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\global\long\def\ilim{\varprojlim}



\title{Modular Curves}


\author{David Loeffler\\
Notes by Florian Bouyer}

\maketitle
\begin{center}
Copyright (C) Bouyer 2014. 
\par\end{center}

\begin{center}
Permission is granted to copy, distribute and/or modify this document 
\par\end{center}

\begin{center}
under the terms of the GNU Free Documentation License, Version 1.3 
\par\end{center}

\begin{center}
or any later version published by the Free Software Foundation; 
\par\end{center}

\begin{center}
with no Invariant Sections, no Front-Cover Texts, and no Back-Cover
Texts. 
\par\end{center}

\begin{center}
A copy of the license can be found at \href{http://www.gnu.org/licenses/fdl.html}{http://www.gnu.org/licenses/fdl.html}
\par\end{center}

\tableofcontents{}

\newpage{}

\textbf{Why study modular curves?}
\begin{enumerate}
\item They give more information on modular forms.
\item They give friendly examples of moduli spaces.
\end{enumerate}

\section{Waffle}

Let $\mathcal{H}=\{z\in\cc|\im z>0\}$. We have that the group $\SL_{2}(\rr)$
acts on $\mathcal{H}$. Take $\Gamma<\SL_{2}(\zz)$ of finite index.
We define $Y(\Gamma)=\Gamma\backslash\mathcal{H}$. We will equip
this with various interesting structures. 


\subsection{Recap of modular forms}

Fix:
\begin{itemize}
\item $\Gamma\leq\SL_{2}(\zz)$ with finite index (which we call the \emph{level})
\item $k\in\zz$ (which we call the \emph{weight})
\end{itemize}
Then there exists a space $M_{K}(\Gamma)$ of \emph{modular forms}
which are functions $f:\mathcal{H}\to\cc$ holomorphic, such that
\[
f\left(\frac{az+b}{cz+d}\right)=(cz+d)^{k}f(z)\,\,\forall\begin{pmatrix}a & b\\
c & d
\end{pmatrix}\in\Gamma,
\]
and a growth condition on the boundary.

There exists a subspace $S_{k}(\Gamma)\subset M_{K}(\Gamma)$ of \emph{cusp
forms}. 
\begin{fact*}[Basic]
 Both $M_{k}$ and $S_{k}$ are finite dimensional over $\cc$. 
\end{fact*}
Any modular form has a $q$\emph{-expansion, }let $h$ be the least
integer such that $\begin{pmatrix}1 & h\\
0 & 1
\end{pmatrix}\in\Gamma$, then 
\[
f(z)=\sum_{n\geq0}a_{n}q_{h}^{n},\, a_{n}\in\cc,\, q_{h}=e^{2\pi iz/h}.
\]


\newpage{}


\section{Modular Curves as Riemann Surfaces}


\subsection{Modular curves as topological spaces}

$\mathcal{H}$ has a topology (obviously) so $Y(\Gamma)$ gets a quotient
topology (i.e., strongest topology such that $\pi:\mathcal{H}\to\Gamma(Y)$
is continuous). Quotient topology can be pretty nasty (for example
$\qq$ acting on $\rr$ by translation) - quotient can even be the
indiscrete topology. 
\begin{prop}
\label{prop:1.1.1}For any $\tau_{1},\tau_{2}\in\mathcal{H}$ there
exists a neighbourhood $U_{1}\ni\tau_{1},U_{2}\ni\tau_{2}$ such that
if $\gamma\in\SL_{2}(\zz)$ satisfies $\gamma(U_{1})\cap U_{2}\neq\emptyset$,
then $\gamma(\tau_{1})=\tau_{2}$. \end{prop}
\begin{proof}
See Proposition 2.1.1 of Diamond and Shurman. 
\end{proof}
We say that $\SL_{2}(\zz)$ acts \emph{properly discontinuously on
$\mathcal{H}$.}
\begin{cor}
\label{cor:1.1.2}$Y(\Gamma)$ is Hausdorff\end{cor}
\begin{proof}
Let $P_{1}\neq P_{2}$ be two points of $Y(\Gamma)$. Choose, $\tau_{1},\tau_{2}\in\mathcal{H}$
be lifting of $P_{1},P_{2}$ respectively. Let $U_{1},U_{2}$ be neighbourhoods
of $\tau_{1},\tau_{2}$ as in Proposition \ref{prop:1.1.1}. We claim
that $V_{i}=\pi(U_{i})$ are open neighbourhoods of $P_{i}$ such
that $V_{1}\cap V_{2}=\emptyset$.

Suppose $V_{1}\cap V_{2}\neq\emptyset$, then $\pi^{-1}(V_{1})\cap\pi^{-1}(V_{2})\neq\emptyset$.
Hence $\cup_{\gamma\in\Gamma}\gamma U_{1}\cap\cup_{\gamma'\in\Gamma}\gamma'U_{2}\neq\emptyset$.
So there exists $\gamma$ and $\gamma'$ such that $\gamma U_{1}\cap\gamma'U_{2}\neq\emptyset$.
This gives $(\gamma')^{-1}\gamma U_{1}\cap U_{2}\neq\emptyset$, which
by Proposition \ref{prop:1.1.1} means $(\gamma')^{-1}\gamma\tau_{1}=\tau_{2}$.
This is a contradiction to the assumption that $P_{1}$ and $P_{2}$
are distinct point (hence $\tau_{1},\tau_{2}$ are in different orbits)
\end{proof}
We are also interested in a slightly larger space $X(\Gamma)$, which
is a compactification of $Y(\Gamma)$. As a set, we have 
\[
X(\Gamma)=Y(\Gamma)\cup\underset{\mathrm{"cusps"\, of\,}\Gamma}{\underbrace{C(\Gamma)}}
\]
\[
C(\Gamma)=\Gamma\backslash\pp^{1}(\qq).
\]
Let $\mathcal{H}^{*}=\mathcal{H}\cup\pp^{1}(\qq)$. Give $\mathcal{H}^{*}$
a topology by extending the topology we have on $\mathcal{H}$. 

\includegraphics[scale=0.5]{Modular_Curves_Pic1}

If we define neighbourhoods of $\infty$ as $\{z:\im z>R\}$ for any
$R\in\nn$. Then so that the topology make sense, we defined neighbourhood
of $x\in\qq$ as circles tangent to $\rr$ at $x.$ Actions on $\mathcal{H}^{*}$
are still properly discontinuous so $X(\Gamma)$ is Hausdorff.
\begin{prop}
\label{prop:1.1.3}$X(\Gamma)$ is compact.\end{prop}
\begin{proof}
It suffices to find a compact subset of $\mathcal{H}^{*}$ mapping
surjectively to $X(\Gamma$). Let $D^{*}=\{\infty\}\cup\{z\in\mathcal{H}:\frac{1}{2}\leq\re z\leq\frac{1}{2},\left|z\right|\geq1\}$.
A standard fact is that $D^{*}$ contains a point of every $\SL_{2}(\zz)$
orbit on $\mathcal{H}^{*}$. So if $\gamma_{1},\dots,\gamma_{n}$
are coset representative for $\Gamma\setminus\SL_{2}(\zz)$, then
$\cup_{i=1}^{n}\gamma_{i}D^{*}$ surjects onto $X(\Gamma)$. It is
easy to check that $D^{*}$ is compact, hence we are done.
\end{proof}

\subsection{Riemann surfaces: recap}
\begin{defn}
\label{1.2.1}A \emph{Riemann Surface} consists of the following data:
\begin{itemize}
\item A topological space $X$, (which is Hausdorff and second - countable)
\item A collection $(U_{i},V_{i},\phi_{i})_{i\in I}$, where $V_{i}\subset X$
are opens forming a cover of $X$, $U_{i}$ are opens in $\cc$ and
$\phi_{i}:U_{i}\to V_{i}$ are homeomorphism, such that if $V_{i}\cap V_{j}\neq\emptyset$,
the map $\phi_{j}^{-1}\circ\phi_{i}:U_{i}\cap\phi_{i}^{-1}(V_{i}\cap V_{j})\to U_{j}\cap\phi_{j}^{-1}(V_{i}\cap V_{j})$
is holomorphic. 
\end{itemize}
\end{defn}
Roughly: A Riemann surface is the least amount of structure on $X$
needed to make sense of a function $X\to\cc$ being holomorphic.

We will now show that $Y(\Gamma)$ and $X(\Gamma)$ have \uline{natural
Riemann surface structures}. 
\begin{defn}
\label{1.2.2}We say $P\in Y(\Gamma)$ is an \emph{elliptic point}
if for some (and hence any) $\tau\in\mathcal{H}$ lifting of $P_{1}$,
$\mbox{Stab}_{\overline{\Gamma}}(\tau)\neq\{1\}$, where $\overline{\Gamma}$
is the image of $\Gamma$ in $\mbox{PSL}_{2}(\zz)$ (that is $\Gamma/\Gamma\cap\{\pm1\}$)
\end{defn}
If $P$ is elliptic for $\Gamma$, then it maps to an elliptic point
of $Y(\SL_{2}(\zz))$. One can see that there are only $2$ of these
(the orbits of $i$ and $\rho=e^{2\pi i/3}$). 

If $P$ is not elliptic or a cusp, we can easily find a chart around
$P$. Let $\tau$ be a lifting of $P$ and apply Proposition \ref{prop:1.1.1}
with $\tau_{1}=\tau_{2}=\tau$. Let $U=U_{1}\cap U_{2}$, then $U$
a neighbourhood of $\tau$ such that $\gamma U\cap U=\emptyset$ for
any $\gamma\neq1\in\overline{\Gamma}$. Let $V$ be the image of $U$
in $Y(\Gamma)$, then $\phi=\pi/U$ is a homeomorphism $U\overset{\sim}{\to}V$. 

If $P$ is a elliptic, need to be a bit cleverer. Proposition \ref{prop:1.1.1}
gives us a $U\ni\tau$ such that $U\cap\gamma U\neq\emptyset$ if
and only if $\gamma\in\mbox{Stab}_{\overline{\Gamma}}(\tau)$ (which
has order $2$ or $3$). Choose $\delta\in\SL_{2}(\cc)$ shifting
$\tau$ to $0$ and $\overline{\tau}$ to $\infty$. Then $\mbox{Stab}_{\overline{\Gamma}}(\tau)$
goes to a finite cyclic group of Mobius transformations fixing $0$
and $\infty$, hence multiplication by $e^{2\pi i/n}$ with $n=2$
or $3$. Define the map $\mbox{Stab}_{\delta\overline{\Gamma}\delta^{-1}}(0)\backslash\delta U\to U'$
by $z\mapsto z^{n}$, then this is a bijection. We define the coordinate
chart to be the map $U'\to V$ as defined in the following map 
\[
\xymatrix{\mbox{Stab}_{\delta\overline{\Gamma}\delta^{-1}}(0)\backslash\delta U\ar[r]\ar[d]^{\sim} & U'\ar[dl]^{\sim}\\
V\subset Y(\Gamma)
}
\]


Lastly, if $P$ is a cusp, we argue similarly. Choose $\delta$ mapping
$P$ to $\infty$, $\mbox{Stab}_{\delta\overline{\Gamma}\delta^{-1}}(\infty)$
is a group of \uline{translation}, and $z\mapsto e^{2\pi iz/h}$
gives a local coordinates. 

We have just proved that there exists a Riemann surface structure
on $Y(\Gamma)$ and $X(\Gamma)$ such that $\pi:\mathcal{H}\to Y(\Gamma)$
is holomorphic. (Clearly the unique such structure).


\subsection{Genus, ramification, Riemann-Hurwitz}
\begin{fact*}
Riemann surfaces are smooth orientable $2$-manifolds, and there are
not many of these. Compacts connected ones all look like doughnuts
with $g$ holes.

\includegraphics[scale=0.5]{Modular_Curves_Pic2}\end{fact*}
\begin{defn}
\label{1.3.1}We define \emph{genus} of a compact $2$-manifold $M$,
as the unique integer $g=g(M)$ such that $H^{1}(M,\zz)\cong\zz^{2g}$. 
\end{defn}
The genus is closely related to the \emph{Euler Characteristic, $\chi(M)=\sum_{i\geq0}(-1)^{i}\rk H^{i}(M,\zz)$.
If $M$ is as in the Fact, then $H^{0}\cong H^{2}\cong\zz$ and $H^{i}=0$
for $i\geq3$, so $\chi(M)=2-2g$. }
\begin{prop}
\label{prop:1.3.2}For $\Gamma=\SL_{2}(\zz)$, the space $X(\Gamma)$
is isomorphic (as a Riemann surface, so in particular as $2$-manifold)
to $\pp^{1}(\cc)\cong S^{2}$. \end{prop}
\begin{proof}
The ``$j$-invariant'' $j(z)=q^{-1}+744+196884q+\dots$ is $\SL_{2}(\zz)$
invariant and descend to a holomorphic map $X(\SL_{2}\zz)\to\pp^{1}(\cc)$.
It is \uline{bijective} (counting zeroes using contour integration)
so it has a holomorphic inverse.
\end{proof}
\textbf{Convention:} Unless otherwise stated all Riemann surfaces
are assumed to be \uline{connected}. 

\textbf{Recap}: We want to find $g(X(\Gamma))$ for all $\Gamma$.
We know that $X(\SL_{2}(\zz))$ has genus $0$. We have that for all
$\Gamma$, there is a map $X(\Gamma)\twoheadrightarrow X(\SL_{2}(\zz))$
\begin{defn}
\label{1.3.3} ~
\begin{enumerate}
\item For $f:X\to Y$ a non-constant morphism, $P\in X$, the \emph{ramification
degree, }$e_{P}(f)$, is the unique integer $e\geq1$ such that $f$
``looks like $z\mapsto z^{e}$ locally''. Note: points such that
$e_{P}(f)>1$ are \emph{isolated.}
\item If $X$ and $Y$ are compact, the sum $\sum_{P\in f^{-1}(Q)}e_{p}(f)$
(which is independent of $Q\in Y$) is called the \emph{degree} of
$f$.
\end{enumerate}
\end{defn}
The degree of $X(\Gamma)\to X(\SL_{2}(\zz))$ is $[\mbox{PSL}_{2}(\zz):\overline{\Gamma}]$
(which is the number of preimages of a generic point of $X(\SL_{2}(\zz))$
\begin{thm}[Riemann - Hurwitz]
\label{thm:1.3.4} For $f:X\to Y$ non-constant of degree $N$ and
$X,Y$ compact, we have $2g(X)-2=N\cdot(2g(Y)-2)+\sum_{P\in X}(e_{P}(f)-1)$.\end{thm}
\begin{cor}
\label{cor:1.3.5} For any $\Gamma$, we have 
\[
g(X(\Gamma))=1+\frac{[\mathrm{PSL}_{2}(\zz):\overline{\Gamma}]}{12}-\frac{\epsilon_{2}}{4}-\frac{\epsilon_{3}}{3}-\frac{\epsilon_{\infty}}{2}
\]
where $\epsilon_{2}$ is the number of elliptic points of order $2$,
$\epsilon_{3}$ the number of elliptic points of order $3$ and $e_{\infty}$
the number of cusps.\end{cor}
\begin{proof}
We need to analyse the ramification of $X(\Gamma)\to X(\SL_{2}(\zz))$
at each $P\in X(\Gamma)$
\begin{itemize}
\item If $P\in Y(\Gamma)$ is not in the $\SL_{2}(\zz)$-orbit of $i$ or
$\rho$.
\[
\xymatrix{\mathcal{H}\ar[d]\ar@(r,r)[dd]^{\alpha}\\
Y(\Gamma)\ar[d]\\
Y(\SL_{2}(\zz))
}
\]
The map $\alpha$ is unramified at any $\tau$lifting a non elliptic
point of $Y(\SL_{2}(\zz))$, so $e_{P}(f)=1$
\item If $P$ maps to $[i]$: all such $P$ are either non-elliptic or elliptic
of order $2$. If $P$ is elliptic of order $2$, then $Y(\Gamma)\to Y(\SL_{2}(\zz)$
is locally an isomorphism at $P$ so $e_{P}(f)=1$. If $P$ is non-elliptic,
then local coordinate for $\SL_{2}(\zz)$ is square of that for $\Gamma$,
so $e_{P}(f)=2$. We use the definition of the degree to count the
number of points above $[i]$. We have $N=1\cdot\epsilon_{2}(\Gamma)+2\cdot($number
of non-elliptic points of $Y[\Gamma]$ above $[i]$). Hence, the number
of non-elliptic points above $[i]$ is $(N-\epsilon_{2})/2$. So 
\[
\sum_{P\in f^{-1}([i])}(e_{P}-1)=\frac{N-\epsilon_{2}}{2}
\]

\item If $P$ maps to $[\rho]$, (where $\rho=e^{2\pi i/3}$). Then $e_{P}(f)=\begin{cases}
1 & P\mathrm{\, elliptic}\\
3 & P\mathrm{\, non-elliptic}
\end{cases}$. We use the same argument as before, using the definition of degree,
to get that the number of non-elliptic points above $[\rho]$ is $(N-\epsilon_{3})/3$.
Hence
\[
\sum_{P\in f^{-1}([\rho])}(e_{P}-1)=\frac{2(N-\epsilon_{3})}{3}
\]

\item If$P$ is a cusp: let $h$ be the width of the cusp $P$ (that is
the integer such that $e^{2\pi iz/h}$ is a local coordinate for $X(\Gamma)$
at $P$). A local coordinate for $X(\SL_{2}(\zz)$ at $\left[\infty\right]$
is $(e^{2\pi iz/h})^{h}$, so $e_{p}(f)=h$. Thus 
\[
\sum_{P\in f^{-1}(\left[\infty\right])}(e_{P}-1)=\left(\sum_{P\in f^{-1}\left(\left[\infty\right]\right)}e_{P}\right)-e_{\infty}=N-e_{\infty}
\]

\end{itemize}
Putting all of this together, we get
\begin{eqnarray*}
2g(X(\Gamma))-2 & = & (-2)N+\frac{N-\epsilon_{2}}{2}+\frac{2(N-\epsilon_{3})}{3}+(N-\epsilon_{\infty})\\
g(X(\Gamma)) & = & 1+\frac{N}{12}-\frac{\epsilon_{2}}{4}-\frac{\epsilon_{3}}{3}-\frac{\epsilon_{\infty}}{2}
\end{eqnarray*}
\end{proof}
\begin{example*}
Let $\Gamma=\Gamma_{0}(11)$. We have that $N=12$, $\epsilon_{\infty}=2$
(they are $\left[0\right]$ and $\left[\infty\right]$), $\epsilon_{2}=\epsilon_{3}=0$
(Exercise, c.f. D\&S). Hence $g(X(\Gamma)=1+\frac{12}{12}-0-0-\frac{2}{2}=1$. \end{example*}
\begin{xca*}
Show that the only primes $p$ such that $g(X(\Gamma_{0}(p)))=0$
are $\{2,3,5,7,13\}$. \end{xca*}
\begin{rem*}
For any $g$, there exists finitely many \uline{congruence} subgroups
$\Gamma$ of $\mbox{PSL}_{2}(\zz)$ of genus $g$. (J.G. Thompson)
\end{rem*}

\subsection{Sheaves and Riemann-Roch}
\begin{conjecture}
\label{1.4.1}Let $X$ be a topological space. Then you already know
what a sheaf on $X$ is.
\end{conjecture}
Now let $X$ be a Riemann surface. We have $\mathcal{O}_{X}$ (``the
structure sheaf'') is defined by $\mathcal{O}_{X}(U)=$holomorphic
functions $U\to\cc$. It is a sheaf of \uline{rings}, so we can
make sense of \uline{sheaves of $\mathcal{O}_{X}$-modules}. 
\begin{defn}
An \emph{invertible sheaf} on $X$ is a sheaf of $\mathcal{O}_{X}$-modules
that is locally free of rank $1$. (They are exactly the ones who
have an inverse with respect to tensor product of $\mathcal{O}_{X}$-modules)\end{defn}
\begin{fact*}
\emph{(For geometers) }Invertible sheaves are in bijection with line
bundles with holomorphic structure.
\end{fact*}
Now we specialise to the case of $X$ is \uline{compact}. We have
a notion of \uline{meromorphic sections} of invertible sheaves
$\mathcal{F}$ (which are sections of $\mathcal{F}\otimes_{\mathcal{O}_{X}}\{$sheaf
of meromorphic functions\})
\begin{thm}[Riemann Existence Theorem]
\label{thm:1.4.2} Any invertible sheaf on a compact Riemann Surface
has a non-zero global meromorphic section.
\end{thm}
This implies that there is a notion of \uline{degree} of an invertible
sheaf, it is defined as the sum of the order of the vanishing of any
non-zero $L$ meromorphic global section. This is wee defined as the
sum of zeroes and poles of a meromorphic function is $0$. 

We have $\deg(\mathcal{F}\otimes\mathcal{G})=\deg\mathcal{F}+\deg\mathcal{G}$
and $\deg(\mathcal{F}^{-1})=-\deg\mathcal{F}$. (Note that invertible
sheaves are a group under $\otimes$ with $\mathcal{O}_{X}$ as identity,
and $\deg$ is an group homomorphism to $\zz$)
\begin{thm}[Riemann - Roch]
\label{thm:1.4.3}  Let $X$ be a compact Riemann Surface, $\mathcal{F}$
an invertible sheaf on $X$. Then
\begin{enumerate}
\item $\mathcal{F}(X)=:H^{0}(X,\mathcal{F})$ is finite dimensional over
$\cc$
\item $\dim H^{0}(X,\mathcal{F})-\dim H^{0}(X,\Omega\otimes\mathcal{F}^{-1})=1-g+\deg\mathcal{F}$,
where $\Omega$ is the sheaf of holomorphic differentials on $X$. 
\end{enumerate}
\end{thm}
Note that if $\deg(\mathcal{F})<0$, then $\mathcal{F}$ has no non-zero
global sections, so if $\deg\mathcal{F}$ is \uline{large}, $H^{0}(X,\Omega\otimes\mathcal{F}^{-1})=0$,
hence we get a formula for $\dim H^{0}(X,\mathcal{F})$. (Note that
$\dim H^{0}(X,\Omega)=g(X)$ by setting $\mathcal{F}=\mathcal{O}_{X}$,
furthermore $\deg\Omega=2g-2$ by taking $\mathcal{F}=\Omega$)

{[}\textbf{Aside}: There exists a cohomological theorem for sheaves
on Riemann surfaces for which $H^{0}(X,\mathcal{F})$ is sections.
the Riemann - Roch is a combination of $2$ things: 
\begin{itemize}
\item a formula for $\chi(\mathcal{F})=\sum_{i\geq0}(-1)^{i}\dim H^{i}(X,\mathcal{F})$
\item Serre duality: $H^{i}(X,\mathcal{F})=H^{1-i}(X,\Omega\otimes\mathcal{F}^{-1})^{*}$.
(This is why sometime $\Omega$ is referred as the ``dualising sheaf''{]}
\end{itemize}

\subsection{The Katz sheaf}

Let $X=X(\Gamma)$ for some $\Gamma$, and choose $k\in\zz$. 
\begin{defn}
\label{1.5.1}Let $\omega_{k}$ be the sheaf defined by $\omega_{k}(V)=\{$holomorphic
functions on $\pi^{-1}(V)\subset\mathcal{H}^{*}$ satisfying $f(\gamma z)=j(\gamma,z)^{k}f(z)$
for all $\gamma\in\Gamma\}$. 
\end{defn}
This is a sheaf of $\mathcal{O}_{X(\Gamma)}$-modules. If $k$ is
odd and $-1\in\Gamma$ it is the zero sheaf, we now assume that we
are not in this case.
\begin{thm}
\label{thm:1.5.2}~
\begin{enumerate}
\item $\omega_{k}$ is invertible
\item $\omega_{2}=\Omega_{X(\Gamma)}($\emph{cusps}$)$
\end{enumerate}
\end{thm}
\begin{rem*}
Here, for $\mathcal{L}$ an $\mathcal{O}_{X}$-module sheaf and $D=\sum n_{i}P_{i}$
is a divisor (formal $\zz$-linear combination of points), we define
$\mathcal{L}(D)(U)=\{$meromorphic sections $x$ of $\mathcal{L}$
over $U$ with $\mbox{div}(x)+D\geq0\}$. For example:
\begin{itemize}
\item $L(P)=$ ``allow simple poles at $P$''
\item $L(-P)=$ ``sections vanishing at $P$''
\end{itemize}
\end{rem*}
\begin{proof}
~
\begin{enumerate}
\item This is a case by case check. We just need to show it on a open neighbourhood
of every $P\in X(\Gamma)$. That is we want to show that for every
$P\in X(\Gamma)$ there exists a neighbourhood $V\ni P$ and $b\in\omega_{k}(V)$
such that $\omega_{k}(V)=\mathcal{O}_{X}(V)\cdot b$.


For $P$ non-elliptic, not cusp, we can find $V\ni P$ open such that
$\pi^{-1}(V)=\sqcup_{\gamma\in\overline{\Gamma}}\gamma U$ and $\omega_{k}(V)\cong\mathcal{O}_{\mathcal{H}}(U)\cong\mathcal{O}_{X}(V)$.


Other cases: Choose $\tau\in\mathcal{H}^{*}$ to be a lifting of $P$,
$U\subset\mathcal{H}^{*}$ open such that $U$ is fixed by $\mbox{stab}_{\Gamma}(\tau)$
and 
\[
\pi^{-1}(V)=\coprod_{\gamma\in\overline{\Gamma}/\mathrm{stab}_{\Gamma}(\tau)}\gamma U
\]
where $V=\pi(U)$. So $\omega_{k}(V)=\{f:U\to\cc$ holomorphic and
weight $k$ invariant under $\mbox{Stab}_{\Gamma}(\tau)$\} while
$\mathcal{O}_{X}(V)=\{f:U\to\cc$ holomorphic and weight $0$ invariant
under $\mbox{Stab}_{\Gamma}(\tau)\}$. So if $\mbox{Stab}_{\Gamma}(\tau)=1$
we are done (take $b=1$), or more generally if the weight $k$ action
of $\mbox{Stab}_{\Gamma}(\tau)$ is trivial. (Happens if $\tau$ is
elliptic and $k$ is divisible by the order of $\mbox{Stab}_{\Gamma}(\tau)$)


For elliptic points: if $\tau$ is elliptic, conjugate it onto $z=0$
as before so $\mbox{Stab}_{\Gamma}(\tau)=$cyclic group of \uline{rotation}
with order $n$, say. A function $U\to\cc$ is weight $k$ invariant
under this group if and only if $f(z)=z^{a}g(z^{n})$ where $a$ is
the least non negative integer congruent to $k$$\mod n$ and some
holomorphic function $g$. So $z\mapsto z^{a}$ is a local basis.


For cusps: if $\tau$ is a cusp, and the cusp is regular or the weight
is even, the action of $\mbox{Stab}_{\Gamma}(\tau)$ in weight $k$
and weight $0$ coincides, so $b=1$ works.


If $\tau$ is an irregular cusps, $k$ odd, (without loss of generality
$\tau=\infty$) then $\mathcal{O}_{X}(V)=\{f:U\to\cc$ holomorphic
such that $f(z+h)=f(z)\}$ while $\omega_{k}(V)=\{f:U\to\cc$ holomorphic
such that $f(z+h)=-f(z)\}$ (where $h$ is the height of the cusp).
Then $z\mapsto e^{i\pi z/h}$ is a local basis. 

\item The isomorphism is $f\mapsto f(z)dz$. Clearly if is a bijection $\mathcal{O}_{\mathcal{H}}(\mathcal{H})\to\Omega_{\mathcal{H}}^{1}(\mathcal{H})$
and it commutes with $\Gamma$-action if we put weight $2$ action
on $\mathcal{O}_{\mathcal{H}}.$ Passing to $\Gamma$-invariants $\omega_{2}|_{Y(\Gamma)}=\Omega_{Y(\Gamma)}^{1}$. 


We need to show that sections of $\omega_{2}$ extending to corresponding
cusps correspond to differentials with simple poles. It suffices to
consider the case $\tau=\infty$ . Sections of $\Omega^{1}$ near
$\infty$ are $f(q)dq$. But $q=e^{2\pi iz/h}$ implies $dq=\frac{2\pi i}{h}e^{2\pi iz/h}dz$,
i.e., $dz$ is a scalar times $\frac{dq}{q}$. Hence ``$\mathcal{O}_{X}dz=\mathcal{O}_{X}\frac{dq}{q}=\mathcal{O}_{X}(\infty)dq$''
(over a neighbourhood $V\ni\infty$)

\end{enumerate}
\end{proof}
Obviously $H^{0}(X(\Gamma),\omega_{k})=M_{k}(\Gamma)$. 
\begin{prop}
\label{prop:1.5.3} 

Let 
\[
r=\lcm\left(\mathrm{order\, of\,}\Gamma\mathrm{-stabaliser\, of\, elliptic\, points},\left\{ \begin{array}{cc}
2 & \mathrm{if\,\exists\, irregular\, cusp}\\
1 & \mathrm{else\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}\right\} \right)
\]


Note that $1\leq r\leq12$. Then $\omega_{k+r}\cong\omega_{k}\otimes\omega_{r}$
for all $k\in\zz$. In particular if $r=1$ then $\omega_{k}=(\omega_{1})^{\otimes k}$
for all $k$.\end{prop}
\begin{proof}
For $k=r$ all the local bases $b$ in Theorem \ref{thm:1.5.2} were
$1$ and bases of $\omega_{k}$ for general $k$ only depended on
$k$ mod $r$. So local basis for $\omega_{k+r}$ is equal to the
product of ones for $\omega_{k}$ and $\omega_{r}$\end{proof}
\begin{defn}
\label{1.5.4}If $r=1$ above (i.e., $\Gamma$ has no elliptic points,
it does not contain $-1$ and all cusps are regular), we say $\Gamma$
is \emph{neat}. Then $\omega_{k}=\omega_{1}^{\otimes k}$, so $\omega=\omega_{1}$
is obviously important. Call this the\emph{ Katz sheaf}. \end{defn}
\begin{cor*}
\label{1.5.5}If $\Gamma$ is neat, then for $k\geq2$ we have $\dim M_{k}(\Gamma)=(k-1)(g-1)+\frac{k}{2}\epsilon_{\infty}$.\end{cor*}
\begin{proof}
We have 
\begin{eqnarray*}
\deg\omega & = & \frac{1}{2}\deg(\omega^{\otimes2})\\
 & = & \frac{1}{2}\left(\deg\Omega+\epsilon_{\infty}\right)\\
 & = & \frac{1}{2}(2g-2+\epsilon_{\infty}).
\end{eqnarray*}
So if $k\geq2$, $\deg\omega^{\otimes k}>2g-2$ and Riemann - Roch
gives 
\begin{eqnarray*}
\dim H^{0}(X(\Gamma),\omega^{k}) & = & k\cdot(g-1+\frac{\epsilon_{\infty}}{2})-g+1\\
 & = & (k-1)(g-1)+\frac{k}{2}\epsilon_{\infty}
\end{eqnarray*}

\end{proof}
There exists similar (but messier) formulae in the non-neat case (c.f.
D-S chapter 3)
\begin{example*}
Let $\Gamma=\Gamma_{1}(5)$, this is neat ($\Gamma_{1}(N)$ is neat
if and only if $N\geq5$), $g=0$ and $\epsilon_{\infty}=4$. So $\dim M_{k}(\Gamma)=k+1$
for $k\geq2$. For $k=1$ we need to worry about $H^{0}(X(\Gamma),\Omega^{1}\otimes\omega^{-1})$,
but $\deg\Omega^{1}=2g-2=-2$, $\deg\omega=1$. So $\Omega^{1}\otimes\omega^{-1}$
had $\deg-1$.

If you try to do this for $\Gamma_{1}(23)$ if fails, $\deg\Omega^{1}\otimes\omega^{-1}$
is $ $$0$. So dimension of Wight $1$ form spaces lie much deeper.
\end{example*}
\newpage{}


\section{Modular Curves as Algebraic Curves}


\subsection{Modular Curves over $\cc$}
\begin{thm}
\label{thm:2.1.1}~
\begin{enumerate}
\item The $\cc$-points of a smooth connected projective algebraic curve
over $\cc$ are canonically a Riemann surface, $X\mapsto X^{\mathrm{an}}$
\item Every compact Riemann surface is $X^{\mathrm{an}}$ for a unique $X$
\item There exists an equivalence of categories between (locally free sheaves
of $\mathcal{O}_{X}$-modules) $\cong$ $($locally free sheaves of
$\mathcal{O}_{X^{\mathrm{an}}}$-modules), preserving global sections
\end{enumerate}
\end{thm}
\begin{rem*}
1.~is basically the implicit function theorem. We'll see later a
bit about the proof of 2.~. 3.~is Serre's ``GAGA'' theorem. 
\end{rem*}
The functors are on one hand $\mathcal{F}\mapsto\mathcal{O}_{X^{\mathrm{an}}}\otimes_{\mathcal{O}_{X}}\mathcal{F}$,
and on the other hand $\mathcal{F}\mapsto($subsheaf of $\mathcal{F}$
whose sections over $U$ are elements of $\mathcal{F}(U)$ extending
to meromorphic sections on $X$$)$.

Hence for any $\Gamma$ there's an algebraic variety $X(\Gamma)_{\cc}$
and invertible sheaves $\omega_{k}$ on it such that $M_{k}(\Gamma)=H^{0}(X(\Gamma)_{\cc},\omega_{k})$. 

Here's an alternative, nicer, construction.
\begin{thm}
\label{thm:2.1.2} 
\[
X(\Gamma)_{\cc}=\mathrm{Proj}\left(\bigoplus_{k\geq0}M_{k}(\Gamma)\right).
\]
\end{thm}
\begin{rem*}
C.f. Hartshorne Algebraic Geometry Chapter 2 for the definition/construction
of Proj.\end{rem*}
\begin{proof}
One knows that for any Noetherian graded $\cc$-algebra $S_{\bullet}$,
with $S_{0}=\cc$, $\mbox{Proj}(S_{\bullet})=\mbox{Proj}(S_{n\bullet})$
for any $n\geq1$. (Where $S_{n\bullet}=$ the subring $\oplus_{k\geq0}S_{nk}$).
Choose $n$ to be $r$ from Proposition \ref{prop:1.5.3}, so 
\[
S_{n\bullet}=\oplus_{k}H^{0}\left(X(\Gamma),\omega_{n}^{\otimes k}\right).
\]
We now quote a standard fact in algebraic geometry: Invertible sheaves
of positive degree on curves are \uline{ample}, so their sections
give an embedding in projective space. \end{proof}
\begin{rem*}
In fact the same argument can be used to prove Theorem \ref{thm:2.1.1}
part 2.~: take any ample invertible sheaf on a Riemann surface and
get an embedding in $\pp^{n}$ for $n\gg0$.
\end{rem*}

\subsection{Descending the base field}

\uline{Question}: Does there exists an algebraic curve over some
number field $ $$K$ such that we get $X(\Gamma)_{\cc}$ by base
extension?

Let's think a bit what this means:
\begin{itemize}
\item Clearly not all varieties over $\cc$ are definable over number fields.
For example $Y^{2}=X^{3}+X+\pi$, this is not defined over any number
field, as its $j$ invariant is $\frac{6192}{27\pi^{2}+4}$. But we
need to be careful as $\pi Y^{2}=X^{3}+X$ is defined over $\qq$
(it is isomorphic to $Y^{2}=X^{3}+X$)
\item Even if descends exists they might not be unique. For example, $\pp_{\qq}^{1}$
and $\{X^{2}+Y^{2}+2Z^{2}=0\}\subset\pp_{\qq}^{2}$ become isomorphic
over $\cc$. 
\end{itemize}
So we need to ask, is there a descend to a number field that ``means
something''?


\subsection*{The curves - fields correspondence}

There is a bijection, for any field $k$, \{smooth geometrically connected
algebraic curves over $k$\}$\leftrightarrow\{$field extensions $K/k$
of transcendence degree $1$ containing no algebraic extensions of
$k$$\}$. The bijection is given by $X\mapsto k(X)$, the field of
rational functions on $X$.

So for a curve$X$ over $\cc$ we have $\{$models of $X$ over $k\subseteq\cc\}\leftrightarrow\{$subfields
$L$ of $\cc(X)$ generating it over $\cc$ but with $L\cap\cc=k\}$. 

So we want to look for nice subfields of $\cc(X(\Gamma)_{\cc})=\{$meromorphic
modular functions of weight $0$ and level $\Gamma\}$. 
\begin{thm}
\label{thm:2.2.1}~Let $N\geq2$
\begin{enumerate}
\item $\cc(X_{0}(N))=\cc\left(j(z),j(Nz)\right)$
\item The minimal monic polynomial of $j(Nz)$ over $\cc(j(z))$ lies in
$\zz[j][Y]\subseteq\cc(j)[Y]$, and is $\Phi_{N}(j,Y)$ for $\Phi_{N}$
symmetric.
\item If $N=p$ is prime, $\Phi_{p}(X,Y)\equiv(Y^{p}-X)(Y-X^{p})\mod p$.
\end{enumerate}
In particular $X_{0}(N)_{\cc}$ has a model over $\qq$ whose function
field is $\qq(j(z),j(Nz))$ \end{thm}
\begin{proof}
(cf. Mine's notes ``MFMF'' pg 90-92)
\begin{enumerate}
\item Clearly $j(Nz)\in\cc(X_{0}(N))$, so $\cc(j(z),j(Nz))\subseteq\cc(X_{0}(N))$.
Over $\cc(j)$, the right hand side has degree $[\mbox{PSL}_{2}(\zz):\overline{\Gamma_{0}(N)}]=[\SL_{2}(\zz):\Gamma_{0}(N)]=:\mu$.
So if we show that $j(Nz)$ has degree $\mu$ over $\cc(j)$, then
part $1.$~follows. 


Let $\gamma_{1},\dots,\gamma_{\mu}$ be such that $\SL_{2}\zz=\coprod_{i=1}^{m}\Gamma_{0}(N)\gamma_{i}$
(without loss of generality $\gamma_{1}=1$). Consider the functions
$j(N\gamma_{i}z)$. All Galois conjugate to $j(Nz)$ over $\cc(j)$
(via the automorphism $z\mapsto\gamma_{i}z$ of $\cc(\mathcal{H})$).
If we can show that they are distinct then part $1.$~follows by
Galois theory. So suppose $j(N\gamma_{i}z)=j(N\gamma_{k}z)$ for some
$1\leq i,k\leq\mu$, for all $z\in\mathcal{H}$. Then 
\[
j\left(\underset{:=A}{\underbrace{\begin{pmatrix}N & 0\\
0 & 1
\end{pmatrix}\gamma_{k}\gamma_{i}^{-1}\begin{pmatrix}N & 0\\
0 & 1
\end{pmatrix}^{-1}}}z\right)=j(z)\,\,\forall z\in\mathcal{H}.
\]
By a question from sheet 1.~this forces $\pm A\in\mbox{PSL}_{2}(\zz)$,
so $\gamma_{k}\gamma_{i}^{-1}\in\Gamma_{0}(N)$, hence $i=k$.

\item From part 1.~we know the monic minimal polynomial of $j(Nz)$ is
$\prod_{i=1}^{n}\left(Y-j(N\gamma_{i}z)\right)$. The coefficients
are symmetric polynomials in $\{j(N\gamma_{i}z)\}$, so they are all
holomorphic on $\mathcal{H}$. Since they are also rational functions
in $j(z)$, they must be polynomials in $j(z)$. 


To control coefficients we use $q$-expansions. We know that $j(z)=q^{-1}+744+\dots\in q^{-1}+\zz[[q]]$.
Moreover, we can choose the $\gamma_{j}$ such that $j(N\gamma_{i}z)=j\left(\frac{az+b}{d}\right)$
for some $a,b,d\in\zz$ such that $ad=N$. So $j(N\gamma_{i}z)\in\zz[\zeta_{N}]((q^{\frac{1}{N}}))$
(where $\zeta_{N}=e^{2\pi i/N}$). So the coefficients of $\Phi_{N}(j,Y)$
(as a polynomial in $Y$) have $q$-expansion in $\zz[\zeta_{N}]((q^{\frac{1}{N}}))\cap\cc((q))=\zz[\zeta_{N}]((q))$.
\begin{claim*}
These coefficients are actually in $\zz[\zeta_{N}][j]$
\end{claim*}

Let $P=\sum_{i=1}^{k}b_{i}j^{i}\in\zz[j]$ have $q$-expansion in
$R((q))$ for some $R\subseteq\cc$ a subring. Inspecting lowest term
of $q$-expansion, we see that $b_{k}\in R$. By induction on degree,
we have $b_{i}\in R$ for all $i$. So $\Phi_{N}(X,Y)\in\zz[\zeta_{N}][X,Y]$.
Write $\Phi_{N}(X,Y)=\sum_{r,s}c_{rs}X^{r}Y^{s}$, substitute in $q$-expansions
of $j(z),j(Nz)$ and equate coefficients. We get equations for $\{c_{rs}\}$,
linear with $\qq$-coefficients. We know $c_{r\mu}=\begin{cases}
1 & r=0\\
0 & r>0
\end{cases}$ and that there is a unique solution over $\cc$. So solution must
be $\mbox{Gal}(\zz[\zeta_{N}]/\zz)$ invariant, i.e., $\Phi_{N}(X,Y)\in\zz[X,Y]$.


To prove symmetry: $\Phi_{N}(j(z),j(Nz))=0$ for all $z$. This implies
$\Phi_{N}\left(j\left(\frac{1}{Nz}\right),j\left(\frac{-N}{Nz}\right)\right)=0\,\forall z\in\mathcal{H}$.
Rearranging, this gives $\Phi_{N}(j(Nz),j(z))=0$. So $\Phi_{N}(Y,X)$
is a multiple of $\Phi_{N}(X,Y)$. Hence $\Phi_{N}(X,Y)=c\Phi_{N}(Y,X)$
for some $c^{2}=1$. Since $N\geq2$, we can't have $c=-1$ as this
would force $\Phi_{N}$ to be a multiple of $X-Y$, hence $c=1$. 

\item Note that $\Phi_{p}(j(z),Y)$ has $q$-expansion 
\[
(Y-j(pz))\left(\prod_{i=1}^{p-1}\left(Y-j\left(\frac{z+i}{p}\right)\right)\right).
\]
Now $q$-expansions of functions $j\left(\frac{z+i}{p}\right)$ are
all congruent modulo a unique prime $\mathfrak{p}|p$ of $\zz[\zeta_{p}]$,
so modulo $\mathfrak{p}$ reduction $\phi_{p}(j,Y)$ is 
\begin{eqnarray*}
(Y-j(z))\left(Y-j\left(\frac{z}{p}\right)\right)^{p} & \equiv & (Y-j(z)^{p})\left(Y^{p}-j\left(\frac{z}{p}\right)^{p}\right)\\
 & \equiv & \left(Y-j(z)^{p}\right)\left(Y^{p}-j(z)\right)\mod\mathfrak{p}.
\end{eqnarray*}
This forces some congruence for $\Phi_{p}$ as a polynomial. 
\end{enumerate}
\end{proof}
\begin{rem*}
~
\begin{enumerate}
\item $Y_{0}(N)$ is \uline{not} the curve in $\aa^{2}$ defined by $\Phi_{N}(X,Y)=0$.
It is \uline{birationally equivalent} to it but not generally isomorphic;
$\{\Phi_{N}(X,Y)=0\}$ is generally \uline{singular}.
\item $\Phi_{N}$'s have \uline{huge} coefficients. For example
\[
\Phi_{2}(X,Y)=X^{3}+Y^{3}-X^{2}Y^{2}+1488XY(X+Y)-162000(X^{2}+Y^{2})
\]
\[
+40733375XY+8748000000(X+Y)-137464000000000
\]

\item Note that mod $p$ relations of $\Phi_{p}$ defines a \uline{reducible}
curve, we get two copies of $\pp^{1}$ intersecting. 
\end{enumerate}
\end{rem*}
We define $X_{0}(N)_{\qq}$ as the unique smooth projective curve
over $\qq$ with function field $\qq(X,Y)/\Phi_{N}(X,Y)$. 
\begin{thm}
\label{thm:2.2.2}There is a sheaf $\omega_{k,\qq}$ on $X_{0}(N)_{\qq}$
whose base-extension to $\cc$ is $\omega_{k}$. \end{thm}
\begin{proof}
As $-1\in\Gamma_{0}(N)$, $\omega_{k}$ is only non-zero for $k\in2\zz$.
We know $\omega_{2}\cong\Omega_{X_{0}(N)_{\cc}}^{1}(\mathrm{cusps)}.$
More generally we always have $\omega_{2k}\cong\left(\Omega_{X_{0}(N)_{\cc}}^{1}\right)^{\otimes k}(D_{k})$
where $D_{k}$ is a $\zz$-linear combination of the divisors: (cusps),
(elliptic points of order 2), and (elliptic points of order 3).
\begin{claim*}
These $3$ divisors descend to $X_{0}(N)_{\qq}$. 

For (cusps) this is clear: the map $X_{0}(N)\to X_{0}(1)$ is defined
over $\qq$ (as $j\in\qq(j(z),j(Nz))$). Cusps are exactly the preimages
of the $\qq$-points $\left(\infty\right)\in X_{0}(1)$. 


For elliptic points we need to be a bit careful: $\{$elliptic points
of $X_{0}(N)_{\cc}$ of order $2$$\}=\{$preimages of $i\in X_{0}(1)$
where the projection map $X_{0}(N)\to X_{0}(1)$ is unramified$\}$.
Ramification degree is \uline{Galois invariant} so we are done.
Note $j(i)=1728$ and $j(\rho)=0$, hence $i,\rho\in X_{0}(1)(\qq)$. 

\end{claim*}
So we can define 
\[
\omega_{2k,\qq}=\left(\Omega_{X_{0}(N)_{\qq}}^{1}\right)^{\otimes k}(D_{k}).
\]
\end{proof}
\begin{cor}
\label{cor:2.2.3}For any $k\geq2$ even, any $N\geq1$, the spaces
$S_{k}(\Gamma_{0}(N))$ and $M_{k}(\Gamma_{0}(N))$ have bases consisting
of forms with $q$-expansions in $\qq[[q]]$. \end{cor}
\begin{proof}
We only give the argument for $M_{k}$, as the argument for $S_{k}$
is similar. We have 
\begin{eqnarray*}
M_{k}(\Gamma_{0}(N)) & = & H^{0}(X_{0}(N)_{\cc},\omega_{k})\\
 & = & \cc\otimes_{\qq}H^{0}(X_{0}(N)_{\qq},\omega_{k,\qq}).
\end{eqnarray*}

\begin{claim*}
The image of $H^{0}(X_{0}(N)_{\qq},\omega_{k,\qq}).$ is a function
of $q$-expansions with in $(2\pi i)^{k/2}\qq[[q]]$. 

Any \uline{meromorphic} section of $\omega_{k}$ is in $\qq(j(z),j(Nz))\cdot(dj)^{\otimes k/2}$.
Now $dj=j'(z)dz$, say $j(z)=J(q)$, then $dj=J'(q)2\pi iqdz$. So
any meromorphic sections of $\omega_{k}$ lands in $(2\pi i)^{k/2}\qq((q))$. 

\end{claim*}
\end{proof}
\newpage{}


\section{Modular Curves as Moduli Spaces}


\subsection{Lattices and Level Structures}

\textbf{Recall}: If $\Lambda$ is a lattice in $\c c$$\cc$ (a discrete
subgroup isomorphic to $\zz^{2}$), $\Lambda$ is homothetic to a
lattice of the form $\Lambda_{\tau}:=\zz+\zz\tau$ for some $\tau\in\mathcal{H}$,
and $\tau$ is uniquely determined modulo $\mbox{PSL}_{2}(\zz)$. 

$Y(\SL_{2}(\zz))=\{$homothety classes of lattices$\}=\{$isomorphism
classes of elliptic curves over $\cc\}$. 
\begin{xca*}
~
\begin{enumerate}
\item For any $N\geq2$ there exists a bijection between $Y_{0}(N)\cong\{$pairs
of $(\Lambda,C)$: $\Lambda$ a lattice, $C$ a cyclic subgroup of
$\cc/\Lambda$ of order $N$, where $(\Lambda,C)\cong(\Lambda',C')$
if there exists an isomorphism $\cc/\Lambda\cong\cc/\Lambda'$ sending
$C$ to $C'$\}. The isomorphism is defined by $[\tau]\mapsto\left(\zz+\zz\tau,\frac{1}{N}\zz\right)$. 
\item For any $N\geq2$ there exists a bijection between $Y_{1}(N)\cong\{$pairs
$(\Lambda,P):$$\Lambda$ a lattice, $P$ a point of exact order $N$,
modulo the equivalence relation $(\Lambda,P)\sim(\Lambda',P')$ if
there exists an isomorphism $\cc/\Lambda\cong\cc/\Lambda'$ sending
$P$ to $P'\}$
\end{enumerate}
\end{xca*}
Note that in part 2.~$(\Lambda,P)$ is always equivalent to $(\Lambda,-P)$.
In part $1.$~we have lots more exceptional cases coming from elliptic
points. 

We have a natural question: if $x\in Y_{0}(N)$ is a $\qq$-point,
does $(E,C)$, $E=\cc/\Lambda$, descend to $\qq$? This is the right
sort of question to ask to understand modular curves over number fields.
(This actual question is vacuous for $N\gg0$, but we don't know that
yet)


\subsection{Moduli spaces and representable functors}

All rings are commutative and unital. We have the categories: \uline{Rings},
$\underline{R}$\uline{-Alg} for a ring $R$, and \uline{Sets}.
(We do not worry about foundational issues). ``Most sets that come
up naturally in algebraic geometry are functors \uline{Ring}$\to$\uline{Set}
(or \uline{$\underline{R}$-Alg$\to$Set})''. 
\begin{example*}
~
\begin{enumerate}
\item Points of varieties over schemes.
\item Classes of varieties or structures on varieties.
\end{enumerate}
\end{example*}
A lot of the fun of algebraic geometry arises from the fact that instances
of 2.~are often 1.~in disguise. These are \uline{moduli spaces}. 


\subsubsection*{Some properties of functors and representable functions}

Let $C$ be a category (``locally small'' - homomorphism between
any two objects are a set). Then $\Hom(X,-)$ is a functor $C\to\underline{\mathrm{Set}}$,
which we denote by $h^{X}$. A functor $\mathcal{F}:C\to\mathrm{\underline{Set}}$
is \emph{representable} if there exists an isomorphism of functors
$\mathcal{F}\cong h^{X}$ for some $X\in\mathrm{Ob}(C)$. How do we
specify the isomorphism $\mathcal{F}\cong h^{X}$?
\begin{note*}
$h^{X}$ has a \uline{canonical element}: namely $\id_{X}$. 
\end{note*}
So we need to know a corresponding element of $\mathcal{F}(X)$, as
$h^{X}(X)\cong\mathcal{F}(X)$. Having specified a $\phi:\mathcal{F}(X)$
corresponding to $\id_{X}$, this determines an element of $\mathcal{F}(Y)$
for every homomorphism $\alpha:X\to Y$, i.e., take $\mathcal{F}(\alpha)(\phi)$. 
\begin{prop}[Yoneda's Lemma]
\label{prop:3.2.1} This construction is a bijection: $\{$\emph{natural
transformations }$h^{X}\to\mathcal{F}\}$ $\overset{\sim}{\to}$$\mathcal{F}(X)$,
for any $\mathcal{F}:C\to\underline{\mathrm{Set}}$ and $X\in\mathrm{Ob}(C)$. \end{prop}
\begin{rem*}
WE have discussed \uline{covariant} functors, but we get the same
for contravariant functors by replacing $C$ with $C^{\mathrm{opp}}$)
\end{rem*}
So if $\mathcal{F}$ is representable, the bijection $\mathcal{F}(X)\cong h^{X}(Y)$
is determined by a $\phi\in\mathcal{F}(X)$; we are saying that for
every $Y\in\mathrm{Ob}(C)$ and every $\psi\in\mathcal{F}(Y)$, there
exists a unique homomorphism $\alpha:X\to Y$ such that $\alpha(\phi)=\psi$.
We say $(X,\phi)$ represents \emph{$\mathcal{F}$. }Hence $\phi$
is essential part of the data.
\begin{example*}
Let $C=\mathrm{\underline{Ring}}$
\begin{itemize}
\item $\mathcal{F}(R)=R$ (the forgetful functor). This is represented by
$(\zz[T],T)$, i.e., for any ring $R$, $r\in R$, there exists a
unique $\alpha:\zz[T]\to R$ such that $\alpha(T)=r$ .
\item $\mathcal{F}(R)=R^{*}$ is represented by $(\zz[T,T^{-1}],T)$
\item $\mathcal{F}(R)=\{n$th root of unity in $R\}$ is represented by
$(\zz[T]/(T^{n}-1),T)$.

\begin{note*}
``primitive $n$th root of unity'' is not a functor on $R$
\end{note*}
\end{itemize}
\end{example*}
\textbf{A Non-Example.} Let $C=\mathrm{\underline{Ring}}$. Consider
$\mathcal{F}(R)=($squares in $R$). This is \uline{not} representable
\begin{proof}
Suppose $\mathcal{F}$ is represented by $(A,a)$, some ring $A$
and $a\in A$ with $a^{2}=b$ for some $b\in A$. Then for any ring
$S$ and element $s\in S$ such that $s$ is a square, there needs
to exists a unique homomorphism $\alpha:A\to S$ such that $\alpha(a)=s$.
But take $S=\zz[T]$, $s=T^{2}$, so there exists unique $\alpha:A\to\zz[T]$
with $\alpha(a)=T^{2}$, hence $\alpha(b)=\left\{ \pm T\right\} $.
Let $\sigma:S\to S$ be $T\mapsto-T$, note that $\sigma(s)=s$. So
$\sigma\circ\alpha\in\Hom(A,S)$ also sends $a$ to $s$, but $\sigma\circ\alpha\neq\alpha$
as $(\sigma\circ\alpha)(b)\neq\alpha(b)$. This contradicts uniqueness
of $\alpha$
\end{proof}
The moral of this example: Automorphism are \uline{bad} for representability.


\subsection{Elliptic curves over general base schemes}

We want to make sense of ``elliptic curves over $S$'', where $S$
a scheme. 
\begin{defn}
\label{3.3.1} Let $S$ be a scheme. An \emph{elliptic curve} over
$S$ is a scheme $\mathcal{E}$ with a morphism $\pi:\mathcal{E}\to S$
(an $S$-scheme) such that $\pi$ is flat and proper, and all fibres
are smooth genus $1$ curves, given with a section ``$O$''$:S\to\mathcal{E}$. \end{defn}
\begin{example*}
In Silverman's book, there is the equation $Y^{2}+XY=X^{3}-\frac{36}{j-1728}X-\frac{1}{j-1728}$.
The associated homogeneous cubic $Y^{2}Z+XYZ=X^{3}-\frac{36}{j-1728}XZ^{2}-\frac{1}{j-1728}Z^{3}$
is a subscheme of $\pp^{2}/R$, where $R$ is the ring $\zz[j,j^{-1},(j-1728)^{-1}]$.
This is an elliptic curve over $\mathrm{Spec}R$. The discriminant
is $\Delta=j^{2}/(j-1728)^{3}$.

Think of this as a \uline{family of elliptic curves}\emph{, }one
for every $j\neq0,1728$, varying in an ``algebraic way''. 
\end{example*}
For $\mathcal{E}$ over $S$ as above, $E(S)=\Hom_{S-\mathrm{sch}}(S,\mathcal{E})=$sections
of $\pi:\mathcal{E}\to S$ picking out a point on each fibre.

\textbf{Warning}: If $P\in\mathcal{E}(S)$ has order $N$, i.e., $N\cdot P=0$
and $M\cdot P\neq0$ for $1\leq M<N$, it is not necessarily true
that $P_{x}$ has order $N$ on $\mathcal{E}_{x}$ for every $x\in S$.
(E.g., if $\mathcal{E}$ over $\Spec(\zz_{p})$ can have points of
order $P$ reduction $\mod p$ to $0$ at closed points of $\Spec\zz_{p}$
).
\begin{prop}
\label{prop:3.3.2} If $\mathcal{E}$ over $S$ is an elliptic curve,
then $\mathcal{E}$ has a Weierstrass equation \uline{locally on
$\underline{S}$}. That is, there exists a covering $\coprod U_{i}\to S$
in Zariski topology such that $\mathcal{E}|_{U_{i}}$ has a Weierstrass
equation for all $i$. 

More precisely, we'll ``show'' the following: $\omega_{\mathcal{E}/S}:=\pi_{*}(\Omega_{\mathcal{E}/S}^{1})$
is an invertible sheaf on $S$, and any local basis $\omega$ of $\omega_{\mathcal{E}/S}$
(over some $U\subset S$ open) determines a Weierstrass equation over
$U$. If $2$ is invertible on $S$, we can do this in such a way
that $\omega=-\frac{dx}{2y}$. \end{prop}
\begin{proof}[Proof (Sketch)]
 The invertibility of $\omega_{\mathcal{E}/S}$ comes from a calculation
in sheaf cohomology, c.f., pg53 of Mumford ``Abelian Varieties''. 

Now, given $U$ and $\omega$ a basis of $\pi_{*}(\Omega_{\mathcal{E}/U}^{1})$,
this gives local parameter on $\mathcal{E}$ at $0$ such that $\omega=dT(1+\mathrm{higher\, order\, terms})$.
$T$ is called a ``local parameter adapted to $\omega$''. Now $\pi_{*}\left(\mathcal{O}_{\mathcal{E}}(2(0))\right)$
is locally free of rank $2$ over $U$. Assume $U=\Spec(A)$ affine,
then $\pi_{*}(\mathcal{O}_{\mathcal{E}}(2(0)))$ is $A\cdot(1,x)$
where $x=\frac{1}{T^{2}}(1+\dots)$. Similarly:
\begin{itemize}
\item $\pi_{*}(\mathcal{O}_{\mathcal{E}}(3(0))$ is $A(1,x,y)$ where $y=\frac{1}{T^{3}}(1+\dots)$,
\item $\pi_{*}(\mathcal{O}_{\mathcal{E}}(4(0))$ is $A(1,x,y,x^{2})$
\item $\pi_{*}(\mathcal{O}_{\mathcal{E}}(5(0))$ is $A(1,x,y,x^{2},xy)$
\end{itemize}
Note $y^{2}-x^{3}\in\pi_{*}(\mathcal{O}_{\mathcal{E}}(5(0))$, so
$y^{2}-x^{3}\in A(1,x,y,x^{2},xy)$ and that's a Weierstrass equation
over $A[x,y]$. Moreover $dx=\frac{-2dT}{T^{3}}+\dots$ and $y=\frac{1}{T^{3}}+\dots$,
so $\frac{-dx}{2Y}=\omega\mod TdT$. (We don't have such a nice characterization
of the Weierstrass equation if $2$ is not invertible on $S$. \end{proof}
\begin{defn}
\label{3.3.3} For $S$ a scheme, $\alpha,\beta\in\Gamma(S,\mathcal{O}_{S})$,
let $E(\alpha,\beta)$ be the subscheme of $\pp_{S}^{2}$ defined
by $Y^{2}Z+\alpha XYZ+\beta YZ^{2}=X^{3}+\beta X^{2}Z$, and let $\Delta(\alpha,\beta)=-\beta^{3}(\alpha^{4}-\alpha^{3}+8\alpha^{2}\beta-36\alpha\beta+16\beta^{2}+27\beta)$
be its discriminant.

If $\Delta(\alpha,\beta)\in\Gamma(S,\mathcal{O}_{S})^{*}$ this is
an elliptic curve over $S$. Note that $P=(0:0:1)\in E(S)$ and we
calculate 
\begin{itemize}
\item $P=(0:0:1)$
\item $2P=(-\beta:\beta(\alpha-1):1)$
\item $3P=(1-\alpha:\alpha-\beta-1:1)$
\item $-P=(0:-\beta:1)$
\item $-2P=(-\beta:0:1)$
\item $-3P=(1-\alpha:(\alpha=1)^{2}:1)$
\end{itemize}
\end{defn}
This means that $P$ does not have order $1,2$ or $3$ in any fibre. 
\begin{prop}
\label{prop:3.3.4} For any scheme $S$, $E$ an elliptic curve over
$S$, and $P\in E(S)$ such that $P,2P,3P\neq0$ on any fibre, there
exists unique $\alpha,\beta\in\Gamma(S,\mathcal{O}_{S})$ such that
$\Delta(\alpha,\beta)\in\Gamma(S,\mathcal{O}_{S})^{*}$ and a unique
isomorphism $E(\alpha,\beta)\overset{\sim}{\to}E$ mapping $(0,0)$
to $P$. \end{prop}
\begin{proof}
First, assume $E$ has a Weierstrass equation over $S$. By a translation
$x\mapsto x+s$, $y\mapsto y+t$ we can assume $P=(0,0)$. Since $P$
does not have order $2$ in any fibre, the gradient of tangent line
at $P$ is in $\Gamma(S,\mathcal{O}_{S})$, so by replacing $y$ with
$y+rx$ for some $r$ we can put equations in the form $Y^{2}+a_{1}XY+a_{3}Y=X^{3}+a_{2}X^{2}$,
with $a_{i}\in\Gamma(S,\mathcal{O}_{S})$. 

Since $P$ does not have order $3$ in any fibre, $(0,0)$ is not
an inflexion point, so $a_{2}\in\Gamma(S,\mathcal{O}_{S})^{*}$. So
by scaling $x\mapsto u^{2}x$ and $y\mapsto u^{3}y$, we have that
$a_{2}=a_{3}$. Then $E$ is $E(\alpha,\beta)$ where $\alpha=a_{1},\beta=a_{2}$.

This gives an isomorphism to a curve in Tate normal form.

Now consider a general $E$ over $S$. We know there exists an affine
covering $S=\cup_{i}U_{i}$ such that $E|_{U_{i}}$ has a Weierstrass
equation over $\Gamma(U_{i},\mathcal{O}_{S})$. So we get $\alpha_{i},\beta_{i}\in\Gamma(U_{i},\mathcal{O}_{S})$
such that $(E|_{U_{i}},P|_{U_{i}})\cong(E(\alpha_{i},\beta_{i}),(0,0))$.
Since $\alpha_{i},\beta_{i}$ are \uline{unique}, they must agree
on $U_{i}\cap U_{j}$. By the sheaf property of $\mathcal{O}_{S}$,
we have that there exists $\alpha,\beta\in\Gamma(S,\mathcal{O}_{S})$
such that $\mbox{res}_{U_{i}}(\alpha)=\alpha_{i}$ and $\mbox{res}_{U_{j}}(\beta)=\beta$
for all $i$. Then $(E,P)\cong(E(\alpha,\beta),(0,0))$.\end{proof}
\begin{rem*}
The last step used in an essential way the \uline{uniqueness} of
$(\alpha,\beta)$: ``local uniqueness give global existence''.\end{rem*}
\begin{cor}
\label{cor:3.3.5}~
\begin{enumerate}
\item The pair $(\Spec\zz[A,b,\Delta(A,B)^{-1}],(E(A,B),(0,0))$ represent
the functor $\underline{\mathrm{Sch}}^{\mathrm{opp}}\to\underline{\mathrm{Set}}$
defined by $S\mapsto\left\{ (E,P)|E/S\mathrm{\, elliptic\, curve\, and\,}P\in E(S)\mathrm{\, not\, of\, order\,}1,2,3\mathrm{\, in\, any\, fibre}\right\} $
\item The pair $(\Spec\zz[B,\Delta(1+B,B)^{-1}],(E(1+B,B),(0,0)))$ represents
$S\mapsto\{(E,P)\big|E/S\mathrm{\, elliptic\, curve\, and\,}P\mathrm{\, point\, of\, exact\, order\,}5\,\mathrm{in\, every\, fibre}\}$
\end{enumerate}
\end{cor}
\begin{proof}
~
\begin{enumerate}
\item This is a restatement of Proposition \ref{prop:3.3.4}
\item Just equate $3P=-2P$. We have $3P=(1-A,A-B-1)$ and $-2P=(-B,0)$. 
\end{enumerate}
\end{proof}
\begin{note*}
$\Delta(1+B,B)=B^{5}(B^{2}+11B-1)$. The discriminant of the quadratic
is $5^{3}$. 
\end{note*}
Is it reasonable to define ``$Y_{1}(5)_{\zz}$'' to be $\Spec\zz[B,\Delta(1+B,B)^{-1}]$.
Sadly \uline{no}: our definition of ``point of exact order $5$''
is too naive in characteristic $5$. If $E$ over $\ff_{5}$ is \uline{supersingular},
$E[5]$ is a single point with multiplicity $25$. So there are no
points of order $5$, even over $\overline{\ff}_{5}$. So this scheme
$\Spec\zz[B,\Delta(1+B,B)^{-1}]$ has empty fibre over supersingular
$j$-invariants. 
\begin{defn}
\label{3.3.6}We set $Y_{1}(5)_{\zz[\frac{1}{5}]}=\Spec\zz\left[\frac{1}{5},B,\Delta(1+B,B)^{-1}\right]$,
and this represents the same functor as before on category of $\zz\left[\frac{1}{5}\right]$-schemes

More generally: For $N\geq4$, let $\mathcal{Y}_{N}=$ closed subscheme
of $\mathcal{Y}=\Spec\zz[A,B,\Delta(A,B)^{-1}]$, where $N\cdot(0,0,1)=(0,1,0)$,
and let 
\[
Y_{1}(N)_{\zz\left[\frac{1}{N}\right]}=\left(\mathcal{Y}_{N}-\bigcup_{d|N,4\leq d<N}\mathcal{Y}_{d}\right)\times_{\Spec\zz}\Spec\zz\left[\frac{1}{N}\right]
\]

\end{defn}
By construction this represents $S\mapsto(\mathrm{elliptic\, curve\,}E/S\,\mathrm{with\, point\, of\, exact\, order\,}N)$
on the category of $\zz[\frac{1}{N}]$-schemes. (More precisely, $Y_{1}(N)_{\zz\left[\frac{1}{N}\right]}$
has a universal elliptic curve over it by restricting $E(A,B)/\mathcal{Y}$,
and this has a point $(0,0)$ and $(Y_{1}(N)_{\zz\left[\frac{1}{N}\right]},(\mathrm{\, this\, curve,\, this\, point}))$
represents the above functor)

\textbf{Two natural questions}:
\begin{enumerate}
\item What does $Y_{1}(N)_{\zz\left[\frac{1}{N}\right]}$ look like? Is
it non-singular?
\item There exists a bijection of \uline{sets} between $Y_{1}(N)_{\zz\left[\frac{1}{N}\right]}(\cc)$
and $\Gamma_{1}(N)\backslash\mathcal{H}$. Is this a map of algebraic
varieties over $\cc$? 
\end{enumerate}

\subsection{Smoothness}
\begin{defn}
\label{3.4.1}A morphism of schemes $\phi:X\to Y$ is \emph{smooth}
if it's locally of finite presentation, flat, and for every point
$y\in Y$, the fibre $\phi^{-1}(y)$ is a smooth variety over $k(y)$. 
\end{defn}
So our definitions of elliptic curves over $S$ required that $\mathcal{E}\to S$
be a smooth morphism.
\begin{lem}
\label{lem:3.4.2}~
\begin{enumerate}
\item The composition of smooth morphism is smooth
\item If $E$ over $S$ is an elliptic curve and $N\geq1$ is invertible
on $S$, then $[N]:E\to E$ is smooth.
\end{enumerate}
\end{lem}
\begin{proof}
~
\begin{enumerate}
\item is standard (see EGA, follow trail of references from Wikipedia)
\item The morphism $\left[N\right]$ multiplies a global differential by
$N$, so it induces an isomorphism on tangent space, i.e., it's on
étale morphism. (and étale morphism are smooth)
\end{enumerate}
\end{proof}
\begin{prop}[Functorial criterion for smoothness]
\label{prop:3.4.3}  Let $X\to\Spec R$ be a scheme of finite type
over $R$, with $R$ Noetherian. The map $X\to\Spec R$ is a smooth
morphism if and only if it's ``formally smooth'', i.e., for every
local $R$-algebra $A$ and nilpotent ideal $I\subset A$, the map
$\Hom_{\mathrm{Scheme}/R}(\Spec A,X)\to\Hom_{\mathrm{Scheme}/R}(\Spec A_{0},X)$
is surjective, where $A_{0}=A/I$. 
\end{prop}
If we replace surjective with bijective, we get a notion of formally
étale
\begin{proof}
See \href{http://stacks.math.columbia.edu/}{Stacks Project}, section
36.9\end{proof}
\begin{thm}
\label{thm:3.4.4}$Y_{1}(N)_{\zz\left[\frac{1}{N}\right]}$ is smooth
over $\zz\left[\frac{1}{N}\right]$.\end{thm}
\begin{proof}
Let $A$ be a local $\zz\left[\frac{1}{N}\right]$-algebra, $I\subset A$
nilpotent. Let $(E_{0},P_{0})\in Y_{1}(N)(A_{0})$. Since $A_{0}$
is local, we have $E_{0}$ has a Weierstrass equation over $\Spec(A_{0})$.
Lift coefficients arbitrarily to $A$ to get an $E/A$ lifting of
$E_{0}$. (The discriminant $\Delta(E)$ is in $A^{*}$ as its image
in $A_{0}$ is in $A_{0}^{*}$). Can we lift $P_{0}$ to an $N$-torsion
point on $E$? In other words, is $E[N]$ smooth? But it is, because
$[N]:E\to E$ is smooth and a composition of smooth morphism is smooth.
($[N]$ composed with the structure map $E\to\Spec A$). So $(E_{0},P_{0})$
lifts to $(E,P)$ and we are done.\end{proof}
\begin{note*}
The schemes $\mathcal{Y}_{N}$ over $\zz$ are very rarely smooth,
this was true for $N=5$ essentially by accident. 
\end{note*}

\subsection{A complex - analytic digression}

Let $\Lambda\subset\cc$ be a lattice. 
\begin{defn}
\label{3.5.1}The Weierstrass $\wp$-function $\wp_{n}(z)$ is the
unique holomorphic function $\cc/\Lambda\to\pp^{1}(\cc)$ such that
$\wp_{n}(z)=\frac{1}{z^{2}}+\mathcal{O}(1)$ at $z=0$, and $\wp_{n}(z)$
is holomorphic away from $\Lambda$. (The $x$-coordinate correspond
to differential $dz$)
\end{defn}
The machinery of Proposition \ref{prop:3.3.2} implies that $\cc/\Lambda$
is isomorphic to $Y^{2}=X^{3}-g_{4}X-g_{6}$ via $z\mapsto(\wp(z),-\frac{1}{2}\wp'(z))$,
where $g_{4}$ and $g_{6}$ are constants depending on $\Lambda$. 

If $\Lambda=\Lambda_{\tau}:=\zz+\zz\tau$, then $g_{4}$ and $g_{6}$
are constant multiples of Eisenstein series $E_{4}$, $E_{6}$ respectively.
Hence they are \uline{holomorphic} functions of $\tau$. 
\begin{prop}
\label{prop:3.5.2} Let $\mathcal{U}=\left\{ \left(\tau,z\right)\in\mathcal{H}\times\cc:z,2z,3z\notin\Lambda_{\tau}\right\} $.
Then there exists a holomorphic map $U\overset{(\alpha,\beta)}{\to}\cc^{2}\setminus\{\alpha,\beta|\Delta(\alpha,\beta)=0\}$
such that $E(\alpha(\tau,z),\beta(\tau,z))\cong\cc/\Lambda_{\tau}$
and $(0,0)\leftarrow z\mod\Lambda_{\tau}$. \end{prop}
\begin{proof}
Start from the pair $(Y^{2}=X^{3}-g_{4}X-g_{6},(\wp(z),-\frac{1}{2}\wp'(z)))$
(with $g_{4},g_{6}$ correspond to $\Lambda_{\tau}$), manipulate
as in Proposition \ref{prop:3.3.4} to put this in Tate normal form.
All coefficients of rescaling and translations are holomorphic on
$U$ as functions of $(\tau,z)$. Hence so are the resulting $\alpha,\beta$. \end{proof}
\begin{cor}
\label{cor:3.5.3}$Y_{1}(N)_{\zz\left[\frac{1}{N}\right]}(\cc)$ is
isomorphic as a Riemann surface to $\Gamma_{1}(N)\backslash\mathcal{H}$. 
\end{cor}
Recall: We characterized $Y_{0}(N)$ over $\qq$ using $q$-expansions. 
\begin{prop}[Siegel, Kato 2004]
\label{prop:3.5.4} Let $E$ be an elliptic curve over $S$, $c>1$
an integer not divisible by $2$ or $3$. There exists a unique element
$_{c}\mathcal{\theta}_{E}\in\mathcal{O}(E\setminus E[c])^{*}$ with
the following properties:
\begin{enumerate}
\item $\mathrm{div}(_{c}\mathcal{\theta}_{E})=c^{2}\cdot\left(0\right)-E[c]$
\item $N_{a}(_{c}\mathcal{\theta}_{E})=c\mbox{\ensuremath{\theta}}_{E}$
for $a$ coprime to $c$, where $N_{a}$ is the norm map $\mathcal{O}(E\setminus E[ac])^{*}\to\mathcal{O}(E\setminus E[c])^{*}$
attached to the $a$-multiplication on $E$. 
\end{enumerate}
Moreover, if $E=\cc/(\zz+\zz\tau)$, $S=\cc$, we have $_{c}\theta_{E}=q^{\frac{c^{2}-1}{12}}(-t)^{-c(c-1)/2}\gamma_{q}(t)^{c^{2}}\gamma_{q}(t^{c})^{-1}$
where $t=e^{2\pi iz}$ (for $z\in\cc/(\zz+\zz\tau)$, $q=e^{2\pi i\tau}$
and $\gamma_{q}(t)=\prod_{n\geq0}(1-q^{n}t)\prod_{n\geq1}(1-q^{n}t^{-1})$.\end{prop}
\begin{proof}
First, note that this unit is unique if it exists. Assume some $f$
satisfying $1.$~and $2.$~exits. Any other $g$ satisfying $1\,$.~and
$2.$~ is $g=uf$ for some $u\in\mathcal{O}(S)^{*}$. Now $N_{3}(g)=g$,
then $N_{3}(uf)=uf$, so $u^{3^{2}}f=uf$, i.e., $u^{8}=1$. Similarly
$N_{2}(g)=g$ implies $u^{3}=1$. Now $u=(u^{3})^{2}(u^{8})^{-1}=1$.
Hence we get uniqueness.

It suffices to show that $c^{2}(0)-E[c]$ is \uline{locally on
$S$} a prime divisor. There exists a theory of ``relative Cartier
Divisor'' and a map $(\deg0\mathrm{\, divisor\, on\,}E)/(\mathrm{pullback\, of\, ones\, on\,}S)\to E(S)$.
Since $c^{2}(0)-E[c]=0$ in $E(S)$. Hence $c^{2}(0)-E[c]$ is the
pullback of a divisor on $S$, hence locally on $S$ principal.

Let $f$ be such that $\mbox{div}(f)=c^{2}(0)-E[c]$. Since $\mbox{div}(f)$
is invertible under $N_{a}$, we must have $N_{a}(f)=u_{a}f$ for
some $u_{a}\in\mathcal{O}(S)^{*}$. Since $N_{a}N_{b}=N_{b}N_{a}$,
$u_{a}^{(b^{2}-1)}=u_{b}^{(a^{2}-1)}$ for all $a,b$ coprime to $c$.
So if we put $g=u_{2}^{-3}u_{3}f$, we have 
\begin{eqnarray*}
N_{a}(g) & = & u_{2}^{-3a^{2}}u_{3}^{a^{2}}u_{a}f\\
 & = & u_{2}^{-3(a^{2}-1)}u_{3}^{(a^{2}-1)}u_{a}g\\
 & = & u_{a}^{-3(2^{2}-1)}u_{a}^{(3^{2}-1)}u_{a}g\\
 & = & u_{a}^{0}g\\
 & = & g.
\end{eqnarray*}
So $_{c}\theta_{E}$ exists locally and by uniqueness it exists globally. 

For the case $S=\cc$ and $E=E_{\tau},$we just check that the given
function has properties $1.$~and $2.$~ (c.f. the $3$rd problem
sheet)\end{proof}
\begin{defn}
\label{3.5.5}For $N\geq4$ and $c>1$ with $\gcd(c,6N)=1$, the \emph{Siegel
unit} $_{c}g_{N}$ is the pullback of $_{c}\theta_{\mathcal{E}}$
along the order $N$ section $Y_{1}(N)\to\mathcal{E}$, where $\mathcal{E}/Y_{1}(N)$
is the universal elliptic curve.\end{defn}
\begin{rem*}
These units are the building blocks of \uline{Euler systems}. (c.f.
Kato's paper (Asterisque 295,2004), and Loefler's paper with Lei and
Zerbes, 2013)\end{rem*}
\begin{namedthm}[Important Corollary]
 $Y_{1}(N)$ is \uline{not} characterised over $\qq$ by having
$q$-expansions of elements of $\qq(Y_{1}\left(N\right))$ in $\qq((q))$. \end{namedthm}
\begin{proof}
Calculate $q$-expansions of $_{c}g_{5}\in\qq(Y_{1}(5))^{*}$. The
order $N$ sections is $z=\frac{1}{5}\mod\zz+\zz\tau$, so $t=e^{2\pi iz}=e^{2\pi i/5}\notin\qq$,
$_{c}g_{5}=q^{(c^{2}-1)/12}(-e^{2\pi i/5})^{-c(c-1)/2}\prod\left(\dots\right)$,
which has $\zeta_{5}$'s everywhere. 
\end{proof}
One can show: $f\in\qq(Y_{1}(N))$ if and only if $f\in\cc(Y_{1}(N))$
and its $q$-expansion lands in $\qq(\zeta_{N})((q))$ and satisfies
$a_{n}(f)^{\sigma}=a_{n}\left(\left\langle \sigma\right\rangle f\right)$
for all $\sigma\in\mbox{Gal}(\qq(\mu_{N})/\qq)\cong(\zz/N)^{*}$. 


\subsection{Quotients and $Y_{0}(N)$}
\begin{prop}
\label{prop:3.6.1}Let $X$ be a quasi-projective $S$-scheme (for
some base scheme $S$) and $G$ a finite group action on $X$ by $S$-automorphisms.
Then there exists a unique $S$-scheme $X/G$ and a morphism $X\to X/G$
representing the functor $Y\mapsto(\mathrm{homomorphism\, of\,}S\mathrm{\, shcemes}\, X\to Y\mathrm{\, commuting\, with\,}G\mathrm{-actions})$.\end{prop}
\begin{proof}
Uniqueness is obvious (representing a functor). So we prove existence:
for $X=\Spec(A)$ affine, then $\Spec(A^{G})$ works; and can show
these patches nicely. (we need quasi-projectivity and finiteness of
$G$ here)\end{proof}
\begin{defn}
\label{3.6.2}For $N\geq4$ let $Y_{0}(N)=Y_{1}(N)/\left(\zz/N\zz\right)^{*}$.
(as a $\zz\left[\frac{1}{N}\right]$-scheme).
\end{defn}
The $\cc$-points of this are $\Gamma_{0}(N)\backslash\mathcal{H}$. 


\subsubsection*{Construction}

Let $S$ be a $\zz\left[\frac{1}{N}\right]$-scheme. There is a map
\begin{alignat*}{2}
\mathrm{} & \left\{ \mathrm{\, isomorphism\, classes\, of\, pairs\,}(E,C)\right\} \\
 & \left\{ E/S\mathrm{\, elliptic\, curve,\,}C\subset E\,\mathrm{subgroups-scheme}\right\}  & \to Y_{0}(N)(S)\\
 & \left\{ \mathrm{\acute{e}tale\, locally\, isomorphic\, to}\,\zz/N\zz\right\} 
\end{alignat*}
defined as follows: Let $(E,C)$ be in the LHS; then there exists
$S'\to S$ étale and $P\in E(S')$ such that $C=\left\langle P\right\rangle $,
and this gives a point of $Y_{1}(N)(S')$. Channing $P$ changes this
by an element of $G=\left(\zz/N\zz\right)^{*}$. So we get a $G$-orbit
of elements of $Y_{1}(N)(S')$. By a scar lemma (``étale descent
of morphism'') this gives an $S$-point of $Y_{0}(N)$. Thus we have
a well-defined map $L_{S}:\left\{ (E,C)/S\right\} \to Y_{0}(N)(S)$.
In general this is neither injective nor surjective, but if $S$ is
$\Spec(\overline{k})$ for some $\overline{k}$ algebraic closed,
it's a bijection. 

\uline{Injectivity}: If $L/K$ is a finite field extension $Y_{0}(N)(K)\to Y_{0}(N)(L)$
is obviously injective, but $((E,C)/K)\to((E,C)/L)$ is not injective
(there exists obsturctions coming from quadratic twists, etc). For
a field $k$, we can check that the image $(L_{k})$ is the set of
pairs $(E,C)$ defined over $k$ modulo isomorphism over $\overline{k}$. 

\uline{Surjectivity}: We can show that for a field $k$, $L_{k}$
is surjective (fairly hard, c.f. Proposition VI.3.2 of Deligne - Rapopok)
but for non-field $S$ surjectivity can also fail. For instance, $S=Y_{0}(N)$
itself; in general there is no $(E,C)$ which correspond to the identity
map. (Can try to use $E/\left(\zz/N\zz\right)^{*}$ but fibres over
points of $Y_{0}(N)$ with nontrivial stabilizers might not be elliptic
curves!). 
\begin{fact*}
$Y_{0}(N)$ is smooth over $\zz\left[\frac{1}{N}\right]$, and it
agrees with our earlier construction over $\qq$. (Sketch of last
point, it suffices to show $j(z)$ and $j(Nz)$ lies in $\qq(Y_{1}(N))^{G}$,
just take $j(E)$ and $j(E/\left\langle P\right\rangle )$.)
\end{fact*}

\subsection{General Modular Curves}

This section is following Katz - Mazur.
\begin{defn}
\label{3.7.1} Let $R$ be a ring:
\begin{enumerate}
\item Let $\underline{\mathrm{Ell}/R}$ be the following category:

\begin{itemize}
\item Objects are diagrams $E\to S$ where $S$ is some $R$-scheme and
$E$ is an elliptic curve
\item Morphisms are squares
\[
\xymatrix{E\ar[r]\ar[d] & E'\ar[d]\\
S\ar[r] & T
}
\]
where $E\cong E'\times_{T}S$. 
\end{itemize}
\item A \emph{moduli problem for elliptic curves over $R$} is a contravariant
functor $\mathcal{P}:\underline{\mathrm{Ell}/S}\to\underline{\mathrm{Set}}$.
\item We say that $\mathcal{P}$ is:

\begin{itemize}
\item \emph{representable} if it is representable. 
\item \emph{relatively representable} if, for every $E/S\in\mathrm{Obj}(\underline{\mathrm{Ell}/R})$,
the functor $\underline{\mathrm{Sch}/S}\to\underline{\mathrm{Sets}}$
defined by $T\mapsto\mathcal{P}(E\times_{S}T/T)$ is representable. 
\end{itemize}
\end{enumerate}
\end{defn}
Aside: The category $\underline{\mathrm{Ell}/R}$ is ``$\underline{\mathrm{Sch}/Y}$
for a $Y$ that doesn't exists''. If functor $S\mapsto\left\{ \mathrm{ell-curve}/S\right\} $
were representable, by some $(Y,E/Y)$, then objects of $\underline{\mathrm{Ell}/R}$
would be maps $S\to Y$. This is the idea of \uline{stacks}. 
\begin{prop}
\label{prop:3.7.2} For $\mathcal{P}$ a moduli problem let $\widetilde{\mathcal{P}}:\underline{\mathrm{Sch}/R}\to\underline{\mathrm{Sets}}$
be defined by $S\mapsto\left(\mathrm{pairs\,}(E,\alpha),\, E/S\mathrm{\, ell\, curve},\alpha\in\mathcal{P}(E,S)\right)$.
If $\mathcal{P}$ is representable on $\underline{\mathrm{Ell}/R}$,
then $\widetilde{P}$ is representable on $\underline{\mathrm{Sch}/R}$. 

(The converse is not quite true).\end{prop}
\begin{proof}
If $(E/S,\alpha)$ represents $\mathcal{P}$, can check $(S,(E,\alpha))$
represents $\widetilde{\mathcal{P}}$. \end{proof}
\begin{defn}
\label{3.7.3} $\mathcal{P}$ is \emph{rigid} if for all all $E/S\in\mathrm{Obj}(\underline{\mathrm{Ell}/R})$,
$\mathrm{Aut}(E/S)$ acts on $\mathcal{P}(E/S)$ without fixed points. \end{defn}
\begin{xca*}
~
\begin{enumerate}
\item A representable functor is rigid.
\item If $\mathcal{P}$ is rigid and $\mathcal{\widetilde{P}}$ is representable,
then $\mathcal{P}$ is representable.
\end{enumerate}
\end{xca*}
\begin{thm}[Katz - Mazur]
\label{thm:3.7.4}  $\mathcal{P}$ is representable if and only if
it is relatively representable and rigid. \end{thm}
\begin{proof}[Sketch of Proof]
 Start from $2$ basic moduli problems:
\begin{itemize}
\item ``naive level $\Gamma(3)$'' over $\zz\left[\frac{1}{3}\right]$. 
\item ``Legendre moduli problem'' $(\Gamma(2)$ and choice of differential)
over $\zz\left[\frac{1}{2}\right]$. 
\end{itemize}
Both have group action ($\GL_{2}(\ff_{3})$ and $\GL_{2}(\ff_{2})\times\left\{ \pm1\right\} $).
Given $\mathcal{P}$ is relatively representable and rigid, construct
one object by taking $\mathcal{E}/Y(3)$ - relative representability
gives us a scheme over $Y(3)$ and this has a $\GL_{2}(\ff_{3})$
action. Take inverse (this is fine since $\mathcal{P}$ is rigid),
and this gives an object $\mathcal{E}/S$ representing $\mathcal{P}$
on $\mathrm{Ell}/R[\frac{1}{3}]$. Legendre gives an object over $R[\frac{1}{2}]$
similarly. By rigidity these agree over $R\left[\frac{1}{6}\right]$,
so we get a representing object over $R$. 
\end{proof}

\subsection{General Level Structure}

Fix $N$ and a subgroup $H\subset\GL_{2}(\zz/N\zz)$. 
\begin{fact}
\label{3.8.1}There exists a moduli problem $\mathcal{P}_{H}$ on
$\mathrm{Ell}/\zz\left[\frac{1}{N}\right]$ such that if $\overline{k}$
is algebraic closed, $E/\overline{k}\in\mathrm{Obj}\left(\mathrm{Ell}/\zz\left[\frac{1}{N}\right]\right)$:
\[
\mathcal{P}_{H}(E/\overline{k})=\left\{ H\mathrm{-orbits\, of\, isomorphisms\,}\left(\zz/\zz N\right)^{2}\overset{\sim}{\to}E[N]\right\} .
\]

\end{fact}
For $H=\left\{ \begin{pmatrix}1 & 0\\
0 & 1
\end{pmatrix}\right\} $, this is $\Gamma(N)$, $E/S\mapsto\left(\mathrm{pairs\, of\, sections\,}P,Q\in E[S]\mathrm{\, generating\,}E[N]\,\mathrm{in\, every\, fibre}\right)$.

For $H=\left\{ \begin{pmatrix}* & *\\
0 & 1
\end{pmatrix}\right\} $, this is the $\Gamma_{1}(N)$.

For $H=\left\{ \begin{pmatrix}* & *\\
0 & *
\end{pmatrix}\right\} $, it's $\Gamma_{0}(N)$. 
\begin{rem*}
If $k$ is a field, $E/k$, then the image of $\mathcal{P}_{H}(E/k)$
in $\mathcal{P}_{H}(E/\overline{k})$ is $\left\{ H\mathrm{-orbit\, of\, bases\, of\,}E[N]\left(\overline{k}\right)\,\mathrm{in\, which\, image\, of\, Gal}(\overline{k}/k)\mathrm{\, lands\, in\,}H\right\} $. \end{rem*}
\begin{prop}
\label{prop:3.8.2} $\mathcal{P}_{H}$ is relative representable and
``étale over $\mathrm{Ell}/\zz\left[\frac{1}{N}\right]$''. (This
means: for all $E/S\in\mathrm{Obj}(\mathrm{Ell}/\zz\left[\frac{1}{N}\right])$,
the functor $T\mapsto\mathcal{P}_{H}(E\times_{S}T)$ is represented
by an étale $S$-scheme.)\end{prop}
\begin{proof}
For $H=\left\{ 1\right\} $, for $E/S\in\mathrm{Obj}(\mathrm{Ell}/\zz\left[\frac{1}{N}\right])$,
we can find an explicit $S$-scheme representing $\mathcal{P}_{H}$
on Schemes over $S$. It's an open subscheme of $E[N]\times_{S}E[N]$
given by non-vanishing of Weil pairing. 

For general $H$ just take the quotient of this by $H$. 
\end{proof}
So it is easier to relatively represent $\mathcal{P}_{H}$ than it
is to define it.
\begin{prop}
\label{prop:3.8.3}$\mathcal{P}_{H}$ is rigid on $\mathrm{Ell}/R\left[\frac{1}{6}\right]$
if and only if the preimage in $\SL_{2}(\zz)$ of $H\cap\SL_{2}(\zz/N\zz)$
contains no elements of finite order (i.e., has no elliptic points
and does not contain $-1$). \end{prop}
\begin{proof}[Sketch of Proof]
 Over $\cc$ this is routine. To prove general statement it suffices
to check it on objects $E/\overline{k}$, where $\overline{k}$ is
algebraic closed. If $\overline{k}$ has characteristic $0$, we can
embed it into $\cc$.

We can show that, if $k$ has finite characteristic $\geq5$, $E/k$
an elliptic curve and $\phi\in\mbox{Aut}(E)$, then the pair $(E,\phi)$
lifts to characteristic $0$. (c.f. somewhere in chapter VI of Deligne
- Rapoport)
\end{proof}
This gives a complete classification of modular curves and their associated
moduli problems.
\begin{rem*}
~
\begin{enumerate}
\item As in the case of $Y_{0}(N)$ for $H$ non-rigid, we can still construct
a $\zz\left[\frac{1}{N}\right]$-scheme which is ``the best approximation''
to representing $\widetilde{\mathcal{P}}_{H}$; we have a map $\widetilde{\mathcal{P}}_{H}(S)\to Y(S)$,
which is surjective for $S$ a field, and a bijection if for $S$
algebraic closed. 
\item If $\Gamma=$ preimage$(H)\subset\SL_{2}(\zz)$, then $Y_{\mathcal{P}_{H}}(\cc)$
is not quite $\Gamma\backslash\mathcal{H}$. It's a \uline{union}
of such things corresponding to quotient $(\zz/N\zz)^{*}/\det H$.
In particular, our version of $Y(N)$ is not geometrically connected.
We can write $Y_{\mathcal{P}_{H}}(\cc)$ more intrinsically as $\GL_{2}(\qq)\backslash\GL_{2}(\aa)/(\rr_{>0}\cdot\mathrm{SO}_{2}(\rr)\cdot U)$
where $U=$ preimage$(H)\subset\GL_{2}(\widehat{\zz})\subset\GL_{2}(\aa_{\mathrm{fin}})$. 
\end{enumerate}
\end{rem*}
\newpage{}


\section{Leftovers}


\subsection{Katz Modular Forms}

Recall we defined, for $E$ an elliptic curve over $S,$$\omega_{E/S}=\pi_{*}(\Omega_{E/S}^{1})$.
\begin{prop}
\label{prop:4.1.1}If $S=Y_{\mathcal{P}_{H}}$ for some $H$ as before,
$E/S$ universal elliptic curve, then $\omega_{E/S}$ is the ``Katz
Sheaf'' from Chapter 2.\end{prop}
\begin{proof}
Just unravel the definitions. We will show that both have the same
pullback to $\mathcal{H}$ and the actions of $\Gamma$ agree.

By definition, the pullback of $\omega_{\mathrm{Katz}}$ is $\cc$.
The pullback of $\omega_{E/S}$ is $\pi_{*}($relative differentials
on $\cc/(\zz+\tau\zz))=\cc\cdot(2\pi idz)$. But the isomorphism $\cc/(\zz+\zz\tau)\cong\cc/(\zz+\gamma\tau\zz)$
for $\gamma\in\SL_{2}(\zz)$ is multiplication by $(c\tau+d)^{-1}$
on $\cc$, so it multiplies $dz$ by this constant. So the action
coincides with the one we defined in the construction of $\omega_{\mathrm{Katz}}$. \end{proof}
\begin{defn}
\label{4.1.2}For $\Gamma$ a torsion free congruent subgroup of level
$N$, $R$ a $\zz\left[\frac{1}{N}\right]$-algebra, we define $KM_{K}(\Gamma,R)=H^{0}(Y(\Gamma)\times R,\omega_{E/Y(\Gamma)\times R}^{k})$.
(This is an $R$-module). 
\end{defn}
Concretely: A Katz Modular form of weight $k$ over $R$ is a rule
attaching to each triple $(E/S,\alpha,\omega)$ - where $S$ is a
$R$-scheme, $E/S$ an elliptic curve, $\alpha\in\mathcal{P}_{H}(E/S)$,
$\omega$ a basis of $\Gamma(E,\Omega_{E/S}^{1})$ - an element of
$\Gamma(S,\mathcal{O}_{S})$ such that 
\begin{itemize}
\item compatible with base change in $S$
\item Homogeneous of weight $k$ in $\omega$. 
\end{itemize}
(C.f. Katz ``$P$-adic properties of modular schemes and modular
forms'', Springer LNM \#330). 

Fun thing: Over $R=\zz\left[\frac{1}{6}\right],$ for any elliptic
curve $E/R$ and $\omega\in\Omega^{1}$, there exists a unique short
Weierstrass equation such that $\omega=\frac{dx}{y}$, and $E_{4}$
(respectively $E_{6})$ are the maps $(E,\omega)\mapsto a_{4}$ coefficient
of this equation (respectively $a_{6}$). 


\subsection{Cups and the Tate curve}

Consider the ring $\zz((q))=\left\{ \sum_{n=-N}^{\infty}a_{n}q^{n}|a_{n}\in\zz\right\} $.
We'll define an elliptic curve over this and a differential, such
that evaluating at this pair gives $q$-expansion of a Katz MF. 
\begin{defn}
\label{4.2.1}Tate$(q)=$ the elliptic curve $y^{2}+ax=x^{3}+a_{4}x+a_{6}$
where $a_{4}=-\sum_{n\geq1}\frac{5n^{3}q^{n}}{1-q^{n}}$ and $a_{6}=-\sum_{n\geq1}\frac{(7n^{5}+5n^{3})/12\cdot q^{n}}{1-q^{n}}$.
(Note $a_{4},a_{6}\in\zz[[q]]$). The discriminant of this curve is
exactly the $q$-expansion of $\Delta(\mathrm{weight\,}12\mathrm{\, cusp\, form})$
in $q+q^{2}\zz[[q]]\subset\zz((q))^{*}$. 
\end{defn}
Hence $\mbox{Tate}(q)$ is an elliptic curve. Tate$(q)$ is ``the
$q$-expansion of $\cc/(\zz+\tau\zz)$'' $=$ ``$\cc^{*}/q^{\zz}$''. 
\begin{prop}
\label{prop:4.2.2}If $\tau\in\mathcal{H}$, then series defining
$\mathrm{Tate}(q)$ converge at $q=e^{2\pi i\tau}$ and define a curve
isomorphic to $\cc/(\zz+\tau\zz)$. 
\end{prop}
Convergence is easy, and we check if $j(\mbox{Tate}(q))$ is the $q$-expansion
of $j(\tau)$. 
\begin{prop}
\label{prop:4.2.3} Let $\oplus$ denote the group law on $\mathrm{Tate}(q)$.
There exists series $X(u,q),Y(u,q)\in\zz\left[u,u^{-1},(1-u)^{-1}\right][[q]]$
such that $(X(u,q),Y(u,q))\oplus(X(v,q),Y(v,q))=(X(uv,q),Y(uv,q))$.
(Interpret $X(u,q),Y(u,q)$ as $\infty$ if $u=1$). \end{prop}
\begin{proof}
Take 
\begin{eqnarray*}
X(u,q) & = & \frac{u}{(1-u)^{2}}+\sum_{d\geq1}\left(\sum_{m|d}m(u^{m}+u^{-m}-2\right)q^{d}\\
Y(u,q) & = & \frac{u^{2}}{(1-u)^{3}}+\sum_{d\geq1}\left(\sum_{m|d}\left\{ \frac{m(m-1)}{2}u^{m}-\frac{m(m+1)}{2}u^{-m}+m\right\} \right)q^{d}.
\end{eqnarray*}
Sneaky part, there exists a straightforward change of coordinates
from $\mathrm{Tate}(q)$ to $y^{2}=4x^{3}-g_{4}(\tau)x-g_{6}(\tau)$,
which is $\cc/(\zz+\tau\zz)$ via $(\wp(z,\tau),\wp'(z,\tau))$. Then
$X$ and $Y$ are just $\wp$ and $\wp'$ as power series in $u=e^{2\pi iz}$,
$q=e^{2\pi i\tau}$. 

So the identity $(X(u,q),Y(u,q))\oplus(X(v,q),Y(v,q))=(X(uv,q),Y(uv,q))$
holds for all $u,q$ in an open subset of $\cc\times\cc$, so it holds
as an identity of power series. \end{proof}
\begin{prop}
\label{prop:4.2.4}Cusps of $Y_{\mathcal{P}_{H}}\leftrightarrow\{\mathcal{P}_{H}$
level structures on $\mathrm{Tate}(q)$ over $\zz[[q^{\frac{1}{N}},\zeta_{N}]],$
modulo automorphism $q^{\frac{1}{N}}\mapsto\zeta_{N}^{a}q^{\pm\frac{1}{N}}\}$. 

And we thus get an action of $\mbox{Gal}(\qq(\mu_{N})/\qq)$.\end{prop}
\begin{example*}
$Y_{1}(5)$.

Points of order $5$ on $\mbox{Tate}(q)$ over $\zz\left[\frac{1}{5},\zeta_{5}\right][[q^{\frac{1}{5}}]]$
are (images of) $q^{a/N}\zeta_{n}^{b}$ , where $a,b\in\left(\zz/5\zz\right)^{2}\setminus\left\{ (0,0)\right\} $. \end{example*}

\end{document}
