Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 137 / 1 / b

Solution

Put

P (τ) = q \prod_{n \geq 1} (1 - q^{n})^{24}, q = e^{2 πi τ} .

The infinite product converges locally uniformly and never vanishes on the complex upper half-plane. With

D = q d / d q = (2 πi)^{- 1} d / d τ

, logarithmic differentiation gives

D lo g P = 1 - 24 \sum_{n \geq 1} \sum_{r \geq 1} n q^{n r} = 1 - 24 \sum_{m \geq 1} σ_{1} (m) q^{m} = E_{2} .

The product is unchanged by

T : τ \mapsto τ + 1

. To study

S : τ \mapsto - 1/ τ

R (τ) = \frac{P ( - 1/ τ )}{τ ^{12} P ( τ )} .

Using the transformation law for the Eisenstein series of weight two,

\frac{d}{d τ} lo g R = \frac{2 πi}{τ ^{2}} E_{2} (- 1/ τ) - \frac{12}{τ} - 2 πi E_{2} (τ) = 0.

Thus

R

is constant. At the fixed point

τ = i

, one has

i^{12} = 1

, so

R (i) = 1

. Hence

P (- 1/ τ) = τ^{12} P (τ) .

The transformations under

S

and

T

, which generate the modular group, show that

P

is a weight-twelve modular form. Its Fourier expansion begins

q + O (q^{2})

, so it is a cusp form. The normalized element of

S_{12} (Γ (1))

is unique by part (a), and therefore

Δ (τ) = q \prod_{n \geq 1} (1 - q^{n})^{24} .

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