Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 137 (Modular Forms)

For an integer $k$ and a congruence subgroup $\Gamma \le S{L}_2(\mathbb{Z})$, the space $M_k(\Gamma )$ consists of holomorphic functions $f:\mathfrak{h}\to \mathbb{C}$ satisfyingfor every $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in \Gamma$, and which are holomorphic at every cusp. This is the space of modular forms of weight $k$ and level $\Gamma$.

For a compact subset $C\subset \mathfrak{h}$, the real-linear maphas inverse norm bounded uniformly for $\tau \in C$. Hence there is $A_C>0$ such thatTherefore the absolute value of the defining series is bounded locally uniformly bywhich converges for $k>2$. The Weierstrass M-test gives locally uniform absolute convergence, so termwise holomorphy proves that $G_k^{(x,y)}$ is holomorphic.

Write $\gamma =\left(\begin{array}{cc}a& b\\ r& s\end{array}\right)$. The automorphy factor identity givesRight multiplication by $\gamma$ bijects ${\mathbb{Z}}^2\setminus \{0\}$ and carries the congruence class $(x,y)$ to $(x,y)\gamma$. Reindexing the absolutely convergent series yields

If $\gamma \in \Gamma (N)$, then $(x,y)\gamma \equiv (x,y)(\mathrm{mod}N)$, so part c gives invariance. It remains to check the cusps. For any $\sigma \in S{L}_2(\mathbb{Z})$,As $\mathrm{Im}\tau \to \infty$, the terms with $c=0$ give a finite constant and the locally uniform estimate for the terms with $c\ne 0$ gives boundedness. A periodic holomorphic function bounded at infinity has a Fourier expansion with no negative powers. Thus every slash transform is holomorphic at infinity, which is holomorphy at every cusp. Hence $G_k^{(x,y)}$ is the congruence-class Eisenstein series in $M_k(\Gamma (N))$.

For fixed $\tau =x+iy$, the set $\{|c\tau +d|:(c,d)\in {\mathbb{Z}}^2\setminus \{0\}\}$ has a positive minimum. Choose a primitive pair $(c,d)$ attaining it and complete it to a matrix $\gamma \in S{L}_2(\mathbb{Z})$. Sincethis point has maximal imaginary part in the orbit. Translate by a power of $T:\tau \mapsto \tau +1$ to arrange $|\mathrm{Re}\tau |\le 1/2$. If now $|\tau |<1$, applying $S:\tau \mapsto -1/\tau$ strictly increases the imaginary part, a contradiction. Thus $|\tau |\ge 1$, proving that every orbit meets the standard fundamental domain of the modular group $\mathcal{F}$.

For a cusp form $f$, the invariant norm of a modular formis modular invariant, bounded on $\mathcal{F}$, and tends to zero at its cusp. Part a therefore makes it bounded throughout $\mathfrak{h}$: $|f(x+iy)|\le C{y}^{-k/2}$.

The correct PDF expansion is $f(\tau )={\sum }_{n\ge 1}{a}_n(f)q^n$. Fourier inversion givesand henceChoosing $y=1/n$ proves the Fourier coefficient bound for a cusp form $|a_n(f)|\le C^{\prime }{n}^{k/2}$.

The corrected PDF integrand isso the integral is a Petersson inner product. On compact subsets it is harmless, while at the cusp the exponential decay of $f$ dominates the polynomial growth of the Eisenstein series; therefore it converges absolutely.

Decompose each nonzero pair uniquely into a positive common divisor times a primitive pair. For even $k$ this writes $G_k$ as $2\zeta (k)$ times the Eisenstein sum over ${\Gamma }_{\infty }\backslash S{L}_2(\mathbb{Z})$. Unfolding the fundamental domain gives a constant multiple ofThe inner integral is the constant Fourier coefficient of the cusp form and is zero. Thus the original integral is zero, expressing the orthogonality of cusp forms and holomorphic Eisenstein series.

The Gamma 1 congruence subgroup is

The point $1/N+{\Lambda }_{\tau }$ has exact order $N$, and changing $\tau$ by ${\Gamma }_1(N)$ changes the pair only by a complex scaling, so the stated map is well-defined.

Conversely, scale a lattice to write it as $\mathbb{Z}\tau \oplus \mathbb{Z}$. A point of exact order $N$ is represented by $(r\tau +s)/N$, where $(r,s)$ is primitive modulo $N$. The group $S{L}_2(\mathbb{Z})$ acts transitively on primitive vectors modulo $N$, so a basis change carries this point to $1/N$. This proves surjectivity. Two resulting normalized pairs are similar precisely when their basis-change matrix fixes $(0,1)$ modulo $N$, namely when it lies in ${\Gamma }_1(N)$. This proves injectivity and the Gamma 1 level structure on a complex lattice bijection

Index-$p$ overlattices ${\Lambda }^{\prime }$ correspond to order-$p$ subgroups of $(p^{-1}\Lambda )/\Lambda \cong {\mathbb{F}}_p^2$, hence to the $p+1$ lines in that vector space.

If $p\nmid N$, the order of $v+{\Lambda }^{\prime }$ cannot decrease: its decrease would have a factor dividing both $p=[{\Lambda }^{\prime }:\Lambda ]$ and $N$. Thus all $p+1$ overlattices are counted.

If $p\mid N$, the element $(N/p)v+\Lambda$ is a nonzero point of order $p$ in $\mathbb{C}/\Lambda$. Exactly one of the $p+1$ overlattices contains it; in that overlattice the image of $v$ has order $N/p$, while in every other one it retains order $N$. ThereforeThis is the Prime-index overlattices preserving a Gamma 1 level structure count.

For a full Euclidean lattice $\Lambda \subseteq {\mathbb{R}}^n$, its dual lattice isFor a Schwartz function, the Poisson summation formula for a Euclidean lattice statesTo prove it, periodize $f$ over $\Lambda$. The resulting function on ${\mathbb{R}}^n/\Lambda$ has Fourier coefficient $m(\Lambda {)}^{-1}\hat{f}(\mu )$ at $\mu \in {\Lambda }^{\vee }$. Evaluating its absolutely convergent Fourier series at zero gives the identity. The same proof applies under the usual weaker hypotheses ensuring convergence of both sides.

Apply part a in ${\mathbb{R}}^2$ toThe supplied identity ${\hat{f}}_k=(-i{)}^k{f}_k$ givesIn the notation of the weighted Gaussian theta sum of a complex lattice, this is

Put $L_{\tau }=y^{-1/2}(\mathbb{Z}\tau +\mathbb{Z})$, a covolume-one lattice, and defineTermwise Mellin transformation in the initial half-plane gives

Split the integral at $t=1$. The integral over $[1,\infty )$ is entire in $s$ because the theta sum decays exponentially. Apply the Poisson summation formula for a Euclidean lattice and the Fourier eigenfunction calculation from part b to the interval $(0,1]$, then substitute $t\mapsto 1/t$. This rewrites the small-time integral as another exponentially convergent integral over $[1,\infty )$ plus explicit elementary Mellin terms. Those terms are meromorphic, but for positive even $k$ their apparent poles are canceled by the zeros of $1/\Gamma (s+k)$. The displayed formula therefore continues holomorphically to every $s\in \mathbb{C}$, proving the analytic continuation of a weight-k real-analytic Eisenstein series.