At weight zero the convention from part (a) is
$$
(T_pF)(\tau)=\frac1p\left(F(p\tau)+\sum_{b=0}^{p-1}F\left(\frac{\tau+b}{p}\right)\right).
$$
Partition the summands of $G$ according to divisibility of the first integer coordinate by $p$. Directly from the definition,
$$
G(p\tau,s)
=p^s\sum_{p\mid m}\frac{y^s}{|m\tau+n|^{2s}}.
$$
In the sum over $b$, the congruence $n\equiv mb\pmod p$ has one solution when $p\nmid m$, has $p$ solutions when $p$ divides both $m$ and $n$, and has none when $p\mid m$ but $p\nmid n$. Therefore
$$
G(p\tau,s)+\sum_{b=0}^{p-1}G\left(\frac{\tau+b}{p},s\right)
=p^sG(\tau,s)+p^{1-s}G(\tau,s),
$$
where the contribution from pairs divisible by $p$ was rescaled by $p^{-2s}$. Dividing by $p$ proves the <Hecke eigenvalue of a nonholomorphic Eisenstein series>:
$$
T_pG(\tau,s)=(p^{s-1}+p^{-s})G(\tau,s).
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 137 / 4 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/iii/paper-137/4/c/solution