Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 23 (Modular Forms)

The central matrices $\pm I$ act identically on the complex upper half-plane, so the effective group is ${\mathrm{PSL}}_2(\mathbb{Z})$. Away from points with nontrivial effective stabilizer, properly discontinuous action supplies ordinary quotient-disc charts. At a fixed point $z_0$, the coordinate $u=(\tau -z_0)/(\tau -{\overline{z}}_0)$ identifies the stabilizer action with a rotation; its invariant coordinate is $u^e$, where $e$ is the effective stabilizer order. These charts give the quotient its Riemann surface structure.

The elliptic stabilizers of the modular group occur only in the orbits of $i$ and $\omega =e^{2\pi i/3}$. They have effective orders two and three. Consequently the analytic ramification indices of the map from the half-plane areThe stabilizers in ${\mathrm{SL}}_2$ have orders four and six, but the central factor does not double these indices.

All rational boundary points, including infinity, lie in one cusp of a modular group: a primitive column $(a,c)$ can be completed to a determinant-one integral matrix taking infinity to $a/c$. The stabilizer of infinity is generated effectively by $T:\tau \mapsto \tau +1$, and the coordinate $q=e^{2\pi i\tau }$ identifies its high horodisc quotient with a punctured disc. Adding $q=0$ fills that disc. This constructs the compactified modular curve from the extended half-plane; it does not use the ordinary subspace topology on the rational boundary.

For compactness use the standard fundamental domain of the modular group. Its part below a fixed height $Y>1$ is compact, since its imaginary part is at least $\sqrt{3}/2$. The part above $Y$, modulo translation and with the modular cusp added, is a closed disc in the $q$-coordinate. Their images cover the quotient, so it is compact. The same argument with finitely many translates proves compactness for every finite-index subgroup.

The weights of $E_4^3$ and $\Delta$ agree, so their ratio is invariant under the modular group. Nonvanishing of the modular discriminant on the half-plane makes the ratio holomorphic there. At an elliptic point, an invariant holomorphic function has a power series in the quotient coordinate, so the ratio descends holomorphically. At the unique modular cusp its expansion isThus it is a meromorphic function on the compact Riemann surface with precisely one pole, of order one. The degree of a nonconstant meromorphic map to the sphere equals its total pole order. It therefore has degree one, hence is a biholomorphism:This is the single-pole criterion for a spherical coordinate.

The matrix $\left(\begin{array}{cc}0& -1\\ 1& 1\end{array}\right)$ fixes $\omega$. Weight-four transformation gives $E_4(\omega )=(\omega +1{)}^4{E}_4(\omega )$, and $(\omega +1{)}^4\ne 1$. Hence $E_4(\omega )=0$ and $j(\omega )=0$. Since $j$ is a degree-one coordinate, this is a simple zero on the quotient surface. Pulled back to the half-plane it has order three, by the analytic ramification index in part (a).

Write $t(\tau )=\Delta (\tau )/\Delta (2\tau )$. For $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in {\Gamma }_0(2)$, the integral matrix ${\gamma }^{\prime }=\left(\begin{array}{cc}a& 2b\\ c/2& d\end{array}\right)$ has determinant one and satisfies $2\gamma \tau ={\gamma }^{\prime }(2\tau )$. Both discriminants acquire the same factor $(c\tau +d{)}^{12}$, so $t$ is invariant. It has neither zeros nor poles in the half-plane.

Reduction modulo two gives index three, with two modular cusp classes, infinity and zero, of cusp widths one and two. These can also be found from the orbits of upper triangular matrices on primitive columns modulo two. Compactness follows from the finite-index argument in part (a).

At infinity $t=q^{-1}(1+O(q))$, so it has a simple pole. At zero use $S\tau =-1/\tau$ and the modular discriminant inversion law:Thus zero is a simple zero, not a pole, in its width-two modular cusp coordinate. The single-pole criterion for a spherical coordinate now givesThis discriminant-ratio coordinate on X0 2 is different from a ratio of two $j$-invariants.

View $j$ as a meromorphic function on $X({\Gamma }_0(2))$, using its coordinate $t=j_2$. At the infinity modular cusp its pole order is one. At the zero modular cusp, $j(S\tau )=j(\tau )=q^{-1}+O(1)=q_0^{-2}+O(1)$, so its pole order is two. Part (c) identifies these poles with $t=\infty$ and $t=0$.

The zero of $j$ on the full modular curve lies at its order-three elliptic point. There are no effective order-three stabilizers in ${\Gamma }_0(2)$: their lifts have trace $\pm 1$, whose characteristic polynomial modulo two is $X^2+X+1$, whereas an upper triangular matrix over ${\mathbb{F}}_2$ has both diagonal entries one. Therefore the degree-three covering has exactly one point over that elliptic point, with analytic ramification index three. Its $t$-coordinate is $a=j_2(\omega )$, and the zero divisor of the pulled-back $j$ is $3[a]$.

A rational function with these zeros and poles must be $C(t-a{)}^3/t^2$. At the infinity modular cusp both $j$ and $t$ have leading coefficient one times $q^{-1}$, giving $C=1$. HenceThe proof fixes the coefficient and the powers using modular cusp cusp widths and elliptic ramification, rather than assuming a formula for the level-two coordinate.

Apply the Poisson summation formula to $h(x)=e^{2\pi i\tau x^2}$, with Fourier kernel $e^{-2\pi ix\xi }$. Its complex Gaussian Fourier transform isChoose the square root holomorphic on the half-plane and positive when $\tau$ is positive imaginary. Summing over integer $\xi$ givesHere the printed theta series of integer squares uses $q=e^{2\pi i\tau }$; the common theta-constant convention instead uses $e^{\pi iz}$.

To keep track of the shifted series, put $z=2\tau$ and defineThus $\varphi (\tau )={\Theta }_4(2\tau )$. The same Poisson calculation with a phase or shifted lattice gives ${\Theta }_4(-1/z)=\sqrt{-iz}{\Theta }_2(z)$ and ${\Theta }_2(-1/z)=\sqrt{-iz}{\Theta }_4(z)$. Translation gives ${\Theta }_4(z+2)={\Theta }_4(z)$ and ${\Theta }_2(z-1)=e^{-\pi i/4}{\Theta }_2(z)$. These are theta-constant inversion and translation laws.

Let $U\tau =\tau /(2\tau +1)$ and put $w=-1/z-1$, so $2U\tau =-1/w$. Raising the preceding identities to the eighth power removes all square-root and phase ambiguities:Also $\varphi (\tau +1{)}^8=\varphi (\tau {)}^8$. The given generators, including their negatives, therefore establish weight four for ${\varphi }^8$ on ${\Gamma }_0(2)$, and weight $4k$ for ${\varphi }^{8k}$.

The defining series is holomorphic and its expansion at infinity has no negative powers. At the other modular cusp,Writing $q_0=e^{\pi i\tau }$ gives ${\Theta }_2(\tau /2)=2q_0^{1/8}{\sum }_{n\ge 0}{q}_0^{n(n+1)/2}$, so the right side begins $16q_0$ and is a holomorphic power series in $q_0$. Taking its $k$th power proves modular cusp holomorphy for every $k>0$. ConsequentlyThe exponent in the shifted series is $n^2$, as printed in the PDF, not the corrupted exponent in the TeX aid.

Use the three left-coset representatives $I,S,ST$ for ${\Gamma }_0(2)\backslash {\mathrm{SL}}_2(\mathbb{Z})$. If a nonzero weight-four cusp form $h$ existed, its coset norm of a modular formwould be a nonzero weight-twelve modular form for the full group: right multiplication permutes its factors. At infinity $h$ has order at least one in $q$. The other two factors correspond to the zero modular cusp of cusp width two and each have order at least $1/2$ in $q$. Thus $H$ has order at least two there.

The ratio $H/\Delta$ is weight zero, holomorphic on the half-plane because $\Delta$ has no zeros there, and holomorphic at the modular cusp with value zero. It descends to a holomorphic function on the compact full modular curve. Such a function is constant, hence zero, contradicting $H\ne 0$. ThereforeConstant terms at the two modular cusps define a linear map $M_4({\Gamma }_0(2))\to {\mathbb{C}}^2$ whose kernel is this modular cusp space. It is injective, so the dimension is at most two. The forms $E_4(\tau )$ and $E_4(2\tau )$ are holomorphic weight-four forms for this group. For the second, the same conjugation used in 1(c) proves transformation, and $E_4(2S\tau )|$ after the weight factor is $2^{-4}{E}_4(\tau /2)$, proving holomorphy at zero. Their constant terms are both one but their $q$ coefficients are respectively $240$ and zero, so they are independent. We obtainThis is the weight-four Eisenstein basis at level two.

The constant and first Fourier coefficients of ${\varphi }^8$ are $1$ and $-16$. In the weight-four Eisenstein basis at level two, these forceFor $n>0$, its $q^n$ coefficient is $-16{\sigma }_3(n)+256{\sigma }_3(n/2)$, where the second divisor sum is zero if $n$ is odd. The even divisors have cube sum $8{\sigma }_3(n/2)$, soOn the other hand, expanding the eighth power of the theta series of integer squares counts ordered integer eight-tuples of square sum $n$. Their sign is $(-1{)}^{{\sum }_j{m}_j}=(-1{)}^n$, since $m_j^2\equiv m_j(\mathrm{mod}2)$. If $r_8(n)$ denotes that count, thenSeparately $r_8(0)=1$, from the all-zero tuple. The positive-divisor formula is not a formula at zero. This is the eight-square representation formula.

The map $X(\Gamma )\to X(\Gamma (1))\cong {\mathbb{P}}^1$ has degree $d$, the effective projective index. Ramification occurs only over the two elliptic points and the modular cusp. Over the order-two elliptic point, $r_2$ points have index one and the other $(d-r_2)/2$ points have index two, contributing $(d-r_2)/2$ to the ramification divisor. The analogous order-three contribution is $2(d-r_3)/3$.

The local degree at a modular cusp is its effective cusp width. The cusp widths sum to $d$, so the modular cusp contribution is $d-r_{\infty }$. The Riemann-Hurwitz formula, $2g(X)-2=d(2g(Y)-2)+{\sum }_P(e_P-1)$, now yieldsTherefore the genus formula for a modular curve isUsing the index in ${\mathrm{SL}}_2$ without accounting for its center would give the wrong degree when $-I\notin \Gamma$.

Reduction modulo $p$ maps ${\mathrm{SL}}_2(\mathbb{Z})$ onto ${\mathrm{SL}}_2({\mathbb{F}}_p)$, of order $p(p^2-1)$. The image of ${\Gamma }_1(p)$ is the upper unipotent subgroup of order $p$. Thus its index in ${\mathrm{SL}}_2$ is $p^2-1$, and, since $-I\notin {\Gamma }_1(p)$ for $p\ge 5$, its effective index isEvery element of ${\Gamma }_1(p)$ has trace congruent to two. An effective elliptic element of order two or three has trace zero or $\pm 1$ in a lift, so neither is possible for $p\ge 5$. Hence $r_2=r_3=0$.

Represent a modular cusp by a primitive column $(a,c)$, modulo sign. Its reduction is a nonzero vector in ${\mathbb{F}}_p^2$ modulo sign, and the unipotent subgroup acts by $(a,c)\mapsto (a+bc,c)$. For $c=0$, the nonzero values of $a$ give $(p-1)/2$ orbits. For $c\ne 0$, $a$ varies freely and $c$ modulo sign gives another $(p-1)/2$ orbits. Thus there are $p-1$ modular cusps. The reduction classification is sufficient as well as necessary: completing two primitive columns to determinant-one matrices and adjusting the second columns by a translation makes congruent columns related by $\Gamma (p)$.

More explicitly, if a determinant-one matrix has first column $(a,c)$, conjugating $T^w$ givesIts least allowable cusp width in ${\Gamma }_1(p)$ is one for $c\equiv 0$ and $p$ otherwise. Both types therefore number $(p-1)/2$, with cusp widths one and $p$; their cusp width sum is $d$. There are no sign-twisted modular cusp periods here, because trace two cannot be congruent to minus two for these primes. Substitution givesThese are the prime Gamma 1 cusp counts and widths.

The effective group is torsion free and every modular cusp is regular in the preceding sense, so orders of a meromorphic weight-$k$ form are integers. At interior points use a local automorphy trivialization; at a modular cusp use the Fourier order of the appropriate slash transform in its cusp width coordinate. Let $C$ be the sum of all modular cusp points, each once, and setA meromorphic function $\phi$ belongs to the Riemann-Roch space $\mathcal{L}(D(f))$ exactly when $\mathrm{div}\phi +D(f)\ge 0$. In the interior this requires $f\phi$ to have no pole; at a modular cusp it requires order at least one. Conversely, the quotient of any weight-$k$ cusp form by $f$ is a meromorphic weight-zero function satisfying precisely those inequalities. This proves the cusp-form divisor presentationTo compute the degree without imposing a valence formula as an extra assumption, use the meromorphic tensor differential $f^{12}(d\tau {)}^{6k}$. Its automorphy factors cancel. Its order at an interior point is $12{\mathrm{ord}}_{P}f$; at a modular cusp it is $12{\mathrm{ord}}_{P}f-6k$, since $d\tau$ is a nonzero constant times $d{q}_c/q_c$. A meromorphic section of the $6k$th tensor power of the canonical bundle has total divisor degree $6k(2g-2)$. The regular-cusp valence formula on a torsion-free modular curve is thereforeFor $k\ge 3$, $\mathrm{deg}D(f)-(2g-2)=(k-2)d/12>0$. The Riemann-Roch theorem says $\ell (D)-\ell (K-D)=\mathrm{deg}D+1-g$, and a divisor of negative degree has no nonzero sections. Thus $\ell (K-D)=0$ andUsing $d=(p^2-1)/2$ and $r_{\infty }=p-1$ givesThe canonical-degree and Riemann-Roch facts used here are general results for compact Riemann surfaces, as permitted.

For a pair $(L,t)$ choose an oriented basis $({\omega }_1,{\omega }_2)$ with $\mathrm{Im}({\omega }_1/{\omega }_2)>0$ and $t={\omega }_2/N$ modulo $L$. Such a basis exists because an exact-order point gives a primitive vector modulo $N$, which can be completed to a determinant-one basis. Define the marked-lattice model of a modular form byChanging to a basis with the same marked point uses a matrix with $c\equiv 0$, $d\equiv 1$ and therefore $a\equiv 1(\mathrm{mod}N)$, exactly ${\Gamma }_1(N)$. The weight-$k$ transformation of $f$ cancels the factor from ${\omega }_2^{-k}$, proving independence of the basis.

The resulting function has homogeneity $F_f(uL,ut)=u^{-k}{F}_f(L,t)$ for $u\in {\mathbb{C}}^{\times }$. Conversely evaluating at $(\mathbb{Z}\tau +\mathbb{Z},1/N)$ recovers $f$. Holomorphy in $\tau$ and holomorphy in the local parameters of degenerating lattices at every modular cusp characterize the functions arising from modular forms, rather than arbitrary homogeneous lattice functions.

Use the following normalizations for Hecke operators on marked lattices and the diamond operator:Multiplication by a unit $d$ preserves the order of the marked point. There are $p+1$ prime-index overlattices; when $p\nmid N$ they all preserve that order. When $p\mid N$, precisely the overlattice killing the order-$p$ subgroup generated by $(N/p)t$ is excluded, leaving $p$ terms. Thus the definition is meaningful at bad primes too.

Both operations preserve homogeneity, since scalar multiplication bijects the indexing lattices and multiplies every summand by the same $u^{-k}$. Changing the marked basis merely permutes the overlattices, giving the required ${\Gamma }_1(N)$ transformation. Locally each term is a modular form evaluated after a rational fractional-linear substitution, multiplied by its appropriate automorphy factor; this preserves holomorphy on the half-plane.

For modular cusp holomorphy, factor any such rational substitution at a rational modular cusp into an integral modular substitution followed by an upper triangular map $z\mapsto Az+B$ with $A>0$. An existing holomorphic modular cusp expansion stays bounded under this map. The finite sum has the positive period supplied by its new level, so boundedness makes its singularity removable in the new modular cusp parameter. This is cusp holomorphy under rational slash operators. It proves that the operators preserve the functions arising from $M_k({\Gamma }_1(N))$.

For clarity, the factor $1/p$ is paired with the homogeneity convention $F(uL)=u^{-k}F(L)$; it gives the Fourier normalization requested in part (c). The diamond action on modular forms equals $f{|}_k\gamma$ for a lift $\gamma \in {\Gamma }_0(N)$ with lower-right entry congruent to $d$, since the transformed marked point is $d/N$ modulo the original lattice.

For $L_{\tau }=\mathbb{Z}\tau +\mathbb{Z}$, the index-$p$ overlattices areThe first $p$ preserve the marked point $1/N$. At a good prime the last equals $p^{-1}{L}_{p\tau }$, with marked point corresponding after scaling to $p/N$. Consequently the lattice formula becomesThe first term has coefficient $a_{pn}$, because the sum of $p$th roots of unity is zero unless its exponent is divisible by $p$. The second has coefficient $\chi (p)p^{k-1}{a}_{n/p}$ when $p\mid n$ and zero otherwise. ThusThis includes $b_0=(1+\chi (p)p^{k-1})a_0$ at a good prime.

At $p\mid N$, the last overlattice loses the required exact order and is excluded. The operator is then $U_p$, with $b_n=a_{pn}$. The formula remains valid for all primes if the Dirichlet character is extended to integers by zero on nonunits, so $\chi (p)=0$ at bad primes. Without that extension, the printed $\chi (p)$ is defined only when $p\nmid N$. This is the good-prime and bad-prime Hecke coefficient formula.

Take a nonzero eigenform of positive weight, and put $h(\tau )=f(p\tau )$ and $C=\chi (p)p^{k-1}$. The original form belongs to level $Np$ by inclusion of groups. For $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in {\Gamma }_0(Np)$, the conjugate $\left(\begin{array}{cc}a& pb\\ c/p& d\end{array}\right)$ lies in ${\Gamma }_0(N)$, soIts Dirichlet character is therefore the specified reduced Dirichlet character; modular cusp holomorphy is permitted in the question. This is the Dirichlet character version of an oldform by argument dilation.

A nonzero positive-weight modular form cannot be constant, because the matrix $\left(\begin{array}{cc}1& 0\\ N& 1\end{array}\right)$ would force a nonzero constant to equal $(N\tau +1{)}^k$ times itself. Let $n_0>0$ be its first nonzero positive Fourier index. In a relation $Af+Bh=0$, the $q^{n_0}$ coefficient forces $A=0$, then $B=0$. Hence their span is two-dimensional, even when the original constant term is nonzero.

At the new level $p$ is a bad prime, so its operator is $U_p$. The good-prime eigenrelation at the old level and part (c) giveThus the matrix in the ordered basis $(f,h)$ iswith characteristic polynomial $X^2-\lambda X+C$. For its distinct roots,This is prime stabilization of an oldform.

The nonzero positive-weight qualification is necessary for the two-dimensional assertion. The zero form gives no such span, and if weight zero is allowed, $f=1$ at trivial Dirichlet character has distinct good-prime roots $1$ and $p^{-1}$, while $f(\tau )=f(p\tau )$ spans only one dimension. The asserted result uses the usual nonzero positive-weight eigenform setting.

For a cusp form, the invariant norm of a modular form $|f(\tau )|(\mathrm{Im}\tau {)}^{k/2}$ is bounded on the entire half-plane. It is invariant under the level group; on a truncated fundamental region of a modular subgroup boundedness is compactness, and near each modular cusp exponential decay of the modular cusp expansion dominates the power of height. With this bound, Fourier inversion on a horizontal interval givesChoosing $y=1/n$ proves the Fourier coefficient bound for a cusp form, $a_n=O(n^{k/2})$. Therefore $\sum |a_n|n^{-\sigma }$ converges when $\sigma >k/2+1$, and uniformly on every compact subset of that half-plane. The locally uniform limit of its holomorphic terms is holomorphic, so

With the determinant-normalized slash operator, let $W_N=\left(\begin{array}{cc}0& -1\\ N& 0\end{array}\right)$ and $Jf=i^k(f{|}_k{W}_N)=g$. Conjugation by $W_N$ preserves ${\Gamma }_1(N)$, and rational slash operators preserve modular cusp holomorphy and vanishing. Hence $g$ is also a cusp form at that level. Direct substitution gives $J^{2}f=f$; the factor $i^k$ cancels the central weight sign.

Put $F(y)=f(iy/\sqrt{N})$ and $G(y)=g(iy/\sqrt{N})$. The defining formula gives the exact relationsIn the initial half-plane of absolute convergence, termwise integration of the Fourier series and the gamma function yield the Mellin transform of a cusp-form L-functionThe scaling of $y$ accounts for $N^{s/2}$ in the completion. At infinity $F$ and $G$ decay exponentially. At zero the boxed relation expresses $F$ as a power times an exponentially decaying function of $1/y$. Thus this integral converges locally uniformly for every complex $s$, including after differentiation in $s$, and defines an entire function.

Splitting at one and changing $y$ to $1/y$ in the lower integral givesApplying the same formula to $g$ interchanges $F,G$, because $J^2=1$. It proves the phase-normalized Fricke functional equationThe entire function here is the completion, despite the apparent poles of the gamma factor in its initial product formula.

Use the product printed in the original PDF, with factors $(1-q^n{)}^2$ and $(1-q^{11n}{)}^2$. The Fricke involution normalizes ${\Gamma }_0(11)$ and preserves its modular cusp space. Therefore $g=-11^{-1}{\tau }^{-2}f(-1/(11\tau ))$ lies in the given one-dimensional space, so $g=cf$ for a constant $c$.

At the Fricke fixed point ${\tau }_{\ast }=i/\sqrt{11}$, the prefactor $-11^{-1}{\tau }_{\ast }^{-2}$ is one. Thus $g({\tau }_{\ast })=f({\tau }_{\ast })$. The product has $0<q_{\ast }<1$, every factor is positive, and its limit is nonzero since ${\sum }_n{q}_{\ast }^n+{\sum }_n{q}_{\ast }^{11n}<\infty$. Hence $f({\tau }_{\ast })>0$, forcing $c=1$. This Fricke sign from a nonvanishing fixed-point value provesPart (b) now gives $\Lambda (f,s)=\Lambda (f,2-s)$. Its Taylor series at one contains only even powers, so its order of vanishing is even. The function is not identically zero, since its first Fourier coefficient is one. In this example the product is positive on the entire positive imaginary axis, and its Mellin integral at $s=1$ is positive. Thus the stronger conclusion isThe TeX aid duplicates and corrupts the product in this part; neither corrupted expression is used.