Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 137 (Modular Forms)

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.

A modular function of weight $k$ and level $\Gamma (1)$ is a meromorphic function $f:\mathfrak{h}\to \mathbb{C}$ satisfyingfor every $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in \Gamma (1)$, and having a meromorphic Fourier expansion at the cusp infinity. Equivalently, $f{|}_k\gamma =f$ for the slash operator for modular forms.

A modular form is a modular function that is holomorphic on $\mathfrak{h}$ and holomorphic at infinity. Its Fourier expansion therefore has the formSince $-I\in \Gamma (1)$, a nonzero level-one form must have even weight.

Extend the slash operator for modular forms to positive-determinant matrices byThe double cosethas left-coset representativesTherefore the sum of the corresponding slashes, multiplied by $p^{k/2-1}$, is exactly

Right multiplication by an element of $\Gamma (1)$ permutes these left cosets. The cocycle law for the slash operator consequently givesso $T_p(f)$ is a weight-$k$ level-one modular function. Each displayed summand is holomorphic on $\mathfrak{h}$, hence so is their finite sum. This is the Hecke operator on modular forms.

Because $f$ is a modular function that is holomorphic on $\mathfrak{h}$, it has a Laurent expansionwith a finite principal part at infinity. The Hecke operator on modular forms acts on this expansion by

If $N>0$ and $a_{-N}\ne 0$, the term $p^{k-1}f(p\tau )$ shows that $T_{p}f$ has pole order $pN$. Inductively, $T_p^{r}f$ has pole order $p^{r}N$ with nonzero leading coefficient. Functions with distinct pole orders are linearly independent, sowould span an infinite-dimensional vector space. This contradicts the hypothesis. Hence $N=0$, and $f$ is holomorphic at infinity. Together with its assumed holomorphy on $\mathfrak{h}$, this proves that $f$ is a modular form, as asserted by the finite Hecke orbit criterion for holomorphy at a cusp.

The cusps are the ${\Gamma }_1(p)$-orbits of primitive columns $(a,c{)}^T$, with simultaneous negation representing the same point of ${\mathbb{P}}^1(\mathbb{Q})$. For the Gamma 1 congruence subgroup, the standard primitive-vector classification separates the orbits according to $d=\gcd (c,p)$. For each divisor $d\mid p$, the two surviving unit coordinates giveorbits when $p>2$. Hence an odd prime gives

For $p=2$, simultaneous negation is already trivial modulo $2$, so the division by two does not apply. In that case ${\Gamma }_1(2)={\Gamma }_0(2)$ has two cusps, represented by infinity and zero. Thus the Number of cusps of Gamma 1 of prime level is

The Klein j-invariantis a weight-zero modular function for $\Gamma (1)$, holomorphic on $\mathfrak{h}$ and meromorphic at infinity. LetSince $c$ is even,The modular invariance of $j$ therefore givesThe quotient is meromorphic on $\mathfrak{h}$ and at the cusps, so it is a weight-zero modular function of level ${\Gamma }_1(2)$. This is the Level-two modular ratio of Klein j-invariants.

The two cusps of ${\Gamma }_1(2)$ are infinity and zero. At infinity,soThus $f$ is holomorphic at infinity and has a simple zero there.

Use the scaling matrixat the cusp zero. Since $j(-1/\tau )=j(\tau )$,The width of zero is two, so its local parameter is $q_0=e^{\pi i\tau }$. As $\mathrm{Im}\tau \to \infty$,and thereforeIt has a simple pole at zero and is not holomorphic there. This uses the width of a cusp to express the two expansions in their correct local parameters.

The left cosets ${\Gamma }_{\infty }\backslash \Gamma (1)$ correspond to primitive bottom rows $(c,d)\in {\mathbb{Z}}^2$, up to simultaneous sign. PutFor $\tau =x+iy$, the identitiesshow that the absolute value of a summand isThe hypothesis $\mathrm{Re}s>(2-k)/2$ says precisely that $r>2$.

If $\tau$ ranges over a compact subset $K\subset \mathfrak{h}$, the positive-definite quadratic form $|c\tau +d{|}^2$ has a uniform lower boundfor some $C_K>0$. The summands are therefore bounded uniformly on $K$ by a constant timesThe corresponding two-dimensional lattice sum converges for $r>2$. The Weierstrass M-test proves absolute and locally uniform convergence. This is the absolute convergence of a weight-k real-analytic Eisenstein series.

For $\delta \in \Gamma (1)$, right multiplication by $\delta$ permutes ${\Gamma }_{\infty }\backslash \Gamma (1)$. The automorphy factor identitytherefore givesThe modular form $f$ obeys the same weight-$k$ transformation law, whileConsequentlyThus the product is invariant under the weight-zero action of $\Gamma (1)$, as described by the invariant product with a weight-k real-analytic Eisenstein series.