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

The cusps are the ${\Gamma }_1(p)$-orbits of primitive columns $(a,c{)}^T$, with simultaneous negation representing the same point of ${\mathbb{P}}^1(\mathbb{Q})$. For the Gamma 1 congruence subgroup, the standard primitive-vector classification separates the orbits according to $d=\gcd (c,p)$. For each divisor $d\mid p$, the two surviving unit coordinates giveorbits when $p>2$. Hence an odd prime gives

For $p=2$, simultaneous negation is already trivial modulo $2$, so the division by two does not apply. In that case ${\Gamma }_1(2)={\Gamma }_0(2)$ has two cusps, represented by infinity and zero. Thus the Number of cusps of Gamma 1 of prime level is

The Klein j-invariantis a weight-zero modular function for $\Gamma (1)$, holomorphic on $\mathfrak{h}$ and meromorphic at infinity. LetSince $c$ is even,The modular invariance of $j$ therefore givesThe quotient is meromorphic on $\mathfrak{h}$ and at the cusps, so it is a weight-zero modular function of level ${\Gamma }_1(2)$. This is the Level-two modular ratio of Klein j-invariants.

The two cusps of ${\Gamma }_1(2)$ are infinity and zero. At infinity,soThus $f$ is holomorphic at infinity and has a simple zero there.

Use the scaling matrixat the cusp zero. Since $j(-1/\tau )=j(\tau )$,The width of zero is two, so its local parameter is $q_0=e^{\pi i\tau }$. As $\mathrm{Im}\tau \to \infty$,and thereforeIt 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.