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\title{Elliptic Curves}
\author{JWR}
\date{Friday November 30, 2001,7:40 AM}


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\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}
