Past exam of the mathematics course of the University of Cambridge / 2006 / iii / Paper 88 / 3 / Solution

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.