Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 23 / 1

The central matrices $\pm I$ act identically on the complex upper half-plane, so the effective group is ${\mathrm{PSL}}_2(\mathbb{Z})$. Away from points with nontrivial effective stabilizer, properly discontinuous action supplies ordinary quotient-disc charts. At a fixed point $z_0$, the coordinate $u=(\tau -z_0)/(\tau -{\overline{z}}_0)$ identifies the stabilizer action with a rotation; its invariant coordinate is $u^e$, where $e$ is the effective stabilizer order. These charts give the quotient its Riemann surface structure.

The elliptic stabilizers of the modular group occur only in the orbits of $i$ and $\omega =e^{2\pi i/3}$. They have effective orders two and three. Consequently the analytic ramification indices of the map from the half-plane areThe stabilizers in ${\mathrm{SL}}_2$ have orders four and six, but the central factor does not double these indices.

All rational boundary points, including infinity, lie in one cusp of a modular group: a primitive column $(a,c)$ can be completed to a determinant-one integral matrix taking infinity to $a/c$. The stabilizer of infinity is generated effectively by $T:\tau \mapsto \tau +1$, and the coordinate $q=e^{2\pi i\tau }$ identifies its high horodisc quotient with a punctured disc. Adding $q=0$ fills that disc. This constructs the compactified modular curve from the extended half-plane; it does not use the ordinary subspace topology on the rational boundary.

For compactness use the standard fundamental domain of the modular group. Its part below a fixed height $Y>1$ is compact, since its imaginary part is at least $\sqrt{3}/2$. The part above $Y$, modulo translation and with the modular cusp added, is a closed disc in the $q$-coordinate. Their images cover the quotient, so it is compact. The same argument with finitely many translates proves compactness for every finite-index subgroup.

The weights of $E_4^3$ and $\Delta$ agree, so their ratio is invariant under the modular group. Nonvanishing of the modular discriminant on the half-plane makes the ratio holomorphic there. At an elliptic point, an invariant holomorphic function has a power series in the quotient coordinate, so the ratio descends holomorphically. At the unique modular cusp its expansion isThus it is a meromorphic function on the compact Riemann surface with precisely one pole, of order one. The degree of a nonconstant meromorphic map to the sphere equals its total pole order. It therefore has degree one, hence is a biholomorphism:This is the single-pole criterion for a spherical coordinate.

The matrix $\left(\begin{array}{cc}0& -1\\ 1& 1\end{array}\right)$ fixes $\omega$. Weight-four transformation gives $E_4(\omega )=(\omega +1{)}^4{E}_4(\omega )$, and $(\omega +1{)}^4\ne 1$. Hence $E_4(\omega )=0$ and $j(\omega )=0$. Since $j$ is a degree-one coordinate, this is a simple zero on the quotient surface. Pulled back to the half-plane it has order three, by the analytic ramification index in part (a).

Write $t(\tau )=\Delta (\tau )/\Delta (2\tau )$. For $\gamma =\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in {\Gamma }_0(2)$, the integral matrix ${\gamma }^{\prime }=\left(\begin{array}{cc}a& 2b\\ c/2& d\end{array}\right)$ has determinant one and satisfies $2\gamma \tau ={\gamma }^{\prime }(2\tau )$. Both discriminants acquire the same factor $(c\tau +d{)}^{12}$, so $t$ is invariant. It has neither zeros nor poles in the half-plane.

Reduction modulo two gives index three, with two modular cusp classes, infinity and zero, of cusp widths one and two. These can also be found from the orbits of upper triangular matrices on primitive columns modulo two. Compactness follows from the finite-index argument in part (a).

At infinity $t=q^{-1}(1+O(q))$, so it has a simple pole. At zero use $S\tau =-1/\tau$ and the modular discriminant inversion law:Thus zero is a simple zero, not a pole, in its width-two modular cusp coordinate. The single-pole criterion for a spherical coordinate now givesThis discriminant-ratio coordinate on X0 2 is different from a ratio of two $j$-invariants.

View $j$ as a meromorphic function on $X({\Gamma }_0(2))$, using its coordinate $t=j_2$. At the infinity modular cusp its pole order is one. At the zero modular cusp, $j(S\tau )=j(\tau )=q^{-1}+O(1)=q_0^{-2}+O(1)$, so its pole order is two. Part (c) identifies these poles with $t=\infty$ and $t=0$.

The zero of $j$ on the full modular curve lies at its order-three elliptic point. There are no effective order-three stabilizers in ${\Gamma }_0(2)$: their lifts have trace $\pm 1$, whose characteristic polynomial modulo two is $X^2+X+1$, whereas an upper triangular matrix over ${\mathbb{F}}_2$ has both diagonal entries one. Therefore the degree-three covering has exactly one point over that elliptic point, with analytic ramification index three. Its $t$-coordinate is $a=j_2(\omega )$, and the zero divisor of the pulled-back $j$ is $3[a]$.

A rational function with these zeros and poles must be $C(t-a{)}^3/t^2$. At the infinity modular cusp both $j$ and $t$ have leading coefficient one times $q^{-1}$, giving $C=1$. HenceThe proof fixes the coefficient and the powers using modular cusp cusp widths and elliptic ramification, rather than assuming a formula for the level-two coordinate.