\chapter{Modular Abelian Varieties Attached to Newforms}
In this chapter, we provide an important construction of two abelian varieties, associated to a
given newform $f \in S_2(\Gamma_0(N))^{\textrm{new}}$. One of the varieties is a quotient of the
modular Jacobian $J_0(N)$ and the other one is a subvariety. In fact, the two abelian varieties are dual
to each other. Shimura first associated such abelian varieties to newforms, although his construction is rather
different from the one presented here. We describe the construction in a much more explicit way, in terms of the
action of the Hecke operator on the modular Jacobian. The subvariety is obtained directly from the dual, using a
more general result about optimal quotients.
\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 noncuspidal points on the modular curve $X_0(N)$ over $\C$ is
as isomorphism classes of pairs $(E, C)$, where $E$ is an elliptic curve and $C$ is a
cyclic subgroup of $E(\C)$ of order $N$ (see \cite[Appendix C \S3]{silverman:aec}). 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 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 $T_p$ on the space of cusp forms $S_2(\Gamma_0(N))$ for all
$p \nmid N$. Moreover, the two operators define compatible actions on the space of
differentials and on the modular Jacobian, because we can consider the modular curve $X_0(N)$ over $\C$
and since the Jacobian is then
$$
J_0(N)(\C) = H^0(X_0(N)(\C), \Omega^1)^* / H_1(X_0(N)(\C), \Z).
$$
But $H^0(X_0(N)(\C), \Omega^1)$ is the space of differentials, so we obtain the compatibility of the actions.
\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. We want to associate an abelian variety to a newform
$f$ of level $N$. 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 $\ds f = \sum_{n = 1}^\infty a_n q^n$ is a newform of level $N$ and weight
2. Consider the ideal $I_f = \textrm{ker}(\phi)$, where $\phi : \mathbb{T} \ra K_f$ sends $T_n \mapsto a_n$. It is
not hard to check that $I_f$ is also the annihilator of $f$.
We saw that the Hecke algebra acts on both $S_2(\Gamma_0(N))$ and $J_0(N)$, so the ideal $I_f$ acts on $J_0(N)$.
\begin{prop}
(i) $I_fJ_0(N)$ is strictly contained in $J_0(N)$, so the
quotient variety
$$
A_f := J_0(N) / I_f J_0(N)
$$
is nonzero. \\
(ii) The variety $A_f$ is an optimal quotient of $J_0(N)$ of dimension $[K_f : \Q]$.
\end{prop}
The key idea for the proof is the following
\begin{lem}
Let $\mathbb{T}_\C = \mathbb{T} \otimes \C$. There exists a perfect
Hecke-compatible pairing
$$
\mathbb{T}_\C \times S_2(\Gamma_0(N))\ra \C.
$$
In particular, Hom$(S_2(\Gamma_0(N)),\C)$ and $\mathbb{T}_\C$ are isomorphic
as $\mathbb{T}_\C$-modules.
\end{lem}
\begin{proof}
Define a pairing
$$
\alpha : \mathbb{T}_\C \times S_2(\Gamma_0(N)) \ra \C
$$
as $\alpha(T, f) := a_1(Tf)$, where $a_1(f)$ returns the first coefficient in the
Fourier expansion of the modular form $f$. We claim that this pairing is nondegenerate.
Suppose that there is a cusp form $f$, such that $\alpha(T, f) = 0$ for any operator $T$
in the Hecke algebra $\mathbb{T}_\C$. In particular, $\alpha(T_n, f) = 0$ for each $n$.
But one can figure out exactly the Hecke action on $q$-expansions, namely if
$\ds f = \sum_{n = 1}^\infty a_nq^n$, then
$$
T_p f = \begin{cases}\displaystyle\sum_{n=1}^\infty a_{pn}q^n + p \sum_{n = 1}^\infty a_nq^{pn} & \text{if }p\nmid N\\
\displaystyle\sum_{n=1}^\infty a_{pn}q^n & \text{if }p\mid N
\end{cases}
$$
so the formula implies that $a_1(T_n, f) = a_n$. Thus, $a_n = 0$ for each $n$, so $f = 0$.
To show nondegeneracy on the right, suppose that $T$ is an operator, for which
$\alpha(T, f) = 0$ for each $f \in S_2(\Gamma_0(N))$. Then for all $n$ and all cusp forms $f$,
$\alpha(T, T_n f) = 0$. But $\alpha(T, T_nf) = a_1(T(T_nf)) = a_1(T_n(Tf))$, because the Hecke
algebra is commutative. By the Hecke action on Fourier expansions, we notice that
$a_1(T_nf) = a_n(f)$ for arbitrary cusp form $f$, so it follows that
$0 = \alpha(T, T_nf) = a_n(Tf)$ for all $n$, i.e. $Tf = 0$. Thus, $T$ kills all cusp forms and
so $T \equiv 0$. Therefore $\alpha$ is a nondegenerate pairing. Finally, we have to prove that
the pairing $\alpha$ is Hecke equivariant, i.e. $\alpha( T_n T, f) = \alpha(T, T_n f)$. But this
follows from the definition, using that the Hecke algebra is commutative.
Finally, it follows from the perfect, Hecke-equivariant pairing $\alpha$ that
$$
\textrm{Hom}(S_2(\Gamma_0(N)),\C) \cong \mathbb{T}_\C
$$
as $\mathbb{T}_\C$-modules.
\end{proof}
\begin{proof}[Proof of Proposition 3.2.2.]
$(i)$ The most difficult part is to show that $I_fJ_0(N) \subsetneq J_0(N)$. To do this,
we consider the variety $J_0(N)$ over $\C$ and then
$$
J_0(N)(\C) \cong H^0(X_0(N), \Omega^1)^* / H_1(X_0(N), \Z).
$$
The idea is to reduce the statement to showing that $I_f \mathbb{T}_\C \subsetneq
\mathbb{T}_\C$, which is easy to prove (in particular, $\mathbb{T}_\C/I_f\mathbb{T}_\C
\cong K_f \otimes \C$, where $K_f$ is the Hecke eigenvalue field). The first step in this
reduction is to notice that it is enough to check that $I_fH^0(X_0(N), \Omega^1)^*
\subsetneq H^0(X_0(N),\Omega^1)^*$ (e.g. by counting dimensions of vector spaces and using
the fact that $J_0(N)(\C)$ and $I_fJ_0(N)(\C)$ are obtained by taking modulo one and the
same lattice $H_1(X_0(N), \Z)$). Next, using the correspondence between differentials on
$X_0(N)$ and cusp forms, explained in the previous section, we have
$H^0(X_0(N),\Omega^1) \cong S_2(\Gamma_0(N))$ as $\mathbb{T}_\C$-modules, so
$$
H^0(X_0(N),\Omega^1)^* \cong \textrm{Hom}(S_2(\Gamma_0(N)),\C),
$$
as $\mathbb{T}_\C$-modules. Finally, by using lemma 3.2.3,
$\textrm{Hom}(S_2(\Gamma_0(N)),\C)$ is isomorphic to $\mathbb{T}_\C$ as $\mathbb{T}_\C$-module,
so we reduced the question to proving that the $\mathbb{T}_\C$-module
$\mathbb{T}_\C / I_f\mathbb{T}_\C$ is nonempty. But $\mathbb{T}_\C / I_f\mathbb{T}_\C \cong K_f \otimes \C$, so
the result follows.
\noindent $(ii)$ To see that $I_fJ_0(N)$ is connected, note that each Hecke operator $T : J_0(N) \ra J_0(N)$
defines a morphism. Thus, the image $T(J_0(N))$ is an abelian variety. To see this, note that the
image of an abelian variety under a morphism is a group variety. Moreover, since connected sets are mapped to
connected sets via continuous maps, then $T(J_0(N))$ is connected and therefore an abelian variety. Finally,
the theorem will follow if we show that the sum of two abelian subvarieties of an abelian variety is an abelian
subvariety. Indeed, let $A$ and $B$ be abelian subvarieties of $J_0(N)$. Consider the map
$$
A \times B \ra J_0(N),
$$
defined by $(a, b) \mapsto a + b$. Since this map is a morphism then its image is an abelian subvariety, i.e.
$A + B$ is an abelian subvariety. Therefore, $I_fJ_0(N)$ is an abelian subvariety of $J_0(N)$ which shows that
$A_f$ is an optimal quotient.
\end{proof}
\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, 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
$$
A_f^{\vee} \xrightarrow{\pi^{\vee}} J_0(N)^\vee
\xrightarrow{\theta_{J_0(N)}} J_0(N).
$$
The fact that the dual morphism is a closed immersion follows from 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}
The key idea in the proof of the statement is that for any abelian variety
$A$, the group $A(\overline{K})$ is divisible.
\begin{defn}
We say that an abelian group $G$ is \emph{divisible}, if for any element
$x \in G$ and any positive integer $n \in \Z$, there exists $y \in G$, such that
$x = ny$.
\end{defn}
One can see that $G$ is divisible, if and only if the homomorphism
$[n] : G \ra G$ is surjective for each $n \in \mathbb{N}$.
\begin{example}
A nontrivial example of a divisible group is the group of $\overline{K}$-rational points
on an elliptic curve $E$, defined over a number field $K$, since for any point
$P \in E(\overline{K})$ and any integer $n$, one can choose a point
$Q \in E(\overline{K})$, such that $nQ = P$ (because one can recover the group
law from the Weierstrass equation of the curve). More generally, for abelian variety
$A$ over a number field $K$, the group $A(\overline{K})$ is a divisible group. This
fact is proved in \cite{mumford:abvars}.
\end{example}
\begin{example}
No nontrivial finite group is divisible. Indeed, if $G$ is a finite
group and $x \in G$ is a nontrivial element of order $n$, then $[n] :
G \ra G$ has a nontrivial kernel. In particular, it is not surjective
and thus $G$ is not divisible.
\end{example}
We will also use the following
\begin{lem}
Let $f : B \ra A$ be a surjective morphism of abelian varieties. The dual morphism
$f^\vee : A^\vee \ra B^\vee$ has finite kernel.
\end{lem}
\begin{proof}
We will use that the ``double dual'' functor is the identity on the category of abelian varieties.
This is a nontrivial result from the theory of abelian varieties. A very ellegant proof, which makes use
of the Poincar\'e sheaves is presented in \cite{}.
Consider the abelian variety $C = f^\vee(A) \subset B^\vee$. We have the composition
$$
A^\vee \ra C \hookrightarrow B^\vee.
$$
We can dualize this composition to get
$$
B \ra C^\vee \ra A.
$$
Since the double dual is the identity, the composition of the two maps is $f$, which is surjective.
Hence $C^\vee \ra A$ is surjective. Hence, $\textrm{dim}(C) = \textrm{dim}(C^\vee) \geq \textrm{dim}(A)$.
But $A^\vee \ra C$ is surjective, since $C$ is the image of $A^\vee$ under $f^\vee$. Therefore,
$\textrm{dim}(A) = \textrm{dim}(A^\vee) \geq \textrm{dim}(C)$. Thus, $\textrm{dim}(A) = \textrm{dim}(C)$,
so $f^\vee$ has finite kernel.
\end{proof}
\begin{proof}[Proof of Proposition 3.3.1.]
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
$\pi^{\vee}$, which is $\pi$ is surjective, then $\pi^{\vee}$ must
have a finite kernel, according to lemma 3.3.5. 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
\begin{equation*}
\xymatrix{
G \ar[r] & A^\vee \ar[r] \ar[rd]^{\pi^\vee} & C \ar[d] \\
& & J
}
\end{equation*}
After dualizing the diagram, we obtain
\begin{equation*}
\xymatrix{
G^\vee & \ar[l] A & \ar[l] C^\vee \\
& & J \ar[u]^{\varphi} \ar[ul]^{\pi}
}
\end{equation*}
where $G^{\vee}$ is the Cartier dual of $G$. Since $G$ is finite,
then $G^{\vee}$ is also finite, so the kernel of the map $\varphi : J \ra
C^{\vee}$ is a finite index subgroup of the kernel of $\pi : J \ra
A$. But ker$(\pi)$ is an abelian variety, so it is a divisible
group. This means that any quotient
of ker$(\pi)$ is divisible as well. Therefore, the finite quotient
ker$(\pi)/\textrm{ker}(\varphi)$ is divisible. But no finite group is
divisible, so $G$ must be trivial. Therefore, $\pi^\vee : A^\vee \ra J^\vee$
is a closed immersion.
\end{proof}