Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 137 (Modular Forms and L-functions)

The Mellin transform iswhere this integral converges. Rapid decay at infinity makes the integral over $[1,\infty )$ an entire function of $s$, but alone gives no control near zero. With the additional expansion, the integral initially converges absolutely for $\mathrm{Re}s>-{\sigma }_1$.

For $\mathrm{Re}s>1$, expand $(e^{\pi y}-1{)}^{-1}={\sum }_{r\ge 1}{e}^{-\pi ry}$. Absolute convergence justifies termwise integration, using the Gamma integral, and gives the scaled Bose integral

There is a missing hypothesis in the general continuation claim: one needs ${\sigma }_j\to +\infty$. The original PDF, like the TeX, only says that the sequence increases. Under the intended additional hypothesis, split the integral at one and subtract $r$ terms of the asymptotic expansion:Continuity of $g_r$ at zero makes it bounded on $[0,1]$. The last integral is holomorphic on $\mathrm{Re}s>-{\sigma }_{r+1}$: on every compact subset, its integrand and all its $s$ derivatives are dominated by an integrable power of $y$ times a power of $|\log y|$. The expressions for successive $r$ agree on their common initial domain and hence everywhere they overlap by the identity theorem. As ${\sigma }_{r+1}\to \infty$, these half-planes cover $\mathbb{C}$. Taking $r\ge j$ with ${\sigma }_{r+1}>{\sigma }_j$ isolates the term $c_j/(s+{\sigma }_j)$, while all other terms are holomorphic near $-{\sigma }_j$. Thus the corrected claim isThis is meromorphic continuation of a Mellin transform from an asymptotic expansion.

To show why the correction matters, set ${\sigma }_j=1-1/j$, $c_j=2^{-j}$, andwhere $\eta$ is continuous, equals one for $0<y\le 1$ and vanishes for $y\ge 2$. On $[0,1]$, the remainder after division by $y^{{\sigma }_{r+1}}$ is a uniformly convergent series of nonnegative powers of $y$, so it extends continuously to zero with value $c_{r+1}$. Away from zero, define $g_r$ by the required remainder quotient; it is continuous on the rest of $[0,\infty )$ as well. Thus all the printed hypotheses hold. Its Mellin transform equalsThe series continues meromorphically on $\mathrm{Re}s>-1$ and has genuine poles at $-1+1/j$, accumulating at $-1$. A meromorphic function on $\mathbb{C}$ cannot have such an accumulation of poles. This is the accumulating asymptotic exponents obstruct Mellin continuation counterexample.

For the Gamma function, the Gamma function recurrence givesAt $s=-j$ the numerator is $\Gamma (1)=1$ and the product of the nonzero denominator factors is $(-1{)}^{j}j!$. Therefore the residues of the Gamma function are

The generating function of the Bernoulli numbers gives the convergent expansion near zeroThe subtraction proof above applies to this expansion, including its zero coefficients, which give no pole. At $s=1-n$, for $n\ge 1$, the residue of its transform is $B_n{\pi }^{n-1}/n!$, while the residue of ${\pi }^{-s}\Gamma (s)$ is ${\pi }^{n-1}(-1{)}^{n-1}/(n-1)!$. Their quotient defines a holomorphic continuation of the Riemann zeta function at this point, including when the first residue is zero. Dividing gives the Bernoulli formula for zeta values at nonpositive integers:In particular the generating-series convention is $B_1=-1/2$, so the formula includes $\zeta (0)=-1/2$.

Write $x=g^{-1}h$ and $S(x)={\sum }_{\chi \in \hat{G}}\chi (x)$, where $\hat{G}$ is the character group of a finite abelian group. The Fundamental theorem of finitely generated abelian groups writes the finite group $G$ as a product of cyclic groups. A character of a cyclic group factor of order $r$ is determined by an arbitrary $r$th root of unity, so $|\hat{G}|=|G|$. This description also shows that the characters separate points: if $x\ne 1$, a nonzero coordinate of $x$ is detected by a character $\psi$ with $\psi (x)\ne 1$.

Multiplication by $\psi$ permutes $\hat{G}$, soFor $x\ne 1$ this forces $S(x)=0$. For $x=1$ every summand is one. Since $\chi (g{)}^{-1}\chi (h)=\chi (g^{-1}h)$, the character-sum cancellation lemma givesAll character values are roots of unity, so inversion here is also complex conjugation. This is the finite-abelian version of character orthogonality.

Extend each Dirichlet character periodically to all integers by setting $\chi (n)=0$ when $\gcd (n,N)>1$. Its Dirichlet L-function is initially the absolutely convergent Dirichlet seriesThe group $(\mathbb{Z}/N\mathbb{Z}{)}^{\times }$ has $\phi (N)$ elements. Part (i) gives, for every integer $n$, including nonunits,For a nonunit both sides are zero, because the fixed class $a$ is a unit. Interchanging the finite character sum and the absolutely convergent Dirichlet series therefore provesThis is Orthogonality of Dirichlet characters applied to a residue-class indicator. Any meromorphic continuations of the two sides agree by uniqueness of analytic continuation.

The original PDF has the summation condition $\gcd (n,N)=1$; the TeX's $\gcd (n,n)=1$ is a transcription error. There is also an actual missing hypothesis in the PDF's coefficient formula: that simplified formula requires a primitive Dirichlet character. We first derive a formula valid for every character, and then show both the primitive specialization and a counterexample to the unrestricted version.

For $k>2$, the character-twisted Eisenstein series converges absolutely and locally uniformly on the complex upper half-plane. On a compact subset, $|mz+n|$ is bounded below by a positive constant times $(m^2+n^2{)}^{1/2}$, and the corresponding two-dimensional lattice sum converges. Changing $n$ to $n+mN$ provesThe condition $\chi (-1)=(-1{)}^k$ makes the terms for $(m,n)$ and $(-m,-n)$ equal. The terms with $m=0$ contribute $2L(\chi ,k)$, and all other terms are twice the sum over $m\ge 1$.

For $\mathrm{Im}w>0$, the cotangent identity andgive, after $k-1$ differentiations,Differentiation is justified by locally uniform convergence. This is the cotangent partial-fraction Fourier kernel.

Write $n=a+Nt$, with $a$ running through the unit classes modulo $N$, and define the finite Fourier transformHere $g(\chi )$ is the Gauss sum of a Dirichlet character.
Applying the kernel with $w=(mz+a)/N$ yieldsThe double series converges absolutely: $|S_{\chi }(r)|\le \phi (N)$ and the exponential decay controls ${\sum }_{m,r}{r}^{k-1}{e}^{-2\pi mr\mathrm{Im}z/N}$. Grouping the terms with $mr=n$ proves the general Fourier expansion of a character-twisted Eisenstein series:

If $r$ is a unit modulo $N$, substitution $a\mapsto r^{-1}a$ gives $S_{\chi }(r)=\chi (r{)}^{-1}g(\chi )$. Suppose now that $\chi$ is primitive and $r$ is a nonunit. Choose a prime $p\mid \gcd (r,N)$. Reduction of units modulo $N$ onto units modulo $N/p$ is surjective. Primitivity means that $\chi$ is nontrivial on its kernel, so there is a unit $u\equiv 1(\mathrm{mod}N/p)$ with $\chi (u)\ne 1$. Since $ru\equiv r(\mathrm{mod}N)$, substitution by $u$ forces $S_{\chi }(r)=\chi (u{)}^{-1}{S}_{\chi }(r)$ and thus $S_{\chi }(r)=0$. This proves the finite Fourier transform of a primitive Dirichlet character identity, and consequentlyThe inverse-character notation in the question is understood to mean $\overline{\chi }(r)$, extended by zero on nonunits; literal inversion of $\chi (r)=0$ would be undefined.

For a counterexample without primitivity, take $N=2$, the principal character, and $k=4$. Then $S_{\chi }(r)=(-1{)}^r$, so $g(\chi )=-1$ but $S_{\chi }(2)=1$. At $n=2$, the general formula gives $c_2=7A_4/8$, whereas the printed simplified formula, interpreted as zero on nonunits, gives $-A_4/8$. Hence the printed coefficient formula is false for arbitrary imprimitive characters; the general boxed formula above supplies the correction.

Let $S=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ and $T=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$. The modular group acts on the complex upper half-plane by fractional linear transformations, andFor fixed $z=x+iy$, the values $|cz+d|$ with primitive integer pairs $(c,d)$ have a positive minimum: the conditions $|cz+d|\le R$ bound both $|c|\le R/y$ and $|d|\le R+|cx|$, leaving finitely many pairs. By Bezout identity, a minimizing pair is the bottom row of a matrix in $S{L}_2(\mathbb{Z})$. Applying that matrix gives a point of maximal imaginary part in its orbit.

Translate it by a power of $T$ into $-1/2<x\le 1/2$. It must have modulus at least one, since otherwise $S$ would increase its imaginary part. If it is on the unit circle with negative real part, apply $S$, which there changes $x$ to $-x$. This gives an element of the specified half-open standard fundamental domain of the modular group. It proves existence by reduction to the standard modular region.

For uniqueness, if $z$ lies in the closed standard region then $y\ge \sqrt{3}/2$ andfor every primitive pair. If $c=0$, this follows from $d=\pm 1$; if $|c|\ge 2$, it follows from $|c|y\ge \sqrt{3}$. If $|c|=1$, then $d=0$ uses $|z|\ge 1$, while $|d|\ge 1$ givesIf both $z$ and $z^{\prime }=\gamma z$ lie in the region, applying this inequality to $\gamma$ and ${\gamma }^{-1}$ forces their imaginary parts to agree, hence $|cz+d|=1$.

The equality cases describe every possible boundary identification. For $c=0$ the map is an integer translation; the half-open vertical strip retains one representative. For $|c|=1,d=0$, equality requires $|z|=1$, and the map is an inversion followed by an integer translation. Inversion exchanges the two halves of the circle, so retaining only the right half removes the duplication; the only translation that returns the exchanged endpoint is the one identifying the two corners. The remaining equality cases have $|c|=|d|=1$, $|z|=1$ and $x=\pm 1/2$, and give only those same corner identifications. Our region retains the right corner alone. Therefore $z^{\prime }=z$, proving

The equality analysis also gives the elliptic stabilizers of the modular group. The central matrices $\pm I$ fix every point. The point $i$ is fixed by $S$, with $S^2=-I$, and the retained corner ${\rho }_{+}=e^{i\pi /3}$ is fixed by $R=TS$, with $R^3=-I$. There are no other noncentral stabilizers, and the possible bottom rows in the equality cases give exactlyThe corresponding stabilizer orders in $PS{L}_2(\mathbb{Z})$ are two, three and one; the requested group is $S{L}_2(\mathbb{Z})$, so the center must be included.

Use the convention that a weak modular form is a holomorphic function $f$ on the complex upper half-plane satisfyingwith no condition at infinity. In particular $f(z+1)=f(z)$, so it has a Laurent series $f(z)={\sum }_{n\in \mathbb{Z}}{a}_n{q}^n$ on $0<|q|<1$, where $q=e^{2\pi iz}$. It is a modular form when it is holomorphic at a cusp, here the unique level-one cusp at infinity, equivalently $a_n=0$ for $n<0$. It is a cusp form when additionally $a_0=0$. If one uses the narrower weakly holomorphic convention, one also requires finitely many negative terms; the derivative argument below holds in that subspace as well. The matrix $-I$ forces odd-weight forms to be zero.

Put $(a{)}_r=a(a+1)\cdots (a+r-1)$, with $(a{)}_0=1$, andWe prove the derivative transformation of a weak modular form in the uniform formFor $\ell =0$ this is the weight-$k$ transformation under $S$. Differentiating the formula and multiplying by $z^2$, since $d(-1/z)/dz=z^{-2}$, gives the recurrencewhere out-of-range coefficients are zero. The displayed explicit coefficients satisfy it. Indeed, for interior $j$, extract the common polynomial $(k+j{)}_{\ell -j}$; the remaining identity isThis follows from Pascal's identity and $j\left(\genfrac{}{}{0ex}{}{\ell }{j}\right)=(\ell -j+1)\left(\genfrac{}{}{0ex}{}{\ell }{j-1}\right)$. The endpoint coefficients satisfy the recurrence directly. This polynomial proof also applies when a factor is zero, avoiding division by that factor. As $C_{\ell ,\ell }=1$, the result isThe letter $l$ in the printed exponent is the same derivative order $\ell$.

For integer $k<0$, take $\ell =1-k$. If $j<\ell$, the product runs from $k+j\le 0$ through $k+\ell -1=0$, and therefore vanishes. Consequently $f^{(1-k)}$ has weight $k+2(1-k)=2-k$ under $S$. It is also invariant under $T$, by differentiating $f(z+1)=f(z)$, and $S,T$ generate the modular group. Its holomorphy is preserved by differentiation. With $D=(2\pi i{)}^{-1}d/dz=qd/dq$, termwise differentiation of the locally convergent Laurent series gives the Bol identity for modular forms:For negative indices the exponent $1-k$ is a positive integer, and the constant term is killed, so every term is unambiguous.

We use left multiplication by $\Gamma =S{L}_2(\mathbb{Z})$. For an integral matrix $A$ of determinant $n>0$, the Bezout identity gives a determinant-one row operation sending its first column to $(a,0{)}^t$, where $a$ is the positive gcd of that column. The resulting matrix is $\left(\begin{array}{cc}a& b\\ 0& d\end{array}\right)$ with $d=n/a>0$. Adding a multiple of the second row to the first makes $0\le b<d$.

These representatives are unique. Left multiplication by a unimodular matrix preserves the gcd of the first column, so two representatives in one orbit have the same $a$ and hence $d$. A matrix taking $(a,0{)}^t$ to itself has the form $\left(\begin{array}{cc}1& t\\ 0& 1\end{array}\right)$, so it changes $b$ by $td$; the prescribed range makes $b$ unique. Thus the determinant-n matrix representatives for Hecke operators are exactly ${\Pi }_n$. In the printed set, $0\le b<d$ already forces $d>0$, and $ad=n>0$ forces $a>0$.

For positive determinant extend the slash operator for modular forms byThe determinant factor makes this a right action: $(f{|}_{k}A){|}_{k}B=f{|}_k(AB)$. Right multiplication by $\delta \in \Gamma$ permutes the left cosets represented by ${\Pi }_n$, since it preserves the set of determinant-$n$ integral matrices. Hence $T_{n}f$ satisfies the modular transformation law. Its summands are holomorphic on the upper half-plane. For an upper-triangular representative,Substitute the Fourier expansion of a modular form. The sum over $b$ is zero unless the original index is divisible by $d$, in which case it equals $d$. Writing that index as $dr$ givesOnly nonnegative powers occur, so $T_{n}f$ is holomorphic at infinity, and all level-one cusps are equivalent to infinity. We have proved $T_{n}f\in M_k(\Gamma )$ and the Fourier coefficients of a composite-index Hecke operator formulaHere $\gcd (n,0)=n$, giving $a_0(T_{n}f)={\sum }_{a\mid n}{a}^{k-1}{a}_0(f)$. In particular cusp forms stay cusp forms. For nonzero level-one cusp forms the weight is an even integer at least twelve, so the powers $a^{k-1}$ are integers; the zero cusp space causes no exception. Thus every $T_n$ preserves integral Fourier coefficients, and closure under addition, multiplication and integer scalars provesThis is the integral Hecke algebra of level-one cusp forms.

We will also need commutativity. The coefficient formula directly givesFor the first identity, the divisors of $u$ and $v$ in the iterated coefficient sum combine uniquely into a divisor of $uv$. For the second, write a coefficient index as $p^t{r}_0$ with $p\nmid r_0$; the formula isApplying $T_p$ splits this into the two sums with shifts $t+1$ and $t-1$; their overlapping terms give exactly the stated recurrence. Since $T_1$ is the identity, induction expresses each $T_{p^r}$ as a polynomial in $T_p$, and coprime multiplicativity expresses every $T_n$ as a product of these polynomials. Distinct prime operators commute by the coprime identity. Hence all Hecke operators commute, and so do all elements of $\mathbb{T}$.

Let $\Lambda =S_k(\Gamma ,\mathbb{Z})$ and $m=\dim S_k(\Gamma )$. In the permitted basis $F_j={\Delta }^j{E}_4^{a_j}{E}_6^{b_j}$, the modular discriminant begins $q+O(q^2)$ and the normalized Eisenstein series begin $1+O(q)$, soThe matrix of $a_1,\dots ,a_m$ on this integral basis is unitriangular and has determinant one. Integer elimination therefore gives an integral basis $f_1,\dots ,f_m$ with $a_i(f_j)={\delta }_{ij}$ for $1\le i,j\le m$. Thus $a_1,\dots ,a_m$ are a $\mathbb{Z}$-basis of the dual module ${\mathrm{Hom}}_{\mathbb{Z}}(\Lambda ,\mathbb{Z})$.

The map $\alpha$ is well-defined by lattice preservation and is $\mathbb{Z}$-linear. Because $\alpha (T_j)=a_j$, it is surjective. For injectivity suppose $\alpha (T)=0$. For every $f\in \Lambda$ and every $n\ge 1$, commutativity givesThe constant term is also zero because $Tf$ is a cusp form. The identity theorem applied to its Fourier expansion of a modular form gives $Tf=0$. The integral basis spans the complex cusp space, so $T=0$ as an endomorphism. This proves the perfect integral Hecke pairing andIf $m=0$, both modules are zero and the asserted basis is empty.

A modular form of integer weight $k$ for a finite-index subgroup $\Gamma \le S{L}_2(\mathbb{Z})$ is a holomorphic function on the complex upper half-plane satisfying $f{|}_k\gamma =f$ for $\gamma \in \Gamma$ and holomorphic at every cusp of a modular group. Explicitly, for each $\sigma \in S{L}_2(\mathbb{Z})$ choose a positive integer $h_{\sigma }$ with $\sigma T^{h_{\sigma }}{\sigma }^{-1}\in \Gamma$. Then $f{|}_k\sigma$ is periodic with period $h_{\sigma }$, and its expansion in $q_{h_{\sigma }}=e^{2\pi iz/h_{\sigma }}$ must have no negative powers. Equivalently it must be bounded as $\mathrm{Im}z\to \infty$, uniformly in one period. This definition applies to noncongruence subgroups too. If $-I\in \Gamma$, nonzero forms require even weight.

Put $d=[S{L}_2(\mathbb{Z}):\Gamma ]$ and let $f\ne 0$. Form the product over left cosetsChanging a representative does not affect a factor. Right multiplication permutes the cosets, so the product is a nonzero level-one modular form of weight $kd$. It is holomorphic on the upper half-plane and at infinity, since every factor is bounded there by cusp holomorphy.

Let $h$ be the least positive integer with $T^h\in \Gamma$, and write $f={\sum }_{n\ge 0}{b}_n{q}_h^n$. If $r$ is its first nonzero index, the $h$ distinct cosets $\Gamma T^j$, $0\le j<h$, give the factors $f(z+j)$. Their product has leading term a nonzero constant times $e^{2\pi irz}=q^r$. All remaining factors are bounded at infinity. Hence $F(z)=O(e^{-2\pi r\mathrm{Im}z})$, so its level-one order satisfies $v_{\infty }(F)\ge r$. Notice that using just the single factor $f$ would only give $r/h$; the entire translation orbit is needed.

The permitted valence formula for the modular group bounds $v_{\infty }(F)\le kd/12$, because all its other weighted zero orders are nonnegative. Thus $r\le kd/12$. With $R=\lfloor kd/12\rfloor$, the map $f\mapsto (b_0,\dots ,b_R)$ is injective: a nonzero form in its kernel would have $r\ge R+1$. This proves the dimension bound for modular forms on a finite-index subgroup:

For the dilation assertion, the original PDF states $N_{1}D\mid N$, a product-divisibility condition, rather than the TeX's chain $N_1\mid D\mid N$. Use the PDF condition. Put $A=\mathrm{diag}(D,1)$. If $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in {\Gamma }_0(N)$, thensince $N_{1}D\mid c$. Therefore, for $g(z)=f(Dz)$, either direct substitution or the slash operator for modular forms gives $g{|}_k\gamma =g$. It is holomorphic on the upper half-plane.

To verify every cusp, take $\sigma \in S{L}_2(\mathbb{Z})$ and choose $\tau \in S{L}_2(\mathbb{Z})$ taking infinity to the rational cusp $A\sigma \infty$. Then $B={\tau }^{-1}A\sigma$ is rational upper triangular, with positive ratio of its two diagonal entries. Consequentlyis a nonzero constant factor times $f{|}_k\tau$ evaluated at an affine map $z\mapsto uz+v$ with $u>0$. The imaginary part tends to infinity, where $f{|}_k\tau$ is bounded. Thus $g{|}_k\sigma$ is bounded too. Together with its periodicity this proves cusp holomorphy by the Riemann removable singularity theorem. Hence the oldform by argument dilation satisfies

Finally ${\Gamma }_0(3)$ has index four: reduction modulo three gives a transitive $S{L}_2({\mathbb{F}}_3)$ action on the four points of ${\mathbb{P}}^1({\mathbb{F}}_3)$, and ${\Gamma }_0(3)$ is the inverse image of the stabilizer of infinity. Reduction is surjective, since elementary determinant-one matrices generate $S{L}_2({\mathbb{F}}_3)$ and lift integrally. The dimension bound gives $\dim M_4({\Gamma }_0(3))\le 2$.

The normalized Eisenstein series $E_4(z)$ and the oldform $E_4(3z)$ both belong to this space. Their Fourier expansion of a normalized Eisenstein series starts asA linear relation between them has zero constant term only if its two scalars sum to zero; its $q$ coefficient then forces both scalars to vanish. They are independent, proving the weight-four modular forms on Gamma 0 3 description