\chapter{Raising the Level of Modular Representations.}
\section{Modular Representation Arising from Maximal Ideals of the Hecke Algebra}
Let $\mathbb{T} := \mathbb{T}_N$ be the Hecke algebra, acting on the complex
vector space $S_2(N)$ of modular forms of weight 2 for $\Gamma_0(N)$. The goal of
this section is to establish a correspondence between maximal ideals of $\mathbb{T}$
and certain semisimple representations
$\rho_{\prid m} : \textrm{Gal}(\overline{\Q}/\Q) \ra \textrm{GL}_2(\mathbb{T}/ \prid m)$.
\begin{prop}
There exists a unique (up to isomorphism) representation
$$
\rho_{\prid m} : \textrm{Gal}(\overline{\Q}/\Q) \ra \textrm{GL}_2(\mathbb{T}/ \prid m),
$$
satisfying
$\textrm{tr }\rho_m(\textrm{Frob}_r) \equiv T_r$ (mod $\prid m$) and
$\textrm{det }\rho_m(\textrm{Frob}_r) \equiv r$ (mod $\prid m$).
\end{prop}
\section{Constructing the Map $\varphi : J_0(N) \times J_0(N) \ra J_0(pN)$.}
The goal of the section is constructing a map $\varphi : J_0(N) \times J_0(N) \ra J_0(N)$, which
will be necessary for proving the existance of \emph{visible} elements of certain order of
$\Sha(A)$, where $A$ is a modular abelian variety. We describe the kernel of
$\varphi$ in terms of the Shimura subgroup and prove that this kernel
is computable.
\subsection{The Degeneracy Maps $\alpha, \beta : X_0(pN) \ra X_0(N)$.}
Suppose that $N$ is a positive integer. Consider the modular curve $X_0(N)$ as a curve
over $\C$. Let $p$ be a prime which does not divide the level $N$. There are two degeneracy
maps $\alpha : X_0(pN) \ra X_0(N)$ and $\beta : X_0(pN) \ra X_0(N)$, which can be defined in
two different ways.
First, $\alpha$ and $\beta$ can be constructed as induced maps from endomorphisms of
the upper half plane $\mathfrak h$. This is achieved in the following proposition:
\begin{prop}
The maps id $: \mathfrak h \ra \mathfrak h$ and $\cdot p : \mathfrak h \ra \mathfrak h$
induce well-defined maps $\alpha, \beta : X_0(pN) \ra X_0(N)$.
\end{prop}
\verb Proof: The identity map id $: \mathfrak h \ra \mathfrak h$ induces a well defined map
$\tilde{\alpha} : Y_0(pN) \ra Y_0(N)$, because $\Gamma_0(pN) \subset \Gamma_0(N)$.
Using the fact that $X_0(N)$ is the compactified curve $Y_0(N)$, it follows that
$\tilde{\alpha}$ extends to a map $\alpha : X_0(pN) \ra X_0(N)$.
The second map is induced by the map $z \mapsto pz$. Indeed, if
$z_1$ and $z_2$ are $\Gamma_0(pN)$-equivalent, then $z_2 = \gamma z_1$ for some $\gamma =
\left [
\begin{array}{cc}
a & b \\
pNc & d
\end{array}
\right ]
\in \Gamma_0(pN)$. But then $pz_1$ and $pz_2$ are $\Gamma_0(N)$-equivalent, since
$pz_2 = \gamma' pz_1$ for
$\gamma' =
\left [
\begin{array}{cc}
a & pb \\
Nc & d
\end{array}
\right ]
\in \Gamma_0(pN)$. This gives us a well defined map $\tilde{\beta} : Y_0(pN) \ra Y_0(N)$, and
so we get a map $\beta : X_0(pN) \ra X_0(N)$, between the compactified curves. $\hfill \Box$ \\
A different way to construct $\alpha$ and $\beta$ is by viewing the elements of
the modular curve $X_0(N)$ as pairs $(E, C_N)$, where $E$ is an elliptic curve and
$C_N$ is a cyclic subgroup of order $N$. For an element $(E, C_{pN})$, the group $C_{pN}$
has a unique subgroup of order $N$, which we call $C_N$. Then
$\alpha : (E, C_{pN}) \mapsto (E, C_N)$. Let $C_p$ be the unique subgroup of $C_{pN}$ of order
$p$. Then $\beta : (E, C_{pN}) \mapsto (E / C_p, C_{pN} / C_p)$ (note that $C_{pN} / C_p$ is
a cyclic group of order $N$).
\begin{prop}
The two definitions of $\alpha$ and $\beta$ are equivalent.
\end{prop}
\verb Proof: *** later *** $\hfill \Box$ \\
We can use these correspondences to
define an action of the Hecke operator on the Jacobian $J_0(N)$.
The maps $\alpha$ and $\beta$ combine to induce a map
$\varphi : J_0(N) \times J_0(N) \ra J_0(pN)$ by $\varphi (x, y) = \alpha^*(x) + \beta^*(y)$.
We first compute the kernel of the map $\varphi$.
\begin{prop}
Let $\Sigma_N$ be the kernel of the map $J_0(N) \ra J_1(N)$ induced from the covering of
modular curves $X_1(N) \ra X_0(N)$. Consider the image $\Sigma$ of $\Sigma_N$ in
$J_0(N) \times J_0(N)$ under the anti-diagonal map $J_0(N) \ra J_0(N) \times J_0(N)$
(i.e. the map $x \mapsto (x, -x)$). Then $\Sigma$ is the kernel of the map
$\varphi : J_0(N) \times J_0(N) \ra J_0(pN)$.
\end{prop}
\verb Proof: ***later*** ...
Since our final goal will be to test whether certain primes divide the order of the
kernel of $\varphi$, we will use the following property of the Shimura subgroup.
\begin{prop}
The Shimura subgroup $\Sigma_N$ of $J_0(N)$ is contained in the kernel of the
endomorphism $\eta_p = T_p - (p + 1)$ of $J_0(N)$ for all primes $p$ which do
not divide $N$.
\end{prop}
\verb Proof: Suppose that $p$ is a prime which does not divide $N$. We have two
degeneracy maps $\alpha^* : J_0(N) \ra J_0(pN)$ and $\beta^* : J_0(N) \ra J_0(pN)$,
which we constructed above. Taking duals and using the self-duality of the Jacobian,
these maps induce maps $(\alpha^*)^{\breve{}} : J_0(pN) \ra J_0(N)$ and
$(\beta^*)^{\breve{}} : J_0(pN) \ra J_0(N)$. It follows from the geometric
definition of the action of $T_p$ on the Jacobian $J_0(N)$ that
$(\alpha^*)^{\breve{}} \circ \beta^* = T_p = (\beta^*)^{\breve{}} \circ \alpha$.
Furthermore, the degree of the map $\alpha^*$ is the same as the degree of $\alpha$.
Since $\alpha : X_0(pN) \ra X_0(N)$ has degree $p + 1$ whenever $p$ is relatively prime to
$N$, then $\alpha^*$ has degree $p+1$ as well, i.e.
$(\alpha^*)^{\breve{}} \circ \alpha^* = p + 1$. We will be done, if we show that
$\alpha^* = \beta^*$ on the Shimura subgroup $\Sigma_N \subset J_0(N)$.