\documentclass{article}
\title{Visibility of Shafarevich-Tate groups of modular
abelian varieties}
\author{William Stein}
\date{July 2000}
\include{macros}
\DeclareMathOperator{\Vis}{Vis}
\DeclareMathOperator{\Res}{Res}
\usepackage[all]{xy}
\begin{document}
\maketitle
\begin{abstract}
These are the notes for my July 2000 Durham talk on visibility
of Shafarevich-Tate groups. I typed them in quickly the night
before my talk, in order to make precise what I would say.
\end{abstract}
\section{Modular abelian varieties}
A central class of abelian varieties of interest in number theory
are the modular abelian varieties of $\GL_2$-type. It is now known
that every elliptic curve over~$\Q$ is a member of this class.
Such abelian varieties are constructed as follows.
Let $f=\sum a_n q^n\in S_2(\Gamma_0(N),\C)$ be a newform. The
Hecke algebra $\T=\Z[\ldots T_n \ldots]$ acts on~$f$, and
there is a surjective homomorphism
$\T\ra \O_f = \Z[\ldots a_n \ldots]$ sending $T_n$ to $a_n$.
The kernel $I_f$ of this homomorphism is the annihilator of~$f$ in $\T$.
Attached to $I_f$ there is a quotient $A_f = J_0(N)/I_f(J_0(N))$,
where $J_0(N)$ is the Jacobian of the modular curve $X_0(N)$.
The abelian variety $A_f$ has dimension $d=[\Q(\ldots a_n\ldots):\Q]$,
and the dual $A_f^{\vee}$ is an abelian subvariety of $J_0(N)$.
There are two groups attached to an abelian variety~$A$ over $\Q$.
One is the Mordell-Weil group $A(\Q)$, which is a finitely generated
abelian group, and the other is the Shafarevich-Tate group
$\Sha(A) = \ker(H^1(\Q,A) \ra \prod_v H^1(\Q_v,A))$,
which is ({\em conjecturally}!) a finite abelian group. This talk
is about a link between these two groups, which is frequently signaled
by the term ``visibility''. Before proceeding further, we state
a fundamental conjecture that both guides and stimulates our work.
\begin{conjecture}[Birch and Swinnerton-Dyer]
Let $A$ be an abelian variety over~$\Q$. Then $\Sha(A)$ is finite,
and the following formula holds:
$$L(A,1) = \frac{\prod c_p \cdot \Omega_A}{\#A(\Q)\cdot \#A^{\vee}(\Q)} \cdot \#\Sha(A),$$
where the right hand side should be interpreted as~$0$ if $A(\Q)$
is infinite.
(There is a refinement when $L(A,1)=0$.)
\end{conjecture}
\section{Mazur: Visualize Sha!}
Let~$A$ be an abelian variety over~$\Q$. To give an elements of $\Sha(A)$
is the same as giving an algebraic variety $X$ over~$\Q$ equipped with
an action $A\cross X \ra X$ satisfying analogues of the usual axioms
for a simply transitive group action. Where can we find these varieties~$X$?
As translates of $A$ inside a bigger abelian variety~$J$!
Fix an embedding of $A$ into an abelian variety $J$, and let $C$ be the
quotient:
$$A \hookrightarrow J \onto C.$$
Taking cohomology, reveals the following diagram:
$$\xymatrix{
& & & {\Vis_J(\Sha(A))}\ar@{^(->}[r] & {\Sha(A)}\ar@{^(->}[d] \ar[r] & {\Sha(J)}\ar@{^(->}[d]\\
0\ar[r] & A(\Q) \ar[r] & J(\Q) \ar[r] & C(\Q) \ar[r] & H^1(\Q,A) \ar[r] & H^1(\Q,J) }$$
\begin{definition}[Visible part]
The \defn{visible part of $\Sha(A)$} with respect to the embedding $A\hookrightarrow J$
is the subgroup
$$\Vis_J(A) = \ker(\Sha(A) \ra \Sha(J)).$$
Likewise, the visible part of $H^1(\Q,A)$ is
$$\Vis_J(H^1(\Q,A)) = \ker(H^1(\Q,A) \ra H^1(\Q,J)).$$
\end{definition}
\begin{question}[Mazur]
Is every element $c\in H^1(\Q,A)$ visible somewhere?
\end{question}
Johan de Jong answered this question in the affirmative, when $\dim A = 1$; however,
his proof, which exploits Azamuya algebras over $\Spec(\Z)$, is not elementary.
Fortunately, Mazur's question has an elementary answer.
\begin{proposition}[Folklore?]
Given $c\in H^1(\Q,A)$ there is an abelian variety $J$ and an embedding $\iota:A\hookrightarrow J$
such that $\iota_*(c) = 0 \in H^1(\Q,J)$.
\end{proposition}
\begin{proof}
Because cochains are continuous, there is a finite extension $K$ of $\Q$
such that $\res_K(c) = 0 \in H^1(K,A)$. The Shapiro lemma implies that
$H^1(K,A)$ is canonically isomorphic to $H^1(\Q,\Res_{K/\Q} A_K)$, where
$\Res_{K/\Q} A_K$ is the Weil restriction of scalars of $A_K$. Furthermore,
there is a canonical embedding $\iota: A\hookrightarrow J=\Res_{K/\Q} A_K$,
and this embedding sends~$c$ to $\res_K(c) = 0$.
\end{proof}
\begin{corollary}
Every $c\in \Sha(E)[2]$ is visible in an abelian surface, and
every $c\in \Sha(E)[3]$ is visible in a three-dimensional abelian variety.
\end{corollary}
Using more sophisticated techniques, Mazur proved that every $c\in \Sha(E)[3]$ is visible
in an abelian surface.
For the rest of this talk, we return to the case where $A$ is a attached to a
modular form.
\begin{question}[Mazur]
Let~$f$ be a newform of level~$N$.
Can one visualize the elements $c\in \Sha(A_f^{\vee})$ inside $J_0(N)$? Inside $J_0(M)$
for some multiple $M$ of $N$?
\end{question}
\section{Some data}
Mazur, Cremona, and Logan collected data on visibility of Sha using only elliptic curves
and congruences between elliptic curves. The data is impressive, but probably only because
$\Sha$ is very small in the range considered. As Frey emphasized in his talk, there
are very few intersections between elliptic curves.
The author made a table of rational numbers $L(A_f,1)/\Omega_{A_f}$ for
all $f \in S_2(\Gamma_0(p))$, for $p\leq 2161$.
To analyze the resulting data, let us momentarily assume the truth of the Birch and Swinnerton-Dyer
conjecture. We find~$38$ of the $A_f$ have the property that an odd prime~$\ell$ divides
$\#\Sha(A_f)$. Of these, the full odd part of $\Sha(A_f)$ is visible in $22$ cases, whereas
it is invisible in all of the remaining $16$ cases. A small {\em selection} of data is
recorded in the table below.
\begin{center}
\begin{tabular}{lcll}
$p$ & $\dim A_f$ & $\sqrt{\#\Sha(A_f)^{\text{odd}}}$ & $\dim B$\\
389 & 20 & 5 (VIS)& 1\\
433 & 16 & 7 (VIS)& 1\\
1061 & 46 & 151 (VIS)& 2 \\
1091 & 62 & 7 (INV, so far) & ---\\
1429 & 64 & 5 (VIS at level $2\cdot 1429$) & 2\\
2111 & 112 & 211 (INV, so far) & ---\\
2333 & 101 & 83341 (VIS) & 4\\
2849 & 1 & 3 (VIS at level $3\cdot 2849$) & 1\\
5389 & 1 & 3 (VIS at level $7\cdot 3849$) & 1\\
\end{tabular}
\end{center}
\section{Criterion for visibility}
\begin{theorem}[Stein]
Let $A,B \subset J$, and let $N$ be divisible exactly by the primes of bad reduction for~$J$.
Suppose that
\begin{itemize}
\item $(A \intersect B)(\Qbar)$ is finite,
\item $A(\Q)$ is finite,
\item $B$ has purely toric reduction at each $\ell\mid N$.
\end{itemize}
Let $p$ be a prime such that
\begin{enumerate}
\item
$$p\nmid 2N \cdot \# (\frac{J(\Q)}{A(\Q)})_{\tor} \cdot B(\Q)_{\tor} \cdot
\prod_{\ell \mid N} \#\Phi_{A,\ell}(\overline{\F})\cdot \#\Phi_{B,\ell}(\F_\ell),$$
\item $B[p]\subset A$ (this condition can be relaxed).
\end{enumerate}
Then
$$B(\Q)/p B(\Q) \hookrightarrow \Vis_J(\Sha(A)).$$
\end{theorem}
\begin{proof}[Sketch]
Consider the diagram
$$\xymatrix{
& {A \intersect B} \ar[r]\ar[d] & B\ar[d] \ar[r]^p& B\ar[d]\\
0 \ar[r] & A \ar[r] & J \ar[r] & C \ar[r]& 0.
}$$
Taking cohomology, we arrive at the following key diagram:
$$\xymatrix{
0\ar[r] & B(\Q)\ar[d] \ar[r]^{p} & B(\Q)\ar[d] \ar[r] & B(\Q)/p B(\Q)\ar[d] \ar[r] & 0\\
0\ar[r] & J(\Q)/A(\Q) \ar[r] & C(\Q) \ar[r] & \Vis_J(H^1(\Q,A)) \ar[r] & 0.
}$$
Upon applying the snake lemma and using the above hypothesis, we
deduce that
$B(\Q)/p B(\Q) \hookrightarrow \Vis_J(H^1(\Q,A))$. The
deduction that $B(\Q)/p B(\Q)\hookrightarrow \Vis_J(\Sha(A))$ involves
a local analysis at each prime $\ell$ of $\Q$.
\end{proof}
\section{Visibility at higher level}
There is a $53$-dimensional abelian subvariety~$A$ of $J_0(1429)$
such that~$5$ divides the conjectural order of $\Sha(A)$. However, this part of
$\Sha$ can not be visible in $J_0(1429)$.
Ribet's level raising theorem supplies us with an infinite collection
of primes $p$ such that an image of $A$ in $J_0(pN)$ meets an abelian
variety $B$ in some $\m$-torsion, where $\m$ is an appropriate maximal ideal
of residue characteristic $5$ in $\T$.
For each of these $B$, {\em probably} (I haven't proved this)
$$L(B,1)\con L(A,1) = 0 \pmod{\m},$$
so either $5\mid \#\Sha(B)$ or $L(B,1)=0$. In the latter case, the Birch
and Swinnerton-Dyer conjecture predicts that $B(\Q)$ is infinite;
the theorem above then implies that
$$5\mid \Vis_{J_0(pN)}(\Sha(A)).$$
There are many similar examples, some of which have $\dim B=1$, and hence don't
rely on any conjectures. Based on this evidence, we make the following conjecture.
\begin{conjecture}[Stein]
Let $N$ be a prime and $A=A_f^{\vee}\subset J_0(N)$.
If $c\in \Sha(A)$, then there is a prime $p$ such that
$\delta_1^*(c)+ \delta_p^*(c) = 0 \in \Sha(J_0(pN))$.
\end{conjecture}
\section{Speculation: constructing points?}
Constructing elements of $\Vis_{J_0(N)}(\Sha(A))$ is equivalent to
constructing points on~$B$. This observation might lead to an
approach to proving the implication $L(B,1)=0$ implies $B(\Q)$ is infinite.
At present, mathematicians seem to have no good construction of points
when $L(B,1)=L'(B,1)=0$. However, we do have Kolyvagin's Euler systems when $L(A,1)\neq 0$!
\end{document}