\documentclass[12pt]{article} \usepackage{amsthm,amsfonts,amsmath,amssymb,latexsym,epsfig} %\usepackage{showkeys} \usepackage{hyperref} \title{Elliptic Curves} \author{JWR} \date{Friday November 30, 2001,7:40 AM} \newcommand{\N}{\mathbb{N}} \newcommand{\Z}{\mathbb{Z}} \newcommand{\R}{\mathbb{R}} \newcommand{\C}{\mathbb{C}} \newcommand{\D}{\mathbb{D}} \newcommand{\E}{\mathbb{E}} \newcommand{\F}{\mathbb{F}} \newcommand{\G}{\mathbb{G}} \renewcommand{\H}{\mathbb{H}} \newcommand{\K}{\mathbb{K}} \renewcommand{\P}{\mathbb{P}} \newcommand{\Q}{\mathbb{Q}} % \newcommand{\fH}{\mathfrak{H}} \newcommand{\fF}{\mathfrak{F}} % \newcommand{\cA}{\mathcal{A}} \newcommand{\cB}{\mathcal{B}} \newcommand{\cBbar}{\overline{\mathcal{B}}} \newcommand{\cC}{\mathcal{C}} \newcommand{\cD}{\mathcal{D}} \newcommand{\cF}{\mathcal{F}} \newcommand{\cE}{\mathcal{E}} \newcommand{\cG}{\mathcal{G}} \newcommand{\cH}{\mathcal{H}} \newcommand{\cI}{\mathcal{I}} \newcommand{\cJ}{\mathcal{J}} \newcommand{\cK}{\mathcal{K}} \newcommand{\cL}{\mathcal{L}} \newcommand{\cM}{\mathcal{M}} \newcommand{\cMbar}{\overline{\mathcal{M}}} \newcommand{\cN}{\mathcal{N}} \newcommand{\cP}{\mathcal{P}} \newcommand{\cO}{\mathcal{O}} \newcommand{\cQ}{\mathcal{Q}} \newcommand{\cR}{\mathcal{R}} \newcommand{\cS}{\mathcal{S}} \newcommand{\cT}{\mathcal{T}} \newcommand{\cU}{\mathcal{U}} \newcommand{\cV}{\mathcal{V}} \newcommand{\cUbar}{\overline{\mathcal{U}}} \newcommand{\cX}{\mathcal{X}} \newcommand{\cZ}{\mathcal{Z}} % \newcommand{\GL}{{\mathrm{GL}}} \newcommand{\SL}{{\mathrm{SL}}} \newcommand{\U}{{\mathrm{U}}} \newcommand{\SU}{{\mathrm{SU}}} \newcommand{\mcg}{\mathrm{mcg}} \newcommand{\id}{\mathrm{id}} \newcommand{\ad}{\mathrm{ad}} \newcommand{\codim}{\mathrm{codim}} \newcommand{\Cusp}{\mathrm{Cusp}} \newcommand{\Perm}{\mathrm{Perm}} \newcommand{\Diff}{\mathrm{Diff}} \newcommand{\Vect}{\mathrm{Vect}} \newcommand{\Stab}{\mathrm{Stab}} \newcommand{\Span}{\mathrm{Span}} \newcommand{\free}{\mathrm{free}} \newcommand{\Aut}{\mathrm{Aut}} \newcommand{\Iso}{\mathrm{Iso}} \newcommand{\Mor}{\mathrm{Mor}} \newcommand{\Fix}{\mathrm{Fix}} \newcommand{\Ord}{\mathrm{Ord}} \newcommand{\Kernel}{\mathrm{Kernel}} \newcommand{\Image}{\mathrm{Image}} \newcommand{\End}{\mathrm{End}} % \newcommand{\Stretch}{\renewcommand{\arraystretch}{1.5}} % \newcommand{\p}{\partial} \newcommand{\Wedge}{{\textstyle\bigwedge\nolimits}} \newcommand{\jhat}{{\hat{\jmath}}} \newcommand{\pbar}{{\bar\partial}} \newcommand{\eps}{\varepsilon} \newcommand{\Mat}[1]{\left(\begin{array}{#1}} \newcommand{\Rix}{\end{array}\right)} % % \newcommand{\inner}[2]{\left\langle #1,#2\right\rangle} %\newcommand{\proof}[1]{\par\smallskip\par\noindent{\bf Proof#1:\ }} %\newcommand{\QEDSYM}{\mbox{$\square$}} %\newcommand{\QED}{\hfill\QEDSYM} \newcommand{\jdef}[1]{{\bf #1}} \newcommand{\mdef}[1]{#1} % \newcommand{\half}{{\textstyle\frac{1}{2}}} \newcommand{\quarter}{{\textstyle\frac{1}{4}}} \newcommand{\ds}{\displaystyle} % \newtheorem{PARA}{}%[section] \newcommand{\para}{\begin{PARA}\rm } \newcommand{\arap}{\end{PARA}} \newtheorem{theorem}[PARA]{Theorem} \newtheorem{corollary}[PARA]{Corollary} \newtheorem{lemma}[PARA]{Lemma} \newtheorem{proposition}[PARA]{Proposition} \newtheorem{definition}[PARA]{Definition} \newcommand{\dfn}{\begin{definition}\rm } \newcommand{\nfd}{\end{definition}} \newtheorem{remark}[PARA]{Remark} \newcommand{\rmk}{\begin{remark}\rm } \newcommand{\kmr}{\end{remark}} \newtheorem{example}[PARA]{Example} \newcommand{\xmpl}{\begin{example}\rm } \newcommand{\lpmx}{\end{example}} \newtheorem{exercise}[PARA]{Exercise} \newcommand{\xrcs}{\begin{exercise}\rm } \newcommand{\scrx}{\end{exercise}} % \newcommand{\note}{[} % These may eventually become end notes. \newcommand{\eton}{] } % \newcommand{\medskipnoindent}{\par\medskip\par\noindent} \newcommand{\Item}[1]{\medskipnoindent\ItemMark{#1}} \newcommand{\ItemMark}[1]{{\bf (#1)}} % \newcommand{\resp}[2]{$\displaystyle \left\{\begin{array}{l}\mbox{#1}\\ \mbox{#2} \end{array}\right\} $} % \newcommand{\mapdown}[1]{\Big\downarrow \rlap{$\vcenter{\hbox{$\scriptstyle#1$}}$}} \newcommand{\mapup}[1]{\Big\uparrow \rlap{$\vcenter{\hbox{$\scriptstyle#1$}}$}} \newcommand{\mapright}[1]{\smash{ \mathop{\longrightarrow}\limits^{#1}}} % \begin{document} \maketitle This is an account of the talk of Cheol-Hyun Cho. The aim is to construct a ``universal elliptic curve''. Tong Hai Yang helped me with this - he told me about~\cite{MUMFORD}. \section{Discrete Groups} A good reference for the material in this section is~\cite{SHIMURA}. \para \label{notation:mobius} Throughout $\P=\C\cup\{\infty\}$ denotes the Riemann sphere, $\H$ denotes the upper half plane, $\C^*$ denotes the multiplicative group of complex numbers, and $\P^n=(\C^{n+1}\setminus\{0\})/\C^*$ denotes $n$ dimensional complex projective space. For $w\in\C^{n+1}\setminus\{0\}$ let $[w]:=w\C^*$ denote the corresponding point of $\P^n$. For $A\in\GL_{n+1}(\C)$ let $M_A$ denote the corresponding automorphism of projective space so that $$ M_A([w])=[Aw]. $$ Identify $\P^1$ and $\P$ via $z=[z,1]$ and $\infty=[1,0]$ so that $$ M_A(z)=\frac{az+b}{cz+d}, \qquad A=\Mat{cc}a&b\\c& d\Rix $$ for $A\in\GL_2(\C)$. A transformation of form $M_A$ is called a \jdef{M\"obius transformation}. \arap \para A matrix $A\in\SL_2(\R)\setminus\{\pm I_2\}$ is called \jdef{hyperbolic} iff its eigenvalues are real and distinct, \jdef{elliptic} iff its eigenvalues are distinct and not real (and therefore complex conjugate), and \jdef{parabolic} otherwise. It is easy to see that a nontrivial M\"obius transformation has either exactly one or exactly two fixed points; Hence a matrix $A\in\SL_2(\R)\setminus\{\pm I_2\}$ is hyperbolic if and only if the corresponding automorphism $M_A$ of $\P$ has two fixed points in $\R$, elliptic if and only if $M_A$ has two fixed points one in $\H$ and the other in $-\H$, and parabolic if and only if it has exactly one fixed point. The fixed point of a parabolic element lies in $\R\cup\{\infty\}$. \arap \para Let $\Gamma\subset\SL_2(\R)$ be a subgroup. A point $z\in\H$ is called a \jdef{regular point} of $\Gamma$ iff the isotropy group $\Gamma_z$ is essentially trivial, i.e. $\Gamma_z=\Gamma\cap\{\pm I_2\}$. A point $z\in\H$ is called an \jdef{elliptic point} of $\Gamma$ iff it is a fixed point of $M_A$ some elliptic element $A\in\Gamma$. A point $z\in\R\cup\{\infty\}$ is called an \jdef{cusp} of $\Gamma$ iff it is a fixed point of $M_A$ for some parabolic element $A\in\Gamma$. Denote by $$ X(\Gamma):=\{\Gamma(z): z\in \H\cup\Cusp(\Gamma)\}, \qquad \Gamma(z):=\{M_A(z): A\in\Gamma\} $$ the orbit space of $\Gamma$ acting on the union of $\H$ with the set of cusps of $\Gamma$. For $z\in \H\cup\Cusp(\Gamma)$ let $$ \Gamma_z:=\{A\in\Gamma: M_A(z)=z\} $$ denote the stabilizer group of $z$. Points on the same orbit have conjugate (in $\Gamma$) stabilizer groups so orbits of $\Gamma$ in $\H\cup\Cusp(\Gamma)$ may be classified as regular, elliptic, or cusp. It is easy to see that the stabilizer group $$ \SL_2(\R)_\infty:=\{A\in\SL_2(\R): M_A(\infty)=\infty\} $$ is the set of all real matrices $A$ of form $A=\pm\Mat{cc}1 & h\\0 & 1\Rix$ where $h\in\R$. \arap \begin{lemma}\label{lem:funnyPoint} Let $\Gamma\subset\SL_2(\R)$ be a discrete group. Then \begin{description} \item[(i)] Let $z_0\in\H$ be an elliptic point of $\Gamma$. Then there is an element $C\in\SL_2(\C)$ with $M_C(z_0)=0$ and $M_C\circ\Gamma_z\circ M_C^{-1}$ a finite cyclic subgroup of the stabilizer subgroup $\C^*\cdot I_2$ of the origin in $\GL_2(\C)$. \item[(ii)] Let $x_0\in\R\cup\{\infty\}$ is a cusp of $\Gamma$. Then is an element $C\in\SL_2(\R)$ with $M_C(x_0)=\infty$ and $M_C\circ\Gamma_z\circ M_C^{-1}$ an infinite cyclic subgroup of $\SL_2(\R)_\infty$. \end{description} \end{lemma} \para Introduce a topology in $\H\cup\Cusp(\Gamma)$ by taking as a basis for the open sets the open sets of $\H$ together with all sets $D\cup\{z_0\}$ where $z_0\in\Cusp(\Gamma)$ and $D$ is an open disk in $\H$ whose boundary is tangent to the $\R\cup\{\infty\}$ at $z_0$. (In case $z_0=\infty$ this means a set of form $\Im(z)> c$.) Since M\"obius transformations map circles to circles it follows that for every $A\in\Gamma$ the map $M_A:\H\cup\Cusp(\Gamma)\to\H\cup\Cusp(\Gamma)$ is a homeomorphism of in this topology. \arap \begin{lemma} For every point $z\in\H\cup\Cusp(\Gamma)$ there is an open neighborhood $U$ of $z$ such that for $A\in\Gamma$ we have $$ A\in\Gamma_z\iff M_A(U)\cap U\ne\emptyset. $$ Such an open set $U$ is called an \jdef{open slice} centered at $z$. \end{lemma} \begin{lemma} \label{cor:localInvariant} Let $\Gamma\subset\SL_2(\R)$ be a discrete group and $z_0\in\H\cup\Cusp(\Gamma)$. Then there is a continuous function $\zeta:U\to\C$ defined in an open slice centered at $z_0$ which is holomorphic on $U\setminus\{z_0\}$ and such that for $z,z'\in U$ we have $$ \zeta(z')=\zeta(z)\iff z'\in\Gamma(z). $$ (In case $z_0\in\H$ the function $\zeta$ is holomorphic on $U$ since the singularity is removable.) \end{lemma} \para\label{def:localInvariant} A function $\zeta:\U\to\C$ as in Corollary~\ref{cor:localInvariant} is called a \jdef{local holomorphic invariant} for $\Gamma$ at $z_0$. The injective map from $U(\Gamma):=\{\Gamma(z):z\in U\}$ to $\C$ induced by $\zeta$ is called a \jdef{holomorphic coordinate} for $X(\Gamma)$ at $\Gamma(z_0)$. \arap \begin{theorem}\label{thm:X(Gamma)} Let $\Gamma\subset\SL_2(\R)$ be a discrete group. Then the various holomorphic coordinates for $X(\Gamma)$ form an atlas. This atlas equips $X(\Gamma)$ with the structure of a (Hausdorff) orbifold Riemann surface. \end{theorem} \xmpl By the Uniformization Theorem a compact Riemann surface of genus greater than one is isomorphic to some $X(\Gamma)$ where every element of $\Gamma$ is hyperbolic. \lpmx \xmpl\label{xmpl:Zh} Let $\Gamma$ be the cyclic subgroup of $\SL_2(\R)$ generated by $\Mat{cc}1&h\\0&1\Rix$ where $h\ne 0$. Then every point of $\H$ is regular, $\Cusp(\Gamma)=\{\infty\}$, and $\zeta(z)=e^{2\pi iz/h}$ is a local holomorphic invariant for $\Gamma$. The holomorphic coordinate induced by $\zeta$ on $X(\Gamma)$ maps $X(\Gamma)$ isomorphically to the unit disk. \lpmx \rmk If $\Gamma$ is a discrete subgroup of $\SL_2(\R)$ so is the group $\Gamma'$ generated by $\Gamma$ and $\pm I_2$. Since as $-I_2$ acts trivially, the group $\Gamma'$ has the same orbits as $\Gamma$ so $X(\Gamma')=X(\Gamma)$. In particular, any discrete subgroup of $\SL_2(\R)_\infty$ has an orbit space identical to an $X(\Gamma)$ as in Example~\ref{xmpl:Zh}. \kmr \xmpl Let $\Gamma=\SL_2(\Z)$. Then $\Cusp(\Gamma)=\Gamma(\infty)=\Q\cup\{\infty\}$. There are two elliptic orbits $\Gamma(i)$ and $\Gamma(e^{\pi i/3})$. The stabilizer subgroups at $\infty$, $i$, and $e^{\pi i/3}$ are generated by $$ A= \Mat{cc} 1& 1\\ 0 & 1\Rix, \qquad B= \Mat{cc} 0& -1\\ 1 & 0\Rix, \qquad AB= \Mat{cc} 1&-1\\ 1& 0\Rix $$ respectively. The element $A$ has infinite order, the element $B$ has order four, and the element $AB$ has order six. The transformation $M_A$ has infinite order, the transformation $M_B$ has order two, and the transformation $M_{AB}$ has order three. The function $z\mapsto e^{2\pi i z}$ is a local invariant at $\infty$, the function $z\mapsto ((z-i)/(z+i))^2$ is a local invariant at $i$, and the function $z\mapsto ((z-e^{\pi i/3})/(z-e^{-\pi i/3}))^3$ is a local invariant at $e^{\pi i/3}$. In Theorem~\ref{thm:X(SL2(Z))} below construct an isomorphism from $X(\Gamma)$ to projective space $\P$; more precisely, to weighted projective space $\P(2,4)$. \lpmx \section{Lattices} \para \label{def:lattice} A \jdef{lattice} is a subgroup of the additive group of $\C$ of form $$ \Lambda=\Z\omega_1+\Z\omega_2 $$ where $\omega_1,\omega_2\in\C$ are independent over $\R$, i.e. one of $\omega_1/\omega_2$ and $\omega_2/\omega_1$ lies in the upper half plane $\H$ and the other in the lower half plane. Choose the indexing so $\tau:=\omega_1/\omega_2\in\H$. Then the automorphism $z\mapsto z/omega_1$ of $\C$ carries the lattice $\Lambda$ to a lattice $$ \Lambda_\tau:=\Z+\Z\tau $$ where $\tau\in\H$. \arap \begin{lemma}\label{lem:lattice} For $\tau,\tau'\in\H$ the following are equivalent: \begin{description} \item[(i)] there exists $A\in\SL_2(\Z)$ with $\tau'=M_A(\tau)$; \item[(ii)] there is an automorphism $z\mapsto \alpha z+\beta$ of $\C$ with $\Lambda_\tau=\alpha\Lambda_{\tau'}+\beta$; \end{description} \end{lemma} \begin{proof} Assume~(i). Then $$ \Z(a\tau+b)+\Z(c\tau+d)\subset \Z+\Z\tau=\Lambda_\tau. $$ The automorphism $z\mapsto z/(c\tau+d)$ sends this $a\tau+b$ to $1$ and $1$ to $\tau'$ and hence $\Lambda_\tau$ to $\Lambda_{\tau'}$. Since $ad-bc=1$ interchanging $\tau$ and $\tau'$ constructs the inverse homomorphism. Conversely assume~(ii). Since $0\in\Lambda_\tau$ we have $\beta\in\Lambda_{\tau'}$ so $\Lambda_\tau=\alpha\Lambda_{\tau'}$ so $\alpha\tau'$ and $\alpha$ generates $\Lambda_\tau$. Hence there exist integers $a,b,c,d$ with $ad-bc=\pm1$, $\alpha\tau'=a\tau+b$, $\alpha=c\tau+d$ and hence $\tau'=(\alpha\tau')/\alpha=M_A(\tau')$. Since $\tau,\tau'\in\H$ we have $ad-bc=1$ (and not $-1$). \end{proof} \para Warning: Lemma~\ref{lem:lattice} says when the lattices are isomorphic, not when they are identical. The condition that $\tau'\in\Lambda_\tau$ implies that $\tau$ satisfies a quadratic equation with integer coefficients. There are only countably many such equations so for most $\tau$ we have $\Lambda_{\tau'}\ne\Lambda_\tau$ for {\em all} $A\in\SL_2(\Z)\setminus\{\pm I\}$. It is not hard to see when $\Lambda_{\tau'}=\Lambda_\tau$. Two lattices $\Lambda=\Z\omega_1+\Z\omega_2$ and $\Lambda'=\Z\omega_1'+\Z\omega_2'$ are equal if and only if there are integers $a,b,c,d\in\Z$ with $ad-bc=\pm1$ and $\omega'_1=a\omega_1+b\omega_2$, $\omega'_2=c\omega_1+d\omega_2$. Taking the cross product gives $\omega'_1\times\omega'_2=\pm\omega_1\times\omega_2$, i.e. the vectors $\omega_1$ and $\omega_2$ determine a parallelogram with the same area as the one determined by $\omega'_1$ and $\omega'_2$. Hence $\Lambda_\tau=\Lambda_{\tau'}$ implies that $\Im(\tau)=\Im(\tau')$. Now $\Lambda_\tau=\Lambda_{\tau'}\implies\Lambda_{M_B(\tau)}=\Lambda_{M_{B(\tau')}}$ for $B\in\SL_2(\Z)$ and $\Lambda_\tau=\Lambda_{t+1}$ so assume that $\tau$ lies in the fundamental region $-\frac12<\Re(\tau)\le\frac12$ and $|\tau|\ge1$ (see~\cite{SHIMURA} page~16) and that $-\frac12<\Re(\tau')\le\frac12$. Since $\Im(\tau)=\Im(\tau')$ it follows that $\tau=\tau'$ and hence that $\tau$ is an elliptic fixed point of $A$. The possibilities are $\tau=i$, $M_A=$ a power of $z\mapsto iz$ and $\tau=e^{\pi/6}$, $M_A=$ a power of $z\mapsto 1-1/z$. This discussion has the following \arap \begin{corollary} The lattice $\Lambda_\tau$ admits a nontrivial automorphism (i.e. one different from the two automorphisms $z\mapsto\pm z$) if and only if $\tau$ lies on one of the two elliptic orbits of $\SL_2(\Z)$. \end{corollary} \section{Elliptic Curves} \begin{theorem} Let $X$ be an \jdef{elliptic curve}, i.e. a compact Riemann surface of genus one. Then $X$ is isomorphic to $\C/\Lambda_\tau$ for some $\tau\in\H$. For $\tau,\tau'\in\H$, the elliptic curves $\C/\Lambda_\tau$ and $\C/\Lambda_{\tau'}$ are isomorphic if and only if $\tau$ and $\tau'$ lie in the same $\SL_2(\Z)$ orbit. \end{theorem} \begin{proof} We first show that the universal cover of $X$ is $\C$ and not the upper half plane $\H$. The group of holomorphic automorphisms of $\H$ is the same as the group of isometries of $\H$ so if $\H$ were the universal cover the group of deck transformations would act by isometries and there would be a metric of negative curvature on $X$. By the Gauss Bonnett theorem, the Euler characteristic of $X$ would be negative contradicting the hypothesis that $X$ has genus one. Hence $X=\C/\Gamma$ where $\Gamma$ is the group of deck transformations of the universal cover $\C\to X$. Every automorphism of $\C$ has the form $t\mapsto at+b$; every element of the subgroup $\Lambda$ is fixed point free and so has the form $t\mapsto t+b$. Thus the orbit $\Lambda$ of $0$ under $\Gamma$ is a subgroup of the additive group of $\C$ and so $X=\C/\Lambda$. We must show that $\Lambda$ has the desired form. By composing with a rotation we may assume w.l.o.g. that $\Lambda\cap\R\ne\{0\}$. Since $\Lambda$ is discrete, $\Lambda\cap\R$ must contain a smallest positive element so by rescaling we may assume w.l.o.g. that $\Lambda\cap\R=\Z$. We cannot have $\Lambda=\Z$, else $X=\C/\Lambda$ would be noncompact. Hence $\Lambda$ contains elements of the upper half plane. As $\Lambda$ is discrete, and as any element of $\Lambda\cap\H$ can be moved to the strip $\Im(\tau)>0$, $-1/2\le \Re(\tau)<1/2$ by translation by an element in $\Z\subset\Lambda$ there must be such a $\tau\in\Lambda\cap\H$ with $\Im(\tau)$ smallest. The parallelogram $P$ with vertices $0,1,\tau,1+\tau$ is a fundamental domain for the lattice $\Lambda_\tau\subset\Lambda$, so for any $t\in\C$ there is an element $\omega\in\Lambda_\tau$ with $t+\omega\in P$. In particular this is so for $t\in\Lambda$. By construction $t+\omega$ cannot lie in the interior of $P$ or in the interior of an edge of $P$. Hence $t+\omega$ is a vertex of $P$ so $t\in\Lambda_\tau$. Thus $\Lambda=\Lambda_\tau$ as required. Any isomorphism $\C/\Lambda_\tau\to\C/\Lambda_{\tau'}$ lifts to an automorphism of $\C$ which carries $\Lambda_\tau$ to $\Lambda_{\tau'}$. Hence, by Lemma~\ref{lem:lattice}, $\C/\Lambda_\tau$ and $\C/\Lambda_{\tau'}$ are isomorphic if and only if there are integers $a,b,c,d$ with $\tau'=(a\tau+b)/(c\tau+d)$ and $ad-bc=1$. \end{proof} \para Define an action of $\Z^2$ on $\H\times\C$ by $$ T_{(m,n)}(\tau,t)=(\tau, t+m\tau+n) $$ for $(m,n)\in\Z^2$. This action maps each fiber of the projection $\H\times\C\to\H$ to itself. Introduce the space $$ W:= (\H\times\C)/\Z^2 $$ and the projection $W\to\H:(\tau,t+\Lambda_\tau)\mapsto\tau$. The fiber over $\tau$ of this projection is the corresponding elliptic curve $$ W_\tau=\C/\Lambda_\tau. $$ \arap \para Define an action of $\SL_2(\Z)$ on $\H\times\C$ by by $$ \Phi_A(\tau,t)=\left(\frac{a\tau+b}{c\tau+d},\frac{t}{c\tau+d}\right), \qquad A=\Mat{cc} a& b\\ c& d\Rix\in\SL_2(\Z). $$ Then the diagram $$ \begin{array}{ccc} \H\times\C& \mapright{\Phi_A}&\H\times\C\\ \mapdown{} && \mapdown{} \\ \H & \mapright{M_A} & \H \end{array} $$ commutes (the vertical arrows are projection on the first factor). Note that the action $A\mapsto\Phi_A$ (unlike the action $A\mapsto M_A$) is effective: $$ \Phi_{-I}(\tau,t) = (\tau, -t). $$ \arap \begin{lemma} For $A\in\SL_2(\Z)$ and $(m,n)\in\Z^2$ we have $$ \Phi_A\circ T_{(m,n)}= T_{\mu(m,n,A)}\circ \Phi_A $$ where $\mu(m,n,A)\in\Z^2$ is given by $$ \mu(m,n,A)= (m,n)A^{-1}=(dm-bn,-cm+an). $$ \end{lemma} \begin{corollary} The action $A\mapsto\Phi_A$ of $\SL_2(\Z)$ on $\H\times\C$ induces an action (denoted by the same symbol) of $\SL_2(\Z)$ on $W$; the projection $W\to\H$ is equivariant. In other words, the diagram $$ \begin{array}{ccc} \H\times\C& \mapright{\Phi_A}&\H\times\C\\ \mapdown{} && \mapdown{} \\ W & \mapright{\Phi_A} & W\\ \mapdown{} && \mapdown{} \\ \H & \mapright{M_A} & \H \end{array} $$ commutes. \end{corollary} \para The stabilizer groups of the action on $\H$ are all finite: a point $\tau\in\H$ has nontrivial stabilizer if and only if it lies on one of the two orbits of elliptic fixed points. \arap \section{Cubic Curves} \para \label{def:cP} For each lattice $\Lambda\subset\C$ define the \jdef{Weierstrass $\cP$ function} by the $$ \cP_\Lambda(t) = \frac{1}{t^2}+\sum_{\omega\in\Lambda\setminus\{0\}} \left\{ \frac{1}{(t+\omega)^2} - \frac{1}{\omega^2}\right\}. $$ In case $\Lambda=\Lambda_\tau$ we write $\cP_\tau$ for $\cP_\Lambda$. Define $$ \cP_\Z(t)= \frac{1}{t^2}+\sum_{m\in\Z\setminus\{0\}} \left\{\frac{1}{(t+m)^2} - \frac{1}{m^2}\right\}. $$ The following facts are not hard to prove: \begin{description} \item[(i)] The series defining $\cP_\Lambda$ and $\cP_\Z$ converge uniformly on compact subsets to holomorphic maps $\C\to\P$. \item[(ii)] Hence the derivatives are given by $$ \cP_\Lambda'(t)= - \sum_{\omega\in\Lambda}\frac{2}{(t+\omega)^3},\qquad \cP_\Z'(t)= - \sum_{m\in\Z}\frac{2}{(t+m)^3}. $$ \item[(iii)] The limit $$ \lim_{\tau\to i\infty} \cP_\tau(t)=\cP_\Z(t) $$ holds uniformly on compact sets, i.e. for every neighborhood $\cU$ of $\cP_\Z$ in the compact open topology on $C^0(\C,\P)$ and every $a>0$ there exists $T>0$ such that $ \cP_\tau\in\cU$ for $\Im(\tau)>T$ and $-a\le\Re(\tau)\le a$. \item[(iv)] The functions $\cP_\Z$ and $\cP_\Lambda$ are respectively periodic and doubly periodic in the sense that $$ \cP_\Z(t+m)=\cP_\Z(t), \qquad \cP_\Lambda(t+\omega)=\cP_\Lambda(t) $$ for $m\in\Z$ and $\omega\in\Lambda$. (It is obvious that the derivatives $\cP_Z'$ and $\cP_\Lambda'$ satisfy these relations.) \end{description} \arap \begin{lemma}\label{lem:cubicEqn} The functions $x=\cP$ and $y=\cP'$ satisfy a cubic equation $$ y^2=4x^3-g_2x-g_3. $$ (Here $\cP$ is either $\cP_\Lambda$ or $\cP_\Z$.) \end{lemma} \begin{proof} By Laurent expansion \begin{eqnarray*} \cP(t)&=&\frac{1}{t^2}+at^2+bt^4+O(t^6)\\ \cP'(t)&=&-\frac{2}{t^3}+2at+4bt^3+O(t^5)\\ \cP(t)^3&=&\frac{1}{t^6}+\frac{3a}{t^2}+3b+O(t^2)\\ \cP'(t)^2&=&\frac{4}{t^6}-\frac{8a}{t^2}-16b+O(t) \end{eqnarray*} so $\cP'(t)^2-4\cP(t)^3+20a\cP+28b=O(t)$. This function is doubly periodic, has no pole, and vanishes at the origin. Hence it vanishes identically. Take $g_2=20a$ and $g_3=28b$. \end{proof} \para \label{g2g3} To evaluate $g_2$ and $g_3$ calculate the Taylor expansion of $F(t)=\cP(t)-t^{-2}$. Then $F''(0)=6\sum_{\omega\in\Lambda\setminus\{0\}}\omega^{-4}$ and $F^{(4)}(0)=120\sum_{\omega\in\Lambda\setminus\{0\}}\omega^{-6}$ so $$ g_2=60\sum_{\omega\in\Lambda\setminus\{0\}}\frac{1}{\omega^4},\qquad g_3=140\sum_{\omega\in\Lambda\setminus\{0\}}\frac{1}{\omega^6}. $$ If $\Lambda$ is replaced by $\Z$ this becomes $g_2=120\,\zeta(4)$ and $g_3=280\,\zeta(6)$ where $\zeta(s):=\sum_{n=1}^\infty n^{-s}$ is the Riemann zeta function. It is easy to see that functions $$ g_2(\tau):=g_2(\Lambda_\tau),\qquad g_3(\tau):=g_3(\Lambda_\tau) $$ are \jdef{modular forms} of weights four and six respectively, i.e. $$ g_2(M_A(\tau))=(c\tau+d)^4g_2(\tau),\qquad g_3(M_A(\tau))=(c\tau+d)^6g_3(\tau) $$ for $\tau\in\H$ and $A\in\SL_2(\Z)$ as in paragraph~\ref{notation:mobius}. We extend $g_2$ and $g_3$ to $\H\cup\{\infty\}$ via $$ g_2(\infty):=g_2(\Z),\qquad g_3(\infty):=g_3(\Z). $$ \arap \para\label{cPZ} An explicit formula for the map $\cP_\Z$ is $$ \cP_\Z(t) = \pi^2\csc^2(\pi t)-\frac{\pi^2}{3}. $$ This follows from termwise differentiation of the series $$ \pi\cot(\pi t) = S(t):=\frac{1}{t} + \sum_{n\in\Z\setminus\{0\}} \left(\frac{1}{t+m}-\frac{1}{m}\right) $$ together with Euler's formula $$ \sum_{m=1}^\infty \frac{1}{m^2} = \frac{\pi^2}{6}. $$ To prove that $\pi\cot(\pi t)=S(t)$ note that both have the same poles with the same residues and that both are odd and have period one. This means that the difference $\pi\cot(\pi t)-S(t)$ is bounded in a strip about the real axis and, as $\cot(w)=i(e^{iw}+e^{-iw})/(e^{iw}-e^{-iw})$, both are bounded on the complement of this strip. Thus the difference $\pi\cot(\pi t)-S(t)=c$ a constant. To see that $c=0$ is zero evaluate at $t=1/2$: We get $S(1/2)=S(1/2-1)=S(-1/2)=-S(1/2)$ by periodicity and oddness so $S(1/2)=0=\cot(\pi/2)$ so $c=0$. To prove Euler's formula calculate the Fourier series $\theta=\sum_n c_ne^{in\theta}$ and use Parseval's equality. \arap \para\label{degree-cP-L} Since $\cP_\Lambda$ has a pole of order two at the origin, the map $\C/\Lambda\to\P$ induced by $\cP_\Lambda$ has degree two and the origin is a critical point. The other critical points are the zeros of the map $\C/\Lambda\to\P$ induced by $\cP_\Lambda'$. This map has degree three (because it has a unique pole of order three) so $\cP_\Lambda'$ has at most three zeros modulo $\Lambda$. Let $\omega_1$ and $\omega_2$ be generators of $\Lambda$. The half periods $\omega_1/2$, $\omega_2/2$ and $(\omega_1+\omega_2)/2$ satisfy $t=-t\mod\Lambda$ and hence are zeros of the odd doubly periodic function $\cP'_\Lambda$. \arap \para \label{degree-cP-Z} Similarly the map $\C/\Z\to\P$ induced by $\cP_\Z$ has degree two. To prove this use the identities $\csc^2(\pi t)=\cot^2(\pi t)+1$ and $$ \cot(\pi t) = -i\frac{e^{\pi i t}+e^{-\pi i t}}{e^{\pi i t}-e^{-\pi i t}} =-i\frac{q+1}{q-1} $$ where $q=e^{2\pi i t}$. By paragraph~\ref{cPZ} we have $$ \cP_\Z(t)= -\pi^2\left(\frac{q+1}{q-1}\right)^2+\frac{2\pi^2}{3}, \qquad \cP_\Z'(t)=2i\pi^3\left(\frac{q+1}{q-1}\right)^2\frac{q+1}{q-1} $$ where we have used $\cP_\Z'(t)=-2\pi^3\csc^2(\pi t)\cot(\pi t)$ in the second formula. The formula for $\cP_\Z'(t)$ shows that $t$ is a critical point of $\cP_\Z$ if and only if $t=n+\frac12$ where $n\in\Z$. \arap \para \label{def:Xtau} For $\tau\in\H\cup\{\infty\}$ denote by $X_\tau\subset\P^2$ the cubic curve $$ X_\tau:=\{[x,y,z]\in\P^2: y^2z=4x^3-g_2(\tau)xz^2-g_3(\tau)z^3\}. $$ Define projections $Q_\tau:\C\to\C/\Lambda_\tau$ and $Q_\infty:\C\to\P\setminus\{0,\infty\}$ by $$ Q_\tau(t)=t+\Lambda_\tau, \qquad Q_\infty(t)=e^{2\pi i t}. $$ For $\tau\ne\infty$ define $F_\tau:\C/\Lambda_\tau\to\P^2$ by $$ F_\tau(Q_\tau(t))=[\cP_\tau(t),\cP_\tau'(t),1] $$ for $t\ne 0$ with $F_\tau(0)=[0,1,0]$. Define $F_\infty:\P\to\P^2$ by $$ F_\infty(Q_\infty(t)) = [\cP_\Z(t),\cP_\Z'(t),1] $$ with $F_\infty(0)=F_\infty(\infty)=[0,1,0]$. \arap \begin{theorem} (i)~For $\tau\ne\infty$ the map $F_\tau:\C/\Lambda_\tau\to\P^2$ is a holomorphic embedding with image $X_\tau$. (ii)~The map $F_\infty:\P\to\P^2$ is a holomorphic immersion, injective except for a single double point, with image $X_\infty$. \end{theorem} \begin{proof} Lemma~\ref{lem:cubicEqn} says that the image of $F_\tau$ is a subset of $X_\tau$. If $F_\tau$ is not an immersion at some point $q=Q_\tau(t_0)$ where $t_0\in\C\setminus\Lambda_\tau$, then $\cP'(t_0)=\cP''(t_0)=0$ so $\cP-\cP(t_0)$ has a zero of order three at $t_0$ contradicting the fact that the map induced by $\cP$ has degree two. Near $0$ the map $F_\tau\circ Q_\tau$ has the form $F_\tau\circ Q_\tau(t)=[t+O(t^2),-2+O(t),t^3]$ which shows that $F_\tau$ is an automorphism at the points $q\in Q_\tau(\Lambda_\tau)$ as well. The only case not covered by these arguments is the case $\tau=\infty$ and $q=\infty$. That $F_\infty$ is an immersion at this point follows from the symmetry $F_\infty(q^{-1})=T\circ F_\infty(q)$ where $T([x,y,z])=[x,-y,z]$. The map $\C/\Lambda_\tau\to\P$ induced by $\cP$ has degree two and its branch points are the (projections of the) half periods of $\Lambda_\tau$ which are not periods (see paragraphs~\ref{degree-cP-L} and~\ref{degree-cP-Z}). In particular, $\cP(t_1)=\cP(t_2)$ implies that $t_1=\pm t_2$ modulo $\Lambda_\tau$. Since the map $\C/\Lambda_\tau\to\P$ is surjective and $\cP'$ is odd it follows from Lemma~\ref{lem:cubicEqn} that the image of $F_\tau$ is $X_\tau$. For the injectivity properties of $F_\tau$ note first that the preimage of $[0,1,0]$ consists of one point if $\tau\in\H$ and exactly two points if $\tau=\infty$ and that moreover $[0,1,0]$ is the only point at which the image of $F_\tau$ intersects the line at infinity. Hence it suffices to prove that the restriction of $F_\tau$ to $\C/\Lambda_\tau\setminus\{0\}$ is injective. Hence we assume that $\cP(t_1)=\cP(t_2)$, $\cP'(t_1)=\cP'(t_2)$, $t_1,t_2\in\C\setminus\Lambda_\tau$ and prove that $t_1=t_2$ modulo $\Lambda_\tau$. If not, then $t_1=-t_2$ modulo $\Lambda_\tau$ so, as $\cP'$ is odd, $\cP'(t_1)=\cP'(t_2)=0$. Hence $t_1$ and $t_2$ are branch points of $\cP$. But the branch points of $\cP$ are half periods of $\Lambda_\tau$ and two half periods $t_1$ and $t_2$ which are distinct modulo $\Lambda_\tau$ cannot be negatives of one another modulo $\Lambda_\tau$. \end{proof} \begin{corollary} For $\tau\in\H\cup\{\infty\}$ the discriminant $g_2(\tau)-27g_3(\tau)$ vanishes if and only if $\tau=\infty$. \end{corollary} \begin{proof} The discriminant vanishes precisely when the polynomial $4x^3-g_2x-g_3$ has a double root and this occurs precisely when $X_\tau$ is not a submanifold. \end{proof} \section{A Projective Embedding} \begin{theorem} \label{thm:X(SL2(Z))} For $\Gamma=\SL_2(\Z)$ the Riemann surface $X(\Gamma)$ defined in Theorem~\ref{thm:X(Gamma)} is isomorphic to the Riemann sphere $\P$. \end{theorem} \begin{lemma}\label{lem:J} There is a holomorphic map $J:\H\to\P$ such that for $\tau,\tau'\in\H$ we have $J(\tau)=J(\tau')$ if and only if $\tau$ and $\tau'$ lie in the same $\SL_2(\Z)$ orbit. Thus $J$ induces a bijection $\H/\SL_2(\Z)\to\C=\P\setminus\{\infty\}$. \end{lemma} \begin{proof} The \jdef{cross ratio} $(e_1,e_2,e_3,e_4)$ of four distinct points $e_1,e_2,e_3,e_4$ of $\C$ is defined by $$ (e_1,e_2,e_3,e_4)= \frac{e_1-e_2}{e_1-e_3}\cdot\frac{e_4-e_2}{e_4-e_3}; $$ the definition extends to four distinct points of $\P=\C\cup\{\infty\}$ by continuity. The symmetric group $\Sigma_4$ permutes the four numbers $e_i$ and permutes their cross ratios accordingly: $$ \Sigma_4(\lambda)= \{\lambda,1/\lambda,1-\lambda,1/(1-\lambda), \lambda/(\lambda-1), (\lambda-1)/\lambda\}. $$ It is easy to check that the polynomial $$ S(\lambda)=\frac{(\lambda^2-\lambda+1)^3}{\lambda^2(1-\lambda)^2} $$ has the property that $S(\lambda)=S(\lambda')$ if and only if $\lambda'\in\Sigma_4(\lambda)$. For $\tau\in\H$ the four branch points of the map $\C/\Lambda_\tau\to\P$ induced by $\cP_\tau$ are $e_4=\infty$ and $$ e_1=\cP_\tau(1/2), \qquad e_2=\cP_\tau(\tau/2),\qquad e_3=\cP_\tau((1+\tau)/2) $$ (see paragraph~\ref{degree-cP-L}). Let $\lambda$ be the cross ratio of the four branch points so $\lambda = (e_1-e_2)/(e_1-e_3)$ and then define $J(\tau)= 4 S(\lambda)/27$. (The factor of $4/27$ is traditional.) \end{proof} \rmk Since $e_1,e_2,e_3$ are the zeros of the polynomial $4x^3-g_2x-g_3$ we have $$ e_1+e_2+e_3=0, \qquad e_1e_2+e_2e_3+e_3e_1=-\frac{g_2}{4},\qquad e_1e_2e_3=\frac{g_3}{4}. $$ It follows easily that $$ J(\tau)=\frac{g_2(\tau)^3}{g_2(\tau)^3-27g_3(\tau)^2}. $$ \kmr \para Recall the cubic curve $X_\tau$ defined in paragraph~\ref{def:Xtau} for $\tau\in\H\cup\{\infty\}$. Define $$ V:=\{([x,y,z],\tau)\in\P^2\times\H\cup\{\infty\}: [x,y,z]\in X_\tau\} $$ and subsets $$ V_1:=\{([x,y,z],\tau)\in V:\tau\in\H\},\qquad V_2:=\{([x,y,z],\tau)\in V:\tau\in U\cup\{\infty\}\} $$ where $U$ is as in Lemma~\ref{lem:U}. \arap \begin{lemma} For $A\in\SL_2(\Z)$ and $([x,y,z],\tau)\in V_1$ the point $$ R_A([x,y,z],\tau):=([(c\tau+d)^2x,(c\tau+d)^3y,z],M_A(\tau)) $$ (see paragraph~\ref{g2g3}) also lies in $V_1$. For $A\in\SL_2(\Z)_\infty$ and $([x,y,z],\tau)\in V_2$ the point $$ R_A([x,y,z],\tau):=([x,ay,z],\tau+n) $$ also lies in $V_2$. (See Lemma~\ref{lem:U}.) \end{lemma} \begin{proof} \end{proof} \section{A Projective Model} \begin{lemma} Let $B=\C^2\setminus\{0\}$ and $W$ be the set of all pairs $([x,y,z],a,b)$ in $\P^2\times B$ such that $$ y^2z=4x^3-axz^2-bz^3. $$ Then \begin{description} \item[(i)] The set $W$ is a complex submanifold of $\P^2\times B$. \item[(ii)] The set of critical values of the projection $W\to B$ onto the second factor is the zero set of the discriminant $a^3-27b^2$ of the polynomial $4x^3-ax-b$. \item[(iii)] Over each critical value $(a,b)$ there is a unique critical point namely the point $[x_0,0,1]$ where $x_0$ is the double root of the polynomial $4x^3-ax-b$. \end{description} \end{lemma} \para Define actions of the complex multiplicative group $\C^*$ on $B$ and $W$, by $$ \lambda_B(a,b)=(\lambda^4a,\lambda^6b), \qquad \lambda_W([x,y,z],(a,b))=([\lambda^2x,\lambda^3y,z], \lambda_B(a,b)). $$ so the projection $W\to B$ is equivariant. It is easy to see that the set of critical values of the projection $W\to B$ form a single orbit of the action of $\C^*$ on $B$. Define $\Phi:V\to W$ by $$ \Phi([x,y,z],\tau)=([x,y,z],(g_2(\tau),g_3(\tau))). $$ \arap \begin{lemma} The map $\Phi|V_1$ induces a bijection from the $\SL_2(\Z)$ orbits of $V_1$ onto the $\C^*$ orbits of the set of regular points of the projection $W\to B$. The map $\Phi|V_2$ induces a bijection from the $\Z$ orbits of $V_2$ onto the $\C^*$ orbits of a neighborhood of the set of critical points of the projection $W\to B$. \end{lemma} \begin{thebibliography}{99} \bibitem{MUMFORD} D. Mumford: {\em Tata lectures on theta}, Progress in mathematics {\bf 28, 43, 97} Birkhäuser, 1983-1991. \bibitem{REYSSAT} E. Reyssat: {\em Quelques Aspects des Surface de Riemann}, Progress in mathematics {\bf 77}, Birkh\"auser, 1989. \bibitem{SHIMURA} G. Shimura: {\em Introduction to the Theory of Automorphic Functions}, Princeton University Press, 1971. \end{thebibliography} \end{document} The quotient $$ \P(4,6)=B/\C^* $$ is called the \jdef{weighted projective space} with weights $4,6$. Let $$ M=W/\C^* $$ denote the orbit space of $W$ under this action. \begin{theorem} Every elliptic curve is isomorphic to a regular fiber $$ W_{a,b}:=\{[x,y,z]\in\P^2: y^2z=4x^3-ax-b\} $$ of the projection $W\to B$. Two fibers $ W_{a,b}$ and $W_{a',b'}$ are isomorphic as elliptic curves if and only if $(a,b)$ and $(a',b')$ lie in the same $\C^*$ orbit. \end{theorem} \begin{lemma}\label{lem:U} Define $U:=\{\tau\in\H:\Im(\tau)>1\}$. Then for $\tau,\tau'\in U$ we have that $\tau$ and $\tau'$ lie in the same $\SL_2(\Z)$ orbit if and only if they lie in the same $\SL_2(\Z)_\infty$ orbit, i.e. if $M_A(\tau)=\tau'$ where $A\in\SL_2(\Z)$ then $A$ has form $$ A = \Mat{cc} a& b\\ 0 & a\Rix $$ where $a=\pm 1$ and $b\in\Z$. \end{lemma} \begin{proof} This follows immediately from the fact that the intersection of the exterior $|z|>1$ of the unit circle with the strip $-\frac12\le\Re(\tau)\le\frac12$ is a fundamental domain for the action of $\SL_2(\Z)$ on $\H$. \end{proof}