Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 88 (Modular Forms)

For a period lattice $\Lambda$, define the Weierstrass elliptic function byThe subtraction is essential: the separate inverse-square lattice sum need not converge. Let $r$ be the distance to the nearest nonzero lattice point. For $|z|<r$, the power series identity $(1-w{)}^{-2}={\sum }_{m\ge 0}(m+1)w^m$ givesThe stipulated convergence allows the sums to be interchanged on smaller disks. Pairing $\omega$ with $-\omega$ cancels odd powers. In terms of the lattice Eisenstein series, this proves the Laurent coefficients of the Weierstrass elliptic function:

For completeness, the function is even and is a periodic function: differentiating gives ${\wp }^{\prime }(z)=-2{\sum }_{\omega \in \Lambda }(z-\omega {)}^{-3}$, which is invariant under translation by a lattice generator. Thus $\wp (z+\omega )-\wp (z)$ is constant. For a primitive generator, evaluating at $z=-\omega /2$ and using evenness makes this constant zero. Two generators establish lattice periodicity. Hence any polynomial in $\wp ,{\wp }^{\prime }$ is an elliptic function.

Set $c_1=3G_4$ and $c_2=5G_6$. The first terms are $\wp =z^{-2}+c_1{z}^2+c_2{z}^4+O(z^6)$. Squaring its derivative and cubing the function giveConsequently $H={\wp }^{\prime 2}-4{\wp }^3+20c_1\wp +28c_2$ has no pole at zero and has value zero there. Lattice periodicity removes the poles at every lattice point; there are no others. Thus $H$ is an entire function and an elliptic function, bounded on a closed fundamental parallelogram and therefore throughout the plane. The Liouville theorem makes it constant, and its value at zero makes it zero. This proves the Weierstrass elliptic differential equation

Differentiating and cancelling ${\wp }^{\prime }$ wherever it is nonzero gives ${\wp }^{\prime \prime }=6{\wp }^2-g_2/2$; the identity extends across its isolated zeros by holomorphy. Write $\wp =z^{-2}+{\sum }_{n\ge 1}{c}_n{z}^{2n}$. For $n\ge 3$, compare coefficients of $z^{2n-2}$:The denominator is positive and rational, and $c_1=3G_4,c_2=5G_6$. Induction therefore gives a polynomial with positive rational coefficients on every monomial that occurs. Since $G_{2k}=c_{k-1}/(2k-1)$, the same holds for every $k>3$. This is the positive polynomial recurrence for lattice Eisenstein sums; for examplePositive coefficients do not assert that the complex values of the lattice sums themselves are positive.

Use the Hecke-normalized rational slash operator from the PDF, writing $f[\alpha {]}_k$ for its right action. Put $A_D=\mathrm{diag}(D,1)$. If $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in {\Gamma }_0(N)$, thenIndeed $N/D$ is an integer divisible by $M$, so $c/D$ is an integer divisible by $M$. The ordinary weight transformation of the original cusp form therefore givesThus $g(\tau )=f(D\tau )$ has the correct transformation for the Gamma 0 congruence subgroup at level $N$, and it is holomorphic in the complex upper half-plane.

It remains to check every modular cusp, not just infinity. Let $\rho \in S{L}_2(\mathbb{Z})$ carry infinity to that cusp, and choose $\sigma \in S{L}_2(\mathbb{Z})$ with $\sigma \infty =A_D\rho \infty$. The matrix $B={\sigma }^{-1}{A}_D\rho$ is upper triangular over the rationals, with positive ratio of diagonal entries. Therefore, up to a nonzero constant weight factor,The original cusp expansion of $f[\sigma {]}_k$ has only positive exponents. The change $\tau \mapsto r\tau +s$ induced by $B$, with $r>0$, preserves exponential decay as $\mathrm{Im}\tau \to \infty$. The transformed function also has a positive integral period because $g$ is modular for a finite-index subgroup. Its bounded periodic holomorphic expansion consequently has no negative exponents and its limiting constant term is zero. This proves cusp holomorphy under rational slash operators and establishes the degeneracy map for cusp forms:

The modular discriminant has the convergent productis a weight-twelve level-one cusp form, and has order one at infinity. Applying the degeneracy map for cusp forms with $M=1,N=D=p$ gives $\boxed{\Delta (p\tau )\in S_{12}({\Gamma }_0(p))}$. The original form $\Delta$ is also a cusp form on this subgroup.

There are precisely two modular cusps for prime level, represented by infinity and zero. To see this directly, write a rational cusp as $a/c$ in lowest terms. If $p\mid c$, completing $(a,c{)}^T$ to a determinant-one matrix gives a member of ${\Gamma }_0(p)$ taking infinity there. If $p\nmid c$, the equation $Ac-aC=1$ has a solution with $p\mid C$: adjust a Bézout solution modulo $p$, using the invertibility of $c$. The matrix $\left(\begin{array}{cc}A& a\\ C& c\end{array}\right)$ then lies in ${\Gamma }_0(p)$ and takes zero there. The two classes are distinct because their denominator divisibility is preserved by the group.

At infinity the cusp width is one, so $q$ is the local parameter. The products start with $q$ and $q^p$, giving orders one and $p$. At zero use $S=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$. Since $S{T}^h{S}^{-1}=\left(\begin{array}{cc}1& 0\\ -h& 1\end{array}\right)$, the cusp width is $p$ and the parameter is $q_0=e^{2\pi i\tau /p}$. The weight-twelve inversion law of the modular discriminant givesHere determinant one makes the printed and ordinary slash operators for modular forms agree. In terms of $q_0$, the leading terms are $q_0^p$ and $p^{-12}{q}_0$. Hence the prime-level cusp orders of discriminant degeneracy forms areThe scalar $p^{-12}$ changes no order of vanishing.

The dimension assumption supplies a nonzero $h\in S_2({\Gamma }_0(11))$. The projective index of the Gamma 0 congruence subgroup is twelve: reduction modulo eleven identifies its cosets with the twelve points of ${\mathbb{P}}^1({\mathbb{F}}_{11})$. There are two modular cusps, by the preceding classification. There are no effective elliptic fixed points. An order-two fixed point would require a root of $x^2+1$ modulo eleven, which does not exist; an order-three fixed point would require a root of $x^2+x+1$, equivalent to a nontrivial cube root of unity in a group of order ten, which also does not exist.

The regular-cusp valence formula on a torsion-free modular curve therefore givesAll orders are nonnegative and both cusp orders are at least one. Thus $h$ has order exactly one at both modular cusps and has no zeros in the half-plane. One way to justify the degree count here is the genus formula for a modular curve: the same index, cusp and elliptic data give genus one. The invariant differential $h(\tau )d\tau$ has cusp order $\mathrm{ord}h-1$, because $d\tau$ is a nonzero constant times $d{q}_c/q_c$. Its total canonical degree $2g-2=0$ therefore makes the sum of the two cusp-adjusted orders and the interior orders zero, giving precisely the count above.

Let $H(\tau )=\Delta (\tau )\Delta (11\tau )$. By the product formula it has no zeros in the half-plane; by the cusp-order table it has order twelve at each cusp. Its weight is twenty-four. Therefore $H/h^{12}$ is a weight-zero modular function, with neither zeros nor poles anywhere on the compactified modular curve. It is constant by the maximum modulus principle on a compact surface. Normalize $h$ so its leading Fourier coefficient at infinity is one. Since $H=q^{12}+O(q^{13})$, that constant becomes one:This fixes the holomorphic twelfth root by its leading term and proves the weight-two discriminant root at level eleven is a nonzero member, hence a generator, of the given space. Taking an arbitrary pointwise twelfth root without this argument would not establish modularity: it could carry an unwanted character.

Put $A_N=\left(\begin{array}{cc}0& -1\\ N& 0\end{array}\right)$. All calculations here use the Hecke-normalized rational slash operator of the PDF. Its automorphy factor identity gives $(f[\alpha {]}_k)[\beta {]}_k=f[\alpha \beta {]}_k$. For $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in {\Gamma }_1(N)$,This proves that $A_N$ normalizes the Gamma 1 congruence subgroup. Right-action associativity shows $f[A_N{]}_k$ has the same transformation law as $f$. It is holomorphic in the complex upper half-plane; at each modular cusp, the rational change of variable preserves cusp decay by the upper-triangular factorization used in 2(i). Consequently $W_N$ maps the cusp form space to itself.

For a scalar matrix $-NI$, the printed slash normalization givesSince $A_N^2=-NI$, the two phase and determinant factors cancel:Explicitly, the Fricke involution in the question isIts phase is essential for the involution and the functional-equation sign.

Suppose $W_{N}f=\epsilon f$. Define $F(y)=f(iy/\sqrt{N})$ for $y>0$. Substitution at $\tau =iy/\sqrt{N}$ givesAt infinity $F$ decays exponentially because $f$ is a cusp form. This identity gives $F(y)=O(y^{-k}{e}^{-c/y})$ near zero for some $c>0$. Thus its Mellin transform converges for every complex $s$, locally uniformly with all $s$ derivatives.

Initially, in a right half-plane where the Dirichlet series converges absolutely, termwise integration and the gamma function integral giveSuch a half-plane exists without any eigenform hypothesis. Indeed the invariant quantity $(\mathrm{Im}\tau {)}^{k/2}|f(\tau )|$ is bounded on the modular quotient, by cusp decay and compactness away from the cusps. Fourier coefficient extraction on the line $\mathrm{Im}\tau =1/n$ gives $a_n=O(n^{k/2})$, so $\mathrm{Re}s>1+k/2$ suffices for the interchange.

Split the integral at one and change $y$ to $1/y$ in its lower part. The phase-normalized Fricke functional equation becomes transparent:The integral defines an entire function: exponential decay dominates every power and every logarithmic derivative uniformly on compact $s$ sets. Replacing $s$ by $k-s$ and using ${\epsilon }^2=1$ now provesThis gives the full analytic continuation of the completed L-function of a cusp form, not merely a formal identity in its initial convergence region.

Use the Fourier transform convention $\hat{g}(\xi )={\int }_{\mathbb{R}}g(x)e^{-2\pi ix\xi }dx$. For $t>0$, completing the square in the Gaussian integral givesBoth functions are rapidly decreasing, so the Poisson summation formula applies:In terms of the Jacobi theta function, this is $\vartheta (i/t)=\sqrt{t}\vartheta (it)$. The theta series converges locally uniformly in the complex upper half-plane, defining a holomorphic function. The branch of $\sqrt{-i\tau }$ with positive value at $\tau =it$ is holomorphic there, because $-i\tau$ lies in the right half-plane. Both sides of the proposed inversion identity are holomorphic; agreement on the positive imaginary axis and the identity theorem giveEquivalently, the complex Gaussian has transform $(-i\tau {)}^{-1/2}\exp (-\pi i{\xi }^2/\tau )$, which produces the same identity directly by Poisson summation.

Put $\psi (t)=(\vartheta (it)-1)/2={\sum }_{n\ge 1}{e}^{-\pi n^{2}t}$. For $\mathrm{Re}s>1$, integrate term by term in its Mellin transform:The theta function inversion givesSplit the integral at one. In the term containing $\psi (1/t)$ put $t=1/u$. The elementary remainder integral isThus the pole-subtracted theta integral for the completed zeta function isThe integral is entire by exponential decay. This supplies a meromorphic continuation with simple poles at zero and one, of residues $-1$ and $1$. The expression is unchanged by $s\mapsto 1-s$, proving the functional equation of the Riemann zeta functionThe reflection and duplication identities $\Gamma (z)\Gamma (1-z)=\pi /\sin \pi z$ and $\Gamma (z)\Gamma (z+1/2)=2^{1-2z}\sqrt{\pi }\Gamma (2z)$ convert it toMultiplying $Z$ by $s(s-1)/2$ removes its two poles and yields the Riemann xi function, which is an entire function, satisfying $\xi (s)=\xi (1-s)$.

Write $A,B$ for the two given congruence subgroups, to distinguish them from the notation ${\Gamma }_1(N)$. Their double coset has a finite decompositionFiniteness follows because $B\cap {\alpha }^{-1}A\alpha$ has finite index in $B$. To check this without assuming integrality of $\alpha$, clear its denominators: a sufficiently deep principal congruence subgroup lies both in $B$ and, after conjugation by $\alpha$, in $A$. Representatives for $(B\cap {\alpha }^{-1}A\alpha )\backslash B$ give the displayed decomposition.

Using the Hecke-normalized rational slash operator from the PDF, define the double-coset operator on modular forms byReplacing ${\alpha }_j$ by $\gamma {\alpha }_j$ for $\gamma \in A$ changes no term because $f[\gamma {]}_k=f$. Right multiplication by $\beta \in B$ permutes these left cosets, so $([A\alpha B]f)[\beta {]}_k=[A\alpha B]f$. Each summand is holomorphic in the complex upper half-plane. At any rational modular cusp, factor ${\alpha }_j\rho =\sigma C$ with $\sigma \in S{L}_2(\mathbb{Z})$ and $C$ upper triangular of positive slope, as in 2(i). The cusp expansion of $f[\sigma {]}_k$ has no negative terms, and substitution by $C$ preserves boundedness. The sum is periodic in a cusp coordinate for $B$, so it has a removable singularity there. This provesIf $f$ vanishes at all its cusps, the same argument gives vanishing for the image too.

Now set $\Gamma ={\Gamma }_1(N)$. For an integer $d$ prime to $N$, choose ${\sigma }_d\in {\Gamma }_0(N)$ with lower-right entry congruent to $d$ modulo $N$. Such a lift exists by choosing $a$ with $ad\equiv 1(\mathrm{mod}N)$ and using $\left(\begin{array}{cc}a& (ad-1)/N\\ N& d\end{array}\right)$. The subgroup $\Gamma$ is normal in ${\Gamma }_0(N)$, and its quotient is $(\mathbb{Z}/N\mathbb{Z}{)}^{\times }$ through the lower-right entry. Define the diamond operator byDifferent lifts differ by a member of $\Gamma$, so this is well defined. The diamond operators satisfy $⟨d⟩⟨e⟩=⟨de⟩$.

For a prime $p\nmid N$, put $A_p=\mathrm{diag}(1,p)$ and define the Hecke operatorChoose $a,b\in \mathbb{Z}$ with $ap-bN=1$ and set ${\sigma }_p=\left(\begin{array}{cc}a& b\\ N& p\end{array}\right)$. A complete set of left-coset representatives isHere $A_j=A_p{T}^j$ and $B_p=A_p\left(\begin{array}{cc}ap& b\\ N& 1\end{array}\right)$, with both right factors in $\Gamma$. To prove completeness, $H=\Gamma \cap A_p^{-1}\Gamma A_p$ consists of the members of $\Gamma$ whose upper-right entry is divisible by $p$. Reduction of $\Gamma$ modulo $p$ is onto $S{L}_2({\mathbb{F}}_p)$: use the Chinese remainder theorem to impose identity modulo $N$ and any chosen determinant-one matrix modulo $p$, then lift to $S{L}_2(\mathbb{Z})$ by elementary matrices. Reduction of $H$ is the lower triangular subgroup. Its left cosets correspond to the $p+1$ lines of top rows in ${\mathbb{F}}_p^2$. The right factors $T^j$ give lines $[1:j]$, while the last factor gives $[0:b]$ with $b\not\equiv 0(\mathrm{mod}p)$. They are all the lines, proving the decomposition.

The printed slash normalization now gives $f[A_j{]}_k=p^{-1}f((\tau +j)/p)$ andThus

To prove commutation, use the marked-lattice model of a modular form. A pair $(L,t)$ consists of a lattice and a point of exact order $N$ modulo $L$, and a weight-$k$ function satisfies $F(uL,ut)=u^{-k}F(L,t)$. In a basis $L=\mathbb{Z}\tau +\mathbb{Z}$ with $t=1/N$, it represents $f(\tau )$. The diamond operator multiplies $t$ by $d$. The same Hecke operator isEvery marked point retains exact order $N$: multiplication by $p$ is invertible on its cyclic group, since $p\nmid N$. This lattice sum agrees with the displayed slash sum. Its $p$ overlattices with basis $((\tau +j)/p,1)$ give the first terms, and the remaining one with basis $(\tau ,1/p)$ has normalized marked point $p/N$, giving $p^{k-1}(⟨p⟩f)(p\tau )$. Multiplying the mark by $d$ commutes with passage to each overlattice and does not change the set of overlattices. Hence

Finally write $f={\sum }_{m\ge 0}{a}_m(f)q^m$. The first term of $T_p$ contains the finite averageIt therefore contributes $a_{np}(f)$ to the coefficient of $q^n$. The second term contributes $p^{k-1}{a}_{n/p}(⟨p⟩f)$ if $p\mid n$, and zero otherwise. This proves the good-prime and bad-prime Hecke coefficient formula at a good prime, without assuming that $f$ has a particular character:where a nonintegral coefficient index means zero. For $n=0$ both terms are retained. On a diamond-character eigenspace the second term becomes $\chi (p)p^{k-1}{a}_{n/p}(f)$.