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\title{The Modular Curves $\X$ and $\XX$}
\author{Tom Weston}

\begin{document}

\maketitle

This paper is intended as a brief introduction to the theory of moduli
spaces through the concrete examples of certain modular curves.  The
motivating question, which we seek to answer, is whether or not there
exist any rational elliptic curves over $\Q$ with a rational point of
order $11$.  This problem, which seems quite hopeless in terms of
explicit polynomials, turns out to have a beautiful solution in terms
of modular curves.

The portions of this paper dealing with explicit computations for
$\X$ are based on a talk given by Matthew Emerton, incorporating some
modifications by Keith Conrad and Robert Pollack.  Throughout the
paper all algebraic curves are assumed to be smooth and projective.
We give very few complete proofs; to give complete details would have
made the paper many times longer, and we will be content to give the
main ideas.

\section{The $j$-line}

For this section we are interested in classifying elliptic curves up
to isomorphism; if $K$ is a field, we will denote by $\Ell(K)$ the set
of isomorphism classes of elliptic curves defined over $K$.  Here we
consider two elliptic curves $E_{1}$ and $E_{2}$ over $K$ to be
isomorphic if there is an isomorphism $E_{1} \to E_{2}$ defined over
$K$.

We will begin with a description of $\Ell(\C)$.
Recall that an elliptic curve over $\C$ can be realized as $\C/\L$ for
some lattice $\L \subseteq \C$; here by lattice we mean a free $\Z$-module of
rank $2$ generated by two $\R$-linearly independent complex numbers.  
If $\omega_{1}$, $\omega_{2}$ are
complex numbers which are linearly independent over $\R$, we will write
$\left< \omega_{1},\omega_{2} \right>$ for the lattice which they generate.
Two complex elliptic curves
$\C/\L_{1}$ and $\C/\L_{2}$ are isomorphic (over $\C$) if and only if the
lattices $\L_{1}$ and $\L_{2}$ are {\it homothetic}; that is, if there
is some $\alpha \in \C^{*}$ such that $\alpha \L_{1} = \L_{2}$.
(To go from the $\C/\L$ form to the usual Weierstrass form
$y^{2}=x^{3}+ax+b$ one uses the Weierstrass functions and its
derivative; for details on all of this, see 
\cite[Chapter 6]{Silverman} and \cite[Chapter 1]{Silverman2}.)

Using this description of complex elliptic curves, we see that in order
to get a description of $\Ell(\C)$ it will
suffice for us to classify lattices up to homothety.  We summarize the
construction; see \cite[Chapter 7, Sections 1 and 2]{Serre} or
\cite[Chapter 1, Sections 1 and 2]{Silverman2} for more details.
The first step is to normalize the lattices somewhat.
Whenever we are dealing with a basis $\left<\omega_{1},\omega_{2}\right>$
of a lattice
$\L$, we will assume that $\omega_{1},\omega_{2}$ are ordered so that
$\im \omega_{2}/\omega_{1} > 0$.  This lattice $\L$ is homothetic to
the lattice $\Lt=\left<1,\tau\right>$, where $\tau=\omega_{2}/\omega_{1}$;
thus we can restrict our attention to lattices of this form $\Lt$ with
$\tau \in \H$, $\H$ denoting the upper half plane in $\C$.

Two such lattices can still be homothetic.  To determine when this
happens, we consider a single lattice $\Lt$.  The possible ordered
bases of $\Lt$ are precisely the bases $\left<c\tau+d,a\tau+b\right>$ with
$\abcd \in \SLZ$.  (Recall that $\SLZ$ is the group of $2 \times 2$ matrices
with integer entries and determinant $1$.
Here the positive determinant condition insures that
our basis will still satisfy $\frac{a\tau+b}{c\tau+d} \in \H$.)  These lattices
are in turn homothetic to the lattices $\L_{\tau'}$ with
$\tau' = \frac{a\tau+b}{c\tau+d}$, and these are the only lattices of this
form which are homothetic to $\Lt$.

We rephrase our results as follows: we define an action of the group
$\SLZ$ on $\H$ by
$$\abcd \tau = \frac{a\tau+b}{c\tau+d}.$$
(One checks easily that this really is a group action.)
We have a map from $\H$ to the set of homothety classes of lattices
given by $\tau \mapsto \Lt$, and two lattices $\Lt$ and $\L_{\tau'}$ are
homothetic if and only if there is some 
$\gamma \in \SLZ$ with $\gamma\tau=\tau'$.  That is,
we can regard isomorphism classes of lattices as parametrized by the
orbit space $\SLZ\bs\H$.  (We write the group in this quotient
on the left since it acts on $\H$ on the left.)
This in turn allows us to identify isomorphism classes
of complex elliptic curves with elements of $\SLZ\bs\H$:
\begin{align*}
\SLZ\bs\H &\bij \Ell(\C) \\
\tau &\longrightarrow \C/\Lambda_{\tau} \\
\frac{\omega_{2}}{\omega_{1}} &\longleftarrow \C/\left<\omega_{1},
\omega_{2}\right>
\end{align*}

One can show (see \cite[Chapter 7, Theorem 2]{Serre} or
\cite[Chapter 1, Corollary 1.6]{Silverman2})
that $\SLZ$ is generated by the two matrices
$$S=\left(\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right) \quad
T=\left(\begin{array}{cc} 1 & 1 \\ 0 & 1 \end{array}\right);$$
these act on $\H$ by $S(\tau)=-1/\tau$ and $T(\tau)=\tau+1$.  It follows easily
(see \cite[Chapter 7, Theorem 1]{Serre} or
\cite[Chapter 1, Proposition 1.5]{Silverman2}; in fact, one usually proves
the last two facts simultaneously)
that a fundamental domain for $\SLZ\bs\H$ is given by the following region:
(For this region to be a fundamental domain just means that every
$\SLZ$-orbit on $\H$ corresponds to a unique point of the region.)

\vspace{0.2in}

\centerline{\psfig{figure=ell1.eps}}

\vspace{0.2in}

The arrows on the boundary lines indicate that these lines
are identified by the given transformations.  $\rho$ is the sixth
root of unity $\frac{1}{2} + \frac{\sqrt{3}}{2}i$.

Note that on identifying the edges of this region one appears to obtain
a sphere with a single point removed (this point being infinitely far up the
imaginary axis).  

\centerline{\psfig{figure=ell5.eps}}

On inserting this point (which is called the {\it cusp at
infinity} and written as $i\infty$)
one obtains (at least topologically) a sphere.  In fact, it is
possible to define a complex structure on this space (this is the obvious
complex structure at most points, although one has to be a little careful at
$\rho$, $i$ and the cusp $i\infty$),
making it into a compact {\it Riemann surface};
see \cite[Chapter 1, Section 2]{Silverman}.
(A Riemann surface is essentially just a topological space for which every
point has a neighborhood isomorphic to an open subset of $\C$.)
When regarding it from this point of view, we will denote this Riemann
surface by $X(1)$; denoting by $\Cs$ the set containing the cusp, we
have
$$X(1) = (\SLZ\bs\H) \cup \Cs.$$
Our description above shows that it is isomorphic
the the {\it Riemann sphere} $\Pl^{1}(\C)$.

The Riemann sphere is usually obtained by adding a single
point at infinity to the complex plane $\C$.  From this point of view it
is easy to determine the meromorphic functions on $\Pl^{1}(\C)$: let
$f(z)$ be such a function, which we regard as a function on $\C$.  Since
$f(z)$ does not have an essential singularity at infinity, it only has
poles in some bounded region of $\C$ and therefore has finitely many
poles.  We can multiply $f(z)$ by polynomials to eliminate these poles, and
we are left with a function $g(z)$ which has a pole at infinity and no
other poles.  Suppose that this pole has order $n$.  The standard function 
with an order $n$ pole at $\infty$ is $z^n$, which is 
holomorphic on the rest of $\Pl^{1}(\C)$.  So subtract from $g(z)$
an appropriate multiple of $z^{n}$ to obtain a function $h(z)$ with
a pole at infinity of order at most $n-1$ and no other poles.
Continuing in this way we find that we can subtract a polynomial from $g(z)$
to obtain a function which has no poles on $\C$ or at infinity; by
Liouville's theorem such a function is a constant.  Thus $g(z)$ is a
polynomial, and $f(z)$ is rational function.  We conclude that the
field of meromorphic functions on $\Pl^{1}(\C)$ is just the field
$\C(z)$ of rational functions in $z$.  (The fundamental reason why there
are so few meromorphic functions on $\Pl^{1}(\C)$ is that it, unlike
$\C$, is compact.)

We conclude that the function field of $X(1)$ is generated
over $\C$ by a single transcendental function.  Regarding points of
$X(1)$ as isomorphism classes of 
elliptic curves $\C/\L$, the standard choice
of such a function is the function $j$ which sends an elliptic curve
to its $j$-invariant, which is a complex number which classifies it up
to isomorphism.  (See \cite[Chapter 3, Section 1]{Silverman}.)
Regarding $j$ as a function of the coordinate $\tau$ on $\H$, we have the
identity
$$j\left(\frac{a\tau+b}{c\tau+d}\right) = j(\tau)$$
for $\abcd \in \SLZ$,
since $\frac{a\tau+b}{c\tau+d}$ and $\tau$ give rise to the same point of
$X(1)$.  This means that $j$ is a {\it modular form of weight $0$},
otherwise known as a modular function.  In particular,
$j(\tau+1)=j(\tau)$, so we can write out a Fourier expansion for $j$ in terms
of $q=e^{2\pi i\tau}$.  It can be shown that
$$j(\tau) = \frac{1}{q} + 744 + 196884q + 21493760q^{2} + \cdots;$$
see \cite[Chapter 1, Remark 7.4]{Silverman2}.

Recall that the algebraic curve $\Pl^{1}_{\C}$ also has a function field
$\C(z)$.  This suggests that we should be able to regard the Riemann
sphere as an algebraic curve.  In fact, both complex 
algebraic curves and compact Riemann surfaces are uniquely determined by
their function fields, which are finitely generated field extensions of $\C$
of transcendence degree $1$.  This sets up a bijection between
complex algebraic curves and Riemann surfaces: we associate to the
(unique)
complex algebraic curve $X$ with function field $K$ the (unique)
Riemann surface $X'$ with function field $K$.  This bijection turns out
to respect all of the other relevant structure as well; in particular,
$X$ and $X'$ have the same sets of points: $X(\C) = X'(\C)$.

In summary, we have seen that the set of isomorphism classes of
complex elliptic curves is in natural bijection with the non-cuspidal
points of a complex algebraic curve $X(1)_{\C}$:
$$X(1)_{\C}(\C) \bij \Ell(\C) \cup \Cs.$$
We have also seen that $X(1)_{\C}$ is isomorphic to the projective line
$\Pl^{1}_{\C}$.

Of course, the curve $\Pl^{1}_{\C}$ can actually be defined over $\Q$ as
the projective line $\Pl^{1}_{\Q}$ with function field $\Q(j)$; the fact
that $\C\Q(j)=\C(j)$ means that $\Pl^{1}_{\Q}(\C) = \Pl^{1}_{\C}(\C)$.
Thus, setting $X(1)_{\Q} = \Pl^{1}_{\Q}$, we have a bijection
$$X(1)_{\Q}(\C) \bij \Ell(\C) \cup \Cs.$$
Since we have defined $X(1)_{\Q}$ over $\Q$, one might hope that we
also obtain a bijection between the $\Q$-valued points of $X(1)_{\Q}$ and
isomorphism classes of elliptic curves over $\Q$.  Unfortunately, this
is not the case: there is a map
$$\Ell(\Q) \to \Pl^{1}_{\Q}(\Q) = \Q \cup \{ \infty \}$$
sending an elliptic curve to its $j$-invariant, but it is not a bijection.
This is because 
if two elliptic curves over $\Q$ have the same $j$-invariant
it does not mean that there exists an isomorphism between them defined over
$\Q$, but only that such an isomorphism exists over $\Qbar$.  Thus the
non-cuspidal points $X(1)_{\Q}(\Q)$ classify elliptic curves over
$\Q$ only up to isomorphism over $\Qbar$.

\section{Modular pairs}

We now seek to redo the construction of the previous section for a slightly
different class of objects.  We define a {\it modular pair} over a field $K$
to be a pair $(E,C)$ of an elliptic curve over $K$ and a cyclic subgroup
$C$ of $E(\Kbar)$ of order $11$.  We further require that every $\sigma$ in
the Galois group $\Gal(\Kbar/K)$ maps $C$ to itself.  This is certainly
the case if $C$ actually lies in $E(K)$, but this is not necessary;
essentially all that we are requiring is that each $\sigma$ sends each
element of $C$ to a multiple of itself.
Two modular pairs $(E_{1},C_{1})$ and
$(E_{2},C_{2})$ over $K$ will be considered to be isomorphic if there is an
isomorphism of $E_{1}$ with $E_{2}$, defined over $K$, which sends
$C_{1}$ to $C_{2}$.

\begin{figure}
\centerline{\psfig{figure=ell2.eps}}
\caption{The $11$-torsion on a complex elliptic curve}
\end{figure}

We begin as before with the situation over the complex numbers.  Here
we can consider modular pairs as pairs $(\C/\L,C)$, and two pairs
$(\C/\L_{1},C_{1})$ and $(\C/\L_{2},C_{2})$ are isomorphic if there
exits a complex number $\alpha$
such that $\alpha\L_{1} = \L_{2}$ and $\alpha C_{1} \equiv C_{2}
\pmod{\L_{2}}$.

If $(\C/\L,C)$ is a modular pair, we can regard $C + \L$ as a complex
lattice in its own right; we have $\L \subseteq C + \L$, and the condition
that $C$ has order $11$ means precisely that $(C+\L)/\L$ has order $11$.
We can therefore think of modular pairs as pairs $\L_{1} \subseteq \L_{2}$ of
complex lattices such that $\L_{2}/\L_{1}$ has order $11$; the associated
modular pair is $(\C/\L_{1},\L_{2}/\L_{1})$.  By the structure theorem
for finitely generated abelian groups we can find a basis
$\left<\omega_{1},\omega_{2}\right>$
of $\L_{1}$ such that $\left<\frac{1}{11}\omega_{1},
\omega_{2}\right>$ is a basis for $\L_{2}$; as usual we can also assume that
$\omega_{2}/\omega_{1}$ has positive imaginary part.  This modular pair
is isomorphic to the modular pair $(\C/\Lt,\frac{1}{11}\Z)$ (corresponding
to the inclusion of lattices $\Lt \subseteq \frac{1}{11}\Z + \Lt$)
with $\tau=\omega_{2}/\omega_{1} \in \H$; the
isomorphism is given by the homothety $\omega_{1}^{-1} : C/\L_{1} \to
\C/\Lt$.

Combining all of this, we can associate to every modular pair 
$(\C/\L,C)$ a complex number $\tau \in \H$ such that $(\C/\L,C)$ is isomorphic
to the $(\C/\Lt,\frac{1}{11}\Z)$.  The difference between this and
the situation of
the previous section is that $\tau_{1},\tau_{2} \in \H$ give rise to
isomorphic modular pairs if and only if 
$(\C/\L_{\tau_{1}},\frac{1}{11}\Z) \cong
(\C/\L_{\tau_{2}},\frac{1}{11}\Z)$ as modular pairs.

To sort out this condition, we proceed as in the previous section and fix
one $\tau$.  The possible bases for $\Lt$ are those of the form
$\left<c\tau+d,a\tau+b\right>$
with $\abcd \in \SLZ$.  These in turn are associated to
lattices $\L_{\tau'}$ with $\tau' = \frac{a\tau+b}{c\tau+d}$.
However, not all
such $\tau'$ correspond to $\tau$ when the extra structure of the cyclic
subgroup of order $11$ is taken into account.  

We must determine for which $\abcd$ the modular pair
$(\C/\Lt,\frac{1}{11}\Z)$ is isomorphic to the modular pair
$(\C/\L_{\tau'},\frac{1}{11}\Z)$.  The homothety of $\Lt$ and
$\L_{\tau'}$ is given by multiplication by $c\tau + d$; this homothety
takes $\frac{1}{11}\Z$ to $\frac{1}{11}\Z$ if and only if
$$(c\tau+d)\frac{1}{11}\Z \equiv \frac{1}{11}\Z \pmod{\Lt}.$$
Since $d$ is an integer and $1$ and $\tau$ are linearly independent this
is the same as requiring that $\frac{c\tau}{11}$ is a multiple of $\tau$.
Of course, this is the case if and only if $11$ divides $c$.  We conclude
that a matrix $\abcd \in \SLZ$ takes $(\C/\Lt,\frac{1}{11}\Z)$
to $(\C/\L_{\tau'},\frac{1}{11}\Z)$ if and only if $11$ divides $c$.

Define $\G \subseteq \SLZ$ to be the subgroup of $\SLZ$ of such matrices:
$$\G = \{ \abcd \in \SLZ \mid c \equiv 0 \pmod{11} \}.$$
Our arguments to this point have shown that we can identify modular pairs
over $\C$
with points of $\H$ modulo the action of $\G$; that is, we can identify them
with the orbit space $\G\bs\H$:
\begin{align*}
\G\bs\H &\bij \Ell_{0}(\C) \\
\tau &\longrightarrow \left(\C/\Lambda_{\tau},\frac{1}{11}\Z\right) \\
\end{align*}

We would like a description of $\G\bs\H$ as a Riemann surface.  To do this
we will need a fundamental domain.  It can be shown that the matrices
$$T=\left(\begin{array}{cc} 1 & 1 \\ 0 & 1\end{array}\right) \qquad
U = \left(\begin{array}{cc} 7 & -2 \\ 11 & -3 \end{array}\right) \qquad
V = \left(\begin{array}{cc} 8 & -3 \\ 11 & -4 \end{array}\right)$$
generate $\G$.  Using this and a little care one can show that the region
below is a fundamental domain for the $\G$-action on $\H$.

\centerline{\psfig{figure=ell3.eps}}

When one identifies the lines $\tau=0$ and $\tau=1$, the region can be
drawn in the following simpler way:

\centerline{\psfig{figure=ell4.eps}}

We see now that $\H\bs\G$ is almost a torus, except that it is missing
two points.  We call the missing interior point the {\it cusp at
infinity} and the missing corner point the {\it cusp at zero},
and we let $\Cs_{0}$ denote the set of these two points.  Once
these points are added, $\G\bs\H$ is a torus.  In fact, as with $\SLZ\bs\H$,
it is possible to give
$\G\bs\H$ the structure of a Riemann surface which we denote
$\X$; see \cite[Chapter 11, Section 2]{Knapp}.
  From our construction, we have a bijection
$$\X \bij \Ell_{0}(\C) \cup \C.$$

It is important to note that we add the cusps only so that $\X$ becomes
compact; they do not correspond to honest elliptic curves.  In particular,
after we have defined $\X$ as an algebraic curve over $\Q$, the cusps
will yield two rational points which don't actually correspond to
elliptic curves.

\section{$\X$ as an elliptic curve over $\Q$}

The first key to using $\X$ to analyze elliptic curves over $\Q$ is to show
that it can actually be realized as a curve defined over $\Q$.  As it is,
we have $\X$ as a Riemann surface.  As we have said before, every Riemann
surface corresponds to a complex algebraic curve, so we
can also regard $\X$ in this way; when
we do so we will write it as $\X_{\C}$.  Let $L$ be its function field; it is
a finitely generated extension of $\C$ of transcendence degree $1$.

To show that $\X_{\C}$ can actually be defined over $\Q$, we must find a
finitely generated field extension $K$ of $\Q$ of transcendence degree $1$
such that $\C K = L$.  Indeed, we can then define $\X_{\Q}$ to be
the algebraic curve corresponding to $K$; the fact that $\C K
= L$ will insure that $\X_{\Q}(\C)$ recovers our previous complex
algebraic curve $\X_{\C}$.

There are several approaches to defining the field $K$.  The first
approach begins by explicitly determining $L$ as the field of
meromorphic functions on $\X$ regarded as a Riemann surface.
To do this, we must exhibit
some rational functions on $\X$.  The first is the function
$j$ we had on $X(1)$; it still makes sense as a function on
$\X$ since $\G \subseteq \SLZ$.

Let $j_{11}(\tau) = j(11\tau)$.  We claim that $j_{11}$ is also a
function on $\X$.  To show this, we must check that
$$j_{11}\left(\frac{a\tau+b}{c\tau+d}\right) = j_{11}(\tau)$$
for all $\abcd \in \G$, for then $j_{11}$ descends from a function on
$\H$ to a function on $\G\bs\H=\X$.  We compute
\begin{align*}
j_{11}\left(\frac{a\tau+b}{c\tau+d}\right) &=
j\left(11\frac{a\tau+b}{c\tau+d}\right) \\
&= j\left(\frac{a(11\tau)+11b}{\frac{c}{11}(11\tau)+d}\right) \\
&= j(11\tau) \\
&= j_{11}(\tau)
\end{align*}
since $\left(\begin{array}{cc} a & 11b \\ c/11 & d \end{array}\right)
\in \SLZ$ and $j$ is $\SLZ$-invariant.

We have now exhibited two meromorphic functions on $\X$.  One can
show that $L=\C(j,j_{11})$; that $j_{11}$ is a root of a polynomial
$\Phi(x)$ in $\Z[j,x]$; and that $\Phi(x)$ is irreducible as a 
polynomial over $\C(j)$.  See \cite[Chapter 11, Section 6]{Knapp}.

One can use the above assertions to realize $\X_{\C}$ as an algebraic
curve over $\Q$.  Indeed, we simply define $K=\Q(j)[j_{11}]/\Phi(j_{11})$,
where we are regarding $j$ as an abstract transcendental element and
$j_{11}$ as a formal variable; let $\X_{\Q}$ be the associated
algebraic curve over $\Q$.  This makes sense, since $\Phi(x)$ has
coefficients in $\Z[j]$.  The fact that $\Phi(x)$ is irreducible over
$\C(j)$ insures us that $\C K$ really is just $L=\C(j)[j_{11}]/\Phi(j_{11})$.

There is a second approach to defining $\X_{\Q}$.  This approach does not
use the Riemann surface description of $\X$ at all.  We only sketch the
main ideas; see \cite[Section 1]{Rohrlich} for details.  We begin with
the elliptic curve $E$
$$y^{2} = 4x^{3} - \frac{27t}{t-1728}x - \frac{27t}{t-1728}$$
defined over the field $\Q(t)$.  This curve has the property that its
$j$-invariant is just $t$.  Let $\Q(t,E[11])$ be the extension of $\Q(t)$
generated by the coordinates of the points of $E$ of order $11$.
Fixing a basis $P,Q$ of $E[11]$, we can regard $E[11]$ as a two dimensional
vector space over $\Z/11\Z$.  $\Gal(\Q(t,E[11])/\Q)$ acts on
$E[11]$ in a natural way, and we therefore obtain a map
$$\Gal(\Q(t,E[11])/\Q) \to \Aut(E[11]) \cong \GL_{2}(\Z/11\Z),$$
(the isomorphism coming from our fixed basis $P,Q$) which one easily
sees is injective.  It is a much deeper fact that it is actually
an isomorphism.  

Let $H$ be the subgroup of $GL_{2}(\Z)$ which maps the cyclic subgroup
generated by $Q$ to itself; one finds that
$$H = \left\{ \left(\begin{array}{cc} a & 0 \\ b & d \end{array}\right) \mid
a,d \in (\Z/11\Z)^{*}, b \in \Z/11\Z \right\}.$$
This corresponds to a subgroup of $\Gal(\Q(t,E[11])/\Q(t))$ by our above
isomorphism, and we let
$K$ be the fixed field of $\Q(t,E[11])$ by this subgroup.  One defines
$\X_{\Q}$ to be the algebraic curve over $\Q$ corresponding to
$K$.  Of course, one now has to show that $K$ really is the same as
the field we constructed before; this amounts to getting a better
understanding of the polynomial $\Phi(x)$.

Using either of the two methods, we have shown that the Riemann
surface $\X$ can be represented by an algebraic curve $\X_{\Q}$ over
$\Q$.  Recall that we found that the Riemann surface $\X$ is a torus;
that is, it has genus $1$.  Since the complex points on $\X_{\Q}$ yield
a space of genus $1$, it follows that $\X_{\Q}$ itself is an elliptic
curve.  $\X_{\Q}$ therefore has a Weierstrass equation; we will determine
what it is in the next section.

Now that we have realized $\X_{\Q}$ as a curve over $\Q$, we can ask if
our description of the points on this curve has the same interpretation
it had over $\C$.  That is, we can ask if the points $\X_{\Q}(\Q)$
correspond to pairs $(E,C)$ of elliptic curves over $\Q$ and cyclic
subgroups of order $11$ stable under $\Gal(\Qbar/\Q)$,
up to isomorphism over $\Q$.  Unfortunately, this
does not quite work out; there is a natural map
$$\{ \text{modular pairs $(E,C)$ over $\Q$ up to isomorphism} \} \to
\X_{\Q}(\Q) - \Cs_{0}$$
but it is a priori not necessarily a surjection or an injection.  We
will have more to say about these rational points later.

\section{An equation for $\X$}

Since $\X_{\Q}$ is an elliptic curve defined over $\Q$ we can hope to find
an equation for it.  This turns out to be a fairly involved exercise in
modular functions.  One could attempt to carry out these calculations with
the modular functions $j$ and $j_{11}$ which we have already introduced,
but it turns out that the coefficients are so large that computation becomes
difficult.  Instead we will introduce some other modular functions.  We
omit most of the details.

Our basic method is as follows: given an abstract genus $1$ curve
$E$ with a given point $O$,
one determines a Weierstrass equation
$$y^{2} + a_{1}xy + a_{3}y = x^{3} + a_{2}x^{2} + a_{4}x + a_{6}$$
by finding functions $x$ and $y$ on $E$ which have poles of order $2$
and $3$ at the point $O$ of $E$, respectively.  The Riemann-Roch
theorem shows that the space of functions on $E$ with poles of order at
most $6$ at $O$ and no other poles has dimension exactly
$6$; since we have the seven functions $y^{2},xy,y,x^{3},x^{2},x,1$ all
with poles of order at most $6$ at $O$ and no other poles, they must
satisfy a linear relation.  Renormalizing $x$ and $y$ one
obtains a Weierstrass equation
as above; see \cite[Chapter 3, Section 3]{Silverman}
for details.  We must find such functions $x$ and $y$
and then determine the linear relation.

We first exhibit some modular forms of weight $2$ for $\G$; recall
that these are functions $f : \H \to \C$ such that
$$f\left(\abcd \tau\right) = (c\tau+d)^{2}f(\tau)$$
for all $\abcd \in \G$.  The first is constructed as a {\it theta
series} (see \cite[Chapter 7, Section 6]{Serre} for a similar
construction): we let $Q(x,y) = x^{2} + xy + 3y^{2}$ be a quadratic form
of discriminant $11$ and define $r_{Q}(n)$ to be the number of integer
solutions $(x,y)$ to the equation $Q(x,y)=n$.  Set
\begin{align*}
\theta_{Q}(\tau) &= \sum r_{Q}(n)q^{n} \\
&= 1 + 2q + 4q^{3} +2q^{4} + 4q^{5} +6q^{9} + 2q^{11} + 4q^{12} + 8q^{15} +
\cdots
\end{align*}
where $q=e^{2\pi i\tau}$ as before.  One uses the Poisson
summation formula to show that
$$\theta_{Q}\left(\abcd\tau\right) = \pm(c\tau+d)\theta_{Q}(\tau)$$
for all $\abcd \in \G$; $\theta_{Q}^{2}(\tau)$ is therefore a modular
form of weight $2$ for $\G$.

There is another well known modular form of weight $2$ for $\G$
\begin{align*}
h(\tau) &= q\prod_{n \geq 1}(1-q^{n})^{2}(1-q^{11n})^{2} \\
&= q -2q^{2}-q^{3}+2q^{4}+q^{5}+2q^{6}-2q^{7} -2q^{9}-2q^{10}+\cdots;
\end{align*}
see \cite[Chapter 8, Corollary 8.9 and Chapter 9, Section 4, Example 5]{Knapp}.
The first modular function we will use is
\begin{align*}
F(\tau) &= \frac{\theta_{Q}^{2}(\tau)}{h(\tau)} \\
&= \frac{1}{q} + 6 + 17q + 46q^{2} + 116q^3+252q^4+533q^5+1034q^6
+1961q^7 + \cdots
\end{align*}
Since both $\theta_{Q}^{2}$ and $h$ have weight $2$, $F$ will have
weight $0$ and thus satisfies
$$F\left(\abcd\tau\right) = F(\tau)$$
for all $\abcd \in \G$.  As we saw before, this means that $F$ defines
a meromorphic function on the Riemann surface $\X$.

To obtain a second modular function, we begin with
$q\frac{dF}{dq}$, which one easily check (from the fact that $F$
is a modular function) is also modular of weight $2$ for $\G$.  We set
\begin{align*}
G(\tau) &= \frac{q\frac{df}{dq}(\tau)}{h(\tau)} \\
&= -\frac{1}{q^{2}} - \frac{2}{q} + 12 + 116q + 597q^{2} + 
2298q^3+7616q^4+22396q^5+ \cdots
\end{align*}
As with $F(\tau)$, $G(\tau)$ defines a meromorphic function on $\X$.

Let us choose to make the cusp $i\infty$ the identity point on the
elliptic curve $\X$.
We must find linear combinations of $F$ and $G$ with the required
poles at $i\infty$ and no other poles.  It is clear from the
Laurent series above that $F$ has a pole of order $1$ and $G$ has
a pole of order $2$ at $\tau=i\infty$ (which corresponds to $q=0$).
It is also not too hard to check
that $F$ and $G$ have no other poles on $\H$.  Unfortunately, it turns
out that they do have poles at the cusp $\tau=0$ (where $q=1$) and we will
have to account for these poles.

We do this using the {\it Atkin-Lehner involution} $w$.  This is an
automorphism of $\H$ given by $\tau \mapsto \frac{-1}{11\tau}$.  One
checks easily that it is compatible with the action of $\G$ on $\H$,
so it yields an involution $w$ of $\G\bs\H=\X$.  Note that $w$
interchanges the cusps $\tau=0$ and $\tau=i\infty$.
It can be shown using a trick involving the Poisson summation formula
that $F \circ w = F$ and $G \circ w = -G$.  To find a candidate
function for $x$, then, we need to find a polynomial $p(u,v)$ such
that $p(F,G)$ starts with $\frac{1}{q^{2}}$ and $p(F,G) \circ w =
p(F,-G)$ has no negative powers of $q$.  This is an exercise in linear
algebra; one of the simplest such polynomials is
$p(u,v) = (u^{2}-v-10v)/2$, which yields
\begin{align*}
x &= \frac{1}{q^{2}}+\frac{2}{q} -1+5q+8q^2+q^3+7q^4-11q^5+10q^6 + \cdots \\
x \circ w &= 22 + 242q+1210q^2+4598q^3+15246q^4+44770q^5+121484q^6
+\cdots
\end{align*}

We find $y$ in a similar way; the polynomial $p(u,v)=(-uv+u^3-10u^2-22u)/2$
works and yields
\begin{align*}
y &= \frac{1}{q^{3}}  + \frac{8}{q^{2}} + \frac{17}{q} + 13 +42q+
66q^2+24q^3+72q^4-70q^5 + \cdots \\
y \circ w &= 121 + 1331q+7986q^2+37268q^3+149072q^4+531069q^5+ \cdots
\end{align*}

To determine the polynomial satisfied by $x$ and $y$ we start with
$$y^{2}-x^{3} = \frac{10}{q^{5}} + \frac{89}{q^{4}} + \frac{287}{q^{3}}
+\frac{506}{q^{2}} + \frac{1111}{q} + 2606 + 3498q + 4729q^{2} +
\cdots$$
and successively add multiples of $xy$, $x^2$, $y$ and $x$ to cancel off
the rest of the poles.  Thus the next terms are
\begin{align*}
y^{2}-x^{3}-10xy &= -\frac{11}{q^{4}} - \frac{33}{q^{3}} + \frac{66}{q^{2}}
+\frac{121}{q} -264+198q+\cdots \\
y^{2}-x^{3}-10xy+11x^{2} &= \frac{11}{q^{3}} + \frac{88}{q^{2}}
+\frac{187}{q} + 143 + 462q + 726q^{2} + \cdots \\
y^{2}-x^{3}-10xy+11x^{2}-11y &= 0
\end{align*}
We have therefore found the Weierstrass equation
$$y^{2} -10xy -11y = x^{3}-11x^{2}.$$

This equation can be written in a different form which is in more common
usage: replacing $y$ by $y+5x-19$ and $x$ by $x-5$ yields the standard
form
$$y^{2}+y=x^{3}-x^{2}-10x-20.$$

\section{Rational points on $\X$}

At this point we have found that $\X$ can be represented by an
elliptic curve over $\Q$, and that an equation for this elliptic curve is
$$y^{2}+y=x^{3}-x^{2}-10x-20.$$
It is easy to determine the torsion subgroup of this curve: it is
the group of order $5$
$$\X_{\Q}(\Q)_{\tors} = \{ O,(5,5),(16,-61),(16,60),(5,-6) \}.$$
The rank of $\X_{\Q}(\Q)$ is more difficult to determine, although
there are methods to compute it. 
One finds (see \cite[p. 110]{Cremona} that $\X_{\Q}(\Q)$ has
rank $0$, so that
$$\X_{\Q}(\Q) = \{ O,(5,5),(16,-61),(16,60),(5,-6) \}.$$

Two of these rational points were expected: they correspond to the
cusps.  Since $\X_{\Q}(\Q)$ is not the same as $\Ell_{0}(\Q)$, we
can not at the moment give good interpretations of the other $3$ points.
Note that even if they do correspond to elements $(E,C)$ of $\Ell_{0}(\Q)$,
we can not be sure whether or not there are any elliptic curves over
$\Q$ with rational $11$-torsion; indeed, all we would know is that
there exists a rational elliptic curve $E$ and a point $P \in E[11]$
such that $\sigma(P)$ is a multiple of $P$ for all $\sigma \in
\Gal(\Qbar/\Q)$.


\section{Refined modular pairs and $\XX$}

Although we developed a good theory of the modular curve $\X$, it ended
up not being quite good enough to answer the question of whether or not
there exist elliptic curves over $\Q$ with rational $11$-torsion.  In
order to determine this, we will consider classifying a different collection
of pairs: we define a {\it refined modular pair} over a field $K$ to be
an elliptic curve $E$ defined over $K$ together with a point
$P \in E(K)$ of exact order $11$.  Two such pairs $(E_{1},P_{1})$ and
$(E_{2},P_{2})$ are said to be isomorphic if there exist an isomorphism
of $E_{1}$ and $E_{2}$, defined over $K$, which sends $P_{1}$ to $P_{2}$.
We will write $\Ell_{1}(K)$ for the set of isomorphism classes of
refined modular pairs $(E,P)$ over $K$.  The question of the existence
of rational elliptic curves with rational $11$-torsion is precisely the
question of whether or not $\Ell_{1}(\Q)$ is nonempty.

As usual, we begin by analyzing $\Ell_{1}(\C)$ using lattices.
First one normalizes the lattices to reduce to considering refined
modular pairs of the form $(\C/\Lt,\frac{1}{11})$ with $\tau \in \H$.
Proceeding as with $\Ell_{0}(\C)$, one finds that two such pairs
$(\C/\Lt,\frac{1}{11})$ and $(\C/\L_{\tau'},\frac{1}{11})$ are isomorphic
if and only if there is a matrix $\abcd \in \SLZ$ with
$$\abcd \tau = \tau',$$
and $a,d \equiv 1 \pmod{11}$ and $c \equiv 0 \pmod{11}$.  We will write
$\GG$ for the subgroup of $\SLZ$ of such matrices; our results to this
point are that there is a bijection
\begin{align*}
\GG\bs\H &\bij \Ell_{1}(\C) \\
\tau &\longrightarrow \left(\C/\Lambda_{\tau},\frac{1}{11}\right) \\
\end{align*}

$\GG$ turns out to be a significantly more complicated group than
$\G$, and there is no particularly simple fundamental domain for
$\GG\bs\H$.  However, it is still possible to do a little
combinatorial algebra and compute what $\GG\bs\H$
should look like: it turns out that we need to adjoin a set
$\Cs_{1}$ of $10$ cusps to $\GG\bs\H$ to obtain a compact Riemann surface
which we denote $\XX$:
$$\XX(\C) \bij \Ell_{1}(\C) \cup \Cs_{1}.$$

In fact, with the proper machinery (primarily involving a fairly well
developed theory of the modular functions for $\GG$;
see \cite[Example 9.1.6]{DI}) it can be shown
that $\XX$ still has genus $1$.  Methods similar to those we discussed
for $\X$ can also be used to show that $\XX$ can be defined as
an algebraic curve $\XX_{\Q}$ over $\Q$: for example, in the notation
of Section 4, $\XX_{\Q}$ can be realized
as the curve associated to the fixed field of $\Q(t,E[11])$ by
$$H_{1} = \left\{ \left(\begin{array}{cc} a & 0 \\ b & \pm 1 
\end{array}\right) \mid a \in (\Z/11\Z)^{*},b \in \Z/11\Z \right\},$$
considered as a subgroup of $\Gal(\Q(t,E[11])/\Q(t))$.
As always, we still have the interpretation
$$\XX_{\Q}(\C) \bij \Ell_{1}(\C) \cup \Cs_{1}.$$

However, there is an unexpected complication in this rational structure:
not all of the cusps actually lie in $\XX_{\Q}(\Q)$.  In fact,
it turns out that $5$ of the cusps lie in this group, while the other $5$
are defined over the maximal real subfield $\Q(\zeta_{11})^{+}$ of
$\Q(\zeta_{11})$; this is a degree $5$ extension of $\Q$.  It follows that
for a general field $K$ containing $\Q$, $\XX_{\Q}(K)$ contains either
$5$ or $10$ cusps, depending on whether or not $K$ contains
$\Q(\zeta_{11})^{+}$.  See \cite[Example 9.3.5]{DI} for a discussion.

\section{Moduli spaces}

The key to determining whether or not there exist elliptic curves over
$\Q$ with rational $11$-torsion is an understanding of the relationship
between the $\Q$-rational points of $\XX_{\Q}$ and $\Ell_{1}(\Q)$.  We
constructed $\XX$ so that its non-cuspidal complex points correspond to
$\Ell_{1}(\C)$, but it is not at all clear if the same interpretation should
hold over $\Q$.

This question is an example of a very important sort of problem in
modern algebraic geometry called a {\it moduli problem}.  Let $F(K)$
be some sort of set of geometric objects over a field $K$; for example,
$F(K)$ could be
\begin{itemize}
\item isomorphism classes of elliptic curves $E$ over $K$;
\item isomorphism classes of modular pairs $(E,C)$ over $K$;
\item isomorphism classes of refined modular pairs $(E,P)$ over $K$;
\item isomorphism classes of algebraic curves of genus $2$ over $K$;
\item algebraic surfaces embedded in $\Pl^{3}_{K}$;
\end{itemize}
or any other similar sort of set.  The key thing is that $F$ should be
defined for every field $K$ (say, of characteristic $0$) and that
for every inclusion of fields $K \inj L$ there should be a
{\it restriction map}
$$F(K) \to F(L)$$
such that for any tower of fields $K \inj L \inj M$, the composition
of the restriction maps $F(K) \to F(L)$ and $F(L) \to F(M)$ is just
the restriction map $F(K) \to F(M)$.  (In all of the examples above the
restriction map is the obvious one reinterpreting an object defined over
$K$ as defined over $L$.)  $F$ is an example of a {\it functor} from
the category of fields of characteristic $0$ to the category of sets.

The goal is to find a nice geometric object (say, a projective
variety defined over $\Q$)
$X$ such that for every field $K$ there is a bijection
$$X(K) \bij F(K).$$
(It should also be compatible with the restriction maps $X(L) \to X(K)$
and $F(L) \to F(K)$.)  We have never actually been able to do this,
even over $\C$; we always had to add a few extra points (the cusps)
in order to
fill in some holes.  This situation turns out to be very common; let us
agree to allow a ``small'' exceptional set of points as well.

If we could find such an $X$, we would suddenly have a powerful tool 
with which to study $F$.  For example, if we could find such an $X$ for
$\Ell_{1}$, then the question of the existence of elliptic curves
with $11$-torsion becomes a question about rational points on a single
variety.  This philosophy of using geometric objects to study whole
families of other geometric objects is fundamental in modern algebraic
geometry; such an $X$ is called a {\it fine moduli space} for $F$.

Unfortunately, fine moduli spaces do not exist for all $F$.  In fact,
they are fairly rare.  One simple necessary condition for the existence
of such a moduli space $X$ for $F$ is that the restriction maps
$F(K) \to F(L)$ be injective for all inclusions $K \inj L$; 
this is because the maps $X(K) \to X(L)$ are obviously injective,
and $F(K) \to F(L)$ must agree with this map.  This immediately shows
that $F=\Ell$ has no fine moduli space: indeed, take $K=\Q$ and
$L=\Qbar$.  There are many elliptic curves over $\Q$ which are not
isomorphic over $\Q$ but which have the same $j$-invariant.  Since they
have the same $j$-invariant, they become isomorphic over $\Qbar$; therefore,
these elliptic curves have the same image under $\Ell(\Q) \to \Ell(\Qbar)$,
so it is not injective and no fine moduli space can exist.

If $F$ does not have a fine moduli space, it still could happen that
it has a {\it coarse moduli space}.  This is a projective variety $X$ 
such that for every $K$ there is a map
$$F(K) \to X(K)$$
which satisfies certain technical properties, but which need not be
an isomorphism.  This is what $X(1)_{\Q}$ and $\X_{\Q}$ are for
$\Ell$ and $\Ell_{0}$.  Coarse moduli spaces are also useful, but it
requires more care to obtain information about $F$ from them.  In the
case of $\Ell_{0}$, the points $\X_{\Q}(K)$ turn out to correspond to
modular pairs $(E,C)$ defined over $K$, but only up to isomorphism
over $\Kbar$.  In particular, the natural map
$$\Ell_{0}(K) \to \X_{\Q}(K) - \Cs_{0}$$
is surjective but need not be injective.  Even given this, the
existence of rational points on $\X_{\Q}$ does not imply the existence
of elliptic curves over $\Q$ with rational $11$-torsion; for the
cyclic subgroup $C$ to be defined over $\Q$ means only that it is
mapped to itself under the action of $\Gal(\Qbar/\Q)$, not that it is
actually pointwise fixed.

Remarkably, $\XX_{\Q}$ turns out to be a fine moduli space for
$\Ell_{1}$ (once we have taken into account the cusps).  That is, for
every field $K$ containing $\Q$ there is a bijection
$$\XX_{\Q}(K) \bij \Ell_{1}(K) \cup \Cs(K)$$
where we remember that $\Cs(K)$ contains either $5$ or $10$ cusps
depending on whether or not $K$ contains $\Q(\zeta_{11})^{+}$.

These facts are quite difficult to prove.  The standard references
are \cite{DR} and \cite{KM}, but they both require a great deal of
algebraic geometric background.  There is a nice summary in
\cite[Part 2]{DI}, although even it requires a fair amount of
algebraic geometry.

We give a brief indication of why $\XX_{\Q}$ is better behaved than
our other examples.  The key fact turns out to be the following
rigidity statement (see \cite{KM}; this particular section requires
only familiarity with the material of \cite[Chapter 3]{Silverman}):
If $(E_{1},P_{1})$ and $(E_{2},P_{2})$ are refined modular pairs
defined over a field $K$, then there is at most one isomorphism
$E_{1} \cong E_{2}$ sending $P_{1}$ to $P_{2}$.

Given this, we can show that the restriction maps
$\Ell_{1}(K) \to \Ell_{1}(L)$ are injective for extensions $K \subseteq L$,
so that at least that obstruction to the existence of a fine moduli space
is avoided.  Unwinding the definitions, we must prove that if
$(E_{1},P_{1})$ and $(E_{2},P_{2})$ are modular pairs over $K$ for which
there exists an isomorphism $\varphi : E_{1} \to E_{2}$ defined over $L$
with $\varphi(P_{1}) = P_{2}$, then $\varphi$ is actually defined over $K$.
Let $\sigma$ be any element of $\Gal(L/K)$.  Acting on everything
by $\sigma$, we obtain an isomorphism $\varphi^{\sigma} : E_{1}^{\sigma}
\to E_{2}^{\sigma}$ sending $\sigma(P_{1}) \to \sigma(P_{2})$; here
by $E_{i}^{\sigma}$ we mean the elliptic curve with equation given by
acting on an equation of $E_{i}$ by $\sigma$.  But
since $E_{1}$ and $E_{2}$
are defined over $K$, $\sigma$ does not effect their defining equations
at all, and $E_{1}^{\sigma} = E_{1}$, $E_{2}^{\sigma} = E_{2}$.  The
same is true of $P_{1}$ and $P_{2}$, so we see that $\varphi^{\sigma}$
yields an isomorphism of modular pairs over $L$
$(E_{1},P_{1}) \cong (E_{2},P_{2})$.  By rigidity this must agree with
$\varphi$, so $\varphi = \varphi^{\sigma}$ for all $\sigma \in \Gal(L/K)$.
But by the main theorem of Galois theory (more or less) this means that
$\varphi$ is defined over $K$, which is what we were trying to show.

\section{The elliptic curve $\XX_{\Q}$}

Given what we said in the last section, to determine whether or not there
exists a rational elliptic curve with rational $11$-torsion is the same
as determining whether or not there are non-cuspidal rational points on
$\XX_{\Q}$.  To do this, we must find an equation for $\XX_{\Q}$.  This
can be done in principal as we did for $\X_{\Q}$, although it is more
complicated; in any event, one finds the equation
$$y^{2}+y = x^{3}-x^{2}.$$
This elliptic curve has $5$ torsion points:
$$\XX_{\Q}(\Q)_{\tors} = \{ O, (0,0), (1,-1), (1,0), (0,-1) \}.$$
The rank of $\XX_{\Q}$ is, as always, more difficult to determine.
By \cite[p. 110]{Cremona}, it can be shown to have rank $0$.
Thus,
$$\XX_{\Q}(\Q) = \{ O, (0,0), (1,-1), (1,0), (0,-1) \}.$$

So $\XX_{\Q}$ has exactly five rational points.  But recall that we
expected five rational 
cusps!  Thus $\XX_{\Q}$ has no non-cuspidal rational
points; $\Ell_{1}(\Q)$ is therefore empty, and we conclude that there
are no elliptic curves over $\Q$ with rational points of order $11$!

\section{Generalizations}

Of course, very little specific to the number $11$ came into our
construction of $\X_{\Q}$ and $\XX_{\Q}$.  One can construct
such curves $X_{0}(N)_{\Q}$ and $X_{1}(N)_{\Q}$ for any positive
integer $N$, corresponding to pairs of elliptic curves and cyclic
subgroups of order $N$ and elliptic curves and points of order $N$,
respectively.  For $N \geq 4$ it turns out that $X_{0}(N)_{\Q}$ is
a coarse moduli space and $X_{1}(N)_{\Q}$ is a fine moduli space.

It follows that to determine whether or not there exist elliptic curves
over $\Q$ with a point of order $N$, ``all'' we have to do is determine
whether or not $X_{1}(N)_{\Q}$ has non-cuspidal rational points.
Unfortunately, as $N$ increases the genus of $X_{1}(N)_{\Q}$ increases
and the methods we used for $N = 11$ fail to work.

It has been known for quite a while that for $N=1,2,3,4,5,6,7,8,9,10,12$,
$X_{1}(N)_{\Q}$ has genus $0$ and therefore is isomorphic to $\Pl^{1}_{\Q}$.
(In general, a curve of genus $0$ over $\Q$ is isomorphic to $\Pl^{1}_{\Q}$
if and only if it has a rational point.  In these cases, the cusps always
give automatic rational points.)  It follows that there exist infinitely
many rational elliptic curves with points of order $1,2,3,4,5,6,7,8,9,10,12$.

For higher $N$ the problem is much more difficult.  Before 1977, results
were known only for a few values of $N$, such as $N=11,13$.  This all
changed when Barry Mazur succeeded in fully analyzing the rational points
on these curves, yielding the following remarkable theorem.

\begin{theorem}[Mazur \cite{Mazur1} \cite{Mazur2}]
Let $E$ be an elliptic curve over $\Q$.  Then the torsion subgroup
of $E(\Q)$ is one of the following $15$ groups:
$$\Z/N\Z\quad \text{(N=1,2,3,4,5,6,7,8,9,10,12);}$$
$$\Z/2\Z \times \Z/N\Z \quad\text{(N=1,2,3,4).}$$
\end{theorem}

More recent work on the points of $X_{1}(N)$ over larger number fields
has yielded the following theorem.

\begin{theorem}[Merel \cite{Merel}]
Let $K$ be a number field of degree $d$.  Then there is an integer $M$,
depending only on $d$ (and not on $K$ itself) such that every elliptic
curve over $K$ has torsion subgroup of order at most $M$.
\end{theorem}

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\end{document}