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

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: