Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 137 / 2

The Gamma 1 congruence subgroup isThe space $M_k({\Gamma }_1(N))$ consists of the holomorphic functions $f:\mathfrak{h}\to \mathbb{C}$ satisfyingfor every $\gamma \in {\Gamma }_1(N)$ and which are holomorphic at a cusp for every cusp. Its subspace $S_k({\Gamma }_1(N))$ consists of the forms that vanish at every cusp.

For matrices of positive determinant, use the determinant-normalized slash operator for modular formsIt satisfies the composition law $(f{|}_k[\alpha ]){|}_k[\beta ]=f{|}_k[\alpha \beta ]$. IfthenThus the Fricke involution normalizes the group, and the composition law proves that $f{|}_k[{\alpha }_N]$ has the required transformation law.

The matrix ${\alpha }_N$ permutes the rational cusps. Applying it to a local Fourier expansion merely transports that expansion to the image cusp, with a nonzero change of local parameter. It therefore preserves holomorphy and vanishing at every cusp. Hence

Put $g=f{|}_k[{\alpha }_N]$. Termwise Mellin transform in the initial half-plane givesso $\Lambda (f,s)=N^{s/2}I(f,s)$.

From the definition of the slash action,or equivalentlySubstituting $y=1/(Nt)$ in the Mellin integral therefore givesMultiplying by $N^{s/2}$ and recognizing the completed function on the right yields

At infinity, both $f(iy)$ and $g(iy)$ decay exponentially because they are cusp forms. The transformation just used converts the behavior of $f(iy)$ near zero into the exponential decay of $g(i/(Ny))$ at infinity. Consequently $I(f,s)$ converges absolutely for every $s\in \mathbb{C}$ after splitting the integral at one and applying that substitution to the part near zero. It is locally uniformly convergent in $s$, hence entire, and agrees with the original Dirichlet series in its initial domain. This proves the analytic continuation and the stated functional equation, as summarized by the Mellin transform of a cusp-form L-function.