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Chapter 14 From Species and Geometric Moduli to Defining Equations
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,
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).
Y
7'
to be (2)
Let
Let
: ti 1/2 K
t
+
s = 2
if
ti/2 if
that
Yslt
and
Y
s
Let
be t h e
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
0,
or
is dianalytically
Y
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
1, and l e t i t b e
or
s = 2
K
+
if
1/2
s = 0.
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
6
Let
then
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. would l i k e t o e x p r e s s 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
cn
then
Lrl,
(3)
and
x
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 ] .
Since
we must look elsewhere f o r
x
p'
and
'p
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
'p'
(( Z - a )
R€L*
~ ( z ) ; hence
is i n
3
Similarly 3
E.
q21Q3
and
(9) E I R " p 1 .
'
is i n
F = C@,v').
Since
By ( 7 . 3 3 : 1 9 ) ,
is i n
The0 r e m
2
and
(9') =475
A ( :g 2
IR
g2'93
3
-
PI.
-
-g$
g3
(7.33)
g3 =
is i n
IR*
.
By Theorem 7 . 4 2 ,
IR ( ' p I p l )
w e see t h a t Since
and
g2
A # 0,
IR. By ( 7 . 3 3 : 4 a n d 2 2 ) ,
.
3
27g32)
E.
c
@,'#I)
[F:E] = 2 ,
(S')2 = W
and
q2
S i n c e (1)
E.
IR, p r o v i n g t h e lemma.
E
Further,
By t h e Lemma,
Proof.
IR, A
g3
E = IR(?\,3'),
Theorem.
W
g2
IR.
(1) w e see t h a t t h i s i s
a2
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
are i n
g3
&
R
I).U s i n g
-
( z - n 2
L L * (
g2
'I))
-
1 2
1
+
and
El
K ~ ( Z=)
a(?) ( z ) E
Proof.
.(;+I
and
9
= E.
g3
are i n
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
ti
on
{m]
IR + ( n + 1 / 2 ) t i ,
w e see t h a t each
n
Since
E
f o r each
-
Since
N.
Z;
E
x
Let
IR.
E
and hence
IR+ t i / 2 ,
Z.
Combining t h e s e r e s u l t s
IRu
{m}
on
IR+ n t i / 2 ,
for
K?)(xi) = ' ! ? ( - x i ) .
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
takes its values i n
thus
n
t h e same i s t r u e o n
L,
c
(14.20:1),
L = L
?(xi); E
n
assumes v a l u e s i n
W
n
{a}
R(x+ti/2) =
i s r e a l - v a l u e d on
~ ? ( x + t i / 2 ) ; thus on
IR+ n t i .
IRu
the latter quantity equals
L,
E
f o r each
= p(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 , ~ ' p ( ~ - t i / 2 ) .S i n c 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
'p
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,
on
{ m l
~ V ' ( x i )= ' p ' ( - x i ) = - y ' ( x i ) ,
for
Thus
'p'
IR+ i n / 2 . 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
0,
avoiding
el
p(z)
+ 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
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
b (x)
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
< 1
for
t > 1
J = g2 3/A,g2(1/2+ti/2)
> 0
J ( p ) = 0 (9.28:7), g2(1/2+ 31i2i/2 = 0.
Since
such that
1 ~ =1
1
and
0 < Re-r < 1/2:
J ( T )E
be in case 3 of 512.3; then
T
257
i.e.,
(9.28).
(0,l)
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 .
14.30
Species 0
Assume now t h a t
t > 0,
with
by u s i n g
9
and
s = 0; 5 K
and
+
1/2.
it i s
9'
t h u s by ( 1 4 . 1 0 : l
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
Since
o (lp) # 'p
.
'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
and
1,
Clearly
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)
with
and x1
xo
( ( 6 . 4 1 ) and ( 6 . 4 2 ) ) .
2
more d e t a i l s . )
Clearly
b
each being simple; t h u s
xl,
(See e . g . ,
(8)
+
( 1 2 . 3 6 ) and ( 1 3 . 1 5 ) ;
{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.
-
Since
IRU
L = L,
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
Indeed, = KQ(z)
I R i + n/2,
,
f o r each
n
of period
1 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. IRu {ml
n
takes i t s values i n
Q
Q(-z)= -Q(z), s i n c e
values i n
where
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'
(15) Q '
on
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
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
Since
=
- ti/2)
0 (1/4 + t i / 2 )
ti
Since
; thus
takes its values i n
Q'
= 0.
= i['p(1/4
'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
in
Q'
P(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
are in
IR. 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
Indeed,
sn'(u) = cn(u)dn(u).
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
(7) Let L
and
I ) ,
= -(l+rX2)(l+k%'2).
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),
0
thus
A s a consequence
Hence
tween Jacobi
.
14.42
5
s
Let
+ ti/2
:1 / 2
( 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
H i s t o r i c a l note.
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
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.
W'/W
'I
= 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 t o (2).
r :: m i n ( 1 , e )
Let
r
radius
@
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).
+
ti/4,
where and where
E(YlIt)
Corollary 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).
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,
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).
Y
7'
to be (2)
Let
Let
: ti 1/2 K
t
+
s = 2
if
ti/2 if
that
Yslt
and
Y
s
Let
be t h e
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
0,
or
is dianalytically
Y
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
1, and l e t i t b e
or
s = 2
K
+
if
1/2
s = 0.
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
6
Let
then
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. would l i k e t o e x p r e s s 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
cn
then
Lrl,
(3)
and
x
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 ] .
Since
we must look elsewhere f o r
x
p'
and
'p
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
'p'
(( Z - a )
R€L*
~ ( z ) ; hence
is i n
3
Similarly 3
E.
q21Q3
and
(9) E I R " p 1 .
'
is i n
F = C@,v').
Since
By ( 7 . 3 3 : 1 9 ) ,
is i n
The0 r e m
2
and
(9') =475
A ( :g 2
IR
g2'93
3
-
PI.
-
-g$
g3
(7.33)
g3 =
is i n
IR*
.
By Theorem 7 . 4 2 ,
IR ( ' p I p l )
w e see t h a t Since
and
g2
A # 0,
IR. By ( 7 . 3 3 : 4 a n d 2 2 ) ,
.
3
27g32)
E.
c
@,'#I)
[F:E] = 2 ,
(S')2 = W
and
q2
S i n c e (1)
E.
IR, p r o v i n g t h e lemma.
E
Further,
By t h e Lemma,
Proof.
IR, A
g3
E = IR(?\,3'),
Theorem.
W
g2
IR.
(1) w e see t h a t t h i s i s
a2
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
are i n
g3
&
R
I).U s i n g
-
( z - n 2
L L * (
g2
'I))
-
1 2
1
+
and
El
K ~ ( Z=)
a(?) ( z ) E
Proof.
.(;+I
and
9
= E.
g3
are i n
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
ti
on
{m]
IR + ( n + 1 / 2 ) t i ,
w e see t h a t each
n
Since
E
f o r each
-
Since
N.
Z;
E
x
Let
IR.
E
and hence
IR+ t i / 2 ,
Z.
Combining t h e s e r e s u l t s
IRu
{m}
on
IR+ n t i / 2 ,
for
K?)(xi) = ' ! ? ( - x i ) .
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
takes its values i n
thus
n
t h e same i s t r u e o n
L,
c
(14.20:1),
L = L
?(xi); E
n
assumes v a l u e s i n
W
n
{a}
R(x+ti/2) =
i s r e a l - v a l u e d on
~ ? ( x + t i / 2 ) ; thus on
IR+ n t i .
IRu
the latter quantity equals
L,
E
f o r each
= p(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 , ~ ' p ( ~ - t i / 2 ) .S i n c 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
'p
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,
on
{ m l
~ V ' ( x i )= ' p ' ( - x i ) = - y ' ( x i ) ,
for
Thus
'p'
IR+ i n / 2 . 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
0,
avoiding
el
p(z)
+ 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
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
b (x)
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
< 1
for
t > 1
J = g2 3/A,g2(1/2+ti/2)
> 0
J ( p ) = 0 (9.28:7), g2(1/2+ 31i2i/2 = 0.
Since
such that
1 ~ =1
1
and
0 < Re-r < 1/2:
J ( T )E
be in case 3 of 512.3; then
T
257
i.e.,
(9.28).
(0,l)
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 .
14.30
Species 0
Assume now t h a t
t > 0,
with
by u s i n g
9
and
s = 0; 5 K
and
+
1/2.
it i s
9'
t h u s by ( 1 4 . 1 0 : l
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
Since
o (lp) # 'p
.
'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
and
1,
Clearly
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)
with
and x1
xo
( ( 6 . 4 1 ) and ( 6 . 4 2 ) ) .
2
more d e t a i l s . )
Clearly
b
each being simple; t h u s
xl,
(See e . g . ,
(8)
+
( 1 2 . 3 6 ) and ( 1 3 . 1 5 ) ;
{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.
-
Since
IRU
L = L,
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
Indeed, = KQ(z)
I R i + n/2,
,
f o r each
n
of period
1 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. IRu {ml
n
takes i t s values i n
Q
Q(-z)= -Q(z), s i n c e
values i n
where
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'
(15) Q '
on
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
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
Since
=
- ti/2)
0 (1/4 + t i / 2 )
ti
Since
; thus
takes its values i n
Q'
= 0.
= i['p(1/4
'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
in
Q'
P(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
are in
IR. 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
Indeed,
sn'(u) = cn(u)dn(u).
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
(7) Let L
and
I ) ,
= -(l+rX2)(l+k%'2).
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),
0
thus
A s a consequence
Hence
tween Jacobi
.
14.42
5
s
Let
+ ti/2
:1 / 2
( 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
H i s t o r i c a l note.
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
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.
W'/W
'I
= 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 t o (2).
r :: m i n ( 1 , e )
Let
r
radius
@
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).
+
ti/4,
where and where
E(YlIt)
Corollary 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).