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

The dimension of level-one cusp forms is zero when $k$ is odd or $k<12$. For even $k\ge 12$ it isEquivalently, multiplication by the modular discriminant gives $S_k(\Gamma (1))=\Delta M_{k-12}(\Gamma (1))$, and the displayed formula follows from the standard dimension formula for $M_{k-12}(\Gamma (1))$.

PutThe infinite product converges locally uniformly and never vanishes on the complex upper half-plane. With $D=qd/dq=(2\pi i{)}^{-1}d/d\tau$, logarithmic differentiation gives

The product is unchanged by $T:\tau \mapsto \tau +1$. To study $S:\tau \mapsto -1/\tau$, defineUsing the transformation law for the Eisenstein series of weight two,Thus $R$ is constant. At the fixed point $\tau =i$, one has $i^{12}=1$, so $R(i)=1$. HenceThe 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

For $t>0$, the product from part (b) has $q=e^{-2\pi t}\in (0,1)$, soThe weight-twelve transformation law givesTogether with exponential decay as $t\to \infty$, this implies rapid decay at both endpoints for the Mellin transformThe integral therefore converges for every real $s$, and its integrand is strictly positive.

In the half-plane where the Dirichlet series may be integrated term by term,Analytic continuation preserves this identity. For real $s>0$, both $(2\pi {)}^s$ and the Gamma function $\Gamma (s)$ are positive, soThis is the positivity statement in the Mellin transform of the modular discriminant, and in particular $L(\Delta ,s)\ne 0$.