\chapter{Modular Abelian Varieties Attached to Newforms}
\section{Hecke Operators as Correspondences}
Consider the modular curve $X_0(N)$ and let $p \nmid N$ be a prime. There are two
degeneracy maps $\alpha, \beta : X_0(pN) \ra X_0(N)$ which can be defined in two different
ways.
One interpretation of the modular curve $X_0(N)$ is as a parametrization of the isomorphism
classes of pairs $(E, C)$, where $E$ is an elliptic curve and $C$ is a cyclic subgroup
of order $N$ *** CITE ***. Consider a pair $(E, C)$, where $C$ is a cyclic subgroup of
order $pN$. Since
$p \nmid N$, then $C \cong C' \oplus D$, where $C'$ is a cyclic subgroup of order $N$ and $D$
is a cyclic subgroup of order $p$. Moreover, the subgroups $C'$ and $D$ are unique. Thus, we
can define the degeneracy maps
$$
\alpha : (E, C) \mapsto (E, C')
$$
$$
\ds \beta : (E, C) \mapsto (E / D, (C + D)/D)
$$
An equivalent construction is can be obtained by viewing $X_0(N)$ and $X_0(pN)$ as
quotients of the upper-half plane. In this case we will not worry about the cusps,
since any rational map between nonsingular curves extends uniquely to a morphism. So,
we look at $\Gamma_0(N) \backslash \mathfrak h$ and $\Gamma_0(pN) \backslash \mathfrak h$.
Since $\Gamma_0(pN) \subset \Gamma_0(N)$, then there is a natural map
$$
\alpha' : \Gamma_0(N) \backslash \mathfrak h \ra \Gamma_0(pN) \backslash \mathfrak h.
$$
It is not hard to verify (by tracing the definitions) that
$\alpha$ and $\alpha'$ represent the same map.
The equivalent way of defining $\beta$ is as follows: note that there is an inclusion
$\Gamma_0(pN) \hookrightarrow \Gamma_0(N)$, defined by
$$
X \mapsto \left[ \begin{array}{cc} p & 0 \\ 0 & 1 \end{array} \right] \cdot X
\cdot \left[ \begin{array}{cc} p & 0 \\ 0 & 1 \end{array} \right]^{-1}.
$$
We can use this inclusion to construct a map between the quotients of the upper half plane:
$$
\Gamma_0(pN) \backslash \mathfrak h \simeq
\left[ \begin{array}{cc} p & 0 \\ 0 & 1 \end{array} \right] \Gamma_0(pN)
\left[ \begin{array}{cc} p & 0 \\ 0 & 1 \end{array} \right]^{-1} \backslash \mathfrak h \ra
\Gamma_0(N).
$$
Again, one checks immediately that this definition agrees with the one given in terms of
the parametrization interpretation.
Thus, we have a correspondence
\begin{equation*}
\xymatrix{
& X_0(pN) \ar[dl]_{\alpha} \ar[dr]^{\beta} & \\
X_0(N) & & X_0(N)
}
\end{equation*}
We can recover out of this correspondence the pullback and push-forward maps on divisors.
Consider the induced map
$$
\alpha^* : \textrm{Div}(X_0(N)) \ra \textrm{Div}(X_0(pN)).
$$
The preimages of the prime divisor $(E, C)$ under $\alpha$ are all $(E, C \oplus D)$,
where $D$ is a cyclic subgroup of $E$ of order $p$. Therefore
$$
\alpha^* : (E, C) \mapsto \sum_{|D|} (E, C \oplus D),
$$
where the above sum runs over all cyclic subgroups of $E$ of order $p$.
Similarly, we recover the push-forward map $\beta_* : X_0(pN) \ra X_0(N)$: if $(E, C')$ is
a point $X_0(pN)$, then $C'$ can be written uniquely as $C \oplus D$, where $C$ and $D$ are
cyclic subgroup of $E$ of orders $N$ and $p$, respectively. Then
$$
\beta_* : (E, C \oplus D) \mapsto (E / D, (C + D)/D).
$$
Consider $\varphi := \beta_* \circ \alpha^*$. Then $\varphi$ is a map
$\varphi : \textrm{Div}(X_0(N)) \ra \textrm{Div}(X_0(N))$. Since $\varphi$
multiplies the degree of a divisor class by the degree of the map $\alpha$, then
it restricts to a map $\textrm{Div}^0(X_0(N)) \ra \textrm{Div}^0(X_0(N))$. This is precisely
how one obtains the Hecke operator $T_p$ on $J_0(N)$. In other words, we define
$T_p := \varphi|_{\textrm{Div}^0(X_0(N))}$.
Notice that it is possible to define the Hecke operator on modular forms from
the above correspondence. Indeed, recall that $S_2(\Gamma_0(N)) \cong H^0(X_0(N), \Omega^1)$,
where the isomorphism is given by $f(z) \mapsto f(z)dz$. Indeed, if $f(z)$ is a modular form
of weight 2 for $\Gamma_0(N)$, then the differential $f(z)dz$ is $\Gamma_0(N)$-invariant.
Since $\alpha$ and $\beta$ induce maps on differentials, we have a composition
$$
H^0(X_0(N), \Omega^1) \xrightarrow{\alpha^*} H^0(X_0(pN), \Omega^1)
\xrightarrow{\beta_*} H^0(X_0(N), \Omega^1).
$$
Thus, we have an operator $\mathfrak T_p$ on the space of cusp forms $S_2(\Gamma_0(N))$ for all
$p \nmid N$. The fact that the two operators define compatible actions on the space of
differentials and on the modular Jacobian, we consider the curve $X_0(N)$ over $\C$ and since
$$
J_0(N)(\C) = H^0(X_0(N)(\C), \C)^* / H_1(X_0(N)(\C), \Z)
$$
\section{Constructing an Abelian Variety $A_f$ as a Quotient of $J_0(N)$}
In the previous section, we defined Hecke operators on the
modular Jacobian $J_0(N)$ and on the space of holomorphic differentials
$H^0(X_0(N), \Omega_1)$ and showed that the two actions are compatible, i.e the two Hecke
algebras are in fact isomorphic. Before proceeding, we need one more definition
\begin{defn}[Optimal Quotient]
Let $J$ be a Jacobian of a curve, $A$ be an abelian variety and $\pi : J \ra A$ be a smooth,
surjective morphism. We say that $A$ is an \emph{optimal quotient} of $J$ if the kernel of
$\pi$ is connected.
\end{defn}
Suppose now that $\ds f = q + \sum_{n = 1}^\infty a_n q^n$ is a newform of level $N$ and weight
2. Consider the annihilator $I_f$ of $f$ as an ideal of $\mathbb{T}$. From the discussion in the
previous section, the ideal $I_f$ acts in a compatible on the modular Jacobian $J_0(N)$
\begin{prop}
(i) $I_fJ_0(N)$ is a proper abelian subvariety of $J_0(N)$, so one can construct the
quotient variety
$$
A_f := J_0(N) / I_f J_0(N).
$$
(ii) The variety $A_f$ is an optimal quotient of $J_0(N)$.
\end{prop}
\section{The Dual Abelian Variety as a Subvariety of $J_0(N)$}
So far, we constructed an optimal quotient $A_f$ of $J_0(N)$. The dual variety
$A_f^{\vee}$ is an interesting abelian variety. The remarkable property it has is that
it can be viewed as a subvariety of $J_0(N)$. Thus, one can associate to each newform $f$
two isogenous abelian varieties - the optimal quotient $A_f$ and the subvariety $A_f^{\vee}$.
The main goal of this chapter is to construct the morphism, which makes $A_f^{\vee}$ a
subvariety of $J_0(N)$.
First of all, we showed in *** *** that Jacobians of curves are self-dual, i.e. $J^{\vee} = J$.
Moreover, they come equipped with a canonical, principal polarization
$\theta_J : J \ra J$, which satisfies $\theta_J^{\vee} = \theta_J$.
Let $\pi : J_0(N) \ra A_f$ be the quotient map. Consider the dual
map $\pi^{\vee} : A_f^{\vee} \ra J_0(N)$ and compose it with $\theta_{J_0(N)}$. Thus, we
get a morphism
$$
A_f^{\vee} \xrightarrow{\pi^{\vee}} J_0(N) \xrightarrow{\theta_{J_0(N)}} J_0(N).
$$
We will show that this morphism is a closed immersion by proving the following more
general statement
\begin{prop}
Let $J / K$ be a Jacobian and $\theta_J$ be the canonical
polarization of $J$. Suppose that $\pi : J \twoheadrightarrow A$
is an optimal quotient. Then the morphism
$$
A^{\vee} \xrightarrow{\pi^{\vee}} J^{\vee} \xrightarrow{\theta_J}
J
$$
is a closed immersion.
\end{prop}
\begin{proof}
It suffices to show that $\pi^{\vee}$ is injective, since
$\theta_J$ is an isomorphism and any monomorphism between smooth
schemes of finite type is a closed immersion. Since the dual of
$\pi^{\vee} = \pi$ is surjective, then $\pi^{\vee}$ must have a
finite kernel. Let $G = \textrm{ker}(\pi^{\vee})$. Let $C =
\textrm{im}(\pi^{\vee})$. Then $A^{\vee} \ra C$ is an isogeny. Let
$G$ be the kernel of this isogeny. One has the following
commutative diagram
After dualizing the diagram, we obtain
where $G^{\vee}$ is the Cartier dual of $G$.
\end{proof}
\begin{proof}
It suffices to check that $A^{\vee} \xrightarrow{\pi^{\vee}} J$ is injective, since
every monomorphism between smooth schemes of finite type is a closed immersion.
Consider the commutative proper group subscheme $G = \textrm{ker}(\pi^{\vee})$ of
$A_f^{\vee}$. It suffices to check that $G$ does not have any torsion elements. Indeed,
if $G$ is finite, there is nothing to check. Otherwise, consider the scheme
$(G / \overline{K})_{\textrm{red}}^0$. It is a complete, commutative, group scheme over $K$,
so it must be an abelian variety, so it must have a torsion (since
$A[n] \cong (\Z / n\Z)^{2d}$ for each $n$; in particular, the torsion is nonempty).
Therefore, to prove injectivity, it suffices to show that the induced morphism
$A^{\vee}[n] \ra J[n]$ is a closed immersion.
\end{proof}
\section{Shimura's Construction}
Suppose that $f$ is a newform for $\Gamma_0(N)$, which is an eigenform for the operators of
the Hecke algebra $\mathbb{T}$. We will associate to $f$ an abelian variety $A_f$ which is
a quotient of $J_0(N)$. Let $K_f$ denote the subfield of $\C$, which is generated by the
Fourier coefficients of the modular form $f$. We also denote by
$\mu : S_2(\Gamma_0(N)) \ra H^(X_0(N), \Omega^1)$ the isomorphism, which identifies
modular forms of weight 2 with differentials on the modular curve $X_0(N)$.
For the purpose of this section, we will denote the Hecke operators on $J_0(N)$ defined
in the previous section by $\xi_n$ (since we keep the notation $T_n$ for the operator on
the space of modular forms).
Also, we need to define more precisely what is meant by a quotient of an abelian variety.
\begin{defn}
A \emph{quotient of an abelian variety $B$ by an abelian subvariety $C$} is a couple
$(A, \nu)$, formed by an abelian variety $A$, which canonically represents $B / C$ and
a natural morphism $\nu : B \ra A$ with $C = \textrm{ker}(\nu)$.
\end{defn}
The main result of this section is the following theorem:
\begin{thm}
There exists a quotient $(A_f, \nu)$ of $J_0(N)$ and isomorphism
$\theta : K \ra \textrm{End}(A) \otimes \Q$, such that dim$(A) = [K : \Q]$ and
$\nu \circ \xi_n = \theta(a_n) \circ \nu$ for all $n$.
\end{thm}
\begin{proof}
Let $\mathfrak T$ be the algebra, generated by the Hecke operators $\xi_n$ on $J_0(N)$.
Let $\mathfrak R$ be the radical of $\mathfrak T$. According to Wedderburn's structure
theorem, $\mathfrak T = \mathfrak R \oplus \mathfrak S$, where $\mathfrak S$ is a commutative
semisimple algebra. Let $\mathfrak S_1, \mathfrak S_2, \dots, \mathfrak S_r$
denote its simple components.
First, we define a homomorphism $\rho : \mathfrak T \ra K$ as follows: since
$\delta \xi_n (\mu(f)) = a_n \cdot \mu(f)$, then one can set $\rho(\xi_n) = a_n$. Since
$\mathfrak S_i$'s and $K$ are simple, Schur's lemma implies that $\rho$ restricts to
an isomorphism $\mathfrak S_1 \ra K$ (after rearranging indices if necessary). Take
the inverse of this isomorphism $\rho' : K \ra \mathfrak S_1$. Consider the ideal
$\mathfrak U = \mathfrak S_2 + \dots + \mathfrak S_r + \mathfrak R$ of $\mathfrak T$.
Let $C$ be the abelian subvariety of $J$, generated by $\alpha(J_0(N))$ for all $\alpha \in
\mathfrak T \cap \textrm{End}(J_0(N))$. We can construct the quotient $J_0(N) / C$,
which is going to be rational, since the endomorphisms in
$\mathfrak T \cap \textrm{End}(J_0(N))$ are rational. Next, notice that $C$ is stable
under the endomorphisms of $\mathfrak S_1 \cap \textrm{End}(J_0(N))$. Thus, one can define
an automorphism $\theta : K \ra \textrm{End}(A) \otimes \Q$, such that
$$
\nu \circ \rho'(a) = \theta(a) \circ \nu.
$$
Note that $\xi_n - \rho'(a_n) \in \mathfrak U$ for every $n$ and therefore
$\nu \circ \xi_n = \theta(a_n) \circ \nu$. We only need to show that $J \ne C$.
\begin{lem}
Let $\mathfrak T_\C = \mathfrak T \otimes_{\Q} \C$. Then
$$
\mathfrak T_\C \simeq S_2(\Gamma_0(N))
$$
as $\mathfrak T_\C$-modules.
\end{lem}
\begin{proof}
This is proved in [Shi-1, \S3.5, Thm 3.51].
\end{proof}
\end{proof}
\section{The Variety $A_f$ as a Subvariety of $J_0(N)$}
\subsection{Modular Abelian Varieties Attached to Newforms}
\begin{defn}
A newform is a modular form $\ds f = q + \sum_{n \geq 2} a_nq^n$ for $S_2(\Gamma_0(N))$,
which is a simultaneous eigenvector for every element of the Hecke algebra $\mathbb{T}$ and
which DOES NOT ARISE from a modular form of lower level (translate the last statement
more rigorously).
\end{defn}
\begin{defn}
The \emph{anemic} Hecke algebra $\mathbb{T}_0$ is the algebra $\Z[T_n\ :\ (n, N) = 1]$.
\end{defn}
This algebra is interesting, since we know precisely how the algebra
$\mathbb{T}_0 \otimes \Q$ decomposes. This is proved in the following
\begin{prop}
The anemic Hecke algebra $\mathbb{T}_0$ is a finite-rank, commutative algebra
of simultaneously diagonalizable operators on the vector space $S_2(\Gamma_0(N))$.
In particular, we obtain a decomposition
$$
\mathbb{T}_0 \otimes \Q \simeq \prod_{f} K_f,
$$
where $f$ ranges over the Galois conjugacy classes of newforms and $K_f = \Q(a_1, a_2, \dots)$,
where $a_1, a_2, \dots$ are the Fourier expansion coefficients for a representative $f$ of the
Galois conjugacy class.
\end{prop}
\verb Proof: Since $\mathbb{T}_0 \subset \textrm{End}(J_0(N))$, then $\mathbb{T}_0$ is
finitely generated. The fact that $\mathbb{T}_0$ is commutative is an immediate consequence of
the definition of the Hecke correspondences. The last statement of the theorem will follow,
provided we prove that the operators in $\mathbb{T}_0$ are simultaneously diagonalizable.
Since $S_0(\Gamma_0(N))$ as a vector space is equipped with the Petersson inner product,
then the last statement follows from the fact that the operators of the anemic algebra are
self-adjoint with respect to that inner product. This is proved in [La-***]. Notice that
this fact is not necessarily true for $T_n$ with $(n, N) \ne 1$. Finally, choose a basis
$\{f_1, \dots, f_n\}$ of eigenforms for all the Hecke operators and look at the map
$T \mapsto (c_1, c_2, \dots, c_n)$, where $c_i$'s are the eigenvalues of $T$ for the
eigenforms $f_1, f_2, \dots, f_n$, respectively. It is easy to check that this map gives an
isomorphism $\ds \mathbb{T}_0 \otimes \Q \ra \prod_{f}K_f$, where $f$ is an eigenform and
$K_f = \Q(a_1, a_2, \dots)$ for $a_i$'s being the Fourier expansion coefficients of $f$.
$\hfill \Box$
\begin{thm}
Let $\ds f = \sum_{n = 1}^\infty a_nq^n$ be a newform of weight 2 for $\Gamma_0(N)$. Let
$K = \Q(a_1, \dots, a_n, \dots)$ be the Hecke eigenvalue field. Then we can associate to $f$
an abelian variety $A_f$, which is defined over $\Q$, such that dim$(A_f) = [K : \Q]$.
\end{thm}
\verb Proof: For each newform $f$, let $\pi_f$ denote the operator, which is a projector
with respect of the coordinate, corresponding to the conjugacy class of $f$. In other words,
$\pi_f = (\dots, 0, 1, 0, \dots) \in \prod_{f}K_f$. The operators $\pi_f$ are idempotent
and their sum is $(\dots, 1, \dots)$. Since $\mathbb{T}_0 \otimes \Q \subset
\mathbb{T} \otimes \Q$ and $\mathbb{T}$ does not have any additive torsion, then
the map $\ds T \mapsto \sum_{f}T \pi_f$ provides a decomposition of the Hecke algebra
$$
\mathbb{T} \otimes \Q \simeq \prod_{f} L_f,
$$
where $L_f$ is an algebra.