5Modular forms of level 1

III Modular Forms and L-functions

5.3 Arithmetic of
Recall that we had
∆ =
X
τ(n)q
n
,
and we knew
τ(1) = 1, τ(n) Q.
In fact, more is true.
Proposition.
(i) τ(n) Z for all n 1.
(ii) τ(n) = σ
11
(n) (mod 691)
The function
τ
satisfies many more equations, some of which are on the
second example sheet.
Proof.
(i) We have
1728∆ = (1 + 240A
3
(q))
3
(1 504A
5
(q))
2
,
where
A
r
=
X
n1
σ
r
(n)q
n
.
We can write this as
1728∆ = 3 · 240A
3
+ 3 · 240
2
A
2
3
+ 240
3
A
3
3
+ 2 · 504A
5
504
2
A
2
5
.
Now recall the deep fact that 1728 = 12
3
and 504 = 21 · 24.
Modulo 1728, this is equal to
720A
3
+ 1008A
5
.
So it suffices to show that
5σ
3
+ 7σ
5
(n) 0 (mod 12).
In other words, we need
5d
3
+ 7d
5
0 (mod 12),
and we can just check this manually for all d.
(ii) Consider
E
3
4
= 1 +
X
n1
b
n
q
n
with b
n
Z. We also have
E
12
= 1 +
65520
691
X
n1
σ
11
(n)q
n
.
Also, we know
E
12
E
3
4
S
12
.
So it is equal to
λ
for some
λ Q
. So we find that for all
n
1, we have
665520
691
σ
11
(n) b
n
= λτ(n).
In other words,
65520σ
11
(n) 691b
n
= µτ(n)
for some τ Q.
Putting
n
= 1, we know
τ
(1) = 1,
σ
11
(1) = 1, and
b
1
Z
. So
µ Z
and
µ 65520 (mod 691). So for all n 1, we have
65520σ
11
(n) 65520τ(n) (mod 691).
Since 691 and 65520 are coprime, we are done.
This proof is elementary, once we had the structure theorem, but doesn’t
really explain why the congruence is true.
The function
τ
(
n
) was studied extensively by Ramanujan. He proved the 691
congruence (and many others), and (experimentally) observed that if (
m, n
) = 1,
then
τ(mn) = τ(m)τ(n).
Also, he observed that for any prime p, we have
|τ(p)| < 2p
11/2
,
which was a rather curious thing to notice. Both of these things are true, and
we will soon prove that the first is true. The second is also true, but it uses deep
algebraic geometry. It was proved by Deligne in 1972, and he got a fields medal
for proving this. So it’s pretty hard.
We will also prove a theorem of Jacobi:
∆ = q
Y
n=1
(1 q
n
)
24
.
The numbers τ (p) are related to Galois representations.
Rationality and integrality
So far, we have many series that have rational coefficients in them. Given any
subring
R C
, we let
M
k
(
R
) =
M
k
(Γ(1)
, R
) be the set of all
f M
k
such that
all
a
n
(
f
)
R
. Likewise, we define
S
k
(
R
). For future convenience, we will prove
a short lemma about them.
Lemma.
(i)
Suppose
dim M
k
=
d
+ 1
1. Then there exists a basis
{g
j
: 0
j d}
for M
k
such that
g
j
M
k
(Z) for all j {0, ··· , d}.
a
n
(g
j
) = δ
nj
for all j, n {0, ··· , d}.
(ii) For any R, M
k
(R)
=
R
d+1
generated by {g
j
}.
Proof.
(i)
We take our previous basis
h
j
= ∆
j
E
b
6
E
(k12j6b)/4
4
M
k
(
Z
). Then we
have a
n
(h
n
) = 1, and a
j
(h
n
) = 0 for all j < n. Then we just row reduce.
(ii) The isomorphism is given by
M
k
(R) R
d+1
f (a
n
(f))
d
X
j=0
c
j
g
j
(c
n
)