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

n≥1

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

n≥1

b

n

q

n

with b

n

∈ Z. We also have

E

12

= 1 +

65520

691

X

n≥1

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

(k−12j−6b)/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

)