The two cusps of $\Gamma_1(2)$ are infinity and zero. At infinity,
$$
j(\tau)=q^{-1}+744+O(q),
\qquad
j(2\tau)=q^{-2}+744+O(q^2),
$$
so
$$
f(\tau)=\frac{j(\tau)}{j(2\tau)}=q+O(q^2).
$$
Thus $f$ is holomorphic at infinity and has a simple zero there.

Use the scaling matrix
$$
S=\begin{pmatrix}0&-1\\1&0\end{pmatrix}
$$
at the cusp zero. Since $j(-1/\tau)=j(\tau)$,
$$
(f|_0S)(\tau)
=\frac{j(-1/\tau)}{j(-2/\tau)}
=\frac{j(\tau)}{j(\tau/2)}.
$$
The width of zero is two, so its local parameter is $q_0=e^{\pi i\tau}$. As $\operatorname{Im}\tau\to\infty$,
$$
j(\tau)=q_0^{-2}+O(1),
\qquad
j(\tau/2)=q_0^{-1}+O(1),
$$
and therefore
$$
(f|_0S)(\tau)=q_0^{-1}+O(1).
$$
It has a simple pole at zero and is not holomorphic there. This uses the <width of a cusp> to express the two expansions in their correct local parameters.

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 137 / 3 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-137/3/c/solution