Cryptography Reference
In-Depth Information
This yields the expression
1 + 240 j =1
1 − q j 3
j 3 q j
q k =1 (1 − q k ) 24
j =
,
which also works very well for computing j .
More generally, if L is a lattice, define
g 2 ( L ) 3
g 2 ( L ) 3
j ( L ) = 1728
27 g 3 ( L ) 2 .
C × , then the definitions of G 4 and G 6 easily imply that
g 2 ( λL )= λ 4 g 2 ( L ) d g 3 ( λL )= λ 6 g 3 ( L ) .
If λ
(9.14)
Therefore
j ( L )= j ( λL ) .
Letting L = Z ω 1 + Z ω 2 and λ = ω 1
,weobtain
2
j ( Z ω 1 + Z ω 2 )= j ( τ ) ,
where τ = ω 1 2 .
Recall that
SL 2 ( Z )= ab
a, b, c, d ∈ Z ,ad− bc =1
cd
acts on the upper half plane H by
ab
cd
τ = + b
+ d
for all τ
∈H
.
PROPOSITION 9.13
Let τ
∈H and let ab
SL 2 ( Z ) .Then
j + b
+ d
cd
= j ( τ ) .
PROOF We first compute what happens with G k :
G k + b
=
1
( m + b
+ d
+ d + n ) k
( m,n )
=(0 , 0)
1
( m ( + b )+ n ( + d )) k
=( + d ) k
( m,n ) =(0 , 0)
1
(( ma + nc ) τ +( mb + nd )) k .
=( + d ) k
( m,n )
=(0 , 0)
 
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