Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 137 / 1

Write $\tau =x+iy\in \mathfrak{h}$. Forthe imaginary part of the Möbius transformation isAmong the primitive integer pairs $(c,d)$, choose one minimizing the nonzero quantity $|c\tau +d|$. Such a minimum exists because only finitely many lattice points lie in a bounded region. Complete $(c,d)$ to a matrix $\gamma \in S{L}_2(\mathbb{Z})$. Then $\gamma \tau$ has maximal imaginary part in its modular group orbit.

Applying an integral translation does not change that imaginary part, so arrangeIf $\mathrm{Im}(\gamma \tau )<\sqrt{3}/2$, then $|\gamma \tau |<1$. The modular inversion $S:z\mapsto -1/z$ would givecontradicting maximality. Hence every orbit meets the stated region. This is the reduction to the standard modular region argument.

Put $\tau =x+iy$. The modular transformation law andshow that the invariant norm of a modular formis invariant under $\Gamma (1)$. On the region from part (a), it is bounded: it is continuous on every truncated region, while the cusp form condition makes it tend to zero as $y\to \infty$. Thusfor all $x\in \mathbb{R}$ and $y>0$.

The Fourier coefficient formula on one period givesand henceChoosing $y=1/n$ yieldsThis proves the Fourier coefficient bound for a cusp form.

Suppose for a contradiction that $b_0\ne 0$. The Eisenstein series in the question has constant term $2\zeta (k)$, sohas zero constant term and is therefore a level-one cusp form of weight $k$. Part (b) gives the coefficient bound $[q^n]h=O(n^{k/2})$. Since $b_n=O(n^{k/2})$ as well, the displayed Fourier expansion of $G_k$ would imply

Take $n=\ell$ through the primes. Thenwhich cannot be $O({\ell }^{k/2})$ because $k-1>k/2$ for $k\ge 4$. Therefore $b_0=0$, so $g$ vanishes at the only cusp of $\Gamma (1)$ and belongs to $S_k(\Gamma (1))$. This is the Fourier coefficient growth criterion for a level-one cusp form.