E_{k} (τ) = 1 + \frac{2}{ζ ( 1 - k )} \sum_{n \geq 1} σ_{k - 1} (n) q^{n} = 1 - \frac{2 k}{B _{k}} \sum_{n \geq 1} σ_{k - 1} (n) q^{n}, q = e^{2 πi τ} .

(1)

Thus

a_{0} (E_{k}) = 1, a_{n} (E_{k}) = \frac{2}{ζ ( 1 - k )} σ_{k - 1} (n) q u a d (n \geq 1) .

(2)

The Bernoulli number

B_{k}

is rational, so every coefficient is rational. Equivalently, the standard Fourier calculation expresses the coefficient as a rational multiple of

(2 πi)^{k} / ζ (k)

, which is rational by the given fact that

ζ (k) / π^{k} \in Q

The supplied values give the familiar expansions

E_{4} = 1 + 240 \sum_{n \geq 1} σ_{3} (n) q^{n}, E_{6} = 1 - 504 \sum_{n \geq 1} σ_{5} (n) q^{n} .

(3)

Since every divisor sum is an integer, all their coefficients are integers.

b

 0  0

Solution

 0  0

Let

N + 1 = dim M_{k} (S L_{2} (Z))

. For each

0 \leq i \leq N

, the dimension formula, equivalently the valence formula for the modular group, ensures that

k - 12 i = 4 a_{i} + 6 b_{i}

(1)

for some nonnegative integers

a_{i}, b_{i}

. Define

h_{i} = Δ^{i} E_{4}^{a_{i}} E_{6}^{b_{i}} .

(2)

The modular discriminant,

E_{4}

, and

E_{6}

have integral Fourier coefficients and leading terms

q

1

, and

1

, respectively. Hence

h_{i} = q^{i} + \sum_{n > i} c_{i, n} q^{n}, c_{i, n} \in Z .

(3)

Their distinct orders of vanishing make the

h_{i}

linearly independent, so they form a basis.

Starting with

f_{N} = h_{N}

, define

f_{i}

downwards by subtracting from

h_{i}

the integral multiples of

f_{i + 1}, \dots, f_{N}

needed to kill the coefficients of

q^{i + 1}, \dots, q^{N}

. This integer Gaussian elimination preserves all integral coefficients and gives

f_{i} = q^{i} + \sum_{n \geq N + 1} a_{n} (f_{i}) q^{n}, a_{n} (f_{i}) \in Z .

(4)

This is the integral echelon basis of level-one modular forms.

c

 0  0

Solution

 0  0

Write

a_{1} (E_{k}) = \frac{2}{ζ ( 1 - k )} = \frac{r}{s}, g cd (r, s) = 1,

(1)

and suppose

p ∣ s

. In the basis from part b, comparison of the constant term and the first

N

nonconstant coefficients gives

s E_{k} = s f_{0} + r \sum_{i = 1}^{N} σ_{k - 1} (i) f_{i} .

(2)

Indeed,

f_{0}

has constant term one and no terms

q, l d o t s, q^{N}

, while

f_{i}

has the sole term

q^{i}

in that range.

Set

f = \sum_{i = 1}^{N} σ_{k - 1} (i) f_{i} .

(3)

Every

f_{i}

with

i \geq 1

vanishes at infinity and is therefore a cusp form; moreover

f

has integral coefficients. Comparing the coefficient of

q^{n}

in the displayed identity gives

r σ_{k - 1} (n) = s, a_{n} (f_{0}) + r, a_{n} (f) .

(4)

Reduction modulo

p

kills the first term on the right. Since

p ∤ r

, cancellation of

r

yields

a_{n} (f) \equiv σ_{k - 1} (n) (mod p)

(5)

for every

n \geq 1

, proving the Eisenstein congruence from a denominator prime.

d

 0  0

Solution

 0  0

The space

M_{8} (S L_{2} (Z))

is one-dimensional. The two normalized weight-eight forms

E_{8}

and

E_{4}^{2}

therefore agree. Their expansions are

E_{8} = 1 + 480 \sum_{n \geq 1} σ_{7} (n) q^{n}

(1)

and

E_{4}^{2} = 1 + 480 \sum_{n \geq 1} σ_{3} (n) q^{n} + 57600 \sum_{n \geq 2} (\sum_{i = 1}^{n - 1} σ_{3} (i) σ_{3} (n - i)) q^{n} .

(2)

Equating the coefficient of

q^{n}

and dividing by

480

proves the divisor-sum convolution identity of weights four and eight

σ_{7} (n) = σ_{3} (n) + 120 \sum_{i = 1}^{n - 1} σ_{3} (i) σ_{3} (n - i) .

(3)

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Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 137 / 1

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