Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 137 (Modular Forms)

The Fourier expansion of a normalized Eisenstein series isThusThe Bernoulli number $B_k$ is rational, so every coefficient is rational. Equivalently, the standard Fourier calculation expresses the coefficient as a rational multiple of $(2\pi i{)}^k/\zeta (k)$, which is rational by the given fact that $\zeta (k)/{\pi }^k\in \mathbb{Q}$.

The supplied values give the familiar expansionsSince every divisor sum is an integer, all their coefficients are integers.

Let $N+1=\dim M_k(S{L}_2(\mathbb{Z}))$. For each $0\le i\le N$, the dimension formula, equivalently the valence formula for the modular group, ensures thatfor some nonnegative integers $a_i,b_i$. DefineThe modular discriminant, $E_4$, and $E_6$ have integral Fourier coefficients and leading terms $q$, $1$, and $1$, respectively. HenceTheir distinct orders of vanishing make the $h_i$ linearly independent, so they form a basis.

Starting with $f_N=h_N$, define $f_i$ downwards by subtracting from $h_i$ the integral multiples of $f_{i+1},\dots ,f_N$ needed to kill the coefficients of $q^{i+1},\dots ,q^N$. This integer Gaussian elimination preserves all integral coefficients and givesThis is the integral echelon basis of level-one modular forms.

Writeand suppose $p\mid s$. In the basis from part b, comparison of the constant term and the first $N$ nonconstant coefficients givesIndeed, $f_0$ has constant term one and no terms $q,ldots,q^N$, while $f_i$ has the sole term $q^i$ in that range.

SetEvery $f_i$ with $i\ge 1$ vanishes at infinity and is therefore a cusp form; moreover $f$ has integral coefficients. Comparing the coefficient of $q^n$ in the displayed identity givesReduction modulo $p$ kills the first term on the right. Since $p\nmid r$, cancellation of $r$ yieldsfor every $n\ge 1$, proving the Eisenstein congruence from a denominator prime.

The space $M_8(S{L}_2(\mathbb{Z}))$ is one-dimensional. The two normalized weight-eight forms $E_8$ and $E_4^2$ therefore agree. Their expansions areandEquating the coefficient of $q^n$ and dividing by $480$ proves the divisor-sum convolution identity of weights four and eight

The Gamma 1 congruence subgroup isThe space $M_k({\Gamma }_1(N))$ consists of the holomorphic functions $f:\mathfrak{h}\to \mathbb{C}$ satisfyingfor every $\gamma \in {\Gamma }_1(N)$ and which are holomorphic at a cusp for every cusp. Its subspace $S_k({\Gamma }_1(N))$ consists of the forms that vanish at every cusp.

For matrices of positive determinant, use the determinant-normalized slash operator for modular formsIt satisfies the composition law $(f{|}_k[\alpha ]){|}_k[\beta ]=f{|}_k[\alpha \beta ]$. IfthenThus the Fricke involution normalizes the group, and the composition law proves that $f{|}_k[{\alpha }_N]$ has the required transformation law.

The matrix ${\alpha }_N$ permutes the rational cusps. Applying it to a local Fourier expansion merely transports that expansion to the image cusp, with a nonzero change of local parameter. It therefore preserves holomorphy and vanishing at every cusp. Hence

Put $g=f{|}_k[{\alpha }_N]$. Termwise Mellin transform in the initial half-plane givesso $\Lambda (f,s)=N^{s/2}I(f,s)$.

From the definition of the slash action,or equivalentlySubstituting $y=1/(Nt)$ in the Mellin integral therefore givesMultiplying by $N^{s/2}$ and recognizing the completed function on the right yields

At infinity, both $f(iy)$ and $g(iy)$ decay exponentially because they are cusp forms. The transformation just used converts the behavior of $f(iy)$ near zero into the exponential decay of $g(i/(Ny))$ at infinity. Consequently $I(f,s)$ converges absolutely for every $s\in \mathbb{C}$ after splitting the integral at one and applying that substitution to the part near zero. It is locally uniformly convergent in $s$, hence entire, and agrees with the original Dirichlet series in its initial domain. This proves the analytic continuation and the stated functional equation, as summarized by the Mellin transform of a cusp-form L-function.

The modular curve isGive it the quotient topology on $\Gamma \backslash \mathfrak{h}$ and adjoin one end for each cusp of a modular group. At a point with trivial effective stabilizer, the quotient map identifies a sufficiently small disk with a chart. At an elliptic point with cyclic stabilizer of order $e$, choose a disk coordinate $z$ centered there; then $z^e$ descends to a quotient coordinate. At a cusp represented by $\sigma \in S{L}_2(\mathbb{Z})$ and having width $h$, a punctured neighborhood is described byand adjoining $q_h=0$ fills in the cusp. These charts make $X(\Gamma )$ a compact Riemann surface.

A weight-zero modular function is invariant under $\Gamma$, so it descends uniquely through the quotient map to a meromorphic function on $\Gamma \backslash \mathfrak{h}$. Its assumed meromorphic Fourier expansion at every cusp makes the descended function meromorphic in each cusp coordinate. A meromorphic function on a compact Riemann surface is equivalently a holomorphic morphism to the Riemann sphere, sending each pole to infinity. Thus there is a morphismwith $f={\varphi }_f{|}_{\Gamma \backslash \mathfrak{h}}\circ \pi$. The open quotient $\Gamma \backslash \mathfrak{h}$ is dense in $X(\Gamma )$, so this identity also proves uniqueness.

The genus formula for a modular curve iswhere $\mu =[PS{L}_2(\mathbb{Z}):\overline{\Gamma }]$, $e_2,e_3$ count elliptic orbits of orders two and three, and $c$ is the number of cusps.

For ${\Gamma }_0(3)$,There are two cusps, represented by infinity and zero. The congruence $x^2+1\equiv 0(\mathrm{mod}3)$ has no solution, so there is no elliptic orbit of order two. The congruence $x^2+x+1\equiv 0(\mathrm{mod}3)$ has the single solution $x=1$, giving one elliptic orbit of order three. Thereforewhich is the Modular curve X0 3 calculation.

LetForthe matrix ${\gamma }^{\prime }=\left(\begin{array}{cc}a& 3b\\ c& d\end{array}\right)$ lies in $S{L}_2(\mathbb{Z})$ and satisfies $3\gamma \tau ={\gamma }^{\prime }(3\tau )$. The weight-twelve transformation law for the modular discriminant gives the same factor $(3c\tau +d{)}^{12}$ in numerator and denominator. Hence $f(\gamma \tau )=f(\tau )$, so $f$ is a weight-zero modular function of level ${\Gamma }_0(3)$.

The discriminant has no zero in the upper half-plane, so $f$ has neither zeros nor poles there. At infinity,and therefore $f=q^{-2}+O(q^{-1})$: it has a pole of order two. The transformation $\Delta (-1/\tau )={\tau }^{12}\Delta (\tau )$ givesThe Fricke involution exchanges infinity and zero, so $f$ has a zero of order two at the cusp zero.

Thus the morphism ${\varphi }_f:X_0(3)\to \hat{\mathbb{C}}$ has degree two, equal to its total pole order. An isomorphism of compact Riemann surfaces has degree one. Although $X_0(3)$ has genus zero, this particular morphism is therefore not an isomorphism.