Put
$$
P(\tau)=q\prod_{n\geq1}(1-q^n)^{24},
\qquad q=e^{2\pi i\tau}.
$$
The <infinite product> converges locally uniformly and never vanishes on the <complex upper half-plane>. With $D=q\,d/dq=(2\pi i)^{-1}d/d\tau$, logarithmic differentiation gives
$$
D\log P
=1-24\sum_{n\geq1}\sum_{r\geq1}nq^{nr}
=1-24\sum_{m\geq1}\sigma_1(m)q^m
=E_2.
$$

The product is unchanged by $T:\tau\mapsto\tau+1$. To study $S:\tau\mapsto-1/\tau$, define
$$
R(\tau)=\frac{P(-1/\tau)}{\tau^{12}P(\tau)}.
$$
Using the transformation law for the <Eisenstein series of weight two>,
$$
\frac{d}{d\tau}\log R
=\frac{2\pi i}{\tau^2}E_2(-1/\tau)-\frac{12}{\tau}-2\pi iE_2(\tau)=0.
$$
Thus $R$ is constant. At the fixed point $\tau=i$, one has $i^{12}=1$, so $R(i)=1$. Hence
$$
P(-1/\tau)=\tau^{12}P(\tau).
$$
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 $S_{12}(\Gamma(1))$ is unique by part (a), and therefore
$$
\Delta(\tau)=q\prod_{n\geq1}(1-q^n)^{24}.
$$

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-137/1/b/solution