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
/ω
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
=
aτ
+
b
cτ
+
d
for all
τ
∈H
.
PROPOSITION 9.13
Let
τ
∈H
and let
ab
∈
SL
2
(
Z
)
.Then
j
aτ
+
b
cτ
+
d
cd
=
j
(
τ
)
.
PROOF
We first compute what happens with
G
k
:
G
k
aτ
+
b
=
1
(
m
aτ
+
b
cτ
+
d
cτ
+
d
+
n
)
k
(
m,n
)
=(0
,
0)
1
(
m
(
aτ
+
b
)+
n
(
cτ
+
d
))
k
=(
cτ
+
d
)
k
(
m,n
)
=(0
,
0)
1
((
ma
+
nc
)
τ
+(
mb
+
nd
))
k
.
=(
cτ
+
d
)
k
(
m,n
)
=(0
,
0)
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