1 4 ) and (16) f o r

and

S1(x)

Lemma, w e see t h a t t h e r e e x i s t

A1,A2,

S2(x),

and u s i n g t h e

B1,

and

and

2.

B2

in

IR*

such t h a t

(17) Sk(x) =Ak(x-al)

2

+Bkr

for

k = l

Now assume t h a t n e i t h e r (18) al thus

- t .3 a 2

J(t.,x)

is zero;

i s a r e a l q u a d r a t i c whose d i s c r i m i n a n t i s z e r o .

7 Hence t h e r e e x i s t

(19) s1 ( x ) for

a1

- tjS2 (x) j = 1

and

and

a2

in

IR

( = J ( t,.x ) 1 = ( a l 7

2.

such t h a t

- t 7. a2 1 ( x - a 7. I 2

Development of Some Discoveries Made Prior to 1 8 2 7 Solving these equations for

S1(x)

and

S2(x)

(20) Sk(x) = A k ( ~ - ~ 1 ) 2 + B k ( ~ - i2i,2 )where in lR*,

for

Note that where each

k,

and

k = l

a1 - a 2 ,

and thus

P(x)

then

Ak

45

gives and

Bk

are

2. Sk

(x)= (Ak + Bk) (x - "1)

would have only one root in

hence would not be admissible: which is absurd.

Thus

for

I

and

@,

a1

#

a2.

Without loss of generality assume that ( 2 1 ) a1 > a2:

proving the Theorem.

3.21

Let us return to consider

(1) dx/P(x)',

:.>

If (3.20:3) of Theorem 3.20 holds let

As Euler did c. 1 7 6 6 . (2)

M

<SL2(R).

If (3.20:4) of Theorem 3 . 2 0 holds let

Set

y 2 =P(x)

and let

2

and

be defined by (3.15:3).

Using (3.15:4) and (3.15:ll) we find that -2

(4) y

=

-2

-2

(A1x + B 1 ) (A2x +B2)

By Lemma 3.15

eIR[x]

and

dx/y=dk/y.

g ( % ) is admissible. This is, essentially, what

Euler asserted [22, vol.XX, pp. 303-3041, c. 1 7 6 6 . Without loss of generality we may assume that (5)

IA1/B1l 5 [A2/B21.

Norman L. Alling

46

then

defined by (6), has determinant

Id,

is in

GLiW).

By ( 5 ) ,

ulvf and

O
w be in

k=h(M-')(%),

and hence

so that

{0,1)

and let $ = d 4 9 ,

as in

By (3.15:4)

(3.15:3). (8)

Let

IA2B1 I-%,

ir2=( - l ) ' ( l -(-l)vG2)( 1 - (-1)wk222 ) .

Note, since this polynomial is admissible (Lemma 3.15) , if v=w

then

k < 1.

Then, by (3.15:ll) (9)

dx =

y

P

= (detM) dft ,

Y

This then motivates the following definitions. w

be in

I0,ll

and let

Let

O
u,v,

v=w,

and

k
Let ( l o )Lu ,v ,w

-

(x,k) :( - l ) u ( l(-1)"~~) (1 - (-1) wk2x2 ) .

This is a real admissible polynomial of degree 4 . The equation 2 y = L u f v l w(x,k) will be said to be in generalized Legendre form. (uIvfwIk)will be called the index of L u l V I w (x,k). Let (")

W u f v f w f kdX/[LU,v,W(x,k) z 1'

.

This differential of the first kind will be said to be in generalized Legendre

form, and

(u,v,w,k) will be called its

index. Theorem.

2 y = P(x) , where

P(x)

is real, admissible and

of degree 4 , can always be transformed, by means of

M

E

GI,:@?)

Development of Some Discoveries Made Prior to 1827

-2

(as indicated above) to

(2,k). Similarly

=LUrVrW

always be transformed to

ccl)uIvIw,kt for some

dx/y

47

can

c en*.

Note that the differential considered by Fagnano was c. 1716.

Some of Legendre's

3.30

Leqendre (1752-1833) wrote a great deal on elliptic integrals.

There is his Exercises de calcul int6qral etc., a 2

volume work that appeared in Paris in 1811 and 1819.

Then he

published his very extensive 3 volume work, Trait6 des fonctions elliptiques, in Paris from 1825 to 1828.

His work on elliptic

integrals contains a summary of much of the work of the 18'th One of his contributions was to define elliptic inte-

century.

grals of the first, the second and third kind, and to show that any elliptic integral could be given in terms of elementary functions and elliptic integrals of these three kinds. The subject of elliptic integrals is a very large one, one which we will not treat at all extensively.

See e.g., [69] for

details on the theoretical aspects of the theory.

In Legendre's Trait6 des fonctions elliptiques

3.31

[49, vol. 1, Chapter 31 he transforms (1) If

x: sine,

df3/[1 (2) If

for

w

~tvtwik

-~r/2<0 <~i/2, then

w

as follows:

OtOtOtk

- k2sin281 % . 0< 8< n ,

then

+ k2]%[1 - kfsin2f314;

where

x:cosO,

-dO/[l

for

w

0 1 0 tlik

kl = k/(l

-

+ k2)'.

--

Norman L. Alling

48

(3)

Clearly

w

which can then be Or1rOtk = k-lwOtO,l,k-l,

transformed by ( 2 ) . (4)

If

X-sece, for

O < e<.ir/2, then

2 2e l 4: k2de/[1 - k2sin (5)

Clearly

w

1t1tOrk

where

w

1rOrltk

-

k2 = (1+ k2)-'.

= k -1w l,otl,k-lt which can be trans-

formed by ( 4 ) . (6) If

x-tane, for

de/[1

-

(k')2sin2el';

(1 - k2)', (7)

If

O < 0
where

then

k'

w

O,l,l,k

denotes as usual

the complementary modules to O < 8 < n/2,

X-sece, for

de/[(k1I2- sin2el 4

-

then

w

k.

1tOrOik

--

.

Legendre gives the following substitution [ 4 9 , vol. 1, Chapter 3, p . 101: let

( 7 ' ) Wl,O,O,k

.

=da/[1

that is let

-

Then de/[ (k'I2 sin2el 'I=

a :arcsin( (sine)/k').

da/[l- (k')2sin2a3 +

sine Ek'sina;

Thus

-

(k'I2sin2a1 %

.

Legendre does not appear to consider the 8'th case, namely w l,ltl,k.

In fact this is exactly the case which has been most

often neglected. (8)

If

xEtan0,

for

O < 8 IT/^,

do/[ (-1)(1 - (k')2sin28 1 4

then

w

1rltltk

-

.

One reason why Legendre takes de/[l

- k2sin2e14

fundamental in his theory is his results (1) - (7').

to be

Development of Some Discoveries Made Prior to 1827 3.32

Using these results of Legendre we can compute the

following "real periods".

K (k2)

-

:

using ( 3 . 3 1 : 3 ) and ( 2 ) .

[ l+k2]' m

(4)

dx 11 [ (-1)(1-x2)( l + k x = klK(kl)/k;

(6)

J

.ir/2

I0

dB [ I - k 2 sin2 01 %

1

using ( 3 . 3 1 : 5 and 4 ) .

Im

dx 0 [(1+x2) (1+k2x2)1'

0

)

'

- -kl

da 2

2

d0 (1-(k'I2sin2 0)%

= K(k').

= K(k').

( l - ( k I ) sin a )'

(1)-(7) are due, with only slight modifications, to

Legendre.

49

Norman L. Alling

50

jm(-1)( 1 + xdx2 ) (l+k2x2)1 %

(8)

= 0

0 [

d0 [(-l) (l-(k')2sin2 a] +

iK(k').

= f

According to Fricke 125, p. 1901, the first result found in which the modulus changes in the course of computing an elliptic integral, is due to J. Landen [47], c. 1775, who showed the following:

3.40

Gauss's a r i t h m e t i c

-

mean

geometric

In 1818 Gauss (1777-1855) published a paper 127, vol. 3 , pp. 333-3551 in which he established the following result. a

and

(1)

where

b

Let

be positive; then

12,

-

d6

2T

0 (a2 cos2 0+b2sin2 0 ) + - M ( a , b ) ,

M(a,b)

is the arithmetic-geometric mean of

(a,b),

which is defined as follows. (2)

Let

a'

y(a,b)

(a+b)/2 and let (a',b'), a(')

-a,

b'

b(O) '.b,

(a(n),b(n)) :y(a(n-l)lb(n-l)), Gauss showed that

(a(n))ncN

common limit, M(a,b) ,

.

(ab)f

and

Then let

and let

for

each

ncN.

(b(n))nrN converge to a

the arithmetic-geometric mean of

(1) is an elliptic integral, which equals

is the eccentricity of the ellipse x2/a2 reason for considering (1) is that

4K(k)/a,

when

.

(a,b) k

+ y2/b2 = 1. Another

(a(n))ncN and

(b(") ) n ~ N

Development of Some Discoveries Made Prior to 1827 converge very rapidly indeed to

In Gauss's Nachlass

3.41

.

M(a,b)

posthumus work) a good deal of

(=

space is devoted to the arithmetic-geometric mean. vol. 3, pp. 361-4031.)

(1) y(a,b) =y(b,a),

(See [27,

First note that for all

a r b> 0.

Thus, without loss of generality, we may assume that Since

y(a,b)

=

(arb) if and only if

attention to the case in which (2)

a=b,

a z b > 0.

let us restrict our

a > b > 0.

2 {(arb)E I R : a > b > 0 ) ;

Let

S :

y

is a

then

51

Cm-map of

S

into

S.

Gauss noted [27, vol. 3,

pp. 361-3631 that (3)

(a' - b') (a' +b') = (a - b)2/4, <

and that

b < b' <

. . .< b (n)

... < a(n) < .. . < a' < a.

He also noted that al-b' -

(4)

a-b

a-b

- 4(a'+b')

=

a-b < 1/2. 2(a+b)+4b1

On the basis of this inequality, Gauss - of course that

(a(n)

and

-

concluded

(b(n)) n € N converged to a common limit

M(a,b) -1-1.

3.42

To prove (3.40:l) it suffices to prove that d0

-

Ti

2M(a,b)

*

Gauss gives the following substitution[27, p. 3521: (2) Sin0

=

2a2sin 8 I (a+b)cos 0'+2asin 8'

Then he shows that

I

for

O< 0'(7~/2.

Norman L. Alling

52

dB (a2cos2e+b2sin2B)'

0

=jy2

dt3

-

dB

-

( p 2 cos2 e + p 2 sin2 el %

Gauss also considers

3.43

(1) 'a-a+ (a2-b2)'

and

noting that they are the roots of y('a,'b) Let (''a

=

x2

- 2ax + b2 ,

and that

(a,b). (Thus we see that y maps S onto - a , ("b-b, let (n)a 1 ( (n-1)a) , and

= 1 ( (n-1)b),

(n)

( a 2 - b2 ) %,

'b-a-

for each

n c N.

S.)

Gauss follows this discus-

sion with some numerical examples.

The fourth of these is given

below [27, vol. 3, p. 3 6 4 1 . Let

a Z '2

and

b :1. Then Gauss givesthe following

Table (4)a

=19.17024 37557 69475 31905 0 ( 4 ) b

=0.00000 00009 32560 02627 6

(3)a

=

9.58512 18783 51017 67266 3 ( 3 ) b

=0.00013 37064 06056 69181 0

(')a

=

4.79262 77923 78537 18223 7 (')b

=0.03579 93323 67652 95745 7

(l)a

= 2.41421 35623 73095 04880 2 ( l ) b

=0.41421 35623 73095 04880 2

1.41421 35623 73095 04880 2

b

a(')=

1.20710 67811 86457 52440 1

b(')=l.l8920

a(')=

1.19815 69480 94634 29555 9

b(2)=1.19812 35214 93120 12260 7

a ( 3 ) = 1.19814 02347 93877 20908 3

b(3)=1.19814 02346 77307 20579 8

a ( 4 ) = 1.19814 02347 35592 20744 1

b(4)=1.19814 02347 35592 20743 9

a

=

=1.00000 00000 00000 00000 0

71150 02721 06671 7

Clearly this table strongly suggests that a very rapidly convergent process is at hand.

Development of Some D i s c o v e r i e s Made P r i o r t o 1 8 2 7

3.44

(1) Let

(2)

6 y ( a , b ) = (a'-

u=a/b;

Hence

b ' )2 / 2 .

y(a,b) =by(a/b,l) =ay(l,b/a).

Further

;)

= $ (a;a J <

P ( c a , c b ) = p ( a ,b )

1/2.

.

u>l.

( a , b ) = b ( u , 1),

y ( a , b ) = by ( u , 1)

.

6y ( a , b ) = b6y ( u , 1) and hence

then

Gy(a,b) : 6 ( y ( a , b ) ) ; then

6y(a,b)/6(a,b)

then

c > 0;

and l e t

S

and l e t

6(a,b) : a - b

Finally

(3)

E

y(ca,cb) = c y ( a , b ) .

p(a,b)

Let

(a,b)

Let

53

6 y ( u , 1) = (u'

1 u'-1 u'+1

p(u,l) =-

- 1)2/2,

.

From t h e l a s t r e m a r k w e s e e t h a t

Lim p ( u , l ) = 1 / 2 ,

showing t h a t

m'U

G a u s s ' s estimate ( 3 . 4 1 : 4 ) i s t h e b e s t p o s s i b l e g l o b a l e s t i m a t e . Let

(4)

then

x Z u - 1;

x > 0.

( a , b ) = b ( u , l ) = b ( l + x , l ) , and

6(a,b) =bx.

L y ( a , b ) = b y ( l + x , l )= b ( l + x / 2 , ( l + x ) * ) ,

and hence

(5)

If

x <1

6y(a,b) = b(1 +x/2 then

1" n= 0

- (1+ x ) 4 ) .

(')xn=1+x/2-x n

i s t h e b i n o m i a l s e r i e s expansion of

( l + x )4 ,

a b s o l u t e l y and i s an a l t e r n a t i n g s e r i e s .

m

(7)

and S O

p(a,b) =

1

n= 2

- (?)xn-l<

2 / 8 + x 3/ 1 6 -

x/8.

Hence

...

which converges

Norman L . A l l i n g

54

Let

for a l l

y(O) ( a r b ) 5 ( a , b )

nE Z,

and l e t

y ( n + l ) ( a , b ) :y ( y ( n ) ' ( a , b ) )

n'o.

y ( a , b ) = b y ( u , l ) = b u f ( ( u4 + u - ' ) / 2 , 1 ) .

(9)

u ( 0 ) E u , u(')

Let

f o r each

Hence

5

(u'

+ u-')/2,

and l e t

u ( n + l ) - ( u ( n ) ) (1)

n EN.

(u ( n ) l n E N

bounded below by

i s a s t r i c t l y d e c r e a s i n g s e q u e n c e that i s 1.

Clearly (9) gives us

(11) y(l) ( a , b ) = b u f ( u ( l ) , 1 ) y ( 2 ) ( a , b ) = b ( u (01, (1)) q U ( 2 ) , 1 )

Let

,(n)

,(n)

- 1, f o r e a c h

T o i l l u s t r a t e how r a p i d l y

u

S

LOg9

n.

( u ( ~) n) c N

converges l e t

and c o n s i d e r t h e f o l l o w i n g t a b l e of a p p r o x i m a t e v a l u e s .

Development of Some D i s c o v e r i e s Made P r i o r t o 1 8 2 7

(12)

n

X

1.58

x

10 4 9

2

1.99

x

1 0 24

3

7.05

x

1 0 11

4

4.20

x

105

5

3.24

x

102

6

9.03

8.03

7

1.669

8 9

x

1 0 49

1.99

x

1 0 24

4.20

x

1 05

3.23

x

10

6.69

x

lo-’

1.03295

3.30

x

1.0001314

1.134

t ( n ) logu‘”)

2

2 x

1+5

II:=o~(n)

1.58

7 . 0 5 x 1 0 11

1.000000002

11

(n)

log9

1

10

(13)

(n)

log9

0

Let

U

5

,

f o r each

n; t h e n

=e

L i m n + m t (n)= 1, a n d h e n c e

m

IIn,ou ( n l i s a b s o l u t e l y c o n v e r g e n t .

From t h e s e r e m a r k s w e c o n c l u d e t h a t t h e f o l l o w i n g holds: m

Theorem.

55

-

M ( a , b) = b ( IInbou ( n ) )

’

= be

l n = ot ( n )/ 2

PART I1 ELLIPTIC FUNCTIONS

This Page Intentionally Left Blank

CHAPTER 4

I N V E R T I N G THE INTEGRAL

Three g r e a t mathematicians i n v e r t e d a n e l l i p t i c i n t e g r a l

t o form a n e l l i p t i c f u n c t i o n :

Gauss, Abel, and J a c o b i .

This

s i m p l e new i d e a r e v o l u t i o n i z e d t h e s u b j e c t a n d p r o v e d t o b e enormously f r u i t f u l .

Although t h e o r d e r of d i s c o v e r y w a s as g i v e n

above, Abel w a s t h e f i r s t t o p u b l i s h a n a c c o u n t of t h e s u b j e c t . Then J a c o b i p u b l i s h e d h i s a c c o u n t , b a s e d t o some d e g r e e o n A b e l ’ s work.

G a u s s d e v e l o p e d t h e s u b j e c t l o n g b e f o r e A b e l ’ s work a p -

peared; b u t he never published h i s r e s u l t s .

T h e s e o n l y became

g e n e r a l l y known l o n g a f t e r G a u s s ’ s d e a t h when h i s c o l l e c t e d w o r k s appeared.

W e h a v e c h o s e n t o p r e s e n t t h e work o f t h e s e t h r e e

g r e a t mathematicians i n t h e order of publication. t h e r e i s n o t a l i t t l e j u s t i c e i n d o i n g so.

F i r s t of a l l ,

Secondly, Abel’s

m a s t e r f u l t r e a t m e n t seems t o b e t h e c l e a r e s t of t h e t h r e e . F i n a l l y , h a v i n g d i g e s t e d A b e l ’ s v e r s i o n , t h e o t h e r s seem e a s i e r

t o follow.

4.10

Abel ‘ s e c h e r c h e s .

..

N i e l s H e n r i k A b e l (1802-29) p u b l i s h e d h i s e p o c m a r k i n g R e c h e r c h e s s u r l e s f o n c t i o n s e l l i p t i q u e s i n volumes 2 a n d 3 o f C r e l l e ‘ s J o u r n a l i n 1827 a n d 1828.

I n t h e f i r s t t w o p a g e s Abel

c i t e s t h e work of “ l ’ i m m o r t e l E u l e r ” a n d o f L e g e n d r e , j u s t a s w e h a v e d o n e v e r y e x t e n s i v e l y i n t h e f i r s t p a r t o f t h i s work. 59

Norman L. A l l i n g

60

( W e have used A b e l ' s c o l l e c t e d w o r k s I1, v o l .

e n c e and w i l l r e f e r t o i t s page numbers.) second page [l, v o l . 1, p.

2641

,

11 a s o u r r e f e r -

A t t h e bottom o f t h e

he w r i t e s

" J e m e p r o p o s e , d a n s ce m6moire, d e c o n s i d 6 r e r l a f o n c t i o n inverse, c'est-&-dire l a fonction

$a, d g t e r m i n g e p a r les

6qua t i o n s

0

=

i

d8

[l-c2sin2e]+

1

s i n 9 = Qa = x." And t h a t i s t h e f i r s t g r e a t i d e a ; i n v e r t t h e i n t e g r a l , j u s t a s w e i n v e r t e d t h e a r c s i n e i n t e g r a l i n 52.21.

I n August of 1823 Abel w r o t e Holmboe (one of h i s t e a c h e r s i n Oslo) from Copenhagen.

H e dated t h e letter a s follows:

"Copenhague, 1I a n (6064321219)1/ 3

( e n comptant l a f r a c t i o n d6cimale.

)I'

[l, v o l . 2, p. 2541.

The l e t t e r c o n t i n u e s i n a l i g h t - h e a r t e d mood.

W r i t i n g of t h e

m a t h e m a t i c i a n Degen, whom h e m e t i n Copenhagen, h e w r i t e s

"You remember t h e l i t t l e p a p e r which t r e a t e d t h e i n v e r s e f u n c t i o n o f t h e e l l i p t i c t r a n s c e n d e n t a l s ; I a s k e d him t o r e a d

i t ; b u t h e c o u l d n o t d i s c o v e r any e r o n e o u s c o n c l u s i o n s o r where t h e m i s t a k e may be h i d d e n . "

152, p. 651.

( S e e a l s o [l, v o 1 . 2 ,

p. 2541 f o r t h e t e x t i n F r e n c h . ) Thus Abel seems t o have had t h e i d e a o f i n v e r t i n g t h e i n t e g r a l , a t least as e a r l y a s 1823.

4.11

I n § I l l of h i s R e c h e r c h e s . . .

Abel i s a l i t t l e more f o r m a l . f o 1lowing

[l, v o l . 1, p. 2661,

H e writes, e s s e n t i a l l y , t h e

Inverting the Integral

(1)

Let

where

c

and

numbers, and (2)

=

a(x)

Let

e e

X

jo

dx 2 [(l-c x) (l+e2x2)I 4

-

are arbitrary

and presumably positive

-

real

.

is not necessarily 2.718281828...

+(a)= x

be the inverse function of (3)

61

a(x)

.

dx w/2 :jYc[(l-c2x2 (l+e2x2)I+

Let

I

and note that (4)

@(O) = 0,

@(w/2) =l/cI

and that

The next idea is to extend

+

@(-a)=-@(a).

to the imaginary axis.

Abel writes [l, vol. 1, p. 2661, "En mettant dans (1) xi an lieu de

.

XI'.

.I'

et dgsignant la valeur de

a

par

Pi.. . ' I

X

(5)

x i = +(Bi)

and

B=!

dx 2 2 2 2 4 0 [(l+c x )(l-e x )]

He then asserts [I, vol. 1, p. 2671 that x=O

and

x=l/e

B

.

is positive between

and he defines

Further, Abel introduces the following auxiliary functions (7)

f(a) : [ l - c24 2 (a)]% and

Finally he notes that

4.12

where

All this can be put into conformity with the present

a(x) :

ji

c

.

+(Gi/2) = i/e.

standards of rigor as follows. (1)

+

F(a) E [ l + e2 2 (a)]4

and

[

e

For

dt (1-c2t2)(1+e2t2)1 4

X E

(-l/c,l/c)

let

1

are arbitrary positive real numbers.

If one

62

Norman L. Alling 0 :arcsin(ct) I

were to let

for all

would see that (1) converges for defined by (1) for all (1) is positive over

function on

x

E

[-l/c,l/c].

Clearly

[-l/c,l/c], differentiable on (2)

2 2

2 2

a ' ( x ) = [(l-c x ) ( l + e x

Thus there exists an inverse [-w/2,w/2]

E

(-l/c,l/c) I

x=+l/c;

(-1/c,1/c),

1-l/c,l/c].

t

thus

is a strictly increasing

ci.

is continuous on and

.

to

a

-

which

of course

41

-

maps

in a strictly increasing fashion onto ,[-l/cr1/c1I

which is continuous, and differentiable on(-w/2,w/2), (3)

can be

a

-f

4

a(x)

Since the integrant of

(-1/c,1/c),

)I

then one

(u) =

[

+

2 2

(1 - c

such that

(u)(1 + e2c+2(u)1%.

Concerning Abel's idea of letting xi

play the role of

in (4.11:1), we could proceed in at least two ways.

We could

define X

dt I [(l-e2t2) ( ~ +2ct2 114

o

and proceed as in 54.11. (5)

a(z)

to be

where

z

is in

fied.

For

C

dC (1-c252)(1+e252)14

ji

and the path of integration is to be speci-

z E [-l/c,l/c]

requiring that

5

a(z)

Let

p(v)

(7)

p'

for all (8)

we could define

run through IR.

(Z{ti: t E [-l/e,l/e] 1 ) I (6)

Or we could regard

and let

Now let

5

be the inverse function to

v

For

E

v

2 2 p

(-l/e,l/e) E

a(xi

2 2

( v ) )( l + c p (v))lfr

.

[-G/2,G/21 I

let

z

E

by ( 5 ) by [-i/e,i/el

run through IRi; then

=iB(z/i).

(v) = [(I - e

a(z)

$(iv) = ip(v

x

Inverting the Integral Clearly (9)

+

63

is continuous on

Do :[-l/c,l/cl u i[-l/e,l/el.

One readily sees that (10

for all z E (-l/cI1/c). Let z = iv. Then d4 dv d+ 1 d ~ ( z =)-(iv) = 7 -(ip (v)) , by (8), which equals dz dv 1 dv 2 2 2 2 P ' (v) = [ (1- e p (v) (1 + c p (v) I+, by ( 7 ) , which equals [(l - c2$ 2 (iv))( l + e24 2 (iv)l4 Thus (10) holds for all z c E0.

.

4.13

In the next section of Abel's Recherches

...

,

51.2,

Abel states an addition theorem

Abel writes of

and

c1

11, vol. 1, p. 2681.

to be

r(a,B).

then

r = $ (a);

that they are "deux ind6termin6es"

He then defines the right'hand side of (1)

He then computes

them to be equal. conclusion that

B

ar/aa

and

ar/aB,

finding

From this he quite naturally is drawn to the

r=$(a+ B ) ,

for some function

thus he concludes that

$.

r = 4 (a+ 8)

,

B= O

For

completing

his argument. Addition theorems for

f

and

F

are also given.

Similar

arguments in their support are supplied [l, vol. 1, pp. 268-2691 Many consequences of Abel's Addition Theorem (1) are derived in the course of the paper.

4.14

One difficulty with Abel's argument is that (4.13:l)

is only defined for

a,B,

and

a+

E

Do

(4.12:9).

In an attempt

to prove a restricted version of Abel's Addition Theorem,

Norman L. A l l i n g

64

let

(1)

Clearly

E~ E c ( ~ , P ) E El

a , ~ , a+

n 2 :

i s a subdomain o f

IR2

P

.

E

(-0/2,w/2)1.

Using A b e l ' s c a l c u l a t i o n ,

which i s now v a l i d , o n e sees t h a t grad r =

(2)

(g)

(l,l), on

El.

Thus, t h e d i r e c t i o n a l d e r i v a t i v e o f ( - l , l ) , is z e r o .

on

Thus, f o r

2

E l n { ( a , @ )EIR :

r(a,B)

i n the direction

k e (-w/2,w/2),

a+P=k}.

r(a,B)

is constant

Thus

r ( a , @ )=$(k) = Q ( a + P ) ,

(3)

f o r some f u n c t i o n

$.

For

a = k

and

@=O,

r ( a , B ) = $ ( a )=$(k);

S i m i l a r l y A b e l ' s A d d i t i o n Theorem c a n be p r o v e d when and

c1

t B

are a l l in

i (-G/2 , G / 2 )

.

Abel n o t e s [l, v o l . 1, pp. 271-2721

that and t h u s t h a t

(5) (6)

$(a+ 2 0 ) = $(a)*

C l e a r l y ( 5 ) and ( 6 ) c a n be used t o e x t e n d t h e d e f i n i t i o n o f t o a l l of

from [ - w / 2 , 0 / 2 ]

(8)

$ ( a+ 2 G i )

Hence

$

a,@ E I R ,

$

Abel a l s o n o t e s t h a t

IR.

= $(a).

c a n be e x t e n d e d t o

$(a+ P i )

IRi,

@.

E x c e p t f o r t h e s e poles o f

Given Thus

$

Abel a s s e r t s [l, v o l . 1, p. 2751 t h a t

$ [ ( m + + ) w +( n + + ) G i l =1/0,

be shown t o h o l d .

u s i n g ( 7 ) and ( 8 ) .

c a n t h e n b e d e f i n e d u s i n g (4.13:l).

c a n be d e f i n e d on a l l

(9)

a,R

$,

for all

(4.12:lO)

m,nE Z.

and (4.13:l)

can

Two o f t h e most i n t e r e s t i n g r e s u l t s a r e

61

Inverting the Integral

and (8); thus

$

65

is doubly periodic.

Much of the rest of Abel's Recherches... representing $,f,

and

F

is devoted to

in terms of other functions.

One

aspect of this work will be described in the next chapter on theta functions.

4.20

Jacobi's Fundamenta

w...

Carl Gustav Jacob Jacobi (1804-1851) sent two letters to Schumacher which were published in his Astronomische Nachrichten, in 1827.

These were essentially research announcements.

(See

[37, pp. 29-48] for extracts of these letters or the September and December numbers of the letters.)

...Nachrichten for

the complete

The first of these extracts is dated June 13, 1827.

It concerned some special transformations of elliptic integrals. By October 4'th, 1827 the first part of Abel's Recherches... reached the Konigsberg Library.

(On that date the University

Librarian wrote a letter protesting that this number had been sent at letter rates, and thus cost extra taler [52, p. 1861.) On November 18'th Jacobi dated his last letter to Schumacher, which appears in the second extract.

In this, the idea of in-

verting the integral becomes explicit.

Neither extract contains

much in the way of proof. Jacobi's Fundamenta Nova Theoriae Functionum Ellipticarum [37, pp. 49-2393 appeared in 1829, and is dated February 1829. Expanding on his letter to Schumacher of November 18'th 1827, Jacobi proceeds as follows [37, p. 81 ff.].

and let

$

be known as the amplitude of

u,

Norman L. Alling

66

(2)

$ -am(u);

then

X

dx 2 2

2

0 [ (1-x ) (1-k x

x = sinam(u)

,

)I4

and

.

Then Jacobi notes that dx 0 [ (1-x2)(1-k

Jacobi defines ~ k'

tary modules

(5)

(

and lets

u to ) be

to

k

K' E j'12 0

am(K-u), defines the complemen-

as Legendre did; d$ [l-(k')

2

2

sin $ I 4

I

and considers (6)

Aam(u)

E

11 - k 2sin2am(u)l s,

.

Jacobi considers the trigonometric functions of coam(u).

am(u)

and

In the next section of the paper, 518, Jacobi states

the following addition Theorem.

Let

am(u)

5

a,

am(v) E b,

and

am(u+v)

5

a;

then

(7)

sina = sinacosbAb+ sinbcosaaa 1 - k 2 sin2 asin2 b

There follow a vast number of formulas.

In 5 9 Jacobi introduces

the following transformation: (8)

sin@ = itan$,

asserts that

cos$=sec$

From this Jacobi deduces that

1

and hence

Inverting the Integral

67

(10) sinam(iu,k) = itanam(u,k')

cosam(iu,k) = secam(u,k') tanam(iu,k) = isinam(u,k'),

etc.

It is interesting to note that Jacobi's mathematical style is quite reminiscent of Legendre's, where as Abel's seems much more like that of our era.

Legendre apparently found Abel's

style radical and Jacobi's more to his liking [52, p. 1871.

Let us try to bring all this up to contemporary

4.21

standards of rigor.

Let

O < k < l , and let

4

E

[-lr/2 ,lT/2] ;

then let

Then

$

E

[-7/2rn/2I +u($)

E

is a strict y increasing con-

[-K,K]

tinuous surjective map that is differentiable on Clearly

u ( - $ ) =-u($) I

for all

$

E

[-n/2,n/2].

(-n/2,n/2).

Let

am

be

its inverse function; thus (2)

T E

[-K,K] +am(.r) E [-n/2,n/21

is a strictly increasing continuous surjection, which is an odd function, and which is differentiable on (3)

v(x)

=

On letting (5)

with

d am(-r)= (1- k2 sin2am(.r))'. d.r Now let

(4)

(-K,K),

v(x) =

X E

(-1,l) and let ds

[(l-s2 ) (1-k2 s2)]+ s Esine

*

, dB [l-k2 sin2 el'

thus ( 4 ) converges for

x = +I,

=

u(arcsinx) ;

and hence ( 4 ) may be used to

Norman L. Alling

68

define

v(x)

on

[-1,1].

x = sinamv (x) and thus

(6)

sinam T

(7)

T =

jo

ds

,

2

[(l-s 1 (1-k s

Differentiating ( 7 ) with respect to (8)

d (z sinam(T))L=

T E

[-K,Kl.

(-K,K)

gives

for all

I1 T

on

2 2 2 (1-sin am(-r))(1-k sin am(,r)).

Jacobi's Addition Theorem (4.20:7) is just Legendre's, in this context; and hence it holds for all

u,v

and

u+v

in

[-K,KI.

The addition theorem can then be used to extend the definition of

sinam(-r) for all

4.22

(1)

'r

~ m .

Now let us consider Jacobi s substitution. 137, p. 851 (4.20:8

sin+=itan$

;

which later was called Jacobi's imaginary transformation. Proceeding formally, cos2 $ = l - sin2 + = l + t a n2 $ = s e c2 IJJ; thus cos$ =sec$b. Differentiating (1) gives cos$d+=isec2$bd$ and hence

d$=id$/cos$b, as Jacobi asserts.

Then we obtain

(4.20:9),

by formal means, much in the spirit of the 18'th

century.

Nevertheless we would like to see (1) put into more

current mathematical form.

An effort in this direction might

be to try to define (2)

IJJ

to be

arctan(sin$/i).

Had the trigonometric functions and their inverses been extended to the complex domain in 1 8 2 9 ?

Gauss' famous letter to Bessel

on complex integration is dated 1811 [ll, p. 311.

Cauchy's

papers on the subject began appearing by 1814 [ll, p. 3 3 ff.]. Nevertheless, Jacobi may have felt comfortable with (3)

arctan z =

1 ,

l + z

an$. thus with

Inverting the Integral

2”n=0

arctan z =

(4)

Of course

(-1)n p + 1

2n+l

-

(5)

1 log l+iz : arctan z = - 21 1-12

(6)

1 $=-log2i

As

l+sin$ 1-sin$

runs through

$

69

thus

’

( - ~ T T / ~ , T TJ1~ ~runs ) , through IRi.

Thus

Jacobi’s transformation (1) can be put on a firm foundation. With

simam(u,k)

defined on IRi, by means of (4.20:10), we

then can extend it to all of C! by using its addition formula (4.20:7).

4.23

Jacobi abbreviated his functions sinus amplitudinis,

cosinus amplitudinis, delta amplitudinis, Aam,

... .

sinam, cosam,

It was apparently Jacobi’s student Gudermann, who

further abbreviated these terms to

sn, cn, dn,

1882, Glaisher introduced the notation l/cn(u) ,

nc(u)

... to

... .

ns(u)

... .

Later, in

l/sn(u) ,

(See e.g., [69, p. 4941 for further de-

tails. ) Much of the rest of Fundamenta E . . is . devoted to the problem of representing elliptic functions in terms of other functions.

One of the most fruitful of these representations

is in terms of, what he called theta functions. troduced in his Fundamenta

=...

,

These were in-

and will be discussed in

the next chapter.

4.30

Gauss’

w o r k on e l l i p t i c f u n c t i o n s .

Carl Friedrick Gauss (1777-1855) investigated elliptic functions extensively, having inverted an elliptic integral of

70

Norman L. Alling

the first kind to form an elliptic function by 1801, before either Abel or Jacobi were born.

Very little of his work on

this subject was published during his lifetime, most coming into other hands only after his death.

(Felix Klein gives a

long and excellent account of Gauss' work in his Vorlesungen [40].)

...

In 1 8 0 1 Gauss published his extremely important and

influencia1 Disquisitiones Arithmeticae [ 2 6 ] .

In Section 3 3 5

of that work he dropped a few hints, in his elegant, withdrawn style, about elliptic functions: namely the following: "The principles of the theory which we are going to explain actually extend much further than we will indicate.

For

they can be applied, not only to circular functions, but just as well to other transcendental functions, e.g., to those which depend on the integral

and also to various types of congruences.

Since, however, we

are preparing a substantial work on transcendental functions, and since we will treat congruences at length as we continue our discussion of arithmetic, we have decided to consider only circular functions here. " In 1818 Gauss published a paper 127, vo1.3, pp. 3 3 2 - 3 5 6 1 on the arithmetic-geometric mean, which we considered in 53.4. What beyond Gauss' meager published work on the subject was known to Abel and Jacobi before they inverted the integral might be a topic of very interesting historical research: however,it goes beyond the scope of the present work.

Inverting the Integral 4.31

71

By 1797 Gauss had considered ([27, vo1.3, pp. 404-

4291 Nachlass), the function sin lemn, which inverts the lemni-

i. e. ,

scate integral:

Gauss then defines

He gives (3)

sin lemn 6 =

$

1 5+ -10 4

+

1 $g 120

'

11 13 -15600

...

211 $171007 318240000 35 3 60 00

[27, vo1.3, p. 4051 (The carrying of numerical calculations to

many many places seems to be one of the hallmarks of Gauss' work.) (4)

(5)

He then gives 2

5-36

9 552 13,

5136

17,

& + m $ 17! 4 Q($) = 1 + 4! 2 $ 4 -8!4 $ + 4 0 8m $ 1 2 + = 16! $I6+ P($) = $

- -5! 4

37 113 + 159667200 '12 2 4151347200 [27, vo1.3, pp. 405-4061.

5146848 $ 2 1 + 21!

...

...

4171

'I4 + 69742632960000

p+..,

Gauss states Fagnano's addition

theorem as follows: (7)

let

X E

sinlemn

$,

y

= sinlemn$,

and z 5 sinlemn($+

then (8)

z=

x(l-y2)% +y(l-x 2 ) 4 l+x2y2

127, vo1.3, p. 4061.

$);

Norman L. Alling

72

4

He clearly allows p. 4 0 7 ff.].

to take on complex values [ 2 7 , vo1.3,

In Gauss' second paper on the lemniscate function

[ 2 7 , vo1.3, pp. 4 1 5 - 4 1 6 1 ,

(9)

he asserts that

sin lemn $ = P ( $ ) / Q ( $ )

(10) cos lemn $ =p($l/q($) I

where

-

a E w/n

and

s

:sin(.rr+/w). Gauss also asserts

[27,

vo1.3, p. 4161 that (15)

2% ( $ + ;/2) 2% ( $

Thus

= p(4)

+ 2 2 ) =q ($1t L

p($ +;/2)

=-Z4P($)

q(+ +;;;/2)

=

P

and

and

L

p

24~($). are periodic of period

are periodic of period

-

Further,

W.

2

a(i+o) = e'TJI

Q(~o),

2

q(+o),

p(i$o) = e"

and

2

q(i$o) = en+ p($w). From (9) and ( 1 5 ) we deduce that

2w,

while

Q

and

q

Inverting the Integral (17) sin lemn(i4)

=

isin lemn 4 ;

thus

(18) sin lemn(4 + 2i0) = sin lemn(4) , Thus

73

for all

4.

sinlemn is doubly periodic of periods

2;

and

2i;.

Gauss' notes on this subject, which appear in his Nachlass, were evidently not prepared for publication.

There is very

little prose in the text and much calculation.

Nevertheless,

it becomes clear that Gauss had within his grip a very great deal of the theory that Abel and Jacobi published.

Perhaps, in

fact, he knew all of it; but he published almost none of it.

4.40

The

guestion

of priority.

It seems evident that Gauss knew a very great deal about elliptic functions by 1801. integral by 1823.

Abel had the idea of inverting the

The first part of his Recherches...

,

complete

with arguments, appeared in the fall of 1827, by which time Jacobi had written to Schumacher.

By October 4'th, 1827 the

first part of Abel's Recherches... had arrived in the University Library in Kbnrgsberg, where Jacobi was in residence. This issue of Crelle's Journal (vol. 2) contained a paper by Jacobi; "...otherwise, most of it was filled by Abel's long memoir, dealing with the topic on which Jacobi was working.

It

is hardly believable that he had not glanced at it and noticed the main idea stated in the first section.

It is impossible to

deny that Jacobi may have found the idea of the inversion independently; Gauss' remarks about Abel's discoveries shows how coincidence may occur.

But under the circumstances, one is in-

clined to believe that the idea had been obtained from Abel. The young Jacobi was at a critical point in his scientific

Norman L. Alling

74

career: for several months his only thought had been to search for a proof of the announcements he had made.

Suddenly he

stood before the goal; the brief paper was probably written in a few hours with only the rapture of victory in his mind, making him forget in the moment that he had, in past, supported himself upon an idea which justly belonged to another."

So writes

Oystein Ore in his excellent biography of Abel [52, p. 1861. This is not to suggest that Jacobi was not a great mathematician. Nevertheless it seems certain that Abel was the first to publish a full account of inverting the integral.

Unfortunately Abel

died very young: in April of 1829, the year Jacobi's Fundamenta

Nova... appeared.

Jacobi lived on until 1851.

He made enormous

contributions to the theory of elliptic functions, developing the theory of theta functions into a very complete and important theory, as we will see in the next chapter.

Gauss died in 1855.

Not until long after his death did the notes and manuscripts he left upon his death

reveal to the mathematical world how

complete and deep his knowledge of elliptic functions was, even before Abel and Jacobi were born.

CHAPTER 5

THETA FUNCTIONS

Having succeeded in defining elliptic functions on the whole complex plane, the problem of how to represent these new and very interesting functions interested Gauss, Abel, Jacobi, Eisenstein, Weierstrass,...

.

One of the most useful ways was

in terms of quotients of certain Fourier series, called theta functions.

5.10

Origins

Except for minor (but annoying) variations in normalization and notation, the theory of theta functions

-

as applied to

the classic case of elliptic functions - is very nearly in its present form in Jacobi's Theorie den eigenschaften Borchardt

der

der

elliptischen Functionen

Thetareihen abgeleitet, notes made by C. W.

of Jacobi's lectures in 1 8 3 8 [ 3 7 , pp. 497-5381.

treatment of theta functions is based on Jacobi's functions and

H,

... .

aus

This 0

which he considered extensively in his Fundamenta Nova

However, theta functions go back further.

They may be

found in Abel's Recherches and in Gauss' work on elliptic functions,since P ( @ ), Q($) , p($) , and q($)

are theta functions.

Jacobi refers to Gauss [ 3 7 , p. 1 5 2 and p. 2731, to Legendre [37,

p. 2021, and to Euler's work on the partition function

[ 3 7 , p. 1 4 5 1 , where Jacobi considers 75

Norman L. A l l i n g

76

J a c o b i used (1) e x t e n s i v e l y 1 3 7 , p . 1 4 4 f f . I [ 2 7 , vo1.3,

p. 436 f f . ] , Gauss u s i n g

[q]

,

as d i d Gauss

t o denote (1) [ 2 7 ,

vo1.3, p. 4 4 6 1 .

5.11

Accordins t o Dickson [18, v01.2, p. 1011,Leibniz

wrote t o J o h . B e r n o u l l i i n 1 6 6 9 a s k i n g him i f he had i n v e s t i g a t e d t h e number of ways a g i v e n number

nc N

c a n be s e p a r a t e d i n t o

two, t h r e e , o r many p a r t s ; remarking t h a t t h e problem seemed d i f f i c u l t and i m p o r t a n t . number of p a r t i t i o n s on 3

Given

n;

n

133, pp. 274-2761

generating function f o r essentially vol.1,

N,

= 1

p(n)

5=5

F(x)

+0=4+1= p ( 5 ) =7 .

is called the

Euler apparently discovered

( S e e 118, vo1.2,

p. 279 f f . ] f o r more d e t a i l s . )

denote t h e

+ 1 + 1 + 1 + 1,

for details.)

( p( n )) ncN.

t h i s i n 1748.

let

t h u s , s i n c e e.g.,

+ 2 = 3 + 1 + 1= 2 + 2 + 1 = 2 3.1 -t 1 + 1

(See e . g . ,

E

p . 1 0 4 1 and [18,

'According t o Whittaker

and Flatson, i n a f a s c i n a t i n g f o o t n o t e [69, p . 4 6 2 1 , t h e p a r t i t i o n function,

c o n s i d e r e d by E u l e r i n 1748, i s t h e f i r s t f u n c t i o n of t h e t a t y p e t o appear i n a n a l y s i s .

S u f f i c e i t t o say t h a t t h e t a f u n c t i o n s

have a l o n g h i s t o r y and wide and v a r i e d a p p l i c a t i o n s .

Further,

t h e t a f u n c t i o n s and t h e i r a p p l i c a t i o n s are s t i l l under a c t i v e investigation.

I t i s a l s o w e l l t o note t h a t t h e r e are a g r e a t

many i n t e r c o n n e c t i o n s between t h e t h e o r y of e l l i p t i c f u n c t i o n s

77

Theta Functions M o s t p r o b a b l y i t i s no c o i n c i d e n c e t h a t

a n d number t h e o r y .

F u l e r , Gauss, Jacobi, and E i s e n s t e i n a l l c o n t r i b u t e d v e r y s i g n i f i c a n t l y t o t h e t h e o r y of e l l i p t i c i n t e g r a l s a n d f u n c t i o n s on t h e o n e h a n d , a n d t o number t h e o r y o n t h e o t h e r .

Definitions

5.20

The a u t h o r h a s f o u n d i t c o n v e n i e n t t o u s e o n e t e x t a s a s t a n d a r d s o u r c e f o r n o t a t i o n , n o r m a l i z a t i o n , and t e r m i n o l o g y o n H e h a s c h o s e n [361 f o r t h i s p u r p o s e .

theta functions.

Unfor-

t u n a t e l y , annoying minor v a r i a t i o n s i n t h e s e c o n v e n t i o n s o c c u r f r o m book t o book.

1361 i s t h e most r e c e n t of t h e t h r e e c l a s s i c s ,

[69] and [ 3 6 ] ; it h a s uone t h r o u q h 4 e d i t i o n s between 1922

[60],

a n d 1 9 6 4 , a n d seems n e a r l y l e t t e r p e r f e c t .

The same m a t e r i a l ,

w i t h o n l y s l i q h t a n d i n e s s e n t i a l v a r i a t i o n s may a l s o b e f o u n d i n [69],

[60]

(1)

Let

,... w

and

T F w'/w

{IR,

be i n

0'

and

(C*(E(C

Im(r)

3

I t is also convenient t o l e t

'r = w / w 2 1'

Also l e t

and hence

s > 0.

(3.12:4);

thus

r

5

r

As u s u a l T

2w

with

and

w2

3

r , s EIR.

20';

thus

Then

Im(7)

Let

hEeinT;

(3)

Let

U E

8a d e n o t e s t h e u p p e r h a l f p l a n e

then

v-u/2w,

h=e-'**e and

z :e

inr

,

inv

inu/2w); (=e

and

-nImv

IzI = e

Let

Z

= s,

€9.

(2)

C,

such t h a t

0.

wl-

+ si,

- {O})

d e n o t e t h e s e t of a l l i n t e g e r s .

Now l e t

Ihl=e-"
Norman L . A l l i n q

78

136, pp. 188-2121,

(See e.g.,

11, C h a p t e r 1111.)

1 6 9 , pp. 462-4901,

o r [ 6 0 1 Tome

(The t y p e - f o n t u s e d f o r t h e " t h e t a " i n ( 4 )

i s n o t t h e c o n v e n t i o n a l o n e , which i s a n open lower c a s e G r e e k Only a c l o s e d lower c a s e " t h e t a " seems t o be a v a i l a b l e

theta.

S e l e c t r i c t y p e w r i t e r being used t o t y p e t h i s

f o r t h e I.B.M. manuscript.

Apologies t o t h e r e a d e r . )

T o see t h a t t h e t h e t a f u n c t i o n s d e f i n e d i n ( 5 . 2 0 : 4 )

5.21

converge c o n s i d e r t h e following:

such t h a t with

Let

la,( 'A,

a z 0;

E 5

0

and l e t

CC

u n i f o r m l y on

and l e t

2

( a n I n c Z be complex numbers

n.

for a l l

Let

f ( x ) :a x 2

+ b x + c cIR[x],

g ( x ) :j x + k ~ I R t x l . C o n s i d e r

and l e t

{T € 3 : I m - r , E l

c

A > 0

Let

Lemma.

yen.

.

Let

K ( ~ , E )Z { V E Q : :I m v L y }

x

Then t h e series (1) c o n v e r g e s a b s o l u t e l y thus

K ( y 1 e );

T(vl T)

i s an a n a l y t i c f u n c t i o n on

X Q .

Let

Proof.

b

n

- a hf(n)zg(n)l for a l l n

n c 2.

then

(2) thus, (3)

lbnl

l/n,Al/ne-v~(an+b+c/n) ,.-~y(j+k/n).

limn+,

-

I

I bn 1 'In = 0.

Further,

- Al/ne-TE

(an-b+c/n)

knIl / n

. e-iiy

(-

j+k/n)

.

I

Let

ncN;

Theta Functions

thus

Limn+, 1 b-n 1 'In

79

proving the Lemma.

= 0,

This lemma is very far from being the best possible result, but it more than suffices to prove the following: Theorem.

ly on

(5.20:4) converqe absolutely uniform-

80,...,83

K ( ~ , E,) hence are analytic functions of

(v,r)

.

Properties of theta functions.

5.30

Theta functions have an enormous number of properties and interrelations. We will only mention a very few of these here. (Pee the references cited for more theta.relations.) In many of the applications

r

is held fixed and thus

the theta functions are functions of a single variable v. i rrv , 0,(v) ,'a3 (v) are Fourier series in v Since z = e

,...

periods

(1)

1,2,2, and

Let

m e h-'z-l

Vote:

1

respectively.

and let

k = exp(-irrr

of

k: h-1z-2.

- 27iiv);

then the following can easily be obtained, essentially by substitution.

[36, p. 1991 :

i.e.,

OO(v+ 1/21

It is also convenient to define

=

e3(v), etc.

Norman L. Alling

80 (3)

e2

E

e3

E €i3(0),

e2(o) and

where differentiation is with respect to any

v.

(2) shows that

could be taken as fundamental, the others being derivj able from it. to choose one, is periodic of period 2, and 8

el,

by ( 2 )

€I1("+ 2.r) = k 2 €I,(v).

Such a function is called guasi-

periodic of guasi-periods 2

and

'r

€I1.)

are also auasi-periods of

5.31

0

i

2.r.

(Note also that

1 and

is periodic, but not doubly periodic; however,

the followina functions are doubly periodic:

(See e.g., 136, pp. 2 1 2 - 2 1 8 1 ,

169, pp. 491-5351, or 160, Tome I11

for more details.) Many authors have based their theory of elliptic functions on theta functions, usinq (1) to define dn(u).

sn(u), cn(u), and

Then virtually every other quantity in the theory of

elliptic functions can be computed in terms of theta functions. This approach can be found in the 1838 lectures of Jacobi 137, pp. 497-5381.

seen,

One advantaqe of doinq this is that, as we have

e i (v) is easily and unanbiquously defined, and is ana-

Theta Functions

u EC

lytic for all

and all

r € 4i

be so defined, and are meromorphic. by Jacobi in his Fundamenta managed to define

sn(u)

81

.

sn(u) ,.. ,dn(u)

thus

can

In the classic case studied

E... , r

=

iy,

for

y > 0.

He

etc. by inverting the integral, using

the addition theorem etc., but only for the special case of .r=iy. Usinq theta functions it is no more difficult to let run throuah all of

Q.

methods of Fundanenta

The author does not know if Jacobi's

Nova...

have ever been riqorously extend-

ed to cover the case of non-real

kcq: - {0,1}.

One of the dis-

advantaaes of usina theta functions to define

sn(u)

etc. is

that, a priori, it is not evident that theta functions have anything at all to do with elliptic integrals. show 136, p. 2161 that

sn(u), as defined by (l), satisfies

(2)

2 2 2 ( s n ' ( ~ ) ) ~(=1 - s n (u))(l-k sn ( u ) ) ,

(3)

k = B'//B' 2

However, one can

where

3'

Of course (2) establishes a very close connection between

sn(u)

and dz

1 (1-z2)(1-k2 z2 ) ] 4

-

One could adopt the point of view, which TQeierstrass seems to have adopted, that

sn(u)

satisfies ( 2 ) , subject to an initial condition: and

sn'(0) > 0.

that

is a meromorphic function on C namely

sn(0) = 0

Viewed in this way theta functions provide us

not only with an existence theorem, but a vast amount of information about the solution of (2) as well. Theorem.

sn(u)

periodic with periods and its poles are at

is meromorphic on C . 40

and

2w'.

2nw+ (2n'tl)w',

It is doubly

Its zeros are at for each

n

and

2n0+ 2n'w' n' E

2.

Norman L . A l l i n g

a2

t36, p . 2 1 5 1 .

5.32

(1)

One e a s i l y sees t h a t t h e f o l l o w i n g h o l d :

e , ( v ) = 1-

-

A

. ..

9

~ ~ C O S ~t T~ V ~ - C O S ~ K 2hV C O S ~ I T Vt

( v ) = 2hk s i n n v 0 2 ( v ) = 2h’cos~v

- 2 h 9 / 4 s i n 3 , ~ vt 2 h 2 5 / 4 s i n 5 ~ v- .. . + 2 h 9 / 4 c o s 3 ~ vt

2 h 2 5 / 4 c o s 5 ~ v+

A 9 O 3 ( v ) = 1 t 2hcos2rrv t 2h-cos4rrv t 2h C O S ~ I T V t

Although

ej(v)

and

. .. .

i s d e f i n e d by a n i n f i n i t e s e r i e s , it c a n

a l s o be g i v e n by a n i n f i n i t e p r o d u c t . and l e t

{2,4,...,2nI...1

.. . ,

v

g

Let

run through

run through

~ ~ , ~ , ~ , . . . , 2 n ~ 1 , . . .

Let

CZn(1-h‘);

(2)

then

9

B o ( v ) = C TI ( 1 - h v z 2 ) ( l - h ” z - 2 ) .

(3)

V

(1 - hqz-2

e 2 ( v ) =Ch’(z

t z-l)

II (1 + h q z 2 (1t hgz-2

I

,

and

9

e 3 ( v ) = C TI (1+ h ” z 2 ) (1+ h v z - 2 ) . v

Note t h a t t h e p r o d u c t s a p p e a r i n g i n ( 2 ) a n d ( 3 ) a r e v e r y c l o s e l y related to the p a r t i t i o n function (5.11:2).

Indeed, each can be

w r i t t e n i n terms o f t h e p a r t i t i o n f u n c t i o n f o r v a r i o u s v a l u e s of

x

and

z.

F r i c k e p o i n t s o u t 125, p. 2271 t h a t

(4)

P ($1 =

z3l4(!)

s

( e - I T I 4 s i n a - e -9T’4sin34

+ e-25n/4sin5+ -

. . . I,

Theta Functions

83

and (5)

(E) 4

= 2-k

O($)

(.1+ 2ecTIcos2$ + 2e-4TIcos4 $

+

... )

,

where

$ = $;/TI.

r = i,

Let (6)

then

thus, referring to (1), we see that

h = e-';

P(@)/el($/z)

and

B($)/e3($/w)

are constant; thus we have the following: Theorem.

F7here

r = i l Jacobi's theta functions are con-

stant multiples of suitably normalized versions of Gauss' "theta" functions, P,Q,p,

and

q.

Very near the end of the part of Abel's Recherches...

5.33

that appeared in volume 2 of Crelle's Journal in 1827 [l, vol.1,

p. 3471, Ahel qives the followins development of his elliptic function

6:

In [l, vol.1, SVIII, p. 3521 of his Recherches... siders the case in which $

e=c=l.

,

Abel con-

In this case Abel's function

is Gauss' sin lemn function, and Abel's constant

to Gauss' constant

-w.

Putting

(W/TT)

(sin(am/w))

product in the numerator of (1), gives Gauss' in (4.31:ll). Q(a),

(2)

w

is equal

with the

P(a) ,

as given

The product in the denumerator of (1) is Gauss'

as siven in (4.31:12).

Thus

$ (a1 = P ( a ) / Q(a)

Gauss wrote to Bessel, in Konigsberg, as follows, after

84

Norman L. Alling

the first part of Abel's gcherches

... had

appeared.

"I shall

most likely not soon prepare my investigation on the transcendental functions which I have had for many years

-

since 1798

-

because I have many other matters which must be cleared up. Herr Abel has now, as I see, anticipated me and relieved me of the burden in regard to one third of these matters, particularly since he has executed all developments with great stringency and elegance.

He followed exactly the same road which I traveled

in 1798; it is no wonder that our results are so similar.

To my

surprise this extended also to the form and even, in part, to the choice of notations, so several of his formulas appeared as if they were copied from mine.

But to avoid every misunderstand-

ing, I must observe that I cannot recall ever having communicated any of these investigations to others." 152, p. 1831.

CHAPTER 6

THE INTRODUCTION OF ANALYTIC FUNCTION THEORY

6.10

Early history.

In 1811 Gauss wrote, in a letter to Bessel, that function theory should be carried out

-

when possible

-

in the complex

plane; sketched what the integral should be; and asserted that what we know as the "Cauchy Integral Theorem" held.

He went on

to mention [27, vo1.8, pp. 90-911 "This is a very beautiful theorem whose proof (not difficult) I shall give at a suitable opportunity.

It is connected with other beautiful truths touch-

ing on expansions in sums".

Gauss waited until 1832 [27, vo1.8,

102 ff.] to publish some of his results about complex function

theory.

6.11

Augustin-Louis Cauchy (1789-1857), one of the found-

ers of analytic function theory, apparently had his Integral Theorem in mind as early as 1814, when he read a paper on the subject to the Paris Academy; however his argument in its support did not appear in print until 1825.

(See [ll, pp. 33-37] for an

English translation of some of this very interesting early paper of Cauchy.)

His Integral Formula appeared in 1841 and his theory

of residues appeared in 1826.

(See e.g.,

111, pp. 31-441 for

exact references.) Cauchy's manuscripts are easily accessible and surprisingly 85

Norman L. Alling

86

contemporary in style.

The underlying topology of the complex

plane had yet to be worked out, and the idea of uniform continuity had not yet emerged. still shaky

The existence of the integral was

by present standards of rigor.

However, the

assertions and arguments are presented in a very plausible way. Clearly it can easily be brought up to present standards of rigor by supplying a supplementary argument here and there.

6.12

Joseph Liouville (1809-1882) stated, in 1844 (see

e.g., 111, p. 32 ff], that every doubly periodic analytic function is constant.

This theorem

-

of course

-

played an important

role subsequently in the theory of elliptic functions, as we will see.

Its generalization to bounded entire functions was

called "Liouville's Theorem'' by Jordan in his Cours d'analyse [38,

p. 3081. The idea of the periodparallelogram emerged in 1847 out

of the work of Liouville and others.

(See [25, pp. 232-2331

for historical details.)

6.13

It is not the purpose of this work to trace the

development of analytic function theory from its origins, in the work of Gauss and Cauchy, to maturity at the hands of e.g., Riemann and Weierstrass.

Suffice it to say that, during the

19'th Century, it developed into the powerful and magnificent theory we know today. We will develop

in this chapter some of the consequences

of Cauchy's theory of residues and of Liouville's Theorem as applied to elliptic functions.

However, first we will make some

remarks about lattices in the complex plane, using contemporary

The Introduction of Analytic Function Theory

87

terminology and methods.

6.20

& I

Lattices

By a lattice

@,

is meant a discrete subgroup L

(1:

that is a free Abelian group of rank

the additive group of C 2:

i.e., it is a free 2-module of rank 2 .

{ w l r o 2 ) of

L=

ZWl@ZW

Given a (free) basis

then

L

will be regarded as a basis of (2)

of

L.

Clearly

2 '

let Q denote the field of all rational numbers. Theorem.

A

necessary and sufficient condition for

to be a basis of some lattice (4)

(01p021

(5)

T :w2/w1

is not in

n=m=O,

Thus if

for

L

n L

[0,1)

is

R

satisfies

is dense in some line in

C

through

mE

2.

called the period parallelogram of Let

R

satisfies ( 4 ) and (5).

satisfies ( 4 ) and ( 5 ) .

It EIR: 0 < t < 11,

has a positive area.

nw1+mw2= 0

Assume that

and

:{xu~ + Y w Z :Xry

P(R) EP(w1,02)

(where

R

and

n.

is a lattice in C r

Conversely, assume that (6)

is that

Clearly ( 4 ) holds if and only if

( 4 ) but not (5); then

0.

in

is linearly independent over Q r

Proof.

implies

L

E

[ O r l ) I ,

as usual).

R.

Let

P(R)

will be

Note that by (5)

P(R)

88

Norman L. A l l i n g

-

-

P(R) z P ( w 1 , w 2 )

(7)

P(R) i s

then

(8)

[0,111;

E

a compact p a r a l l e l o g r a m a n d h a s v e r t i c e s a t

and

0,w1f~2f

{XUl +yw2: x , y

3

Clearly

w1+02.

(A+P(R))XEL

i s a p a r t i t i o n of

From t h e p i c t u r e i m p l i c i t i n ( 8 ) o n e sees

C.

immediately t h a t

i s a d i s c r e t e s u b g r o u p of

L

proving t h e

C,

theorem.

( 5 ), w i l l be d e n o t e d by

of

T,

R.

t h e q u o t i e n t of

~

(9)

(:) u

=

L.

R

and l e t

@,

B E GL2(Z);

( i i ) Given two b a s e s of

R'.

t h e e n t r i e s of GL2 ( Z )

Clearly C

t h e i d e n t i t y e l e m e n t of t h e s u b g r o u p of L" c L ' c L.

be a b a s i s

E

Since

R"

L.

L,

there exists

B E M

B E GL2(Z).

212

over

Clearly

then t h e r e

R' = B R . C g e n e r a t e d by

such t h a t

R"

Let

R;

thus

(Z)

IR, B

so t h a t R=CR';

R";

L" = L ,

CB=I.

thus

being be

L"

then

proving t h a t

R'

are i n

Similarly there Since

R=CBR.

Similarly

I

and l e t

CR',

R' = B R .

Since

L.

CB = I ,

( i i ) S i n c e t h e e n t r i e s of

C E M ~ , ~ ( Z )so t h a t C

=

is a basis

i s a s u b g r o u p of

GL2 ( Z )

GL2 ( Z ) .

CB = I ,

i s a b a s i s of

i s a b a s i s of

L'

R'

R',

and

g e n e r a t e d by t h e e n t r i e s of

C

R'

exists

R

L,

BRZ

b e t h e s u b g r o u p of

L'

there exists

then

such t h a t

B E GL2(Z)

(i) L e t

Proof.

E

be a l a t t i c e i n

(i) L e t

e x i s t s a unique

B

.r=q(~).

L. Theorem.

of

and w i l l be c a l l e d

Note t h a t

~where

L

Let

6.21

of

,

q(R),

BC=I;

i s u n i q u e l y d e t e r m i n e d by

R

R

hence and

R',

89

The Introduction of Analytic Function Theory proving the theorem.

R

Let

6.22

be a basis of a lattice

quotient (6.20) is

By (6.20:5),

T.

L

in C

Im(.r)# O .

R

whose

will be

called positive or negative according as Im(.r) > 0 or Clearly if fi( = (w1w2)t ) is negative then (ol - w2)' positive basis of B :

Let

(Ei: Eii) q(R');

T'

(2)

T' =

Since

E

L;

thus

GL2(Z)

L

R' EBR;

and let

. B

is non-singular,

It is easy to see that

Im(r') = (Im(-r)detB)/Ibll+b12?I 2

(3)

then

then

(6.20:s) , and since

b l l + b 1 2 ~# O .

is a

always has a positive basis. Let

(b21+ b22T)/ (bll+ b12T)

r LIR

Im(r) < 0.

.

Thus we have the following from Theorem 6.21. Theorem.

B E SL2(Z),

then

positive basis such that

6.23

Assume that

R'

R

is positive.

is a positive basis of

BR

of

L

Given

(1)

L.

(ii) Given a

then there exists a unique

B E SL2(Z)

R' = B R .

Two lattices

L

equivalent if there exists

and

L'

c1 E ~ * ( : Q : -

in c

{Ol)

will be said to be such that

L' =nL.

Clearly this is an equivalence relation between lattices in Let (1)

R

be a positive basis of Let

RT

1

5 (T)

L

and let

let E M ~ , ~ ( C and )

T :q ( R )

L :Z$rZ. T

.

C.

Norman L. Alling

90

L

and

LT

are equivalent, since

Let

6.24

WILT = L.

be a lattice in a: and consider the follow-

L

ing exact sequence in the category of Abelian groupsl where

x-

qyL:

(1)

R

O-M-X~O:

(i.e.l .9 X

a: onto X whose kernal is L).

is a homomorphism of

L

may be topologized by requiring that

be an open mapping

X havethe weakest topology making' L

and that

is easy to see that

continuous.

It

is a compact topological group, which is

X

homeomorphic to a torus. Let lent. (2)

and let

a EC*

Let

X'

f

C/L'

be exact.

Let

5

aL;

then

L'

and

clzl for all

f( z )

z

E

Clearly

C.

onto

X'

Proof.

flL

denotes L

onto

f

restricted to

L.

L'.

making ( 3 ) commutative (1.e. , such that Let

x

E

since (1) is exact. other preimage L.

@.

There exists a homeomorphic isomorphism g

Theorem.

E

is an

R

It is clearly an isomorphism of

h

f

O---,L---*C~X-+O

is, of course, row exact.

X

are equiva-

'-0

analytic autohomeomorphism of the additive group (3)

L

and let

R'

o--;rL'--,B:-x

L'

z1

X.

z

of

f(z1 ) = a z + c c h .

There exists

z EC

x

in

Since

Q:

is of the form

L' =aL,

f(zl)

gR = L'f)

and

x.

z+A, f(z)

.

R(z) =x,

such that

is not uniquely determined by

of

Any for some

differ

The Introduction of Analytic Function Theory L' ,

by an element in

is well defined.

g

hence

R ' (f (2,))

91

.

Thus

g

has

R ' (f ( 2 ) ) 5 g ( x )

=

The reader can easily check that

the requisite properties. As stated this theorem i s not very strong or in-

Remark.

teresting, for groups to

W/Z)

X

and

X'

are isomorphic

as topological

and hence isomorphic to each other.

@ CtR/Z)

The importance of this theorem emerges only when we define analytic structures for is also analytic. and

X

and

then

X';

it turns out

that

g

Analytic equivalence between the surfaces X

is much stronger than topological equivalence, as we

X'

will see.

F i e l d s o f elliptic functions ~-

6.30

Let

(1)

L

be a lattice in

f

F(L) !If:

a meromorphic function on C

f ( z + A ) =f(z),

f E F(L)

Let

C.

for all

z EC

and all

is said to be invariant under

respect

L.

Let

f?:

(w

w2)t

L

function, if

6.31

lytic; then Proof.

L

f

L.

f E F(L) f

w2.

is

may

L-elliptic function or merely an elliptic

is known to be the lattice in question.

is constant.

Let

continuous map of pact, f

L).

and

w1

Liouville's Theorem (1844). Let f

E

such that

or automorphic with

be a basis of

also called doubly periodic with periods also be called an

X

,

R

is bounded on

is bounded on C.

L

and note that

(6.20:7) into C .

P(Q)

-

P(Q).

be ana-

(See (6.12) for historical details.)

be a basis of

-

f E F(L)

Since

Since

-

P(f?)

u XEL ( X + P ( R ) )= C

f

is a is com(6.20:8),

By the classic theorem, usually called

92

Norman L. Alling

"Liouville's Theorem", f

is constant, proving the Theorem.

Since the reader is hopefully very familiar with the classic theorem called "Liouville's Theorem" we have used it to prove the original theorem of Liouville.

6.32

Liouville's theorem has other immediate important

consequences, when applied to elliptic functions. Let

Theorem.

f,g E F(L).

have the same poles on that each pole

of

has

f

corresponding pole of that

f

and

h = f/g.

Then

Theorem, h

6.40

-

on

P(R);

f-g

then

In case (i), let

Proof.

of

and

g

L,

and

the same principle parts as the then

g;

R

for some basis

f

is constant.

(ii) Assume

have the same zeros and same poles - each to

g

the same order

P(R),

(i) Assume that

h

E

F(L)

and

h

f/g

h

G

f

is constant.

- g.

In case (ii), let

is analytic.

By Liouville's

is constant.

Some

a p p l i c a t i o n s of C a u c h y ' s

and

Liouville's

work.

Cauchy's theory of residues can also be applied with considerable effect. of

(1)

L,

f E F(L) t R = (ol w 2 ) Let

and let

R

be a position basis

.

with

A (R) :8 (P(R))

thus it is

Let

,

be positively oriented;

[O,wll u [wl,wl + w21 u [wl + 02,w2] u [w,, 0 1 , oriented

by reading from left to right.

Since

f c F(L)

can have only a

finite number of poles on

F ( R ) , there exists z o

could be chosen to lie in

P(R)) such that

A(n)

f

+zo. Residue

T h e o r e m for e l l i p t i c f u n c t i o n s .

E

C

(which

has no poles on

The Introduction of Analytic Function Theory

(2)

JA(W

93

f (z)dz = 0; +zo

thus the sum of the residues of

f

inside

A(Q) + zo is zero.

Proof.

z +w f (z)dz = A(Q) + zo zO

fzo+w2 f

+w +w f(z)dz

f(z)dz+j20+w1

zO

f (z)dz

Z0+W1+W2

+

Since

! +!

2

f (z)dz. z +w 0 2

is doubly periodic of periods

w1

and

w2,

the right

hand side of (3) equals z +w

z +w f(z)dz+!

(4)

2O

zO

Z

f(z)dz+jZ0 f(z)dz+j f(z)dz, z0+w 1 z +w2 0

which clearly is zero, proving the theorem. Given a function f z

in CC

(5)

let

Resz(f)

(see e.g.,

meromorphic in some neighborhood of

denote the residue of

f

at

z

[69, p. 1111 for the definition); thus ( 2 ) is equiv-

alent to Resz (f) = 0. Note:

( 2 ) is dependent on

f

having no poles on

however the statement of (6) is not. holds for all

(7)

Let

S

zo

A (Q) + zo;

One easily sees that (6)

C.

be a set of coset representatives of

d L

in

then (8)

l z e s Resz(f) = O . Corollary.

Given

f c F(L) - @ ,

then

f

can not have a

@;

94

Norman L. Alling

single simple pole on

S;

thus the degree of poles of

f

on

is at least 2.

S

Note:

(9)

one can prove a version of Theorem 6.32 for

MZ

Let

6.41

S.

denote the set of all functions meromor0

phic in some neighborhood of extension of C .

For

in C.

zo

-

ft-M Z

Clearly it is a field

let

{O},

0

(1)

vz (f) be the unique n

f(z) (z - z ~ ) - is ~

such that

E Z

0

bounded and bounded away from zero in some deleted neighborhood of

v

in

zo

(f) is the order

of

C.

f

at

It is easy to see that

zo.

zO

for all

- {O}.

f , g M~Z

ation on

MZ 0

.

Thus

0

is a discrete rank one valu-

vz 0

(See e.q., [ 7 0 , v01.21 for a discussion of valu-

ation theory. ) Let

Theorem.

(4)

FZES

f E F(L)*;

vz(f) = 0 . Let

Proof.

(5)

ilD(f) :f1/2.rrif;

then

ilD(f),

and

vz(f)

then

the logarithmic derivative of is the residue of

kD(f)

at

f,

is in

z. Hence (4) follows

from (6.40:8). Corollary.

Let

f E F(L)

Let

-

C;

f E F(L) - C.

then

F(L);

For all

a

E

C

The Introduction of Analytic Function Theory

(7)

L

-

S

vz

(f)

N.

is in

(f)


Clearly it is independent of the choice of

will be called the order - _ of By Corollary 6.40 , (8)

95

f

ord (f)> 2.

S.

and will be denoted by

It

ord(f).

Clearly

ord(f) =ord(f - a ) =ord(l/f)

for all

a

C.

E

From these remarks we have proved the following.

6.42

Let

Theorem.

f EF(L - C

be of order

n.

For all

aeC L

vz (f - a) = n.

S

vz(f-a) > 0 If

fcC

6.43

R (:(wl f

we will define

Let

Lemma.

w2) t )

f E F(L) - C

f

and

bl,...,b n

0.

be of order

be a positive basis of

has no zeros or poles on

zeros of

ordf to be

A(Q) + z o .

L.

Let

Let

be the poles of

n.

Let such that

zo E C

all...,a n

f

in

be the

P(Q)+zO,

each appearing to its multiplicity: then (1)

- bj) E

r;=,(aj Proof.

L.

Consider the complex number

First let us see that sideration.

A(R) + z o I (3)

IJ

IJ

is related to the problem under con-

Since, by assumption, is well defined.

g ( z ) :zf' (z)/2iiif ( 2 )

f

has no zeros or poles on

The only possible poles of

Norman L. Alling

96

A(Q) + z o

inside

occur at the points

By Cauchy's residue theorem, 1-1

al,

A(R)+zO.

Let

inside

A(n) +

It is easy to see that

zo.

ResC ( g ) =cvc(f);

thus

(5)

P = Ij=l(aj n - bj).

Clearly

f'/2nif

is in

F(L)

.

To help simplify

To help to simplify 1 2 ,

13,

let

5 :z - w 2 ;

(
Let these integrals be called compute J1

where

let

and

Jl

let

f ( z 0 + w 2 t) :h(t) r(t)

G

then

Hence

5: z 0 +02t, for all no poles or zeros on A (n) + z o l (10)

respectively. Note

then

5E2-w1;

(8)

f

zo+w2

Let us call these integrals 11,...,14 that

g(z)

be one of the zeros or poles of

(4)

J

blI...,bn.

is the sum of residues of

inside

c

... ,an'

J2,

t

E

respectively. To [0,11. Since

f

has

r(t)e i0 (t)I

Ih(t) I > 0, for all

t r [0,1]; and

e(t) r I R

has

The Introduction of Analytic Function Theory been chosen to be continuous on (0,l).

97

[0,1] and differentiable on

Hence

W1 [logr(t) + ie (t)]1 (11) J1 = 2ni

0 -

,

f ( z 0 + 02)= f (2,)

Since some

me

J = m u E L . (Note: J1/ol is just a wind1 1 Similarly, J 2 E L; thus u E L, proving the Lemma.

Let

Theorem.

f E

F(L) - @

set of coset representatives of be the zeros of

j

f

in

R

Choose

(resp. bj)

a .+ A 1 1

and

- b i. - u i. )

j the Theorem.

6.50

Let

is in

Let

in C.

Let

S

be a

al,...,a

(a) = S

W E

\cc

and let

(60).

rrs

f(u+u)

i.e., that Let

as in the Lemma.

P(Q) + z o .

L,

n f

f

=

f(u),

in

L

For each so that

By the Lemma,

proving (12), and thus proving

Theta f u n c t i o n s t r e a t e d

with

analytic function theory

- -m < r <s ( + m .

Let

r < Im(u/w) .< s } .

srIs(w) :{ u E C :

"r,s such that

(3)

c/L

(resp. p j )

j

is in

This is just a strip in

(2)

n.

blr... ,bn be the poles of

zo

A

there is a

(resp. b . + u . ) 7 7

(aj+ A

(1)

and

S

be of order

each appearing to its multiplicity; then

S,

Proof.

a

for

Hence

2.

ing number.)

in

6 (1)- 6 ( 0 ) = 2rmr

r(1) = r(0) , and

Let

C.

f

Note that if

then

be an analytic function on

for all

U E S

for all

Srrs(o)

(w):

rIs

is periodic of period

h(5) E f(gw),

6 EC*r

5 E

w

on the strip

Srrs(l)i

Sr,s(~).

Norman L. Alling

98

then h

(4)

is periodic of period

Let

Clearly S-m

,

+m

g(5) :e

g

2~ic

,

on the strip

1

for all

5 EC.

is periodic of period

(1)). Given

1 on all of C

logw

WE^*,

is

number defined uniquely only modulo

el

0 2 0

2

, any

~

branch of

-

of course

-

(i.e., on a complex

2 ~ i , Further, given any

logw can be defined to be a single

- { peie: p

valued analytic function on C * 8.

SrlS(l).

>

01 ,

for any fixed

Let 2~ic 1;

(5)

w:g(c)(=e

(6)

then there exists n

Let

Z

such that

logw = 2nic + 2ain;

(logw)/2ni = 5 t n.

hence (7)

E

H(w) :h((logw)/2~i),

and note that, a priori,

H

may not be well defined.

is periodic of period

since h

h( ( logw)/2ni + n) = h ( (logw)/2ni)

However,

1, ;

thus

H

is a well defined

analytic function on the annulus (8)

where

A

so'ro

E

{w E C :

ro :e-2nr and

s o < IwI < ro),

so :e-2ns.

As such H

has a Laurent

expansion (9)

I;=-,

n anw , which converges to

uniformly on compact subsets of

AsOl'O are uniquely determined by H

(an)nc z determine

H.

H,

.

Note:

the coefficients

and, conversely, uniquely

Thus

the convergence being uniform on compact subsets of Finally,

Sr,s(l).

The Introduction of Analytic Function Theory

the convergence being uniform on compact subsets of The series (10) and (11) are the coefficients series.

-

of course

-

99

SrlS(w).

Fourier series and

(an)nEZ are the Fourier coefficients of these

As remarked above, they are unique.

6.51

Recall that 1

periodic of period pansion.

6

or

(v) is an entire function that is 2;

thus it has a Fourier series ex-

Indeed, as defined (5.20:4), e

its Fourier series expansion. of translating the argument by Theorem.

f(v+1) = f (v),

(2)

f(v+T)=kf(v),

then there exists

for all forall

a. E C

Since

f

is given by giving

(5.30:2) gives the effect on

8

j

T.

Given an entire function

(1)

Proof.

j

f

such that

V E C , and V E C , where

-1 - 2 k-h z

(5.30:l);

such that

is periodic of period

1

(l), it has a

In=-, m

ane2ninv ( 1 6 . 5 0 ) . m f (v)= ln=-manz2n. Let v1 E v +

Fourier series expansion z - eiav (5.20:3),

z1

E

Since T

and let

einvl; then z1 = zh, since h E ein T (5.20:2). Thus m h2nanz2n. kf (v) = ln=-mh-1an+lz2n. m Since ( 2 )

f(V+T)

=In=-,

holds , (4)

an+l=

h2n+la n’

for all

n

E

Z.

Using induction one may show that ( 3 ) holds, proving the Theorem.

Norman L. A l l i n g

100

Given a n e n t i r e f u n c t i o n

Theorem.

6.52

(1)

f ( v + l ) = f ( v )I

(2)

f (v +

T) =

there exists

for a l l

- k ( f (v)),

E

such t h a t

and

VEC,

for a l l v

f

C;

then

such t h a t

a. E C

L

(3)

f ( v ) = a o ~ ~ = - m ( - l ) n zh 2n n = a0 6 0 ( v ) . The p r o o f i s v i r t u a l l y t h e same a s t h a t o f Theorem 6 . 5 1 .

,

(1)

f ( v + l )=-f(v)

(2)

f ( v + T )= k f ( v )

cEC

there exists

for a l l

V E C ~ and

for a l l

VEC;

such t h a t

2 (1)I t h u s i t h a s a nliiv= m n f ( v ) = cn=-mane Cn,-,a z Since m

F o u r i e r series e x p a n s i o n

n

(1) h o l d s ,

(1+ (-l)n)an=O

all n

Hence

Let m

Cn=-m

(5)

(6)

E

Z.

,

V ~ - V + ' I ,

h 2 n - l b 22n-1 n

bn+l = h Then

theorem.

2n

then

is periodic of period

f

Proof.

be an e n t i r e f u n c t i o n such t h a t

f

Let

Theorem.

6.53

bn

.

Since

for all

z1 = hz

then k - h

for all

b n = h ((2 n -1 )/2 )' c

n

-1 -2

E

z

thus

nE Z ;

I

.

a2n=0,

and t h u s

for

f (v+T ) =

m -1 2n-1 k f ( v ) = l n = - m h bn+lz

Z.

f o r some

c

E

cI

proving t h e

The Introduction of Analytic Function Theory

6.54

Let

Theorem.

(1)

f ( v + 1) =-f(v),

(2)

f (v+T)

there exists

=

-kf ( v ) ,

cEC

f

be an entire function such that

for all for all

C,

V E

v

E

C;

and then

such that

The proof is only a slight variation of that of Theorem 6.53.

101

This Page Intentionally Left Blank

CHAPTER 7

WEIERSTRASS'S WORK ON ELLIPTIC FUNCTIONS

7.10

Introduction

Karl Weierstrass (1815-1897) worked on elliptic functions during much of his mathematical career, at least from 1840 into the 1890's.

He was a student of Christof Gudermann (1798-1852),

who was a student of Jacobi. Although Gudermann suggested a research problem to him, Weierstrass left Munster without receiving a doctors degree.

Instead he received a certificate that

allowed him to teach in secondary schools. 1856, when he went to Berlin.

This he did until

Although many of his early papers

- unpublished until his collected works began to appear at the very end of his life

-

dealt

with elliptic functions, much of

his work on the subject is contained in Volume 5 of his collected works under the title Vorlesungen uber die Theorie d s elliptischen Functionen, which we will frequently refer to in this monograph as his Vorlesungen. his Berlin lectures of 1863.

.. .

Much of it comes from

As explained, lectures from other

years found their way into his Vorlesungen 1864-65 and 1874-75.

..., e.g.,

the years

In general it is difficult to date many of

Weierstrass's discoveries since

-

in many cases - publication

was so long delayed. Weierstrass's great importance comes not only from his discoveries but from his methods. 103

For example, many of his

104

Norman L. Alling

results appeared in the work of Eisenstein (1823-18521, as Professor Andre Weil pointed out in his review [66] of Elsenstein's collected works [ZO]. book [65]).

(See also Weil's interesting

Weierstrass uses power series, infinite products,

and analytic continuation very skillfully, and very much as we do today.

We are treated to many proofs in Weierstrass's works

that meet currect standards of rigor.

This is not

-

perhaps -

surprising, for Weierstrass helped to shape these standards. For biographic details see e.g.,Klein's

...

...

([49, pp. 276-2951), and Behnke and Kopferman Festschrift

7.20

Weierstrass's V o r l e s u n g e n . .

Weierstrass's Vorlesungen

...

...

Entwicklung

[9].

.

is an imposing work which

comprises 34 chapters and runs to 327 pages in length in [64, In Chapter 1 he deals with the transformation of

vol.V]. (1) where

dx/P(x)Si, P(x)

E

c[x]

is admissible (3.10), by means of a linear

fractional transformation, f ( z )

=

h(M) ( 2 )

(3.13) , much as Euler

did [22, vol.XX, pp. 302-3071. He shows that (1) equals ds/W(g2/g3) 4 ( s ) I

is admissible. Weierstrass's that

P(x)

where

(This polynomial will be said to be in

form.)

To see that this can be done, assume first

is of degree

n=4.

Let

p

be a root of

P(x)

c and let

then

M

is in

SL2(C).

Let

y2

f

P(x),

and let

2,

9,

and

in

105

Weierstrass's Work on Elliptic Functions

g(k)

be defined as in (3.15:3) and (3.15:4); then, by Lemma

p ( % ) is admissible. Further P ( % ) is of degree 3 and

3.15,

p2 = ? ( % ) . n=3.

then

Thus, without loss of generality, we may assume that

Now let

M

E

SL2 (C),

coefficient of

$(%)

%2

is admissible of degree

in it is zero.

ality, we may assume that

then

M E SL2 ( C ) ,

and

C=

-2

and the

Thus, without loss of gener-

Now let

0.

(2).

y =W

3,

Thus

'2 "3

(6)

dx/P (x)% = ds/W

(s)

%

'2"3 can be effected by a linear fractional change of variable, from

x

to

s.

In the third Chapter of his Vorlesungen

7.21

...

Weierstrass introduces his famous ?-function as the solution of the differential equation (1)

(Z)

L

3 =4s - g 2 s - g 3 ,

subject to the initial condition (2)

P(U) =

+ c 0 + c 1u 2 + c 2 u4 + 2 U

... .

He then gives several variants of an addition theorem for

'p,

and gives some information on the coefficients of (2): namely (3)

c o = 0,

cl=g2/20,

and

c2=g3/28.

In Chapter 2 , Weierstrass gives an argument, using power

Norman L. Alling

106

series, to establish the existence of a solution of ( 1 ) satisfying (2). In the fourth chapter, devoted to the sigma function

u (u),

he defines it by means of 2

(4)

10gO(u) 2

P(u) = -

such that

I

du

At the end of the chapter he gives the familiar version of the Weierstrass Addition Theorem 'p(u+v)= -I 4

(6)

(W) -

Nu)

- P(V).

In the fifth chapter he also introduces the roots and e3

(x) :

W

of

elIe2'

thus

'2Ig3

(7)

-

g2'g3

(8)

-

(x) = 4 (x el) (x - e2) (x e3) , and hence

w

e1 + e2 + e3 = 0, e2e3 + e3e 1 + ele2 = -g2/4,

and

ele2e3 = g3/4.

It is not until Chapter 14 [64, vol.V, p . 1201 that Weierstrass asserts that

(9)

where L

a

( u ) = un; ((1 - u/R)exp(u/e + u2/2R 2)

R

of 'p.

(11)

1;

)

,

and

ranges over non-zero elements of the period lattice This follows a discussion of the fact that l/t3

is absolutely convergent. Further, Weierstrass shows [64, vol.V, p. 1211 that (12) g2 = 6 0

1;

l/R 4 ,

and

g3 =140

1;

1/11 6

.

Weierstrass's Work on Elliptic Functions

7.22

There are

-

-

of course

Weierstrass's Vorlesungen

... .

107

a vast number of results in

When the author first read it

he was very surprised to find that the product expansion of u(u)

(7.21:9) and the series expansion of

were not used to define occurred

-

u(u)

and

(7.21:lO)

respectively, but

for the first time - at least a third of the way

through Weierstrass's Vorlesungen

7.30

'p(u)

'p(u)

Weierstrass's

... .

Theory Reordered

In 1893 the first volume of Tannery and Molk's four volume work, h6ments de la th6orie des fonctions elliptiques [601 began to appear in Paris. Weierstrass ' s functions

In it the authors consider ~ ( u ) and

u (u)

among others.

than proceeding as Weierstrass did in his Vorlesungen defined

u(u)

I

they

from its product formula (7.21:9) and derived the

Weierstrass zeta function

(2)

...

Rather

5 (u),

d2 p ( u ) = -5' ( u ) = - logo (u)1 du2

and his ?-function as follows:

[601 ~01.1,pp. 153-1561.

This is, of course, pretty much the way most 20'th century texts present Weierstrass's functions. Many excellent texts were written after Weierstrass began giving his lectures on elliptic functions in Berlin in 1857. (Note: Chapter 12 of [64, vol.V] consists of notes taken in 1863.)

Briot and Bouquet wrote Th6orie

[12] c. 1859.

des

fonctions elliptiques

The second edition is dated 1875.

In it they

mention Weierstrass's functions Al, All, AlZI and

A13

(named

in honor of Abel) [12, p. 519 ff.]; however,they do not seem to

Norman L. Alling

108

p

mention Weierstrass' functions

and

G.-H. Halphen pub-

u.

lished his 3 volume Trait6 des fonctions elliptiques et de leurs applications [31], in 1886, 1888, and 1891. Weierstrass's functions r ) , ~ , and

5

at length; however

is defined in terms of Jacobi's function p. 241.

p(u) , and

p(u)

[31, vol.1, p. 1681.

sn(u)

r)(u)

[31, vol.1,

~ ( u ) is defined [31, vol.1,

Weierstrass's function

p . 1351 in terms of

Halphen discusses

u(u)

is defined in terms of

Thinking that the Franco-Prussian

war might have inhibited the flow of mathematical ideas between Germany and France, the author saught a German text of this era, in the Rush Rhees Library at the University of Rochester, and found Elliptische Functionen

und

algebraische Zahlen [62] by

H. Weber of Marburg, which was published in 1891. is quite unlike Tannery and Molk's.

Weber's work

Thus, it may be that Tannery

and Molk introduced the method of defining the Weierstrass functions by defining

u(u)

by an infinite product. Whittaker and

Watson [69, p. 4341 define sion.

first by the usual series expan-

Of course, the facts about convergence one needs for this

reordering of the treatment of Weierstrass's functions are to be found in Weierstrass's work.

Following Whittaker and Watson we

will begin our development by considering (3)

1;

l/L3,

7.31

[64,

vo1.5, p. 1161.

Theorem ( W e i e r s t r a s s ) .

and let L* :L- { O l ;

Let

L

be a lattice in

then

We will establish this by proving that, for real

a,

c

Weierstrass's Work on Elliptic Functions

(2)

converges if and only if

1/1 I l l a ,

Proof (of (2)).

Choose a basis

each

kEN,

(3)

Lk(Q) : {mul + nu2:

has

8k

Let

h

> 2.

2)t

w

of

L.

For

let

In1 L k ,

or

n=+k

(Lk( n ) IkEN

Note that

R: ( w 1

CL

109

m,n

E

and

Z

and either

and

Iml (kl.

is a partition of

points in it.

m=+k

L*,

and that

Lk(Q)

Let

be the smallest of the two altitudes of the period

-

parallelogram P ( f i ) , diagonals of (5)

k

(6)

8 -

E

Lk(Q) *

and let

P(Q).

H

be the largest of the two

One easily sees that

implies, h k z

1 -

8

lkl

(Hk;

thus

1

Ha

Since

lk=l l/k"-'

converges if and only if

a - 1 > 1 , (2) is

established,proving the Theorem. Apparently by 1736 Euler knew that

([43, pp. 237-2381).

Since

k=m

1/k2 >

1-

From this and (6) we obtain

dx/x 2 21/3m 3 ,

m

obtain m

(')

lk=m

0) > 1/3m3.

Note also that

for each

m

E

N

we also

Norman L. Alling

110

L k ( ~ ) has

(10): : : u

Let

7.32

(1)

zo

Br(zo) : I z

E

E

C:

4m(m-l)

points in it.

r > 0.

C and let Iz

- zoI

<

Let

r

This set will be called the open -ball about

Iz

Clearly its closure ==

B,o

E

c:

Iz

zo

- z 0 I -< r).

of radius

r.

Further,

is always compact. For all

Lemma.

1ilcL* I z/il I

r > 0,

is uniformly conver-

B,o.

gent on

Using this Lemma we know, from the classical theory of Weierstrass products (see e.g., [60, vol.1, p . 114 ffl), that

defines an entire function having as zeros exactly the set

L,

each zero being simple. As remarked before, Tannery and Molk [60] take (2) to be the definition of

u

[60, vol.1, p. 1551;

whereas Weierstrass gave (2) [64, vol.V, p. 1201 as an expansion of a known function, a . Since (3

u

(-2) =

il

E

-u

L*

-+

-il E

L*

is a bijection,

(2):

.,

u

is an odd function.

Since

u

is entire and is non-constant, it cannot be doubly

i.

periodic. We want to define Weierstrass’s ?-function as follows.

To make sure that the expression on the right of (4) defines a

111

Weierstrass's Work on Elliptic Functions meromorphic function, let

C-L,

and let

-

of course

a.

E

K

be a non-empty compact subset of

r: supZEK/zI. Let

L'

is finite.

Let

L'

L" 5 L

E

I.L E L: ILI 22r3; then

- L'.

For

z

E

Br(0)

-

and

L",

1

1

1

(5)

1 2- 7 1

=

(2Q-z)z < 3 1 ~ / r - 12r (z-L)2Y21 - (le1/2)21a.12 -

*

Using (7.31:l), together with (5), we see that the right hand side of ( 4 ) converges absolutely uniformly on

K;

thus 'p

having as its set of poles

a meromorphic function on C ,

is L,

each of these poles being of multiplicity 2 and of residue 0. It is clear that (6)

p(-z) = p ( z )

thus

P

zrC;

is an even function.

directly that 'p take

for all

mEL

It is tempting to try to prove

is invariant under

L.

To do this one can

and note that

Rearranging the right hand side of (7) to equal the right side of ( 4 ) would seem to be blocked by the fact that

CLEL*1/E l L The standard way around this diff culty

is divergent (7.31:2).

PI,

is to consider

and to note that

and that the expansion on the right converges absolutely uniformly on

K.

'p'

is

-

of course - meromorphic on

set of poles the set

L,

Clearly

Further, one easily sees that

9' ( z - m )

=

P'

(z),

has as its

each of these poles being of multiplic-

ity 3, and of residue zero.

(9)

@,

'p'

is an odd function.

Norman L. Allins

112

L E L

since

is a bijection, and since the risht hand

L

I l + m E

+

side of ( 8 ) is absolutelv converqent on F(L) - C

'p

E

Let

Proof.

F

(L) -C.

R

( E (wl

j = 1 and 2 .

Let

g(z t

f .( z )

regular at

be a b a s i s of

- ?(z)

for all

f! ( z ) 3

for each thus

L:

Since

j.

and

.p(ztwj)

is a basis

Q

are

'p(z)

Since

-'p(-wj/2).

I

and each

z

and for

z EC,

j i 3 is an even function (6), this number is zero: thus 3

Clearly it

and note that

c = f ( - w./2) ='p(wj/2)

w./2.

L.

3

I

is not in

7

t)

0.)

3 f ( z ) = c . E C, 7 3

is zero: thus w./2

w2)

p ( z + w j ) = p ( z ) for all

suffices to show that

Lr

is in

'p'

(6.30:l).

Theorem.

Of

Thus

K.

.p

f (z) =0 I

7

j, Provins the theorem.

Thus we see that 'p

and

are in

'p'

F(L).

The alqebraic

relation between these functions is of qreat importance to us, and will be established in the next section.

pole in

R :(wl

Let

7.33

at

p(R),

be a basis of

w2)t

0,

0'

The zeros of and

(w,+021/2,

(1)

Let

(w1/2,

Proof.

Let

,

on

E

countinq multiplication. P(Q)

are at

w1/2,

be called the half-periods

(w,+w2)/2,w2/2)

R. be a half-period; then

is an odd function

9' ( - E ) = - P I ( € 1 3

'p'

P(Q),

Each of these zeros is simple.

w2/2.

associated with

Since 'p'

has, bv the same argument, order

has 3 zeros on

Lemma.

has onlv one

and that Dole is of multiplicitv 2; thus

7, is of order 2 (6.41). 'p' 3; hence

L. 'p

p'

(-E)

proving that 9'( E )

it has at most 3 zeros on

P(R).

=-TI = 0.

and

E{L

(E).

'pl

(E)

Since '0'

2~

EL.

= ' p ' ( E - 2 ~ )=

is of order

Since we have found all

three, each must be simple, provinq the lemma.

Weierstrass's Work on Flliptic Functions

113

It is sometimes convenient to define

w - w1/2

(2)

w' !w2/2;

and

Clearly the set of zeros of Let

?\(w1/2)

E

p'

is the set

p( (wl + w 2 ) / 2 )

ell

= (w

w') t ,

and it is a

1 zL.

basis of the lattice

(3)

71 Q

then

Note that the indexing in (3) is not

E

-

e2,

1

L.

zL-

and

p(w2/2)

unfortunately

-

It is in conformity with [ 3 6 ] , but not with all texts.

E

standard. One

justification for this indexing is that with the boundary -

of

e3.

A(R)

positively oriented,the half-periods occur in the order

P(Q)

given in (1) and that the order is compatible with (3), provided

R

is positive.

Of greater importance is that the indexing of

R

(3) is dependent on the choice of basis (4)

The numbers

Indeed,

p'

el,e2, and

is zero at

p(z-w1/2) - e l the value

el

w1/2,

e3

not assume the value

el

01/2.

L.

are distinct.

the zero being simple: thus

has a double zero at doubly at

of

w1/2.

Since

again on

p

P(Q).

That is,

p

assumes

is of order 2 it canA similar statement

holds for the other half periods, estahlishing ( 4 ) . Let (5)

ncN,

sn(L)

with

zPEL*P-"

nZ3,

and let

.

By (7.31:2), the series on the riqht in ( 5 ) is absolutely convergent: thus

s,(L)

complex number.

(or

sn

for short) is a well defined

The right hand side of (5) is known as an

Eisenstein series. (6)

s n = 0,

Indeed, let

CREL*

for all odd n

n L 3 , and let it be odd. - S n = -n thus sn = 0. since -I,* = L*; = s n' (-a)-" = zmEL*m E

N,

n.

114

Norman L. Allinq Let

r:q.l.b.{ILI

REL*);

then

Consider now the Laurent expansion of and let

Q E

L*;

then

lz/.Q

r>O. 'p

Let

about

U-Br(0).

0.

Let

z

E

U

< 1 ; hence

As a conseauence, on rearranqinq we find that

(See e.q., the main rearranqement theorem in [43, PD. 143-1441 Usinq (6) we see that (10) simplifies to

for details.)

the converqence beinq absolutely uniform on compact subsets

u-

(01. Theorem.

(12)

('PI

(z,L))2=4'p3 (Z,L) - 60s4 (L)'p(z,L) - 140s6 (L),

z

\c.

E

Proof.

(13)

'p(z) = z

(14) 'p' (15)

(2)

We have seen (11) that on -2

+3s z

=

4

4 +5s6z +

... .

+ 6s4z + 20s6z3 +

['p' (z)I2 = 4z-6

(16) 4'p3 ( z ) = 4z-6

2

U-

fo)

Thus

... ,

- 24s4z-2 - 80s6 + . . .

+ 36s4z-2 + 6 0 s 6 +

. ..;

hence

for all

W e i e r s t r a s s I s Work on E l l i D t i c F u n c t i o n s

(17)

115

[ p ' ( 2 ) ] - 4p3 ( 2 ) + 6 0 s 4 p ( z ) = - 1 4 0 s 6 + . . . .

Since both s i d e s of

( 1 7 ) a r e e l l i p t i c f u n c t i o n s and s i n c e t h e

f u n c t i o n on t h e r i g h t h a s no p o l e s , w e mav a p p l y L i o u v i l l e ' s Theorem ( 6 . 3 1 ) and c o n c l u d e t h a t t h e r i g h t hand s i d - e o f constant: thus it i s

provinq ( 1 2 ) ,

-140s6,

(17) is

and h e n c e t h e

I t is convenient t o d e f i n e

Theorem.

(18) g 2 ( L )

t o be

and

60s4(L),

q3(L)

t o be

140s6(L);

From t h e Lemma a t t h e b e g i n n i n g of t h i s s e c t i o n , and u s i n g

( 3 1 , w e know t h a t (20)

[P

I

(2)

I

= 4

CP ( 2 ) - e l ) CP( 2 ) - e 2 ) CP( 2 ) - e 3 ) .

On m u l t i p l y i n g o u t t h e r i g h t hand s i d e o f

(201, a n d u s i n g

(20)

and (191, w e see t h a t

(21)

e l + e 2 + e 3 = 0 , e l e 2 + e2 e 3 + e 3 e l = - g 2 / 4 ,

and

ele2e3 = g 3 / 4 . Further, the discriminant, s i d e of (22)

g

o f t h e c u b i c o n t h e r i g h t hand

A,

( 1 9 ) and ( 2 0 ) c a n h e g i v e n a s follows: 3 2

-

2 27g3 = A = 1 6 (e l

(See e . g . ,

By ( 4 1 1

- e 2 )2 ( e 2 - e,)

[ 3 6 , p.

2

(e,

- el) 2 .

1681.)

A # 0.

W e may o b t a i n more i n f o r m a t i o n a b o u t t h e c o e f f i c i e n t s o f t h e Laurent expansion of follows. (23)

let then

'p

about

(ll), by p r o c e e d i n g a s

0,

First t o reduce t h e n o t a t i o n a l burden s l i g h t l y , b ( n ) E ( 2 n + 1)s b(1) Eg2/20

~ and

,

~f o r + n

E

~ N;

b ( 2 ) =g3/28.

Norman L . Alling

116

(24) Let

T(z) =

b ( - l ) 5 1;

and let

h(0) - 0

In--- lb(n)z2n, m

on

then

TJ- {O}.

Differentiating (19), with respect to

z

gives

~ P ’ ( z ) ’ ~ ” (=12p z ) 2 ( z ) ~ ’ ( z-g2P’(z); ) thus

From (24) we obtain m (26) $ o ” ( z ) = ln,-12n(2nl)b(n)z 2n-2 ,

on

IJ- {O}.

The left hand side of (25) is then (27) 6z-4 + 12b(l) + r:=22n(2n-l)h(n)z

2n-2

.

The right hand side of (25) is

(29) thus, For

(29)gives

n = 2,

Note that

n(2n-l)h(n)

(30) b(n)

=

31y=-1b(j)b(n-1-j),

6b(2) = 6h(2),

1123,

3(l~~:b(j)b(n-l-j))/(n-2) (2n+3).

b(n) ,

n 2 3.

for each

expression concerning the (31)

~

for

m24.

Let

L

and

L’

g . (1,) = g . ( L ’ ) ,

3 I Proposition.

b(2),

s

(2k-1)(2m-2k-1)~

j= 2

Assume that

and

Converting (30) back into an

be lattices in for

b(1)

~ gives ~ ‘

(2m + 1) (m-3)(2m-1)s~~ = 31:1;

implies

n22.

and thus is uninteresting.

Clearly (30) is an algorithm which, given determines

for each

+ b(O)h(n-l) + b(n-l)h(O) + b(n)b(-1) = 2b(n) ;

b(-l)b(n)

thus for each

=

and

g .

3

Clearly

@.

~

L=L’

3.

(L) = g .( L ’ ) 7

for

j=2

,

~

Weierstrass's Work on Elliptic Functions and 3; then

117

L=L'. Using (30) we see that the Laurent expansion of

Proof.

'p(z,L) and of

at

T(z,L')

0

are identical. Hence

rp(z,L) =

'p(z,L'). As a consequence their period lattices are equal, proving the proposition.

7.34

Weierstrass introduced the function, that was later

known as the Weierstrass

zeta function

<(z)

as follows:

(Halphen [31, ~01.1, p. 134 ff.] refers to this function as

< (z),

c. 1886.)

Weierstrass asserts that

[64, vol.V, p. 1201.

To see that the series on the right hand

side of (2) does indeed converge, let subset of

C-L.

Let

K

be a non-empty compact

r-sup{lz/: Z E K ) ;

then

r > O . Let

and let

Since

converges (7.31:l) , we see that the series

lREL* I R

on the right hand side of (2) converges absolutely uniformly on K;

thus it can be used to define

meromorphic function on Q: at the points of

L.

P(z)

=-

d <(z)

.

Clearly

5(z)

is a

whose poles are all simple, being

Further from (2) we can conclude that

~ ( z ) is an odd function.

(4)

5(z)

Differentiating (2) gives

164, V O ~ . V ,p. 321.

Norman L. Alling

118

One can also prove easily that (5)

(z)/o(z) = 5(z).

0'

R

be a fixed positive basis of

< (z)

has exactly one pole on

Let see that

R;

parallelogram of

< (z)

is not in

(6)

R E L,

Let

m

and

2

P ( Q ),

From ( 2 ) we the period

that pole being a simple one at

F(L)

0.

Thus

(Corollary 6.40).

fL(z) :< ( z + k ) -

let

L.

be in

L;

then

<

(z).

c ~ + <~( z=+ I ! + m ) - < ( z ) =

< ( z + R + m ) - < ( z t m ) + < ( z + m ) - < ( z ) = c R + c m . Hence

C: L

(8)

is a homomorphism of

L+cR

E

L

into the additive

group of @ . (9)

Let

cw

5 ri

1

Clearly

{cR: il

{mnl+nn2: 5

E

j'

L},

m , n E 2).

would be in

for

j = 1 and 2.

the range of the homomorphism I.e., it is

Tlj

c(L) = I 0 1

then

is

qj

= 25 ( O j / 2 ) .

Indeed, 5

Where

is

Since this is not the case, c(L) # {Ol.

F(L).

Another convenient expression for (10)

c(L).

c,

n

j

:<(-w./2+w.)

-<(-w./2) 1

3

1

=<(w./2) 3

-<(-wj/2).

Since

is an odd function, (10) is established. We now come to an equation usually called Legendre's

equation. (11) qlw2

- q2w1

Proof.

A(Q)

+zo

=

2ni.

Choose

(A(G)

being

Z

~

E

@

so that

a P ( R ) ) . Then

<

has no poles on has one pole inside

Weierstrass's Work on Elliptic Functions

A(Q)

+ zol

119

and that pole is a simple one with residue 1; thus z +w

5 (z)dz=

(12) 2 ~ i =

!

'L(z)dz+

zO

Izo+w2

5 (z)dz+

z +w1+w2 0

1

z +w

jzo

i

z +w +w

2<(z)dz+ zo+wl

i (z)dz =

z +w 0 2

1

(r(z+w1)-

zO

c

dz -

zO

n19-

r\*y

c(z))dz-

zO

- C(z))dz= /zo+u2

'(r,(z+w,)

z +w 0 2

z +o 'c

dz= w2

zO

I

proving (11). Let

Theorem.

f

be a meromorphic function on C

as its set of poles the set

L,

each pole being simple.

having Assume

also that (13) f(z+R) =f(z) + c R , for each Limz+o(f (z)- l/z)

(14) and that then

ZEC

and

R E L ~

= 0;

f = 5. Proof.

Let

(13), g E F(L).

g(z) :f(z)-<(z),

Since

g

ZE@.

is analytic on all of C

Liouville's Theorem and conclude that k = 0,

for all

g(z) = k c C .

By (7) and we may apply By (14)I

proving the theorem.

7.35

(7.34:5),

How is

u(z+R)

related to

a(z),

for

R E L?

By

o'(z)/u(z) = ~ ( z ) . Using this and (7.34:7) we see

that U l (z+R)60-m+cR

(1) Let

F(C)

Given

UI

(z) I

for

Z E C and

EEL.

denote the field of all meromorphic functions on @ .

f E F(C)*

(-F(C) -

{o})

let

Norman L. Alling

120

(2)

6(f) :f'/f; and

6(f) E F(c), all

then 6(f) = O

if and only if

fE C .

Further, for

f,g E F(C)*,

(3)

6(fg) = 6(f) + 6 ( g ) ;

thus

b is a homomorphism of the multiplicative group of F (c) * the additive group of

6( u ( z ) ) + 6 (exp( cRz))

z E C ,

that has kernel

@*.

Using (1)

E ( a ( z + L ) )=D(o(z)) + c R =

and (2) we see that

for all

F(C),

into

=

(exp( cRz)u ( z ) )

R E L.

and for each

;

hence

Let

k and

R

be in

L.

Then

.

= AkAQexp(ckR)

Clearly ( 4 ) gives (6)

A o = 1.

Using induction and (5) we obtain (7)

n Ant =ARexp(cRLn(n-1)/2) I

for all

nEN.

for all

n

E

Using (6) we see that A_& = AQlexp (c,R)

Hence we see that (7) holds for all

For all m

and

n

in

2

then

k 1? (~~hm(rn-1) / 2 + cRRn(n-l)/ 2 + mnckR) Amk+nR = AmAnexp Since

thus

N, A-nn.=An(-R) = AnRexp (c-R-in(n-1)/2) =

Ainexp(cRa(-n) (-n-1)/2). n E 2.

;

-

for all

k,R E L,

( 5 ) gives us

.

Weierstrass's Work on Elliptic Functions

CkR

(9)

- C,k

121

2.iriz.

E

Recall (7.34:9) that

c w , is defined to be

rl

;

thus

7

(9) is a generalization of Legendre's equation (7.34:11). (10)

, for each j.

:Aw

cij

Let

j 0

Let

z=-w./2; then 7

a

Since

c1

j

=

(-w,/2 J

+ w,)

J

exp(nj (-wj/2) )u(-wj/2)

is an odd function (7.32:3), we obtain

(11) a. = -exp(q .w./2). j 3 7

Let

m

and

(12) u(z

n

+ R)

=

again be in

2

and let

m n exp(q w m(m- 1)/2 1 2 1 1

ci ci

R - m u +nu2;

then

1

+ n2w2n(n - 1)/2

+ rnnnlu2 + (mnl+ nn2)z)a (z) =

(-l)m+nexp(m2n1w1/2

+ n2n2w2/2

+mnq 1o2 + (mnl+nn2)z)u(z).

By Lagrange's equation (7.34:11) , (13)

U(Z

+ a)

(14) Let Clearly if

E(,) E

R E 2L.

= (-1)m+n+mnexp (c, ( z

qlw2

+ 11/21

-

n2(.!jl =

27i;

thus

.

u (x)

m+n+mn

: (-1)

maps

L

into

To see that

E

the multiplicative group j = 1 , 2 , and that

{+1}.

Further, € ( a ) = 1

is not a homomorphism of

{kl}, note that

E(w.)

3

if and only L

onto

= - 1 for

E ( w ~ + w ~=-1. )

Bibliographic note.

the transformations of

a

While the approach of dealing with under

L

at first without using a

basis may be novel, many of the results presented in this section are well known:

e.g., (13)-(14) appear in 136, p. 1811.

122

Norman L. Alling

Theorem.

7.36

Let

f

be an entire function such that

the following hold: (1)

f(z+R) = ~ ( R ) e x p ( c ~ ( z + R / 2 ) ) f ( x ) ,

for all

zE C ,

and each

(2)

Limz,of ( z ) / z = 1;

then

€ = u.

Let

Proof.

(7.35:13),

g

R E L;

and

g(z) -f(z)/u(z),

for all

is doubly periodic.

Liouville's Theorem

g(z)

By

ZE@.

By (2) g

is analytic.

is a constant function

c.

By

By (2)

c = l , proving the theorem.

7.40

Representation o f doubly periodic f u n c t i o n s

In sections 7.40, 7.41, 7.42, and 7.43 we will show how 'p, u ,

and Let

5

can be used to represent elements of

f E F(L) - C

(6.30:1), and let

By Corollary 6.40, n 2 2 . in

P(Q)

Let

al,

n=ord(f)

...,a n

and let bl,... ,bn be the poles of

1.n3 = 1 (a.-b.) E R 3 3

is in

all to

a complete set of zeros and poles of

(2)

f

in

P(Q) ,

By Theorem 6.43,

al- R

we can obtain

f, mod L, which we will

Ial,...,an, bl,...,bn3,

such that

1.n3 = 1 (a.-b.) =O. 3 3

(Note: this set of zeros and poles need not be a subset of Now consider

P ( n ) .)

(3)

g(')

f

L.

By changing one of the zeros, say

again denote by

(6.41:7).

be the zeros of

each appearing according to its multiplicity. (1)

F(L).

u ( z-al) . . . . * u ( z-an) ' u(z-bl). ... .o(z-b,)

I

for a l l

z EC.

Weierstrass's Work on Elliptic Functions

Let

R EL

and recall ( 7 . 3 5 : 4 )

123

u ( z + R) =ARexp(cRz)u(z);

that

thus g(z+ a) =g(z)exp(c~l~=l((z-a.) - (z-b.1))= g ( z ) ,

(4)

7

7

proving that

g

E

F(L).

Clearly

f/g

lytic; thus it is a non-zero constant

is in c,

F(L)

and is ana-

and we have proved

the following 0

=

Theorem.

7.41

(z-a,)

. . . - 0 (z-an)

*

.. .

u (z-bl)

-0

(2-bn)

We can extract more from .this line of argument,

namely the following. Let

Theorem.

.

bl,.. ,bn

(1)

each

be in c

and let

all...,an and

such that

a . $ bk(modL) ,

and

7

- b . )= 0.

(2) Let

n c N, n 2 2 ,

3

cEC*

and let

....u(z-an) . . (z-bn )

(3)

a(z-al). f(z) :c o(z-bl) * .

Then

f

is in

F(L) I

and its poles at

-0

f

bl+L,

*

has its zeros at

...,b n + L ,

al+L,

...,a n + L ,

each according to its mul-

tiplicity. (Note: Were we to try to state this theorem when then (2) would imply

7.42

(1)

Let

f E F(L)

a(f)(z) :f(-z),

Clearly

a(f)

E

al = bll which would violate (11.)

F(L) ,

and let

for a l l

z c @ .

and clearly

n=l

124

Norman L. Alling

(2)

each point of C (3)

F(L) ,

is an automorphism of

a

Let

Fe(L)

{f

of period 2, that leaves

fixed. F(L): a(f) =f).

E

This then is the subfield of all even functions of (7.32:6)

= Fe(L). f E F(L)

Note that

is odd if and only if

is odd (7.32) , we see that F(L). (5)

For any

(6)

= fa(f),

the

of

Since

7'

is a proper subfield of

f,

and

f, are in

Fe(L).

f = (f+a(f))/2+ (f-a(f))/2-fe+fo f

into the sum of an even function, fe)

and an odd function, fo.

We then have

[F(L): Fe(L)] = 2:

i.e. , the dimension of

F(L)

2.

Now assume that

and

c.

Assume that

regular at (8)

a(f) =-f.

f E F (L),

is a decomposition of

(7)

Fe(L)

Tr(f) :f +a(f) , the trace of N(f)

By

is even; thus

'p

C (lp)

(4)

F(L).

f

a.

f

f

as a vector space over g

Fe (L) is

are meromorphic functions on

is even and let

a E C ; then

f' (-a) =

Hence

even implies

f'

odd.

A similar calculation shows that (9)

f

odd implies

f'

even.

It is also obvious that (10) if

of

f

is even or odd and

f, then

(11) Assume that

-a f

aEC

is a zero (resp. pole)

is a zero (resp. pole) of

f.

is non-zero and is either even or odd;

125

Weierstrass's Work on Elliptic Functions then

a (f) =u_,(f),

u

Let

Proof.

f

by

nL0.

n - va(f).

acC. n < 0,

Were

then we could replace

thus we may assume, without l o s s of generality,that

l/f;

Assume now that

By (10) f

n=l;

has a zero at

odd or even. that

for all

Since

holds for all

g,

-a.

g

f

has a simple zero at

By (8) and (9) f'

a.

is either

we may apply (10) and conclude

f'(a) # O r

thus

f'(-a) # O ;

then

v a (f) = l = v -a (f). Now assume that (11)

either even or odd such that

By (8) and ( 9 ) ,

f' :g

is either odd or even.

va (9)= n - 1,

(g) = va (9).

Oiva(g)

n.

Since

We conclude that

v -a (f) = v-,(g) + 1 = va(g) + 1 = va(f) , proving (11). Let

(12)

be non-zero and either even or odd.

f

even or odd according as

at

is

is even or odd.

This may be seen by expanding

f

in a Laurent expansion

Assume now that

We want to investigate the

0.

behavior of

f

(wl+02)/2,

and

R

f

vo(f)

f E Fe(L) -C.

at the half-periods of w2/2.

(13) The half-periods of

1.

- (01

p'

g

E

F(L) - C

that

g(E)

P(R)

(Lemma 7 . 3 3 ) .

are exactly the elements

vE(Ip-l)(~))

E

of

(EnE) is greater than

n E = 2.

be an odd function and let

g

is regular at

E

then

Indeed, by definition, g(-E) period of

01/2,

be a half

E

R.

period of (14) If

at which

R

in

Further, at these points

Let

i.e., at

We have seen that the half periods of

are exactly the zeros of

P(R)

a:

g, =0,

g(-E)

= g (2e

-

E)

proving (14).

,

g(E) = 0 . =-g(E).

Since

which equals

2~

g(E) ;

is a showing

126

Norman L. Alling

(15) Let then

and let

f E Fe(L) - C ;

is even.

vE(f)

Let

Proof.

n

vE(f).

E

If

n < 0, we may replace

f

by

thus, without loss of generality, we may assume that

l/f; n> -O . in

R

be a half-period of

E

Assume that

F(L) -C.

n>2

and

n>O.

By (14)

By (8) f'

f'(E) = O f

< n , then

Clearly

f" P C .

vE(g)

g

Fe(L) -C

E

Clearly it is Assume that

such that

By (8) and (9),

is even.

- 2;

0< v E (f")= n

Further,

odd,

n > 1.

hence

assume that, for each

O(vE(g)

is

f" E Fe(L).

n-2

thus

is

even, proving ( 1 5 ) . f E Fe(L) - C ;

(16) Let

1z E P ( R )

Indeed, ordf E

then

ordf

-vZ(f) 1

is in

2N.

(6.41:7).

v z (f1 < o By (15) if

z

is in

Let

2N.

is a half period of Z E P(R)

z1f z

to a unique

vz(f) = vzl (f),

proving ( 1 6 ) .

E

R.

-vz(f) -z

is

By (11)

P(G).

By (16), ordf = 2m,

for

N.

A point

an A-point. R

in

f E Fe(L) -C.

Assume now that

then each such

be a non half-period of

congruent modL

some m

R

a

E

P(G)

- {0,w1/2,

Other points in

(wl

P(R)

will be called

+w2)/2, w2/2)

are either half-periods of

or zero.

(17)

Let of

a P(R)

be an A-point in

that is congruent to 'p

The zeros of

-

are both simple.

-

P(R).

of course

-

Indeed, since

?(a)

in

P(R)

-a,

a'

0.

be the point

modL;

are at

The only pole of

a double one at a

Let

'p

then

a

and

- ?(a)

is not a half period of

a'

in

Further,

R,

afa'.

P(R)

and is

va(f) =va,(f). afa'.

Weierstrass's Work on Elliptic Functions By (11)

va(f)

=

v-,(f).

-a E a' (modL),

Since

127

v-,(f)

=

v a l (f),

establishing (17). (18) If

f

has a pole at an A-point

(9- ?(a) )

g :f

or

(fl

0

If

vo(f) > 0,

ordg = (ordf) - 2 = 2m - 2, f

( ~ - P ( E ) )

v E (f)+ 2

=

is either

P(Q)

then

2m

E

v o ( g ) = vo(f) - 2.

and

vo(f) ( 0 ,

then

polynomial of degree Let

Proof.

0.

Let

g-f-

possible pole since

of

ordf=2m,

m

f

in

p,

in

0

v E (9) and

ordg=ordf= 2m.

ordg= (ordf)- 2 = 2m- 2,

is the only pole of

0

?-a(€)

(where it has a double zero) and at

then

ordg

2m-2.

E

(20) If

then let

Finally,

are at

vo(f) > 0,

R,

gEFe(L) -C.

it has a double pole) , the statements about

at

of

Since the only zeros and poles of

If

is

thus

Proof.

are obvious.

f

proving (18).

and note that

or

Since

then

vo(f) '2;

has a pole at a half-period

g-f v,(g)

vo(f) < 0,

is even (12). If

ordg=ordf =2m.

If

gt-Fe(L)-C, v ( g ) = v ( f ) + l = a a is either 2m Finally, ordg -2

This follows, in part, from (17).

vo(f)

(19)

P ( Q ) , let

2m- 2.

Proof.

even,

in

then

;

= v

vai (g)r vo(g)

a

in (where vo(g)

If

establishing ( 1 9 ) .

P(R),

then

f

is a

with complex coefficients.

m

ln=-manz2n be the Laurent expansion of a_,? m : then g E Fe(L) , 0 is the only g

in

a #O. -m Assume that

P(R),

If

and m=l,

ordgz2m- 2.

f

Note that

g = c O E C I and hence

f = c + a 'p. m > l and assume that for all 0 -1 h c Fe(L), which has no poles on P ( n ) - { O } , and for which ordh< 2m,

then

h

is a polynomial of degree

(ordh)/2 in

'p

128

Norman L. Alling

with complex coefficients. for some

... ,cm- 1

cor

...

in m- 1

g = c 0 + c 1T +

Hence

. . . +C,-~T m- 1,

Thus m

C.

f = c0 +el?+ + ~ ~ - ~ r+ pa-,p , proving (20). Having set up what could be thought subalgorithms, let us assemble them into the full algorithm. alr...,akE P(R) f

in

{al,air...,ak,ai}

are the poles of

that are A-points, each indexed according to mul-

P(Q)

(N.b. , (11) holds.)

tiplicity. poles of

so that

First choose

f

in

..

Next let

be the

E~,.

that are half-periods, each occurring

P(R)

to half the multiplicity of the corresponding pole of P(C2).

in

(N.b., (15) holds.)

(21) Let

h: lI!=l(fQ

-fQ(aj))IIjZl(T n -

? ( E ~ ) )

h f 0.

Thus

of

in

g

is in

P(C2)

Fe(L).

is at

0.

,

and let

1; then

Here we take an empty product to be

g

f

g

5

ht c[?],

fh. and

Note that the only possible pole

By (20) g

is in C [ ? ] .

Thus we

have proved the following: Fe(L) = C ( T I ;

Theorem.

thus

F(L) = C

(?,?I).

More can be concluded from this argument; namely the following.

Clearly

(i) vo(f) '0, then

then

m=k+n

and

k+nI m

v o ( g ) = v o (f) - 2 (k + n)

and

ordg(!q)

is

vo(f) (m;

thus

2m.

.

If

If (ii) v o ( f ) > 0,

q=2m-v0(f).

Hence, in

general , (22) ordg(:q)

is

of degree Note that

q

q

2m - max(vo (f),0),

and

g

is a polynomial

in 'p.

can very well be

h=T-elr g=1,

and

q=O.

If

0.

For example, if

q = O , then

f=l/('p -el),

g ~ c O ~ Cand *

Weierstrass's Work on Elliptic Functions

q > 0;

Assume that and

g

then

w o ( f ) < 2m.

129

By construction

have exactly the same zeros to the same order on

P(Q) - I O l .

Let

bl,... ,br

be A-points of

so that

P(Q)

{bl,bi,...,br,b~~ is the set of all zeros of

f

in

P(Q)

are A-points, each indexed according to multiplicity. again that (11) holds.) P(Q)

Let

in

f

that are half-periods, each occurring to half the mul-

constant.

Since, by construction, q

it must have a pole at on

P(Q) - { O l Hence

f

in

has no poles on

thus the zeros of

0;

f,

and

listed above are all the zeros of

2(r+s) =q.

By construction q

is not P(Q)

- {O}

hence of q

on

By (20)

. . . + cqpq

(24) q = co + clp +

(Note

P(Q).

Since, by assumption, q > 0, q

again that (15) holds.)

P(R).

that

(Note

be the zeros of

61,... , 6 s

tiplicity of the corresponding zero of

g,

f

E

c "p1 , with

cq # 0 .

and

have the same zeros and the same poles to the same orders on P(Q);

thus by Liouville's Theorem c

(26) f =

g = c t; 9

hence

nj,lr ('P - P (b.1 1 fl Sj = l ( P - 'P (6j ) 1 k n njz1 ('P - ?(aj) ) I ' I ~ , ~( P - ' P ( c j ) )

(Note that (23) may be regarded as a special case of (26) in which

r = O = s.)

By construction, the rational function given

in (26) is reduced.

In terms of

k,nrrr and

s,

w0(f) =

2(k+n) -2(r+s). Bibliographic note.

The Theorem above is asserted by

Weierstrass [64, vol.V, pp. 141-1521.

A

formula quite similar

to (26) can be found in Whittaker and Watson [69, p . 4491.

A

Norman L. Alling

130

more detailed treatment may be found in [36, pp. 171-1721.

Let

7.43

f E F(L)

have only simple poles, let

.

bll.. ,bn be the poles of residue of

f

at

b

f

P ( Q ),

151 'n.

for

j'

in

and let

r

j

be the

We have seen ( 6 . 4 0 : E )

that

(2)

Let

then

q

g(z

+ R)

=

q(z)

: I n1=1r.c(z-bj); 3

is a meromorphic function on @ . Given R E L, n equals lj=lrjc(z + R - b . ) , which by ( 7 . 3 4 : 7 ) 3

l;,l.rjc(z

- bj) + I;,lrjcR

g E F(L).

Since

C,

f-q

which, using (1), equals is in

it is a constant c.

(3)

F(L)

g(z) ;

thus

and is analytic on all of

Hence

f(z) =c+lj,lrj5(z-bj). n

This method of representing

f

can be used to prove an existence

theorem, namely the following Let

Theorem.

tinct points in

(5)

Let

nENI n22,

P (fl)

.

Let

r

j

EQ:

and let bl,...,b n be dis*, with 15 j 'n, such that

f(z) ~ ~ + ~ ~ = ~ r ~ < ( zfor - ball , ) Z, E C , 3

where

c

E @ .

its poles in

Then

f E F(L) ,

f

P(fl)

being at

bl,...,bnI

has only simple poles on C I

.

rll.. ,rn'

An a d d i t i o n t h e o r e m

7.50

Let

u

E

c

-

1 -L 2

for

and consider

'p

with residues

Weierstrass's Work on Elliptic Functions

f (z) 5 p ( z )

(1)

-

P(u),

for all

131

zE C .

f

has a double pole at each point of L, and has a root at df each point of u + L. Since u P T1L , =(u) = 7' (u)# 0, showing that each point in

u+L

is a simple root of

f.

Since

an even function, f (-u)= p(-u) - p ( u ) = I)(u) - I)(u) = 0; f(z) and u(z-u)u(z+u)/u 2 (z) are both elements of

p

is

thus F(L)

that have the same zeros and same poles to the same orders. From Liouville's Theorem we conclude that there exists

cEC*

such that (2)

f(z) =ca(z-u)u(z+u)/a

(3)

Let

2

(2).

g(z) -cz2a(z-u)u(z+u)/u 2 ( z ) ,

for all z E C . g is an entire function and g ( 0 ) = cu ( - u ) u (u)= 2 -CIS (u) (since u is an odd function). From (1) we see that the principal part of (4)

c = -l/a

(5)

Hence

2

f(z)

at

0

is

z - ~ ; thus

.

(u)

p(z) - p(u) = - u ( 2 - u ) 0 2(z+u) 02(z)o (u)

Note that ( 5 ) was established for all

zEC

and all

u

E

C

1 - 2L.

By the identity theorem for meromorphic functions, ( 5 ) holds for all in (6)

z

z

and

and

u E C.

Let

h(z,u)

u.

ah 6Z(h(z,u)) --(z,u)/h(z,u) az ah 6U (h(z,u)) :-(z,u)/h(z,u). au Let

Applying

be analytic and non-zero

E Z to both sides of

(Note that ( 7 . 3 5 : 3 ) holds for

I

and let

( 5 ) gives

-

Dz.)

Applying

-

DU

to both sides

132

Norman L. Alling

of (5) gives

Adding (7) and ( 8 1 together, and dividing by 2 , gives

Differentiating (9) with respect to

z

gives

The left hand side of (10) is

thus

Interchanging

z

and

u

in (12) gives

Adding ( 1 2 ) and (13) together and dividing by 2 gives

(7')2

Since

=4?

3 -g2'p - g 3 ,

(7.33:19),

r ) " p p " = (1r)2 - g 2 ) ? ' ,

and hence

Using (15) to simplify (14), we obtain Weierstrass's Addition Theorem

Let

u

be a point in s: - L

and let

z

approach it;

then (16)

Weierstrass's Work on Elliptic Functions

133

gives

(16) may be found in Weierstrass's

Bibliographic note.

lectures [64, vol.V, p. 381. p. 2181.

(17) appears in [64, VOl.Vr

The treatment given in this section was found by the

author in [54, pp. 384-3861, and modified slightly.

A

7.60

(1)

relation between Weierstrass's

2 f (v) E exp (-nlwlv /2) u (w,v)

Let

simple zero at 0(

+w.

z

for all (3)

function

and

,

v ~ c . Clearly f is an entire function, which has a

for all

(2)

0

=

7

z

0.

Recall (7.35:13) that

-exp ( q . ( z 3

and

E @ ,

j =1

+ w 7./2)1 0 ( 2 ), and

2.

f (v + 1) = -f (v), for all

.

E

C.

w 2/ w 1 (5.20:l) , and that k = exp(-in.r - 2niv) Using Legendre's equation (7.34:11) one can show

Recall that (5.30:1)

v

Then

T 5

that (4)

f(v+T) =-kf(v),

By Theorem 6.54, c

for all

f (v) = cel (v),

VE@.

for some

cei(0) =f'(O) =w,a'(O).

note that

c E C.

To evaluate

We can compute

5'(0)

directly, by differentiating the product (7.32:2) used to define it, and evaluating 0' ( 0 )

= 1;

thus

u'(v)

at

c = w 1/ e l1 ( 0 ) .

( u l / O i ( 0 ) )exp(nlwlvL/2)e,(v).

(5.20:3), let (5)

0 (u) =

u:olv.

0.

On doing so we find that

Hence

0 (w

1v) =

Using the standard notation,

Then

.

2 (wl/O; ( 0 ) 1 exp (qlwlv /2) el (v)

This Page Intentionally Left Blank

CHAPTER 8 RIEMANN SURFACES

Introduction

8.10

Bernhard Riemann (1826-1866) Galois,

... m a t u r e d

,

l i k e Gauss, A b e l , E i s e n s t e i n ,

mathematically a t an e a r l y age.

H e worked a t

t h e v e r y h i g h e s t l e v e l from a b o u t 1 8 5 0 , u n t i l s h o r t l y b e f o r e h i s d e a t h , a t t h e a g e o f 4 0 i n 1866.

H i s main work o n w h a t came t o

der

be known a s Riemann s u r f a c e s was h i s T h e o r i e

Abel'schen

F u n c t i o n e n , which a p p e a r e d i n C r e l l e ' s J o u r n a l i n 1857 [ 5 3 , pp. 88-1421.

While t h i s work w a s v e r y i n f l u e n t i a l , i t w a s d i f f i c u l t

t o r e a d and absorb.

A g r e a t d e a l of e f f o r t h a s been devoted t o

e x p l o i t i n g t h e p o t e n t i a l t h a t l a y i n t h e s e i d e a s o f Riemann. During t h e 1880-1881 academic y e a r a t G o t t i n g e n F e l i x K l e i n l e c t u r e d o n Riemann's Theory o f a l g e b r a i c f u n c t i o n s a n d t h e i r F o r t u n a t e l y t h e s e l e c t u r e s were p u b l i s h e d .

integrals.

There

i s e v e n a n e x c e l l e n t E n g l i s h t r a n s l a t i o n of K l e i n ' s l e c t u r e s [39].

D u r i n g 1 9 1 1 - 1 9 1 2 a t G g t t i n g e n Herman Weyl a l s o deThese a l s o a p p e a r -

l i v e r e d a series o f l e c t u r e s on t h e s u b j e c t . e d i n p r i n t i n German a n d i n E n g l i s h

[67].

For a d i s c u s s i o n

o f t h e v a s t c o n t r i b u t i o n s o f Riemann see e . g . ,

Klein's lectures

[401. 8.11

Let

(7.32:4) ;

A n Example

L

be a l a t t i c e i n

C

t h e n a s w e have s e e n , 135

(6.20).

F

Let

(-F(L))

9(z)

:v ( z , L )

(6.30:l)

equals

136

Norman L . A l l i n g

,

(7.42)

@((n,ll)

3

= 4p

(1)

and -g21\-g3,

(7.33:12 y :p ' ;

and l e t

x 5 'p

(2)

Let

(3)

3 y 2 = 4x - g 2 x - g 3 : f ( x ) ,

and 1 9 ) .

then F = @(x,y).

and

n

Further, y

L

are d i s t i n c t complex numbers t h e r e are complex numbers

w?(z) = f ( z ) , 3 Some

k,

,

= 4 ( x - e l ) ( x - e 2 ) ( x - e,)

(7.33:4).

w,(z)

w (z) = 0 = w 2

t h e Riemann s u r f a c e of

y

and

and

z

e3

C

E

such t h a t

W,(Z)

z = ekl

If

for

The c l a s s i c way t o c o n s t r u c t

(2).

1

= f (x)

elIe2,

For every

w2(z) = -w,(z).

and s u c h t h a t

then

where

,

which might be c a l l e d t h e

" c u t and p a s t e method", i s t o l e t

b e t h e Riemann s p h e r e

@'

minus n o n - i n t e r s e c t i n g c l o s e d J o r d a n arcs j o i n i n g el t o

C

e2

e3,

to

e3

and

to

@'

Over

compact s u r f a c e .

then

m;

e 2'

i s a c o n n e c t e d non-

C'

it i s p o s s i b l e t o choose s i n g l e

v a l u e d a n a l y t i c b r a n c h e s of

w1 ( z )

and

and

j = 1

w,(z) ;

t h e n one c a n

form (4)

W' :{ ( z , w . ( z ) ) : z

Let

vl

3

be t h e f i r s t p r o j e c t i o n of

W',

maps

(5)

Let

then

W'

i n a two-to-one

w0 c

W

0

,

and

z

E

0 C

. -

Let

IT

1

injective.

0

T ~ ( W) Pro

and l e t

W

z

@;

c

C

2

T

1.

vl

then

Ct.

Let mW

W

be t h e one p o i n t

denote t h e element i n

by mapping

mW

to

m

t h e r e a r e e x a c t l y two p o i n t s o f under

.

= f ( a ) )I ;

= C.

extend t o

{elle21e31m)

p r o j e c t down t o

B~

and 2) onto

C2

fashion, onto

c2:

:{ ( a , ~ )E

compactification of W-W

@'

E

Over

E

C.

If

W

that

l e l , e 2 , e 3 , ~ ) , -ir1\w

I t i s n o t d i f f i c u l t t o see t h a t

is

Riemann S u r f a c e s (6)

i s a s u r f a c e and t h a t

W

137

i s a n open c o n t i n u o u s map o f

n, I

onto

PJ

C.

C l e a r l y t h e l o c a l behavior of

el,e2,e3,

and

a t t h e p o i n t s above 1 i s e q u i v a l e n t t o t h e behavior of z

m

7~

The E u l e r c h a r a c t e r i s t i c o f

0.

follows.

Consider a t e t r a h e d r o n

el,e2,e3,

and

Clearly

a.

h a s 6 e d g e s and 4 f a c e s : t h u s

T

genus of

is

g(T),

TI

0.

x(T) = 4

(See e . g . ,

ence t o t h e topology of s u r f a c e s . )

w e w i l l b u i l d a geometric f i g u r e t i n u o u s map morphic.

let

p

of

onto

S

T

Over e a c h v e r t e x of

-6+4

Further,

C.

Thus t h e

= 2.

[51] f o r a g e n e r a l r e f e r -

I n o r d e r t o compute

x(W)

over

T

and a n o p e n con-

such t h a t

S

and

S

a r e homeo-

W

c o n s t r u c t a v e r t e x of

T

and

S

map t h e s e new v e r t i c e s t o t h e c o r r e s p o n d i n g o l d o n e s .

p

Over e a c h e d g e o f

T

c o n s t r u c t two new e d g e s of

S

and l e t

map t h e s e new e d g e s down t o t h e c o r r e s p o n d i n g o l d e d g e s o f Over e a c h f a c e o f

T

c o n s t r u c t two new f a c e s o f

ma P t h e s e down t o t h e c o r r e s p o n d i n g o l d f a c e s o f wa Y t h a t a t t h e v e r t i c e s o f

x(S) W

X : @/L

Let

?\

Since

(8)

= 4 - 1 2 + 8 = 0;

a r e homeomorphic,

(7)

X

S,

p

thus g

g ( s ) = 1.

(W) = 1:

S

i.e.

,

T.

p

i n such a

z

z2

+

at

W L > Z . )

Finally, since

W

p

and l e t

TI

is locally l i k e

(To see t h a t t h i s c a n b e d o n e c o n s i d e r

0.

at

whose v e r t i c e s a r e l a b e l e d

T

i s homeomorphic t o

T

z2

may b e computed a s

x(W),

W,

+

and

S

is a torus.

(6.24).

i s i n v a r i a n t under

L

( 7 . 3 2 ) i t i n d u c e s a map

p

of

X.

onto

7T

7\ X->C

is topologically equivalent t o

W-

>C

I n s e c t i o n 8.20 w e w i l l p u t a n " a n a l y t i c s t r u c t u r e " on and o n

C

so t h a t w e may d o a n a l y t i c f u n c t i o n t h e o r y on t h e s e

X

138

Norman L. A l l i n g

surfaces. 8.12.

Let

f (X,Y)

(1) L e t

and

X

be a n i r r e d u c i b l e p o l y n o m i a l i n

n > 0

degree

(in

C l e a r l y we can f i n d (2)

a(X)f(X,Y)

g(X,Y)

( W e can even choose Let

then

I

such t h a t

i s of degree

in

n

Y,

and i s i n

where

g(X,Y) = i jn= o p j ( X ) Y j ,

a(X)

i s a non-zero,

pj(X)

E

@[XI,

x :h ( X )

Let

t o b e monic and o f minimal d e u r e e . )

proper i d e a l i n

i s maximal.

I

@ ( X ) [Yl

morphism of

onto

and

F = @ ( x , y ),

then

where

f (X)@(X)[Yl;

1

is irreducible,

(4)

of

@ ( X ) [Y]

pn(X) # 0 .

and

(3)

@[XI

E

QI'.

n f(X,Y) = l j = o r j ( X ) Y J r

thus

Y);

a(X)

Hence

@[XI [Y].

be two i n d e t e r m i n a n t s over

Y

and

C(X) [Yl/I

h

f (X,Y)

be t h e c a n o n i c a l homo-

: F.

y :h ( Y ) ; f (x,y) = 0 = g(x,y)

how t h e Riemann s u r f a c e

of t h e equation

W

g(x,y) = 0)

equivalently of

Let

Since

@(XI [Y].

.

W e w i l l now s u g g e s t f(x,y) = 0

(or

c a n b e c o n s t r u c t e d by t h e " c u t and

p a s t e method". T h e "cut and p a s t e -

8.13

surface

(1) L e t

E

(2)

F = @ ( x , y ) by r e a s o n i n g a s f o l l o w s .

of

c0

{A€@:

pn(X) # 0

Since A

W

m e t h o d " c o n s t r u c t s t h e Riemann

pn(A)#ol.

(8.12:2),

cc -

C0

is a f i n i t e set.

For a l l

Co

i s a polynomial of degree

l;=opj ( A ) t J

As such it has

n

roots

tl(A),

occurring t o its multiplicity.

. .. , t n ( A )

E

n C,

in

@[tl.

each r o o t

Riemann Surfaces

Let

(3)

Il

D(A) Z

j
This is - of course but since

f(X,Y)

(tj(A) - t .(A))

-

3

Let

.

the discriminant of ( 2 ) .

is irreducible (8.12:1),

finite number of zeros in (4)

L

139

@O

D

It may be zero; has only a

*

and let

C1 E { X c C 0 : D ( X ) # O } ,

S : (@-Cl) u b).

Using the implicit function theorem, one can show that each tj(X)

can be continued analytic over C1. Let X o E C1 and be a piece-wise C 1-path in C1 from A. back to X o .

r

let

Continuing some j

=

k,

k.

t .( A o )

I

1 < k < n.

By letting

r

along If

r

tk(Xo),

will lead back to

r

for

is homotopically trivial, then

go around each point in

S,

we will

obtain ramification data. Now connect the points of intersecting piece-wise C1-paths

E E

(@u{m})

set the

-

A

together with a non-

S

A,

such that,in

no path goes around any point in

S.

Over this

t.(A)'s can be defined to be single-valued analytic 3

functions such that for each

A,

t.(X) = tk(A) 3

implies

j

=

k.

Let (5)

W

Wo

{(X,tj(X)): A

is then formed from

points over

S,

E

Wo

E

and

1 < j < n]

by adjoining a finite number of

and pasting the sheets together above

A

according to the ramification data. Remark:

The treatment above is very sketchy; however it

can be made to work.

(See e.g., [ 3 5 , vol. 11, Chapter 1 2 1 for

a recently written classical treatment.) Many of the classic texts have wonderful and ingenious drawings to assist the reader in visualizing the Riemann surface

140 W

Norman L. Alling

so constructed. Nevertheless, the author found the following

remarks by Klein very instructive 139, p.xl. "I am not sure that I should ever have reached a well-

defined conception of the whole subject, had not Herr Prym, many years ago (1874), in the course of an opportune conversation, made a communication which has increased in importance to me the longer I have thought over the matter.

He told me that

Riemann surfaces originally are not necessarily many-sheeted surfaces over the plane, but that, on the contrary, complex functions

of position

surfaces plane.

can be studied

on

arbitrarily given curved

exactly the same wax _ as _ on -the surfaces over the

It

These ideas of the 19'th century were then translated into the following in the 20'th century. D e f i n i ti o n s

8.20.

Let

X

be a connected topological space for which a

family

has been given; where each

z

then

X

each

U

each

z

1

j j

will be called a coordinate map on

C :c

all open sets of all sets C

U

c

Let

C

C

C

u

and

U

will be called a coordinate neighborhood in

i.e., let

Then

U

is a surface.

X,

onto an open subset of @; j will be called an Atlas of X, and

is a homeomorphism of

E x a m p l e 0. @:

( U j ) j c J is an open covering of

U

U,

and

1'

be the one point compactification of To topologize

{m}.

in the open sets of

such that

C

-

U

C.

1,

we will include To these adjoin

is a compact subset of

@.

is a compact Hausdorff space which is homeomorphic to

141

Riemann S u r f a c e s

t h e two-sphere. of

into

U1

( u ) : l/u,

Z

i s a s u r f a c e and

and

(2)

101.

0.

b e t h e i d e n t i t y map

U

Let

is closed i n

Q:

R :

(See(6.2) for d e t a i l s . )

L; then

(3)

Let

Each

V

V1

=

v2

: v1 + w1/2,

Ixwl+yw2:

+ (wl + w 2 )

v4

+

o n t o a n open s e t and l e t X;

thus

X

zj

Let

(

~

~

b e0 a ~b a s )i s

~

U

(-1/8,5/81},

E

and

/2,

w2/2.

j

in

J E {1,2,3,4}.

C

X.

Then

and Let

hlV

z

j

U : (U

i s a homeomorphism

: (hIV.)-’, 1 j ”j) j c J

f o r each

j,

i s an A t l a s of

is a surface.

is called analytic i f

U

(4)

x,y

V3 : V1

: v1

Recall t h a t s i n c e

denotes t h e period parallelogram associated

P(R)

i s a n open set i n

j

and l e t

i s a l o c a l homeomor-

phism.

R.

(6.20)

n a t u r a l l y i n h e r i t s t h e s t r u c t u r e of a

X

@,

h

with

thus

0

-f

t o p o l o g i c a l g r o u p i n s u c h a way t h a t

of

let

U2

C.

i s e x a c t , i n t h e c a t e a o r y of A b e l i a n g r o u p s . L

E

(Uj l z j ) j E { 1 , 2 1 ;

be a l a t t i c e i n

L

u

For each

is an A t l a s of

U

Let

h L “@--->X

+

-

z1

b e d e f i n e d so t h a t

X

0

l/m

where

Example 1 .

and l e t

U2 : Z

Let

@.

z2

h

U1 : @

Let

0

Zk

-1

I z7.( U7. ) n

zk(Uk)

i s a n a l y t i c , f o r each

(Here t h e empty mapping i s t a k e n t o b e a n a l y t i c . ) ( 4 ) a r e c a l l e d t h e t r a n s i t i o n f u n c t i o n s of Lemma.

analytic.

j ,k

E

J.

The maps of

U.

The A t l a s e s g i v e n i n Example 0 a n d Example 1 a r e

Norman L . A l l i n g

142

U : ( Vk , w k ) kcK

and

U

Let

b e A t l a s e s on

They w i l l

X.

be c a l l e d a n a l y t i c a l l y e q u i v a l e n t i f

j'

(5)

0

-1 Wk Iz. ( u . ) n 3 7

and a l l

k

is analytic, for a l l

J

E

A n a l y t i c e q u i v a l e n c e between a n a l y t i c A t l a s e s

i s an equivalence r e l a t i o n .

X

X

An e q u i v a l e n c e c l a s s l y t i c Atlases of

of a n a l y t i c a l l y e q u i v a l e n t ana-

i s c a l l e d a n a n a l y t i c s t r u c t u r e on

X

pair

(X,X)

c o n s i s t i n g of a s u r f a c e

ture

X

X

on

i s c a l l e d a Riemann s u r f a c e .

n o t e d by

Clearly

C.

Although i t i s a n

i n s t e a d of

(X,X).

i s i n an a n a l y t i c s t r u c t u r e

S

i s c a l l e d t h e Riemann s p h e r e , and i s u s u a l l y de-

(1,s)

1.

U

(Cont.)

Example 0 .

X

The

X.

and a n a n a l y t i c s t r u c -

X

abuse of n o t a t i o n w e f r e q u e n t l y use

on

j

K.

E

Theorem.

of

wk(vk)

i s a compact s i m p l y c o n n e c t e d Riernann

C

surface.

X.

ture

U

(Cont.)

Example 1.

i s contained i n an a n a l y t i c s t r u c -

is an a n a l y t i c t o r u s .

(X,X)

S i n c e a n a l y t i c maps p r e s e r v e o r i e n t a t i o n and s i n c e

C

is

o r i e n t e d w e have t h e following: Riemann s u r f a c e s a r e o r i e n t e d .

Proposition.

Let

8.21.

let

f:X

U E ( U

x

E

+

z )

j f j jcJ

E

X

there exists

X

f (x)c V k f

(1) wk

b e Riemann s u r f a c e s and

(Y,Y)

w i l l be c a l l e d a n a l y t i c i f t h e r e e x i s t s

f

Y.

and

(X,X)

U z (Vk f W k ) k E KE Y

and

j

E

and

J

k

K

E

such t h a t f o r a l l

such t h a t

x

E

U

jf

and o

f

Lemma.

0

z

-1 j

is analytic a t

Assume t h a t

f

z.(x) 3

E

@.

i s a n a l y t i c ; t h e n f o r any c h o i c e

Riemann S u r f a c e s

of

U

E

X

and

V

Y

E

143

(1)holds f o r a l l

j

and a l l

J

E

k

K.

E

This holds s i n c e t h e composition of t w o a n a l y t i c

Proof.

f u n c t i o n s i s a n a l y t i c and t h e i n v e r s e o f a n i n j e c t i v e a n a l y t i c function is analytic. d e n o t e t h e s e t o f a l l a n a l y t i c maps o f

F(X)

Let

o t h e r t h a n t h e map t h a t t a k e s e v e r y

C,

Clearly

F(X)

contains

@.

Wk( t )

(2)

E

m

E

1.

I t w i l l b e c a l l e d t h e f i e l d of a l l meromorphic

X.

x

E

U

0

f

0

j (1) z . J

n U

and

j( 2 )

f(x)

E

Vk(l)

its derivation

for

g;,

t = 1

-1 wk(2)

91 = w k ( l )

n V

Since

k(2)*

-1 = ( t ) - 9t

i s a n a n a l y t i c f u n c t i o n on a n open s u b s e t o f

(3)

to

X

into

i s a f i e l d , under point-wise o p e r a t i o n s , t h a t

f u n c t i o n s on Let

x

X

g2

and

Clearly

2. 0

'j(2)

2

w e can take

@,

-1 j (1)'

Is t h e r e a n y r e a s o n why (4)

gi

o

zj(l)

should equal Let

Example.

U : (U

Let

z,(X)

wl(A)

and

I

A

E

= X I

21 # 9;

0

where

X

X

E

o

z ~ ( ~ ) ?

Let

f(X) :

U1

E @ :U 2 ,

V

Let

@.

for all

for all

g 2 ( A ) = X/2, 0

X :@ E Y .

j l z j )j r 1 1 , 2 1

for all

= 21,

g;

E

@;

@.

(Vltwl),

gl(X)

thus

=

A,

A

for all

z,(A)

=

E

X,

and

=

v1

where for all X

gi = 1 # 1 / 2 = g ; .

E

@.

@,

@.

Hence

22-

Thus t h i s s t a b ( 4 ) a t d e f i n i n g a d e r i v a t i v e o f f a i l s to define an invariant object.

f

on

x

It t u r n s o u t t h a t deriva-

t i v e s a r e n o t d e f i n a b l e o n Riemann s u r f a c e s ; however d i f f e r e n -

t i a l s are, as w e w i l l see i n t h e n e x t s e c t i o n . 8.22.

Let

(X,x)

b e a Riemann s u r f a c e a n d l e t

144

u

Norman L. Alling

(U.,Z.)

f

I

I jEJ

(1) Tjk where

x.

E

D(zj

o

For all

zkl)

z k,

o

J,

E

on

U . n Uk,

1

denotes differentiation in

D

let

j,k

Let

@.

(&j)j E

J

:6u

be

such that

Aj

(2)

is a meromorphic function on

U

(i.e.,

j

6.:U

i

C

-f

j

is analytic and not the infinite constant map), and 6k = 6 . T 1 jk'

(3)

6u

on

is then a meromorphic differential relative ( C o n t . f r o m 98.20).

Example 0. (E @ - ( 0 ) ) .

Let

T21(X)

:1

cS1

on

=

- 1/X 2

2

Note that

L S ~ ( X ) :-A

-2

on

U.

n U2

U1

T12(X) = - X 2 ,

and

and let

U1

(Sj) j E 1 1 , 2 1

sees that

= @*

X

for all

E

@.

One easily

U2.

is a meromorphic differential relative

u.

to

Example 1.

j

U . n Uk. 3

E

-1 z k (1) = X

J.

D(zj

(Cont. from 5 8 . 2 0 ) .

Since z j o zk-1) = 1, and so

o

T

jk

+ =

wjk,

1,

Let 6 1, for all j where 2 w j k E L,

for all

j,k

is a meromorphic differential with respect to In general

6u

E

J.

Thus

6u

U.

is called analytic if each

sj: u j

@.

-P

Note that the differential constructed in Example 1 above is analytic, whereas the differential constructed in Example 0 is not.

Let

yu

be a meromorphic differential relative to

.

Let

f

(4)

(6 + Y . ) I i jEJ

E

F(X)

a

and

with respect to

(f&j)jEJ are meromorphic differentials U.

These differentials will be denoted by spectively.

U.

Note also that

AU 'yu

and

f6U re-

6 j/yj IUj n Uk = 6 /y I U , n Uk; k k l

Riemann Surfaces

145

thus is a meromorphic function

(5) ujEJ6j/yj

is non-zero.

yu

g

X I provided

on

g

will be written as

(6) dfj

D(f

z-1 .)

o

Finally let

GU/yu.

for each

z . 3,

o

3

j

J.

E

is a meromorphic differ(dfj)jE which will be denoted by dfU.

It is easy to verify that ential relative to

U,

Let

8.23.

V

X

E

(1)

and

6u

and

V

and

U

respectively. are equivalent if

yy

X.

D(X)

Let

X.

entials on

is a meromorphic

yy

X

is an equivalence class

6 u with respect to

of meromorphic differentials E

6u u

U u V.

A meromorphic differential on

U

and

be meromorphic differentials rela-

yu

differential relative to

6

U

(X,X) be a Riemann surface, let

6u

and let

tive to

b U = dzU.

(Cont.)

Example 0 .

U,

with

denote the set of all meromorphic differ-

It is easy to see that (8.22:4) induces the

structure of a vector space on

D(X)

over

F(X).

From (8.22:5)

we see that (2) D(X)

is of dimension at mo.st one

From (8.22:6) one (3)

d:f

E

F(X)

kernel is Theorem.

sion of

D(X)

let

bU

E

6

6.

df

i

D(X)

is a C-linear derivation, whose

C.

If there exists

0 # df E

F(X).

easily sees that

over

Proof.

Let

-+

over

D(X)

F(X) E

is

U(X).

with

6

f

E

F(X) - C ,

then the dimen-

1.

Using (2) the theorem is proved.

# 0.

Let

U

( Uj IZj’j E J E X

and

Norman L . A l l i n g

146

Lemma. ____

No

Proof.

Since

6

i s i d e n t i c a l l y zero.

j

6 # 0

i s not i d e n t i c a l l y zero.

there exists

x

Since

k'

i s , by assumption, c o n n e c t e d ,

and s i n c e (8.22:3) h o l d s , and s i n c e e a c h zero, each

6

k

Let

Z.

T

(8.22:l) i s n e v e r

jk

i s n o t i d e n t i c a l l y z e r o , p r o v i n g t h e Lemma.

j

The o r d e r o f in

6k,

such t h a t

J

E

6

a t some x E U j l ~ ~ ( j s u c h t h a t x E Uk. S i n c e T

J

E

6( 6 .~4 1 :) 1 ) ,

is

is analytic

jk

and w i t h o u t z e r o s o r p o l e s ,

(4)

vx(6.) =

Let

~ ~ ( 6d e) n o t e t h e i n t e g e r i n ( 4 ) .

Vx(6) n <

VX(").

3

for a l l

m,

for a l l

m

n

x

E

X.

(Here l e t

6,y

E

let

6 = 0

If

b e o r d e r e d so t h a t

m

Z.)

E

Let

Proposition.

x

D(X),

E

and l e t

X,

f

E

F(X);

then

(5)

vX(6

f

vx(6) # (6)

n

E

vx(f6) = vx(f)

z u

vx(S) > 0

equality occurring i f

then

+

vx(6)

,

where

n

+

x

x

then

i s c a l l e d a pole o f

is called a

of 6 .

6

E

x

E

X.

X.

t i a l s on

Let

V1(X)

for

and i f

is called

~ ~ ( >6 0)

for

b e t h e s e t of a l l a n a l y t i c d i f f e r e n -

C l e a r l y it i s a v e c t o r s p a c e o v e r

Example 1.

6,

D(X)

a n a l y t i c , o r a l t e r n a t e l y of t h e f i r s t kind, i f all

: a,

m

Cml.

vx(6) < 0

If

,

V,(Y).

Further, all

m i n ( v x ( 6 ), v x ( y ) )

y)

(Cont. f r o m $8.22.)

0

(Cont. f r o m 98.22.)

dz

# 6

E

@.

D(x);

thus

V 1 ( X ) # 0. Example 0 .

pole a t

m

8.24.

E

V(C)

and h a s no o t h e r z e r o s o r p o l e s on Let

6

E

V(X)

and l e t

r

has a double C.

be an o r i e n t e d piece-

Riemann S u r f a c e s

2 C -Jordan arc, o r curve, i n

wise on

r.

x

such t h a t

W e want t o d e f i n e a n i n t e g r a l o f

this let

U : (U

j “ j ’ jcJ

that

r

(1)

zj ( r ) j

f o r some

c U.

3,

6

Assume t h a t

r

o

z-ldz

j

E

x

j

E

and l e t J.

147

6u

6

along

6.

E

h a s no p o l e s

6

r.

To do

Assume f i r s t

Clearly

is w e l l d e f i n e d .

is a l s o i n

Uk,

f o r some

k

E

J.

Let

y

Lemma.

(4)

dr; = D ( z

-1

j

ozk )dw; (3) is

t h u s t h e r i g h t hand s i d e o f

w h i c h , by (8.22:l a n d 3 ) , is

U s i n g t h e Lemma w e c a n d e f i n e

no poles o n

r

and l e t

c

E

@;

then

lr6.

E

D(X)

have

Norman L . A l l i n g

148

r

If

rl

i s t h e sum of two o r i e n t e d J o r d a n a r c s ,

and

r2,

then 6 f j

(9)

6=!6.

r

r

Assume i n a d d i t i o n t h a t

cauchy's Theorem.

is the

boundary of a n open c o n n e c t e d , s i m p l y c o n n e c t e d , r e l a t i v e l y compact s u b s e t Xo :

of

Xo

and assume t h a t

X,

h a s no p o l e s o n

6

then

jr 6

(101

= 0.

L e t us sketch a proof.

Since

Xo

i s compact i t may b e

c o v e r e d by a f i n i t e number o f c o o r d i n a t e n e i g h b o r h o o d s

ul,...,un .

Xo

since

is oriented,

r

i s t h e sum i n t h e s e n s e

of homology of a f i n i t e number o f p i e c e - w i s e CL-Jordan c u r v e s

r

j

for

c

U

1'

such t h a t t h e hypotheses concerning

r

and

6

1'

Let

X.

Let

g = 1 then

generations

a,

(1) a b a - l b - l (2)

-nl(X)

g

2

X.

C, p r o v i n g t h e t h e o r e m .

If

g = 0

i s t h e g e n u s of

then

X

X;

then

= 1;

Z @ Z

b.

Further,

thus

H1(X,Z).

g

is said t o

i s simply connected.

I T ~ ( X ) , t h e fundamental group of

and

X.

i s a t o p o l o g i c a l i n v a r i a n t of

g

b e a n a n a l y t i c s t r u c t u r e on

X

hold

i s homeomorphic t o a s p h e r e t o which

[51] f o r d e t a i l s . )

be t h e genus of If

X

h a n d l e s have b e e n a t t a c h e d , where

(See e . g . ,

r

b e a compact ( c o n n e c t e d ) o r i e n t a b l e s u r f a c e .

X

I t i s w e l l known t h a t

g

and

On a c o o r d i n a t e neighborhood C a u c h y ' s t h e o r e m

c a n be p r o v e d u s i n g C a u c h y ' s t h e o r e m i n 8.25.

6

X I h a s two

149

Riemann S u r f a c e s

We c a n g e n e r a l i z e L i o u v i l l e ' s Theorem a s f o l l o w s .

8.26.

and l e t

Y

and

b e ( c o n n e c t e d ) Riemann s u r f a c e s X

b e a n o n - c o n s t a n t a n a l y t i c map o f

f

i s compact t h e n

X

x

Let

Theorem.

maps

f

onto

X

thus

Y;

into

Y.

If

i s a l s o com-

Y

pact.

i s n o n - c o n s t a n t and

X

i s z e r o o n l y on a d i s c r e t e c l o s e d s u b s e t o f

X.

Since

Proof.

f

H u r w i t z ' s Theorem t h a t

sets i n

i s open.)

(See e . g . ,

Y ) .

Since

i s compact and

ous,

i s c o m p a c t , and h e n c e i t i s c l o s e d i n

i s open

f(X)

i s open.

i s open and c l o s e d on

Since Y,

p. 1621 f o r a

[54,

f

f(X)

continu-

f

Since

Y.

i s c o n n e c t e d and

Y

maps o p e n

f

proof t h a t

X

df

We know from

i s a n open map ( i . e . ,

f

t o open sets i n

X

connected,

a

f

# f(X)

proving t h e theorem.

f(X) = Y,

L e t u s see t h a t t h i s t h e o r e m i m p l i e s t h e t h e o r e m w e c a l l

L i o u v i l l e ' s Theorem nowadays.

f

E

be a bounded e n t i r e f u n c t i o n ; t h e n

f

Let

Corollary. C.

Since

Proof.

f

i s a bounded a n a l y t i c f u n c t i o n i t c a n

X

b e r e g a r d e d as a n a n a l y t i c map o f movable s i n g u l a r i t y a t of

f

into

1

Thus

00.

Since

C.

f

-

into

{a}

w i t h a re-

e x t e n d s t o a n a n a l y t i c map

i s compact and

C

C

u s e t h e theorem above t o conclude t h a t

f

i s n o t w e may

C

i s c o n s t a n t , proving

t h e Corollary. 8.30. -

Some p r o p e r t i e s -

The Riemann s p h e r e Let z

E

z

F(C)

C

of t h e Riemann s p h e r e .

w a s s t u d i e d i n $ 8 . 2 Example 0 .

d e n o t e t h e i d e n t i t y mapping o f (8.21);

C

onto i t s e l f .

Clearly

thus

(1) C ( z ) = F ( C ) . Now l e t

f

E

F(C)

-

C.

Since

C

i s compact,

a f i n i t e number of z e r o s and p o l e s o n

C.

Let

f

can have only

k

5

vm(f).

Let

150

Norman L. A l l i n g

g : z-kf;

then

q

m.

There e x i s t s

of

q

r > 0

and

C

h a s no z e r o s or p o l e s a t

q

s u c h t h a t a l l o f t h e zeros a n d p o l e s

are i n s i d e .

(2)

c,(o)

Let

Cr(0)

:IX

E@:~xI

=r)

be positively oriented i n

view o f maps

z2.

and

U2

Cr(0)

,

h(w) :g ( l / w )

to

,

C1/,(0)

w

for all

E

multiplicity.

is i n

Let

and it

U2,

then

h

C

(0).

Thus ( 3 ) e q u a l s

{A

Br(0)

Let

for definitions.)

negatively oriented.

l/r

h a s a s many z e r o s a s p o l e s i n s i d e

q

W e c a n compute

( S e e Example 0 , 5 8 . 2 0

h a s no zeros o r p o l e s i n s i d e

Thus

@.

F i r s t l e t u s view i t from t h e p o i n t of

i n t w o d i f f e r e n t ways.

z2

-

F(C)

E

E

@:

/XI < r}

F(C)

Cr(0),

counting

and l e t

(5)

al,...,an

Let

of

b e t h e zeros a n d l e t

inside

q

Cr(0).

(z-al)*

(6)

P(Z)

Clearly

- bl)

(Z

p(z)

o r p o l e s on

E

C.

0

be t h e poles

Let

. . . a (

.

bl,...,bn

.

z-an)

.. ( Z - b n )

*

@ ( z ) . By c o n s t r u c t i o n By L i o u v i l l e ' s Theorem,

g(z)/p(z)

h a s no zeros

q(z)/p(z) = c

E

C*;

thus (7)

k

f (z) = z cp(z) Theorem. 8.31.

E

C(z)

,

proving t h e following

C ( z ) = F(C).

Having d e s c r i b e d

F(C)

we can describe

U(C)

and

Riemann S u r f a c e s

D,(Z)

U s i n g Theorem 8 . 2 3 w e see t h a t

(8.23).

(1) D ( C )

C(z)dz.

=

Now l e t

y

D,(C).

E

§8.23), t h a t

up)

by ( 1 ) . W e h a v e s e e n (Example 0

y = fdz,

has a double pole a t

dz

o r zeros i n

(2)

151

proving t h a t

1;

f = 0

m

a n d no o t h e r p o l e s

and t h u s t h a t

= 0.

Some

8.40.

Let

p r o p e r t i e s of

be a l a t t i c e i n

L

C/L.

(6.20).

C

h

Let

and

be

X

defined i n t h e category of Abelian aroups such t h a t

->c-

(1) O + L X

>X

i s exact..

0,

+

c a n b e made i n t o a compact s u r f a c e so t h a t

homeomorphism.

c

Since

w h i c h makes

X

h

i s a compact Riemann s u r f a c e o f g e n u s 1.

X

thus

is the

I n Example 1 ( 8 . 2 0 ) , a n a n a -

X.

w a s d e f i n e d on

X

lytic structure

h CC ->X

i s simply connected,

u n i v e r s a l covering space of

is a local

h

analytic;

Let

F(X)

( 8 . 2 1 ) d e n o t e , as u s u a l , i t s f i e l d o f meromorphic f u n c t i o n s o n

X. Given

(2)

h*(f)

then

h*(f)

f

E

F(X)

f

o

h;

E

let

F(L) ( 6 . 3 0 : l ) . h*

Theorem.

i s a C - l i n e a r isomorphism o f

F(X)

onto

F(L). Proof.

It is clear that

homomorphism o f

F(X)

g(z+R) = g(z) c o s e t s of

C

L

such t h a t

since

h

,

f

i .e.

into

for a l l in

thus

C;

g = h

z

0

f.

h*

.

Let

and

R

F(L)

c

E

g

Since

i s an i n j e c t i v e C-linear

F (X)

.

Clearly

E

E

L,

i n d u c e s a map g

F(L)

.

Since

i s c o n s t a n t on

g f

of

i s meromorphic o n

i s a l o c a l homeomorphism, E

g

h* ( f ) = g ,

f

X

into (I:

and

i s meromorphic o n p r o v i n g t h e Theorem.

X:

152

(3)

Norman L . A l l i n g

Let

serve t o identify

h*

F(X)

and

F(L) ,

whenever t h i s may p r o v e c o n v e n i e n t . 8.41.

then

g

E

(1) L e t

Let

f

-

C.

F(L)

cxEx

ord(f) 5

-

C

-

Vx(f);

F(X)

E

and l e t

g 5 h * ( f ) (8.41:2);

v x ( f l
ord(f) = ord(g)

6.40,

and ( 7 . 3 2 ) w e c o n c l u d e t h a t t h e f o l l o w i n g i s t r u e .

a

For a l l

Theorem.

(2)

(6.41:7).

From Theorem 6 . 4 2 ,

Corollary

C

E

v (f-a) = n. X

IXEX

vx ( f - a ) >O Further,

n > 2

for all

f

E

6

I n Example 1 ( 8 . 2 2 )

8.42.

-

F(X)

C,

(8.23).

Although dz.

w e see t h a t

(1) D ( X 1 = I f d z : f

E

F(X)}.

D1(X)

Theorem.

= {cdz: c

Clearly each

Proof.

D1(X)

is frequently defined to be

an abuse of notation, 6

,

I n 58.23 i t

was d e f i n e d .

was shown t h a t i t i s n o n - z e r o a n d i n

Using (8.23:2)

ordh*(?) = 2.

and

By (1) t h e r e e x i s t s

f

C).

is i n

cdz

F(X)

E

E

D1(X).

5

D1(X).

E

Since

dz

h a s no p o l e s ,

By L i o u v i l l e ' s Theorem ( 6 . 3 )

c a n h a v e no p o l e s .

i

5 = fdz.

such t h a t

h a s no z e r o s o r p o l e s ( 8 . 2 2 ) , and s i n c e

Let

,

f = c

f C,

E

proving t h e theorem.

xo 5 h ( 0 )

Let

8.43.

x.

i s A b e l i a n and

-

T ~ ( X , X ~ )a n d

H1(X,Z)

be a b a s i s o f

L

down, v i a

to a closed path

A. 7

E

IT.

1

h,

(X,x 1 . 0

indeed

-

E

P(R)

X

i s isomorpnic t o

are i s o m o r p h i c .

(6.20).

Let

Since

Let

Let

I 5 [0,1];

is a torus Thus

2 @ Z.

R then

T ~ ( X , X ~ )

( 5 (wl

Iw

j

w2)

maps

i n X. Clearly j be t h e p e r i o d p a r a l l e l o g r a m o f A

t

)

Riemann S u r f a c e s

R

Since

(6.20).

aP3)

[ A ~ , A ~E I

(1)

i s homotopically t r i v i a l i n

i s a basis of

F u r t h e r , g i v e n any b a s i s

(3)

exists a basis

R'

of

{Ai,A;j

of

such t h a t

L

IT~(X,X~). T I ~ ( X , X ~ )t h e r e

{Ai,A;} = h ( Q ' ) . h

T h i s a l l f o l l o w s from t h e f a c t t h a t

Note:

C,

= 1.

{A1,A21 :h ( R )

(2) C l e a r l y

153

t h e u n i v e r s a l covering space of

and t h a t

X

g r o u p o f a l l deck t r a n s f o r m a t i o n s o f

h.

is

C->X

acts as the

L

[51] o r

(See e . g .

[3] f o r d e t a i l s . )

Given

r

E

TI

5

r

(X,xo). 1

w e want t o d e f i n e

V1(X),

E

L

i, f o r a l l

need n o t b e a r e c t i f i a b l e p a t h .

Although

t h i s d o e s n o t r e a l l y p o s e any p r o b l e m i t i s p e r h a p s c o n v e n i e n t

-

-

conceptually

some

m

and

E

r

in

w i t h o u t loss o f g e n e r a l i t y w e

2.

Thus

i s homotopic t o

€or

A?;

i s a p i e c e - w i s e C 1- p a t h .

r

may assume

(4)

n

t o note t h a t

plw

x

i s a 2 - b i l i n e a r map.

p l x o )

+

1

ir

5

E

cl

Further j r c b = c j 5

r

for all

c

E

C.

I n view o f Theorem 8.42, i t i s s u f f i c i e n t t o r e s t r i c t o u r

<

attention to

(6)

P(Q)

I n general

= dz.

= Q.

5 = cdz,

f o r unique

c c @,

and

r

= AmAn

1 2

in

Norman L . A l l i n g

154

IT^ ( X I xo

thus

;

dz

8.44.

V1(X)

E

c a n b e w r i t t e n i n a n o t h e r wayl namely

as follows. dz = q f l ' .

Theorem.

A s w e h a v e seen ( 8 . 2 3 : 3 )

Proof.

'p'

E

-

F(X)

is i n

d'pfil

C,

D(X).

U

Let

V(X)

.

Since

b e d e f i n e d as i n

(dp.h\')

One e a s i l y sees t h a t

Example 1 ( 8 . 2 0 ) .

is i n

@

= (dz)

(see

Example 1 ( 8 . 2 2 ) ) ; p r o v i n g t h e t h e o r e m .

(1)

q = P

d'p

f

(4P

By ( 7 . 3 3 : 2 0 )

X

h((ul+u2)/2),

- 92T - g 3 ) 4

('p')'

I

only pole on

3

at and

= 4('p - e l )

xo

( p - e,)

h(0)).

(

('p

v1

- e,).

It has

h(w1/2),

as i t s z e r o s ( 7 . 3 3 : 3 ) .

h(w2/2)

has i t s

I f these

p o i n t s are n o t i n t h e i n t e r i o r o f a p a t h o f i n t e g r a t i o n o f (1)

(4p 3 - g 2 y - g 3 ) 4

then a continuous branch of W e have s e e n (8.43) t h a t

w

=

j

then

A./2 J

i s a C1-Jordan

A

dz.

Let

X

from

can be chosen. A./2 7

:h ( [ O , S l w . ) ; 3

j

arc i n

xo

to

Clearly J A3J 2

dz = u . / 2 . 3

U s i n g t h e t h e o r e m a b o v e w e see t h a t

Transforming t h i s i n t e g r a l back t o t h e p l a n e g i v e s (3)

u./2 = 7

h(uj/2).

p

A (4T

3

- g2'D - g 3 P

.

155

Riemann Surfaces Since

has a pole at

'I\

(7.33:3), and since

(4) w . / 2

(4s

O3

ds 3 -g2s - g 3 ) 4

(Cf. [64, vol. V, p . 781.)

Q(02/2) = e 3

[0,410

is injective on

'I\

= ?

1

%(wl/2) = el,

0,

we see that

j

and

( 4 ) also goes back to the beginnings

of the study of elliptic integrals, to the notion of the period In the context of

of an elliptic integral of the first kind. integrating

dz on

X

against a basis .{A1,A21

of

~l(X,xo),

much that may have seemed ad hoc and possibly ambiguous, in the early work on periods, in the 18'th and 19'th century

--

is cleared up. Note, in conclusion, that

If

is a positive basis of

fi

L

then

T

E

R,

the upper half

plane, 8.45.

Two Riemann surfaces will be called analytically

equivalent if there exists an analytic homeomorphism between Recall (6.23) that two lattices

them.

said to be equivalent if there exists L'

=

L

and

c1 E

C*

L'

in

C

are

such that

aL. Theorem. C/L(- X )

and

equivalent if and only if Proof.

a e C*

Assume that

so that

L ' = aL.

C/L' L

L

and and

Let

(E X ' )

L' L'

are analytically are equivalent.

are equivalent.

f(z) : a z ,

for all

Let z e C.

Norman L. Alling

156

Consider the following sequences of maps, in the category of Abelian groups, that is row exact: (1) 0-

0-

Since

L'

of

onto

X

->x-h

>L ->

c

>L' ->

11 c>X'-->O.

aL,

=

f(L)

>O

L';

=

thus

Conversely, let

g

is an analytic map of

analytic map over h' g

f

0

0

= g

h(ztR)

(3)

f(z+P.)

Fix

R

into z

E

in (4)

of

XI.

0

h.

=

g

-

0

Let

R

E

h(z)

=

h'

L

L'

z

C

E

by

h',

and let 0

z

E

lation of

XI,

f(z) = az

and so

8.46.

then

C;

is in

L'.

+

for some

p(z,R) = p ( E )

a,B

a, E

f(z+!L)=

0

is an analytic map of

p ( z ,R)

a # 0.

df/dz

E

L',

CC

for all

is entire and

and hence C.

Modifying

g

by a trans-

if necessary, we may assume that L' = aL,

h'

As a consequence

is discrete

Since ( 2 ) is commutative

such that

f(z).

thus it is a constant

f(z) = a z + 8,

thus there is an

XI;

Using this and ( 3 ) , we see that

F(L);

X

the universal covering space

C,

X'

f(z1 : u ( z , ! L )

Since

into

into

C

and note that L'.

C.

f

as mapped down to

X,

is

>O

I

I

h

g

be an analytic homeomorphism of

>L L > C - > X - h

0-

0

g

onto

X

Clearly

g

Then we have the following row exact diagram.

XI.

(2)

induces a homomorphism

which makes (1) commutative.

X'

an analytic homomorphism of

onto

f

proving the Theorem.

$ = 0.

Thus

Riemann S u r f a c e s Any Riemann s u r f a c e o f g e n u s

Theorem.

equivalent t o

is analytically

0

1.

(See e . g . ,

[58,

Surfaces

8.50.

157

C h a p t e r 91 f o r a p r o o f . )

of

g e n u s one

I n 58.4 w e s t u d i e d

and n o t e d t h a t e a c h o f t h e s e

@/L,

We w i l l

s u r f a c e s i s a compact Riemann s u r f a c e o f g e n u s 1. s k e t c h t h e p r o o f of t h e f o l l o w i n g c o n v e r s e . Theorem. L e t

g e n u s 1.

b e a compact Riemann s u r f a c e o f

(X,X)

There e x i s t s a l a t t i c e

in

L

f o r which

@

@/L

( X IX I

are analytically equivalent.

8.51.

The p r o o f o f t h e g e n e r a l i z a t i o n o f Theorem 8.50 t o

and

a r b i t r a r y ( f i n i t e ) g e n u s o c c u p i e s much s p a c e i n t h e l i t e r a t u r e . (See e . g . ,

[58].)

followings

(1)

The s t e p s i n v o l v e d i n p r o v i n g 8.50 a r e t h e Since

A b e l i a n g r o u p of r a n k 2 . space X.

of

U

Then

X

nl(X)

(See e . g . ,

i s of genus

X

ism.) or

L.

(U,U)

on (5)

(4)

and t h e u n i v e r s a l c o v e r i n g map

p

l i f t s t o t h e covering group

of

[3, C h a p t e r 1 1 f o r d e t a i l s . )

c

U

on

U

(This i s e a s i l y seen s i n c e (U,U)

is a free

nl(X)

Construct t h e universal covering

(2)

a unique a n a l y t i c s t r u c t u r e analytic.

1,

(3)

G

p

of

u n d e r which p

p: U

and

onto

U

X

+

X.

induce

is

p

i s a l o c a l homeomorph-

is analytically equivalent to e i t h e r

@,g,

I n t h e s p e c i a l case u n d e r c o n s i d e r a t i o n i n Theorem 8 . 5 0 ,

i s analytically equivalent t o

C,

and t h e a c t i o n o f

i s t h e same as t h e a c t i o n o f some l a t t i c e Thus 8.52.

(X,X)

and

C/L

L

on

G

C.

a r e analytically equivalent.

I n a d d i t i o n t o b e i n g a compact Riemann s u r f a c e o f

g e n u s 1, @/L E X

i s a compact L i e g r o u p .

wonder how t h e g r o u p s t r u c t u r e o n

X

I t is n a t u r a l t o

is reflected i n the

158

Norman L . A l l i n g

f u n c t i o n t h o e r y on L-automorphic

or

XI

- e q u i v a l e n t l y - t h e t h e o r y of t h e

f u n c t i o n s on

i n p a r t i c u l a r how i t i s re-

@;

f l e c t e d i n t h e p r o p e r t i e s of t h e % - f u n c t i o n . For

(11

u

+(u)

thus

let

X

E

c

b ' (ul)

( b ( u )I

2

;

i s a n a n a l y t i c map of

$

X

into

X

be t h o u g h t of a s t h e g r a p h o f compact, and t h u s i s c l o s e d i n

in C

2

Z

C

2

2

.

.

Let

r

$(X)

f

r

Clearly

is

.

Lemma.

$

i s i n j e c t i v e ; t h u s i t i s a n a n a l y t i c homeo-

morphism o f

X

onto

Let

Proof.

uo

W e must show t h a t

denote

X

(having kernel

n

and

u1

uo = u l .

-

- a s usual

h

of

r.

L)

.

h(z.1 = u 1 1' and P ' i n

for

wo

E

w1

+

w1

z

t h e period parallelogram

and

thus i f

and

0

z1

zo

E

P(R)

P I

i s a pole of

X

such t h a t

i s t h e only pole of

0

1.

1)

then

uo - u l . and

p

are r e g u l a r a t

be i n

-

P(R)

z

such t h a t

{O)

w

and

0

i t s two p o l e s i n 2.0

wo

then, of course,

then

-

onto

Q:

0

and

zl;

9 (w,)

=

c.

wo = w1

P(R) ,

and l e t

L

# 0 # zl.

zo

and

wo # wl;

w0

P(R);

and hence

p(wll :x If

j = 0

Assume t h a t 9

Let

be a b a s i s o f

P(R) I

Since

w e may c h o o s e

then

$(ud=$(ul).

t h e c a n o n i c a l homomorphism o f

mod L ;

(2)

R

Let

such t h a t

i s a set of c o s e t r e p r e s e n t a t i v e s of

(6.20:6),

z o = z1

X

be i n

Z

R

E

L;

w1

(mod L ) .

a r e t h e two z e r o s o f

P(R) thus

w1

being a t

w1

Z

-

wo

0.

Assume t h a t p

- X

in

By Theorem 6 . 4 3

(mod L ) .

Thus w e

159

Riemann S u r f a c e s

h a v e shown t h a t

wo

given

(3)

wo : ?r w1

then Since If

w

and

z1

= ~ ( z , ) .

BY ( 3 )

(mod L),

then

z o = zl,

proving t h a t

-

z1 (mod L ) .

P' (z,)

are h a l f p e r i o d s of z

hence

j

sequence

:- z

j z o : z1

is injective.

4

space,

= 'p' ( z , )

:

L:

i.e.,

(mod L )

,

for

(mod L);

z . .E 0 7 j = 0 and

hence

1.

uo = ul,

i s a homeomorphism o n t o

T,

$

i s a n a n a l y t i c homomorphism o f

kernel Clearly E

0

$(O)

r

Ip' ( z , ) , z

z1

and

0

and

A s a con-

4

i s a Hausdorff

p r o v i n g t h e Lemma.

Let

C

4

induces t h e

$ E $

onto

r

0

h;

then

having

2

i s t h e i d e n t i t y element of

(m,m)

R;

in

r.

i s of t h i s f o r m .

k = 1 , 2 , o r 3. (5)

=

be a h a l f - p e r i o d o f

in

-

L.

element o f o r d e r 2

Since

proving t h a t

EL

r.

s t r u c t u r e o f a compact L i e g r o u p o n

1 (mod L ) . uo = ul. z

(mod L ) ,

U s i n g t h e Lemma j u s t p r o v e d w e see t h a t

(4)

=

(z,)

+

j

2

(2,).

By Lemma 7 . 3 3 ,

i s compact a n d

X

Since

z

9'

thus

s

zo

p ' (-zl) = - p '

=

' p ' ( z o ) = 0 = 9'(z1).

proving t h a t

I

i s a n odd

9'

Since

b ' (z,)

L-automorphic f u n c t i o n , $(uo) = 4 (u,),

l)(Wl)

(mod L ) . P(z,)

zo :

Assume t h a t

'D(w0) =

such t h a t

P(n)

,

$(uol = $(ul)

zo E

in

1

then

2~

E

L.

Hence

$(E)

k = 1 , 2 , o r 31.

r

Let

i s an

C o n v e r s e l y , a n y e l e m e n t of o r d e r ?'(E)

= 0

and

p ( ~ =) e k ,

Thus

t h e elements i n

r.

of o r d e r

2

are

{ (ek,O):

for

Norman--L. A l l i n g

160

z1

Indeed, t h e r e e x i s t s

-

- zl)

P1 = qi(

(X1f

- Y1)

(8)

Let

= ( Q ( - z ) , 1\

PI

and

such t h a t then

+

P1

P

P2

1

+

r

$2.)

= P

(9)

Assume f i r s t t h a t

z

z1

(12) b

z2

i

n

@

2

P1

-9'( z 1 ) = 1

i s odd, p r o v i n g

9'

r

E

,

(7).

n C 2 - {(ek,O): k = 1 , 2 , 3 ) ,

. +

Our a i m i s t o d e s c r i b e t h e P2.

Choose

z

j

E

such t h a t

C

?)

x1 # x 2 :

# v ( z 2 ) . Using ( 3 ) w e see t h a t

(z,)

(mod L ) . I\

m

(11) L e t

(13)

= (WZl)

j'

i . e . , assume t h a t (10)

(-zl))

P2 # 0 ;

Cartesian c o o r d i n a t e s of 3

6 ( z l ) = P1.

such t h a t

L

(x2,y2))

(:

is i n

P2

I

-

1 i s even and

s i n c e 'p

I

@

E

-

y1

5

mxl(

-

= q'(z,)

A i {(x,y)

Let

' (z,) - 3 '

E

(2,)

and l e t

E @,

mv(zl)

E

C!).

C2: y = mx+bl. n

A

i s t h e n a (complex) l i n e i n

and

are i n

P2

(14)

on

X

x (E h ( 0 ) ),

at

0

ord Z = 3.

zeros of

-

(rnq(z) + b )

i s a n L-automorphic

Q

X.

on

a n o t h e r zero i n

X

Since

of

z1

+

z2

+

z

3 -

P1

0

for all

z

E

Q;

h(zl) Q

C.

f u n c t i o n which h a s i t s p o l e s

and

h(z2)

i s of o r d e r

c a l l it

By Theorem 6 . 4 3 ,

(15)

,

t h a t pole being three-fold:

By c o n s t r u c t i o n Q

One e a s i l y sees t h a t

r.

p(z) :v'(z)

Let

Clearly

A n

CL.

(mod L ) .

h(z3),

thus

are d i s t i n c t

3 there exists f o r some

z3

E

@.

161

Riemann S u r f a c e s

$ ( z 3 ) :P 3

Let

C

2

.

(x3,y3).

+

P1

P2

+

t h u s , by ( 8 ) ,

P3

c ( x ) : 4x3

(17)

+

P1

Clearly

thus

P3

is i n

c (x);

r;

A n

-

P2 =

P3;

w i l l b e a c c o m p l i s h e d by f i n d i n g t h o s e

P2

Now c o n s i d e r t h e p o l y n o m i a l

(7)).

-

g2x

+

P1

Our a i m o f f i n d i n g t h e C a r t e s i a n

-

g3

i s of d e g r e e

c(x)

is i n

-

.

n C2

r

E

( i n view o f

P3

i.e.,

P3 = 0 ;

c o o r d i n a t e s of

P3

# 0;

P3

From ( 1 5 ) w e see t h a t

(16)

of

By ( 8 )

thus

(mx+bI2 3.

@[XI.

E

Since

x1,x2,

and

z

3

x3

i s a z e r o of

Q,

a r e t h e r o o t s of

thus

(18)

C(X)

E q u a t i n g t h e q u a d r a t i c terms o f

x1 + x 2 + x3

(19)

Note t h a t

x

3

=

2

m /4,

.

( x - x,)

= 4 ( x - x,) ( x - x 2 )

or

= 9 ( z 3 ) = 9(-(z1

( 1 7 ) a n d ( 1 8 ) g i v e s us

x

3

-

=

+ 2,))

- x 2 + m2/ 4 .

x1

= l) ( z l

+

thus (19)

z2) ;

translates t o

a ( z1 + z21

(20)

This is

-

=

-

of course

- a(z2)

co(Z1)

-

+

1 9 '(z2)

(

assume t h a t

x1 # ek ,

By ( 8 )

x1

# x2,

for

k = 1 , 2 , o r 3;

(22)

m Z I \ " ( z )/cD'(z ) 1 1

(23)

b

(24)

Let

y1

-

mxl

P(Z,)

j..

and

x1 = x 2 ' thus

let

2

-

(2,)

'(zl)

W e i e r s t r a s s ' A d d i t i o n Theorem ( 7 . 5 0 : 1 6 ) .

L e t u s now d r o p a s s u m p t i o n ( 9 ) t h a t

(21)

P

-p

C,

E

(=PI (z,) -

and l e t

m'R(zl)

E@)

.

n

A :{(x,y)

E

L

C :

y = mx

+

b}.

a'(z,)

# 0.

Now

162

Norman L . A l l i n g

Pl = P2

Clearly (25)

Let

Clearly

r.

-

V'(z)

Q(z) 5

(mn(z) + b ) ,

i s a n L-automorphic

Q

-

3" ( z )

A n

E

m7)' ( z ) ;

z 1 , z 2 and z 3

least

2.

Since

Q

(26)

2z1

+

z

(27)

2P1

+

P3 = 0 ,

Let

:0

(mod L); or

Q(z)

be t h e z e r o s o f

has a t r i p l e pole a t

3

E

@.

f u n c t i o n of o r d e r 3 .

i s a z e r o of

z1

thus

for all z

Q'(z)

=

of order a t Q(z),

mod L .

0, thus

P 3 = -2P1.

As beforelwe reach t h e following

This

-

-

of course

[ 6 4 , vol. V , p .

i s (7.50:17)

,

d u e t o Weierstrass

2181, and i s d i s t a n t l y r e l a t e d t o F a g n a n o ' s

r e s u l t s on doubling t h e arc In general,

-

P3 =

P1

l e n g t h of t h e lemniscate $2.3.

-

P2;

t h u s , using (71,

Then ( 2 0 o r ( 2 8 ) may b e u s e d t o e v a l u a t e t h e r i g h t hand s i d e of

(30). A g e o m e t r i c i n t e r p r e t a t i o n o f what h a s been done i s t h e

following.

Given

P1

and

P2

A

c o n s t r u c t a (complex) l i n e which then

P1

then

P

of

A

and

P1

+

1 and

P2 and TI

P2

+

lie.

P3 = 0 , P2 and

A

in

r

n @

in

,

s a t i s f y i n g (81,

( b y ( 1 3 ) o r ( 2 4 ) ) on

r

intersects

i n t h e group

2

r.

a t one p o i n t If

x1 f x 2

p3 ( 9 ),

a r e d i s t i n c t s i m p l e p o i n t s of i n t e r s e c t i o n

A

i s t h e l i n e d e t e r m i n e d by

P1

and

P2.

Riemann S u r f a c e s If

x1 - x2

A

( 2 1 ) then

The

let

divX,

+

Elements of

x

Given

a > b

X

X

divX

E

if

be

X.

Clearly

2

b(x),

for a l l

x

E

{Xx:

x

supp(a),

XI

E

i s a f r e e b a s i s of

t h e s u p p o r t of

deg(a)

Let

Xx(y) = 6 x y l

is

a,

ix

for a l l

divX Given

X;

then

x,y

a

divX.

Given

X:

a ( x ) # 01.

E

deg.

divX

Divisors i n

Clearly

divX/div X

for a l l

x

0

E

X.

E

divX

Ixcsupp(a a )( x ) I

i s a homomorphism of

k e r n e l of

X.

E

is finite.

where t h e sum o v e r t h e empty s e t i s d e f i n e d t o b e z e r o . deg

XI.

i . e . , i t i s a Z-module.

a(x)

such t h a t

By d e f i n i t i o n , s u p p ( a ) (4)

and

X

E

let

Clearly (3)

XI

g,

i s a p a r t i a l l y o r d e r e d ( e v e n a l a t t i c e - o r d e r e d ) group.

divX

(2)

let

divX,

r.

and

w i l l be c a l l e d d i v i s o r 0”

divX

i s a t o r s i o n f r e e Abelian group: E

Pl

i s z e r o e x c e p t on a f i n i t e s u b s e t of

a

Z:

and

P1,

d i v i s o r class g r o u p

t h e group of d i v i s o r s on

(1) { a : X

a,b

A

at

b e a compact Riemann s u r f a c e o f genus

X

Let

r

i s tangent t o

i s a d o u b l e p o i n t of i n t e r s e c t i o n of 8.60.

163

X

Since

Let

2.

divOX b e t h e

divOX w i l l be c a l l e d homogeneous.

Given

Z.

2

onto

Clearly

f

E

i s compact

F(X)* f

let

( f ) (x)

f

vx(f),

can have o n l y a f i n i t e

number of z e r o s o r p o l e s ; t h u s

(5)

f

E

F(X)*

+

(f)

divX

E

i s a homomorphism of t h e m u l t i p l i c a t i v e group a d d i t i v e group (6)

into the

divX.

divX/(F(X)*) 5 C ( X ) Lemma. _-

F(X)*

(F(X)*)

c

i s t h e d i v i s o r c l a s s group of divoX.

X.

164

Normal L . A l l i n g

F7e w i l l p r o v e t h i s lemma i n c a s e

or

g = 0

The

1.

p r o o f i n t h e g e n e r a l c a s e may b e found i n t h e l i t e r a t u r e on t h e subject (7)

.

d i v o X / ( F ( X )* )

i s t h e homogeneous d i v i s o r c l a s s

E Co(X)

We w i l l compute

i n case

Co(X)

g = 0

and

1

i n the

n e x t two s e c t i o n s . See e . g . , X

g r o u p on

[29]

f o r c o n n e c t i o n s between t h e d i v i s o r c l a s s

and c e r t a i n s h e a f cohomoloay g r o u p s on Assume

8.61.

then w e have seen t h a t

g = 0;

a r e analytically equivalent (8.46).

Co(C)

Theorern.

=

Let

a

that

a # 0.

Let

S :s u p p ( a ) .

then

g

and

S'

point i n

E

and

F(Z)*

supp(b)

the points i n

d i v 1. 0

If

Let

b

then

a = ( 1 ) . Assume

n :a(m)

b :a

does n o t c o n t a i n

a.

+

and l e t

is i n

(g)

b

g + zn;

div C 0

xl,...,x

Let

be t h e

k yl, . . . , y k

i s p o s i t i v e , and l e t

o n which

S'

C

( F ( L ) * ) c divoC.

a = 0

( 9 ) = nXm.

on which

S'

and

COI.

Proof.

E

X

W e have seen ( 8 . 3 0 ) t h a t

I n ( 8 . 3 0 ) w e a l s o see t h a t

F(C) = C ( Z ) .

X.

be

i s n e g a t i v e , each o c c u r r i n g

according t o i t s m u l t i p l i c i t y . (z-xl)

(1) L e t then

f (z) :

( f ) = b, 8.62.

Let

(z-y,)

.. . . ..

and h e n c e L

(2-x,)

(z-y,)

E

F(C)*;

(f/g) = a ,

be a l a t t i c e i n

p r o v i n g t h e Theorem.

C

and l e t

complete w i t h t h e induced group s t r u c t u r e . o f genus 1.

By Theorem 6 . 4 1

X

is

X

2

-

of c o u r s e

C/L,

-

165

Riemann S u r f a c e s

(1)

(F(X)*)

(2)

Given

(3)

Define

divOX

c

a

divoX,

E

sum(a)

a = Cxcsupp ( a a ( x ) X x -

t o be

CxEsupp(a

C l e a r l y sum i s a homomorphism o f

a ( x )x .

divOX o n t o

X.

By Theorem

6.43 (4)

(F(X)*)

k e r sum.

c

By Theorem 7 . 4 1 , (5)

given

a

( f ) = a;

E

k e r sum,

i.e.,

there exists

f

E

F(X)*,

such t h a t

( F ( X ) * ) = k e r sum.

Thus w e h a v e p r o v e d t h e f o l l o w i n g . Theorem.

onto

X.

sum i n d u c e s a n isomorphism, Sum, of

c,(X)

This Page Intentionally Left Blank

CHAPTER 9

THE ELLIPTIC MODULAR FUNCTION

Introduction

9.10

We have seen that parameters occur in elliptic integrals. For example, Legendre's modulus Chapter 1. functions

Two parameters, c 54.1.

k

occurs in some integrals in

and

e l occur in Abel's elliptic

Legendre's modulus again occurs in Jacobi's

elliptic functions 94.2.

Clearly

(5.20:l) is a parameter

T

that plays a role in defining theta functions Further, a lattice and

w2,

and

L

- in

L

in C

in Chapter 5.

is determined by parameters

turn - determines

1\

and

O1

7'.

One way to describe the problem at hand is the following: find a complex number associaked with lent to

p(X),

X -C/L,

X'(x/L')

the modulus of

such that

if and only if

X

XI

which can be

is analytically equiva-

p(X) =p(X').

This problem

is addressed and solved by using the elliptic modular function J,

and is the subject of this chapter.

eventually, that

J

We will also see,

is intimately connected with the parameters

mentioned in the paragraph above. Klein traces the origins of the elliptic modular function back to work of Gauss.

(See [40, p. 43 ff.].)

Klein himself

did a great deal of work on the elliptic modular function.

(See

the third volume of his collected works and the huge two-volume work, written with Fricke, on the elliptic modular function 1411.) 167

168

Norman L. Alling Finally note that "modular" was used in one sense prior

to about 1850 and in another after about that date. Weierstrass remarked [64, vol.1, p. 501

As

in a note in his col-

lected works written near the end of his lifeton his first papers, the term "modular function" c. 1840 referred to what we now call "elliptic functions". Let

9.11

be a lattice in c ,

L

let

n e N,

with

n> 3,

and recall (7.33:5) that the Eisenstein series (1)

L&-n

Zt

converges absolutely and is defined to be

sn (L) Let

a

E

c*

and let

lattices in c (2)

L' :aL; then

(8.45).

are equivalent

Clearly

g2(L) = 6 0 s4(L)

and that

g3(L) =

thus -4

g2(L') = a

g2(l)

-12

(4)

A(L')

=a

(5)

Let

J(L)

(6)

J(L') = J(L).

Now let

equivalent to Proof.

g3(L') = a

-6 g3(L).

2 - 27g3(L)

(7.33:22); thus

A(L). 3 g2(L)/A(L);

L

and

X ' :C/L'.

Theorem.

and

3 A(L) :g2(L)

Recall also that

and let

L'

.

Recall (7.33:18) that

(31

and

= u-ns, (L)

Sn(L'

140 s6(L)

L

be any lattices in C ,

let

x i C/L

Then we have L

If XI)

L'

then

,

is equivalent to

then

J (L)= J (L')

L'

(resp. X

is

.

( 6 ) gives the non-parenthetical result.

Theorem 8.45, the parenthetical result is proved.

Using

The Elliptic Modular Function

169

Finally note that

Let

9.12

L

L

be a lattice in C .

is a discrete subset of iI:

Since

by definition

there exists

E

L* (EL - C O I )

such that (1)

Clearly

for

1

E

L*.

has the same property.

-wl

Among all elements of

(2)

1 1 1,

is minimal among all

lull

L-Zwl,

choose

so that

w2

Iw2)

is minimal. Let

(3)

T

Note that, without loss of generality,

:w2/w1.

we may assume that Lemma.

R

Proof.

Let

Since

T EIR.

subset of

€6,

) is a basis of L. 1 2 L' :Zol+Zw2. Assume, for a moment, that

5 (w w

L

is a discrete subset of C ,

mul.

By the choice of

TPE.

absurd, proving that

R

T

is a basis of

L'.

L'

is a discrete

wl(l), L' = Zol; which is

Hence

L'

is a lattice in C

and

Assume, for a moment, that there exists

mcL-L'. ( P ( f l ) + u 1 L'

(4)

is a partition of C ;

thus, there exists a unique m o E m - 1 ; then

for

x

and

x , y ~(0,l).

y

mo

in

E

P(R).

roll).

By (1) and ( 2 1 ,

1 E L' As

such that

and w 2 .

Since lull

mPL,

P(R) + 1 . Let

zIw21

L' = L ,

Hence

mo#O.

(Imol;

thus

mo

can-

whose vertices are

As a consequence m l - w1 + w 2 - m o

is absurd; proving that

E

a consequence mo=xwl+ywZ,

not lie in the (closed) triangle T 0, w,l

m

E

LnTI

proving the lemma.

which

170

Norman L. Alling

R of

satisfying (1) and ( 2 ) will be called a minimal basis

L.

If, in addition, it satisfies ( 3 ) it will be called a t minimal positive basis of L. Let L: Z + i Z ; then (1 i) (i - 1)

and

are both minimal positive bases of

a minimal positive basis then so also is of

Let

9.13

L

be a lattice in

be a minimal positive basis of T

Let

E & .

L' 5wi1L

minimal positive basis of a basis of

and

L

L'

Proof.

Since ( 9 . 1 2 : l and 2 ) hold for

1' 1 1;

T

-1

would be in

Re T = t 1 / 2

are equivalent

L'

and

a',

' 1

IT-11

]TI.

Were

would be

then

(1

T

-1/2 < ReT.

7 l ) t is also a normalized

L';

can always be chosen so that

1 / 2 < Re

Note:

is such

which is prohibited by ( 9 . 1 2 : 2 ) , proving that

minimal positive basis of

-

R'

T.

Similarly we can show that

T

is

-1/2 < R e 'I < 1/2.

then

thus

R'

.

and

If

then

clearly

wl=l;

12 I T [

T >1/2,

t)

will be called a normalized

The following hold for

Re T <1/2.

(3)

if

L

w )

1 2

as usual;

T

Lemma.

less than

(2)

Let

R

L'.

R(-(w

rw2/w1, R' : (1 T ) t ;

L.

Note also that

L'.

lattices ( 6 . 2 3 )

Re

is

thus such a 'basis

and let

C

and let

a minimal positive basis of

(1)

R

If

is not unique.

L

then

-R;

L.

T

5

1/2.

we will adopt this convention.

Although it is not

standard it seems to give slightly more appealing results for real elliptic curves. (4)

If

T

€ 4 with

[TI

=1

and

-1/2

< Re T < 0

let

171

The Elliptic Modular Function L" :?L'

and let

:-?

T"

E

4; then R"

normalized minimal positive basis of

IT"I

Note: (5)

and

=1

L"

and

L'

5 1/2

-

are

of course

Corollary.

to a lattice

L'

1 ~ 1

and

.

L"

T " ) ~is

a

Further

< 1/2.

T"

D Z { T cf2t- 1/2 < Re T < 0

Let

0 < Re T

T E

0 < Re

(1

5

-

equivalent.

IT(

and

> 1,

or

1).

Given any lattice

L

which has a basis

R'

in C =

it is equivalent t (1 T ) , for some

D.

Definition

9.20

(1)

For

then

LT

(2)

Let

T E

0,

let

and

elementary properties.

LT:

2 +TZ,

is a lattice in

and

and let QT

(1

QT

T)~;

is a positive basis of

J ( T ):J ( L T ) .

J ( T ) is the elliptic modular function, one of the most inter-

esting €unctions in analysis.

For

9.21

(: :)

ME

E

GL2(c)

and

z E C,

recall

(3.13 :2) that

(1)

h(M) ( z ) z (az+b)/(cz+d) Let

Lemma.

(2)

J(T)

=

and let

Q' 5 M ' R T

Let

MI

5

t

(: .

z).

M E SL2 (Z);

LT.

5 (w'

(By (9.11:6),

then

M I is in SL~(Z). -1 I," :( u i ) LT, and let Clearly

') Let 1 w2 z o'/w' 2 1; then L" is a lattice in

Let

to

CQ

J(h(M) ( T I 1 .

Proof.

T'I

T

E C.

J(L") =J(LT).

which is equivalent Since

detM'(=1) > 0,

172

R'

Norman L. Alling

is positive (Theorem 6.22) )

L" = LTll. Clearly

that

hlSL2(Z)

{+I1 ; (4)

Q

and hence

proving the Lemma.

We have seen (3.14)

be called the modular group.

?I

and let

h (m)(T) ,

E

TI'

r :h(SL2 (2)) ,

Let

(3)

T" =

thus

;

is a homomorphism into

conQ

having kernel

thus

r

is a subgroup of S L (~z ) / { t 1 1

=

conQ

that is isomorphic to

P S L (2) ~ ,

the projective special linear group of rank 2

(2)

2.

may be rephrased as follows (5)

J ( T )= J ( g ( T ) )

thus

J

,

for all

T

E Q

and all

g

E

r;

is invariant under the action of the modular group on

63.

9.22

Let

Lemma.

n

E

N,

with

n13;

then

T

E

Q > -

s

n (L7 )

is analytic. Recall ( 7 . 3 3 : 5 ) that

Proof.

(a,b)# (0,O) ,

such that

a,b E 2 (1)

T

Let

E > O

E

p, >-

(atb?)-"

and let

QE:

1;

(GT)

5 8/Enkn'll

then clearly

is analytic. C T E Q : Im

notation of Theorem 7.31, h)~, (2)

sn(L) E CLEL* l-n. Given

T > E ~ .

Let

T E Q ~ .

and hence, by (7.31:6),

for all

k

E

N;

(3)

(4)

Since

Q = u

E>O

Using the

QE

we can combine (1), (3), and ( 4 ) to prove the Lemma.

173

The Elliptic Modular Function Since

g (L)= 140 s (L), we may use 3 6

g2 (L)= 60 s4 (L) and

the lemma to prove the following. Theorem.

analytic, for

t

gi (LT), A (LT

Q-->

E

j= 2

and

Let

t(z) s z + 1 ,

each

n E Z.

(2)

Let

then

h(T) = t,

(3)

T ~ =1

(o

n) 1

,

(4)

Let

S

(y

):-

(5)

(6)

1>

(o1

T-

S E

1

E

thus

SL2(Z),

t

1

(9.21:3).

E

T'.

nE

tn(z) = z

+ n,

for

z.

and let 2 S =-If

8

and

Clearly

Clearly

for all

Let

r

4.

of

SL~(Z);

for all

z

s : h(S).

and

s

€4,

and

be real, with

E

r. s

2 =l.

r>O

and

O < 8
reie

and

s (reie)= r-lei

Note that

t

€4.

1 0 ) = -r-'eie; s (re'

across C1 0 ) Clearly

r

~ € 4 ) .

for all

s ( z ) =-l/z,

is in s)

J ( T ) are

3.

is just translation by

Clearly

and hence

Now let us study the modu ar group,

9.23

(1)

T

n

8.

hence

(Note: that

has no fixed points.

-s

is reflection of s)

C1(0) = { z E C :

i

121

=l}.)

is the only fixed point of

S.

(7)

-ST-lS = (11

(8)

Clearly

0 1) :U

U m = (m1

E

0 1)

SL2(Z) ,

,

and let

u = st-ls.

for each m e Z.

174

Norman L. Alling

(9)

Note that Um =-ST-mS, and

for each m c Z.

.

T

generate S 1 2 (Z)

Lemma.

S

Proof.

Let

G

and

T.

(We will show that

G.

Hence A E G

erated by

S

S

2 =-I, -I

U

is in

is in Let

G.

(11) For each

be the subgroup of

kcN

nE

2,

and let Ak

SL2(Z)

that is gen-

G = SL2 (Z). )

implies -A be in

Since

is in

thus

SL2(Z).

b +ndk k dk

TnA k =

G;

SL2(Z) I

Consider the following statements: ( 1 3 1 k ) akCk # 0 1 (141’)

Iakl

2 lCkl,

and (15,k)

lakl < Ickl.

Clearly if (13,k) holds then either (14,k) or (15,k) holds. (16,k)

If r

E

(13,k) does not hold then there exists a unique

Z

such that Ak

is either

Indeed, since det Ak = 1 , Ak

(i f)

or

f

(y

-:)

,

t

Tr

E

G

or

*

STr E G.

is either establishing (16,k).

(17,k) Assume that (13,k) and (14,k) hold.

algorithm we can find unique n E 2 n T Ak :Ak+ll then 0 2 ak+l < ICk I

Using the Euclidean such that if

.

The Elliptic Modular Function

Note that

c

175

~ = ck; + ~ thus on applying (17,k), (15,k+l) holds.

(18,k) Assume that (13,k) and ( 1 5 , k ) hold.

Using the Euclidean

algorithm we can find unique n E Z such that if n U Ak:Ak+lr then 0 5 c ~ '+ lakl. ~ Note that

ak+l = ak;

thus on applying (18,k)I

(14,k+l) holds.

Having set up the several reductions of the algorithm, let us let it run.

Let

A1c SL2(Z).

(We must show that

Assume first that (13,l) holds.

G.

A1

is in

If (15,l) holds, replace

by SA1, for which (14,l) holds. Thus we may assume A1 wi hout l o s s of generality - that (14,l) holds for A1. Applying (17,l) we arrive at

to

steps

Ak

E

G

la21 > Ic31. Ak

I al I 2

such that

If (13,2) holds, apply (18 2 ) to define

thus (15,2) holds. such that

A2

By finite induction we come (in 5 /all

for which (13,k) does not hold.

thus Ale G I

A3

By ( 1 6 , k ) ,

proving the Lemma.

From this we see immediately that the following holds. s

Theorem.

and

t

r.

generate

(19) then

Q

is of order 3.

Let

and it also is of order 3.

0

q: st-'.

Then

q(z) =-l/(z-l) ,

Clearly the fixed point of

q

is

(20) Since

p :eiTl3,

a cube root of

SQ = - T - ,~ sq = t-',

and we have the following.

and

q

generate

Two elements

z

and

Corollary.

9.24

s

-1.

congruent mod

r

if

w

r.

in !$

will be called

in

176

Norman L. Alling there exists

(1)

r

Since

g

E

r

such that

g (z) = w.

is a group, this is an equivalence relation. Recall

( 9 . 1 3 : 5) that (2)

D:

-1/2 < Re T < 0

{TEQ:

0 < Re T < 1/2

For each

Theorem.

congruent to Proof.

n(z(w

w ) 2

1

t)

Theorem 6 . 2 2 , T :w2/w1;

QT ( 9 . 1 3 )

then

mod

T

Let

TI

E Q there exists

of

R' : (1

and

E

T I )

thus

TI

and

T

T ~ = ~ ( T ) ; then

If

-1/2

T

that is

and let

L' ( 9 . 1 2 ) .

thus

E

By

n=AQ'.

Let

clcD.

S(T) : T

-

r.

If

-r0

are

= 1,

then i(T-6) D. ~= e

proving the Theorem.

TI

G L 2 ( C ) ->

It would

and

T'

IT^

but

above (3) is an autornorphism of order 2 of SL2(Z).

E

T ~ E D , and

are congruent mod

Note, in passing, that A

an automorphism of

D

are congruent mod

< Re T < 0,

~ / 2< 6 < 2 1 ~ / 3 ; T~

t,

such that

SL2(Z)

By Lemma 9 . 1 3 ,

(w1)-'L.

r.

E

is a normalized minimal positive basisl

(w1)-'O

let

-r0

I'.

there exists A

congruent mod T = ei6 , with Clearly

11.

L' - L T l l let

T = h(B) (T') ;

Re T = - 1 / 2

T'

1, or

be a minimal positive basis of

then

,

IT^ 2

and

1 ~ >1

and

A' :B ,

that induces

GL2(C)

of course

as defined

-

be pleasant

if it was not necessary to employ this automorphism. This could and writing a normalized be accomplished by letting T E w1/w2 basis as ( T 1)t The reason we have not done this is that it

.

is at variance with most of the classical literature. We have continued to use the classical conventions

-

in most cases

-

that e.g., [36] can be used virtually without modification.

so

The E l l i p t i c Modular F u n c t i o n

rr

Clearly

Let

(3)

then

A -

+

C T ~

ri

implies

f

cx

2

+

g = s,

-bE

(d-alx

(9.23);

rp

thus

Let

s

E

If

c#O

may assume

c#O.

a =d

thus

i s then a root

i and

a=O.

b = -c.

Hence

Since

A=*S,

p :e iT/3

and

qL

i s t h e f i x e d p o i n t of

rP.

are in

g c

-

f ( x ) Zcx thus

{lj; 2

+

then

l=ad-bc,

+

c#O.

Note t h a t

(d-a)x-bEZ[x];

a-d=c=-b.

w i t h o u t l o s s of g e n e r a l i t y

Since

a 2 - ac

rP -

then

Since

-

that

p

and

h(-A) = g , c>O;

thus

w e see t h a t

(c2-1) = 0 .

Solving ( 6 ) , u s i n g t h e q u a d r a t i c formula g i v e s

Since

a

then

ri.

c = * l and

that

l,q,

f ( x ) = 0;

satisfy

(6)

rT:

= {ilq,q2}.

p2-p+l=O.

a >d.

!g E

establishing ( 4 ) .

Indeed, l e t

-p

g=1.

then

Z[xl :

c#O,

Recall (9.23:20)

(5)

and hence

Clearly

r i - 11);

gc

and s i n c e

det A = l

q

h(A)

= 0.

A = + I,

Q.

such t h a t

= {~,sI.

f (x)

and so

-b

(d-a)r

Indeed, l e t of

SL2 ( 2 )

E

i s quadratic over

(4)

It i s called the isotropy

T.

c=O

Clearly T

at (: :)

r

subgroup of (2)

r.

i s a subgroup o f

177

i s real

c2 < 1: i . e .

I

c = 1.

Hence

b = -1 I

and

we

178

Norman L. Alling

ad=0.

If

a=O

then d = - 1 and g = q . 2 g=q establishing (5).

a = l ; hence

9.26

then

I

Since

is analytic on

J

variant under the action of

0

d=O,

~f

is the strip

TI

and since it is in-

it is periodic of period

(1) (6.50:l).

So,+m

Q

Thus

J(T)

1.

has a Fourier

series expansion

J(.r)

which converges to

uniformly on compacta of

Q.

(See

6.5 for details.) Let (3)

then

.

w - e2 n i ~I

a wn

m

ln=-m

converges uniformly on compacta to the

analytic function J(w), AOl1 Z { z Clearly

E

0 <

Q;r:

J (w) = J ( T)

<

1).

.

J(w)

Theorem.

IzI

on the annulus

has a simple pole at

w = 0.

Further,

a-l = (12)-3 (=1/1728). Proofs of this very well known theoreml whose methods will not be of further use to us, can be found in the literature. (See e.g.,

[36, p.2233 or [54, pp.393-3941.)

An important consequence of this theorem for us is the following. The limit of

Corollary.

is

J(r),

as

goes to

+m,

as imbedded in

1.

m.

It is now convenient to think of The

Im(T)

.o se

{T

E

@:

ReT=-1/2]

point in common in

C,

and

namely

D

{ T E 55: m.

J

Rer=1/2}

have a

may be extended to

The Elliptic Modular Function clCD by letting

J(m)

be defined to be

tended, is continuous on

9.27

onto

r

Theorem.

E

179

then

w;

so ex-

J,

clCD.

D u

{m)

->

J(T)

is a one-to-one map

C.

As remarked in 99.26, valued.

Let

CEC.

J(m) = m .

Clearly

JID

is finite

The theorem is proved by showing that

Very clear proofs of (1) can be found in the literature.

(See

e.g., 154, pp.396-3991, or 136, pp.223-2271). The following is a strengthening of Theorem 9.24. For each

Corollary.

that is congruent to Proof.

Let g

E

T

r

~

DE that is congruent to

such that

g(To) = T

A subset

'I

T

JID

E

D

.

I

T~

was assured in 59.24.

mod .'I

By Lemma 9.21,

Hence there exists J(.ro)= J( r l ) .

is one-to-one: thus

T o = Tl.

of 6 will be called a fundamental domain

S

if for all

congruent to

~

ro

r.

mod

T

there exists a unique

The existence of such

By Theorem 9.27

for

T E Q

T E Q

mod

there exists a unique

.r0

E

S

which is

r.

The corollary then proves that (2)

D

is a fundamental domain for

9.28

Let

(1)

r(z) = - z ;

then

r

z E Q

transforms Q

r.

and let

by rotating it about the y-axis.

continuous and antianalytic:

i.e.,

z E Q

->

r

is

is analytic.

Norman L . A l l i n g

180

-

Lemma.

J(r(?)= ) J ( T ) , for a l l

Proof.

Let

T

J(r(r))=J(r(LT)).

for a l l

(3)

n> 2.

r c Q .

@. C l e a r l y r ( LT ) = Lr ( T ) ’ L e t L S L . R e c a l l (7.33:5) T

thus

E

-r -2n ~( L ) ) =~CLeL* ( (-L)

~

Clearly

g j ( r ( L ) )= g j ( L ) ,

that

j =2

for

and

3,

and

A(r(L))

p r o v i n g t h e Lemma.

i s r e a l - v a l u e d on

C o r o l l a r y 1. J

+

(4)

{n/2

J(T)

= J (T) ;

iy: y > 0 1 ,

p e r i o d 1, then

and l e t

proving t h a t

+

J(n/2

+

J(1/2

y > 0

Let

Proof.

proving t h a t

Since

Since

n

for a l l

E

22.

J

r ( T ) =

r,

i s p e r i o d i c of

r ’ -= 1 / 2

Let

+

iy;

+ iy) = J(r(1/2 + i y ) ) = J ( 1 / 2 + iy)

iy) = J(-1/2

+

T !i y .

J ( r ) EIR.

i y ) clR

J(1/2

n c Z.

f o r each

iy) E D .

Since

i s p e r i o d i c of p e r i o d

J

1, ( 4 ) h o l d s . Recall t h a t

i s r e a l - v a l u e d on

C o r o l l a r y 2. J

(5)

C1(0) n

Q.

Proof.

Let

Clearly

r s ( T ) = ~ .

i s t h e u n i t c i r c l e w i t h c e n t e r 0.

C1(0)

T

E

C1(0) n 0 ;

Since

J

J ( T )= J ( r s ( T ) ) = J ( s ( T ) = )

‘r = e i e ,

with

is i n v a r i a n t under

m, p r o v i n g

a (clcD).

i s r e a l - v a l u e d on

4.

J

i s r e a l - v a l u e d on e v e r y image of

r.

g3(Li) = 0 ;

thus

J ( i ) = 1.

T.

the Corollary.

J

( 5 ) under

0 < 0 <

S,

C o r o l l a r y 3. Corollary

(6)

then

( 4 ) and

The Elliptic Modular Function

Indeed, since

iLi = Li,

s

181

6 6 (L.) 1 = s 6 (iL.) 1 = i s6 ( L ~ )

6 (Li) , proving that s6(Li) = 0, and hence that 3 (Li), and so J(i) = 1. As a consequence, A (Li)= g2 = -s

(7)

g2(Lc)

thus

0;

=

J ( p ) = 0.

Indeed,recall (9.23:20) that s (L

4

P

1 =s4(PLp) =

P -4s 4 (L ) = p

p=

e

2s (L 1 . 4

As

Corollary 5.

(L

4

P

goes around

T

C1(0) n D+,

(ii) around

1

s

s 4 ( ~ p+)o ,

Were

c

pLp = Lc' p2

proving (7).

) = O r

D+Z{TEQ: O< -R e . r-< l / 2 and l r l ~ l } .

Let

(i) J

thus

;

would be 1, which it is not: thus (8)

g3(Li) = 0.

aD+

(i) down

{iy: y > 11,

{1/2+ti: t > 31/2/2};

and (iii) up

is strictly decreasing from arbitrarily large values to

at i,

(ii) to 0 at

p,

and

+ J(aD )

negative values.

Thus

valued on all of

+ D - aD .

=

(iii) then to arbitrarily small

By Corollary 1 and 2,

Proof.

By Theorem 9.27,

JlaD'

and hence

IR,

J

J

is non-real-

is real-valued on

is one-to-one.

aD+.

By Corollary 9.26,

Lim J(iy) = + m and Lim J(1/2 + ti) = f a. By (6), J(i) = 1 Y+= t+m and by (7) J(p) = 0; proving the Corollary.

9.29

Let

C/L

Theorem.

and only if Proof.

equivalent. Theorem 9.11,

L

and

L'

and

be lattices in

C/L'

are analytically equivalent if

J(L) = J(L'). First assume that By Theorem 8.45, J(L) = J(L')

.

L

@/L

is equivalent to

to

T

and

where

r'

L'

C/L'

are in

LT

D;

are analytically

are equivalent.

Assume now that

L

I

and

and

Corollary 9.13, LT'

C.

and thus

J(L) = J(L'). L' J(r)

By By

is equivalent =

J(LT)

=

Norman L. Alling

182

J(L) = J(L') = J(LTl)= J ( T ' ) . By Theorem 9.27, and

L'

are equivalent. By Theorem 8.45

T = T ' :

C/L

and

thus @/L'

L

are

equivalent, proving the Theorem.

X

Let

XI

and

be compact Riemann surfaces of genus 1.

By Theorem 8.50, there exists a lattice L

X

and C/L

(1)

Let

in C

such that

are analytically equivalent. J(X)

J(L).

Using Theorem 9.29, we see that this definition is independent

L.

of the choice of

X

Corollary.

and only

X'

and

are analytically equivalent if

J(X) = J ( X ' ) .

of

Reflection

9.30

Since

J

across

is real valued on

J

Schwarzian reflection.

Since

D

8clCD.

acl D

c

it can be extended by

is a fundamental domain for

the elliptic modular function, J

so extended covers all of

0. Of particular interest in Part

111 of this monograph is the

behaviour of

J

on IRi n 0 and on

(1/2

+ IRi)

and

1 ~ >1 11:

n

0.

We will

obtain some such information now.

9.31

(1)

Let

+

D- u D = D D

in

D-:

{T E

and

(9.24:2).)

-

D

4: -1/2 < R e T < 0 n

D+=$.

(D'

then

is defined in (9.28:8) and

183

The Elliptic Modular Function

D

-

D+

2

P

4 + ti/2

4 + ti/ -3

+ ti/2 34

3

-4 (3)

of (2) in

is real-valued on all the lines and curves

0.

Let under

J

y > 0.

Note that

s(iy) =i/y.

Since

J

is invariant

s,

(4)

J(iy) = J(i/y) ,

(5)

thus

Y E

Y E

[ I r a ) ->

(0,1] ->

J(iy)

analytic map onto (7)

J(z) -1

(8)

Hence

y > 0;

for all

J(iy)

ing analytic map onto (6)

34

4

0

Note that

+ ti/2

is a monotone strictly increas-

[ I r a ),

is a monotone strictly decreasing [l,m).

From (2 we see that

has a double zero at

y=l

and

i.

is a relative minimum of

J(iy).

Norman L. Alling

184

Now consider relection across

9.32

of radius

1

1.

and center

C1(l),

the circle

This reflection is accomplished

by (1)

€ C ->

2

Under the map maps to 1/2

+

(1/2

Z / ( E - 1). C1(0),

+ IRi) ,

i/2 (3)

{1/2

(2)

+

Y E

i

maps to

+

1/2

i/2,

1

and center

and

p2

Since we know about the behaviour of

C1(0) n s) between on

the circle of radius

i

and

J(1/2

+

iy)

on J

is an unbounded, strictly

decreasing analytic map onto We know (9.28:7) that

J

we have the following:

ti: t > 31'2/21r

[1/2,m) ->

to

and about the behaviour of

p,

0,

J(p) = O .

J(z)

has a triple zero at

(3)

J' ( p )

(-m,ll.

From (9.31:2) it is clear that thus

p;

= 0 =J"(p).

Now consider reflection about

ClI2(1/2).

This can be

effected by z E C ->

(4) Let

- 1).

y > 0, and note that under (4)

(5) Since (6)

Z/(2'i

1/2 J

+

yi/2 ->

1/2

+

i/2y.

is real-valued on

J(1/2

+

yi/2)

=

J(1/2

ClI2(1/2)

+

,

(9.31:3) ,

i/2y).

From (2) we see that

(7)

Y E (0,1] ->

J(1/2

+

iy/2)

increasing analytic map onto (8)

Further, 1 for

is an unbounded strictly (-all].

is the absolute maximum of

y > 0, and

J'(1/2

+

i/2) = O .

J(1/2

+

iy/2),

The Elliptic Modular Function

Having raised questions about isotropy subgroups

9.33

r

of (1)

in 5 9 . 2 5 , Let

and

T

I

let us give a complete analysis of them.

Indeed, for

4

be in

and let r T I = g T T-1 g

T'

f E TT,

g-lf ' q :f;

q

r

E

such that

.

then

T ' = ~ ( T ) ;

and let

185

gfq-l E F T l .

then

then

f E TT

and

f'

Let

qfq-'=f',

E

rT,

establishing

(1).

Let

(2)

and let

T

Do

be the interior of

D0

E

.

rT= i l l .

(3)

Indeed, suppose

for a moment

There exists

is not 1. Do

that is transformed to

T~

E

U

Let

(4)

T

aD- {i,pj; then

E

Indeed, assume E

TT

-

{l}.

Since

real locus of a sub arc

orientation of

g,

aD

A.

But

J

aD,

U

about

T

that in

# T ~ ;

but

establishing (3). T T = {l}

that there exists

is invariant under

into

rT

where -r1

is mapped to itself under

J

of

A

by

for a moment

J

gc

there exists

in an open set

T~

this contradicts Corollary 9 . 2 7 ,

g

D,

fixing

T,

is monotone on

r

(9.21:2),

g.

Hence

the g

maps

and reversing the

A;

which is absurd,

establishing ( 4 ) . Thus we have the following, using ( 9 . 2 5 : 4 Theorem. p.

T i = {l,s}

9.40

Let

For and

TED,

TT={l}

T p = {l,q,q

L

and 5 ) .

if and only if

Tfi

or

1.

M o d u l a r functions

f

be a monomorphic function on

0

that is invariant

186

Norman L. Alling

r

under the action of

8.

on

there exist constants c

(1)

for all Let

F(r)

y

im(z)

is a modular function if

f and

k

such that

greater than some

If

(2)

yo EIR.

be the set of all modular functions. Clearly such an

.

F(r)

element in

C

(2)

is a simple transcendental extension of @ .

C(J)

(J) is in

Let

Proof.

f E ~ ( r ) . f maps

invariant under the action of D,

Further, clearly

F ( T ) = C (J).

Theorem.

on

r,

a fundamental domain for

function F

on C

such that

8

and since J

r,

.

f = F (J)

F

thus

f

is

is injective

(1) and Theorem 9.26,

Theorem 8.30,

Q: ( z ) ;

Since

there exists a monomorphic

extends to an analytic map of E

Z.

into

imply that F

into

C

1.

By

f E C (J), proving the Theorem.

i n v e r s i o n problem

9.50

Let

-< cekY

a

and

b

be in C

such that

2 a3 - 27b # 0.

The

Weierstrass inversion problem is the following: (1)

Find a lattice

L

in Q:

such that

g2(L) = a

and

93 (L)= b* Theorem.

The Weierstrass inversion problem always has

a solution. Proof.

Assume first that

afofb.

If

L

satisfies (1)

then (2)

J(L)

=

a3/(a3-27b2)

Conversely, assume that see that (1) holds. fies (2).

and L

g2 (L)/g2 (L) = a/b.

satisfies (2); then one can easily

Thus it suffices to find

By Theorem 9.27 there exists

T E

D

L

which satis-

such that

The E l l i p t i c Modular F u n c t i o n

3

3

21 .

J ( T )= a / ( a -27b

(9.11:6)

J ( L ) = J ( - r ) . By ( 9 . 1 1 : 3 )

2

cx g 2 ( L T ) / g 3 ( L T ) . C l e a r l y

let

T:

then

p;

# 0, g3 (LT) # 0.

A(LT)

g2(L,) that

a3

-

i;

T

# 0.

then

g 2 ( L ) = cx

g2(L) = a r

a # 0 # b.

(L ) ;

3 . r L a s t l y assume t h a t

(9.28:6).

g 2 ( L T ),

Since

g 3 ( L ) = a-6g

g3(L) = b .

If

Since

thus

ci

b=O.

A(LT) # O r

thus there exists

cx E C

such

p r o v i n g t h e Theorem.

Now assume t h a t a and b a r e i n IR s u c h t h a t 2 27b # O . Assume f i r s t t h a t a # O # b ; t h e n (1) and ( 2 ) a r e

again equivalent.

If

A

unique

J ( r ) = a3/(a3-27b 2 )

such t h a t (3)

g2(L)/g3(L) =

g 2 ( L T )= 0 ( 9 . 2 8 : 7 ) .

g 3 ( L T )= b -4

t h e n by

may be c h o s e n s u c h t h a t

ct

Note t h a t

may be c h o s e n so t h a t Let

,

L : aLT;

p r o v i n g t h e theorem i n c a s e

g2 (L)/g3 (L) = a/b, a =0

and l e t

a€@

Let

187

may b e found i n

.r

aD+ ( 9 . 2 8 : 8 )

.

g 2 ( L T ) / g 3 ( L T ) and

have t h e same ( r e s p . o p p o s i t e )

a/b

s i g n , t h e r e e x i s t s a unique

ct

> 0

such t h a t

ctTT

( r e s p . icxT ) s a t i s f i e s ( 1 ) . T

Assume now t h a t

(4)

If

a=O;

g3(LT)

and

then

b

t h e r e e x i s t s unique

e

in/6

cxLT)

(5)

If

b = 0.

A(LT) # O r

g 2 ( L T ) and

-

b#O.

Let

T:

p.

have t h e same ( r e s p . o p p o s i t e ) s i g n , cx> 0

such t h a t

cxLT

(resp.

Let

.r=i;

then

g3(L,) = b .

g2(L,) # O . a

t h e r e e x i s t s unique

ein’4cx~.r)

of cours e

s a t i s f i e s (1).

L a s t l y assume t h a t Since

-

have t h e same ( r e s p . o p p o s i t e ) s i g n ,

a >0

s a t i s f i e s (I),.

such t h a t

aLT

(resp.

This Page Intentionally Left Blank

CHAPTER 10

ALGEBRAIC FUNCTION FIELDS

Definitions

10.10

Let that

L

F(L)

and

Introduction

be a lattice in

@.

We have seen in Chapter 7

is a field extension of

as follows. p

which can be described

@

is transcendental over

finite algebraic extension of

@(?).

C

and

F(L)

is a

This will serve as a model

for the definition to follow. Let

K

be a subfield of a field

F.

KcF

will be called

an algebraic function field of one variable, or merely an algebraic function field, if there exists over

K,

such that

F

x

E

F,

transcendental

is an algebraic extension of

K(x)

of

finite degree. Example.

@ c

F(L)

is an algebraic function field.

The theory of algebraic function fields is a large one. We will give only a very brief sketch of it in this chapter, giving frequent references to the literature.

Hopefully we will

write enough to motivate what has been done when

K

is IR

or

C, and thus make Part I11 of this monograph more accessible to the reader not knowledgeable in these matters. The standard reference to this subject is Chevalley's concise and extensive Introduction to the theory of algebraic functions of one variable [16]. A more classical reference to the state of the art c. 1900 is Hensel and Landsberg's Theorie 189

Norman L. Alling

190

der - algebraischen Funktionen einer Veriabeln 1341.

Let

10.11

(1)

Let

E

XI

KcF

F

be an algebraic function field (10.10).

be transcendental over

K;

Indeed, since the transcendence degree of definition, 1, x

is algebraic over

[F:K(x)], we see that

[F:K(x,x')]

then [F:K(x')] <

F

over

K(x');

thus

thus

<m;

K

m.

is, by

[F:K(x')l

< m l

establishing (1). Thus the definition of an algebraic function field does not depend on the choice of Let

K

be called a field of constants and

of functions (2)

Let

then

It

of f

{a

F :a

E

function field. 2 E x a m p l e 1.

algebraic over F,

and

KI;

EcF

is an algebraic

IRc@(x)

KcF.

is an algebraic function field.

The

@.

will be called a complex (resp. real) algebraic

function field if Let

the field --

is called the field of constants of

field of constants of it is

(3)

F

KcF.

is a subfield of

KcF

x.

FcF1

K = @ (resp. K=IR). be a finite algebraic extension; then

KcF1

is an algebraic function field. We have seen in 18.13 that a compact Riemann surface

x

may be associated with a complex algebraic function field, using the "cut and paste method". structing

X.

Many choices were made in s o con-

That these arbitrary choices do not influence

the final surface

X

is true, but was not manifest in 58.13.

To motivate the way we will presently construct

sider the following example.

x,

let us con-

Algebraic Function Fields Let a:

Example 2.

over

(4)

z ( 2 )

If

Oa

Let

Clearly

=F(C);

thus the

corresponds to an object of analysis

with some algebraic subobject of

C

is transcendental

Our problem is initially to associate a point

and geometry, 1. of

where

(z)

We saw in Theorem 8.30 that a:

@.

algebraic object @ ( z )

u

c@

191

is regular at

f

E C ( Z ) :

~ ( 2 ) .

ul.

is a proper subring of ~ ( z ) that contains C I

-

f E ~ : ( z ) U0

such that for all

I

l/f

E

0,.

Subrings of this

sort occur elsewhere in algebra; they are called valuation rings. A proper subring

(5)

f t - F-L) implies

l / f E

L) L)

Historical note:

of

F

that .contains K

such that

is called a valuation ring of

KcF.

in this form the idea of a valuation

ring goes back to Krull's classic Allgemeine Bewertungstheorie 1461 I written c. 1931. (6)

Let

RiemKF

{ L ) : L)

is a valuation ring of ,Kc F).

We will think of this set as the formal Riemann surface of K

c

F. It is not difficult to show that

Rie%C

( z ) = {Ou:u

E

11.

In order to work with valuation rings we must know more about them.

10.12 (7

E

Let

KcF

be an algebraic function field and let

RiemKF.

(1)

Let

ul

U(L))

(or U M(O)

and let

for short) be the group of units of (or M

for short) be

0 - U.

It may be useful to refer back to Example 2 in 510.11. Clearly

03;

U

thus

(OD) M((7,)

= {f

E

a: ( 2 ) : f has neither a zero or a pole at

is the maximal ideal of

0,.

Further

192 f

E

Norman L. Alling f(a) E C

Oa ->

M(O,).

as kernel

M

af

fc M

M;

for a moment

that

then it is in

that b a f = l ,

Now let g e M;

(3)

f-g

then

Assume that

is a valuation ring

f

and

or

f/g

M;

by (2) is in

g/f f/g

is zero then clearly

g

are non-zero.

g

proving (3).

is an ideal in

or

f

generality we may assume that

0,

which is absurd; proving

f E U,

+ 1 U.~ If

Indeed, note that

of

such

M.

E

(3) holds.

0, is

af, which must be in

Thus there exists b e 0

U.

showing that

(2)

M

a E 0; then

and let

M.

E

Assume not in

0.

is the maximal ideal of

Let

Proof.

having

These facts suggest general results.

Theorem.

(2)

is a C-linear homomorphism onto d:

is in E

(1.

0

Without loss of

f - g = g (f/g

(1.

Since

- 1) ,

which

From ( 2 ) and (3) we see that

0. Since 0 - M = U,M

is the maximal ideal

proving the Theorem. Clearly

is a subgroup of

K*(:K-{Ol)

U,

thus

0.

K n M = (01.

FO

Clearly K

maps injectively, under the canonical homomorphism

pO,

O/M

is called the residue -class field of

onto a subfield of

to a map

of all of

p

point not in

FO.

Let

0. Given

place of

F.

let U

Let

(5) of

F*

(or p

F

by sending any

Po

EFO u

{a}.

for short) extend

fcF-0

So extended

to po

00,

a

is the

h

f(O)

E pO(f)

E

-

of course

vO

(or v

is

pO

f6 F

h

(4)

Let

onto

F*/U:

GO

FO.

-

a subgroup of

F*.

for short) be the canonical homomorphism (or G

for short).

193

Algebraic Function Fields The Abelian group

0.

is called the -value group of

G

It will

be written additively, and can be ordered as follows:

(6)

let

v ( 0 nF*).

G'Z

0 = v (1)E G+ ,

Clearly

and clearly

-

What is - at first glance given

(7)

gc

then

then

G,

g

or

f E F*

f

a valuation ring

or

-g

is, accordingly, in

in

G',

-g

0: i.e. ,

and so

g = 0, For

f

l/f G

is in

+.

and

g

Let

G.

v(f)

0

0

g. Since

g

and

so that

a,b E U

are in

=

(10.11:5); hence

Assume that

l/f

and

h

(=

let

E G

0.

Thus

-g fa f

is

g

or

are both and

b/€

is a unit

g,h

if

g-h

is in

linearly) ordered Abelian group.

be an element not in E

If both,

G'.

establishing (7).

is a totally

g

is in

such that

then there exist units

are in

all

a bit surprising is that

g = 0.

Indeed, there exists

G

is closed under addition.

G+

G, ordered to be greater than

G+;

then

Let

m

g

for

v(0) E m .

(8)

For all

a,b

(9)

Further, if

E

F, v(a

k

b) Lmin(v(a) , v ( b ) ) .

v(a) # v(b)

equality holds in (8).

Indeed, without l o s s of generality we may assume that v(a) Iv(b).

If

b#O;

v(b/a),O,

then

b=O

then (8) and (9) hold. andso

clearly (8) holds and, since

b/aEOnF*.

-1 E U , v(a)

=

Assume that If

a + b = 0 then

w (b). Assume that

aib#O.

v(a+b) =v(a(l+b/a)) =v(a) +v(li-b/a) 'v(a),

ing (8).

Assume now that

hence

b/a

E

M.

As a consequence

v (a 2 b) = v (a) + v

The map

v(a) < v(b).

v

(1 f b/a) of

F*

=v

(a),

onto

lkb/a

Then E

U,

v(b/a) > 0

provand

and so

showing that (9) holds. G

is known as the valuation

Norman L. Alling

194

with

associated

0.

Let us return to

E x a m p l e 2 (510.11

f E Q : ( z ) (=F(C)); let Clearly

(86.41).

with

{n E Z: n 2 0 )

0,,

and

Let

A

C

u

Let

be the order of

vu

Q:

RE

u

at

onto

(z)*

2

vU(Uu n @ ( z ) * ) =

is the valuation associated

is its value group.

be an integral domain and let

~

f

Further

U ( O u ) nq: ( z ) * .

thus

and let

C

E

is a homomorphism of

;

Z

quotient field. element

vu(f)

vu

having as its kernel Z+(-

continued).

R

Let

denote its

A

be a field extension of

is called integral

A

A.

An

if there exists a

monic polynomial

... + mn-l

(10) m(x) E m 0 + m1 x +

as a root.

A

xn-l + xn E A[x]

is integrally closed in

is integral over

A

Lemma.

Each

Proof.

Let

is in 0

a

E

E

if each

is integrally closed in

be integral over

be a monic polynomial with coefficients in a root. hence

Assume, for a moment, that v(an)

<

v(an-l) <

that

a{O;

F.

Let m(x)

I).

that has

I)

then

(10)

as

a

u(a) < 0

and

. . . < v(a) < 0.

Since n1 mn-l a r nv(a) ,min(v(mo)

... + w (a), . . . , v (mn-l)+ (n-l)v (a)) 2 (n-l)v (a);

a n = - m o - m1 c1-

v (m,)

a~ A

c1

A.

RiemKF

F

A

which has

which is

absurd, proving the Lemma.

10.13

Valuation theory can also be used to good effect

in number theory.

Let

a prime ideal in A. (1)

Let Pe

A

be an integral domain and let P

Let

A

A p - {a/bE A: a E A {a/b

E

A: a E P

and

be the field of quotients of and b

E

bEA-PI

A -PI;

be A.

and let

then

Ap, which is

Algebraic Function Fields

called the ring

A

localized at P,

contains A, and Pe Pe n A = P . Further (2)

Pe

Indeed, give

- Pe.

Lemma.

Let

p

E

- Pel f = a/b with a,b E A - P; thus Hence Ap - Pe = u (Ap), establishing (2).

A

is a prime ideal in

in

A

be a unique factorization domain and let

Let

A.

Ap

A.

Let

P ZAP;

f e A-A,.

Then

f=a/b,

with

thus

plb.

Hence

a

is not in

P,

a

Since

and so

then

P

A.

is a valuation ring of

having no common irreducible factors.

b e P;

such that

Ap.

denote an irreducible element in

Proof.

Ap

that

f E Ap

Ap

f-l = b/a

A

is a subring of

is a prime ideal in

is the maximal ideal in

195

and

b

fpAp, b/aEAp,

proving the Lemma. We can apply this lemma to the following examples. E x a m p l e 1.

0 - z P (PI numbers.

is

Let

p

be a prime number; then by the Lemma,

is a valuation ring of the field Q1

of rational

It is easy to see that the residue class field of

Z

PI

the field having

p

elements.

easy to see that the value group of E x a m p l e 2.

Let

K

reducible polynomial in

Up

is

Further it is Z.

be a field and let K[x]

.

m(x)

be an ir-

By the lemma,

0 - 0

:K[x] is a valuation ring in the field K(x). m (x) (m(x)1 It is again easy to see the residue class field of 0 is

K-isomorphic to

K[x]/(m(x))

and that the value group is

Z.

That these two examples seem so similar, one drawn from number theory and the other from algebraic function theory, might be superficial; however it is not.

The development and

interrelations between these two fields lies outside the scope

Norman L. Alling

196

of this monograph; however before leaving this fascinating topic in comparative mathematics, let us point out an essential difference between these two topics. It is not difficult to show that given

Example 1 ( c o n t . ) .

any valuation ring number

p

see that

such that 2

Q, then there exists a unique prime

of

I)

o=I)

P'

Further it is not difficult to

is the intersection of all valuation subrings of

Q. Example 2

(cont.)

om 5 K [l/x] (l/x)

.

K[l/x]

is a subring of

is a valuation subring of

valuation ring at

(Note: if

co.

K(x),

I)

E

that is

0=0

(m(x))

*

the

here defined

It is not difficult

o = om

RiemKK(x), then either

there exists an irreducible polynomial m(x)

and

called

om

K = @ , then

and as defined in (10.11:4) are identical.) to show that given any

K(x)

in

K[x]

Clearly the intersection of all

0

E

or

such RiemKK(x)

K. Historical note.

The names of Hansel and Hasse are

associated with early applications of valuation theoretic methods to algebraic number theory.

See Endler's excellent

monograph on valuation theory [21] for additional details. Finally let us consider an example that will prove useful in Part I11 of this monograph. Example

3.

In studying R i e w ( x )

Example 2 above, that we must study maximal ideal of IR[x]

.

Given

m

E

we have seen, in

SpecmIR[xl, the set of

Specm lR[x] ,

it is, of course,

a principal ideal generated by an irreducible monic polynomial m(x) x-r,

in IR[x]. for some

m(x)

is of one of two types:

r E ~ R ; or it is

it is either

2

x + b x + c , with

Thus it is natural to associate SpecmIR[x]

with

b 2 - 4 c < 0.

Algebraic Function Fields

Q;+ E { z

(3)

c c : Im(z) > 03.

Hence it is natural to associate

with c + u

RieW(x)

{a}.

Extensions

10.20

Let

197

be an algebraic function field and let

KcF

be a finite algebraic extension of degree

n.

FcF

(As a general

reference on valuations and extensions see e.g., [ 7 0 , vol.11, Chapter VI].

This theory is not trivial.

Most of the results

in this section will be stated, but not proved.

Proofs may be

found e.g., in [ 7 0 , ~01.111. ) Let (1)

U

0E X

8 be in RiemK $:?

0 : ~ ( 8 )-

and let

8

nF;

then

E Riem F.

K

5

will be said to be under

6

and

extension theorem (see e.g. [16, p . 6 1 )

0. The place has many consequences.

One of them is the following: (2)

given Let

0 E X,

M

there exists

8 2

such that

E

0 and let

be the maximal ideal of

8;

maximal ideal of

then

k

n

0 = M.

~ ( 8 =) 0. be the

Then the following dia-

gram is commutative and row exact:

O ->

2

->

8

0 ->

M ->

*

-

->

U / M ->

0

0

->

U / M ->

0

A unique K-isomorphism

g

(3)

of

I

0, into ?J/G

t

of

U/M,

the residue class field

the residue class field of

fined so that ( 3 ) is commutative.

fication. (4)

Let

g

8,

serve as an identi-

Then

[a/i: O / M l

:f,

is finite.

It is called the relative degree of

-

can be de-

0 over

0.

198

Norman L. Alling

fi

Let

-

U.

0=

U n

i j L > F*

-’ G-

0

ih

.

13

F* ->

U->

->

G - > O

A unique group isomorphism h making (5) commutative.

of

Let

h

e

is of finite index

into

G

can be defined

serve as an identification.

Then G

be the value

V

1 2

0 ->

(6)

Let

U.

The following diagram is commutative and row exact:

0 ->

(5)

-

0.

valuation associated with group of

8 and let 5 be the

be the group of units of

e

in

is called ramification index of

-

G.

8

over

0.

As a corollary of (6) we have the following

(7)

is isomorphic to

2.

Indeed, let

x

then

is finite. We have noted (Example 2 (10.13))

G

[F:F]

be transcendental over

is isomorphic to

2.

K

and let

F - K(x);

that

Using (6) we see that (7) holds.

Let us consider some examples to illustrate these

10.21

phenomena. E x a m p l e 1.

U FlR[x]

Let

is IR

and let (XI and that of

of

is 2 .

-

f

Example 2 .

0 E@ [ z ]

(XI

10.22

a :Q: [XI

8

Let

over

and

EC(x).

Let

-

The residue class field of 0 (XI is @ ; thus the relative degree (10.20:4)

8

and let

(10.20:6) of

F -lR(x)

K-IR,

-

F

K-Q:,

0 EQ: [ z

2

I

is

Assume now that

- @ ( Z ) ~

(,2)

-

and let

2 F - @ ( z ).

Let

then the ramification index

2.

-

FcF

is a separable extension.

Algebraic Function Fields

-

(This will - of course

be assured if

K

199

is of characteristic

zero, which it will be in all the central applications in this monograph.)

over

0. For

over

0

1 1 1k'

and let

e

let

f

-

*

0 E X , let

Given any

Theorem.

O1,...,Ok

E

-

X

aj

be the relative degree of

j

be the ramification index of

j

be the points

aj

over

0;

(1)

... + ekfk =n.

e1f1 +

then

(It takes a fair amount of space and care to prove this. A proof may be found as Theorem 20 [ 7 0 , vol.11, pp.60-611.

See

also the remark in the penultimate paragraph of 170, vol.11,

-

P.631 1 There are many applications of this beautiful theorem. Let

E x a m p l e 1.

numbers and let

elle2, and

y = 4(x-el) (x-e2)(x-e,).

-U E X-

Let

0.

be over

is

@.

Since

of

6

is also

or 2 points.

Finally

E

E

f =l.

n-l(o)

0 :

(10.11:4). 0

Over each

*-'((I) consists of 1

contains 1 point if and only if

Let

K=IR,

F EIR(x)

0

E

e=l=f.

X-

and let

F :@ (x) (as in

3 x 5 { 0 E X : the residue class field of

Let

nl. ax = m[xI (x-r):r c I R 1

each point

and let

Hence

u

ax there is only one point in

f=2.

C

00.

Example 1, 110.21).

0

u

(x,y), where

Clearly the residue class field of

thus

@;

Example 2.

is

Let

F SQ:

is algebraically closed the residue class field

Q:

u=el,e2,e3, or

0

be distinct complex

F -a: (x) and let

Let

K -@.

L

e3

ax

'&[[l/xl over

1.

For each

0, since n = 2

and

there exist two points, since for

Norman L. Alling

200

The

10.30

d complex a l g e b r a i c f u n c t i o n

Riemann s u r f a c e

field

Let C c F and 10.11)

.

be a complex algebraic function field (10.10

X

Let

5

Riem F

C

Since C

(10.11:6).

ically closed,each 0 E X has C

is algebra-

as its residue class field

h

F O = Z,

(10.12); thus C

the Riemann sphere ( ( 1 0 . 1 2 ) and ( 8 . 3 0 ) ) .

has, of course, a topology on it which makes it a compact

surface.

0 E X ->

(1)

Let

fc F

gives rise to a map

?(0) E

C

(10.12:4).

have the weak topology making (1) continuous, for a1 L

X

f E F. (2)

Each

Chevalley [16, p.133 ff.] has proved the following. X

is a compact space.

As remarked before ( 1 0 . 2 0 : 7 ) , thus there exists

2;

fE F

the value group of

such that

0

is

vO(f) = 1.

Chevalley has also shown that (3)

f

is a local uniformizer at

0; thus X

is a compact

surface. Finally Chevalley has shown that the Atlas so defined is analytic ( 8 . 2 0 ) (4)

X

;

thus

becomes a compact Riemann surface X

is a meromorphic function on

X;

and

such that each

f E F ->

2

E

F(X)

is

a surjective @-isomorphism. Remark.

structing X

In the author's opinion,Chevalley's method of confrom C c F

is a great improvement over the "cut

and paste method" described in 58.13.

4 theorem

10.40

Let

C

of

coeguivalence

be the category whose objects are complex alge-

Algebraic Function Fields

201

braic function fieldsand whose morphisms are @-linear isomorphisms of one such object into another.

S

Let

be the category

whose objects are compact (connected) Riemann surfaces and whose morphisms are analytic surjections of one such object onto (See,e.g., 1501 for the revelent definitions in cate-

another.

gory theory.) let

f

Let

@

cF

and let

of generality, we may take @

@ cF

and

cF

of the morphism

f.

Let

into

be objects in

F.

[F : F].

Riem F

5

X

and

@

n

(Riem f) ( ? ) E

then

Riem f

@

C F is a finite

-

,..

X :Riem F ;

c

then,

S.

a

Given

let

E

2 n F;

Riem@

Lemma 1.

F

Since

are compact Riemann surfaces:

is an analytic map of

a:

-

F.

is called the degree

and let

each is an object in the category

(1)

and

C

Then, without l o s s

to be a subfield of

n: X

Let

as we noted in 510.30, X

,

F

-

cF

are algebraic function fields

algebraic extension.

i.e.

F

be a @-isomorphism of

@

2 onto

X.

Hence

is a contravariant function of

c

into

S. Y -> a

Now let object in

Let

S.

Y.

tions on

for details.) onto

g

is an

be a triple of object, morphism, and

F(Y)

be the field of a11 meromorphic func-

Using the fundamental existence theorem on Riemann

surfaces, F(Y),

Y

Y

is a proper extension of

Let

g

E

F(Y)

then

-@;

(2)

and

is an analytic map of

It may be shown that there exists

C.

n - to - 1 map of

Y

onto

Z

Then it is not difficult to show that F(Y)

f

(See e.g. , [ 5 8 1

@.

F(V)

are objects in

F(a) (h) ha,

to show that

for all

h

6

c. F(Y)

nc N

such that

(counting rami€ication). [F(Y):

@

(g)]

=n;

thus

Let

.

Then it is not difficult

Norman L. Alling

202

Theorem of c o e q u i v a l e n c e .

Riem

and

Q:

s

is a contravariant functor of

F

Lemma 2 .

into

c.

The contravariant functions

establish a coequivalence of the categories

F

c

S.

and

(See [5, pp.95-1041 for details.) f is a triple of object, morphism, and Thus if F1 > F2

C;

object in (3)

F1

L.

If

X 2 ->

in

S,

then

@ F(Rie% F1)

a

F (Riem@f )

> F(Riemc F 2 ) = @ F 2 .

is a triple of object, morphism, and object

X1

then Rie%F (a)

(4)

> Riem F(X1) =

X 2 = Riem F ( X 2 )

c

c

Let

Theorem.

x :C/L

and

let F '

X':@/L'.

F(L')

L

and Let

L'

F

xl.

be lattices in

@.

Let

F(L) (or equivalently F ( X ) )

(or equivalently F ( X ' ) ) .

and

The following are

equivalent: (i)

L

and

L'

are equivalent lattices ( 6 . 2 3 ) .

(ii) X

and

X'

are analytically equivalent.

(iii) F

and

F'

are @-isomorphic.

(iv) J (L) = J (L') Proof.

.

By Theorem 8.45, (i) and (ii) are equivalent.

Theorem 10.40, (ii) and (iii) are equivalent.

By

By Theorem 9.29,

(ii) and (iv) are equivalent; proving the Theorem.

The Riemann-Roch Theorem

10.50

Let

X

be a compact Riemann surface, and let

the underlying topological space.

Since

X

X

denote

is a compact

orientable surface,it is homeomorphic to a sphere to which

g

Algebraic Function Fields handles have been adjoined. X

called its genus.

g

203

is a topological invariant of

(See e.g., [51] as a general reference on

the topology of surfaces.)

Let

a

be a divisor on

X.

(See

58.60 for a definition.)

(1)

Let

L(a)

then

L(a)

is a complex subspace of

(2)

L(a)

is of finite complex dimension

Riemann's

(3)

L(a)1'

(4)

Let

{f

E

F(X): (f) >a]; F(X)

. L(a).

Theorem

- g - deg(a) .

i(a),

the index - of specialty of

be the non-negative integer

a,

be defined to

-1 + L (a) + deg (a) + g;

then we have

the following. The Riemann-Roch

Theorem.

For all

a

E

divX,

the follow-

ing holds :

(5)

O=l-.l(a) -deg(a) -g+i(a). The Serre D u a l i t y Theorem.

For all

aEdivX

c U (X) (-a), where

(6)

i (a) = dim

(7)

U(X) (-a) :(6

(Note: D(X)

E

D(X):

(6) > -a}.

was defined in 1 8 . 2 3 . )

The literature on the Riemann-Roch and the Serre Duality Theorems is extensive.

Nevertheless, Gunning's excellent

Lectures 0" Riemann surface [29] would serve to give complete proofs of these beautiful and important results.

10.51

The Riemann-Roch theorem can be stated and proved

for any algebraic function field 113.)

KcF.

In so doing a non-negative integer

the genus of

KcF.

(See e . g . , g

[16, Chapter

emerges, called

One of the great accomplishments of contem-

204

Norman L. Alling

porary algebraic geometry is that it has been able to carry over much algebraic and analytic geometry, that was first discovered and investigated over

@,

to arbitrary ground fields.

PART I11 REAL ELLIPTIC CURVES

This Page Intentionally Left Blank

CHAPTER 11

REAL ALGEBRAIC FUNCTION FIELDS AND COMPACT KLEIN SURFACES

Real

11.10 IRc K

Let

algebraic function

be an a l g e b r a i c extension,

a l g e b r a i c a l l y closed; then to

fields.

-

of course

f o r which

-

K

is

K

is

IR-isomorphic

@.

(1) T h e r e e x i s t e x a c t l y two IR-isomorphisms, onto

K

R

Let

Further

@.

CY

=

j

CY 1-j'

for

c1

0

j = 0

and

all

and

1.

d e n o t e t h e c a t e g o r y whose o b j e c t s a r e r e a l a l g e b r a i c

f u n c t i o n f i e l d s ( 1 0 . 1 1 ) and whose morphisms a r e

c

F

be an o b j e c t i n t h e c a t e g o r y

C

IR c

CI

Since

function f i e l d s (10.40).

R.

@ c

R.

Let

F

+

lRcE

R

be an o b j e c t i n

Let

of complex a l g e b r a i c

IR c F i s a n o b j e c t i n

is then a covariant functor

IRc F

in-

IR-linear

j e c t i v e homomorphisms o f one s u c h o b j e c t i n t o a n o t h e r .

C

of

and l e t

5

of

C

into

b e i t s f i e l d of

constants (10.11). (2)

If

-

E i t h e r ( i ) IR i s

IR,

o r (ii) it i s

( i i ) h o l d s t h e n (1)

IR

h a s two d i s t i n c t IR-isomorphisms

IR-isomorphic t o

@.

-

onto

@.

Using e i t h e r o f t h e s e w e c a n make

i n t o an o b j e c t i n

c.

( W e w i l l see l a t e r t h a t t h e s e two o b j e c t s

aO

and

a1

need n o t b e e q u i v a l e n t i n Assume now t h a t

f i e l d of

x2

+

1

C.)

( i )h o l d s i n ( 2 ) ; t h e n

i r r e d u c i b l e polynomial i n over

E;

IRc E

E[x]. then

207

Let IRc F

F

x2 + 1

is an

be t h e s p l i t t i n g

i s an o b j e c t i n

R

208

Norman L . A l l i n g

-

whose f i e l d o f c o n s t a n t s s a t i s f i e s ( i i )o f

IRc F

in

IR

F.

Let

(2).

-

renders

denote a r o o t of

and

t h a t takes

F

h a s a genus

g.

-

of

cil

+

i

to

with

IR

@.

x2

C

This o b j e c t

IR c E .

i s t h e f i x e d f i e l d of

E

E

IRc

cxo

Hence

@.

and w i l l be c a l l e d

@ c F

t h e complexification of

Clearly

i

Let

i n t o an equivalnet o b j e c t i n

IRc F

w i l l be d e n o t e d by (3)

IR-isomorphic t o

b e t h e E-automorphism o f

0

Thus e a c h IR- isomorphism

-1.

is

A s noted i n ( 1 0 . 5 0 )

0.

,

C h e v a l l e y [16, p . 991 h a s p r o v e d t h e

following: Theorem.

i s a l s o o f genus

@ c F

g.

11.11

constants i s

is

tion

c

c

whose f i e l d o f

Its complexifica-

IR, and whose g e n u s i s z e r o . ~ ( x ) .

Example 2 .

IRc

@ ( x ) i s an o b j e c t i n

-

IR i s IR-isomorphic t o

constants

R

is an object i n

IRc IR(x)

Example 1 .

R

whose f i e l d o f

and whose g e n u s i s

@,

zero. The f o l l o w i n g i s a more i n t e r e s t i n g e x a m p l e , which may b e found e . g . ,

i n [16, p . 231.

of

F

F :@ ( z ) . L e t

Let

Example 3 .

b e t h e IR-isomorphism

0

t h a t e x t e n d s complex c o n j u g a t i o n and maps

z

to

-l/z.

( T h a t s u c h a n automorphism e x i s t s f o l l o w s from t h e f a c t t h a t

-l/z

generates

t r a c e of

f,

(1) L e t

x :T ( z ) / 2

then

be

over

F

f

x = ( z - 1/2)/2

+

C.)

For

f

E

F

a(f). and l e t and

y 5 T(iz)/2;

y = i(z+l/z)/2.

let

Tr(f), the

1

209

Real Algebraic Function Fields

(2)

Hence

Let

E

IRc

E;

thus

E

-1.

=

R.

is in

IRc E

Since

0.

Clearly

x + y/i

Since

z,

=

is

CJ

x

R-linear,

and

y

E;

are in

IR(x,y) (i) = F ;

thus

IR(x,y)

=

R

is an object in

IRc E

= @(z).

(4)

+ y2

be the fixed field of

Clearly CC

2

IR(x,y) c E.

thus (3)

x

whose complexification is

Using Theorem 11.10, we see that

E

IRc

is of genus

0.

It has been shown [ 6 1 that (5)

IRc E

If

R

is an object in

of genus

0, then it is

equivalent to the object considered in one of the three examples above.

Now let us consider examples of objects in

11.12.

Let

Example 1 .

- 27g32

2 and let

y

in

R

q2

Let

be in

x

g2x

=

of genus

w

such that

v

g3.

then by Theorem 11.10, IRc E

Let

be in =

such that

is an object

1.

E x a m p l e 2.

u, v, and

-

IR such that

be transcendental over IR

be algebraic over IR(x)

IR(x,y);

E

and

g1

is non-zero.

(1) y2 = 4x3

Let

of

1.

genus

A 'g3

R

w

x

IR.

be transcendental over

{O,l};

implies

and let k < 1.

(2)

Lu ,v w (x,k) : (-l)'(l-(-l)vx2)

Let

y

be algebraic over

(3)

y2

=

Let

E : IR(x,y)

I

IR(x) ,

k

E

IR,

with

Let 0 < k < 1,

Following (3.21:10), let w 2 2

( l - ( - l )k x 1 .

and let

LUIVIW (x,k).

.

By Theorem 11.10, I R c E

is an object in

2 10

R

Norman L . A l l i n g

1.

of genus

I t w i l l b e shown i n t h i s monograph t h a t Examples 1 and 2

R

g e n e r a t e a l l examples of o b j e c t s i n

1.

of genus

Further,

e l e m e n t a r y c r i t e r i a w i l l b e g i v e n which w i l l a l l o w u s t o

R.

d e t e r m i n e when any two s u c h o b j e c t s a r e e q u i v a l e n t i n

11 . 2 0 .

Klein Surfaces

R,

be a n o b j e c t i n

IRc E

Let

and l e t

IR

denote i t s

f i e l d of c o n s t a n t s ( 1 0 . 1 1 ) .

(1) L e t Given

(I

IR

let

X,

E

By ( 1 0 . 2 0 : 4 ) ,

(2)

Thus

(3)

Let

Let

f

E

ax

(4) (I E

Let

x w

is e i t h e r :

(I E

X:

and l e t

(10.12:4);

thus X

denote i t s r e s i d u e c l a s s f i e l d ( 1 0 . 1 2 ) .

EO

is a f i n i t e a l g e b r a i c extension of

El)

EO

E

(10.11) :6)

Riem E

X

If

X.

f If

IR.

E

E

f

0) t h e n

?((I) E

li((I) I

,! 0 l e t

I ; ~ C O )I

continuous, f o r a l l

89 f f ]

.

X

f

E

f

m.

E.

i s a compact c o n n e c t e d s p a c e .

L e t u s s k e t c h a p r o o f o f t h i s t h e o r e m , which

w i l l prove u s e f u l .

IR

and

C

X

is an o b j e c t i n

E

X

i s i n t e g r a l l y closed (10.12),

C.

Let

choose an

be i d e n t i f i e d using it; then

c

are identical.

in itself

i s I R - i s o m o r p h i c t o C,

If

i s o m o r p h i s m and l e t

and Y

EO

b e g i v e n t h e w e a k e s t t o p o l o u y making

Theorem [6, p .

(I

C.

IR}.

EO =

(I E

IZ(0) I

or i s IR-isomorphic t o

IR

IR:

Y

f

Since each

RiemlRE. -

IRc

thus, as sets,

(I;

As n o t e d i n (10.30),

Y

can be given

t h e s t r u c t u r e o f a compact Riemann s u r f a c e i n a v e r y n a t u r a l

X

Real Algebraic Function Fields way.

ing that

Y

and

is identical, show-

is a compact connected surface (without boundary).

X

-

IR = IR.

Assume now that

of

X

Clearly the topology on

211

Let

C

be the complexification

Riemann surface structure

Riem F be endowed with its C Y (10.30). Let u be the non-

trivial E-automorphism of

F

E (11.10:3) and let

IRc

a map

u*

Y

of

u*

that

Y

F

c

onto

f

of order

Y

is a homeomorphism of

Y

onto

X

to see that

Y.

Indeed, one

i.e., that it is

(See, e.g., 1 6 , p. 1 ff].)

2/32.

induces

It is easy to see

2.

easily shows that it is anti-analytic: annihilated by

u

(11.10). Clearly

It is easy

is homeomorphic to the quotient space

Y/u*;

completing our sketch of the proof of the Theorem.

Returning to the examples of $11.11, it is easily

11.21.

seen (for example by consulting Chapter 10)

that

(1) RiemRC (x) is homeomorphic to the Riemann sphere

C;

and that RiemnIR(x)

(2)

(where C+

{a

5

Example

3

is homeomorphic to

+ bi: a,b

(§11.111.

cendental over

IR

is

C

E(i); thus

RiemC@(z) C(z)

takes Thus

and

Let 2

E Z IR(x,y) where

+ y2

.=

u*

C*

to

-

l/h

E

C*,

is fixed point free.

is then homeomorphic to

x

is transF

is the complexification of

c C(z)

is homeomorphic to

E

I:,

-1. Recall that

C.

of

and which permeates The quotient space

RiemnE.

C(z)

IRc E.

The E-automorphism u

induces an anti-analytic involution o*

X

in

{m}

> 01). IR, b -

E

x

C+ u

X

of

I: which 0

and

C/u* : X

is easily seen to be

m

212

Norman L. Alling

homeomorphic to the real projective plane. orientable and has no boundary.

Thus

X

This suggests that

is non-

ax

(11.20:3) should be empty.

It should here be remarked that all of this is worked out in greater detail in [ 6 ] .

Y : Riem IRIR(x).

Since

IR(x) c IR(x,y) O

there is a map which takes

ax,

0'

For

0 to be in

M',

the maximal ideal of

r

IR.

E

Let

pO

then we see that

0

of

Thus

ax

To see why

E

X

4,

let

is a morphism in

0' : 0

to

must be in

o',

=

aY.

n

If

IR(x,y) E Y. x

E

0' then

(x-r)U', for some

must be

be the place associated with pO(x) = r.

R,

0

(10.12);

In the residue class field

EO

(10.12) we see that

cannot be the field

Eo

0' then

is not in

l/x

is.

IR, and hence must be Thus

0'

=

it is the point at infinity in Y. Then equation x 2 + y2 = -1 then gives us (4) 1 + (y/x)

Applying

po

2

= -l/x

2

@.

If

-

IR(l/x)

i.e., (l/x) * po(l/x) = 0. The

.

to (4) gives us

1 + (Po(Y/x))2 = 0;

(5)

thus

EO

cannot be the field

IR, and hence must be

@.

We

have thus shown that (6)

ax

=

This example was cardinal in motivating the research of Newcomb Greenleaf and the author which lead to [ 5 1 and [ 6 ] . It became clear that if X : Riem E

IR

IRc E

x

is an object in

R,

then

might be orientable or non-orientable; and that

213

Real A l g e b r a i c Function F i e l d s

ax

I n [ 6 ] w e showed t h a t

m i g h t o r m i g h t n o t b e empty.

(7)

X

was a l w a y s a c o n n e c t e d s u r f a c e w i t h ( p o s s i b l y empty)

ax,

boundary and

t h a t any s u c h s u r f a c e c o u l d a r i s e i n t h i s way.

(8)

One o f t h e main t h e o r e m s i n [ 6 1 was t o p u t a s t r u c t u r e s u c h t h a t i t s s e t o f "meromorphic" f u n c t i o n s would b e

on

X

E.

The s t r u c t u r e w e p u t on

ture.

we called a dianalytic struc-

X

T h i s i d e a g o e s back i n e s s e n c e t o K l e i n ' s 1882 monograph

[ 3 9 ] , f o r i n t h e c l o s i n g p a g e s of i t h e c o n s i d e r s "Riemann surfaces" t h a t a r e non-orientable.

S c h i f f e r and S p e n c e r [ 5 6 ]

c a r r y o u t t h i s program i n a c o n t e m p o r a r y way.

The i n t e r e s t s

o f G r e e n l e a f and t h e a u t h o r w e r e more i n c l i n e d t o a l g e b r a i c g e o m e t r y t h a n S c h i f f e r and S p e n c e r seem

t o have been; t h u s

o u r e m p h a s i s and o u r r e s u l t s w e r e i n many e s s e n t i a l s q u i t e different.

11.22.

The d e t a i l s on K l e i n s u r f a c e s may b e found i n [ 6 ] .

I n C h a p t e r 1, e n t i t l e d K l e i n S u r f a c e s , a n a l y t i c p r e l i m i n a r i e s may b e found i n $1.

A t h o r o u g h d e s c r i p t i o n of what a d i a n a l y t i c

s t r u c t u r e i s may b e found i n 5 2 . "functions".

5 3 d e a l s w i t h meromorphic

This occupies 1 6 pages o f t e x t and c o n s t i t u t e s

the kernal of the ideas.

less e s s e n t i a l .

§§4-8 a r e u s e f u l b u t a r e a l i t t l e

$ 9 on s u r f a c e s o f g e n u s 0 and 1 i s e s s e n t i a l

f o r t h e reader t o understand.

I n d e e d , t h i s monograph may b e

viewed a s a n e x t e n s i o n o f C h a p t e r 1, $ 9 o f

11.23.

[61.

The h i s t o r y o f K l e i n s u r f a c e s , a s i t was known by

Norman L. Allinq

214

Greenleaf and the author c. 1971 is sketched in [ 6 1 .

The two

additional references which have come to the author's attention

ijber symmetrische Periodicitatsmodulu der

in the intervening decade are Guido Weichold' s Riemann'sche Flachen und die

zugehorigen Abel'schen Normalintegrale erster Gettung [63], published in 1883.

See also

W. - D. Geyer's report given at

Oberwolfach in 1964, in which he uses Galois cohomology to obtain many of Weichold's results [281.

Symmetric Riemann s u r f a c e s

11.30.

X

Let

be a Riemann surface and let

analytic involution of

(1) Then

:Y

X/5

be an anti-

X;

is a symmetric Riemann surface.

(X,5)

(x,o)

Assume that

5

is a symmetric Riemann surface; then

is a Klein surface.

Y

is compact if and only if

X

is. Conversely, given a Klein surface orientable or for which X

[6, $1.61; then X/O

that

and

Y

to work on out on

X.

11.40.

Y.

let

which is non-

X

be its complex double

has an anti-analytic involution are dianalytically equivalent.

have preferred to work on differentials,

8Y # $,

Y

... .

X

o

such

Some authors

with symmetric functions,

The author and Newcomb Greenleaf preferred

Nevertheless, many of our proofs were carried

This point of view is continued in this monograph.

A_ t h e o r e m o f c o e q u i v a l e n c e

A theorem of coequivalence (similar to Theorem 10.40) for a category of real algebraic function fields and iR-linear

Real A l g e b r a i c F u n c t i o n F i e l d s

215

monomorphisms,and a c a t e g o r y o f compact K l e i n s u r f a c e s and K l e i n morphisms i s proved i n [ 6 , pp. 9 5 - 1 0 4 1 . one o f t h e main theorems o f t h a t monograph.

This i s

This Page Intentionally Left Blank

CHAPTER 1 2 THE SPECIES AND GEOMETRIC MODULI OF A REAL ELLIPTIC CURVE

The

12.10

Let lRcE Y

then

1

extended modular yroup

be an object in

R

of genus 1.

Let

Y Z

Rie%E;

is a compact Klein surface whose (algebraic) genus is

[4, p.241. D e f i n i t io n

(1)

lRcE,

Y,

or equivalently

will be called a

real elliptic

curve. Let lRcE

be a real elliptic curve with

As

Rie%E.

noted in 511.10 the constant field, (2)

-

is either (i) lR

R,

Assume that (ii) holds. phisms of % 1.

and

Let

and

(11.1O:l); then

a1

and

a:;

then a: c E

@.

be the twoIR-isomor-

-

aj =alejI for

j=O c1

1'

to

is an object, of genus 1, in

(10.40).

(3)

x 3.

5

Riem E

a:

is a complex algebraic curve of genus 1.

We have seen (10.30:4) that of genus 1.

a:

a0

Having chosen one of these isomorphisms, say

identify

C

onto a:

or (ii) it is Ill-isomorphic to

such that

X

is a compact Riemann surface j By Theorem 8.50 there exists a lattice L(j) in

x

and @/L(j) are analytically equivalent. In j Chapter 9 we saw that there exists a unique T E D (9.13:5), a j

fundamental domain of the elliptic modular function, such that (c/L( j)

and

@/LT

( 9 . 2 0 :1) are analytically equivalent.

j 217

By

218

Norman L. Alling

Theorem 7.42 (4)

E=@(P(',LT

)

i

j

P'(*iLT ) I * j

we have also seen (7.33:19) that

Now, of course, there is no reason to choose opposed to (6)

Since

ctl-j.

-

1'

gk(LTlej) =-,

=

as

j

we see that

'1-j

for

c1

k=2

and

3

and

1.

and

j=O

j

Recall (9.28:l) that a l s o that

gk(r(L)) =

r(z)

=-;,

v,

for all

z

E

fi.

Recall (9.28:3) By Theorem 9.27

for

k= 2

and

3.

for

j= O

and

1.

we see that :r(rj) - ~

(mod

r),

(7)

T

(8)

Let the extended modular group

T

lytic autohomeomorphisms of 9

generated by

~

Bibliographic note.

be the group of diana-

r

and

r.

[Klein-Fricke for example, considers

such maps [41, vol.1, p.196 ff.].

We have adopted the terminol-

ogy of DUVal [19, p.44 and p.2461.

Clearly

12.11

(1)

.. r=r

%

(2)

(9.28:l) is of order two, hence

u r r = r u rr; thus

T/l' =

and Clearly

r

r

Given

Thus

is a normal subgroup of

z2.

is the set of all elements in f c rr, there exist

f(z) = (aE+b)/(ci+d), (3)

r

D+

-r

a,b,c,dE 2

with

-T I

that are analytic. such that

ad-bc=-1.

(9.28:8) is a fundamental domain for

?.

Hence we have proved the following: Theorem.

Let l R c E

be a real elliptic curve, and let

219

The Species and Geometric Moduli Y

Rie

%E.

Assume that the field of constants

morphic to

i.e., that Y

@:

There exists a unique

T E

is IR-iso-

is orientable and has no boundary.

D+

Y

such that

and @/LT

are dian-

alytically equivalent.

12.20

Species

Let IRcE

be a real elliptic curve, let

assume that the field of constants assume that

or that

aY#B

out in (11.101, x L + l

of IRcE

splitting field of

x +1

of all elements of

F

and

is 1R:

i.e.,

is non-orientable.

Y

is irreducible in 2

Rie%E

Y E

over

E[xl.

Let 5

E.

As pointed

Let

F

be the

now denote the set

that are algebraic over IR:

i.e., the

field of constants of IRcF; then 5

(1)

is IR-isomorphic to

By Theorem 11.10, I R c F X:Rie%F.

is an object in

and @/LT

morphism

u*

alent to

x/a*.

of

equals

x (X) = 0;

thus

(2)

=o.

Let

X(Y)

s

of genus

are dianalytically equivalent.

the non-trivial E-automorphism of

x(X),

R

X

onto

X;

F.

then

Let

Let

0

such be

induces an anti-analytic

(5

Y

1.

+ TED

By Theorem 12.11, there exists a unique

X

that

@.

is dianalytically equiv-

It is easy to see that the Euler characteristic, 2x(Y).

Since

x

is of topological genus 1,

be the number of components of

Y.

Using the classifica-

tion of compact surfaces (see e.g., [51] for details), we find that the following holds: Theorem.

~f

Y

s=2,

Mobius strip, and if

s=

0

is an annulus, if it is a Klein bottle.

s=l

it is a

Norman L. Alling

220

The species of

(3)

is the integer s

Y

Y

Note that the species of

.

s( Y )

determines the homeomorphism type

of the underlying space Y.

Geometric moduli

12.30

Let I R c E

be a real elliptic curve and let

be the field of constants of IR c E .

Let

i.e. I that Y

is IR-isomorphic to C : that

aY =

a.

(1)

Now assume that z=IR: i.e., that orientable. Let

F

Let

X

(2)

y

of

a

E @ *

basis

Q

E

Letting

(4)

z =T

over

F; Y

then and

E

x. u*

is

X/u*

are

b

E

of

Z E C ,

P(~,T), the period parallelogram of the L .: LT implies

(6.20:6).

y

One easily sees that

(2) Z y ( z l )

and

z' = O

in ( 3 ) gives

and

z' = 0

in ( 3 ) gives

a+b

a? E L .

Note that

2 x +1

lifts to an anti-analytic automor-

forall

(mod L)

z=1

y

is non-

has an anti-analytic

a E L.

Letting (5)

and

(1 T)

z: z '

D+

which must have the form

Y(z).:ai+b,

where

(3)

Clearly

@,

Y

such that

dianalytically equivalent. Thus C/LT involution y.

E

be the geometric modulus of

T

ananti-analyticautomorphism of

phism

or

a Y f a

be the non-trivial E-automorphism of

CJ

T

Y.

of

be the splitting field of

and let X:Rie%F. Let

is orientable and

are dianalytically equivalent.

is the geometric modulus m(Y),

T

Rie%E.

Assume first that

By Theorem 12.11 there exists a unique

Y and C/LT

such that

Y

y2 = l x I

the identity map of

X.

(mod L). 5

b

(mod L);

hence

The Species and Geometric Moduli

(6)

y 2 (z)= a a z + a E + b , for all

-

aa=l;

la1 = l .

thus

y 2 = l X , y2(0)

Since (8)

is in

L;

thus

ag+bcL. ([6, (1.9.6)]).

Proposition.

either

Re(T)

is

is a basis of T

either

+; 0

or

0

,

BY ( 7 )

Proof.

thus

z c @ .

y 2 = lX,

Since

(7)

221

L,

a

1/2,

Corollary.

Let

12.31

1'11

>1;

Since

(1

then

1/2.

BY ( 4 )

= 1.

,

must either be. 1

(which is

or

la1

Assume that

2Re(r))

is in

a

E

L.

or

L.

By ( 5 )

-1.

Thus

Re(r)

T)

t

7 E L; is

proving the proposition.

T

is always in

T E

aD+.

aD+.

It will be convenient to consider

5 separate cases.

(1)

Case 1.

~ = u i , with

Case 2.

~ = i .

Case 3.

T=

Case 4.

T

Case 5 . Note that if which

-

eiel with

u > 1.

1r/3 < 8 < ~ / 2 .

+ i3&/2. ~ = p + v i = 1 / 2 +(34/2+v)i, =e

E p =

-r=i then

of course

-

L

1/2

with

v>O.

is the set of all Gaussian integers,

has many symmetries.

again has quite a few symmetries.

If

r = p

then

L

Cases 1,3, and 5 will be

called the general cases; cases 2 and 4 will be called the special cases.

12.32

By construction in (12.30)

222

Norman L. Alling Y

(1)

is dianalytically equivalent to

Let these two spaces be identified. lytic involution of @/LT

Let [6,

Let

(C/LT)/y.

be another diana-

y1

and let

be the field of all meromorphic "functions" on

El

p.12 ff.].

Y

Assume that

Yl

and

are dianalytically

equivalent; then there exists an IR-isomorphism h

E.

Since E(i) =F=El(i),

automorphism

h

of

F

F.

(yf

yr(f) = f

if and only if

is then-automorphism of

Thus, for all

El

h(i)=i: i.e.,

h

onto

is a

Since, by hypothesis, h

onto E l we have, for each

(3)

of

can be extended to an IR-linear

such that

@-linear automorphism of El

h

Y1

maps

fEF, y*(h(f)) =h(f). induced by

F

f € E l l yf(f) =f=h''y*h(f).

yl.) Since

y*(i) =-i= 1

h-ly*h (i), (4)

y f = h-'y*h.

Since h

is a C-linear automorphism of

analytic automorphism Riem@ (5)

6

of @/L,

F, there exists an

such that

h= 6*.

Since

is a contravariant functor (see e.g., [ 6 , p.99 ff.l),

Y1 = 6ys-I.

Conversely, given such an automorphism of @ / L T l Y

a dianalytic equivalence between anti-analytic automorphism p,

of

and @

Y1.

it engenders

y1

lifts to an

which must be of the

form (6)

where

p.,(z)

al E

= a l E + b1' @*

and

bl

for all

Z E @ ,

may be chosen to be in

.

P ( 1 , ~ ) (See

The Species and Geometric Moduli

223

(12.30) for more details.) (7)

p and y1 will be said to be equivalent if

Y

and

Y1

are dianalytically equivalent. Thus we have proved the following: and

Lemma.

exists

-6 ( z )

c

and

E @ *

cz + d,

take

dEC

for all

Assume that

are equivalent if and only if there

p

z

such that (5) holds: where E @ .

and

T1

are equivalent.

Since

onto another fundamental domain for

P(1,r)

L

6

must

in C ,

c

must satisfy the following. (8)

In case 1, c=+l. In case 2,

c=+ll

In case 3,

c = +-l .

In case 4,

c =+I,

In case 5,

c = +-l .

Note also (9)

or

+i.

or

+P,

or

2

f~

.

that in each case,

c/c=c 2 :

and that in the general cases (i.e., cases 1,3, and 51, 2 (10) c =l. Clearly 2 (11) 6y6 '(z) = c a z + c b + (d-ac2a) E Y , ( z ) = a 1z + b l , mod L, I _ _ -

for all

z EC.

It will be convenient to have established the following.

X - e i e l with (12)

z

(13) z

+ :A

=

8 EIR,

and let

-i8/2 2eiel2Re(ze 1I

z

E @ ;

Let

then

and

- X z = 2ieie/21rn(ze-i8/2)

Indeed, z

+ As =

-ie/2+zeie/2 i8/2-- (ze-i8/2+ (ze-ie/2) eie/2 (ze )e -

224

Norman L. A l l i n g

w,

Given a complex number

wt

w = 2 R e (w)

w

and

- w = 2 i im(w) :

p r o v i n g ( 1 2 ) and (13).

12.33

i s e q u i v a l e n t t o o n e and o n l y o n e of t h e

Theorem.

following:

1.1

Value of b -

Value of - T

Case ui,

with

I1

I1

I1

1.3

II

11

I1

11

I1

1.4

'I

1

2

-1

2

u > l

1.2

species

0

-1

0

2.1

i

1 ,L -1

2

2.2

If

i

-i

1

2.3

'I

1 'L -1

0

3.1

eiBl

3.2

I'

4.1

p :1 / 2 t i3'/2

with I1

5.1

p

1 , 3 , and 5 ) hence

a

v > 0

1

2 3

'Lp4

1

'LP5

1

1

1

-1

11

,

c2 = 1

(12:32:10); t h u s

is invariant.

(1)

i n cases 1 and 5 ,

(2)

i n case 3 ,

al=a

By (12.30:4 and 5 1 , ( a (= 1 ; a=tl;

hence and

a = +-. r .

Note a l s o t h a t (3)

if

1

0

Note t h a t i n t h e g e n e r a l c a s e s ( i . e . , i n c a s e s

By ( 1 2 . 3 0 : 7 ) ,

L.

T

P'LP

I1

Proof.

1

l'Lp

+viI with

5.2

-T

11

I1

4.2

in

n/3 < B < ?r/2

%

ac2 = 1,

then

d

- ac2a = 2 i I m ( d ) ,

(12.32:11) a

and

aT

,

and are

The Species and Geometric Moduli

(4)

if

ac2=-1,

then

225

d - a c 2d=2Re(d).

Having made these general observations, let us consider the several cases separately (in order of increasing complexity). Let

C a s e 1.

c = +-l .

(12.32:8), d

~ F u i ,with

- ac2d = 2iIm(d).

If

a=l,

Hence

d,

1 E L,

we may assume

and since

-

then

ac = 1 ;

is in

L;

1.1 and 1 . 3 . By (4),

Assume now that

d - ac2d = 2Re(d).

a=-1;

Hence

d,

our disposal, can be chosen so that is in L:

[O,ui). By (12.30:8),

i.e., either

values of

a

or

b=O

and

-

a

then

(12.30),

is in

is either b

0

in cases

acL=-l

(12.32:8).

which is completely at blr or equivalently b,

a6+b

(which is

b=ui/2.

b-6)

is in

This establishes the

in cases 1.2 and 1.4.

b

b

b

and

is real.

P(1,T)

that

hence

This establishes the values of

or 1/2.

By

thus by ( 3 1 ,

was chosen to be in

g+b

a = +-l .

blr or equivalently, b

without loss of generality

By (12.30:8),

[0,1).

b

By (11, L

which is completely at our dis-

posal, can be chosen so that Since

u > l .

Using ( 3 ) and (4),

one easily sees that cases 1.1, 1.2, 1 . 3 , and 1.4 are equivalent. As to the species [ui/2, 1+ui/2] s=2.

s

of

in case 1.1, clearly

[0,1] and

map down to two disjoint circles in

In case 1.2,

[O,ui] and

two disjoint circles in the species of

Y,

Y

period, and hence

aY;

[1/2, 1/2+uil

hence

s

is again

has no fixed points in

X.

thus

map down to 2.

in cases 1 . 3 and 1.4, note that y

aY;

Concerning b

is a half

Thus

s=O

in each case. C a s e 5.

By (12.32:8), hence

a

Let

T

c = +-l .

p+vi

(=1/2+ (3+/2+v)i),

By (l),

is an invariant of

y;

a = +-1 .

Since

with

c2=l,

v > 0.

al=a;

showing that cases 5.1 and

226

Norman L. A l l i n g

5.2 a r e i n e q u i v a l e n t .

Assume f i r s t t h a t

case 5.1 f i r s t ) .

By ( 3 ) ,

t i r e l y a t our d i s p o s a l , taken t o be real.

Since

b

thus

L;

is in

1

is either

b

c ( 35/ 4 + v / 2 ) i ;

-Z + C T . Let

p,

Thus

c=-1;

c=+1.

,

may

b,

i s en-

d

may be

(which i s

E+b

d

-

- further

be

2b)

b=1/2.

b e d e f i n e d t o be

= i + c / 2 + c ( 34 / 2 ) i + c v i = z + c ( p + v i ) =

T,

i s equivalent t o

hence we may t a k e

b

a s given i n c a s e 5.1.

t o be

showing t h a t i f

0;

i s e q u i v a l e n t t o a map covered by case 5.1. 2Now assume t h a t a = -1. By ( 4 ) , d - a c d = 2Re(d) ; t h u s b a=l,

then

be t a k e n t o be pure-imaginary.

Since

[O, ( 3 % +2 v ) i )

(since

t a k e n t o be i n

2 ~ - 1 = (4 3+2v)i.

By ( 1 2 . 3 0 : 8 ) ,

thus e i t h e r

or

b=0

is in

T

2.r

af;+b

b = (3$/2 + v ) i .

-1

b = ( 34 / 2 + v ) i .

A s noted above,

ac2 = -1.

=

p,(z)

d=-1/4;

b

may be

is i n

L

and

Let

c=l;

Since

then

pI ,(z)

d

d=-1/4;

then

T,(z)

a l e n t t o a map covered by c a s e 5.2. c a s e 5 . 1 i s indeed domain f o r (5)

Let

@/LT

1, in

C.

is

= -z - T ,

then

is entirely

=-;+T,

hence i t i s e q u i v a l e n t t o a map covered by c a s e 5.2. and l e t

7

then

L;

Assume t h a t

-; + ( 34/ Z + v ) i + 2 R e ( d ) .

a t o u r d i s p o s a l w e may l e t

L,

b = 0,

If

c=+1.

may

(=2iIm(b)) is in

e q u i v a l e n t t o a map covered by c a s e 5.2.

c=-1

is

Zero i s t h e v a l u e of

1/2.

A s s u m e , f o r a moment, t h a t

Let

p,(z)

then

b

L,

or

0

prescribed i n c a s e 5.1.

A s noted above,

Since

and hence e q u a l l y

By (12.30:8)

t a k e n t o b e i n [0,1). in

d-ac2z=2iIm(d).

bl,

( i . e . , consider

a = l

and

Now l e t

which i s equiv-

To s e e t h a t t h e s p e c i e s i n

let u s c o n s t r u c t a new fundamental Let

m - 3'+2v.

P l ( l , ~ ):{ x + i y : 0 < x < 1 and

max(-mx,mx-m) < y < m i n ( m x ,- m x + m ) I u

COI u

( 0 , ~ )u ( O , ~ - T ) .

The Species and Geometric Moduli

Clearly

is the set of fixed points of

[0,1)

-

z, and clearly

sidering the action of If

s=l.

Modify

on

y

C/LT

Y.

By con-

one easily verifies that

y.

then the imaginary axis is left fixed by

?=-z

to

P1(l,?)

P 1 ( l , ~ ) under

maps down to a circle in

[0,1)

-

227

so that

P2(1,.r)

[0,(3'+2v)i)

is in

P ( 1 , ~ )and ~ so that P2(l,~) is a rhombus that is a funda2 mental domain for @/L,. Then proceed as above to show that s is again

1. Let

Case 3 .

By (12.32:8) , showing that

T

:eiel

c = +-I ,

thus

a

T-

If

(7)

a = -T,

a=-.r; then

Since

and that

.

d - ac2i = 2eie/2Re(de-ie/21 .

then

= 0.

Re

As a consequence, al = a

One easily sees that

(2 + 2c0sB)'e~~/~, l=i(2-2cose)4eie/2

Assume that

gclR.

cL = 1.

+ 1=

T

71/3 < 8 < .rr/2. By (21, a=+T. -

is invariant, and thus that cases 3.1 and 3.2

are inequivalent. (6)

with

may be chosen so that

d

Having so chosen

T-1

is in

L,

d,

b = igei8/21

we may use (6) to show that,

without l o s s of generality, g

may be chosen to be in

[Ol(2-2cosB)4 ) .

By (12.30:8),

ac+b

(which is

this case) is in

L;

is in

L.

either

g=

g= 0

or

hence

2b

(2- 2cosB)'/2.

If

equivalent to a map covered by case 3.1. that

b f 0.

Let

by (12.32:11),

Now let ?,(z)

y1 ( z )

=

and let

-TZ + T ,

c = -1 and let

=-T~-T,

case 3.1.

c=l

d

5

for some

-Tg+b

in

As a consequence,

b = 0,

then

is

Assume, for a moment,

d - (2+2~ose)~e~'/~/4. , then, which is equivalent to case 3.1.

- (2 + 2~ose)'e~'/~/4- ,

then

which again is equivalent to a map covered by

Norman L. A l l i n g

228

(8)

If

a=T

Now assume t h a t thus

a=T.

g

r+l

is in

may b e c h o s e n t o be i n

[O, ( 2

r6 + b i s i n b = e i e / 2 ( 2 + 2cos9)'/2.

3.2. b

d-ac2a=2iei0/21m(de

may b e c h o s e n s o t h a t

d

(12.30:8), or

then

f o r some

b=gei9l2,

by ( 6 ) ; t h u s

2

c =1,

and

Assume t h a t

gt-IR.

It e q u a l s

L.

b = 0,

If

2b;

-i0/2

+2~0~9)').

)

=o;

By

hence e i t h e r

then

1.

and i s g i v e n

L,

b=0

f a l l s into case

t h e n w e see ( 6 ) t h a t

bfO;

Im(be

-i8/2

b= (~+1)/2.

t h e n i s a t t h e i n t e r s e c t i o n o f t h e d i a g o n a l s of t h e rhombus

whose v e r t i c e s a r e

and

0,1,~+1,

Let

T.

d r c i ( 2 - 2 ~ 0 ~ 9 ) ' e ~ ' / ~ (/ 4= c ( ~ - 1 ) / 4 ) ; t h e n by ( 1 2 . 3 2 : 1 1 )

,

( 6 ) , and ( 8 ) .

y

3 . 1 ( r e s p . c a s e 3.2), n o t e t h a t 0

r -1

and

=T;+CT;

i s e q u i v a l e n t t o a map

To see t h a t t h e s p e c i e s i s

c o v e r e d by c a s e 3 . 2 .

t h a t contains

y,

Hence

y,(z)

1

i n case

i s r e f l e c t i o n about t h e l i n e t h u s w e may p r o c e e d

.r+1);

(resp.

as w e d i d above f o r c a s e 5. Case 4 .

-

course then

in

G

Let

T ?

p ( - e in/3=1/2+i3'/2).

a p r i m i t i v e s i x t h r o o t of u n i t y .

i s a c y c l i c g r o u p of o r d e r 6 . By ( 1 2 . 3 0 : 7 and 4 ) ,

G.

is in

a

t h e n it i s a s u b g r o u p o f i n d e x phic t o t h e t w o element group be any e l e m e n t of d e t e r m i n e d by are

given

G,

a , only

2 2 4 G ( = { l , p ,P 1 )

mod G 2 and

Assume f i r s t t h a t

t o be

c L

al=c a

G

pG2

n

pn:

G/G2

Z);

E

c

is

2 2 ={g : g

E

GI;

i s isomor-

can be chosen t o (12.32:ll) i s

The two cosets of

i f it i s i n

of

G

a € G2

If

mod G

2

then

it i s a s cover-

C l e a r l y c a s e 4 . 1 and 4.2 a r e i n e q u i v a l e n t .

e d by case 4 . 2 .

a

Let

G.

Since

aE G,

:c

G

-

By ( 1 2 . 3 2 : 8 ) ,

and t h u s

Z2.

.

Let

2 pG ( = { p l p 3 , p 5 } ) .

i t i s a s ' c o v e r e d by case 4 . 1 ;

take

2,

is

p

1.

a

E

GL.

Since

By ( 1 2 . 3 0 : 8 )

c

i s a t o u r d i s p o s a l w e may

ag+b

is i n

L.

By c h o i c e

229

The S p e c i e s and Geometric Moduli

is in

b

is either

0

can choose

or b

i n case 4.1. t o be

a g + b = E + b = 2 R e ( b ) ; thus

P ( ~ , T ) (12.30).

y1

giving u s

Now assume t h a t

a

l o s s of g e n e r a l i t y .

Fix

2iein/61m(de-i'/6).

Since

g.

t h u s we may choose

g=O

If

but

L;

4 g=3/2.

s = l r let

y

Lastly let

a

5

1.

Since

-

G/G2

Since

(12.30:8) h o l d s , or

G.

then

or

b=1/2.

1E L ,

aE+b If

y

7

then

g = 3'/2.

yl(z)=pz+p,

i s complex con-

i s r e f l e c t i o n about

A s w e can see

p + l .

from t h e

1 i n e a c h case.

i.

G:{*l,+i}.

Let

By ( 1 2 . 3 2 : 8 ) , G 2 ( :cg2:

ac2 = 1 and so

C E

(which i s then

2b)

y

Clearly

- ac2a = d -

is i n

-

and

of

a l = %a. L e t

thus d

G

By

G.

g e G I ) ={+1}

w e may assume t h a t

b=O

a6+b

Assume t h a t

then

i s a t o u r d i s p o s a l w e may assume t h a t

(12.32:ll).

b=O

g= 0

I

34 e i v / 6 . ,

is

L,

iv/6

To see t h a t i n e a c h case

is T 5

ge

By (12.30:E)

i s t h e two element group;

c = +1;

Let

d

is i n

a

2ac d =

is

is i n case 4.2.

and

s

without

p,

-

d

b

i71/6 d=ie /4;

i s a c y c l i c group of o r d e r 4 .

course

+

[0,3 1.

a=p;

0

demonstration i n c a s e 5 ,

(12.30:4 and 7),

t o be

i n c a s e 4 . 1 and n o t e t h a t

let

may be chosen

By ( 8 )

Hence e i t h e r

andlet

the l i n e t h a t contains

Case 2.

1.

a

which i s i n

p+1,

and s o

I n case 4.2

jugation.

c2

Since

or equivalently

is i n class 4.2. a = l

may be chosen

i s e n t i r e l y a t our d i s p o s a l , w e

d

t o be i n

b=O

c-1

y

and hence

g

.

t o be

ag+ b=2b.

then Let

c

bl;

f o r some r e a l number

2

d

which i s e q u i v a l e n t t o

w e may choose

GLI

may choose it s o t h a t

pG

E

then

z + pr

then we a r e

b=O

If

1/2.

b=1/2;

(z) =

2d - a c d=2iIm(d), we

Since

or

0

Assume now t h a t

t o be any element i n

is in

c = l .

Let

t o be e i t h e r

i3'/4,

case 4 . 1 .

1/2.

Re(b)

b b

E

L;

= 2iIm(d).

i s real [0,1).

Since

hence e i t h e r

is i n case 2.1.

Since

230

c

Norman L. A l l i n g runs through

2 c =?l;

G,

a r e equivalent.

t h u s t h e cases

Assume now t h a t

i s as d e s c r i b e d by case 2.3.

a = l

Since

d

c a s e s 2 . 1 and 2.3 a r e i n e q u i v a l e n t .

c2 = il Let

and

a = + l and

i s pure imaginary,

a c{*iI.

Now assume

w e may assume, w i t h o u t loss of g e n e r a l i t y , t h a t

c 2 = 1;

t h e n by ( 8 )

may be c h o s e n t o be

'

( = 2 e i7T/4)

is i n

(12.30:8)

a6 + b

and h e n c e

g

i n c a s e 2.2. d E i2'eP'/4/4;

E

d - a c 2 z = 2 i e i T i l 4 I m ( d e- i ~ i / 4) ;

g e iTi/4, g

L,

L.

with

is either

0

or is g # 0.

Y1 ( z )

a 6 t b = 2b; 2'/2.

If

c = l

Let

2b

c l o s e indeed t o c a s e 1.3.

b

By E

L

7

is

and l e t

ig +i f and

which

To see t h a t

b=O

t h e a c t i o n i s t h e same a s c o v e r e d i n cases 1.1; t h u s Case 2 . 2 i s v e r y much l i k e c a s e 3.

thus

then

g=0

i s e q u i v a l e n t t o t h e map d e s c r i b e d i n cases 2 . 2 . a = l

a = i.

[0,2').

thus

(12.32: 11) e q u a l s

t h e s p e c i e s i s as c l a i m e d , n o t e t h a t i f

Since

1+i

may be c h o s e n t o be i n

Assume t h a t then

g E R . Since

In t h i s case

7

i.e.,

b=1/2;

- ac2J

b=O

then

s=2.

F i n a l l y c a s e 2.3 i s very

Thus t h e a r g u m e n t s above c a n b e u s e d

t o e s t a b l i s h t h a t t h e s p e c i e s a r e a s claimed i n c a s e 2 , proving t h e theorem. B i b l i o g r a p h i c note.

Most of t h e c o n t e n t s of t h i s c h a p t e r ,

up t o t h i s p o i n t , i s a n e x p a n s i o n o f

[6, pp.60-661.

W e have

s u p p l i e d a r g u m e n t s , m o d i f i e d n o t a t i o n , and made s l i g h t c h a n g e s . On p. 6 6 o f [ 6 ] t h e a u t h o r and Newcomb G r e e n l e a f made a n i n i t i a l s t a b a t g i v i n g a c o h e r e n t d e s c r i p t i o n of t h e moduli s p a c e s of r e a l e l l i p t i c curves

Y

whose c o n s t a n t f i e l d i s lR.

The i n i t i a l

a i m o f t h i s r e s e a r c h was t o g i v e a s i m p l e r p a r a m e t r i z a t i o n o f these spaces.

T h i s w e w i l l now do.

The Species and Geometric Moduli

231

12.34

(1)

Let

(2)

Let lR+s{tcn: t > O I .

For

denote complex conjugation.

K

a

in the upper half plane

T

(3)

let

XT :@/L~.

(4)

Let

Y

be a real elliptic curve whose field of constant

,

i.e.

is lR:

which has a non-empty boundary

aY

or which

is non-orientable. Let

Theorem.

exists a unique

5

t:t(Y)

3

~

cIR

.

Y of species 2 there

For each

+

such that

Y is dianalytically

equivalent to (5)

Xti/5 5

where

Y2,t

1

is the anti-analytic involution of

<. Every

has species 2.

modulus m(Xti)

(12.30:l); then

Let

T I

S

ti;

1.1; also

Let T E

T

aD+

then the following holds: (ii) t = l

T = T ~ .

(iii) O < t < 1

Xti

be the geometric (Corollary 12.30). (i) t > 1

in case 2.1; also

in case 1.2; also

T=

induced by

in cases

T=i=T'.

i/t=-l/T'.

T o prove this theorem one need only look at the action

induced on

LT

in Theorem 12.33, for cases 1.1, 1.2, 2.1; and

shift to the action priate element in (6)

K

on

Lti, by multiplying by an appro-

@*.

A_ fundamental domain for

Let

ten+

0 - y - t/21.

and let

FD(Y

Y2,t. 2,t

)

-{x+iy: O <- x < 1

and

Norman L. Alling

232

1 + ti/2

ti/2

Identification

f

1 + yi

,,

Yi

Clearly

Let

5

c r ( z 0 ): C z

Let

exists a unique

thus it maps down to

[ti/2, 1+ti/2]

in

Xti

and thus maps down to the other component

€a:

and

( z - z o I =rl. ..

EC:

Theorem.

zo

5,

The image of

aY2,t.

is left fixed by

-

is left fixed by

[0,1]

a component of

12.35

9

5

:ic.

let r > O recall that

Y of species

For each

+ t-t(Y) cIR

such that

Y

there

1

is dianalytically

equivalent to

+' + ti/2/ <:Y l,t '

('1

5

where

i.

by

is the anti-analytic involution of Every

modulus T I

also

1.

Let

(12.3O:l); then

The following hold:

T I

T E

T

aD+

(i) t > 3%

(ii) t = 3 4 in case 4.1; also (34/2)i). (iii) 3 4 > t > l in case 3.1;

T=.r'.

( = + +

and

has species

m(X++ti/2)

ti/2.

f)z+

YlIt

=

r ; ( . r )I

where

<(z)

about the circle C1 (1). )

E/(z

!

- 1).

(iv) t = 1

X++ti/2

induced

be the geometric (12.30).

Let

in case 5.1; T = T '

also

(Note: r;

.r=r;(.r')

is reflection

in case 2.2; also

The Species and Geometric Moduli T =

i= 5

and

(T')

3.2;also

T' =

T=<(-?')

case 4.2; also 1/3$ > t > 0

T = % +

(1+ i)/2

(v)

= 5 (T).

and

T'=C(-?).

(3'/2)i

and

in case 5.2; also

T =

1 > t > '3

in case

(vi) t=1/3' =++i/2(3 s, ) .

T I

1/2

233

+ i/2t

and

in (vii)

T I

?/(2?

=

- 1).

Before discussing the proof of this theorem let us give

Ylltt

two fundamental domains for (2)

& fundamental domain for Let

t EIR+

O
and let or

each of which has its usages.

Yl,t

FD1(YlIt) - E x + iy: 0 < x

1 / 2-< x < 1 and

<

1/2

and

<

1/2

and

O
4 + ti/2

=

a+

T'

5 + ti/2

Identification 0

(3)

Another fundamental -domain for Let

1

Boundary

t E R + and let

Yl,t.

FD2 (Yl,t) { x + iy: 0 < x

0< y< t/2).

ti/2

-

1/2 +ti/2

5 T I

Boundary (continued)1/2 A sketch of the proof.

Using (3) it is clear that studying the action of

9,

YlIt

is of species

1.

On

as given in Theorem 12.33 in cases

2.2, 3.1, 3.2, 4.1, 4.2, 5.1, and 5.2,one easily sees that each

234

Norman L. Alling

Y corresponds to a unique Ylrtr the correspondence between action of

on

LT

for

and

T

one need only study the

TI,

and see that, using a little conformal

5

geometry, it corresponds to the action of that

* z/ (22 - 1) is reflection about

z

To establish

t EIR'.

on

LTl. (Note

C + ( + ) .)

12.36

Theorem.

Now let

there exists a unique

f K

+ 1/2.

For each

t :t(Y) EIR+

such that

Y

of species

Y

0

is dianalyti-

cally equivalent to (1) Xti/S

where

2.

5

Let

species

5

is the anti-analytic involution of T

Fm(Xti) , which is in

0. Let

(ii) t = l

:ti.

T=

i/t

Each

YoIt

induced by is of

(i) t > l in case 1.3; a l s o

incase2.3;also

case 1.4; also (2)

TI

aD+.

Xti

T = ~ = T ' .

(iii) O < t < l in

= 1/?1.

A - fundamental domain --for and

Yort.

-t/2

<

FD(YOIt):

Let

y< t/23.

ti/2

1/2

+ ti/2

-a + 1/2 0

1/2

a

-ti/2

1/2 @-ti

T=T'.

- ti/2

The Species and Geometric Moduli

235

Clearly

is a Klein bottle, hence it is of species Yo,t The rest of the theorem follows by studying the action of

0.

in cases 1 . 3 ,

1.4, and 2.3

From these results we obtain the following main

12.37

Let

Theorem.

Y

be a real elliptic curve.

Y

If

is

orientable and has no boundary,then it is characterized up to dianalytic equivalence by its geometric modulus Assume now that

m(Y)

E

D+.

Y has a non-empty boundary or that is is non-

orientable. Then it is characterized u p to dianalytic equivalence by its species t(Y)

( E

s(Y)

(E

{

0,1,2})

and its geometric modulus

IR+).

12.38

Let

A , M,

and

K

denote the sets of dianalytic equivalence

classes of real elliptic curves of species

2, 1, and 0 :

of dianalytic annuli, Mobius strips, and Klein bottles.

Y t+ t (Y)

these sets be topologized so that phism;

EIR+

then each of these sets is homeomorphic

i.e., Let

is a homeomorto .'RI

Thus

these sets are then spaces of moduli of the several species.

12.39

Let

s~{2,1,0}

and let

+,

t,t' E I R

with

tft'.

Ys,t and YsrtI may or may not have dianalytically equivalent complex doubles (YSrt)@ and (YsIt,)@ [ 6 , 1.61. If they do, we will call them complementary.

From the theorems above we

see that we have the following: Theorem.

if

t'=l/t

and Ys,t and t#t'.

YsrtI

It is convenient to call

are complementary if and only

YsIl

__ self

complementary. Clearly

236

Norman L. Alling

m((Ys,l)c) = i ;

thus

Y211

I

YlI1

I

and

yo,l will be referred

to occasionally as the Gaussian annulus, the Gaussian Mobius strip, and the Gaussian Klein bottle.

12.40

Real

A lattice [19], if

Functions

z=L.

lattices

L

in

Q:

will be called

&,

following DuVal

In DuVal's excellent monograph Elliptic

Elliptic Curves [19], he makes an extensive study

of real lattices. Many of his results go back at least to Weierstrass [64, pp.264-2751.

The author has learned much from

DuVal's treatment, and we will have occasion to refer to it frequently.

(The author is also indebted to Dr. Carl S. Weisman

for bringing DuVal's book to his attention).

CHAPTER 13 AUTOMORPHISMS OF REAL ELLIPTIC CURVES

T h e automorphism g r ou p of

13.10

ys,t be a Klein surface 16, 1.21. A dianalytic auto-

u

Let

U.

homeomorphism will be calledanautomorphism of (1) Let

U

If

AutU

If

Aut'U

51

f

f E Autu:

is non-orientable let

always a subgroup of exists thus

U.

is orientable let

(2)

be the automorphism group of

f E AutU- Aut

Aut'U

Autu.

+u.

is orientation preserving). Aut'U

AutU/Aut+U

Clearly

Aut+U

is

Assume, for a moment, that there

For all

is a subgroup of

normal and

E Autu.

g

E

AutU

+

Autu- Aut u ,

fg

of index 2.

E

+

Aut u ;

Hence it is

is the two element group.

Hence, in

general (3) Autu/Aut+U (4)

13.11

The complex plane

in

fixes E

2.

is always a normal subgroup of

Complex conjugation

Autq:/Aut

2

or

1

Aut+U

face.

f^

is of order

Aut'C

+@

^I

Aut'C, a,

Z2,

Each

is in

K

? E Aut+@

which fixes

unique

-

of course - a Klein sur-

Aut+@ - A u W ;

thus

extends uniquely to an element

-.

when restricted to there exist

is

@

Autu.

Conversely, each is in

@

a

E @ *

237

+ Aut @ .

and

bc@

E

Aut'C

For each such that

which

Norman L. Alling

238

(1) ?(z) = a z + b ,

for all

zcc.

is just the affine group.

Aut+C

(?,z) E (Aut'C)

group structure on it such that analytic.

For

b

E

C

,

gb(z)

Let

Zb be called translation of gb

+.

z + b , for all

Aut'C

E

xC+?(z) E C

is

let

(2)

g:b EII:

Clearly it has a natural Lie

z EC.

&

@

b.

Clearly

is an analytic injective homomorphism of the

addition group of C

into

Aut'c,

whose image will be denoted

by (3) TransE,

and will be called the translation subgroup Transq: acts transitively on C .

(4) da(z) = a z ,

for each

will be called dilation

::a

E

+

za

E

Aut'C

a

Aut'C

.

Dil@

Clearly

let

E @ * ,

of

C

&

a.

Clearly

is an analytic injective homomorphism of

the multiplicative group @ * into Aut'C denoted by

+

Aut C .

ZEC.

za

@*

For

of

,

whose image will be

and will be called the dilation subgroup of

Clearly the following holds

..,-

5

-.-.

-

(5)

f=sbd,=d asb/a'

(6)

thus

--1-

da sbaa=GbIar

for all

aEC*

and

bcC.

Hence we have proved the following: Theorem.

Aut'C

Trans@

is a normal subgroup of

Aut+C.

is a semi-direct (or normal product) product of

Further, Tram

by D i X . (See e.g., [30, p.88 ff.] for details on semi-direct products.)

Automorphisms of Real Elliptic Curves

x

Let

13.12

constant

239

be a real elliptic curve whose field of

is IR-isomorphic to

or, equivalently, has no

@;

boundary and is orientable. We have seen that there exists

X

such that

TEQ)

(12.11).

Let

X,

and

X

X,

and

lytic homomorphism of a:

are dianalytically equivalent be identified.

XT

onto

Let

having kernel

duces the structure of a compact Lie group on X,

XT

acts on itself by translation; thus

as a subgroup of (1) Let

E

2 - 2 '

E

and let :G

LT},

is a subgroup of

Proof.

Clearly the identity element

and

z'=G(w').

(2)

w - w'

Since

E

h

and

let w - g - l ( z )

be in

G,.

w' E 9 - I

and

Since

L,

in-

As such

XT.

may be thought of

= GT n Aut+C.

G,

5

q

if and only if

L,

Lemma.

Let

GT,

L,.

Aut+X,.

GT={$~Au=:

G(z) -G(z')

be the ana-

g

Aua.

Let

and

z

then

(2') ;

1

-

of

Aut@

z'

be in

of course

-

is in @,

z =

and

(w)

GEG,

if and only if

- z'

z

E

L,.

--1 --1 --I--Ig (z) - g (2') = g g(w) - g g(w') = w - w ' ,

(3) G - l ( z ) - G - ' ( z ' )

E

LT

if and only if

Combining ( 2 ) and (3) proves that

G-'

w-w' is in

E

LT.

G,.

Since

G E GT (4)

hg(z) -hg(z')

E

LT

if and only if

Since, by definition, ?j is in that

hi

is in

G,,

(Recall that Theorem.

f

G,

4(z) -<(z')

E

L,.

we can use ( 4 ) to show

proving the Lemma.

-f(z)

is in

was defined to be GT

if and only if

a z + b (13.11:1).) aL,

=

L,;

thus

Norman L. Alling

240

?<

-

$ ( z ' ) = a(z

aLT=LT

z

Let

Proof. ?(z)

la1 = I .

implies

G~

- z')

implies

a,

=

aA;

be in C

A :z

and let

- z'.

proving the first statement.

Clearly

la/= 1 .

?

Note that for placed on

z'

and

to be in

GT,

a strong restriction is

but that no restriction at all is placed on

b;

thus we have Corollary.

TransC cGT.

Given

13.13

G

G,,

E

constant on each coset of a:

mod L,;

element

such that

g -q*(G)

in

G

then by definition ( 1 3 . 1 2 : l )

AutXT

iS

thus it induces a unique

(1) qG=gq. (i) q*

Theorem.

G C Autq:,

(ii) Given

is a homomorphism of

GT

f o r which there exists

g

that (1) holds, then

G

AutXT

Aut+XT.

and

: G

onto

is in

GT.

(iii) q*

E

m(XT) E D -{i,p}

m(X ) = i ; T

m(X

T

and

) = p

Proof.

and let

(12.30:l);

a = pnr

(y)

(Ee (i) Let

h gq*(h).

g (hq)= (gh)q = q*

obvious.

9

Then

CS)q* (5)q.

q* (gh) = q* (<)q, (h) (iii) Let

such

GT

AutfXT,

(B)

a = + l and

for some

n

in

E

E

LT}.

by

(a) a = + 1

+i

if

IOr1r2r3r4r511

(vii) ker n DilC is trivial. q* and 6 be in GT, let g q * (G) q*(GL)q=q(gfi) = (q<)fi=(gq)fi=g(qfi)= Since

q

maps

C

onto

proving the first assertion. g

onto

(vi) GTnDilC

such that the following hold:

+

if

if

da

AutxT

(iv) kerq, n TransC = { gb: b

virtue of it acting on itself by translation. is the set of all

AutXT.

maps

(v) q,(TransC) = X T r considered as a subgroup of

-

into

AutX,T. Since C

xT

I

(ii) is

is the universal

Automorphisms of Real Elliptic Curves

covering space of

XT

lifts to

such that (1) holds.

@ E

AuW.

and

its universal covering map,

q

is in

is orientation preserving if and only if

third assertion of the Theorem. to (vi), by 12.11, we may let

T

be in ,'D

da

By Theorem 13.12,

Clearly

il

have this property.

elements

a

than

+1,

E

is, proving the As

without loss of

if and only if

GT

aLT = L T .

By the same theorem such For such

1.

must clearly be either

T

g

Note that

GT.

(iv) and (v) are obvious.

generality.

must have modulus

g

Using the second asser-

G

tion of the Theorem, we conclude that

G

241

proved the penultimate assertion.

i

or

a

to exist other thus we have

p;

The last follows easily from

the earlier assertions, proving the Theorem.

Theorem 13.13 gives a detailed description of

13.14

+ Aut xT.

We will now investigate AutXT.

+. + g tAutXT -Aut XT

erality (12.11), we may assume that There exists

Lemma.

Without loss of gen-

T E

D

if and only if

+

TtaD. By Theorem 13.13 such

Proof.

there exists

G

+ GT-GT.

in

By Theorem 12.33 such

G

exists.

zt

@ ,

Since

where

g

Assume first that

exists.

and

bE@.

Since

9.

GTf

a = + l , L T = z T . By the choice of

T

or

If

must either 1/2;

-T

or

- ~ + 1 . Accordingly

hence in each case

T E

then by (13.13) we conclude that in

aD+,

i.e.,

proving the Lemma.

+.

aD

- z'

E

LT =aLT.

la/=l.

-

(13.12:l):

z

-

a(z-2')

LT

f

aD

.

for all

G ( z ) =aZ+b,

and only if

E

is in

T

Conversely, assume that such

KG is in Aut'c,

at@*

exists if and only if

ReT

if

Clearly

(9.12:2),

T

is either

Assume now that

1 ~ =1 1

LT

and hence

T

0

aftl; is again

242

Norman L. Alling

Let

be in

T

aD+.

As noted above (13.10:4),

is always a normal subgroup of

XT

that

AutfXT

In Theorem 12.33 we saw

AutXT.

always has an anti-analytic involution.

Thus we have

the following: For

Theorem.

product of

T E

Aut+XT

-

constants, lR,

is 1R:

X

1.63: i.e.,

tion of

and

X,

is always a semi-direct AutX,.

i.e., which has a non-empty boundary or Let

(X,p,<) be its complex double

is a complex elliptic curve (which we can

XT

identify with

AutXT

be a real elliptic curve whose field of

which is non-orientable. [6,

+,

by a two element subgroup of

Y

Let

13.15

aD

for p

T

E

aD+)

5

I

is an anti-analytic involu-

is the quotient map of Y.

which is dianalytically equivalent to

X

onto

Note:

p

X/<,

is diana-

lytic. (1) Let

Aut X E { f ~ A u t x :fE;=
5

Aut+X

Aut'X

5

n

Aut X .

5

(2) Aut5X

is a subgroup of

AutX

Aut+X

is a subgroup of

Aut X .

5

Indeed:

then

Aut5X

and as a consequence

<

Let

f

E

Aut X

5

be in

f(z.) = w 3

j'

- Aut'X;5 Aut5X

for

AutX

is clearly in

is closed under composition.

by definition

(3)

5

the identity element in

and clearly

and

j=O

fE=
Hence

f-I5 = Ef -1 thus

5

f EAut x; 5 < = f - l € < =f-1
Let

establishing (2).

Aut X/Aut +X

5

and let and

I

Aut X

zo E X.

1; then

5

2

Let

Z2. zlE 5(zo)

and let

5(wj) =
3

Automorphisms of Real Elliptic Curves f(zl-j) = w ~ - ~ Hence . there exists a unique

p,(f)

243

E

AutY

such

that (4)

pf =p*(f)p. Lemma.

p,

is a homomorphism of

Proof.

Let

Aut X

be in

Aut X;

p(fg) = ( p f ) g = p , ( f ) p g = p , ( f ) p , ( g ) p .

Since

Y

this shows that

of

g

then

5

p

is a homomorphism.

=c

1,5} ,

and

p,

p,(fg)p=

X

maps

onto

The surjectivity

X, f (x) = x

or

C(x,)

x1 fx,.

3

dianalytic

+ Aut5X

maps

isomor-

AutY.

Clearly {1,<]ckerp,.

Proof.

then

p,

kerp,

phically onto

E

and

AutY.

was established in [ 6 , 1.9.21, proving the Lemma.

p,

Theorem.

x

f

onto

5

f (x) =

<(XI.

Let

There exists

Assume ( i ) that

f(x) = x

f E kerp,. x,

X

E

f (x,) =xo.

in some neighborhood of

x,

-

For each p-1 (aY);

Since in

X.

f

is

Using

analytic continuation, together with Schwarzian reflection (see e.g., (12.34:6) , (12.35:3) , and (12.36:2)) for all

x

E

X:

i.e.,

then

( s f ) (x,) =xo.

that

kerp,c{l,5}.

5

generated by that

By (i),

p, IAutiX AutY;

dianalytically) on

13.16

Let

5f = 1,

and hence

f (x,) = x1;

f = 5,

proving

Y

5

proving the Theorem.

take the topology on then

AutY

Aut'X

and induce a

5

acts continuously (and even

Y.

be as in (13.15) and let

As we have seen (12.34, 12.35, and 12.36) equivalent with

f (x) = x

As to the rest of the Theorem, Aut 5 x is and Aut+X; thus using the Lemma above we see

5

topology on

we see that

f = 1. Now assume (ii) that

+ p,(Aut X) =AutY, Let

I

YSltl

for a unique

t

Y

En+.

s

be its species.

is dianalytically

244

Norman L. Tilling

(1) Let

-ti

TI

be

if

1/2+ti/2

(2) Let

if

if

:K

s=2

or

and let it be defined to

0,

s=l.

s=2

1, and let it be

or

~ + 1 / 2 if

s=o.

Thus

Y s l t Z X,,/~,

X,,

of

where ..

5.

induced by

5X

translation

into

-cs_

X

Aut'X,

E

-+

Aut'X,,

x EIR.

Let

,

As we have seen,each (Corollary 13.12) : thus

G,,

x EIR

(13.13) for each

is a continuous (even a

whose kernel is the group

+ x = gx< (z),

( z ) =:

is the anti-analytic involution

(13.11:2) is in

+ sx-q*(gx) ~ A u tX T l x E I R sx

5

(3) csx = sxc

x

since

and hence

sx

E

(13.13). C (m)

Further,

homomorphism

)

Clearly

2.

is real; thus

+ AutSXT,.

Using Theorem 13.15 we see that the following is proved. Proposition.

(even Cm) image,

homomorphism of IR,

ima,

is a continuous s,t whose kernel is 2 ; thus its

a : x c I R - + a x E p*(sx)

-IR/Z).

S'(

induce an orientation

imn.

(5) Let (6) Then

and

AutY

is isomorphic to the circle group

(4) Let the usual orientation on IR on

E

H T l E q;'Aut HTl

+

=c f

5 E

xP I

and let

H,:

( z i - s?) ( z )

GTl:

+

E

:H,,

LTl

I

( z t - i F ) ( z ) EL^,,

HT,={feGT,:

n

Aut'C.

zEC}

for all for all

Z E C } .

By Theorem 13.13 and 13.15, (7)

p*q,

(8) Let

is a homomorphism of

a :p*(s ti/2)

and let

then, by ( 7 1 , each is in Theorem.

(i) H,

I

n

onto

H:,

AutYsIt.

6 :p*(d-l) ;

AutYsIt. Trans

={

gbb: b = x + int/2 ,

where

x

245

Automorphisms o f R e a l E l l i p t i c C u r v e s

in

i S

--

5

and

IR

+

a

HTl - H T t .

E

(iv)

2.

and

6

(ii) H T l n D i l @ = { d a : a = * l } .

n c 2).

If

6

and

are i n

(i.e., i f

s = 2

The s e t

s = 2 : then

Let

by

and

{l,y},

S , t

d i r e c t product of (x)

a 6 = 6a :y = p* ( d

+

for a l l

yaX ~ = a - ~f o, r a l l

Aut'Y

AutY

{ 1,u } .

and

(v)

-lSti/2) (vi)

s.

xcx.

i s a group t h a t i s isomorphic t o

{l,u,6,y~

(viii)

and

and a r e e a c h o f o r d e r s,t is o r i e n t a b l e ) , then a

Aut Y s I t ,

in

2

6ax6 = a- X I

u aX u = a

Aut'Y

ys,t

are orientation reversing.

i s an e l e m e n t of o r d e r

ima

AutY

(iii)

ima

by

Z2@Z2.

i s a semi-direct product of s,t i s t h e d i r e c t product of

s, t

(ix)

(vii)

is (the semis,t d i r e c t product with Cl,al.

I n g e n e r a l , AutY {l,y}),

I n g e n e r a l , t h e c o n n e c t e d component o f t h e i d e n t i t y e l e m e n t

of

AutY

is

s,t

ima,

f(z) Eaz+b

Let

Proof.

a group isomorphic t o be i n

HTt.

EIR/Z).

S1(

If

or

s = 2

1,

then

(9) If

(rg - g?Z) ( z ) =

a); + b -

z

for all

that

(i.e., let

a = l

2 i Im(b),

?

be i n '

of course) i s i n

k > 0

and minimal i s

with

x EIR

and

is i n

n

Since HTl

LTl,

i s r e a l i n e a c h case.

a

with

la1 = I .

Z E @ .

E @ .

One sees, a t o n c e , t h a t

-E;

LTl,

( ~ g - f ? Z ) ( ~ ) = ( a - a ) z + ( b - i j + ( a - 1 ) / i2s) i n

for a l l

is

is in

then

s = 0,

(10)

(a-

a

E

Z,

is i n

but not

ti,

Trans); LT,.

then

The e l e m e n t

.

a = t1;

HTt.

are in

AutYSlt.

(which

kie LTlr b = x + int/2,

By Theorem 1 3 . 1 3 ( v i ) proving ( i i ) . C l e a r l y

H +T l . By ( i ) and ( i i ) gtiI2

By d e f i n i t i o n ( 5 ) and Lemma 1 3 . 1 5 ,

are in

b-5

i n each c a s e : hence

proving (i)

IR,

Assume f i r s t

The r e s t o f ( i i i ) i s o b v i o u s .

a (iv)

-

and and

-

d-l

6

(vii)

246

Norman L . A l l i n g

a r e a l s o obvious.

(viii)

-

(x) f o l l o w e a s i l y from t h e r e s u l t s

j u s t proved. Thus w e have o b t a i n e d a d e t a i l e d d e s c r i p t i o n o f

AutY stt

i n a l l cases.

The o r b i t subspaces

13.20

of

ys,t.

Ys,tr

W e w i l l now c o n s i d e r t h e o r b i t s u b s p a c e s o f

+ Aut Y s , t l

ima,

and

A s noted before

AutYs,t,

and

s ~{2,1,0)

(Theorem 1 3 . 1 6 (x)) I

component of t h e i d e n t i t y o f

ima

AutYsIt.

tclR

+.

i s t h e connected

Since

h a s been

imci

i t s o r b i t s a c q u i r e a n o r i e n t a t i o n from i t .

oriented (13.16:4), (Let

where

under

be d e f i n e d a s i n ( 1 3 . 1 6 : 1 ) . )

T I

y ~ n ,l e t

(1) F o r

U

Y

b e t h e image o f

IR+yi

pq

under

in

ys,t.

i s homeomorphic t o S Y a c q u i r e s a n o r i e n t a t i o n from IR. C l e a r l y Y s r t Since

l € L T I

each

U

U Is.

j o i n t union of

Y

(3)

6(U 1

= u - Y'

(4)

Y(U y )

= u- y - t / 2 -- U - y + t / 2 '

y

Y

U

Y

is the dis-

and t h u s

Theorem.

13.21

(Uy) 0 2 y 9 2 ' Let

uo

and Y E

YZrt

The o r b i t s o f

imci

[O,t/2];

and of t h e

ima.

under

are

a r e t h e two b o u n d a r y components

%/2

a ( U ) = Ut / 2 - y '

then

s e r v e s t h e o r i e n t a t i o n of t h e s e o r b i t s . t i o n of

Note that

= uY + t / 2 '

O(U

YZtt.

.

One e a s i l y sees t h a t

(2)

of

1

U 's

Y reverse t h e o r i e n t a t i o n of t h e

Y

pre-

0

Further t h e orienta-

are c o m p a t a b l e . U Is.

and

6

The o r b i t s of

and

Y2rt

y

Automorphisms of Real Elliptic Curves

under

+

Aut Y2,tr or under

(Uy Ut/2-y) O~y
AutY21t, are Ut/4

I

being the only orbit of

these two groups that is connected. guished orbit of

Thus

U

is a distin-

t/ 4

Y2,t.

gtii2 (IR+yi)=IR+ (y + t/2)i.

Proof.

247

takes this set to

IR- (y + t/2)i.

This latter set and IR+ (t/2 - y)i

same image in

YZrs, namely

Ut/2-y.

-

the orientation of the

u

Clearly

have the preserves

U ' s . d-l(IEl+yi)=IR-yi, reversing Y orientation. 5 mpas IR-yi to lR+yi; thus 6(U ) = U and Y Y' orientation is reversed. The rest of the Theorem follows easily. (l) Ut/4

will be called the -center orbit of

Since it is the only connected orbit of analytic invariant of

13.22

( ' y )

Theorem.

l~y~t/ ' 4 Let

damental domain [

(t/2 - y)i, 1/2

YZIt.

AutY21t, it is a di-

YaIt.

The orbits of

0< y
FD2(YlIt)

Ylrt

under

imcl

are

The preimage of

(12.35:3)

U in the funY is [yi, 1/2 +yi) u

+ (t/2 - y)i) . Uo is the boundary of Y.

u(U ) = U and the action of u on U is the same as the Y Y' Y action of at on U thus u preserves the orientation of -5 Y; U The action of u (or equivalently c14) on U is trivial Y' Y is distinguished. 6 and if and only if y=t/4; thus U t/4 Y both map U to itself but are orientation reversing. For Y 0< y
Clear, on making some pictures and doing a few

very elementary calculations.

Norman L. Alling

248

Let

(Uy)O5yct/ 2

preimage of

Y

If

FD(YOIt)

The preimage of

[-yi, 4-yi). and of

0 5y(t/2.

in

U

is

U

t/2 trivial on U

y# 0

under

or, t/2

(12.36:2) is in

Uo

[ti/2, S, + ti/2).

if and only if

Y

Yoft

The orbits of

Theorem.

13.23

ima

then the

[yi, 4 + yi) u

[of%)

FD(YOIt) is

The action of

y=O

or

are

is

y=t/2;

thus these

and u preserves t/2-y the orientation of these orbits. o ( U ) = U if and only if Y Y and the y=t/4; thus this orbit is distinguished. 6 ( U ) = U Y Y' action of 6 on U reverses orientation. y ( U ) =Ut/2-y and Y Y y reverses the orientation of U For y E (O,t/2), U is Y Y contained in a tubular annular neighborhood. This, however, is

two orbits are distinguished.

u(U ) =U

y

.

not true of Proof.

Uo

and

UtI2.

Clear, on making some pictures and doing a few

very easy calculations.

In general

13.24

In fact the orbits of

YSrt imcl

is the union of orbits of constitute a foliation of

ima.

Ysrt.

Orthogonal t r a j e c t o r i e s

13.30

Since a Klein surface has

-

by definit on

-

a dianalytic

structure on it, it makes sense to say that two curves in a Klein surface cross at right angles.

We will now investigate

U Is. Y is a vertical line in a:.

the family of orthogonal trajectories to the For each

x EIR,

(1) Let

Vx

then the

Vxls

x' €IR

5

IRi + x

pq(JRi+x); are orthogonal trajectories to the

U 's. Y

For

Automorphisms of Real Elliptic Curves

(2) ax' (Vx) = V x + x l ;

thus

ima

249

acts transitively on the

VXk.

(3)

u(Vx) =Vx,

Theorem.

and

6(Vx) =V-x,

Let

s=2;

v(Vx)

then each

Vx

=v-~. is a closed dianalytic

arc that runs from a point on one component of unique point on the other. arcs. Vx

constitute all of these

(Vx)ox
Now let

In general, V x + l = V x .

aYZIt to a

s=l;

then again,each

i s a closed dianalytic arc that runs from one point on

to a unique other point. arcs.

V X + + --V x .

morphic to

S1

Lastly, and

dianalytic curves.

(Vx)ox<4

let

(Vx)o<x<4 -

s=O;

aY

constitute all of these then each

Vx

is homeo-

constitutes all of these closed

CHAPTER 1 4

FROM SPECIES AND GEOMETRIC M O D U L I TO DEFINING EQUATIONS

Introduction

14.10

Y

Let

-

is

IR,

b e a r e a l e l l i p t i c c u r v e whose f i e l d of c o n s t a n t s

IR;

e q u i v a l e n t l y , assume t h a t

boundary,

Y

of

or t h a t it i s non-orientable.

aY,

s p e c i e s of

( 1 2 . 2 0 ) and l e t

Y

(512.3).

7'

to be (2)

Let

K

+

Y

Let

: ti 1/2

t

ti/2

0,

or

be i d e n t i f i e d . and l e t i t b e d e f i n e d

s = 1.

if

s = 2

if

be t h e

is dianalytically

Y

that

Yslt

and

s = 2

if

s

Let

b e t h e g e o m e t r i c modulus

W e have s e e n (512.3)

YsIt.

equivalent t o

(1) L e t

h a s a non-empty

Y

1, and l e t i t b e

or

K

+

1/2

if

s = 0.

6

Let

then

b e t h e a n t i - a n a l y t i c i n v o l u t i o n of

YsIt Let

i s d e f i n e d t o be E(YsIt)

(or

for short)

Ys

,

(512.3).

Let

[6,1.3].

F(XT1) (or

F

b e t h e f i e l d o f a l l meromorphic f u n c t i o n s o n

We h a v e s e e n [ 6 , p p . 9 5 - 1 0 4 1 ,

is an a l g e b r a i c function IR.

,/i

XTl.

and r e s t a t e d i n (11.401, t h a t

f i e l d , o f a l g e b r a i c g e n u s 1,

Thus t h e r e e x i s t e l e m e n t s

x

and

y

in

E

such t h a t

IR, y

E = IR(x,y).

s a t i s f i e s an a l g e b r a i c equation with

coefficients i n

y

IR(x).

i s algebraic over

IR(x),

Such a n e q u a t i o n w i l l be c a l l e d a 251

E

over

is transcendental over Hence

-5 ;

i n d u c e d by

f o r s h o r t ) b e t h e f i e l d o f a l l mero-

E

morphic " f u n c t i o n s " on

XT

XTl

and

x

252

Norman L. A l l i n q

defining equation f o r

R c E.

There are

-

-

of course

an

i n f i n i t e number o f s u c h d e f i n i n g e q u a t i o n s ; t h u s w e s e e k a deFurther, we

f i n i n g e q u a t i o n o f a p a r t i c u l a r l y " n i c e " form. x

would l i k e t o e x p r e s s

cn

i n Chapter 7 t h a t i f lattice

=

y

analytically.

W e have s e e n

i s t h e Weierstrass 3 - f u n c t i o n f o r t h e

then

Lrl,

(3)

and

4p

3

- g2r)

-

and t h a t

g3,

F =

C(piq').

T h i s w i l l s e r v e a s a model f o r t h e c o r r e s p o n d i n g p r o b l e m f o r E

A s w e w i l l see ( 3 ) q i v e s u s a s o l u t i o n t o

i n t h i s chapter.

s = 2 o r 1,

t h e p r o b l e m i n case

-

essence

t o Weierstrass.

a r e s u l t t h a t g o e s back

The case o f s p e c i e s

0

-

in

does not

seem t o h a v e b e e n t r e a t e d b e f o r e i n t h e l i t e r a t u r e , beyond what i s u i v e n i n [ 6 ] .

'p

Since

we must look elsewhere f o r

x

p'

and and

y.

are not i n

E(YOIt),

This search lead the

a u t h o r b a c k t o J a c o b i e l l i p t i c f u n c t i o n s , a s w e w i l l see i n

514.3.

I n general, given (4) U ( f ) then

a

f

Kfc;

i s a n I R - l i n e a r automorphisrn o f

V i r t u a l l y by d e f i n i t i o n , (5)

F(LTl) let

E

Let

L

7

Let

of o r d e r 2.

is t h e fixed f i e l d of

a.

LTl.

Species

14.20

E

F,

2 and 1

b e t h e Weierstrass p - f u n c t i o n f o r

L

(7.32).

First note that

(1)

E

=

Indeed i f

L s = 2,

-

T' =

-

ti = - t i .

If

s

-

= 1, T '

= 1/2

-

ti/2.

253

S p e c i e s , Geometric Moduli, Defininq Equations

- 1) i s

t i (= 2 ( 1 / 2 + t i / 2 )

Since

in

-

L,

is i n

T'

L;

e s t a b l i s h i n g (1). Lemma.

Z

_ -1 Z

are i n

(( Z - a ) 1

+

~ ( z ) ; hence

is i n

3

(1) w e see t h a t t h i s i s

a2

Similarly 3

E.

'

is i n

h o l d s w e may u s e i t a n d t h e d e f i n i t i o n o f

t o prove t h a t

q21Q3

and

g3

(9) E I R " p 1 .

F = C@,v').

Since

By ( 7 . 3 3 : 1 9 ) ,

2

(9') =475

and

A ( :g 2

Further,

IR

By t h e Lemma,

Proof.

and

q2

S i n c e (1)

E.

g3

(7.33)

IR, p r o v i n g t h e lemma.

E

E = IR(?\,3'),

Theorem.

W

g2

IR.

are i n

g3

&

R

I).U s i n g

-

( z - n 2

L L * (

g2

'I))

-

1 2

R€L*

and

El

K ~ ( Z=)

a(?) ( z ) E

Proof.

.(;+I

'p'

and

9

-

-g$

27g32)

g3 =

is i n

IR*

.

By Theorem 7 . 4 2 ,

E.

IR ( ' p I p l )

w e see t h a t

PI.

(S')2 = W

-

c

@,'#I)

[F:E] = 2 ,

3

3

Since

and

g2

= E.

g3

are i n

g2'93

IR, A

is i n

The0 r e m

.

A # 0,

IR. By ( 7 . 3 3 : 4 a n d 2 2 ) ,

proving t h e

As n o t e d i n ( 7 . 3 3 : 2 0 a n d 21) ,

(3)

thus

e

1

+

e2

+

e 3 = 0 , ele2 + e2e3 + e 3 e l = - g 2 / 4 ,

and

ele2e3 = g 3 / 4 .

14.21.

IR+ n t i / 2

(i)

Theorem.

and

its values i n

IRi IRu

+

n/2,

{a}

on

rp

takes its values i n

for each

IR+nti/2

n

E

2.

IRu

(ii) p '

{a}

on

takes

and it t a k e s i t s v a l u e s

254

in

Norman L . A l l i n g

Riu

p

thus

f o r each

E, 9

is i n

ti

Since

1R.

I R i + n/2,

Since

Proof.

on

on

{m)

E

IR u

= p(z),

on

{m]

Using t h e Schwarz r e f l e c t i o n p r i n c i p l e ,

ti

~ ' p ( ~ - t i / 2 ) .S i n c e

IR + ( n + 1 / 2 ) t i ,

on

w e see t h a t each

n

E

-

Since

N.

'p

thus

n

t h e same i s t r u e o n

L,

c

Z;

E

x

Let

IR.

E

R(x+ti/2) =

and hence

IR+ t i / 2 ,

Z.

Combining t h e s e r e s u l t s

IRu

{m}

takes its values i n

?(xi); E

n

on

IR+ n t i / 2 ,

for

K?)(xi) = ' ! ? ( - x i ) .

(14.20:1),

L = L

n

{a}

i s an even f u n c t i o n ( 7 . 3 2 : 6 ) , t h e l a s t q u a n t i t y i s

9

Since

f o r each

assumes v a l u e s i n

W

IR+ n t i .

i s r e a l - v a l u e d on

~ ? ( x + t i / 2 ) ; thus

f o r each

IRu

the latter quantity equals

L,

E

Z

E

takes its values i n

L, I n ( z + n t i )

t a k e s on v a l u e s i n

n

IF.i

+

IRu

on

{m)

n, f o r e a c h

n

Since

IRi. E

Z.

Using

t h e Schwarz r e f l e c t i o n p r i n c i p l e , as w e d i d a b o v e , o n e sees that

n assumes i t s

each

n

E

x

E

3'

{m}

IR; t h e n

on

IRi.

y'

above, t h a t f o r each

14.22

n

E

IRi+

Z;

IRu

n/2,

for

Thus

'p'

IR+ i n / 2 .

on

{ m l

~ V ' ( x i )= ' p ' ( - x i ) = - y ' ( x i ) ,

since

9'

is

assumes i t s v a l u e s i n

From t h i s i t f o l l o w s , from t h e a r g u m e n t takes its values i n

IRiu

{m}

on

I R i + n/2,

p r o v i n g t h e Theorem.

Assume t h a t

Theorem.

on

Cm)

take its values i n

a n odd f u n c t i o n ( 7 . 3 2 ) . IRi u

IRu

e s t a b l i s h i n g ( i ) . The same a r g u m e n t c a n b e u s e d

Z;

t o show t h a t Let

values i n

( i ) el,e2,

s = 2.

a r e real and d i s t i n c t . 3 is p o s i t i v e , zero, or neaative and

e

(ii) A > 0 .

(iii) g3

according as

t > 1, t = 1, o r t < 1.

(iv)

g2 > 0.

(v)

'p

maps t h e p e r i m e t e r o f t h e r e c t a n g l e whose v e r t i c e s a r e 0 , 1/2,

S p e c i e s , Geometric Moduli, Defining Equations

+

1/2

ti/2

and

ti/2,

injectively onto

255

(vi)

[-m,ml.

goes through t h i s s e t , counterclockwise s t a r t i n g a t

el

p(z)

0,

avoiding

+ ti/2)

but

> e3 ( i n ( t i / 2 ) ) .

)

e l , e 2 , and e 3

By Theorem 1 4 . 2 1 ,

Proof.

z

is s t r i c t l y decreasing; thus

p ( 1 / 2 ) ) > e2 ( :l p ( 1 / 2

(

0,

As

are real.

By

(7.33:4) t h e y are d i s t i n c t , proving ( i ) . Since 2

A = 16(e -e ) 1 2

( i i ) . For

i

-6

s

= -s

6

zero.

(e2-e,)

t = 1,

Recall t h a t

Hence

3

thus

g3(ti)

s

, A

(7.33:5)

6

and hence

s6, L

1 = 27g3/A

t > 1

> 0,

equals (7.33:18)

q6

(9.11:7)

,

proving

is

and t h a t

( C o r o l l a r y 5 , 99.28); t h u s

t > 1.

for a l l

t > 0,

-

(7.33:22)

1

i L = L;

J

for a l l

g3(ti) # 0,

(1) F o r

( e -e ) 2

showing t h a t

6'

J(ti) > 1

2

(i/t)Lti

= Lilt;

t = 1.

changes s i g n a t

Since

T

CD

I+

g3(LT)

i s a n a l y t i c (Theorem 9 . 2 2 1 ,

g3(ti)

shown t h a t

e x i s t s a n d i s p o s i t i v e [ 1 9 , p p . 6-

Limt++m

s6(Lti)

i s continuous.

7, a n d p p . 3 3 - 3 4 ] ; p r o v i n g ( i i i ) . S i n c e t > 1

(Corollary 5, 99.28))

for a l l

(3)

t > 1.

g2(Li,t)

proving ( i v ) .

,

and s i n c e

J(ti)> 1,

U s i n g (1) w e see t h a t 4

= t q2(Lti)

I

To see t h a t ( v ) a n d ( v i ) h o l d ,

Note i n p a s s i n g t h a t ( 2 ) and ( 3 ) i m p l y A(Li,t) 14.23

for a l l

3 J = g 2 /A, g 2 ( Lt l- 1 > 0 ,

F i g u r e 11, p r o v i n g t h e Theorem.

(4)

I t can be

= t

12

A(Lti)

Assume t h a t

,

for a l l

s = 1.

t > 0.

see [ 1 9 , p . 38

256

Norman L . A l l i n g

( i ) el

Theorem.

is real.

r e a l complex c o n j u g a t e s .

are distinct,non-

is positive,

( i i i ) g3

t > 1, t = 1, o r

t < 1.

(iv)

i s p o s i t i v e , zero, negative, zero, o r p o s i t i v e according

a s t > 3'12,

x

(v)

(O,l),

E

,

t = 3

> t > 0.

3-1/2, for

e3

and

2

(ii) A < 0 .

zero, o r negative according a s g2

e

t = 3- 1 / 2

3112 > t > 3- 1 1 2 ,

I

or b (x)

i s t h e r e l a t i v e minimum of

1/2

-

t h e r e l a t i v e minimum v a l u e b e i n g

of c o u r s e

-

e 1' el E ? ( 1 / 2 ) .

Proof.

el

By Theorem 1 4 . 2 1 ,

is real.

fi p ( 1 / 4 + t i / 4 ) . (Note: w e have 3 a d o p t e d t h e c o n v e n t i o n of numbering t h e e I s u s e d by [ 3 6 1 . )

e2

15 ( 3 / 4 + t i / 4 )

and

p

By Theorem 1 4 . 2 1 ,

e

i s r e a l - v a l u e d on

of r e f l e c t i o n about t h i s l i n e , a r e symmetric p o i n t s ; t h u s

3/4

+

-

e3 = e 2 ;

7Ri

ti/4

+

1/2.

and

I n terms

1/4

+

ti/4

p r o v i n g ( i ) . By ( 7 . 3 3 : 2 2 ) ,

2 2 2 A = 1 6 ( e -e 1 ( e 2 - e 3 ) (e3-el) , a n d by ( 7 . 3 3 : 4 ) t h e e I s a r e 1 2 j U s i n g (i) o n e e a s i l y sees t h a t A 5 0 , d i s t i n c t ; thus A # 0. proving ( i i ) . C l e a r l y L(

LT ,) ;

thus

(1)

( i / t ) L T I = L1/2

(2)

Hence

(3)

92 ( ~ 1 / 2t i / 2 t )

If

t = 1

'3 ( L 1 / 2

then

g3(L1/2+i/21

+ti/2)

equals

2 3 27g3 / ( 9 2

i/2tI

for a l l

= t 4 g 2 ('112

+

-

i/2

# 0.

< 1;

27g3)

t > 0.

6 = -t g 3 ( L 1 / 2 t t i / 2 )

+i/2t)

thus

,

+ti/2)

= T'.

As w e saw,

= 0.

J(1/2

+ti/2

+

1/2

t > 1,

93 (L1/2

is a basis of

{ti, 1/2-ti/21

I

and

*

( 2 ) t h e n shows t h a t

i n C o r o l l a r y 5 , 59.28, f o r J(LlI2

+

ti/2)

- 1,

0;

a n d so

i s less than

Using ( 2 ) w e see t h a t

which

t = 1 is t h e only

Species, Geometric Moduli, Defining Equations 93 (L1/2 + ti/2)

zero of

the rest of (iii).

DuVal [19, p. 33 ff.] establishes

Since

J(1/2+ti/2)

(Corollary 5, §9.28)), and since for all

t > 1.

Now let

T E Q

let

Since

< 1

for

t > 1 > 0

J = g2 3/A,g2(1/2+ti/2)

J ( p ) = 0 (9.28:7), g2(1/2+ 31i2i/2 = 0.

such that

1 ~ =1

1

and

be in case 3 of 512.3; then

T

257

0 < Re-r < 1/2:

(9.28).

(0,l)

J ( T )E

i.e.,

Using the Schwarz reflection principle, reflecting across the circle of radius J(1/2+ti/2)

1

and center

(0,lI

E

[19, p. 451.)

for all

1, we know that

31i2 > t > 3

(see e.9.

Using (3), we can establish the rest of (iv).

Using [19, pp. 38-391, (v) can be established, proving the Theorem. 14.24

Vorlesungen

Bibliographic note.

...

[64,

In Chapter 30 of Weierstrass's

pp. 264-2751 he applies some of his earlier

derived results and formulas to the case in which the numbers q2

and

q3

are real.

Many of the results presented thus far

in this Chpater can be found there.

For example, Weierstrass

notes that the study naturally breaks into two cases:

A

>

0,

el > e

and (11) A < 0. > e3.

In case (I) he noted that

is pure imaginary.

one of the

R

Further he noted that a basis

the period lattice could be chosen so that w2

e.'s 3

(I)

w1

5

(wp 2 )

t

of

is real and

In case (111, Weierstrass noted that

is real and that the others are a pair of

conjugate complex numbers. chosen so that it was

He further noted that Q could be (1 1/2 + ti/2) t , for some t > 0.

Most of the function theory theorems thus far presented in this chapter are well known.

See e.g., Chapter 2 and 3 of

DuVal's very useful little book [191.

Norman L . A l l i n g

258

- a s elsewhere i n P a r t

What may be n o v e l h e r e

-

monograph

I11 o f t h i s

i s t h e a s s o c i a t i o n of t h e a n a l y t i c function theory

o f r e a l e l l i p t i c f u n c t i o n s , a s found by E u l e r , L e g e n d r e , Gauss,

-

Abel, J a c o b i , W e i e r s t r a s s , Klein e t a l l w i t h t h e a l g e b r a i c

g e o m e t r i c t h e o r y o f K l e i n s u r f a c e s a s d e v e l o p e d by K l e i n , Witt, S c h i f f e r and S p e n c e r , A l l i n g and G r e e n l e a f , e t a l .

Species 0

14.30

Assume now t h a t

t > 0,

with

by u s i n g

9

and

s = 0; 5 K

9'

and

+

t h u s by ( 1 4 . 1 0 : l

1/2.

it i s

and 2 )

T I

5

ti,

Having m e t w i t h s u c h s u c c e s s

perhaps

a l i t t l e surprising to

learn that (1) 9

Indeed,

p

and

o ( ' p )( 2 )

a r e not i n 5

p e r i o d i c of period t r u e of

9I

,

( z +1 / 2 )

~'p

1/2,

E. = 'p ( z

+ 1/2).

w e see t h a t

'p

Since

o (lp) # 'p

.

is not

The same i s

e s t a b l i s h i n g (1).

Pedagogical

note.

I t seems t h a t o n e o f t h e g r e a t d i f f i -

c u l t i e s s t u d e n t s have when t h e y f i r s t b e g i n t o t r y t o do research i s t h a t mathematics, a s i t appears i n t e x t s , i n lect u r e s , and even i n r e s e a r c h j o u r n a l s , i s n o t o n l y p o l i s h e d ; b u t t h a t t h e i n v e n t o r s ( o r d i s c o v e r e r s ) o f t h e mathematics have been so t h o r o u g h a b o u t c o v e r i n g up t h e way i n which t h e i d e a s came t o them.

I n a n e f f o r t t o s h e d a l i t t l e l i g h t on how

some o f t h e r e s e a r c h was c o n d u c t e d i n a r r i v i n g a t t h e r e s u l t s o f t h e t h i r d p a r t o f t h i s monouraph, which may be of u s e t o s t u d e n t s and may a l s o b e of i n t e r e s t t o o t h e r s , 5914.31

-

14.33

a p p e a r s h e r e u s i n g t h e methods and t h e o r d e r o f t o p i c s a s t h e y a p p e a r e d i n t h e f i r s t d r a f t o f t h i s monouraph.

(Of c o u r s e

t h e y were much messier and more c o n f u s e d t h e r e , b u t t h i s i s

S p e c i e s , Geometric Moduli, Defining Equations

259

how t h e i d e a s e v o l v e d . )

One o f t h e s t a n d a r d a p p l i c a t i o n s o f t h e Riemann-

14.31

Roch Theorem i s t o show t h a t c e r t a i n e l e m e n t s e x i s t i n a n a l gebraic function f i e l d .

510.50 f o r a s t a t e m e n t of

(See e . g . ,

t h e Riemann-Roch Theorem i n t h e complex c a s e a n d e . g . , f o r t h e g e n e r a l t r e a t m e n t i n t h e complex c a s e .

[ 2 9 , 571

See e . g . ,

[ 1 6 , C h a p t e r 111 f o r t h e Riemann-Roch Theorem f o r g e n e r a l a l -

[4, 531 f o r t h e t h e o r e m i n

gebraic function f i e l d s , o r e.g.,

Our n o t a t i o n w i l l b e c o m p a t i b l e w i t h [ 4 ] . )

the r e a l case. yo

Let

be a p o i n t i n

y t i c Klein b o t t e ; thus f i e l d of c o n s t a n t s , field a t

yo

(1) L e t

b

then at

b yo

is not i n

yo

-

of

R,

is

E

i s IR- i s o m o r p h i c t o :-

i s of d e g r e e

ord b,

(3

The g e n u s

aY.

Even t h o u g h t h e

IR,the residue c l a s s (11.20).

C

X! y o 1 ;

i s a d i v i s o r on

(2

w h i c h w e know i s a d i a n a l -

Yo,t'

2

over

t h e o r d e r of g

b,

Yo,t

of

Since t h e residue c l a s s f i e l d

Yo,t'

IR, [ 4 , p.26 1 .

i s -2

i s , by d e f i n i t i o n , i (b)

To compute t h e i n d e x o f s p e c i a l t y

of

1.

b

(see e . g .

,

[4, p . 31]), w e may u s e t h e u s u a l d e v i c e , The S e r r e D u a l i t y Theorem (see e . g . ,

Yo,t (4)

such t h a t

[4,

3.91).

Given a d i f f e r e n t i a l

-

b > 0,

then

(w)

w

w

on

i s zero: t h u s

i ( b ) = 0.

By t h e Riemann-Roch

(5)

k(b) = 2:

(6)

i.e.,

L(b)

{f

Theorem (see e . g . ,

E

E(Y):

(f)

+

b

2

[ 4 , 3.81)

0)

, w e see

that

i s of d i m e n s i o n

260

Norman L . A l l i n g

2 over

IR.

Clearly

1 is i n

Clearly

{l,fj

element

h

and

b

(7)

h

7

poles a t

x

p

Let

-1

for

=

and

0

Clearly

and

xo

1,

with

and x1

Xti,

a

E

being

i s a map

h

A s a consequence,

( 1 0 . 4 0 ) o r [ 6 , p p . 95-1041,

[ C ( h ) : IR(h) ]

= 2

and

for

[F:E] = 2 ;

IR(h)] = 2 .

[E:

k

Let k

14.32

E

E - IR(h);

then

E = IR(h,k)

i s algebraic over

.

xo,

L e t us choose

and l e t

yo

i t s poles i n where

Thus w e h a v e

IR(h) o f d e g r e e 2 .

in

Yo,t

.

E (Yo, t)

t o be

let

pg(O),

q ( 1 / 2 ) = xl.

(1) W e want t o d e f i n e a n e l l i p t i c f u n c t i o n P(R) simple poles a t

R : (1 t i )t

0

Q

in

having

E

and 1 / 2 ,

.

I n 514.31 w e saw t h a t s u c h f u n c t i o n s e x i s t .

Fleierstrass zeta

f u n c t i o n s ( ( 7 . 3 0 : l ) and ( 7 . 3 4 ) ) a r e p a r t i c u l a r l y w e l l s u i t e d f o r t h i s p u r p o s e (Theorem 7 . 4 3 ) . (2)

Let

IR*

has its only

h

l e a r n e d s o m e t h i n g a b o u t d e f i n i n g e q u a t i o n s of

q(0)

b

( 1 2 . 3 6 ) and ( 1 3 . 1 5 ) ;

( ( 6 . 4 1 ) and ( 6 . 4 2 ) ) .

2

more d e t a i l s . )

Clearly

+

each being simple; t h u s

xl,

(See e . g . ,

(8)

{xo,x,}

(yo) =

j = 0

[F: C ( h ) 1 = 2 .

thus

Hence a n y o t h e r

IR.

af

L ( b ) - IR.

0'

of o r d e r

Xti

E

t h a t pole being a simple

Yo,t'

As a meromorphic f u n c t i o n o n

distinct.

of

y

F(Xti).

S(x.1

over

i s of t h e form

h a s o n l y o n e p o l e on

is i n

then

L ( b ) - IR

L(b)

f

Clearly

pole a t h

i s a b a s i s of

in

7R.

E

thus there exists

L(b);

Q ( z ) : i [ < ( z )- 5 ( 2

- 1/2)

- q1/2I,

for a l l

z

E

C,

S p e c i e s , Geometric M o d u l i , D e f i n i n o E q u a t i o n s

<

where

nl

and

261

is Weierstrass's zeta function f o r t h e lattice = 2<(1/2)

zero, i t s pole i n poles on

is

P(Q),

(7.34:2).

1

these being a t

P(Q),

i

these pointsare

-i

and

0

and

Lti

c(z) at

The r e s i d u e of

(7.34:9 and 1 0 ) .

has only simple

Q

The r e s i d u e s a t

1/2.

r e s p e c t i v e l y ; t h u s by Theorem

7.43,

is i n

(3)

Q

Let

L :L

(4)

S(X)

ti ' E

F.

-

L = L,

Since

IRU

for all

Cml,

x

IR.

E

I n d e e d , u s e t h e e x p a n s i o n ( 7 . 3 4 : 2 ) t o show t h i s . that

(14.10:4)

i s t h e IR-automorphism o f

o

a ( Q ) ( z ) = K Q ( Z +1 / 2 )

its fixed field. 11,/2I)

- i ( Z + 1/21 +

= i[T(Z)

is real.

+

5 ( z - 1 / 2 + 1) = (5)

is i n

Q

(z

-

+

1/2)

-

Since (6)

Q(-z)

= -Q(z)

,

q1/2].

1

By ( 7 . 3 2 : 7 )

,

<(2+1/2) =

proving t h a t

ql,

for all

(14.31),

is i n

Q

E

z

@:

E

i.e.,

i s a n odd

Q

U s i n g ( 4 ) o n e sees t h a t for all

Q(K(z)) = -KQ(z),

Q

( = 25 (1/2)

rll

(z)-

E.

values i n

(8)

5

i s a n odd f u n c t i o n ( 7 . 3 4 : 2 )

5

n

since

-

IR.

function.

(7)

= ~ ( i [ ( cZ + 1 / 2 )

as

E

C ( z ) = ~ ( z; ) t h u s

I n d e e d , i n terms of t h e n o t a t i o n o f L(b)

having

_ _

U s i n g ( 4 ) w e see t h a t

a ( Q )( z ) = i [ < ( z ) - < ( 2 + 1 / 2 )

,

n1/2]

F

Recall

IRi u

{m}

on

z

E

@;

thus

IR, a n d h e n c e o n

Q

takes its IR+ n t i ,

for

Z.

takes its values i n

Indeed, l e t

x

E

IR; t h e n

IRi

on

Q ( x +t i / 2 )

IR+ t i / 2 . = Q(K (x- ti/2))

equals,

262

Norman L . A l l i n g

by ( 7 )

- ~ Q ( ~ - t i / 2 ) . Since

t h i s l a s t quantity equals Q

is p e r i o d i c of period

(9)

each

n

E

+

-KQ(x

proving ( 8 ) .

ti/2),

ti,

Since

w e t h e n have

ti

takes i t s values i n

Q

i s p e r i o d i c of period

Q

IRi u

{m)

on

IR+ n t i / 2 ,

for

2.

Combining ( 6 ) and ( 7 ) w e f i n d t h a t (101

Q(-z)= on

I R i + n/2,

,

f o r each

n

IRu {ml

of period

1

n

2.

E

IRu

i t i s odd.

o n lRi i s o b v i o u s .

~Q(n+1/2+iy= )

y

That Since

IR and l e t

be i n

KQ(c(n-iy)) =

a q u a n t i t y which i s i n

IR u

-Q(F)

Using ( 7 1 ,

assumes

Q Q

i s periodic

i t assumes v a l u e s i n t h i s s e t on e a c h Let

{m}

Z.

E

establishing the f i r s t result.

values i n

where

takes i t s values i n

Q

Q(-z)= -Q(z), s i n c e

Indeed, = KQ(z)

thus

KQ(z);

n

be i n

IRi+n, 2.

o(Q)(n-iy) = Q(n-iy), establishing (10).

{a};

By

construction,

(11) t h e p o l e s o f 1/2

Thus

+

L,

Q

on

C

a r e a t t h e p o i n t s of

and

each being simple.

h a s two s i m p l e p o l e s i n

Q

L

P(Q),

U s i n g ( 9 ) and (10) w e

see t h a t (12)

t h e zeros of

Q

+ ti/2 +

L,

1/2 Since

(13)

5' = -'p.

ti/2

+

L

and o f

each being simple.

(7.34:4),

Q ' ( z ) = i " p ( z - 1/21 -!p(z)l

(14) t h e p o l e s of

Q'

t h e p o i n t s of Since

a r e a t t h e p o i n t s of

Q'

(15) Q '

on

C

and

L

a r e e a c h d o u b l e and t h e y a r e a t 1/2

+

h a s two d o u b l e p o l e s o n

i s of o r d e r

4

L.

P(Q),

( ( 6 . 4 1 ) and ( 6 . 4 2 ) ) .

S p e c i e s , Geometric Moduli, D e f i n i n g E q u a t i o n s

h a s 4 z e r o s on

As a c o n s e q u e n c e , Q'

1/41]

-

1/41

I,

9

since

=

= 0.

- ti/2)

= i['p(1/4

0 (1/4 + t i / 2 )

ti

Since

; thus

takes its values i n

Q'

Since

'p(1/4 + t i / 2 1

i s a n even f u n c t i o n .

+ (1/4 - t i / 2 )

= 0.

9,

p(3/4)] = i [ ~ ( 1 / 4 )- ?(-1/4)1

+ti/2) -

p,

Q'(1/4) =

1 i s a period of

Since

:i[+(-1/4

is a period of Q(l/4 +ti/2)

0.

=

L e t u s f i n d them.

P(fi).

i s an even f u n c t i o n ( 7 . 3 2 : 6 ) ,

Since

26 3

on

Wi

[19, p . 3 8 , F i g u r e 1 1 , w e c a n u s e t h e Schwarz re-

IRi+ 1/2

flection principle t o reflect doing s o w e f i n d t h a t

1/4,

shown t h a t z e r o s of

Q'

across t h i s line.

Q'(3/4 + t i / 2 )

+

1/4

3/4,

in

Q'

ti/2,

Since

P(fi).

found a l l o f t h e z e r o s o f

in

Q'

each of t h e s e z e r o s i s simple.

Hence w e h a v e

= 0.

a n d 3/4

+

P(Q)

are a l l

ti/2

has order 4

Q'

On

(15) w e have

a n d w e know t h a t

Thus w e h a v e e s t a b l i s h e d t h e

following. (16)

The z e r o s o f t h e p o i n t s of

u (17)

(3/4

(1/4

+ t i / 2 ) + L)

The numbers Q3/4)

on

Q'

c

+ L)

a r e a l l s i m p l e and t h e y a r e a t

u

+ L)

u

(1/4

+ t i / 2 + L)

.

Q(1/4), Q(1/4

are a l l i n

(3/4

IRi.

+ti/2) ,

Q(3/4 + t i / 2 ) ,

and

F u r t h e r , t h e f i r s t and f o u r t h

are c o n j u g a t e , a s a r e t h e s e c o n d and t h i r d . I n d e e d , by ( 9 ) t h e y a r e a l l i n v a l u e d on

I R i + 1/2.

IRi.

By (10)

Q

i s real-

U s i n g t h e Schwarz r e f l e c t i o n p r i n c i p l e

w e o b t a i n t h e rest. (18) Since P(Q)

The f o u r numbers g i v e n i n ( 1 7 ) a r e d i s t i n c t . Q

i s o f o r d e r 2 (11) i t c a n assume t h e s a m e v a l u e i n

only t w i c e , counting m u l t i p l i c i t y .

Since

1/4,

264

Norman L . A l l i n g

1/4+ti/2, Q

3/4+ti/2,

and

3/4

a r e t h e zeros of

P(Q),

in

Q'

a s s u m e s e a c h o f t h e s e p o i n t s d o u b l y ; h e n c e t h e v a l u e s of

Q

a t those four points a r e d i s t i n c t .

(i) (Q')

Theorem

2

= -(Q-Q(1/4))

(Q-Q(3/4+ t i / 2 ) ) ( Q - Q ( 3 / 4 ) ) :g(Q) in

-(x

IR[Q].

E

i s not

(ii) Q '

(iii) E ( Y O I t ) = I R ( Q , Q ' ) . (iv) g(x) = 2 + b ) , where a = Q ( 1 / 4 ) / i and b = Q ( 1 / 4

IR(Q).

2

(Q-Q(1/4+ti/2))

2

2 + a ) (x

IR.

are in

Since

Proof.

p o l e s on

and of

(18) ,

map o f hence

1, Q'

Xti

g(Q)

and t o t h e same o r d e r ,

U:

must b e

and

(Q')2

z e r o complex number

(Q')2

+ti/2)/i

(6.31).

g(Q)

at

0

onto

C

( Q ' ) 2/g (Q)

each s t a r t o u t with g(Q)

IR(Q), proving

o f o r d e r 2 (11),

[ E ( Y O I t ) : IR(Q) ] = 2 .

:X

i s a non-

Since t h e Laurent expansion of

proving ( i ) . Since

cannot be i n

h a v e t h e same z e r o s and

-l/z

4

, X

has 4 d i s t i n c t r o o t s i i ) . Since

Q

is a

[F(Xti) : @ ( Q )1 = 2 ;

U s i n g ( i i ) w e see t h a t ( i i i ) h o l d s .

( i v ) f o l l o w s d i r e c t l y from ( 1 7 ) .

14.33

W h i t t a k e r and Watson w r i t e a s f o l l o w s a t t h e

b e g i n n i n g of C h a p t e r X X I I , t h e i r c h a p t e r on J a c o b i ' s e l l i p t i c functions:

" I n t h e course of proving g e n e r a l theoremconcern-

i n g e l l i p t i c f u n c t i o n s a t t h e b e g i n n i n g o f C h a p t e r X X , i t was shown t h a t two c l a s s e s o f e l l i p t i c f u n c t i o n s were s i m p l e r t h a n any o t h e r s s o f a r a s t h e i r s i n g u l a r i t i e s were c o n c e r n e d , namely t h e e l l i p t i c functions of order 2 .

The f i r s t c l a s s c o n s i s t s o f

those w i t h a s i n g l e double p o l e (with zero r e s i d u e ) i n each

c e l l , t h e s e c o n d c o n s i s t s of t h o s e w i t h t w o s i m p l e p o l e s i n e a c h c e l l , t h e sum o f t h e r e s i d u e s a t t h e s e p o l e s b e i n g z e r o .

S p e c i e s , Geometric Moduli, Defining Equations

"An example o f t h e f i r s t c l a s s , namely

cl? ( z ) ,

265

was d i s -

cussed a t l e n g t h i n Chapter XX; i n t h e p r e s e n t c h a p t e r w e s h a l l d i s c u s s v a r i o u s examples o f t h e s e c o n d c l a s s , known a s J a c o b i a n e l l i p t i c functions."

p . 4911

[69,

Clearly

is i n t h e

Q

s e c o n d c l a s s o f t h e s e f u n c t i o n s ; t h u s i t is n a t u r a l t o t r y t o

write

i n terms o f s t a n d a r d J a c o b i a n e l l i p t i c f u n c t i o n s .

Q

W e d e s c r i b e d J a c o b i ' s work on h i s e l l i p t i c f u n c t i o n s i n I t w i l l b e h i s s i n a m f u n c t i o n ( 4 . 2 0 : 3 ) t h a t w i l l b e most

94.2.

concerned. W e w i l l u s e Gudermann's n o t a t i o n , function (4.23).

sn,

t o denote t h i s

The most c o n v e n i e n t way t o d e f i n e t h e J a c o b i a n

e l l i p t i c f u n c t i o n s f o r a g e n e r a l complex modulus i s by means of t h e t a functions ( 5 . 3 1 : l ) . [361 l e t

F o l l o w i n g most of t h e n o t a t i o n a l c o n v e n t i o n s of

(1) w then

5

1/4

and l e t

w'

Z

is defined ( 5 . 3 1 : l ) .

sn

then

ti/2;

T

= 2ti

By Theorem 5 . 3 1 ,

[36,

sn

p 190 f f . ] ;

is i n

F (Xti) has zeros a t t h e p o i n t s of

(2)

L

and

1/2

+

L , each being

si m p l e ; (3)

and 1/2

sn

+

h a s i t s p o l e s a t t h e p o i n t s of

ti/2

+

ti/2

+

and o f

L

L, each being simple.

Comparing ( 2 ) and ( 3 ) , on t h e o n e h a n d , w i t h t h e d e s c r i p t i o n o f t h e p o l e s and z e r o s o f Qsn = A ,

f o r some

A

in

has a simple pole a t

0

which h a s r e s i d u e

(4) Q

( 1 4 . 3 2 : 1 1 and 1 2 ) w e see t h a t

Q

simple zero a t

0.

W e have s e e n ( 5 . 3 1 : 2 )

(5)

@*.

L e t u s compute

sn'(0).

that

2 2 2 ( s n ' ) 2 = ( 1 - s n ) (1-k s n ) ,

where

i.

sn

has a

Norman L . A l l i n g

266

k

i s L e g e n d r e ' s modulus ( 5 . 3 1 : 3 ) ;

(6)

s n ' ( 0 ) = i- 1.

thus

sn'(0) = 1 sn'(u) = cn(u)dn(u).

Indeed,

169, p. 4921.)

Since

[36, p. 2161 o r

(See e . u . ,

( 6 ) is established.

cn(0) = 1 = dn(O),

Thus w e h a v e proved t h e f o l l o w i n g :

Q

Theorem.

-

i/sn.

Having d i s c o v e r e d t h a t

Remark.

course

=

have d e f i n e d

Q

t o be

i/sn

Then w e c o u l d have d e r i v e d ( 1 4 . 3 2 : 2 ) . the species

0

w e could

Q = i/sn,

-

of

i n the f i r s t place. Note a l s o t h a t t o t r e a t

case w e a r e n a t u r a l l y l e a d t o s t u d y t h e

Jacobian e l l i p t i c functions.

14.34

From Theorem 1 4 . 3 3 and ( 1 4 . 3 3 : 5 ) w e o b t a i n

Reca 11 ( 5 .3 1:3 ) t h a t (2)

2

k

02/0,

2

e2

One c a n see d i r e c t l y from t h e d e f i n i t i o n o f

and

[ 3 6 , p . 1961 t h a t t h e y a r e r e a l and p o s i t i v e ; t h u s

O3 0 < k.

However, i n g e n e r a l (3)

k2 = ( e 2 - e 3 ) / ( e l - e 3 )

W e have s e e n i n Theorem 1 4 . 2 2 k'

> 1

(4)

[ 3 6 , p . 2281. (vi) that

thus

and h e n c e ,

0 < k < 1,

i n t h e case u n d e r c o n s i d e r a t i o n .

I n Theorem 1 4 . 3 2 w e f o u n d t h a t where

e1 > e2 > e3;

a

w a s d e f i n e d t o be

Q(1/4)/i

Using Theorem 1 4 . 3 3 w e know t h a t b = l/sn(1/4+ti/2).

(Q') and

=

- ( Q2 + a 2

(Q

2

+ b2

b E Q(1/4+ti/2)/i.

a = l/sn(1/4)

and

T h e s e q u a n t i t i e s a r e w e l l known (see e . g . ,

I

S p e c i e s , Geometric M o d u l i , D e f i n i n g E q u a t i o n s

[ 6 9 , p . 498 f f . a n d p . 502 f f . ] w h e r e

K' = t / 4 ) : (5)

K = w = l j 2

267

and

namely

a = 1 and

b = k.

I t w i l l be convenient t o d e f i n e a n o t h e r f u n c t i o n i n E(Yort) 1 (6)

,

namely

-1/Q = i s n .

3

Clearly w e have t h e following: T h e o r e m . E = l R ( 3 ,3

= -(l+rX2)(l+k%'2).

(7) Let L

and

I ) ,

urvl

UIVIW

and

w

be i n

and r e c a l l (3.21:lO)

{0,1)

w 2 2 ( x , k ) 5 ( - l ) u ( lk-1 ) v x 2 ) (1- (-1) k x ) .

that

Thus ( 7 ) i s i n

t h e f o l l o w i n g g e n e r a l i z e d L e g e n d r e form

14.40

Other q u a r t i c d e f i n i n g equations

Having f o u n d t h a t o u r s e a r c h f o r a d e f i n i n g e q u a t i o n f o r species

0

l e a d s us q u i t e n a t u r a l l y t o q u a r t i c d e f i n i n g equations

(Theorem 1 4 . 3 2 a n d ( 1 4 . 3 4 : 7 a n d 8 ) ) , a n d t o J a c o b i a n e l l i p t i c f u n c t i o n s , it i s n a t u r a l t o look f o r q u a r t i c d e f i n i n u e q u a t i o n s f o r the other species.

let

-5

14.41.

:K;

Let

then

s = 2

Y2,t

and l e t

:X

t h i s Chapter f o r d e t a i l s . )

t > 0.

ti/s. Let

Let

L

z

L

ti

and

(See t h e e a r l y s e c t i o n s of sn

be d e f i n e d as i n ( 1 4 . 3 3 ) ;

then

(1)

2 2 2 (sn')2 = (1-sn ) ( l - k sn ) = L

(See (14.33:5) and ( 3 . 2 1 : 1 0 ) . )

0 ,0 ,0 ( s n , k l

As n o t e d i n 514.33,

sn

is i n

268

Norman L . A l l i n g

F(X~-)

(2)

=

a: ( s n , s n ' ) .

I n d e e d , o n e may a r g u e a s w e d i d i n t h e p r o o f o f Theorem 1 4 . 3 2

(iii); e s t a b l i s h i n g ( 2 ) . (3)

sn(u)

IR, f o r a l l

E

u

E

IR.

I n d e e d , t h i s may b e c h e c k e d ( e . g . ,

i n [ 3 6 , pp. 1 9 0 - 2 1 3 1 ) , j u s t

by l o o k i n g a t t h e v a r i o u s d e f i n i t i o n s ; s i n c e

are a l l r e a l [36, p. 1901.

h , u , and v

20, @ o f eo(v), @,(v),

thus

A s a consequence

Hence

tween Jacobi

( s n ' ) 2= L O , O , O ( ~ n , k )

The case i n w h i c h

k

i s r e a l and be-

1 i s t h e c l a s s i c a l case t h a t i n i t i a l l y c o n c e r n e d

and

0

where

1.

k

0

H i s t o r i c a l note.

. 14.42

Yl,t

2ti;

a l l r e a l [36, p.1961.

= IR(sn, s n ' ) ,

E(Y2,t)

Theorem.

T'

is

i s r e a l [ 3 6 , p . 2131.

sn(u)

and

@ia r e

and

T

+ ti/2

:1 / 2 5

s

Let

xT,./5.

and

= 1

and l e t

-

L :L

as always T"

-

-

let

5 :K ;

Let

t > 0.

Let

then

( S e e t h e e a r l y s e c t i o n s o f t h i s c h a p t e r f o r more

details.

'I

W'/W

= 1

136, p . 1901 (2)

and

w 5 1/4

(1) L e t

sn

is i n

+

w'

: ~ ' / 2= 1 / 4

h = -e -llt ,

ti,

sn(u)

E

IR, f o r a l l

m

1 + 2

u

E

u (sn) (u) : csn(u) = sn(u),

W e have s e e n (5.31:3) t h a t

=

then

a n d z = ei n v = e 2 r i u

L e t u s now c o n s i d e r L e g e n d r e ' s m o d u l u s

e3

ti/4;

iR.

E(Yl,t);

s i n c e (1) h o l d s ,

(3)

+

In=1 hn

2

k

f

2 2 R2/e3.

[36,p. 1961.

k.

establishing (2).

S p e c i e s , G e o m e t r i c Moduli, D e f i n i n g E q u a t i o n s

o3

(1),

h = -e -'t

Since

269

is real.

C l e a r l y t h i s can be w r i t t e n a s follows:

e2

(5)

= 2h

h = ei r

e- t r

thus

;

e-tr/4f

( 6 ) h1l4 = ei'/l (7)

k

E

2 hn -n

m

1/4

and h e n c e

lRi

proving t h e following:

k2 < 0.

k L = ( e 2- e 3 ) / ( e l - e 3 )

Since

are d i s t i n c t ( 7 . 3 3 : 4 )

, we

(14.34:3)

,

and s i n c e t h e

e 's j

have proved t h e following;

k2 < 0.

Lemma.

I t i s very easy t o prove t h a t

(8)

t h u s w e have

F(X . ) = C ( s n , s n ' ) ;

tl

E = IR(sn,sn'),

Theorem.

where

It is interesting t o notice t h a t t h i s differential

equation i s a s p e c i a l c a s e of Abel's d i f f e r e n t i a l equation. (10) where

2 2

c

e

and

14.43.

(1) c

2 2

( w ' ) ~= ( 1 - c w ) ( l + e w )

1

are

(4.12:3),

non-zero r e a l numbers.

Let

and

2 1/2

e : (-k )

;

t h e n A b e l ' s d i f f e r e n t i a l e q u a t i o n ( 4 . 1 2 : 3 ) becomes

Norman L . A l l i n g

270

( a s d e f i n e d i n ( 1 4 . 4 2 ) ) i s a meromorphic s o l u t i o n o f

sn (u

Abel s e l l i p t i c f u n c t i o n

r

radius

r :: m i n ( 1 , e )

Let

t o (2).

@

about

In

0.

(94.1)

i s another global solution

and l e t V

s i n g l e valued square r o o t .

be t h e open d i s c o f

V

t h e r i g h t hand s i d e o f ( 2 ) h a s a 2 2 2 1/2 be Let [ ( l - w ) ( l + ew ) I

t h e s q u a r e r o o t t h a t i s p o s i t i v e on

(0,r).

e q u a t i o n t h a t Abel c o n s i d e r e d on

is

$

s a t i s f i e s ( 3 ) [l, V o l .

sn'(u) = cn(u)dn(u)

[36, p. 2181; t h u s (3).

Since

w'

= 1/4

V

The d i f f e r e n t i a l

1, p . 2 6 8 1 .

[ 6 9 , p . 4921 and s n ' ( 0 ) = 1;

cn(0) = 1 = dn(0)

hence

sn

also satisfies

$ ( 0 ) = 0 = s n ( O ) , w e have p r o v e d t h e f o l l o w i n g :

r$ = s n ,

Theorem.

and

(2).

+

Corollary

ti/4,

where and where

E(YlIt)

Historical

note.

=

sn

is defined f o r

r$

w = 1/4

c = 1

i s defined f o r

n(@i@')

Since A b e l ' s paper i s w r i t t e n i n a

s t y l e t h a t i s n o t i n conformity with p r e s e n t standards of r i g o r i t c o u l d p e r h a p s b e a r g u e d t h a t Abel d i d n o t p r o v e t h a t h e had found a g l o b a l meromorphic s o l u t i o n o f (2).

After a l l

t h e n o t i o n o f a meromorphic f u n c t i o n h a s n o t b e e n f u l l y f o r m a l i z e d by 1 8 2 7 .

A t t h e very l e a s t it can be a s s e r t e d t h a t

sn,

as d e f i n e d h e r e w i t h t h e t a f u n c t i o n s , i s a s o l u t i o n o f ( 2 ) which e n j o y s a l l t h e p r o p e r t i e s t h a t Abel a s s e r t e d t h a t

$

had.

a u t h o r i s i n c l i n e d t o f e e l t h a t v i r t u a l l y e v e r y t h i n g on t h i s s u b j e c t a s s e r t e d by Abel and Gauss c a n b e p r o v e d w i t h o n l y a few a d d i t i o n a l comments.

A t t h e t i m e of p u b l i c a t i o n Abel's

The

S p e c i e s , Geometric Moduli, D e f i n i n g E q u a t i o n s

... was

Recherches

regarded,

271

for example by G a u s s , a s b e i n g

w r i t t e n a t a v e r y h i g h l e v e l of r i g o r .

(See O r e [ 5 2 ]

for

details. )

14.44

I n SVIII o f A b e l ' s R e c h e r c h e s

....

he t u r n s h i s

a t t e n t i o n t o t h e l e m n i s c a t e i n t e g r a l , 51.3, which Gauss h a d s t u d i e d e x t e n s i v e l y by 1 7 9 7 ( 4 . 3 1 ) .

lets

e = c = 1

function

9

To do t h i s A b e l m e r e l y

[l, V o l . I , p . 352 f f ] .

On d o i n g t h i s A b e l ' s

e q u a l s G a u s s ' s s i n l e m n (4.31:l).

This Page Intentionally Left Blank

CHAPTER 15

THE DIVISOR CLASS GROUP OF

Ys,t

Introduction

15.10

The divisor class group was considered for compact Riemann surfaces of genus

g

equals

in 558.61 and 8.62.

and

0

1

in 18.6, and computed explicitly when

g

Given a compact Klein

surface,one can define its divisor class group. 5.71 it was computed, using sheaf cohomology.

In [ 4 , 5.6 and

We will now make

an entirely independent computation of the divisor class group for

without the use of sheaf theory, which is both more

Ysltt

elementary and more explicit.

(1)

Let be

(2)

s = 2 , 1 , or 0, and let

Let

15.11 T '

= ti

if

1/2+ti/2,

Let

iL

K

is

s=2 if

or

0,

tclR,

with

t > 0.

and let it be defined to

s=l.

s = 2 or

1, and let it be

~ + 1 / 2 if

s=o.

Let

5

XT,

be the anti-analytic involution of

then

YsIt

(3)

5 {b: Y + Z such that s,t s,t on a finite subset of Yslt1.

Let

Let

is defined to be

XT,/(

(or E

for short)

b

is zero except

be the field of all meromorphic Given

let

associated with the point

Y

273

f

E

E*

and

YSlt,

"functions" on v

[6, 1.31. ys,t be the valuation of E

5;

(§12.3).

div Y

E(Yslt)

induced by

Y E

y.

274

Norman L. Alling

(Since y

is

contains IR, (4)

Let

(5)

then

u

by definition - a valuation ring of Y

group

E

F*-+ (f)

E*

that

is just its valuation.)

(f)(y) E uy(f),

f

E

for all

Y E

Ysrt;

is a homomorphism of the multiplication

into the (additive) group

div(Ysrt), whose

kernel is R*. A divisor in

(E*) is called a principal divisor.

The most fundamental question is to find necessary and sufficient conditions for

b

div Y

in

s,t

to be principal.

A

closely related question is to compute (6)

C(Ysrt) div Y srt/(E(Ysrt)*)r

To do this it will be useful Ys,t. to define the notion of the degree of b. Roughly this measures the divisor class group of

the number of zeros of

b

minus the number of poles of

However, the degree of

b

must also take into account the degree

of the residue class field of each [16, 1.71 for details.)

is either IR

where

2 cy

E

(See e.g.,

Since the residue class field of

or is IR-isomorphic to

correct definition of the degree of (7)

over IR.

y

b.

supp (b) - ay b(y) + ‘y

supp(b), the support

of

E

cr

y

the following is the

b:

supp (b) n ay b(Y) r

b,

is

{ Y E Y: b(y) f 0 1 ,

and

we understand that the sum over the empty set in (7) is the integer zero. onto

2.

(8)

Let

Clearly

deg

is a homomorphism of

div Y

srt

divO Ysrt: ker deg.

A divisor in shortly that

div0Ysrt is called homogeneous.

We will see

The Divisor Class Group of

(91

275

YsIt

(E(YsIt)*)cdivoYsIt. divoYs,t/(E(Ys,t)*) z

(10) Let

c 0 (Ys,t)'

This will be called the homogeneous divisor class group of ys,t. C(Ys,t)/Co(Yss,t) = Z.

(11) Clearly

In [ 4 , 5 . 7 1 it was shown that (12)

CO(YSIt) is isomorphic to

@/Z)

that it is isomorphic to I R / Z

Calculations gg

15.20

~EJZ 2

if

s= 2,

if

s=l

or

and

0.

xTl

Having made definitions and stated results (15.11:9 and 1 2 ) directly on

YsIt, let us proceed up to the covering space

X T , to prove them. and

p

c*(a) 5 a5 5*

(2)

is again in

div X T l

is an involution of

involution of

YsIt.

XT1

(See (13.15) for de-

a € div X T I ,

Given

Clearly

XT1

is an anti-analytic involution of

is its quotient map onto

tails.)

(1)

5

divO X T I

.

.

div X T ,

which induces an

(See §8.6 for details on divisors on

-1

X T l :{a E div E X i 1 : c*(a) = a } 5 sym divoXTl:sym div X T , n divoXTl 5 5

Let

sym div

and let

.

Clearly each of these sets is an additive group.

X T I onto (3)

and

p*(b) p*

Ysrtl

= bp

given is in

b

E

Since

div Y srt

div X T l ,

is an injective homomorphism.

Lemma.

(i) p*

maps

div Y srt

injectively onto

p

maps

Norman L. Alling

276

.

(ii) For all b sym div XT , 5 (iii) p* is an injection of

is injective. I

/5

5

Since

Ys , ,

div Ysft

into

a=p*(b); let

is in

aY;

point

x

a=bp.

then

p* (b),

Y E

Let

b

P*

X T l onto p*

proving that

maps

j= 0

3

deg a = 2.

Y

b c div Ysft; thus and, for the moment,

s,t

is in

X T l . Clearly

S(x.1 = x ~ - ~ for ,

p*(b)

=

xfxl

5

a,

and thus

then (15.11:7) deg b = 2,

j and

is

in

1.

X T l , xO#xl,

Thus

deg a = l .

and

but

p*(b) = x I x o 3 + x f x l l!a ,

Since elements of the form

div Ys,t,

basis of

sym divOX 5 is surjective,

div Ys,t. Assume first that y -1 deg b = 1 and p (y) consists of a single

then

in

=

Clearly

Assume now that y E Y - aY; -1 p (y) = Cxo,xll, where x

and

onto

is the quotient map of

proving (i).

x{YP

b:

deg b = d e g p*

sym div X T , . Let a E sym div X T l ; then 5 5 a = a 5 ; thus a induces a map b of XT,/C

such that

2,

p

S*p* (b) = bpg = bp

a : X T , + Z and into

,

div Ys,

divO ys,t We have noted that since p

Proof.

xT

E

(ii) is proved.

form a free XIYl (iii) then follows,

proving the Lemma.

E(

15.21

3

E(Y

may

))

t F( 5 F(XT , ) ) . Sf

subfield of

(1) o(f)

to be

K

f 5,

having

E

(F*) c divOXT (2)

(E*)

c

Lemma. Proof.

of course - be regarded as a

In fact, if we define for each

as we did in 514.10, then F

-

a

f

E

F,

is an IR-linear automorphism of

as its fixed field.

By Theorem 6.41,

(as noted again in (8.62:l) )

divoXTI

;

thus

.

(E*) = (F*) n sym divOXT,. 5 Let g E E* and let a: (g).

Since

g

E

F*, a

is

The Divisor Class Group of in

Since

(F*).

E* c (F*) n sym divOXTI . 5 (3)

let

f

be in

Let

u(f)/f - h ; then

such that

F*

h

=

Ehf;

hence

Let

1 hl

h = ei8

such that (4)

h

=

1.

and

Clearly

is in

sym divoX,,.

5

(h)=5*(d) - d = O .

By

is in C * .

Clearly a(f) =hf. 2 f = u (f)= u(hf) = g u ( f )

F,

Thus there exists a unique

then

g

E

E*

and

Indeed, u (9)= e-ie/20(f)=eiel2f = g ,

E*.

(15.20:l); and so

8,

02

6 <

(g)=

d,

(9)= d .

proving that

g

is in

proving the Lemma. Some of the ideas in Hilbert's Theorem

Bibliographic note.

90 (see e.g., [48, p.2131) were used in developing this proof.

Note in passing that (5) g

is uniquely determined by

d

up to multiplication by

a non-zero real number. Using Lemma 15.20 and Lemma 15.21 we have proved the following: Theorem.

c o ( Ys, t)

is isomorphic to

sym divoXTI / ( (F*) n

5

sym divoXTI 1 . 5

15.22

divoXT that

sum, defined in (8.62:3), is a homomorphism of

onto the group

XT

I

( 5 C/LT I )

.

We have seen (8.62:5)

ker sum= (F*).

(1) Let

sum 5

be the restriction of sum to

sym divoXT,.

5

Using (8.62:5), Lemma 15.21, and Theorem 15.21 we have proved the following:

271

.

eie'2f;

g

(f) E d

F*

is in

is the identity map of

u2

277

Conversely,

Liouville's Theorem (6.31), Since

S*(a) = a

(l),

u(g) = g

Ys,t

278

Norman L. Alling CO(YsIt) is isomorphic to

Theorem.

(2) Let, q

denote the homomorphism of Q:

kernel (3)

im sum

5'

onto

X T l having

LTl.

Recall also that

p

is the quotient map of

X T l onto

X T I / 5 Z Ys,t.

For

XT1,

X E

a(x)

is in

div X T l and

ao(x)

is in

divoXTl.

is a basis of the free Abelian group div XT Since divoXTl is a subgroup of

div X T l , it too is a free

Abelian group. (6)

(ao(x))x E X T I - {O}

is a basis of

divoXTl.

What will be of greater concern to us is a set of generators of

sym divoXTl.

5

(7) x

E

p-l(aY)

sym

5

ao(x)

d(l)

supp d(l) ,

By (7) and (8)I

be in

G

sym5divOX, I .

is a subset of

sym divoXTl. 5 and recall (8.60:3 ) that

sym divoXTl 5 the support of d(l) I

is

Assume that there exists some point If (i) x

is in

divoXT

Proof.

Let

if and only if

is in

p-1 (aY), then

x

Ix E XT I : d(l) (x) # 0 1 . in

d(2) :a(')

supp d(l) - 10,5 ( 0 ) 1 . -d(')

(x)ao(x)

is

The Divisor Class Group of

279

ys,t

-

(7) again in sym5 div0 X T I r and supp d(2) c (supp d I f (ii) x is not in p-1 (aY), then d(2) Ed(') -d(') is in

sym5 divoXTl (81, and

{O,t;(O)

1.

i.e.,

group generated by

(supp d

G,

d(k) = 0.

two components.

is a subgroup of

Further, q(ti/2)

such that

d(l)

is in the

proving the Lemma.

X T I , which is isomorphic to I R / Z .

is of order 2 in

(1) p-l (aY) is a subgroup of W Z ) @

Hence

kc N

- { X I S(XI 1) u

s = 2 ; then - by definition (512.2) - aY has p-1 (aY) consists of q m ) u q m + ti/2). Clearly

Let

15.23

,,

XT

XT,;

thus

which is isomorphic to

z2. (i) im sym5 = p-1 (aY); thus (ii) Co(Yzrt) is

Theorem.

isomorphic to

m/Z) @ Z 2 .

If

Proof.

in

c

Thus, by induction, there exists

supp d(k) c { 0 , 5 ( 0 ) 1 :

qp)

supp d(2)

u {OI. (x)bo(x) {XI)

x

p-1 (aY) then (15.22:7) ,

is in

ao(x)

is

sym divoXT,. Further,

5

(2) sum a (x)= x ; 5 0

(3)

thus

p-'(aY)

By (15.22:8), (4)

c

im sum

bo(x)

5'

is in

sym divoXT,. Clearly 5

sum b (x) = x + ~ ( x E)p-l(aY). 5 0

Using (2), (4) and Lemma 15.22, we see that

im sim 5

c

p-l( aY) ,

proving the Theorem.

15.24

Clearly

Let

s=l;

then

aY

and

p-l (aY) = q @ , Iand ) hence

p

-1

(aY) are connected.

Norman L. Alling

280

(1) p-l (aY) is a subgroup of

XT I , which is isomorphic to

IR/Z.

(i) im sum5 =p-’(aY);

Theorem.

thus

(ii) co(YlIt) is

isomorphic to I R / Z . The proof is virtually the same as that of Theorem 1 5 . 2 3 .

Let

15.25

s=O.

the cases in which Let

x

E

XT I

Since

s= 2

or

aY=$, p 1,

-1

im sum

5

(aY)= $ ;

thus, unlike

cannot be

p-l(aY).

then

;

(1) sum b (XI = x + ~ ( x -)0 - 5 ( 0 ) = x + E ( x )- q ( 1 / 2 ) . 5 0

Let

z

such that q ( z ) = x ;

E @

then

(2) sum b (x) = q ( z + z ) = q ( 2 R e ( z ) ) e q ( I R ) . 5 0

Indeed, sum b (x) = q ( z + i ( z ) - O - E ( O ) ) 5 0

= q(z

+ z ) , establishing

(2).

=q(z+z+1/2-1/2)

Hence we have proved the follow-

ing : (i)

Theorem.

im surn5=q@);

thus (ii) Co(b’o,t) is

isomorphic to I R / Z .

15.30

Let

Applications

b

be in

div Ys,t.

We will now make explicit the

necessary and sufficient conditions, arrived at in 515.2, for b

to be in

Let

(E*):

arp*(b)

i.e., for

5

divoYs,t

(I) for

T’,

and (Lemma 1 5 . 2 0 ) that p*

injectively onto b

to be a principal divisor.

We saw, Lemma 15.21, that

(15.20:3).

( E * ) = (F*) n sym divOX

b

maps

sym divoXT,; thus 5

to be principal it must be homogeneous (15.11).

By Theorem 6 . 4 3 ,

The Divisor Class Group of (2)

for

b

to be principal, sum

a

Ys,t

281

must be zero.

Conversely, using Theorem 7.41 and Lemma 15.21, we have the following: Theorem.

homogeneous and

b

E

div Y

s,t

is in

sum p* (b) = 0.

(E*)

if and only if

b

is

This Page Intentionally Left Blank

CHAPTER 16 ANALYTIC DIFFERENTIALS

Introduction

16.10

Meromorphic and analytic differentials were discussed, for a Riemann surface X,

in Chapter 8 (8.22 ff.).

differentials was discussed in 1 8 . 2 4 . of genus 1. X

Let

now be compact

The integration of an analytic differential

along an element in the fundamental group of We found that

sed in 18.43.

T

and

X

5

on

was discus-

€ 6 could be computed by means

of periods of a non-zero differential on X

X

Integration of

X

(8.44:5)

Further

are analytically equivalent.

XT

The purpose of this chapter is to try to obtain analogous

YsIt.

results on

16.11

The notion of a meromorphic differential was defined

for Klein surfaces in [6, 1.101, let

r

Let

Y be a Klein surface,

be an oriented arc or curve on

ferential on

Y

that has no poles on

Y .'I

and let

5

be a dif-

The following was

established [6, 1.10.4 and 1.10.51: (a) The real part of

Theorem.

(b) If borhood in

r

Y,

lr<

is always well defined.

is an arc, or a curve, with an orientable neighthen

Ir<

is well defined up to complex conju-

gation. (c) If

Y

is a Riemann surface, then 283

j r < is a well

284

Norman L. Alling

defined complex number.

r=

(d) If

Z,

I,<

then

is well defined and is real

valued. (1) Let

D1(YS,,)

denote the space of all analytic differen-

Ysft.

tials on

In 14, 3.3 (ill it was shown that (2) P1(Ysft) is of dimension 1 over El.

Computations

16.20

Let

s=2.

Since

7 and

(Lemma 14.201, and since

drp

p'

are in

E( !E(Y

))

t is a meromorphic differential on Sf

Ys,t'

(1)

3 Edz

is in

P

and is non-zero.

was defined in Example 1, Chapter 8, and was shown

(Note, dz to be in

U1(Y2,t)f

Ul(c/L).)

Thus

{dzl

is a basis of

U1(Ys,,).

is, of course, an annulus: thus the fundamental group of

Y Y

is

infinite cyclic.) Let

Y E [O,t/21

and recall (13.20:l) that U

Y

(

-pqWtyi))

is an orbit space under the connected component of the automorphism group of

YZtt

(13.21).

From Theorem 16.11 we see

that (21

adz

is well defined and is

a,

for all

a EIR.

J" Y Let

X E

[0,1) and recall (13.30:l) that

A priori, only the real part of (3)

adz vx

Vx Gpq@?i+x).

Analytic Differentials is defined.

Of course it is

285

On orienting

0.

imaginary part of (3) is well defined (16.11). (12.34:6) one easily sees that it is ing on which orientation we put on

then the

Using

FD(Y~,~)

iati/2, the sign depend-

Y.

Of course, this number

is contractable.

is not a period, since Vx

Y

Following the

terminology found in the classical literature on Schottky differentials, which period of

dz

on

dz

Y.

is,

2

ati/2

(See e.g.,

could be called a

[3, V.141 for details on

to the double of

Schottky differentials. When one lifts Vx

YZft, Xtil

and integrate

which is period of

adz

adz

on

half

about it,one obtains

?

ati,

Xti; hence the terminology.)

Combining the results obtained above, we have proved the following: Let

Theorem.

(4)

211m

for all

5

E

D1 ( Y2 It)- 01 ;

then

I, X c l / l I UY 5 1 = t ,

X E

[0,1) and all

Y E

[O,t/2].

Note the U ' s are orbits of the connected component of Y the identity of the automorphism group of YZIt (513.21, and the

Vxls

are the set of orthogonal trajectories to these

orbits (813.3).

Note also that (4) is very much like (8.44:5) ,

the analogous result for complex elliptic curves.

16.21 E(

Let

s=l.

Since

p

and

fo'

are again in

-E(Y1,t) )

(1)

3 =dz 7 -

is in

D1(Yllt)- { O l .

U ' s are the orbits of YlIt under Y the connected component of the automorphism group of Y l I t

Let

Y E [O,t/4];

then the

286

Norman L. Alling

< E Q1(Yllt) - l o } .

(13.22). Let such that

There exists a unique

a EIR*

<=adz. Consider

By Theorem 16.11 the real part of (2) is well defined. 5 = ia

(3) Re U

or' ta/2

according as

y

E

FD2 ( Y1 I t)

.

[O,t/4)

or

Y

Y E Ct/4).

Indeed I consult (12.35:3 ) to see Let

X E

[0,1/2);

then, Theorem 13.30, the

orthogonal trajectories to the

U 's.

Y

Each

Clearly it is

is contained in an orientable neighborhood; thus

Vx

the imaginary part of (4) is well defined up to sign. FD2(YlIt) (5)

Im

(12.35:3).

j

are the

Consider

A priori, only the real part of (4) is defined. 0.

Vx's

c=f

Recall

It is clear that

at/2.

vX

Thus we have the following: Let

Theorem.

5

E

Ql

Yl,t

- {O},

X E

[0,1/2), and let

Y E [O,t/41.

(6) 21Im

I

c

vX

y

is in

[O,t/4)

or

are orbits of YlIt and the Y are their orthogonal trajectories; and that (6) is very

Again note that the

VX ' s

{t/4}.

much like (8.44:5).

U 's

Analytic Differentials

Let

16.22

9

s = 0.

(14.30:l); however,

%(

and

'p'

287

are not in

Eisn) and 3 '

E( ZE(YOrt)

are in

E

(Theorem

14.34). (1) Hence Let

y

[O,t/2];

E

- {O}.

g = d z is in 3 then the

U 's

are the orbits of

Y

Y O r t under

the connected component of the automorphism group of (13.23).

5

Let

such that

D1(Yort)- { O ) .

E

c=adz.

YOlt

There exists a unique

a EIR*

Consider

Using Theorem 16.11 again, we know that the real part of (2) is well defined.

Referring to

FD(YOrt) (12.36:2), one easily

sees that (3)

Re

!

dg

is

a

or

a/2,

according as

y

is in

(O,t/2),

U

Y

or For

Y E

y

is in

(O,t/2)

U

CO,t/21.

Y

is contained in an orientable tubular

neighborhood (13.23); thus the imaginary part of (2) is well defined up to sign.

Clearly it is zero.

For

y=O

or

t/2

this is not the case (13.23); thus the imaginary part of (2) is not defined for such Let

x

E

y.

[0,1/2) ; then, Theorem 13.30, the

orthogonal trajectories to the (4)

U 's.

Y

Vx's

are the

Consider

I 5.

The real part of (4) is well defined and is

0.

Since each

is contained in an orientable tubular neighborhood (see Vx e.g., FD(Y ) (12.36:2)), the imaginary part of (4) is well Olt

Norman L. Alling

28 8

defined up to sign. (5)

Im

Clearly

5 = t at. vX

Thus we have proved the following: Theorem.

Let

5

E

Dl(Yo,t)

- Col,

X E

[0,1/2),

and let

Y E [O,t/21.

(6)

J

51/lRe "X

(O,t/2)

51 = t

U

or

Y

or is in

CO,t/21.

2t

according as

y

is in

CHAPTER 17

FROM DEFINING EQUATION TO SPECIES AND MODULI

17.10

Let

(1)

INTRODUCTION

P(x) :Ax 4 +4Bx 3 +6Cx 2 + 4 D x + E ~ B [ x l

be of degree

n,

where

n

is either

has no multiple roots in a::

P(x)

3

4,

or

i.e. , let

such that

P(x)

be a real

admissible polynomial 9 3 . 1 . IR(x,y) , where

(2)

Let

E

(3)

Let

F EC(x,y);

then

F

E

Y

=P(x).

for a unique

s

F=E(i),

and thus

E

(See Chapters 10 and 11 for details.)

be the Klein surface Rie%E

seen in Chapter 12 that

by

L

is a complex elliptic field,

is a real elliptic field. Let

y

Y

and

(Chapter 11). We have

is dianalytically equivalent to t;

thus

s

and

t

are determined

Pk). The aim of this chapter is to compute

from

s and

t

directly

P(x). Let

X-Riem F;

a:

then

X

is a compact Riemann surface of

genus 1. We saw in Chapters8 and 9 that there exists a unique T

ED

(9.24:2),

function J ,

a fundamental domain for the elliptic modular such that

X

is analytically equivalent to

XT ( -c/IJ,). (4)

Let

T I

-ti

if

s=2

or 289

0,

and let it be

1/2+ti/2

290

Norman L . A l l i n g

s = l

if

I

(5)

5

Let

if

:K

s = 2

and l e t it be

1,

or

~ + 1 / 2 if

s=o.

i s a l s o a n a l y t i c a l l y equivalent t o

X

Note

:XTl/E

YsIt

XTl:

further

(912.3).

Much t h a t f o l l o w s i n t h i s c h a p t e r i s h i g h l y d e p e n d e n t on C h a p t e r 3 , i n p a r t i c u l a r 9 3 . 1 and li3.2.

P ( x ) EIR[x] and

p1,p2,p3 p4=m

If

(2)

Let

Since

r

2.

m

n,

and

p1,p2

t h e degree of

or

2,

r = O ;

[E : I R ( x ) ] = 2

Proof.

a meromorphic f u n c t i o n on

c l C+

( :{a+ bi:

i d e n t i f y with Given

c1 E

c l C

Rie%

+

s,

of

then

P(x)

Ys,t’ and

r/2,

thus

s = 2

2,

i s a two-to-one b> 01 u

{m])

,

map o f i t which w e c a n

( S e e C h a p t e r 11, and [ 6 ] f o r d e t a i l s . )

Gl,...,Gk

G

for a l l

j,

t h e n , by Theorem 1 0 . 2 2 ,

e f + 1 1

... + e k f k = 2.

0.

considered a s

be t h e r a m i f i c a t i o n i n d e x and r e l a t i v e d e g r e e o f

(4)

.

r = 4

if

or

e

(910.2);

P(x)

is e i t h e r p o s i t i v e

Let

such t h a t

3.

P(x).

is

Accordingly,

(17.10:2);

a,b EIR

IR(x).

YsltI

w e know t h a t t h e r e e x i s t

P(6.)=aI I

is

P(x),

and

0.

The s p e c i e s

Assume t h a t

C

,

d e f i n i t e or n e g a t i v e d e f i n i t e .

onto

are in

p3

w i l l be r e g a r d e d a s a r e a l r o o t o f

i s i n IR[xl

Theorem.

or

where

be t h e number of r e a l r o o t s of

P(X)

r=4,

(3)

p4,

n = 3,

species

( ( 1 7 . 1 0 : l ) and (3.10:l)) h a s f o u r r o o t s ,

i f and o n l y i f

(1)

the

Determining

17.20

1< jck.

j j

in

Ysrt

and

f

over

j c1

From Defining Equation to Species and Moduli

291

and f are in N, (4) is a very strong condition. j j Clearly k = 2 or 1. If k = 2, then e l = e 2 = f = f = l . If 1 2 k = l , then either e l = l and f1 = 2 , or e 1 = 2 and f l = l . Since

e

We are here concerned with

If

aY.

El€

then

aY

c1

E I R U{ m } .

f1 = 1 then ?tl E a Y . el > 1 if and only is a root of P(x). (In each case e l = 2.) Let a EX?.

If

a

if

c1

If

P(a)> O

and

CIRU{ m >

then the

E 's

f = 2 and E 1 # a Y . Let 1 f1 = 1 ; thus ?tl E a Y . If on either side of r > 0.

m,

are in

j

as

If

a=m.

n=4

a'

aY.

If

n=3

then

P(a)

then

P(a')

r/2

0,

e1=2

then and

has the same sign

ranges through IR.

Thus there exist exactly

<

Assume that

disjoint closed intervals

.

over which each point is in a Y . Further, over Ill.. ,I r/2 mu { a } ) - (I1u u Ir/2 ) there are no points in a Y . Over the end points of the Im's ramification takes place. Thus

...

s = r/2. Now assume that

r = 0.

Clearly

definite or negative definite.

P(x)

In the first case each of the

two points over the points in IR u no ramification occurs over IRu

{m}

{m);

are in thus

negative definite then every point over R u point; thus

s = 0,

is either positive

aY.

s=2. {m}

Since If

r = 0,

P(x)

is

is a complex

proving the Theorem.

Bibliographic note.

See [6, 2 9 4 1 for related results

and arguments.

17.21

Theorem 17.20 can very easily be generalized as

follows: (1)

Let

Q(x)

be in IR[x] ,

of degree

has no multiple roots in C.

m > 4,

such that it

292

Norman L. Alling

(2)

Let

E' -IR(x,y)I

Let

Y ' z Rie%E',

Let

n Ern

if

and let

m

2

where

y =Q(x).

g

Y'.

be the (algebraic) genus of

is even, and let it be

mtl

if

m

is odd;

then g = (n/2) - 1.

(3)

Let

be regarded as a real root of

m

is odd.

Let

0< r' In.

Q(x);

be the number of components of

s'

If

Proposition.

r' = 0; then

if and only if

be the number of real roots of

r'

Let

P(x)

Q(x)

r' > 0

then

s'=r'/2.

n

then

aY.

Assume that

is either positive definite or negative

definite; accordingly

s'=2

or is

0.

The proof is essentially that given in 517.20. Y'

T h e o r e m (Alling-Greenleaf)

if

r' = n ,

or

Let

Proof.

on which

Cm}

and if

Q(x)

over IR(x)

Y'

is positive definite. Y'

then

can be obtained by gluing

together along the

Q(x)'O;

hence

occurs over

Y'

Y'

intervals in If

r' = O

Q:

-IR.

Since

consequence Y '

E'

is orien-

Riemc E' (i) :X I

can be regarded as a closed subspace of

0 < rl

occurs over C -IR

<

n;

r' = 0

and

then ramification of

Q(x)

E'

and gluing occurs over IR u { - I .

is non-orientable.

Transformations

17.30

n/2

is orientable.

is orientable. Assume now that

negative or that IR(x)

Q(x)

is positive definite than all ramification of

table and since XI,

r' = n ;

cl c f

two copies of IRu

and

r' = O

is orientable if and only

of

(1)

Let

G2(P) :AE - 4BD+3C2 ,

(2)

let

G3

(P) :ACE t 2BCD - AD2

defining equations

and

- B2E - C3 .

is

over As a

From Defining Equation to Species and Moduli G2

H i s t o r i c a l note.

appears to be due to Cayley c. 1845

[15], where Cayley attributes

and

G3

293

G3

to Boole.

the Cayley-Boole invariants of

P.

We will call G2 Presently we will

show that they are indeed invariants, in the classical meaning of the word, as used by Cayley, Sylvester et al. (3)

Let

3 2 = G2(P) - 27G3(P).

A(P)

Note that if

(x) = 4 (x - e,) (x - e2) (x - e,)

P(X) = W

,

then

92 “ 3 for

j=2

and

g2=-4(e 1e2 + e 2 e 3 + e3e 1)

3,

.

g3 = 4e1e2e3’

and

A(P) =

L

(ej -ek)

16Il

1<j
.

Returning to ideas that can be found in the works of [221 r let

M :

(z ,”)

E

GL~(c), 2 p=y(ck+d) ;

Let

%zh(M-’)(x), -2 y=y(c%+d)

then

x=h(m)(%)

and

.

Let

F ( % ) 5x24 + 4g% 3 + 6&

+ 65

2

+ 46%

4

(c% + d ) (P(h (M)(2) 1) E C 121

Then

72 = @ ( % ) ,

and

6 (2)

.

is admissible.

(See (3.13:2) and 9 3 . 1 5 for details.)

It will on occasion be

useful to denote (9)

y2 =$(:)

as

(10) Clearly E

2 y =P(x)

F = @ ( % , Y ),

transformed by

and if

M

is in

=3?(%,7).

17.31

Let

MZS

(E

(i

(3.13:9);

M. GL2(IR)

then

r

Norman L. Alling

294

(1) (2) (3)

then

I=-l/x

and

Y=y?

2

.

y2 = Es4 - 4D23 + 6Cg2 - 4B2 + A Clearly G . (6) = G .(P), for j = 2 I 3 Hence

:P(I) e I R [ 2 1 and

3;

.

and hence

.

A ( $ ) = A (P)

y2 = P ( x )

This transformation of inversion.

Note:

17.32

Let

(1)

then

(2)

Clearly

F2 = $ ( ? )

to

will be called

det S = l .

M-Tb

% = x - b,

(- (o1

and

b

for some bea:

(3.13:6);

? = y.

K=A g=B+Ab

2 = C t 2Bb + AbL a = D + 3 C b + 3 B b 2 + A b3 , 2 3 4 g = E + 4 D b + 6 C b +4Bb + A b

and

.

The following can be checked directly: (3)

Gj($) = G . (P), for

(4)

If

7

b

is in lR

then

17.33

b.

Let

and

g(?)

y 2 = P(x)

The transformation of translation by

j= 2

Note:

M=Da(:

?=x/a' and

3;

thus

A($) =A(P).

is in l R [ x ] . to

7 2 = $(I) will be called

det T b = l .

(: !))

for some

aEC*;

(1)

then

(2)

Clearly

3 2 x = A a 4 , 6 = B a , e = C a , a = D a , and E = E .

(3)

Further

G2 (5)

=

F=y.

G2a4, G3 ( $ 1

=

G3a6 ,

and hence

A(F) = A(P)a 12 (4)

If

a

is. in IR*

then

$(2)

is in IR[2].

From Defining Equation to Species and Moduli y2 = P ( x )

The transformation of be called dilation

.

(x)

:h ( M - l )

(det M ) 4 G 2 ( P )I A

(5)=

=

6(%)

given by (1) will

& a. Note: det Da =a.

T h e o r e m (Cayley-Boole)

let 2

y2

to

295

Given

$ :y(cx

and

(ii) G 3 ( P )

=

M

+ d)2 ;

E

GL2 ( C )

,

then (i) G2 (?)

=

and thus (iii)

(det M ) 6 G 3 ( P ) I

(det M ) 1 2 A ( P I . We have seen in 5 3 . 1 3 that

Proof.

{Tb:b E C } , {Da:a E C * } , and and ( 1 7 . 3 3 : 3 )

S.

GL2(C) is generated by

Using ( 1 7 . 3 1 : 3 )

(17.32:3)

I

,

then gives the result. Since there has been a revival of

Bibliographic note.

interest in invariance theory, and even its history, the author did not do much bibliographic research on it.

Rather he used

Salmon's Modern Higher Algebra [55] c. 1 8 7 6 as a convenient ~referenceto the work of Cayley, Sylvester et al. Salmon [55, p.1041 calls

G2' GjI

and

invariants

A

because of the Theorem above. Corollary.

If

M E G L 2 D )I

then

and

A($)

have

A(P)

the same sign.

17.34

(1) Let

ii

let

Let E

x

9 2 E G(2) c c

~ E C * and

and

?=ay.

[%I.

(2)

then

(3)

G2 (6) = G2 ( P ) a 4 , G3 ( P ) = G3 ( P ) c1

(4)

If

i=Ac121

a EIR*

6 =Da2 ,

B = B a 2 1 ?=Ca2,

then

$(%)

6

and

I

is in IR[x]

and A

and

E=Ea

2

.

( 5 ) = A ( P ) a1 2 .

A(6)

and

A(P)

have the same sign. The transformation of

y

2

= P(x)

to

-2 y

= P ( % ) given by (1) will

Norman L. Alling

296

be called

y

dilation of

by

a.

We will now consider various transformations of -2 to y = 6 ( % ,) using transformations developed in

17.35

2

y = P(x)

5517.31

-

We will write this in a quasi-algorithmic

17.34.

manner. (1)

If

E = 0,

then

let the transformation be inversion (17.31);

p ( % ) is of degree 3.

It will prove convenient, since we will perform a finite sequence of these steps,to (2)

-2 replace y = 6 ( % ) with

2 y =P(x).

(In this way we won't have to make up a whole string of new notation.) If

E # 0,

F(%) If

let the transformation be inversion (17.31);

is of degree 4. b

is a root of

translation by real root let If

A # 0,

b;

b

let

translation by If

A=O,

E=

then

-b;

or 4,

A=O

Note:

then

B f 0.

P(x)

implies B f 0.

if

P(x)

has a

P(x).

and let the transformation be

g = 0.

Let

formation be translation by Indeed, notethat since

0.

be a real root of

b : B/A,

then

let the transformation be

P(x),

is

b = -C/2B, b;

-

then

and let the trans-

E=

0.

by definition

-

of degree 3

The rest of (6) follows from

(17.32:2).

(7)

If 4/B

A=O,

then

BfO.

Let

a

be the real cube root of

if it has one, and a cube root of

let the transformation be dilation by

4/B

a;

if not; and then A = 0

From Defining Equation to Species and Moduli

297

g = 4.

and

We will now use these transformations to put the defining equation into Weierstrassian form. (8)

If

is of degree 3, go to (10). Assume that is of degree 4. Using (4), transform y 2 =P(x) to 2

(9)

P(x)

=@(% Note I . g = 0.

Apply (1); thus

@(%)

P(x)

Apply (2) and then go to (9). is of degree 3.

Apply (2) and then

go to (10). ( 1 0 ) Apply (6). Note

e=O.

(11) Apply (7). Note

A = O , 6 = 4 r and ? = O .

g 2 :-4D,

and

By Lemma 3.15, A (W

)

and (6).

-E.

E

Then

Apply (2), let

y2 = P (x)= 4x3

- g2x - g3

(x) is admissible; thus its discriminant, g2'93 is not zero. Note also that G . (F) = G . (P)I for

,

91'42 j = 2 and

g3

Apply (2) and go to (11).

W

7

3

3,

and hence

A ( P ) =A(P),

for (11, (31, ( 4 1 1 (511

In (7) it transforms according to (17.33:3); thus we

have proved the followings A (P) # 0.

Lemma

(13) Let

M

GL2 ( c ),

be any element in

transformed by

M

and let

(17.30:6 and 7); then

q2 = ( 2 ) be

J(6) = J(P).

Indeed, this follows from (17.31:3), (17.32:3) and (17.33:3). Historical

In the classical theory of invariants J

note.

is called an absolute invariant.

(See e.g.

I

[55, p.1051 . I

We have seen in 99.50 that there exists a lattice

c

such that

g .(L) = g 3

j r

for

j =2

in (17.10:3)) is @-isomorphic to

and

F(L)

3;

thus

F

L

in

(defined

(which was defined in

Norman L. Alling

298

We have seen in Chapter 9 (Corollary 9 . 1 3 and

(6.30:l)).

Corollary 9 . 2 7 ) that T It

in

be in

D. D

L

is equivalent to

Bv Theorem 10.41,

J ( L ) = J ( T " ) .T

such that F

(17.10)

LTl,, for a unique

was chosen to

is @-isomorphic to

bY Theorem 10.41, J ( T " )= J ( T ) . By Theorem 9.27,

F(LT); T

thus Thus

= T".

we have proved the following: J(P) = J(r);

Theorem.

thus

complex elliptic function field

is an invariant of the

J(P)

@ c

F.

Clearly (14)

J(P)

- 1=27Gf(P)/A(P).

We will see that

is slightly more convenient to use

J-1

for real elliptic curves than is Note that if of degree 4 . ficients.

17.36

P(x)

J.

has no real roots then it is necessarily

When we apply ( 4 ) ,

P(2) need not have real coef-

In this event when we apply ( 7 ) ,

Assume that P ( x )

a

need not be real.

has a real root; then all the

polynomials (resp. matrices) involved in ( 1 7 . 3 5 : 8 , 9 , 1 0 , have real coefficients (resp. entries). there exists

and 11)

Hence

a E D * such that g 2 = a4 G 2 ( P ) I g 3 = a6 G3 ( P )

and and

As a consequence, gj j = 2

Let

and r(P)

3.

Gj ( P )

have the same sign, for

The same is true of

A

and

A(P).

denote the number of real roots of

P(x)

.

(Recall the conventions given in ( 1 7 . 2 0 ) . ) Thus

r(P)

=

r (W 92 I g 3

Theorem.

1.

The species s

of

Rie%E

is either

2

or

1.

From Defining Equation to Species and Moduli s= 2

(resp. 1) if and only if

(resp. A(P)

By Theorem 17.20,

the finite roots of

s=r/2.

W

Let

<

0).

it must be either 4

Since, by assumption, r > 0,

Proof.

or 2.

A(P) > 0

299

el,e2, and e3

denote

(x). We have seen (7.33:22) that 91'42

16(el- e2)2 (e2- e 3 ) 2 ( e 3 -el)2

A =

(5)

and by ( 2 )

If

r=4

clearly

A,

If

r=2

let

be real; then,

el

.

A(P),

are strictly positive.

since W

(x) €IR[XII 92- g3 -

e2;

e3 = e2 - e3

thus

e l - e2

and

is pure imaginary.

el-e3 Thus

are complex conjugates and and by (2)

A,

A

(P), are

strictly negative, proving the Theorem.

Assume now only that

17.37

polynomial 13.1. If

a root of

P(x)

.

then let

b;

then

ality we may assume that

4.

(1)

P(x)

(2)

and

b

I?,# 0.

E # 0.

Thus without loss of gener-

Now use (17.35:3); then

is a real admissible polynomial of degree

We saw in 593.20 and 3.21 that there exists U,V,WE {O,ll,

k

1,

is of

be a real number that is not Let the transformation of y2 = P(x) (17.32)

be translation by

p2 = $ ( i t ),

is a real admissible

Assume, for a moment, that

E=O

degree 3.

P(x)

and

such that

M E GL2(IR).

k E (0,1], for which if 2 = F(2) = L (x,k)

v = w then

UIVIW

-

( : ( - l ) u ( l(-1)vx2) (1

Using the' fact that IRu {-I,

(2)

h(M)

-

wk2x2 ) (-1)

(3.21:lO)).

(3.13:2) induces a homeomorphism of

we know that the number of real roots of

r(P1 =r(L

(*,k)), which is

P(x),

2(2-v-w).

UlVlW

Combining (2) with Theorem 17.20 gives us the following:

(4)

If

r > O then

s=2-v-w.

If

r=O

then

w=l=v,

300

Norman L. Alling and

or

s=2

0 according as

Let us compute Gj($), = 0=

Note first that

and

1.

and

3,

A($).

Then note that

e = (-l)l+u((-l)v+ (-1Iwk2)/6,

L = k2 (-1)u+v+w

(5)

6.

j= 2

for

or

u=O

and

g = (-1)U . (6)

Then, =

G2(6) =%+3E2,

G(AE

G3(F)

=ie2-?3,

and

A(F)

- 9E2 ) 2 .

Combining ( 5 ) and (6) gives us the following: (1 + 14k2

(F) =

+ k4)/12.

(7)

G2

(8)

G3 ( 6 ) = (-1Iu[( - l ) v + (-1Iwk2][l - 34k2 (-l)vtw t k4]/216.

(9)

A(P)

=

v+w- 2 4 k2(-1)vtw[ (-1) k ] /16;

thus

A(B) # O .

From these results we immediately conclude the following: (10) J($) =

(1t 14k2 (-l)v+w t k4) 108k2 (-1)v+w [ (-1)v+w - k21 [

(11) J($) - 1 =

A(6) <

0

17.38

+ k4]

Finally note that

if and only if

v

+ w :1

We have seen that if

(det M)I2A(P)

and

w 2 2 ( - 1 I v + (-1) k 3 [l - 34k2 (-l)v+w 108k2 (-l)v+w [ (-1)v+w - k214

(cf. 140, p.15 ff.].) (12)

,

(mod 2 ) .

M E GL2QR) then

(Theorem 17.33) , and that

A(P) # 0

A($)

=

(Lemma 1 7 . 3 5 ) ;

thus (1)

for all Theorem.

of

A(P)

M

E

GL2 QR)

A(P)

< 0

,

s

A ($)

if and only if

is determined by

by the residue of

A (P) and

s,

modulo 2 .

and is

-

have the same sign. s = 1;

in turn

thus the sign

-

determined

From Defining Equation to Species and Moduli

301

Using (1) and (17.37:l) we may assume, without l o s s

Proof.

P(2) = Lu tvrw (2,k). Using (17.37:4) the

of generality, that Theorem is proved.

Note this is in some ways an improvement of

Comment.

Theorem 17.20, for there - to determine r

(the number of real roots of

P(x))

.

s

- we had to compute

Now we see that

is computed knowing only the coefficients of then

know that

s

is either

(Theorem 17.36).

s= 2 either P(x)

Assume that

s=l.

4

or

0

A(P)

2

or

0.

(Theorem 17.20). r

>

0.

If

P(x)

17.40

(1)

(P) < 0

is of degree 3 then r

If any changes in sign of

If none are found then one can P(x)

by considering

T E

can be solved in terms of radicals.

X

a

h(T)

PI,

(or h

the upper half plane (3.12:4), and let for short) be defined to be

(5.20:2)); then

h

einT, (as usual

is analytic and non-zero on

Q1

and

h(r +I) =-h(T). Clearly all (3)

is

Of coursel since the symmetric group on 4 letters

is solvable, P(x) = 0

Let

A

is of degree 4; then

consider relative maxima (resp. minima) of P'(x), etc.

If

(P)

is strictly positive; then we

Assume P(x)

can be found then

.

P (x)

A

t

>

Further

h (ti)= e-nt

and

h(z+ 1 ti/2)

=

ie-nt/2 I

for

0.

Ih(T) I

<

1,

f o r all

T E

4.

It is well known that

(See e.g., 136, p.2051.) To assure oneself that (41, and related infinite products, do

302

Norman L . A l l i n g

a n € @,

indeed converge, l e t

for a l l

nc N

and c o n s i d e r

nnn=l (1 + a n ) . m

(5)

I t i s c a l l e d a b s o l u t e l y Convergent i f

(1+ I a n I )

m

(6)

converges.

I t i s w e l l known (see e . g . , (7)

n;=,

1 5 4 , p.288 f f . ] ) t h a t

i s a b s o l u t e l y c o n v e r g e n t i f and o n l y i f

(l+an)

m

~ ~ = ~ il s ac o~n v el r g e n t . Thus w e see t h a t , f o r e x a m p l e , (8)

nn,l (1- h 2 n ) m

s u b s e t s of (9)

Let

i s a b s o l u t e l y u n i f o r m l y c o n v e r g e n t on compact

$.

12

6 m 6 ( ~ ) ( 2 ~ )h I I n = l ( l - h 2 n )

.

Using ( 7 ) o n e e a s i l y sees t h a t t h e e x p r e s s i o n on t h e r i g h t hand s i d e of ( 9 ) c o n v e r g e s a b s o l u t e l y u n i f o r m l y on compact s u b s e t s o f

a;

thus

(10) 6

i s a n a l y t i c on

(T)

G.

Using (4) w e see t h a t (11) 6

2

( T )

=A(T);

Lemma 1.

Proof.

hence

(i) 6

(T

6 ( ~ f) 0 .

+ 1) = -6

(T)

,

for a l l

Using (1) and ( 9 ) ( i ) i s p r o v e d .

T E

5.

( 2 ) and ( 9 ) p r o v e

( i i ) and ( i i i ) ,p r o v i n g t h e Lemma.

Recall t h a t that

e

3

( T )

el

(T)

5

9(1/2,LT)

I

:1 ) ( ~ / 2 , L ~ ) . ( R e c a l l t h a t

t i o n s found i n [ 3 6 ] t o i n d e x t h e

e2 (

+

:9((1 T ) / 2 1 L T )

,

and

w e a r e u s i n g t h e conven-

e Is.) j

T )

From Defining Equation to Species and Moduli

(12) ej(T)

is analytic in

303

T .

Indeed, this may be seen directly from the definition of

ej ( T )

using results given in 57.32.

(14)

Proof.

Using (11), (13), we see that

6 ( ~ )=

4

&

n

- ek(T)).

(ej( T I

1<j
Since each function is analytic on

Q((10) and ( 1 2 ) ) , we see that

either the plus sign holds for all

T E

holds for all

4.

T E

Let

t>0

and let

el(?) >e,(~)> e 3 ( ~ ) ; thus

14.22,

4,

or the minus sign T

=ti.

By Theorem

(ej(r)-e,(~)) > 0. 12j < k53 6 ( ~ )> 0, proving that the plus sign in ( 1 4 )

By Lemma 1 (ii),

holds for all such

4

Using the identity theorem €or analytic

T .

functions (see e.g., 135, p.199]), we see that the plus sign in

(14) must hold for all

T E

0.

Consider now the transformation anti-analytic involution of in IR;

then -iar h(-.r) = e

(15)

s > *

-TS

- h(-T) =h(T) ,

-T

,

6

-T

E

0;

T

z

-

it is an r

and

s

and

showing that for all

T

€a.

From this, and the definition of (16)

@ *

r + is, for iar -ilrs , h(~) =e e

Let

- =-r+is.

0.

e

4.

T E

6(T)

(9), we see that

(-;)'=s(.r)

17.41

g3(r)

to be analytic on Lemma.

(i)

(

b

-g3(LT))

(7.33:18

and 5 ) , has been shown

(Theorem 9 . 2 2 ) .

g 3 ( T + 1 )

=g3('),

for a l l

T

EB.

Let

t > 0.

3 04

Norman L. Alling

(ii) g3(ti) E D , and it is positive, zero, or negative according as

t

>

1, t = 1, or t < 1.

(iii) g3(1/2

+ ti/2)

is in IR,

and it is positive, zero, or negative according as or

(iv) g 3 ( T I =4e1(~)e2(~)e3(-r). Proof. Since g , ( ~ )= 140s6(LT) (7.33:18) and since

t < 1.

L~ = LT+l,

and

t > 1, t = l ,

(i) is proved.

By (Lemma 14.20) we know that

are in IR.

g3(1/2+ti/2)

g3(ti)

The rest of (ii) and (iii) was

.

established in Theorems 14.22 (iii) and 14.23 (iii)

(iv) was

established in (7.33:21), proving the Lemma. g,(r)

also has an expansion in terms of

(See e.g., 136, p.2101.) (2)

Let

a ( T )

-

and

for all

a(ti)

t=l,

t

<

Now for the definition of

,

33’2g3(T)/6 (i)

Theorem.

J ( T ) 1,

=

T E

T E

4.

h, namely:

Q

t+

for all

€4.

is analytic.

a ( T )

Let

T

t

>

0.

a.

(ii) a2(-c)=

(iii) a(ti)

is in IR,

is positive, zero, or negative according as 1.

(iv) ia(1/2+ti/2)

is in IR,

and it is positive,

t > 1, t = l , or

zero, or negative according as

t > 1,

t < 1.

Further,

(v), the following hold:

Finally, (vi), Proof.

proving (i).

a

(

+~1) = - a

(T)

,

for a l l

T E

b.

By (17.40:lO) and Theorem 9.22, a ( ~ ) is analytic, Since (17.40:11) holds, c ~ ~ (=27g3(T)/A(T). T) 2

From Defining Equation to Species and Moduli

By ( 9 . 1 1 : 7 ) ,

J ( T )- 1 , proving (ii).

this quantity is

Lemma 1 (ii) of 5 1 7 . 4 0

6(ti) >

and by the

( 3 ) follows from ( 1 7 . 4 0 : 9 )

Lemma, part (iii), above, (iv) holds. and (1).

By

and by the Lemma, part (ii),

O r

By Lemma 1 (iii) of 5 1 7 . 4 0 ,

above, (iii) holds.

305

( 4 ) follows from Lemma 2 of 5 1 7 . 4 0 and the Lemma, part

(iv), above, proving (v). To establish (vi), note that it follows immediately from Lemma 1 (i) of 5 1 7 . 4 0 and the Lemma, part (i), above; proving the Theorem. Since

CL(T

+1) = - c L ( T ) ,

is not invariant under the

ci

r,

action of the modular group,

on

0;

thus it is not a mod-

ular function for the full modular group ( 9 . 4 0 ) . c1 ( T

+2)

=

thus

ci ( T ) ;

where recall ( 1 7 . 4 0 )

However

may be regarded as a function CL (h), in h=e T Since c1 is analytic on 0 it ci ( T )

.

admits a Fourier series expansion m

(5)

bnhnr

Cn=-,

which converges absolutely uniformly on compacta on A : {h E @:

at

0 < Ihl

of

<

c i ( ~ )

11.

(See 5 6 . 5 for details.)

The "behaviour

(expressed in condition ( 9 . 4 0 : l ) ) is reflected

by the behaviour of ( 5 ) at

h=O.

From ( 3 ) we see what that

behaviour is, namely; (6)

a (h) has a simple pole at

(7)

Hence

c i ( ~ )

0,

which has residue 31'2/72.

has a simple pole at

A s a consequence we see that

a ( T )

satisfies ( 9 . 4 O : l ) .

Under the anti-analytic involution seen ( 1 7 . 4 0 : 1 5 ) (8)

CJ~(-T)

that

=g3(T)

,

h(-y)

=h(rl.From

for all

T

a(-;) =T('Tj', for all

T

-

T +-T

EQ.

of 4

(1) we see that

~ a .

Using this together with ( 1 7 . 4 0 : 1 6 ) , (9)

m.

we see that

we have

306

Norman L. Alling

A closely related function, which has been exten-

17.42

sively studied, is : (e2(T)-e3(~))/(el(~) -e3(T)),

(1)

defined for all A

T

€9.

is well defined.

Since the

Since the

e Is j

are distinct (7.33:4),

e.(.r)'s are analytic on b 7

(17.40:12) , (2)

h

is analytic and non-zero on 0 .

(T)

Bibliographic n o t e .

is a very well known and much

A(T)

See e.g., [ 2 4 , pp.157-1631.

studied function.

the uses we have for

Since some of

seem too novel, our study of it will

h

be detailed and will bring out some of the algebraic facts surrounding

in an explicit

A

form.

In the next section we will study the action of the modular group the

on

T

r

on

e Is. j

and let m: h(M) m(r)

now we will study the action of

A;

5

(3.13:2); then m

(ar+b)/(c.r td).

basis of Let M

E

Rr

LT

(:

Z fB TZ

M'

5

(bd

SL2 (Z) + MI

( E

T

)

r

and

~ (6.23:l))

is of course a

(6.23:l)).

c) ; then a E

(1

is in

is in

MI

SL2(Z), and

SL2 (Z) is an automorphism of

SL2(Z)

period 2 . Let

ol:c-r+d

M I R T(=(ol w 2 )

and let E

R)

o2 E

ar

+ b;

then

is again a positive basis

(Theorem 6.22). Clearly

olRm

=a;

thus

o L 1 m(r)

-LT*

LT

of

From D e f i n i n g E q u a t i o n t o S p e c i e s a n d M o d u l i

307

e . ( ~ ) ,j = l , 2 , a n d 3 , a r e t h e v a l u e s o f 3 a t t h e z e r o s of ' p ' ( * , L T ) o n a f u n d a m e n t a l domain o f

As n o t e d i n 57.33 'p(- I L T 1

mod L T .

@,

Since

f u n d a m e n t a l domains o f 'pl

( - ,L

)

and

P ( ~ , T ) (6.20:6)

are both

P(wl1w2)

and s i n c e t h e zeros o f

C l mod L T r

are a t t h e h a l f periods i n t h e s e sets ( 7 . 3 3 : l )

I

we

see t h a t

(8)

I

wi2el ( m

-

-

(T)

1 , w12e2

( m ( T 1 1 , u12e3 (m ( T 1 1 I .

Thus t h e r e i s a p e r m u t a t i o n o f t h e symmetric group -2

(9)

= w1

e .1( T )

Let

=

LT+l

see t h a t

n(T) = (2

3)

E

S3.

Now l e t and

S=

(y

-:)

Then

w l = ~

and

( 1 / 2 T ) tTL-l/T)

check t h a t

for

j = 1 , 2 , and 3.

(9.23:2);

M E T

2.

(10)

( s i n c e 'p

i.e., a n element

thus

rn(.r)=.rtl.

Since

t h i s c a l c u l a t i o n i s p a r t i c u l a r l y s i m p l e , and w e e a s i l y

Example

p ( T

{1,2,3I:

on t h r e e letters, such t h a t

S3

e n ( M( )j , ( m ( - r ) ) ,

Example 1 . LT

of

n(M)

=

T

M

S' =

m(T)

-

3.

Recall

q ( ~=) - l / ( ~ - lw )l ,=

~ - l

E

3).

= - l / ~ e,(.r) . =p(1/2, LT) =

2p(1/2T~L-l/T) =

n ( S ) = (1 3 )

= (2

:) .

(-O1

is a n even function) =

Example

n(T-')

then

E S;

and

w2=-1

-

Similarly

T

T

2'p( ( 1 / 2 )

(-1/T)

t

L-l/T)

2e3( - l / ~ )T.h u s o n e c a n e a s i l y

S3. (9.23:19) and

ing through t h e details, t h a t

that

w2=-1.

n(Q)

Q E ST-';

thus

One e a s i l y s e e s , o n c h e c k -

= (1 3

2).

Note that

Norman L. Alling

308

n(S)II(T-’)

is, by Examples 1 and 2, equal to

(1 3) (2

3).

This in turn is equal to

(1 3

on elements of

as functions usually do, on the left);

{1,2,31,

(letting permutations act

.

thus

II (ST-’) = Il ( S ) II (T-l)

Let

Example 4 . wl=

2)

and

T

Note

- 1.

w2=-’

therefore that

now be defined to be

M

One then finds that

note that

T-lS;

(1 2 3).

II(M) =

Il CT-lS) = (1 2 3) = ( 2 3) (1 3) = Il (T-’)II

In 99.23 we saw that

S

and

T-l

generate

SL2 ( 2 )

(S)

;

.

thus

using the results obtained in Examples 1-4, we have the following :

n

Lemma 1 .

is a homomorphism of

SL2(Z)

It remains only to show that

Proof.

II

I I , Q , Q2 1

maps

(11)

Since

(S) =

(1 3 )

E

isomorphically onto

S3 - A3,

II

maps

s3 is surjective.

(1 3 2);

Q, an element of order 3, mapsto a 3-cycle

SL2(2)

-

onto

thus

A3’ onto

S3, proving

the Lemma. (12) Since the Weierstrass p-function is an even function, I I ( - I ) = l e S3;

thus

modular group

r

Il

onto

induces a homomorphism

TI

of the

s3’

Now let us investigate ker Il. Since II(T) = (2 3) (Example l), II(T 2) =I. Recall (9.23:8 and 9) that Um=-ST-mS

2

showing that

II (U ) = Il(-I)II ( S ) I l (T-2)II ( S ) = (1 3 ) (1 3) = 1;

and of

U‘ Z

(141

are in onto

Z2

ker

n.

By (12) -I E ker iI.

The homomorphism

having kernel (2) induces

a homomorphism

h

of

SL2(z)

onto

T2

sL2(z2).

From D e f i n i n g Equation t o S p e c i e s and Moduli

ker h :

(15)

b

0

i

I(:

a',b'c', c=2c',

e , b ) = ?(1/2, using (3)

-

d (mod 2 ) ,

and

k e r II.

d' E Z

and

thus there e x i s t such t h a t

a=2a'+1, b=2b',

d=2d'+1.

L T ) = P(w1(1/2w1)

t

wlLm(T)

1

-2

= w1

P(1/2w11

Lm(,) 1

(9).

1 / 2 0 1 :1 / 2

(17)

c

M E k e r h;

and

5 1 5

I.

ker h Let

Proof.

SL2(Z): a

E

c (mod 2 )

E

Lemma 2 .

(16)

):

309

(mod L m ( T ) ) .

Indeed, t o show t h i s it s u f f i c e s t o show t h a t t h e r e e x i s t s and

V E

Z

(18) 1 / 2 W l

u

such t h a t

- 1/2

=U

+Vm(T).

This equation i s e q u i v a l e n t t o (19)

2uc+2va=-c

and

2ud+ 2 v b = l - d .

Using ( 1 6 ) t h i s r e d u c e s t o (20)

Since

va

+ u c = -c'

det(E

z)

=1,

t h u s ( 1 7 ) holds.

(21) e 3 ( T )

e,(r)

and

vb

+ ud

= -d'.

( 2 0 1 , and t h u s (18), h a s a s o l u t i o n i n

Hence

=w12e1(m(T)).

= p(T/2,

LT) =

-2

iWILm(T))= w 1

p(U1(T/2Wl)

P ( T / ~ W ~ L I

using (3) - ( 9 ) .

(22)

T/2Wl

=m(~)/2

(mod L ~ ( - , ) ) .

To prove ( 2 2 ) , w e must f i n d (23)

T/2Wl

U,VE

Z

- m(r)/2 = u + vm(r).

This equation i s equivalent t o (24)

va+uc=-a'

and

vb+ud=-b',

such t h a t

m ( T ) )I

2;

310

Norman L. Alling

which has a solution in

just as (20) above does.

2,

Thus

(25) e3(T) = ~ ; ~ e ~ ( m ( r ) ) . IT(M) =1,

We conclude that

and hence

M E ker

n,

proving the

Lemma. Using Lemma 2 , we know that (26) n h - l

is a homomorphism of

(27) SL2(Z2)

(=

GL2(Z2))

SL2 ( Z 2 )

onto

S3.

has 6 elements in it.

Indeed, there are 16 2 by 2 matrices with entries in of these are non-singular.

The rest are singular. (29) n h - l

Six

z2.

They are

Thus

is an isomorphism of

SL2(Z2)

onto

S3.

Thus we

have proved the followingr Lemma

ker h

3.

Recall (3.13: 2 )

(9.21:3).

h(ker n )

E

Z,

h(ker h )

(which is also

r2.

denoted by a,b,c,d

n.

h I SL2 (Z) is the natural homomorphism of

that

r

SL~(Z) onto (30) Let

,

ker

=

It is then

ad-bc=l,

{T

E

4

+

by Lemma 3) be

(a-c+ b)/(cr + d ) :

a - 1 - d (mod 2 ) ,

and

b - 0 - c (mod 2 ) ) . Thus we have proved the following: Theorem.

17.43

For'all

m

E

r2

and all

r

Having studied the action of

817.42, let us study the action of

r

on

A(m(-r)) is given in the following table.

4,

T E

A

A

(m(.r)) = A

on the

(T). For

e

's

( T ) .

in

j M E SL2(Z) ,

The top line repre-

From D e f i n i n g E q u a t i o n t o S p e c i e s a n d M o d u l i

(1)

1

(1 2 3 )

(1 3 2 )

x

1/(1-x)

(x-l)/x

for a l l

T

(1 3 )

(1 2 )

( 2 31

1-h

1/x

~ / ( h - l )

311

€5.

U s i n g (1) w e see t h a t , f o r a l l

m e

r,

and f o r a l l

T E

0,

F(m(r)) = F(T).

(3)

F

B i b l i o g r a p h i c note.

v e r y w e l l known.

(See e.g.,

and t h e ideas s u r r o u n d i n g it a r e [54, p.392]..)

I t w i l l be c o n v e n i e n t t o c o n s i d e r a new f u n c t i o n ,

(4)

G(T)

(h+l)

(=+ 1

1)

(++

1)

r

t h e p r o d u c t d e r i v e d from t h e f i r s t t h r e e f a c t o r s i n ( 2 ) .

(5)

G ( T ) = ( h + l ) (2-A) ( 2 A - l ) / A ( l - A ) ,

all

G ( T ) i n terms of t h e

_ -

( T )

= F ( T ),

e 's. j

3 -3 e e e 1 2 3;

(7.33:21), t h e numerator of

thus 3

G ( T ) = 3 e 1 e 2e 3/(e3-el) (e2-el) ( e3 -e 2 ) .

From ( 1 7 . 4 1 : 4 ) w e see t h a t

(8)

for

(e l + e2 - 2e3) ( e 2+ e 3 - 2 e l ) ( e l + e 3 - 2 e 2 ) ( e 3- e l ) ( e 2- e l ) ( e 3 - e 2 )

e l + e 2 + e 3= 0

Since

(7)

2

G(T) =

-

to

-G

€4.

T

L e t u s compute

(6)

and

Clearly

a ( ~ =)- 3 - 3 / 2 G ( ~ ) , a

(T)

=

and t h u s

- ( ~ + l ()2 - A ) ( 2 ~ - 1 ) / 3 ~(I-A) / ~ ~

( 6 ) reduces

Norman L. A l l i n g

312

=

a2 = J

Since

- (-

2x

-1

3

+ 3~ 2 + 31 - 2 ) / 3 3 / 2 ~( l - ~ ) .

(Theorem 1 7 . 4 1 ) , w e o b t a i n t h e f o l l o w i n g w e l l

known e x p r e s s i o n s ( s e e e . g . , (9)

J ( T )- 1 = ( 2 h 3 - 3 h

(10)

J ( T )= 4 ( A 2

2

[ 4 1 , v o l . 1, p . 1 5 1 ) :

2 2 2 - 3 1 + 2 ) /27A ( 1 - A ) ,

2

- A + 1)3/27A 2 ( 1 - A )

and

.

Now l e t u s examine t h e b e h a v i o r o f

"at infinity.''

X(T)

The f o l l o w i n g i s a w e l l known f u n c t i o n (see e . g . , (11)

K ( T )

2

h:e

2

:e 2 / e 3 .

Recall t h a t i nT

O2

and

El3,

Many a u t h o r s ( e . g . ,

t o d e n o t e t h i s (11) q u a n t i t y .

and u s e

and

k

4.

p h i c on K

Further

[ 6 9 , p.4911 u s e

VJe have c h o s e n t o f o l l o w [ 3 6 ]

L

U,VrW

(x,k).

k

to

P r e s e n t l y , w e w i l l see t h a t

are intimately related.

F r o m t h e way

(12)

f u n c t i o n s of

t o d e n o t e t h i s f u n c t i o n s i n c e w e have u s e d

K

d e n o t e t h e modulus i n K

defined in' (5.30:3),are

.

Notational comment.

k

[ 3 6 , p.212 f f ] ) :

K ( T )

From (11) and ( 5 . 2 0 : 4 ) w e o b t a i n

(

2 (h1l4 + h 9 l 4 + 1+2h+

=

K

i s d e f i n e d w e know t h a t it i s meromor-

L

...

= ( e -e

2

3

. . .)

) / ( e1 -e 3 )

>'

[ 3 6 , p.2181.

[ 3 6 , p.2281;

thus

proving t h a t (14)

A(T)

has a simple zero a t i n f i n i t y .

Using t h i s , t h e f a c t t h a t the fact that for a l l A

m c

A

r2r

i s a n a l y t i c on X (m(-r)) = A

w i l l b e c a l l e d a u t o m o r p h i c 0" 9)

with

(T)

4

( 1 7 . 4 2 : 2 ) , and

(Theorem 1 7 . 4 2 )

respect

to r 2 .

(A

,

From D e f i n i n g E q u a t i o n t o S p e c i e s and Moduli

Q

i s sometimes c a l l e d a modular f u n c t i o n on

r2.)

a subgroup Since

(See e . g . ,

transcendental extension of

@(J)

i s a s u b f i e l d of

(15)

@ ( j )c C(T)

b u t o n l y under

[251 f o r d e t a i l s . )

c(A)

i s n o t a c o n s t a n t (13),

A(T)

even though it i s

r,

n o t i n v a r i a n t u n d e r t h e f u l l modular g r o u p

313

i s a pure

By ( 9 ) o r ( 1 0 ) w e see t h a t

@.

and t h a t

@(A)

i s an a l g e b r a i c e x t e n s i o n of d e g r e e a t most

6.

J = a2 + 1

I n t h i s c o n t e x t ( 8 ) and t h e f a c t t h a t

c a n be g i v e n

t h e following i n t e r p r e t a t i o n :

(16)

C(J)

Since

a2

c

@(a)

c

+ 1- J = 0 ,

2 o r 1.

@(A).

c ( a ) over

t h e d e g r e e of

By Theorem 17.41

(vi),

a ( . r + l ) =-a(?);

r;

i n v a r i a n t u n d e r t h e a c t i o n of

hence

is either

@(J) thus

a

is not

As a c o n s e -

a k @(J).

quence [@(a): @(J) I = 2 .

(17)

r

W e have s e e n t h a t t h e modular g r o u p i n g t o (l), and t h a t t h e s u b g r o u p o f

is let

r2

(Theorem 17.42).

f(A)

(18)

Let

thus

r

(19)

Let

Let

m c

r

be a r a t i o n a l f u n c t i o n i n

r

t h a t leaves

and l e t A

a c t s on

f(A)

E

A

accord-

A

fixed

@(A):

i.e.,

w i t h complex c o e f f i c i e n t s

m(f ( A ) ) :f ( m ( A ) ;

acts a s a group of G

C-automorphisms o f

d e n o t e t h e g r o u p o f c-automorphisms o f

i n d u c e d by t h e a c t i o n o f isomorphic t o

r2;

r;

thus

G

c(A)

is naturally

r/r2.

I n 517.42, w e saw t h a t having k e r n e l

@(A).

thus

T

i s a homomorphism o f

r

onto

S3

Norman L. A l l i n g

314

G

(20)

and

on

a r e n a t u r a l l y i s o m o r p h i c , t h e a c t i o n of

S3

b e i n g g i v e n by ( 1 ) . L e t

C(A)

G

and

s3

be

S3

identified. Let

F

be t h e f i x e d f i e l d of

t o conclude t h a t [@(A):

(21)

C(J)] C(J) =

5

~ ( a )is

Indeed, s i n c e

But w e saw i n ( 1 9 ) t h a t

thus [ C ( A ) : C ( J ) l = 6.

and

By c o n s t r u c t i o n ( 4 ) (22)

G(T)

i s f i x e d under

t h e f i x e d f i e l d of a=-3-3/2Gl

a

A3.

A3.

is i n the fixed f i e l d

Using a l i t t l e G a l o i s t h e o r y w e know t h a t a l s o know t h a t

C(a) c

[c(A): C ( J ) ] = 6 Since

A,

that

( 2 1 ) ; hence

C ( a ) [XI

d u c i b l e polynomial i n

-

2x3

Thus

C(X)

( 3 - 33/2a1x2

[ C ( A ) : A ] = 3.

[ C ( a ) : @ ( J )=]2

C(a)

A,

equals

(17),

A3.

W e

and t h a t

establishing (22).

which

i s a r o o t (8):

X

- ( 3 + 33/2a)x + 2;

i s t h e s p l i t t i n g f i e l d of (23) over

The subgroup o f

of

A

(8) g i v e s r i s e t o t h e following irre-

[ C ( A ) : @ ( a ) ]= 6 ,

(23)

@ ( J c) F .

i s 6 w e may u s e s t a n d a r d G a l o i s Theory

S3

[ C ( h ) : F] = 6 .

6;

F

J(.r)-1=a2(-c) =

a s w e have s e e n i n t h i s s e c t i o n ,

G L ( - c ) / 2 7 = -F(-c)/27,

Since t h e o r d e r of

Since

S3.

r

@(a).

t h a t induces t h e a c t i o n

on

A3

@(A)

is (24)

T - ~ ( A ~E )

r;/r3

a r2.

=A ~ .

r2 r;

r:

i s a s u b g r o u p of i s a subgroup of

r

o f i n d e x 3 and o f i n d e x 2.

Thus w e have p r o v e d t h e f o l l o w i n g r Theorem.

a

i s a u t o m o r p h i c on &

I n Example 3 o f 117.42, w e saw t h a t a(q),

e q u a l s (1 3 2 ) :

hence

a(q)

with respect t o

n(Q),

crenerates

ri.

and t h u s A3.

A s a con-

From D e f i n i n g E q u a t i o n t o S p e c i e s and Moduli

315

sequence

r2

(25)

and

r;

Since

q

g

r,

i s o f i n d e x two i n

g i v e n any

(26)

r;.

generate

r

g c

- r;,

rz

t h e g r o u p g e n e r a t e d by

and

r.

is

F i n a l l y w e have: for all

(27)

g

E

r

and a l l

T E

4,

a(

= +a (

g ( T ) )

T )

,

the sign

~ ( g )i s a n e v e n p e r m u t a t i o n i n

being p l u s i f

minus i f i t i s a n odd p e r m u t a t i o n i n

and

S3

s3'

I n d e e d , t h i s f o l l o w s from ( 1 7 . 4 1 : 4 ) .

17.44

v a l u e of

t

Let

We w i l l now c o n c e r n o u r s e l v e s w i t h t h e

> 0.

a ( t i ) and

a (1/2

The map

Theorem 1.

t EIR'

properties:

(i) it h a s r a n g e IR,

increasing,

( i i i l it i s a

(v) a ( i / t ) = - a ( t i ) , Proof.

g3(ti)

.

+ ti/2)

a (ti) has t h e following

+

( i i ) it i s s t r i c t l y monotone

C(")-map,

for a l l

( i v ) it t a k e s 1 t o 0 , and

+.

t EIR

& ( t i> )0.

By Lemma 1 of § 1 7 . 4 0

By Lemma 1 7 . 4 1 ,

is p o s i t i v e , zero, o r negative according as

or

t < l ; thus

as

t >1, t = l , o r

a(ti)

t

<

i s p o s i t i v e , zero, o r negative according

1.

By Theorem 1 7 . 4 1

A c c o r d i n g t o C o r o l l a r y 5 , J9.28,

t

E

[l,")

t-+

a

L

I n Example 2 , § 1 7 . 4 2 , w e see t h a t

Hence

(Recall t h a t

a ( i / t ) = a ( s ( t i ) )= - a ( t i ) ,

From t h e a b o v e w e see t h a t

S(T)

-1

is a

[ o f-)

.

J(ti)

~ ( s =) (1 3 )

= -l/~:

- 1.

(T) = J ( T )

s t r i c t l y monotone i n c r e a s i n g c o n t i n u o u s map o n t o

odd p e r m u t a t i o n .

t >1, t = l ,

thus

E

S3,

an

s ( t i )= i / t . )

by ( 1 7 . 4 3 : 2 7 ) , p r o v i n g ( v ) .

316

Norman L . A l l i n g

t z l , a ( t i ) = ( J ( t i )-1)’I2 and f o r

for

(1)

t c

(O,l),

-

a ( t i ) = - ( J ( t i )- 1 )1/ 2 = - ( J ( i / t ) 1 y 2 . Since

i s a n a l y t i c (Theorem 1 7 . 4 1 ) , w e s e e t h a t t h e Theorem

a(r)

i s proved.

t elR+

The map

Theorem 2 .

( i ) i t has

ing properties:

i a (1/2

H

By Lemma 1 of 517.40,

Proof.

Lemma 1 7 . 4 1 ,

t

or

<

1;

accordingly

around

{1/2

from

Cl(0) n D+

+ ti/2:

t

and maps o n t o

2

i

to

p

-

J(T) 1

31’2}, (-m,0].

Let

of 512.31.

Reflection across

E

T E

[1/2+i/2,p].

it i s clear t h a t

Let

Since

By Lemma 9.28,

C o r o l l a r y 5, 1 9 . 2 8 , (2)

for

t

2

map

t

E

U E

r

passes

T

and t h e n up

C1(l)

( c i r c l e o f r a d i u s 1 and

~ ( z E) E

/(Z-l)

(Theorem

injectively onto

be d e f i n e d a s i n ( 9 . 2 3 : 7 ) ; t h e n

,

f o r all

z

E

C.

i s i n v a r i a n t under

J(-T) = J ( T ) ; t h u s

Thus

J ( T ’ )=

I’,

J ( T ’ )=

J ( T ’= ) J ( T ) . Using

one sees t h a t

1, i a ( 1 / 2 [ l , m )

J

,

as

i s c o v e r e d by c a s e s 2 , 3 , o r 4

T

+ nD

~ ( z =)u ( - z )

J ( c ( T )=J(u(-;)). ) J(-T) ,

C1 ( 0 )

eial3)

(5

c e n t e r 11, i s e f f e c t e d by t h e map

T I

By Theorem 1 7 . 4 1 ,

l i e i n the closed c i r c u l a r a r c

assume t h a t

i.e.,

+ ti/2)

i s s t r i c t l y d e c r e a s i n g from 0

T

C1(0) n D+:

12.35) ; i f mapping

By

> 0.

i a (1/2

According t o C o r o l l a r y 5 , 1 9 . 2 8 ,

= J ( r ) -1.

+.

t E ~ R

for a l l

6(1/2+ti/2)/i

i s positive, zero, o r i s negative. ti2(.,)

+ ti/2) ,

i s p o s i t i v e , z e r o , o r n e g a t i v e ac-

g3(1/2 + t i / 2 )

t > 1, t = 1,

cording a s

( i v ) it takes 1

C(”)-map,

t o 0 , and ( v ) i a ( 1 / 2 + i / 2 t ) = - i a ( 1 / 2

h a s t h e follow-

( i i )i t i s s t r i c t l y

a s i t s range,

lR

( i i i ) ,i t i s a

monotone i n c r e a s i n g ,

+ti/2)

+ ti/2)

-

= (1 J ( 1 / 2

* ia(1/2 + ti/2)

c r e a s i n g c o n t i n u o u s map o n t o

+ ti/2))li2;

thus the

i s a monotone s t r i c t l y i n [O,m).

From D e f i n i n g E q u a t i o n t o S p e c i e s a n d M o d u l i

R e f l e c t i o n of and c e n t e r Clearly

m,

1/2r

c a n be e f f e c t e d by

0

M : (2

then

i

t h e image o f

in

M

r2

c

Let

2

0

<

r;

(Theorem 1 7 . 4 3 )

a(m(z))= a ( z ) .

t <1,

1 < l/t,

n o t e d above.

i a (1/2

Thus

Since

Since

,

a

is

and s i n c e

m(-z) =w(z).

1/2+ti/2=0(1/2 +i/2t),

and

+ti/2)

Note that

= i a (w (1/2

+ i/2t)

as

)=

By ( 1 7 . 4 1 : 9 ) t h i s i s

ia(m(-(1/2+i/2t))) =ia(-(1/2+i/2t)).

iz1/2+i/2t).

r2.

is clearly i n

I',

- 1).

SL2(Z).

(17.43:25), then

z*w(z) E 2/(22

1/2

Let

is i n

M

automorphic with r e s p e c t to

ra

t h e c i r c l e of r a d i u s

(1/2),

ClI2

0(1/2 + i / 2 t ) =1/2 + t i / 2 .

1

(3)

about

C

317

is i n

a(1/2+i/2t)

IRi,

Theorem 1 7 . 4 1

( i v ), we obtain t h e following, (4)

i a (1/2

+ ti/2)

+ i/2t)

= -ia (1/2

:

thus u s i n g ( 4 ) and ( 2 ) w e o b t a i n (5)

for

0 < t

<

1,

ia (1/2

+ti/2)

=

- (1 - J ( 1 / 2 + i / 2 t ) ) 1 / 2 ;

p r o v i n g t h e Theorem.

17.50

The i n v a r i a n t

6.

W e have s e e n i n 517.3 t h a t are defined given

# 0

a n d A(P) 7 ( 1 7 . 1 0 : 1 ) , t h a t e a c h number i s r e a l , a n d

P

G . (P) ( j = 2 and 3)

that

A (P)

(Lemma 1 7 . 3 5 ) .

(1)

If

A (P) >

0,

let

B (P) :33/2G3 ( P ) / ( A ( P ) )'I2;

(2)

if

A(P)

0,

let

B(P) :3 3 / 2 G 3 ( P ) / i ( - A ( P ) ) I/ 2 .

<

and

From t h e C a y l e y - B o o l e Theorem ( 1 7 . 3 3 1 , w e see t h a t t h e following holds: Theorem.

-2 y =P(;)

Let

M

E

(53.15): then

GL2 (IR)

transform

B(P) = B ( P ) .

y

L

=P

(x) t o

Norman L. A l l i n g

318

w e know t h a t

From ( 1 7 . 3 5 : 1 4 ) f?

(3) (T

2

- 1 = J ( T )- 1,

(P) = J ( P )

b e i n g d e f i n e d i n 517.10.)

J ( T ' )

-1,

being defined i n (17.10:4)).

( T '

J ( T )- 1 =

Since

a

L ( T I )

=

(Theorem 1 7 . 4 1 ) , w e see t h a t

J ( T ' )- 1

(4)

By C o r o l l a r y 9 . 2 9

8(P)

(TI

)

-

W e w i l l bend o u r e f f o r t s now t o d e t e r m i n i n g t h e s i g n i n ( 4 ) , i n

a l l cases.

(1)

be a l a t t i c e i n

L

Let

17.51

Let

wL (x) = 4x 3 - g 2 ( L ) x - g 3 (L)I

let

W,(x)

and l e t

T

€8.

and

(x).

WL

E

5:

T

L e t us consider

f? (Wtx)

I

t

for

e . ( t i )E I R , j = 1 , 2 , 3 , A 3 el ( t i )> e 2 ( t i )> e3 ( t i ); t h u s

Since

7

(\47ti)

By ( 1 7 . 3 0 : 4 ) ,

0.

>

0.

By Theorem 1 4 . 2 2 ,

hence w e have proved t h e f o l l o w i n g ; one one r e c a l l s ( 1 7 . 4 1 : 4 ) . L e m m a 1.

Now l e t

real, t h a t

For T 2

t

0, f?(W . ) = a ( t i ) .

tl Recall ( 1 4 . 2 3 ) t h a t

1/2+ t i / 2 .

e2,e3 E @-IR,

that

e3=e2,

and t h a t

e , ( ~ ) is A(F7,)

< 0.

n e e d t o d e t e r m i n e t h e s i g n of

Clearly

(el

- e 2 ) (el - e3) =

(el - e 2 (el

- e2

>O.

2 4 e2-e3=T 0 2

We

From D e f i n i n g Equation t o S p e c i e s and Moduli

[ 3 6 , p.2131.

h

i s d e f i n e d [ 3 6 , p.1901 t o be

e i'/2-e-nt/2.

equals

einT

319

which

s i n c e [ 3 6 , p.1961

w e can compute it.

Notice t h a t

thus it i s

n2 - n = n ( n - 1 ) ; 2

congruent t o

all

n

E

j

( e 2 ( r )-

(7)

pro;ed

is real f o r

)

) / i 2 0.

are d i s t i n c t (7.33:4)

e 3 ( T ) ) / i

, we

conclude t h a t

0;

>

t h e following:

R e c a l l ( 8 . 4 5 ) t h a t two l a t t i c e s

L

are s a i d t o b e e q u i v a l e n t i f t h e r e e x i s t s

a

17.52

C

hn -n

Hence

hn2-n 4

m

= (16h(I n=l

e 's

Since t h e

n.

thus

N;

e 42 / i

(6)

€or a l l

0 , mod 2 ,

and E @ *

L'

in

such t h a t

L ' = aL.

(1)

L

and

w i l l be c a l l e d

L'

a EIR*

such t h a t

IR-equivalent

i f there exists

L' =aL.

Recall ( 1 2 . 4 0 ) t h a t , w i t h D u V a l [ 1 9 r p . 2 1 , w e call

L

real i f

L=L.

Clearly i f

L

and

a r e Ill-equivalent t h e n

(2)

and o n l y i f L = L'

L'

L'

i s real.

i f and o n l y i f DuVal [19, p.21

L

i s real i f

R e c a l l a l s o , P r o p o s i t i o n 7.33,

gj ( L ) = g i ( L ' ) ,

for

that

j = 2 and 3 .

has c l a s s i f i e d real l a t t i c e s as follows:

Norman L. A l l i n g

320

i s rectangular i f it has a b a s i s

L

(3)

that

Iw1,w2 1

i s rhombic i f i t h a s a b a s i s

L

(4)

w 2 clR* i .

and

wlclR*

such

Iwlro21

such t h a t

DuVal [19, p . 2 1 a s s e r t s t h e f o l l o w i n g ( w i t h o u t p r o o f ) :

Assume t h a t

(DuVal).

Lemma

i s r e a l ; t h e n it i s e i t h e r

L

r e c t a n g u l a r o r rhombic. Let

Proof.

L

Q(=(wl

By c o n s t r u c t i o n (9.12:l and 2 1 ,

(9.12).

(A) f i r s t t h a t ( A . 2 ) -wl;

accordingly

If

t wl.

l2It

i s a b a s i s of

L,

and t h u s

A s s u m e now t h a t ( A . 2 ) h o l d s : t h e n

w1 E I R ~ .

then

w1

(w2

and

-w 2 = w 2

would b o t h b e i n

w2

then

z 2 # w2,

and s o

L:

S

E {.E E L:

I II I

6 points.

If

{en'i/2:

L

n=0,1,2,3},

l a s t l y (B.2)

that

{en'i/3:

I n e a c h case L

n=Or..

L

S n=0,

is true

(A.12)

Were

-w 2

=--w

thus

(w2

then

2'

If

(A.21)

If

is a basis

;2)t

lull = I w 2 1 .

Now assume ( B ) t h a t

4 points, o r (B.2)

equals e i t h e r (B.11)

S

i n which c a s e

.,3 ) ,

or it i s ( A . 1 2 )

is r e c t a n g u l a r .

L

Assume

i s rhombic.

L

contains e i t h e r (B.1)

( B . l ) holds,

{en'i/2-ei'/4:

(B.21)

lull 1

=

w 2 f. w;l

i s rhombic.

If

or

w,l

IRi.

which i s a b s u r d .

i s r e a l and hence

w2

t h e n it i s ( A . 2 2 )

of

IRi,

or

-02

is rectangular,

L

(A.1)

IR

is either ( A . l l )

w2

holds

(A.ll)

is e i t h e r i n the

w1

-

Assume

\wl\
is either

then

< Iw21:

lull

t h a t ( A . 1) h o l d s ; t h e n -w2

be a minimal p o s i t i v e b a s i s o f

w,))

i s r e c t a n g u l a r , o r (B.12)

L

i n which c a s e

L

i s rhombic.

Assume

c o n s i s t s of 6 p o i n t s ; then it i s e i t h e r

...,6 1

i s rhombic.

i s r e c t a n g u l a r o r rhombic.

both: t h u s , t h e r e e x i s t

bases

or (B.22)

ten'i/3-ei'/6:

n=0,

...,6).

Thus w e have shown t h a t e a c h r e a l Assume f o r a moment t h a t i t i s

n(:(w,

w2)

t

)

and

n'

( ~ ( 0 'w

1

')

2

t

)

From Defining Equation to Species and Moduli

of

L

such that

w1 € D l

there exists M E G L 2 ( Z )

w2 cDi,

and

such that

R'

-

=MR.

Let

-

Then

cu1+du2 = w '2 = w i = a w l + b o 2 =aol-bw2;

d=-b.

We conclude that

By Theorem 6.22

=wi.

w;

-2ab=det M = f l ,

321

M

thus

E

(ca db ).

c=a

and

which is absurd,

proving the Lemma. (i) Let

L

be rectangular; then there exist

unique positive numbers

c

and

Theorem.

u

such that

c L = Lui'

(ii)

Let

L

be rhombic; then there exist unique positive numbers

and

u

such that Proof.

such that

and

may assume that let

u

c L = L1/2+ui/2'

(i) By definition there exists a basis

w1elR

w1

: (W2/i)/wl;

exists a basis

n

>

and that

0

then

of

of course - equals

Re

wi=2Reol> 0.

W;

:2Irn(w$/oi) ;

Let

then

w

w

2

such that

2)

:ol.

/ i > 0.

-

U ~ E Q I

-

c :l/ulI

and

Flithout loss of

and that

is positive. Let

Let

w2-w1.

c :l/ui

Let

Real

(which then

w i :w 1 + w 2 ;

and let

proving the Theorem.

cL = L1/2+ui/2'

Bibliographic n o t e .

L

(ii) By definition there

c L = Lui'

L

of

R

Without l o s s of generality we

ED^.

w2

generality we may assume that

u

c

DuVal 1191 goes on to extend his

classification of real lattices

L

on pp. 2-3 and p. 33.

Using

the Theorem above we obtain the following versions of DuVal's definitions as follows: (5)

L

is vertical or horizontal according as

O
Note:

L

is square if

exactly these values of

determining the sign of is rhombic; then

g3(ui)

u > l

or

u=l. u

are cardinal in Chapter 14 in and

g3(1/2+ui/2).

Assume

L

322

Norman L. A l l i n g

Note a l s o t h a t t h e s e v a l u e s p l a y a c e n t r a l r o l e i n d e t e r m i n i n g

t h e v a l u e of

g2(1/2+ui/2)

(Theorem 1 4 . 2 3 ) .

Thus D u V a l ' s

t h e o r y and o u r s d o v e t a i l v e r y n i c e l y .

Assume t h a t

17.53

has a real root: i . e . ,

P

( t h e number o f r e a l r o o t s of s=r/2;

Theorem 1 7 . 2 0 ,

E EIR(x,y),

where

y

L

Y

,

i s e i t h e r 2 o r 1.

i s such t h a t r

such t h a t

M E GL2@R)

M

M

Hence

i s diana-

t

0

(12.34 and

a n d 11) t h a t t h e r e e x i s t s 2 y =P(x)

transforms

has real c o e f f i c i e n t s ,

By

Y -Rie%E

f o r a unique

W e have s e e n (17.35:8,9,10,

12.35).

Since

s

thus

=P(x),

l y t i c a l l y equivalent t o

i s e i t h e r 4 or 2.

P(x))

r

that

E =IR(%,y).

to

W e have s e e n

(517.3) t h a t

- 279:

) ) # 0, w e know t h a t t h e Weierstrass '2"3 i n v e r s i o n problem h a s a s o l u t i o n ( 9 . 5 0 ) ; t h u s t h e r e e x i s t s a

Since

g2

lattice (3)

L

(=A(W

c

in

gj = g j ( L ) t

such t h a t

By P r o p o s i t i o n 7 . 3 3 , thus g. 7

L

(L)

3. (4)

.

L

Since

'j By P r o p o s i t i o n 7 . 3 3

-L = L :

i.e.,

i s u n i q u e l y d e t e r m i n e d by

L

is real,

g . (L) = g ,

3

3

( c ),

CL

is

o f rhombic.

or

L~~

Let

p

and

2

g3;

for

j = 2 and

is a real lattice (12.40).

By Theorem 1 7 . 5 2 t h e r e e x i s t u n i q u e p o s i t i v e that

g

From ( 7 . 3 3 : 5 a n d 1 8 ) w e see t h a t

i s unique.

= gj ( L )

j = 2 and 3.

for

L1/2+ui/2

according as

be d e f i n e d t o b e

ui

or

and

c L

u

such

is rectangular

1/2+ui/2

From D e f i n i n g Equation t o S p e c i e s and Moduli

323

accordingly: thus i n e i t h e r case, (5)

cL=L

LJ

.

From ( 7 . 3 3 : 5 and 1 8 ) w e see t h a t g . ( L ) = c2 1 g . ( p ) , for I 3 thus A(L) = c l 2 A ( p ) .

(6)

j = 2 and 3:

By Lemmas 1 and 2 of 517.51 w e know t h a t

P ( I h 7

P

) =LY(P);

thus,

u s i n g ( 2 ) and ( 7 ) , w e have

(8)

6 (PI =

fi

LY

(P).

is i n

GL20R). -2

,u

Let

(10)

M

transform

y 2 = w P (2);

Y

=wg

thus

to

(f;)

2lg3

E =IR(G,$).

W e have s e e n t h a t

(11) 0 ' ( . r L P ) 2 = W u ( p ( * l L P ) ) ; t h u s I R ( p ( * , L P ),TI ( - , L L J ) ) ) a r e

E

and

IR-isomorphic.

By 12.34, 1 2 . 3 5 , and t h e Theorem of Coequivalence (11.40), w e know that

Ys,t

and

(C/LP)/~

are d i a n a l y t i c a l l y equivalent.

Using

( 8 ) w e t h e n have proved t h e f o l l o w i n g : Theorem.

6 (P)

that

= LY ( t i ) . I f

Assume t h a t

has a r e a l r o o t .

P

s = 1 then

17.54

Assume

r = O .

By Theorem 1 7 . 2 0 ,

B ( P ) = c1(1/2

lastly that

P

s

53.20 and 53.21 t h a t t h e r e e x i s t s

If

s = 2

then

+ti/2).

h a s no r e a l r o o t s :

i s e i t h e r 2 o r 0.

i.e.,

W e saw i n

M E G L ~ O R ) ,U , V , W E ~ 0 , 1 ~and 1

324

Norman L. Alling

M

k G (0,1], such that v = w implies k < l , for which -2 forms y2 = P (x) to y =LUrVrW(Pfk). Since r = O , L

trans-

UrViW

has no real roots; thus

v=l=w.

By Theorem 17.20,

implies

s=O.

u=0

(Prk)

Hence

s = 2,

and

u=1

implies

By Theorem 17.50, and (17.37:8 and 9), 31/2(-1)1-u(l+k2) (I-34k2+k4 ) @ ( P I =B(LUrl,l(*rk))=

(2)

lSk(1-k2) Note:

Since Let

(3)

e2

V=Wr

a > 0,

:2ai,

T

e3

and

k c (0,l).

let

and let

K ( T )

2

:e2/e3

2

being defined in terms of

T

((5.20:2,3 and 4) and

0

such that

.

(5.30:3) )

Lemma.

There exists a unique

Proof.

It is easier to consider

modulus to Legendre modulus

K

a

>

k =K(T).

the complementary

K ' ( ; ) ~

(See e.y. , (1.10:3) and

(;).

(3.31:6) for other references to this modulus.) (4)

K~

+ ( K ' )= ~1 and

K '

= Oo/e3 2 2

Using the product development of

@

[36, p.2141. and

0

e3

[36, p.2041 we

find that

Since K '

(G)

ii is a function of a, namely e-2na., thus is a function, F(a) 4 , of a. Clearly it is continuous =2ai,

and differentiable on IR+ always positive.

.

Since

a

>

0, 0 < h

<

1; thus

F(a)

is

From D e f i n i n g E q u a t i o n t o S p e c i e s a n d Moduli

then 0

<

F ( a ) =II"

fi

<

1,

a.

(7)

hence

EIR

+

F ' ( a ) / F ( a ) = ~ ~ = l ~ ~ ( a ) / u n (S ai n)c .e uA(a) > 0;

thus

i s a monotone s t r i c t l y i n c r e a s i n g f u n c t i o n

F

l i m F(a)= O a+O+

Clearly a

and

One e a s i l y sees t h a t

u n ( a ) > 0.

F' ( a ) > 0; of

u (a),

n=l n

325

c ( 0 ~ 1 i)s

P K '

l i m F(a) = l ; thus a++-

and

a monotone s t r i c t l y i n c r e a s i n g

continuous s u r j e c t i o n . U s i n g ( 4 ) w e see t h a t

(8)

+

a EIR

H

K(;)

E

i s a monotone s t r i c t l y d e c r e a s i n g

( O r l )

continuous s u r j e c t i o n , p r o v i n q t h e Lemma. T h i s p r o o f may b e f o u n d , i n a l l

Bibliographic n o t e .

e s s e n t i a l s , i n [ 6 9 r p.4811. w :1/4

Let wl(-

2 ~ =)1 / 2

and

and

w ' =- a i / 2 ;

w2(-

then

2w') = a i

n o t t h e u s u a l c h o i c e s of

136, p . 1 9 0 1 ;

and

w1

and

w

i s p e r i o d i c of p e r i o d s 1 a n d (9)

ai.

thus these a r e

W e have s e e n 15.31

i s d e f i n e d and t h a t it

V7e h a v e a l s o s e e n t h a t

2 2 2 ( s n ' I 2 = ( 1 - m I ( 1 - k s n ) ( = L O , O , O ( ~ n , k )()5 . 3 1 : 2 ) , and t h a t

F(Xai) = @ ( s n , s n ' )

---

Assume f i r s t t h a t that

'3: i s n ,

(10)

('3')

2

E

s = 0;

and

(14.41:2).

thus

u=l.

Recall (14.34:6)

that

2 = - ( 1 + R 2 ) ( 1 + k % ) (= L

and t h a t Thus

sn(u)

w',

(Note:

f o r t h e Weierstrass theory.

w2

They a r e , however, f o r t h e J a c o b i t h e o r y . ) t h a t having chosen

;.

W'/U

E(Y

) =IR(s9,'3')

Ora

E(YO,a)

are

1 r 1 r 1

(tRik))

(14.34:7)

r

(Theorem 1 4 . 3 4 ) .

IR-isomorphic.

By t h e Theorem o f c o -

326

Norman L. Alling

equivalence (11.40),

YOft

equivalent; thus by Theorem 12.36,

y2

Now transform (17.33) to G 2 =@(;).

=

Yofa

(17.10) and a=t.

(1) by dilation by

L1,1, (%, k)

Since

are dianalytically

s=ik

and

?=?

i

(17.33:1),

P(k) = L

(k,k). Further, G3(@) =-G3(Llflf1(*,k)) and A(?) 1, O f 0 A(Llflf1(-,k)) (17.33:3). Using this and (2) we see that

y 2 =1,~

Now transform -2

- -

O f

y =P(x)

0

y 2 = Lo,0,

-

and

by

i

to

n

y = iy.

B (Lo, o ( - ,k)1 =

Since

0 (;,kl.

*

x=x

(17.34); then

y

(x,k) by dilation of -

=

-a

As a consequence

(L1,

(.

,k)1

(17.34:3), we see that (12)

B (PI = B (Lo,

(.

tk)1

-

By Theorem 14.41, (13) E(Y2,t) =IR(sn,sn') , where (sn'12=Lo,0,0(snfk). (This differential equation is just Jacobi's classical equation.) (x,k) has 4 real roots, thus it is treated by the methods Lo,0,o used in 817.53. By Theorem 17.53 and (7) we see that I?(Lo,0,

(14)

0

(.,k)) =a(ti).

B (P) =

c1

Hence

(ti).

Lastly assume that E -IR(x,y). Rie%E.

Y

thus

s=2;

Y

By (1) E = I R ( % , F ) . is

-

u=O.

Recall (17.10) that

is defined (17.10) to be

by definition (17.10)

-

dianalytically equivalent

f o r a unique t > 0 (12.34). By the Theorem of coto Y2,t' equivalence (11.40), E and E(YZft) are =-isomorphic.

F(EC(x,y) identify

(17.10)) F

with

= C(%,?),

F(Xti).

by (1). It is convenient to Let

Eo

be used to denote

E.

Let

From D e f i n i n g E q u a t i o n t o S p e c i e s a n d M o d u l i

y2 - L o , l r l ( ~ , k ),

us transform

i

by d i l a t i o n by

$=v

(17.33)

,

to

327

a defining equation f o r

Eo

y2 =;(;).

and

E 7 = i;

Since

(11,

(17.33:1), Let

Note t h a t for

Eo(i) = F =El(i),

and t h a t s i n c e t h e d e f i n i n g e q u a t i o n

-

(15) h a s 4 real r o o t s , t h e n

El

-

by Theorem 1 7 . 2 0

h a s s p e c i e s 2. (16)

R-isomorphic to

El

a

(12.34).

>

By ( 2 ) , -B

is

Thus 0

6 i = - 1 , B (Lo,l,l( * t k ) 1 =

B(P) = B ( L o r l l l ( ~ r k ) ) . S i n c e

(Lo, 0,o ( . , k ) ) . U s i n g (Lo, 0,o ( a r k ) ) = a ( a i ) .

f o r a unique

E(b’2,a)r

( 1 5 ) a n d Theorem 1 7 . 5 3 ,

w e know t h a t

By Theorem 1 ( v ) o f 1 7 . 4 4 ,

w e know t h a t

a ( a i ) = -a (i/a) ;

(17) Since

B (P) =

Thus

x

is i n

(i/a).

CY

but is not i n

El,

t i n c t s u b f i e l d s of

j = O a n d 1.

volutions

Xti.

(12.32:7))

I n t u r n , each is induced on

Since

+K.

7,

t a k e n t o be

by

.z

Y2,1/t

to

K

y.

or t o

Eo

-K

(where

K

d e n o t e s complex

yj

c a n be t a k e n t o be t h e f i x e d f i e l d o f

c a n be t a k e n t o be -K.

(16) also holds.

and

Xti

Thus, w i t h o u t loss o f g e n e r a l i t y w e c a n t a k e

conjugation).

(12.34),

are dis-

(2) F a + b j’ f o r a l l z E 7 7 By Theorem 1 2 . 3 3 e a c h o f t h e s e i s e q u i v a l e n t ( i n t h e

(12.30:2).

t o be

El

u

d i s t i n c t a n t i - a n a l y t i c map

s e n s e of

and

of F having E as i t s f i x e d f i e l d , j j Each g i v e s r i s e t o d i s t i n c t a n t i - a n a l y t i c i n -

of

u)r 7

Eo

thus t h e r e exist two distinct=-automor-

F;

phisms, of p e r i o d 2 , for

Eo,

Thus

El

is

K.

Since

u i f u f ,

IR-isomorphic t o

C,

E (Y2,1/t)

By t h e Theorem o f c o e q u i v a l e n c e ( 1 1 . 4 0 ) ,

are d i a n a l y t i c a l l y e q u i v a l e n t .

K

can be But

*

Y2,a

By Theorem 1 2 . 3 4

328

Norman L . A l l i n g

(18)

a=l/t.

Combining (16), ( 1 7 ) and ( 1 8 ) w e t h e n have t h e f o l l o w i n g : B(P) = a ( t i ) ,

Theorem.

f o r all

P

which have no r e a l

roots.

Theorems 1 7 . 5 3 and 17.54 combine t o g i v e t h e f o l l o w -

17.55

ing.

B (I?) = a ( t i ),

s = 2 o r 0, Let

let

and

Po

i n a l l cases.

b(P) = a(.r')

T h e o r e m 1.

and i f

s = 1, t h e n

In particular i f B ( P I = a (1/2 + t i / 2 ) .

be r e a l a d m i s s i b l e p o l y n o m i a l s ( 3 . 1 0 ) ,

P1

E . :IR(x

, y . ) , where y ? = P . ( x . ) , f o r j = O and 1. L e t 1 j i 3 3 7 r be t h e number o f r e a l z e r o s o f P I n Theorem 1 7 . 2 0 w e j j' saw how t o d e t e r m i n e t h e s p e c i e s s of E d i r e c t l y from P j j j' The f o l l o w i n g i s t h e main theorem o f t h i s c h a p t e r .

Y

E

Y

and

Rie%Eo

i f and o n l y i f

and

Eo

T h e o r e m 2.

are IR-isomorphic

El

:Rie%E1

and

sl=sl

a r e dianalytically equivalent)

B(Po) = B ( P ~ ) .

Using t h e Theorem o f c o e q u i v a l e n c e ( 1 1 . 4 0 ) w e know

Proof.

and

a r e Ill-isomorphic

that

Eo

Y

are d i a n a l y t i c a l l y e q u i v a l e n t .

Chapter 1 2 Y S j l t j

are

(or equivalently

El

w e know t h a t

for

unique

IR-isomorphic;

t. 7 then

Y'j)

>

i f and only i f

is dianalytically equivalent t o

and

Y

and

Using r e s u l t s o b t a i n e d i n

Assume, f i r s t , t h a t

0.

Y(O)

Y

Eo

and

El

are dianalytically

SO'

Sl'tl Using Theorem 1 2 . 3 7 w e know t h a t

equivalent.

tO=t 1' so=sl

By Theorem 1,

and

b ( P 0 ) = B(P1).

B(Po) = B(P1).

know t h a t

t =tl.

know t h a t

Y (O)

0

and

p r o v i n g t h e Theorem.

so=sl

and

C o n v e r s e l y , assume t h a t

Using Theorem 1 and 2 of 5 1 7 . 4 4 ,

we

Using t h e r e s u l t s o f C h a p t e r 1 2 ( 1 2 . 3 7 ) w e Y

a r e dianalytically equivalent;

From Defining Equation to Species and Moduli

In view of Theorem 2, B ( P ) modu1.u~of

(resp. Y

E

5

329

will be called the algebraic

Rie%E).

Because of the values of

(17.44) we know the

CL(T')

following:

(1)

(resp. n*i)

~ ( p E )~ R *

Since

2

B

(P) = B 2 ( T I )

implies

(resp. s = 1).

s = 2 or 0

-1

= J ( T ' )- 1 = J ( T )

(17.41), we have

the following: Eo(i)

~ h ~ o r e3. m

and

El(i)

are

C-isomorphic (or equiv-

RiemSO (i) and x :RiemcEL (i) are analytically 2 2 5quivalent) if and only if B ( P o ) = B (P1).

alently

x(O)

= -

Recall (12.39) that complement-

equivalent.

are called

(Ys,t)s

and

(Ys,tl)C are analytically

It was shown (Theorem 12.39) that this is so if and

tft'

and

complementary. E'

Ysrtr

and

if they are not dianalytically equivalent, but

their complex doubles

only if

Ys ,

t' =l/t.

Let

Further

E-E(Ysrt) and

was called S,l

E ' :E(Ysrt,).

Ysrt

will be called complementary if

plementary.

Y

and

Further, E(Ysrl) will be called

self

Then

E

and

Ys,tl are com-

self complementary.

Clearly (2)

E

and

are

E'

are not IR-isomorphic but

C-isomorphic if and only if

E

E(i)

and

and

E'

E' (i)

are comple-

mentary. Assume that

C o r o l l a r y 1.

and that if

s o = sl.

g3( O ) + g 3

tary if

g3( 0 )

=

0

Y(O) and

and

Y # 0#

9:')

j = O and 1,

are complementary if and only g:').

Y ( O ) is self complemen-

= 0.

Assume now that j= 0

and

for

P.=W ( 1 ) (j) 1 92 '93

1; then

s

P I

1

0

L

O,O,O(*rkj)'

= 2 = sl.

with

kj

E

(O,l),

for

We have seen (Lemma 17.54) that

Norman L. Alling

330

there exist unique Let

(3)

kc 3

t. > 0

such that

3

~(i/t.): thus

5

3

Y(O)

complementary if and only if for

j = 0 and 1, and

3

3

and

Y(l)

are

k . = kC 7 1-j

t. fl#tl-j. 3

2 4 (l+k2) (1-34k.+k.) (17.37:8 and 9) 18k. (1-ki)

2 ' 1 3

Since

(4)

k. = ~(t,i).

B(P.) = 3

3

We have the following: and

Y(O)

C o r o l l a r y 2.

Y(l)

are complementary if and

only if ( l + k 2o ) (1-34kO+k0) 2 4

(5)

ko ( l - k o l 5 (Po)

#

0

-

# B (P,)

Equivalently, Y (6)

k.=ky-j 3

0

=

3

- Z3I2 ( = Proof.

t

0

=

+

*

Y

are complementary if and only if

j = O or 1, and

for

and

kl (1-k:)

and

y (O)

Corollary 3 .

k

( l + k 2l ) (1-34kl+kl) 2 4 = 0,

2 2

B(Po)#O#B(P1).

is self complementary if and only if

approx. 0.17157288). By definition, Y ( O )

1. By Theorem 17.41,

CY

is self complementary if By Theorem 1,

(i) = 0.

The numerator of (4) factors into

3l/'(l+k?) 3

The only root of this in (0,l) is

3

- 23/2,

B ( P o ) = 0.

(1-6k.+k?)(1+6k.+k?). 3

1

3

3

proving the

Corollary. Assume now that k. 3

E

(0,l).

P. 3

5

Ll , l , l ( * , k j ) l

By Theorem 17.20,

so=O=sl.

we see that (7)

4 3ll2 (l+k?)(1-34k2.+k.) B(Pj) = 3 3 3 18k. (1-k2) 3

3

for

j = 0 and 1, where

Using (17.37:8 and 9)

From Defining Equation to Species and Moduli

Corollary

and

Y(O)

4.

331

are complementary i f and

Y(')

only if (5), or equivalently (6) holds.

Assume now that

Po

5

L o l o , o ( ~ l k o and ) that

P l ~ L o I l l l ( ~ I k l ) l where the 17.20,

s

0

= 2 = s

1'

k . ' s are in (0,l). 7 By ( 1 7 . 3 7 : 8 and 9)

By Theorem

j1l2 ( l + k i ) (1-34kO 2 + k40 ) (8)

31/2

Em1) = Corollary

(l+k:)

and

I

1 8 k 0 ( l - k o2 ) 2

B(Po) =

(1-34kl+kl) 2 4

1 8 k , (1-k,2 1 2 5.

Y(O)

Y

and

are dianalytically equiv-

alent (resp. complementary) if and only if k. =k

I

k. = k

and 6 ( p 0 ) # 0 # B (Pl))I for j = 1-1 Let us turn our attention to species 1.

v + w = 1 (mod 2); then

L

( a

UIVIW

lk)

gives rise t o ,a curve of Let

By (17.37:8 and 9 ) , 2 4 3 l I 2 ( 1 - k 2 ) ( 1 + 3 4 k +k )

B(Lgl0,1(*ik)) =

I

18ik ( l + k 2 )

B(Lllo,l('tk))

=

B ( L o l l l O ( ' i k ) )=

-

31'2

2 4 ( 1 - k 2 ) ( 1 + 3 4 k +k ) I

18ik ( l + k 2 ) 2 4 31i2 ( 1 - k 2 ) ( 1 + 3 4 k +k ) I

18ik(l+k2)

2 4 3 l / * ( 1 - k 2 ) ( 1 + 3 4 k +k )

B (L1I

( I

i

k) 1 =

Clearly the following hold:

(resp.

Assume that

species 1; thus there are 4 cases to consider.

(9)

C

I 1-1 0 and 1.

18ik(l+k2)

k

6

( O l l ] .

332

Norman L

Alling

(10) i B ( L U r V f W ( a r k ) ) i s i n n , i B ( L u f v r w( - , k ) ) = O if

k=l,

i B ( L o , O , l ( ~ , k ) = i B ( L l , l , o ( ~ , k ) )2 0

(.,k)) = i B ( L (.,k)), i B (L1, 0, 1 OI1,O

-

Assume t h a t

Corollary 6 .

(mod 2 )

,

and

t h e s i g n b e i n g o b t a i n e d from ( 9 ) .

+al

and

i f and o n l y

k

i f and o n l y i f

E

(0,1];

then

P = L UfVlW

l i m iB(LufVlW(.ik)) k-. 0

(.,k),

where

v + w - l

i s s e l f complementary

Y

s = 1.

2

k=l. A s e a r l y a s 1797 G a u s s

H i s t o r i c a l note.

ed t h e l e m n i s c a t e i n t e g r a l ( 1 . 3 0 ) .

t h i s function is

(y')

2

( 4 . 3 0 ) had i n v e r t -

The d e f i n i n g e q u a t i o n f o r

= L o , v , w ( y , l ) = (1-y

4

).

(y')

2

= Lo,v,w(y,k)

i s a l s o a n e q u a t i o n t h a t A b e l c o n s i d e r e d i n 1827 ( 4 . 1 0 ) .

A b e l [l, p . 2 6 5 f f . ] c o n s i d e r e d t h e f o l l o w i n g d e f i n -

17.56

ing equation: (1)

y 2 = (1-c 2 x 2 ) ( l + e2x 2 ) -= P ( x ) ,

L e t us c h o o s e

c

e

and

in

+ W .

when

c2

a n d ( i i ) c > e.

form (1) by d i l a t i o n by (2)

Let

G2

Erc/e;

= (l-k2H2)

B(P) = -

then

l/e

"e".)

"h"

to

There are

A s s u m e ( i )h o l d s .

(17.33); then

O < k s 1,

e 2 > 0.

(Note: A b e l u s e d

denote t h e base of t h e n a t u r a l logarithm, not

t w o cases ( i ) c s e

and

% = e x and

Transp=y.

and

(1+z2)= Lo , , , o ~ x " , ~ ~ .

Then,

3 l l 2 (1-k2) ( 1 + 3 4 k Z + k 4 ) 18ik(l+k2)

Assume now t h a t ( i i ) h o l d s . (17.33); then (3)

x=cx

and

T r a n s f o r m (1) by d i l a t i o n by y=y.

k:

e/c; t h e n 0 < k c 1, and -2 -2-2 y = (1-x ) ( l + k x ) = L 0 , 0 , 1 (&,I;) 2 4 3'12 ( 1 - k 2 ) (1+34k +k ) Then @ ( P ) = 18ik (l+k2) Let -2

-

l/c

From D e f i n i n g E q u a t i o n t o S p e c i e s and Moduli

I t i s i n t e r e s t i n g t o s e t down

333

B(LUIVIW), for general

u l v l and w.

(4)

a

Let

be d e f i n e d t o be 0 o r 1 a c c o r d i n g as

u+v

is

c o n g r u e n t t o 0 o r 1, mod 2 ; (5)

t h e n B ( L u I v I w) =

3 l l 2 (-l)u ([ - l ) v(+ - 1 ) w k 2 ] [1-34k2 (-1)V+W+k4] v+w- 2 2 1 8 i a k [ (-1) k l

a c c o r d i n g t o ( 1 7 . 3 / : 8 and 9 ) and ( 1 7 . 5 0 : l and 2 ) . I n 1 9 7 1 , i t w a s n o t e d , w i t h r e f e r e n c e t o Theorem 1 2 . 3 3 , " a n n u l i a r e p a i r e d w i t h K l e i n b o t t l e s w h i l e Mobius s t r i p s s t a n d alone" P1 = L

[6, p.651. (-,k),

1,1,1

Assume l a s t l y t h a t with

k c (0,l);

P o = L o , O , O( - , k )

and

so=2

and

t h e n (17.20)

s1 = 0 .

VJe w i l l c a l l

YZlIt

and

YoIt

s p e c i e s 2 and 0 r e s p e c t i v e l y .

p a i r e d r e a l e l l i p t i c c u r v e s of From Theorem 1 o f 317.55 w e o b t a i n

t h e following: Corollary.

Y(O)

and

Y(l)

are paired r e a l e l l i p t i c

c u r v e s of s p e c i e s 2 and 0 r e s p e c t i v e l y .

This Page Intentionally Left Blank

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Gunning, R.C., Lectures on Riemann Surfaces (Princeton Univ. Press, Princeton, 1966).

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Hardy, G.H. and Wright, E.M., An Introduction to the Theory of Numbers, 3rd ed. (Oxford Univ. Press, 1956).

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Hensel, K. and Landsberg, G., Theorie der algebraischen Funktionen einer Variabeln (Chelsea, New York, 1965).

35

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Hurwitz, A. and Courant, R., Vorlesungen uber allgemeine Funktionentheorie und elliptische Funktionen, 4th ed. (Springer-Verlag, Berlin, 1964).

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Jacobi, C.G.J., Gesammelte Werke (Reimer, Berlin, 1881) , vol. 1.

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Klein, F., On Riemann's Theory of Algebraic Functions and Their Integrals (Dover, New York, 1963).

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Klein, F., Vorlesungen uber die Entwicklung der Mathematik im 19 Jahrhundert (Springer-Verlag, Berlin, 1979).

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Klein, F. and Fricke, R., Vorlesungen uber die theorie der elliptischen Modulfunctionen, 2 vols. (Johnson Reprints, New York, 1966).

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Klein, M., Mathematical Thought from Ancient to Modern Times (Oxford Univ. Press, New York, 1972).

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Knopp, K., Theory and Application of Infinite Series (Blackie, London, 1928).

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Koenigsberger, L., Zur Geschichte der Theorie der elliptischen Transcendenten in Jahren 1826-1829 (Teubner, Leipzig, 1879).

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Koenigsberger, L., Carl Gustav Jacob Jacobi (Teubner, Leipzig, 1904).

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MacLane, S., Categories for the Working Mathematician (Springer-Verlag, New York, 1 9 7 1 ) .

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Massey, W.S., Algebraic Topology: An Introduction (Harcourt, Brace and World, New York, 1 9 6 7 ) .

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Ore, O., Niels Hinrik Abel, Mathematician Extrodinary (Univ. of Minnesota Press, 1 9 5 7 ) .

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Riemann, G.F.B., Mathematische Werke (Teubner, Leipzig, 1892).

54

Saks, S. and Zygmund, A., Analytic Functions, 2nd ed. (Pafistwowe Wydawnictwo Naukowe, Warsaw, 1 9 6 5 ) .

55

Salmon, G., Lessons Introductory to Modern Higher Algebra (Hodges-Foster, Dublin, 1 8 7 6 )

56

Schiffer, M. and Spencer, D.C., Functionals of Finite Riemann Surfaces (Princeton Univ. Press, Princeton, 1 9 5 4 ) .

57

Siegel, C.L., Topics in Complex Function Theory (Wiley, New York, 1 9 6 9 1 , vol. I.

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Tannery, J. and Molk, J. , 6lbments de la thborie des fonctions elliptiques, 2 vols. (Chelsea, New York, 1 9 7 2 ) ; originally published in vols. I, 11, 111, and IV (Paris,

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van der Waerden, B.L., Modern Algebra (Frederick Ungar, New York, 1 9 4 9 1 , vol. I.

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64

Weierstrass, K., Mathematische Werke (Berlin, 1 8 9 4 , 1 9 1 5 , 1 9 0 3 ) vols. I, V, and VI.

65

Weil, A., Elliptic Functions According to Eisenstein and Kronecker (Springer-Verlag, Berlin, 1 9 7 6 ) .

66

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Whittaker, E.T., A Treatise on the Analytical Dynamics of Particles and Rigid Bodies, 4th ed. (Dover, New York, 1944).

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70

Zariski, 0. and Samuel, P., Commutative Algebra, 2 vols. (Van Nostrand, Princeton, 1958 and 1960).

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INDEX

a , §57.41:2 and 13.16

Abel's Addition Theorem, 54.13:l Abel's elliptic function, $, 94.1 Addition Theorem for 9 , 557.50:16 and 17 Admissible polynomials, 53.10 Affine group, Aff2(k) , 53.13:4 Algebraic function field (of one variable), 510.10 Algebraic modulus, B ( P ) , of E, §17.50:1 and 2 Amplitude, am(u), of u, 54.20 Analytic Atlas, 58.20 Analytic differential on a Riemann surface, 558.22 and 8 23 Analytic differentials U1(Yslt) on Yslt, Chapter 16 Analytic map between Riemann surfaces, 58.21 Analytic structure, 58.20 Analytic torusI 58.20 Analytically equivalent Riemann surfaces, 58.45 Ap, A localized at P, 510.13:l A-point, 57.42 Arcsine integral, 52.20 Arithmetic-geometric mean, 53.4 A, space of annuli, 512.38 Atlas of X, 58.20 Automorphic with respect to L, 96.30 Automorphism group of Ys,t, Chapter 13 Aut U and Aut'U, 513.10:l and 2 + Aut u and AutSX, §13.15:1 and 2 5 a(x) and ao(x), 515.22:4 341

Norman L. Alling

342

algebraic modulus, 917.5O:l and 2 Br(z0), the open ball about z o of radius bo(x), §15.22:8 B(P),

@,

r, §7.32:1

the field of all complex numbers

@*-CXEc:X # O l

and cl&+, 93.12:4 the category of all complex algebraic function fields, 910.40 C(YsIt), the divisor class group of YsIt, Chapter 15 CL' 97.34:7 x(W), the Euler characteristic of W Coequivalence Theorems, 910.40 and 911.40 Complementary ellipse, 51.10 Complementary modulus, k', of Legendre, 51.11 Complementary period (of the pendulum), 91.22 Complementary real elliptic curves, 912.39 Complete elliptic integral (of the first kind), K=K(k), 91.11 Complex algebraic function field, 910.11 Complexification, §11.11:3 Commutator [A1,A21 of A1 and A2, 98.43:l Conformal group of C, conX, 93.12:l Conb, §3.12:5 C O ~ X ,93.12:3 con C a t con C a , B , and con C 93.12 :2 a , BlYl Coordinate map, 18.20 Coordinate neighborhood, §8.20

@+,

C,

I A - z ~ I = r) the divisor class group of X I §8.60:6 Co(X), the homogeneous divisor class group of

cr (2,)

EEXEC:

C(X),

D, §9.13:5 D+ and D-, 99.28:8 and 99.31:l Da , dilation by a, §3.13:7 Defining equation, § 14.10 Degree of a morphism, 110.40 A ( A ( L ) ) , §7.33:22 6 and u , 913.16:8 3 2 A(P) GZ(P) . - 27G3(P), 917.30:3

(z y)

X,

98.60:7

Index

343

a square root of A ( T ) , 517.40:9 det(B) , the determinant of the matrix B, 53.11 Differential of the first kind on a Riemann surface (= an analytic differential ) , 58.23 Q1(Ys,t), the space of all analytic differentials on Yslt, Chapter 16 , 53.13:7 and 517.33 Dilation by a, Da : Dilation, da, of @ by a, §13.11:4 Dilation of y by a, 517.34 D i a l the dilation group of @ , 513.11 6(.r),

(z y )

%

Divisor class group C(Yslt) of Yslt, Chapter 15 Divisor class group of X, c ( X ) , 58.60:6 div X, the space of divisors on X, 58.60 divoX, the space of homogeneous divisors on X, 58.60 Doubly periodic, with periods w1 and w 2' 56.30 ax, the boundary of X D(X), the space of meromorphic differentials on X, 58.23 D(X)l, the space of differentials of the first kind (i.e., analytic differentials) on X, 58.23 el,e2, and e3, §7.33:3 Eccentricity, 51.10 Eisenstein series, 57.33:5 Elliptic functions, 56.30 Elliptic integral, 51.11 Elliptic modular function, J ( T ) ,59.20:2 E ( L ), 57.35:14 Equivalent lattices, 56.23 nj : cw , §7.34:9 j

-

Euler's Addition Theorem, 52.40 Extended modular group T I §12.10:8 Fagnano's Theorem, 52.30 Field of elliptic functions, F(L), 56.30:l Field of constants, 510.11 Field of functions, 510.11 F(L), the field of elliptic functions, 56.30:l Fe(L) , 57.42:3

344

Norman L. Alling

0, 510.12 Fu U 510.12 Formal Riemann surface, RiemKF, 510.11:6 Fractional linear substitutions, 93.10 Fundamental domain FD(YOrt), for Y O r t , §12.36:2 Fundamental domain for YlIt, 912.35; FD1(Yl,t) §12.35:2, and FD2 (YlIt) I 512.35: 3 Fundamental domain, FD(Y2,t), for Y2,t, §12.34:6 Fundamental domain for r , 59.27

Fu, the residue class field of

f?,

E

G+, 510.12:6 g2 and g3, §§7.21:12, 7.33:18 2 G2(P) AE 4BD + 3C , §17.30:1 2 C3, §17.30:2 G3(P) Z ACE + 2BCD - AD2 - B E GLn(A), the general linear group of rank n over A, §3.11:2 GO (or G for short), the value group of 0, §10.12:5 GT, §13.12:'1 r 5 h(SL2(Z)), the modular group, §9.21:3

-

-

the isotropy subgroup of r at T , 99.25 ?, the extended modular group, §12.10:8 y, 513.16 (Theorem (v)) Gaussian annulus, Mobius strip and Klein bottle, 512.39 General linear group of rank n over A, GLn(A), 53.11 Generalized Legendre €orm, LUIVIW(xlk),§3.21:10 Genus (topological), §§8.25 and 10.50 Geometric modulus of a real elliptic curve, 512.30 and §17.50:1 and 2 Group of units U(A) , 53.11

rT,

Q, the upper half plane, §3.12:4 h(M) , §3.13:2 H1(X,A), the first homology group of X with coefficients in A H,, and HTlt 513.16 :5 Half periods associated with Q , 57.33:l Historical and bibliographic notes 50.20 Homogeneous divisor class group of X I C o ( X ) , §8.60:7 +

Index

345

Homogeneous divisors on X , divoX, Li8.60 Horizontal lattice, 9 1 7 . 5 2 : 5 Im(z), the imaginary part of z, § 3 . 1 2 : 4 Index, of a generalized Legendre form, 5 3 . 2 1 Indexing, 00.40 Integral elements over A, 5 1 0 . 1 2 Integrally closed, 5 1 0 . 1 2 Invariant under L, 5 6 . 3 0 Inversion, S : , 5 § 3 . 1 3 : 9 and 1 7 . 3 1

(y

Jacobi's imaginary transformation, 5 4 . 2 2 J(L) , § 9 . 1 1 : 5 J(T), the elliptic modular function, 5 9 . 2 0 : 2 k K,

e;/e3,2

§5.31:3

complex conjugation extended to

~0

K , the space of Klein bottles, 5 1 2 . 3 8

Klein surface, 5 1 1 . 2 0 L, a lattice, in @ , 9 6 . 2 L" E L - I 0 1 L-elliptic function, 9 6 . 3 0 LUIVIW(x,k), generalized Legendre form, 5 3 . 2 1 : l O A(T), § § 1 7 . 4 2 : 1 and 1 7 . 4 Lattice, L, in @ , 9 6 . 2 Legendre's Addition Theorem, 5 2 . 5 0 Legendre's equation, q 1 0 2 - n 2 w 1 = 2ni, 5 7 . 3 4 : 1 1 Legendre's modulus, k , 5 1 . 1 1 Lemniscate integral, 91.3 Linear fractional substitution, 9 3 . 1 0 Liouville's Theorem (c. 1 8 4 4 ) , 5 6 . 3 1 Localized at P , 9 1 0 . 1 3 Logarithmetic derivative, k ? D ( f ) , 9 6 . 4 1 : 5

346

Norman L. Alling

M , the maximal ideal of a valuation ring M I the space of Mobius strips, 912.38

0, §10.12:1

MmIn(A), the set of all m by n matrices with coefficients in the ring A, 93.11 Meromorphic differential on a Riemann surface, 958.22 and 8.23 Meromorphic function on a Riemann surface, 98.21 Minimal basis, and minimal positive basis of L, 59.12 Modular, meaning prior to c. 1840, 59.10 Modular angle, §1.20:9 Modular functions, 59.40 Modular group, r : h(SL2(Z)) , §9.21:3 Modulus, F! ( X ) , 99.10 N : {112,...,n,...3 N(f), the norm of f, 57.42:5 Negative lattices, 56.22 Norm, N(f), of f, 57.42:5 Normalized minimal positive basis of L, 59.12 v0 (or v for short), the valuation of 0, 910.12:5 the order of f at z o l 56.41:l Z (f)'

0

0, a valuation ring, §10.11:5

O, the valuation ring at m 'Open ball, Br(z0), about zo of radius 913.2 Orbit space of yS,t' ord(f), the order of f, 98.41:l Order of f at z o , v (f), 56.41:l zO Orthogonal trajectories, 513.3

r, 57.32:l

p, the map of the complex double X of Y onto p-function of Weierstrass, 997.21, 7 . 3 2 ~ 4 Pel 910.13:l p*, §13.15:4 P(n), the period parallelogram of f i t 56.20:6 Paired real algebraic curves, 117.56 Partition function, 55.11 Pendulum equation, 91.20:3

Y,

913.15

Index

347

Pendulum equation (linearized), §1.20:4 Pendulum, simple, 91.2 Period (of the pendulum) , 51.20 Period parallelogram, P(R), of R , §6.20:6 4 , Abel's elliptic function, 54.1 X ( M ) , 517.42:9 n l ( X ) , the fundamental group of X Positive lattice, 56.22 Prerequisites and exposition, 50.30 Principal divisor, 515.11 ,Projective special linear group of rank 2 over A , PSL2(A) , 53.13 Ax4 + 4Bx3 + 6Cx2 + 4Dx + E , §17.10:1 P(x) q, the quotient map of C onto C / L T q* 1 513.13 Q ( z ) , §14.32:2 Quasi-angle (of a pendulum), 51.20 Quasi-periodic, 95.30 Quasi-periods, 15.30 Quotient, T , of R , 56.20:5 r, the number of real roots of

XT

P(x) cIR[x], 517.20:l and 2

1R, the field of all real numbers

1R-equivalent lattices, 517.52:l IE -l/Q = isn, §14.34:6 Re(z) , the real part of z , §3.12:4 Ramification index, 510.20:6 Real algebraic function field, 5910.11 and 11.10 Real elliptic curve, 512.10 Real lattice, 1512.40 and 17.52:2 Rectangular lattice, 517.52:3 Relative degree, §10.20:4 Residue class field of 0, O/M E F O , 510.12 Residue Theorem for elliptic functions, 56.40:2 Rhombic lattices, 517.52:4 Rie% , a contravariant functor, 510.40 Riem F, the formal Riemman surface of F over K, §10.1:6 K Riemann-Roch Theorem, 510.50

Norman L. Alling

348

Riemann sphere, 1, 9 3 . 1 2 Riemann surface, 5 8 . 2 0 Riemann surface of a complex algebraic function field, 910.30 S :

(! ):-

, inversion,

§3.13:9

s 2 h(S), 5 3 . 1 3 S, the category of all Riemann surfaces, 5 1 0 . 4 0 gb, translation of a: by b, § 1 3 . 1 1 : 2 SLn(A), the special linear group of rank n over A , 53 ,11:3 Sn, the symmetric group on n letters sn(u) , cn(u) , and dn(u) , § 5 . 3 1 : 1 S1 :: E/Z, the circle group Srls(o), 5 6 . 5 0 : l D and 6, §13.16:8 a-function of Weierstrass, § § 7 . 2 1 : 4 and 5, and 7 . 3 2 : 2 Self complementary real elliptic curves, 9 1 2 . 3 9 Serre Duality Theorem, § 1 0 . 5 0 : 6 and 7 sinlemn, § 4 . 3 1 : 1 Special linear group of rank n over A , SLn(A) , § 3 . 1 1 : 3 Species of a real elliptic curve, 5 1 2 . 2 0 : 3 Square lattice, § 1 7 . 5 2 : 5 sum, 5 8 . 6 2 : 3 Surface, 5 8 . 2 0 symgdiv X T l and symg divO X T , , § 1 5 . 2 0 : 2 Symmetric Riemann surface, 911.30

(i 11)

T

,

§9.23:2

(0' );

Tb : , translation by b, §3.13:6 t > 0, the geometric modulus of Ys,t' Chapter 1 2 , 5 1 2 . 3 t E h(T), 5 9 . 2 3 T E

W'/W

5 W2/W1,

§5.20:1

ti if s = 2 or 0, and 1 / 2 + ti/2 if Theta functions, Chapter 5 and 3, 5 5 . 2 0 : 4 ej(v) , for j = O , 1 , 2 , e;,e2,e3, and e4, § 5 . 3 0 : 3 Trace, T,(f), of f, § § 7 . 4 2 : 5 and 11.11 Tr(f), the trace of f, § § 7 . 4 2 : 5 and 11.11 T'

f

s = l , §13.16:1

Index

349

Transition functions, §8.20:4 Trans @ , the translation subgroup of @ , §13.11:3 Translation by b , Tb : , §§3.13:6 and 17.32 Translation of c by b, gb, 513.11:2 Transpose Mt of a matrix M Triangular, extreme, and median lattices, 517.52:6

(i y )

u

('1 y)

,

and u h(U), §9.23:7 U(A) , the group of units of the ring A, 53.11 U(0) (or U for short), the group of units of the valuation ring 0, §10.12:1 U §13.20:1 Y' Upper half plane, Q, §3.12:4 f

Vx, §13.30:1 Value group GO (or G for short) of 0, §10.11:5 Valuation ring, 510.11:5 Valuation ring at m, om, 510.13, Example 2 (cont.) Vertical lattices, §17.52:5 W ( s ) f 4s3 - g2s - g3, Weierstrass form, §7.20:2 2' "3 WL(X) f 4x3 - g2 (L)x - g 3 (L)I 517.51:l Weierstrass's Addition Theorem, 58.52:20 Weierstrass's inversion problem, 99.50 Weierstrasslsl)-function, 997.21 and 7.32:4 Weierstrass's o-function, §§7.21 and 7.32:2 Weierstrass's t-function, §§7.30:1 and 7.34:l (X,p,S), the complex double of Y, 513.15 6 , 913.15 5 : K if s = 2 or 1 and K + 1/2 if s = O , §13.16:2

-

Yo,t' 512.36: 1 Yl,t' 512.35 :1 512.34:5 Y2,t' Z, the ring of integers or its addition group <(u), function of Weierstrass, 5§7.30:1 and 7.34:l

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