Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 23 / 5

For a cusp form, the invariant norm of a modular form $|f(\tau )|(\mathrm{Im}\tau {)}^{k/2}$ is bounded on the entire half-plane. It is invariant under the level group; on a truncated fundamental region of a modular subgroup boundedness is compactness, and near each modular cusp exponential decay of the modular cusp expansion dominates the power of height. With this bound, Fourier inversion on a horizontal interval givesChoosing $y=1/n$ proves the Fourier coefficient bound for a cusp form, $a_n=O(n^{k/2})$. Therefore $\sum |a_n|n^{-\sigma }$ converges when $\sigma >k/2+1$, and uniformly on every compact subset of that half-plane. The locally uniform limit of its holomorphic terms is holomorphic, so

With the determinant-normalized slash operator, let $W_N=\left(\begin{array}{cc}0& -1\\ N& 0\end{array}\right)$ and $Jf=i^k(f{|}_k{W}_N)=g$. Conjugation by $W_N$ preserves ${\Gamma }_1(N)$, and rational slash operators preserve modular cusp holomorphy and vanishing. Hence $g$ is also a cusp form at that level. Direct substitution gives $J^{2}f=f$; the factor $i^k$ cancels the central weight sign.

Put $F(y)=f(iy/\sqrt{N})$ and $G(y)=g(iy/\sqrt{N})$. The defining formula gives the exact relationsIn the initial half-plane of absolute convergence, termwise integration of the Fourier series and the gamma function yield the Mellin transform of a cusp-form L-functionThe scaling of $y$ accounts for $N^{s/2}$ in the completion. At infinity $F$ and $G$ decay exponentially. At zero the boxed relation expresses $F$ as a power times an exponentially decaying function of $1/y$. Thus this integral converges locally uniformly for every complex $s$, including after differentiation in $s$, and defines an entire function.

Splitting at one and changing $y$ to $1/y$ in the lower integral givesApplying the same formula to $g$ interchanges $F,G$, because $J^2=1$. It proves the phase-normalized Fricke functional equationThe entire function here is the completion, despite the apparent poles of the gamma factor in its initial product formula.

Use the product printed in the original PDF, with factors $(1-q^n{)}^2$ and $(1-q^{11n}{)}^2$. The Fricke involution normalizes ${\Gamma }_0(11)$ and preserves its modular cusp space. Therefore $g=-11^{-1}{\tau }^{-2}f(-1/(11\tau ))$ lies in the given one-dimensional space, so $g=cf$ for a constant $c$.

At the Fricke fixed point ${\tau }_{\ast }=i/\sqrt{11}$, the prefactor $-11^{-1}{\tau }_{\ast }^{-2}$ is one. Thus $g({\tau }_{\ast })=f({\tau }_{\ast })$. The product has $0<q_{\ast }<1$, every factor is positive, and its limit is nonzero since ${\sum }_n{q}_{\ast }^n+{\sum }_n{q}_{\ast }^{11n}<\infty$. Hence $f({\tau }_{\ast })>0$, forcing $c=1$. This Fricke sign from a nonvanishing fixed-point value provesPart (b) now gives $\Lambda (f,s)=\Lambda (f,2-s)$. Its Taylor series at one contains only even powers, so its order of vanishing is even. The function is not identically zero, since its first Fourier coefficient is one. In this example the product is positive on the entire positive imaginary axis, and its Mellin integral at $s=1$ is positive. Thus the stronger conclusion isThe TeX aid duplicates and corrupts the product in this part; neither corrupted expression is used.