Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 137 (Modular Forms and L-functions)

Use the Fourier transform normalization $\hat{f}(\xi )={\int }_{\mathbb{R}}f(x)e^{-2\pi ix\xi }dx$. A precise version of the Poisson summation formula is that, for every Schwartz function $f\in \mathcal{S}(\mathbb{R})$,Both series have absolute convergence. Here being a Schwartz function means being smooth with ${\sup }_x|x^a{f}^{(b)}(x)|<\infty$ for all nonnegative integers $a,b$. This hypothesis makes the sums and the termwise operations below legitimate; the formula is not being asserted for arbitrary integrable functions.

Form the periodization of a Schwartz function $P_f(x)={\sum }_{n}f(x+n)$. Every differentiated series has uniform convergence on $[0,1]$, by rapid decay, so $P_f$ is a smooth periodic function of period one. Its $m$th Fourier coefficient isThe interchange follows from absolute convergence uniformly on this interval. Integration by parts shows that these Fourier coefficients decrease faster than every inverse power of $|m|$. Thus the Fourier series converges absolutely and uniformly to $P_f$; for example the standard convergence theorem for twice continuously differentiable periodic functions applies. Evaluating at zero proves the Poisson summation formula.

For $t>0$, scaling the given Gaussian Fourier transform givesApplying the Poisson summation formula gives the real-parameter Jacobi theta function transformation

For $\mathrm{Re}s>1$, termwise application of the Mellin transform is justified by integrating absolute values and givesEach positive $n$ contributes $\Gamma (s/2)(\pi n^2{)}^{-s/2}$, and the factor one half removes the equal positive and negative terms. This is the Mellin representation of the completed Riemann zeta function.

Split the integral at one. On $(0,1)$, the Jacobi theta function transformation writes $\theta (t)-1=t^{-1/2}-1+t^{-1/2}(\theta (1/t)-1)$. Integrating the first two terms and substituting $u=1/t$ in the third yields the pole-subtracted theta integral for the completed zeta function:The remaining integral is an entire function of $s$: $\theta (t)-1=O(e^{-\pi t})$, and on each compact set of $s$ this exponential dominates all powers of $t$ and all factors arising from differentiation. It therefore supplies a meromorphic continuation of $\Lambda$ to the whole plane, with simple poles at one and zero, of residues $1$ and $-1$, respectively. Its expression is unchanged by $s\mapsto 1-s$.

The reciprocal Gamma function is entire, with simple zeros at its nonpositive integer arguments. Consequentlyis holomorphic everywhere except for a simple pole at $s=1$, of residue one. The apparent pole at $s=0$ cancels; in fact $\Gamma (s/2{)}^{-1}\sim s/2$ gives $\zeta (0)=-1/2$. This proves the analytic continuation of the Riemann zeta function. The completed functional equation of the Riemann zeta function isThese are equalities of meromorphic functions; at apparent singularities they are interpreted by continuation. Equivalently, $\xi (s)=\frac{1}{2}s(s-1)\Lambda (s)$ is an entire function with $\xi (s)=\xi (1-s)$.

Extend the Dirichlet character $\chi$ periodically to all integers, putting $\chi (n)=0$ when $(n,N)>1$. Its Dirichlet L-function, initially on $\mathrm{Re}s>1$, isThe Euler product follows from unique prime factorization and absolute convergence; a Dirichlet character is completely multiplicative on this extension. Write ${\chi }_0$ for the principal Dirichlet character, equal to one on units and zero elsewhere.

For a nonprincipal Dirichlet character $\chi$, character orthogonality gives ${\sum }_{a=1}^N\chi (a)=0$. Explicitly, multiplication by a unit $u$ with $\chi (u)\ne 1$ permutes the unit residues and multiplies this sum by $\chi (u)$, forcing it to vanish. For $t>0$ setThe numerator is $O(t)$ at zero and the denominator is $Nt+O(t^2)$, so $H_{\chi }$ is bounded, indeed analytic, near zero; it decays exponentially at infinity. Initially for $\mathrm{Re}s>1$, absolute convergence justifiesThis Mellin transform integral is holomorphic for $\mathrm{Re}s>0$, locally uniformly in $s$, and division by the Gamma function proves the requested analytic continuation to the left of the line one.

In fact, the same argument proves Mellin continuation of a nonprincipal Dirichlet L-function to the entire plane. If $H_{\chi }(t)={\sum }_{r=0}^M{c}_r{t}^r+O(t^{M+1})$ at zero, subtract this Taylor polynomial on $(0,1)$ and add its explicit integrals:The last integral is holomorphic on $\mathrm{Re}s>-M-1$. Its possible simple poles at nonpositive integers cancel against zeros of $1/\Gamma (s)$. Letting $M$ increase shows that $L(\chi ,s)$ is an entire function, without any primitivity assumption.

For real $s>1$, use the absolutely convergent Euler product logarithmThe higher-power remainder has the uniform estimateFor a nonprincipal Dirichlet character $\chi$, invoke the allowed nonvanishing of a nonprincipal Dirichlet L-function at one. Its holomorphy and nonvanishing give a holomorphic logarithm on a small disk about one. On the connected real interval $1<s<1+\eta$, $B_{\chi }(s)$ differs from this logarithm by a fixed element of $2\pi i\mathbb{Z}$: the difference is continuous with exponential one. Thus the prime-character sum near one is bounded. This branch argument is needed for complex-valued Dirichlet characters.

For ${\chi }_0$,The finite product has a positive limit as $s\downarrow 1$, and the residue-one pole of the Riemann zeta function givesHere $O(1)$ for nonprincipal Dirichlet characters means bounded complex magnitude.

Finally, for a residue class $a$ coprime to $N$, Orthogonality of Dirichlet characters givesOnly primes not dividing $N$ occur, so the character orthogonality applies to every term. This sum diverges as $s\downarrow 1$. A finite collection of primes would give a bounded sum, a contradiction. Every reduced residue class contains infinitely many primes. This is the Dirichlet theorem on primes in arithmetic progressions; the coprimality hypothesis is essential.

Use the valence formula for the modular group: for a nonzero modular form of weight $k$ on $S{L}_2(\mathbb{Z})$,The sum contains one representative of each other modular group orbit of zeros. The orders are nonnegative because $f$ is holomorphic on the complex upper half-plane and at infinity. The half and third weights account for the elliptic stabilizers of the modular group.

In weight two, the transformation under $S=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ at its fixed point gives $f(i)=i^{2}f(i)=-f(i)$. Thus any nonzero $f$ would have $v_i(f)\ge 1$. The valence formula for the modular group would give a left side at least $1/2$ and a right side $1/6$, which is impossible. Therefore the unheaded request is answered by vanishing of weight-two level-one modular forms:

The displayed definition uses iterated Eisenstein summation in weight two: the sum in $n$ is evaluated before the sum in $m$, with the single $(0,0)$ term omitted. This order is essential. The two-dimensional lattice series does not have absolute convergence, so arbitrary rearrangement would not be justified.

For noninteger $w$, the cosecant partial-fraction identity isFor completeness, apply the residue theorem to $\pi \cot (\pi \zeta )/(\zeta -w{)}^2$ on squares with large half-integer sides. The cotangent is bounded on the contours and the integral is $O(R^{-1})$. Its residues at the integers are $(n-w{)}^{-2}$ and its residue at $w$ is $-{\pi }^2{\csc }^2(\pi w)$, proving the formula. If $\mathrm{Im}w>0$, the geometric-series expression $\cot (\pi w)=-i(1+2{\sum }_{r\ge 1}{e}^{2\pi irw})$, differentiated termwise, gives the cotangent partial-fraction Fourier kernel

Put $q=e^{2\pi iz}$ with $z$ in the complex upper half-plane. For positive $m$, this gives $-4{\pi }^2{\sum }_{r\ge 1}r{q}^{mr}$. Negative $m$ gives the same value, by replacing $n$ with $-n$ in its inner sum. The $m=0$ row is ${\sum }_{n\ne 0}{n}^{-2}={\pi }^2/3$, by the Basel problem. The resulting series in $m,r$ does have absolute convergence, locally uniformly in $z$, so collecting the coefficient at $q^{\ell }$ is legitimate:The coefficient is the sum-of-divisors function, since $r$ runs over the positive divisors of $\ell$. ThusThere is no conflict with vanishing of weight-two level-one modular forms: the Eisenstein series of weight two has an anomalous transformation term, so it is not a weight-two modular form.

Write $D=(2\pi i{)}^{-1}d/dz=qd/dq$ and use the Serre derivative $D_{k}f=Df-(k/12)E_{2}f$. Translation invariance of $f$ and the Eisenstein series of weight two gives $D_{k}f(z+1)=D_{k}f(z)$.

Differentiate the weight-$k$ transformation $f(-1/z)=z^{k}f(z)$. The chain rule yieldsThe assumed transformation of the Eisenstein series of weight two givesSubtracting cancels the extra term. Hence $D_{k}f(-1/z)=z^{k+2}{D}_{k}f(z)$. Since $S$ and $T=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$ generate the modular group, these two transformations prove the weight-$k+2$ law for all its elements. Holomorphy on the complex upper half-plane follows from the formula.

For the Fourier expansion of a modular form $f={\sum }_{n\ge 0}{a}_n{q}^n$, differentiation gives $Df={\sum }_{n\ge 1}n{a}_n{q}^n$. The product of the convergent series for $E_2$ and $f$ likewise has no negative powers. Thus $D_{k}f$ is holomorphic at a cusp at infinity, and therefore at every cusp of the modular group. Its constant coefficient is $-k{a}_0/12$. Since $k>0$, it vanishes exactly when $a_0=0$. Consequently

Apply the Serre derivative in weight twelve to the modular discriminant. The preceding result makes $D_{12}\Delta =D\Delta -E_2\Delta$ a weight-fourteen cusp form. The vanishing of weight-fourteen level-one cusp forms follows directly from the valence formula for the modular group: a nonzero such form has order at infinity at least one, and $f(i)=i^{14}f(i)=-f(i)$ forces order at $i$ at least one. Thus its weighted number of zeros would be at least $1+1/2>14/12$, a contradiction. Therefore $D\Delta =E_2\Delta$.

Compare the Fourier coefficients at positive indices using $\Delta ={\sum }_{n\ge 1}\tau (n)q^n$ and $E_2=1-24{\sum }_{r\ge 1}{\sigma }_1(r)q^r$. This gives $n\tau (n)=\tau (n)-24{\sum }_{r=1}^{n-1}{\sigma }_1(r)\tau (n-r)$. For the Ramanujan tau function, the concise recurrence isAt $n=1$ the sum is empty and both sides are zero; normalization supplies $\tau (1)=1$. For instance the next coefficient is $\tau (2)=-24$.

First establish rational conjugation of finite-index modular subgroups without assuming that $\Gamma$ is a congruence subgroup. Multiply $\gamma$ by a positive integer to obtain an integral matrix $A$, and let $D=\det A>0$. Conjugation is unchanged by this scalar. If $h=I+DB\in \Gamma (D)$, thenThus the principal congruence subgroup $\Gamma (D)$ is contained in $H=S{L}_2(\mathbb{Z})\cap {\gamma }^{-1}S{L}_2(\mathbb{Z})\gamma$. It has finite index because reduction modulo $D$ has finite image. Inside $H$, pullback under conjugation of $\Gamma$ has relative index at most $[S{L}_2(\mathbb{Z}):\Gamma ]$. ConsequentlyThis argument does not assert that an arbitrary finite-index subgroup contains a principal congruence subgroup.

Use the determinant-normalized slash operatorThe positive real power of the determinant is used; on $S{L}_2(\mathbb{Z})$ this reduces to the usual slash operator for modular forms. The automorphy factor identity gives the right-action rule $(f{|}_k\gamma ){|}_k\eta =f{|}_k(\gamma \eta )$.

A modular form on a finite-index subgroup of integer weight $k$ is a holomorphic function on the complex upper half-plane, invariant under this weight-$k$ action of $\Gamma$, and holomorphic at a cusp at each of its cusps. A cusp of a modular group is a $\Gamma$ orbit in ${\mathbb{P}}^1(\mathbb{Q})$. If $\sigma \in S{L}_2(\mathbb{Z})$ carries infinity to its representative, choose a positive integer $h$ with $\sigma T^h{\sigma }^{-1}\in \Gamma$. Such $h$ exists by finite index. Then $f{|}_k\sigma$ is periodic and has a convergent expansion in $q_h=e^{2\pi iz/h}$ near zero; holomorphy means no negative exponents, and being a cusp form means zero constant term. Using an actual translation period avoids possible signs if a smaller width of a cusp is defined only modulo the center, particularly in odd weights.

For cusp holomorphy under rational slash operators, choose $\sigma \in S{L}_2(\mathbb{Z})$ with $\sigma \infty =\gamma \infty$, possible by completing a primitive integer pair to a determinant-one matrix. Then ${\sigma }^{-1}\gamma =\left(\begin{array}{cc}a& b\\ 0& d\end{array}\right)$, with $a/d>0$. Up to a nonzero constant factor,The imaginary part of the argument tends to infinity with that of $z$, so this remains bounded by the cusp expansion of $f{|}_k\sigma$. It tends to zero if $f$ is a cusp form. Moreover $f{|}_k\gamma$ is invariant under ${\Gamma }^{\prime }$: for $\eta \in {\Gamma }^{\prime }$, $\gamma \eta {\gamma }^{-1}\in \Gamma$ and the right-action rule applies. Finite index gives a translation period for ${\Gamma }^{\prime }$, so boundedness is a removable singularity at zero in that periodic parameter. This proves holomorphy at infinity. For every other cusp, apply the same argument to the rational matrix $\gamma \rho$, with $\rho \in S{L}_2(\mathbb{Z})$. Thus all cusp conditions hold, and

For the character twist by rational translations of a cusp form, put $B_j=\left(\begin{array}{cc}1& j/N\\ 0& 1\end{array}\right)$ and $K={\Gamma }_1(N)\cap {\Gamma }_0(N^2)$. For $\eta =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in K$, direct conjugation givesIndeed $N^2\mid c$ and $a\equiv d\equiv 1(\mathrm{mod}N)$. Every $f{|}_k{B}_j=f(z+j/N)$ is therefore $K$-invariant and vanishes at all its cusps by the preceding rational-translate argument. Their finite weighted sum is a cusp form, for every Dirichlet character:

There is, however, a missing primitivity hypothesis in the printed final expansion claim. The exact Fourier expansion of a modular form is alwaysValues of a Dirichlet character on units have modulus one, so $\chi (j{)}^{-1}=\overline{\chi (j)}$. For unit $n$, substitution gives $S_{\overline{\chi }}(n)=\chi (n)g(\overline{\chi })$, where $g$ is the Gauss sum of a Dirichlet character. For nonunit $n$, this vanishing formula requires a primitive Dirichlet character.

Here is its proof in that case. Choose a prime $p\mid (n,N)$. Primitivity supplies a unit $u\equiv 1(\mathrm{mod}N/p)$ with $\chi (u)\ne 1$: otherwise the character would factor through the surjective reduction to units modulo $N/p$. Surjectivity follows by lifting a unit and, if needed, adjusting the lift to avoid the additional prime $p$, using the Chinese remainder theorem. Multiplication by $u$ fixes $e^{2\pi inj/N}$ because $p\mid n$, but multiplies the character factor by a nontrivial constant. Hence $S_{\overline{\chi }}(n)=0$. The finite Fourier transform of a primitive Dirichlet character now gives the corrected formulaThe constant is nonzero: finite exponential orthogonality gives ${\sum }_{n\mathrm{mod}N}|S_{\overline{\chi }}(n){|}^2=N\phi (N)$, whereas the proved formula makes this $\phi (N)|g(\overline{\chi }){|}^2$. Thus $|g(\overline{\chi })|=\sqrt{N}$.

For a concrete counterexample to the printed unrestricted claim, take $N=2$, the principal Dirichlet character, and $f=\Delta$. The translation sum is $\Delta (z+1/2)={\sum }_n(-1{)}^n\tau (n)q^n$, whose $q^2$ coefficient is $-24$. Any constant multiple of the proposed odd-index-only series has $q^2$ coefficient zero. Thus the general modularity conclusion is proved, while the claimed simplification is false without the stated extra hypothesis.

For a weight-$k$ cusp form, the invariant norm of a modular form $y^{k/2}|f(z)|$, with $y=\mathrm{Im}z$, is unchanged by the modular group: $\mathrm{Im}(\gamma z)=y/|cz+d{|}^2$ and $|f(\gamma z)|=|cz+d{|}^k|f(z)|$ cancel exactly.

Every orbit meets the standard fundamental domain of the modular group $\mathcal{F}=\{z:|\mathrm{Re}z|\le 1/2,|z|\ge 1\}$, where $y\ge \sqrt{3}/2$. The Fourier expansion of a modular form at the cusp has $f(q)=qh(q)$, with $h$ holomorphic near zero, since $f$ is a cusp form. Thus $|f(x+iy)|=O(e^{-2\pi y})$ uniformly in $x$ as $y\to \infty$, and $y^{k/2}|f|$ tends to zero there. On the remaining compact part of $\mathcal{F}$ it is bounded by continuity. Invariance therefore gives

For every $y>0$, Fourier coefficient extraction givesFor $n\ge 1$, minimize the last factor by taking $y=k/(4\pi n)$. This proves the Fourier coefficient bound for a cusp formThe assertion concerns positive indices; the constant coefficient vanishes because $f$ is a cusp form.

Write $\Gamma =S{L}_2(\mathbb{Z})$ and $G=G{L}_2(\mathbb{Q}{)}^{+}$. The double-coset Hecke algebra consists of $\Gamma$-bi-invariant complex functions on $G$ supported on finitely many double cosets, with convolutionEach double coset has finitely many orbits under left multiplication by $\Gamma$, by rational conjugation of finite-index modular subgroups. Thus the sum is finite and independent of representatives. The characteristic functions of the double cosets form its basis, and its right action on invariant modular forms is $f\ast h={\sum }_{\Gamma x}h(x)f{|}_{k}x$, using the determinant-normalized slash operator.

For a positive integer $n$, let ${\mathcal{M}}_n$ be all integral two-by-two matrices of positive determinant $n$. It is $\Gamma$-bi-invariant. Define $T(n)$ to be its indicator function, equivalently the sum of the distinct double cosets it contains, each with coefficient one. This is the convention consistent with the requested formula. For composite $n$, it need not be the single double coset of $\mathrm{diag}(1,n)$: for instance $2I\in {\mathcal{M}}_4$ cannot lie in that double coset, since multiplying by unimodular integral matrices preserves the greatest common divisor of the entries.

Prove all the needed subgroup facts directly. For a finite-index subgroup $L\subseteq {\mathbb{Z}}^2$, take $a$ to be the least positive first coordinate appearing in $L$ and $d$ the least positive second coordinate on its intersection with the second axis. Euclidean division then shows that the first-coordinate projection is $a\mathbb{Z}$ and $L\cap (\{0\}\times \mathbb{Z})=\{0\}\times d\mathbb{Z}$. Positivity follows, for example, because the finite quotient group kills a nonzero multiple of each coordinate vector. Choose $(a,b)\in L$ and reduce $b$ modulo $d$ to $0\le b<d$. Every vector of $L$ has first coordinate a multiple of $a$, and subtracting that multiple of $(a,b)$ leaves a multiple of $(0,d)$. Therefore these two vectors form a basis of $L$. The parameters $a,d,b$ are unique. Reducing the first coordinate modulo $a$ and then the second modulo $d$ gives precisely $ad$ quotient representatives, so $[{\mathbb{Z}}^2:L]=ad$.

Apply this row Hermite normal form in rank two to the row lattice of an integral matrix $A$ with $\det A=n$. Its row lattice contains $n{\mathbb{Z}}^2$, since $\mathrm{adj}(A)A=nI$, so it has finite index. Its two rows and the displayed two rows are bases of the same row lattice. The two inverse change-of-basis matrices have integer entries, so their determinants are integers whose product is one. Thus the change-of-basis matrix has determinant $\pm 1$; since both orientations are positive, its determinant is one. Thus each orbit under left multiplication by $\Gamma$ has exactly one of the determinant-n matrix representatives for Hecke operatorsThere are ${\sum }_{d\mid n}d$ such representatives, proving finiteness as well as the formula. The subgroup argument is a proof of the relevant Hermite normal form, not an invocation of an unproved lattice classification.

Consequently the normalized Hecke operator isIt preserves $M_k(\Gamma )$: right multiplication by $\Gamma$ permutes the left-multiplication orbits in ${\mathcal{M}}_n$, giving invariance, and cusp holomorphy under rational slash operators gives the holomorphy of each term at every cusp.

For a triangular representative the determinant-normalized slash operator is $f{|}_{k}A=n^{k/2}{d}^{-k}f((az+b)/d)$. Summing the Fourier expansion of a modular form over $b$ kills every index not divisible by $d$, by finite exponential orthogonality. ThusThe Fourier coefficients of a composite-index Hecke operator are thereforeIn particular $a_1(T_{n}f)=a_n(f)$. If $T_{n}f=\lambda f$, comparison of the $q$ coefficients gives

Finally suppose $a_0(f)\ne 0$ and $f$ is a simultaneous eigenfunction. Constant coefficients force ${\lambda }_n={\sigma }_{k-1}(n)$, so $a_n(f)=a_1(f){\sigma }_{k-1}(n)$ for all $n\ge 1$. Weight two cannot occur, by vanishing of weight-two level-one modular forms. For even $k\ge 4$, use the normalized Eisenstein series with its Fourier expansion of a normalized Eisenstein serieswhere $B_k$ is the Bernoulli number. Then $h=f-a_0(f)E_k$ is a cusp form with coefficients $(a_1(f)-a_0(f)c_k){\sigma }_{k-1}(n)$. The Fourier coefficient bound for a cusp form bounds these by $O(n^{k/2})$, but at arbitrarily large primes $p$ their magnitude is $|a_1-a_0{c}_k|(1+p^{k-1})$. Since $k-1>k/2$, the coefficient factor must vanish. All coefficients of $h$ then vanish, including its constant coefficient, so $h=0$ by its cusp expansion and the identity theorem. This proves the noncuspidal level-one Hecke eigenform characterization: