Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 88 / 2

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.