Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 137 / 3

The modular curve isGive it the quotient topology on $\Gamma \backslash \mathfrak{h}$ and adjoin one end for each cusp of a modular group. At a point with trivial effective stabilizer, the quotient map identifies a sufficiently small disk with a chart. At an elliptic point with cyclic stabilizer of order $e$, choose a disk coordinate $z$ centered there; then $z^e$ descends to a quotient coordinate. At a cusp represented by $\sigma \in S{L}_2(\mathbb{Z})$ and having width $h$, a punctured neighborhood is described byand adjoining $q_h=0$ fills in the cusp. These charts make $X(\Gamma )$ a compact Riemann surface.

A weight-zero modular function is invariant under $\Gamma$, so it descends uniquely through the quotient map to a meromorphic function on $\Gamma \backslash \mathfrak{h}$. Its assumed meromorphic Fourier expansion at every cusp makes the descended function meromorphic in each cusp coordinate. A meromorphic function on a compact Riemann surface is equivalently a holomorphic morphism to the Riemann sphere, sending each pole to infinity. Thus there is a morphismwith $f={\varphi }_f{|}_{\Gamma \backslash \mathfrak{h}}\circ \pi$. The open quotient $\Gamma \backslash \mathfrak{h}$ is dense in $X(\Gamma )$, so this identity also proves uniqueness.

The genus formula for a modular curve iswhere $\mu =[PS{L}_2(\mathbb{Z}):\overline{\Gamma }]$, $e_2,e_3$ count elliptic orbits of orders two and three, and $c$ is the number of cusps.

For ${\Gamma }_0(3)$,There are two cusps, represented by infinity and zero. The congruence $x^2+1\equiv 0(\mathrm{mod}3)$ has no solution, so there is no elliptic orbit of order two. The congruence $x^2+x+1\equiv 0(\mathrm{mod}3)$ has the single solution $x=1$, giving one elliptic orbit of order three. Thereforewhich is the Modular curve X0 3 calculation.

LetForthe matrix ${\gamma }^{\prime }=\left(\begin{array}{cc}a& 3b\\ c& d\end{array}\right)$ lies in $S{L}_2(\mathbb{Z})$ and satisfies $3\gamma \tau ={\gamma }^{\prime }(3\tau )$. The weight-twelve transformation law for the modular discriminant gives the same factor $(3c\tau +d{)}^{12}$ in numerator and denominator. Hence $f(\gamma \tau )=f(\tau )$, so $f$ is a weight-zero modular function of level ${\Gamma }_0(3)$.

The discriminant has no zero in the upper half-plane, so $f$ has neither zeros nor poles there. At infinity,and therefore $f=q^{-2}+O(q^{-1})$: it has a pole of order two. The transformation $\Delta (-1/\tau )={\tau }^{12}\Delta (\tau )$ givesThe Fricke involution exchanges infinity and zero, so $f$ has a zero of order two at the cusp zero.

Thus the morphism ${\varphi }_f:X_0(3)\to \hat{\mathbb{C}}$ has degree two, equal to its total pole order. An isomorphism of compact Riemann surfaces has degree one. Although $X_0(3)$ has genus zero, this particular morphism is therefore not an isomorphism.