Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 75 (Modular Forms)

The closed standard fundamental domain of the modular group isIts vertical sides are paired by $T:z\mapsto z+1$, and its circular sides by $S:z\mapsto -1/z$. It is a fundamental region with boundary identifications. To obtain exactly one representative of each orbit, use $-1/2<x\le 1/2$ and, on $|z|=1$, retain the half with $x\ge 0$. We will include both vertices when discussing the closure.

To prove existence of a representative, fix $z\in \mathbb{H}$. Among primitive integer pairs $(c,d)$ choose one minimizing $|cz+d|$. The minimum exists: below any fixed bound, $|c|y\le |cz+d|$ bounds $c$, and then the real part bounds $d$, so only finitely many pairs occur. Extend this primitive pair to the bottom row of a determinant-one integer matrix $\gamma$. Sincethis makes the imaginary part maximal over the orbit. Translate by an integer to put the real part in $[-1/2,1/2]$. If the resulting point had modulus below one, inversion by $S$ would increase its imaginary part, a contradiction. Thus every orbit meets $\mathcal{F}$.

For uniqueness in the interior, note that $y\ge \sqrt{3}/2$ throughout $\mathcal{F}$ and that $|cz+d|\ge 1$ for every primitive pair. If $|c|\ge 2$, then $|cz+d|\ge |c|y\ge \sqrt{3}>1$. For $|c|=1$, the minimum over integer $d$ is attained at $d=0$ or, at an endpoint, at a neighboring integer: $|z|\ge 1$ and $|z\pm 1{|}^2=|z{|}^2\pm 2x+1\ge 1$. All larger $|d|$ give a strict inequality. If $c=0$, primitivity gives $d=\pm 1$.

If two points in $\mathcal{F}$ are equivalent, applying this inequality to the transformation and its inverse shows their imaginary parts are equal. For two interior points equality forces $c=0,d=\pm 1$, hence an integer translation. Their real parts lie in an interval of length one, so the translation is zero. The only matrices acting trivially are $\pm I$. On the boundary, the equality cases give precisely the stated vertical and circular identifications. This establishes the fundamental-domain claim, including its boundary convention.

An elliptic point of a modular curve means a fixed point of a nonidentity element of the effective modular group $PS{L}_2(\mathbb{Z})$; the central matrix $-I$ is not counted, since it acts trivially everywhere. The equality cases above show that the only such points in the closed region areTheir stabilizer subgroups in $S{L}_2(\mathbb{Z})$ arewhereIndeed $S^2=-I$, $(ST{)}^3=(TS{)}^3=-I$, and their fractional-linear fixed-point equations give the listed points. To see that these are the full stabilizer subgroups, equality in $|cz+d|\ge 1$ leaves only bottom rows $(0,\pm 1),(\pm 1,0)$, with the extra neighboring rows at the vertices. The fixed-point equation then fixes the top row. This yields exactly four matrices at $i$, six at each vertex, and just $\{\pm I\}$ elsewhere. These are the elliptic stabilizers of the modular group. After passing to $PS{L}_2(\mathbb{Z})$ their effective orders are two and three. The two vertices represent the same elliptic orbit.

Normalize the weight-four and weight-six Eisenstein series to have constant term one. Their Fourier expansions beginTheir absolutely convergent lattice sums transform by modular weights four and six under reindexing by the modular group, and their Fourier expansions prove holomorphy at its modular cusp. Thus they are elements of $M_4,M_6$.

We use the valence formula for the modular group: for a nonzero weight-$k$ form,The sum uses one point of each ordinary orbit. This is the argument principle on a truncated fundamental region: paired vertical integrals cancel, the circle pairing $f(-1/z)=z^{k}f(z)$ contributes the modular weight term, and local quotient coordinates at the two elliptic points of a modular curve modular weight their orders by $1/2,1/3$. All orders are nonnegative for a holomorphic modular form.

Define the modular discriminant byIt is a weight-twelve cusp form. Its order at infinity is one, exhausting the valence total $12/12=1$. Hence $\Delta$ has no zeros in $\mathbb{H}$. Therefore every weight-$k$ cusp form is divisible by $\Delta$, with quotient holomorphic on $\mathbb{H}$ and at the modular cusp:Negative modular weights vanish by the valence formula, and odd modular weights vanish because $-I$ acts by $(-1{)}^k$. Weight-zero forms are constant: they descend holomorphically to the compactified modular curve, and the maximum modulus principle applies. There are no weight-two forms: $f(i)=i^{2}f(i)=-f(i)$ forces a zero at $i$, whose valence contribution $1/2$ exceeds $2/12$.

Every nonnegative even modular weight other than two is $4a+6b$ with $a,b\ge 0$. If $k\equiv 0(\mathrm{mod}4)$, take $b=0$; if $k\equiv 2(\mathrm{mod}4)$ and $k\ge 6$, take $b=1$. For $f\in M_k$, choose such a monomial $h=E_4^a{E}_6^b$, whose constant term is one. Subtracting the constant term $c$ of $f$ gives $f-ch\in S_k$, soInduction on modular weight, together with $1728\Delta =E_4^3-E_6^2$, proves that every modular form is a polynomial in $E_4,E_6$.

For algebraic independence, first fix one modular weight $w$. All solutions of $4a+6b=w$ have the same residue of $a$ modulo three and the same parity of $b$. Consequently their monomials can be writtenfor suitable $a_0,b_0,L$. On an open set where the common factor and $E_6$ are nonzero, a linear relation would give a polynomial relation inThis is nonconstant. By the open mapping theorem, a polynomial vanishing on all its values must be the zero polynomial. Thus same-weight monomials are linearly independent.

Different modular weights cannot cancel either. Suppose ${\sum }_w{P}_w(E_4,E_6)=0$ with each $P_w$ weighted homogeneous. Transform by ${\gamma }_n=\left(\begin{array}{cc}1& 0\\ n& 1\end{array}\right)$. At any fixed $z\in \mathbb{H}$ this givesThe numbers $nz+1$ are infinitely many distinct values, so the polynomial in that number vanishes identically, and every $P_w(z)=0$. The same-weight independence then sets every coefficient to zero. This is the graded weights separate analytic polynomial relations argument. Hencewith algebraically independent generators. Multiplying the two Eisenstein series by their nonzero lattice-normalization constants does not alter either generation or independence.

Use the determinant-normalized slash operator and let ${\mathcal{M}}_n$ be all integral matrices of positive determinant $n$. Their left $S{L}_2(\mathbb{Z})$ cosets have the unique determinant-n matrix representatives for Hecke operatorsInteger row operations reduce the first column to $(a,0{)}^T$, and a row shear then reduces $b$ modulo $d$. DefineChanging representatives does not affect the sum, while right multiplication by the modular group permutes the cosets. Hence $T_{n}f$ has the same modular transformation law. Rational positive-determinant slash transforms are holomorphic on $\mathbb{H}$ and preserve modular cusp vanishing: choose an integral matrix carrying infinity to their rational image modular cusp, after which the remaining matrix is upper triangular. Its positive dilation sends the original decaying modular cusp expansion to another decaying expansion. Thus these are well-defined Hecke operators on $S_k$.

For the Fourier expansion, insert $f(z)={\sum }_{m\ge 1}a(m)e^{2\pi imz}$. The sum over $b$ vanishes unless $d\mid m$, in which case it is $d$. Writing $m=d\ell$, the resulting exponential is $q^{a\ell }$, and its factor is $n^{k-1}{d}^{1-k}=a^{k-1}$. ThereforeIn particular, the first coefficient of $T_{n}f$ is $a(n;f)$.

Assume $f\ne 0$ is a simultaneous Hecke eigenform. The first-coefficient identity gives $a(n)=\lambda (n)a(1)$. If $a(1)=0$, all coefficients vanish, contradicting $f\ne 0$. After dividing by $a(1)$, we may therefore suppose $a(1)=1$, and then $\lambda (n)=a(n)$.

We derive the relations needed for the Euler product. For coprime $m,n$, applying the coefficient formula twice splits the divisors into their disjoint prime supports and yields $T_m{T}_n=T_{mn}$. For one prime define formal Fourier operators $U_p(\sum a(r)q^r)=\sum a(pr)q^r$ and $V_{p}f(z)=f(pz)$. The coefficient formula givesMultiplying this finite sum by $T_p=U_p+p^{k-1}{V}_p$ shows thatThese auxiliary operators need not separately preserve level one; the identities concern their action on Fourier series. Applied to the eigenform, they giveThe prime-power generating function consequently satisfiesMultiplicativity and unique prime factorization now prove the Euler product of a Hecke eigenform:For the original unnormalized form, the right side is multiplied by $a(1)$ and $a(p)$ in the factors is replaced by $\lambda (p)=a(p)/a(1)$. The zero form has $L=0$ and is excluded from the nonzero-eigenform product assertion.

These manipulations initially hold in an absolutely convergent right half-plane. For example, $y^{k/2}|f(z)|$ is bounded by modular invariance and modular cusp decay. The coefficient integral at $y=1/n$ gives $|a(n)|\le C{n}^{k/2}$, so $\mathrm{Re}s>k/2+1$ is sufficient. No analytic continuation is needed to justify the initial product.

Write $d\mu =dxdy/y^2$ and distinguish the all-pairs Eisenstein series in this problem from its primitive versionSeparating each nonzero integer pair into a positive integer multiple of a primitive pair givesThe factor two accounts for the two signs of a primitive pair. It must not be omitted in the Rankin–Selberg unfolding.

Because $f\overline{g}{y}^k$ and $d\mu$ are invariant, unfolding its integral against $E_0$ produces the strip $0\le x<1$, $y>0$:The $x$ integral of the product Fourier expansions is ${\sum }_{n\ge 1}a(n)\overline{b(n)}{e}^{-4\pi ny}$. The Gamma function integral then givesThese steps first follow by absolute convergence sufficiently far right. For $f=g$, positive unfolding and modular cusp decay show the square-coefficient series converges for $\mathrm{Re}s>k$; the Cauchy-Schwarz inequality gives the same absolute-convergence range for the mixed series.

Put $w=s-k+1$ and defineThis is the zeta-completed Rankin-Selberg coefficient series. Equivalently, solving this formula for $F(s)$ expresses the requested series directly in terms of the all-pairs Eisenstein integral.

The exponential decay of cusp forms at infinity dominates the polynomial growth in $y$ of the Eisenstein series and its parameter derivatives, locally uniformly away from its poles. Thus its stated continuation can be passed through the integral. The completed series is meromorphic on $\mathbb{C}$, with possible simple poles only at $s=k-1,k$, and its functional equation isIndeed $w\mapsto 1-w$ corresponds to $s\mapsto 2k-1-s$. If $⟨f,g⟩=0$, both pole residues vanish and the completion is entire.

The normalization of the poles can also be made explicit. The constant term of the given Eisenstein series isThe second term has residue one at $w=1$, since the pole of $\zeta (2w-1)$ has residue $1/2$; the functional equation gives residue minus one at $w=0$. Hence the residues of $\mathcal{R}$ at $k$ and $k-1$ are respectively $⟨f,g⟩$ and $-⟨f,g⟩$, the Petersson inner product.

It is important to separate this completed statement from the raw $F$. Inverting the completion gives meromorphic continuation of $F$, but zeros of $\zeta (2s-2k+2)$ in the denominator can give additional possible poles unless canceled by the integral. Nor does $F$ itself satisfy the unweighted symmetry $s\leftrightarrow 2k-1-s$. At the two displayed completion poles, for a nontrivial modular cusp modular weight,The latter is finite because the pole of $\Gamma (w)$ cancels that of the integral at $w=0$, using $\zeta (0)=-1/2$. In particular, taking $f=g\ne 0$ gives a raw-series pole at $k$ and regularity at $k-1$. The direct analogue of the stipulated two-pole Eisenstein properties is therefore the completed function $\mathcal{R}$, not the raw series with all completion factors suppressed.

Finally suppose the two $L$-series have Euler products. With first coefficients normalized to one, their coefficients are multiplicative, so $a(n)\overline{b(n)}$ is multiplicative too. Thereforeinitially in its absolute-convergence half-plane. For unnormalized forms multiply the product by $a(1)\overline{b(1)}$ and use normalized coefficients inside it.

For the usual degree-two Hecke products of Question 3, take roots ${\alpha }_p,{\beta }_p$ with sum $a(p)$ and product $p^{k-1}$, and ${\gamma }_p,{\delta }_p$ with sum $b(p)$ and the same product. The prime-power recurrence gives $a(p^r)=({\alpha }_p^{r+1}-{\beta }_p^{r+1})/({\alpha }_p-{\beta }_p)$ and the analogous expression for $b$. Multiplying and summing four geometric series proves the Euler factor of a coefficientwise product of Hecke eigenforms:Repeated roots are handled by continuity, or directly by the recurrence. Multiplication by $\zeta (2s-2k+2)$ cancels the local numerators and yields the degree-four convolution product. This also explains why the zeta factor belongs in the natural analytic completion.

Let ${\mu }_N=[PS{L}_2(\mathbb{Z}):\overline{\Gamma (N)}]$, where the bar denotes the effective projective image. This is the degree of the modular curve projection, rather than the full $S{L}_2$ index when $-I$ is not in $\Gamma (N)$.

Reduction modulo $N$ is onto $S{L}_2(\mathbb{Z}/N\mathbb{Z})$. One proof uses the Chinese remainder theorem and elementary matrices: over $\mathbb{Z}/p^r\mathbb{Z}$, a unimodular first column has a unit entry, so elementary row operations reduce a determinant-one matrix to the identity. Each elementary matrix lifts integrally. Over ${\mathbb{F}}_p$, the first column has $p^2-1$ choices and the second has $p$ choices with prescribed determinant, giving $p(p^2-1)$. Each subsequent prime-power lift has $p^3$ choices, because the determinant-one linearized condition is one trace equation on four entries. HenceFor $N=1,2$, the subgroup contains $-I$; for $N\ge 3$ it does not. ConsequentlyFor $N\ge 2$ there are no effective elliptic stabilizer subgroups in the subgroup. Indeed, write $\gamma =I+NB\in \Gamma (N)$. The determinant condition gives $\mathrm{tr}\gamma =2-N^2\det B$. A noncentral elliptic element in $S{L}_2(\mathbb{Z})$ has trace $-1,0$ or $1$, none congruent to $2$ modulo $N^2$ for $N\ge 2$. Thus the projective action is torsion-free.

All cusp widths are $N$. At infinity, $T^h$ lies in the effective subgroup exactly when $h$ is divisible by $N$: the alternative $T^h\equiv -I(\mathrm{mod}N)$ can only occur for $N=1,2$, and for $N=2$ it imposes the same condition. Every rational modular cusp is a modular translate of infinity, and the subgroup is normal, so the same width holds there. The sum of modular cusp analytic ramification indices equals ${\mu }_N$, giving ${\mu }_N/N$ modular cusps.

The base $X(1)$ is a sphere. For instance $j=E_4^3/\Delta$ is a meromorphic function on it with a single simple pole at the modular cusp and no other poles, using the nonvanishing of $\Delta$ proved above. It therefore defines a degree-one map to the Riemann sphere, making the base genus zero.

The only branch values of $X(N)\to X(1)$ are the two elliptic orbits and the modular cusp. For $N\ge 2$, the numbers of points above them and their analytic ramification indices are respectively ${\mu }_N/2$ with index two, ${\mu }_N/3$ with index three, and ${\mu }_N/N$ with index $N$. Applying the Riemann-Hurwitz formula directly,This is the elliptic and cusp ramification for principal level calculation. Level one is the identity projection. The complete result isFor example the genera for $N=3,4,5,6,7$ are $0,0,0,1,3$. Keeping the exceptional projective index at level two is essential; applying the factor one-half there would give a nonintegral answer. This agrees with the principal congruence modular curve genus formula obtained from these branch counts.

Write $U_u=\left(\begin{array}{cc}1& u/D\\ 0& 1\end{array}\right)$ and $\tau (\chi )={\sum }_{u\mathrm{mod}D}\chi (u)e^{2\pi iu/D}$. The finite Fourier transform of a primitive Dirichlet character givesFor $(n,D)=1$ this follows by multiplying residues by $n$. For a nonunit $n$, primitivity supplies a unit $v\equiv 1(\mathrm{mod}D/p)$ with nontrivial character value, for a prime $p\mid (n,D)$; multiplication by $v$ leaves the exponential unchanged and forces the sum to vanish. Orthogonality of the finite exponentials gives ${\sum }_{n\mathrm{mod}D}|{\sum }_u\chi (u)e^{2\pi inu/D}{|}^2=D{\sum }_u|\chi (u){|}^2=D\phi (D)$. The primitive transform identity makes the left side $\phi (D)|\tau (\chi ){|}^2$. Thus the primitive Gauss sum of a Dirichlet character is nonzero, with $|\tau (\chi ){|}^2=D$. Consequently the twist is the finite rational-translation sumThis representation will prove both the level statement and the Fricke formula.

Take $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in {\Gamma }_0(N{D}^2)$, and choose $v\equiv d^{2}u(\mathrm{mod}D)$. Since $ad\equiv 1(\mathrm{mod}D)$, this means $av\equiv ud(\mathrm{mod}D)$. Direct multiplication givesEvery entry is integral, its determinant is one, and its lower-left entry is divisible by $N$. Thus ${\gamma }_u\in {\Gamma }_0(N)$. Its lower-right entry is congruent to $d$ modulo $N$, so $f|{\gamma }_u=\psi (d)f$ by its nebentypus character. Using the right-action property of the slash operator,The map $u\mapsto v=d^{2}u$ permutes the residue classes. Reindexing gives $\overline{\chi }(u)=\chi (d{)}^2\overline{\chi }(v)$, soThis proves the required modular weight and character on ${\Gamma }_0(N{D}^2)$.

Cuspidality must also be checked. For any modular cusp matrix $\sigma \in S{L}_2(\mathbb{Z})$ and any rational translation $U_u$, choose $\rho \in S{L}_2(\mathbb{Z})$ with $\rho \infty =U_u\sigma \infty$. Then ${\rho }^{-1}{U}_u\sigma$ is upper triangular with positive dilation. Since $f|\rho$ decays exponentially at infinity, so does $f|U_u\sigma$. The finite sum defining $f_{\chi }$ therefore tends to zero at every modular cusp; its transformation law supplies a modular cusp period, so its holomorphic Fourier expansion has zero constant term there. Hence the primitive twist at coprime level is

Now put $W_M=\left(\begin{array}{cc}0& -1\\ M& 0\end{array}\right)$ and $g=f{|}_k{W}_N$, always using the determinant factor specified in the question. For each unit $u$ choose a unit $v$ with $Nuv\equiv -1(\mathrm{mod}D)$, possible because $(N,D)=1$. The crucial integral matrix factorization isThe right side multiplies out to $\left(\begin{array}{cc}uND& -1\\ N{D}^2& 0\end{array}\right)$, so the identity is exact. Its determinant-one matrix has lower-right entry $D$, and therefore $f|{\gamma }_u=\psi (D)f$. A positive scalar matrix $DI$ acts trivially under the determinant-normalized slash operator, since its determinant contributes $D^k$ and its denominator contributes $D^{-k}$. ThusThe inverse-residue substitution is $u\equiv -(Nv{)}^{-1}(\mathrm{mod}D)$, giving $\overline{\chi }(u)=\chi (-N)\chi (v)$. The sum over $v$ is $\tau (\chi )g_{\overline{\chi }}$. Hence the Fricke transform of a primitive coprime twist isThe second version uses $\tau (\chi )\tau (\overline{\chi })=\chi (-1)D$, which follows from $\overline{\tau (\chi )}=\chi (-1)\tau (\overline{\chi })$ and $|\tau (\chi ){|}^2=D$. Here $g_{\overline{\chi }}=\sum b(n)\overline{\chi }(n)q^n$ if $g=\sum b(n)q^n$; it is not the complex conjugate function $\overline{g}$. The coprimality of $N,D$ is used both in the matrix factorization and in evaluating $\psi (D)$. For $D=1$ the twist is trivial and the formula simply reads $f|W_N=g$.