Cryptography Reference
In-Depth Information
This allows us to define a formal series
g
=
∞
a
n
q
n
.
n
=1
We refer to
g
as the
potential modular form
attached to
ρ
.Ofcourse,
some conditions must be imposed on the
a
r
in order for this to represent a
complex function (for example, the numbers
a
n
∈O
must be identified with
complex numbers), but we will not discuss this general problem here.
Let
N
be a positive integer. Recall that a modular form
f
of weight 2 and
level
N
is a function analytic in the upper half plane satisfying
f
aτ
+
b
cτ
+
d
=(
cτ
+
d
)
2
f
(
τ
)
(15.5)
for all
ab
cd
∈
Γ
0
(
N
)
(where Γ
0
(
N
) is the group of integral matrices of determinant 1 such that
c ≡
0(mod
N
)). There are also technical conditions that we won't discuss
for the behavior of
f
at the cusps. The
cusp forms
of weight 2 and level
N
,
which we'll denote by
S
(
N
), are those modular forms that take the value 0 at
all the cusps.
S
(
N
) is a finite dimensional vector space over
C
. We represent
cusp forms by their Fourier expansions:
f
(
τ
)=
∞
b
n
q
n
,
n
=1
where
q
=
e
2
πiτ
.
If
M
Γ
0
(
M
), so a modular form of level
M
can be re-
garded as a modular form of level
N
. More generally, if
d
|
N
,thenΓ
0
(
N
)
⊆
(
N/M
)and
f
(
τ
)
isacuspformoflevel
M
, then it can be shown that
f
(
dτ
) is a cusp form of
level
N
. The subspace of
S
(
N
) generated by such
f
,where
M
ranges through
proper divisors of
N
and
d
ranges through divisors of
N/M
, is called the
subspace of
oldforms
. There is a naturally defined inner product on
S
(
N
),
called the Petersson inner product. The space of
newforms
of level
N
is the
perpendicular complement of the space of oldforms. Intuitively, the newforms
are those that do not come from levels lower than
N
.
We now need to introduce the
Hecke operators
.Let
r
be a prime. Define
T
r
∞
|
⎧
⎨
n
=1
b
rn
q
n
+
n
=1
rb
n
q
rn
,
b
n
q
n
=
if
r
N
(15.6)
n
=1
b
rn
q
n
,
⎩
if
r | N.
n
=1
It can be shown that
T
r
maps
S
(
N
)into
S
(
N
) and that the
T
r
's commute
with each other. Define the
Hecke algebra
T
=
T
N
⊆
End(
S
(
N
))
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