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\title{Visible Shafarevich-Tate groups
of modular abelian varieties\footnote{Thanks
for inviting me!}\vspace{1em}\\
{\small (Utrecht Arithmetic Geometry Workshop)}}
\author{William A. Stein}
\date{Tuesday, 27 June 2000}
\newcommand{\head}[1]{\begin{center}\bf #1\end{center}}
\begin{document}
\maketitle

\begin{slide}
\head{Modular abelian varieties}
Let $f=\sum a_n q^n\in S_2(\Gamma_0(N);\C)$ be a newform.\\
\mbox{}$\qquad f \leadsto$ optimal quotient $A_f$ of $J_0(N)$
$$\xymatrix{ A_f^{\vee}\[email protected]{^(->}[r] & J_0(N)\[email protected]{->>}[d]\\ & A_f }$$
Shafarevich-Tate group of $A$:
$$\Sha(A) := \ker\left(H^1(\Q,A) \ra \prod_v H^1(\Q_v,A)\right)$$
{\bf Conjecture:} (Birch, Swinnerton-Dyer, Tate)\vspace{-3ex}
\begin{enumerate}
\item {\em Formula:} $\Sha(A_f)$ is finite and
$$L(A_f,1) = \frac{\Omega_{A_f}\cdot \prod_{p\mid N} c_p}{\#A_f(\Q)\cdot \#A_f^{\vee}(\Q)}\cdot \#\Sha(A_f).$$
\vspace{-5ex}
\item {\em Rank:} Order of vanishing of $L(A_f,s)$ at $1$ equals
rank of $A_f(\Q)$.  (Regulators, etc.)

\end{enumerate}

\end{slide}

\begin{slide}
\head{Mazur: Visualize $\Sha$!}

Given $A \hra J$, let $B=J/A$.  We have:
$$\xymatrix{ 0\ar[r] &{A(\Q)} \ar[r] & {J(\Q)} \ar[r] & {B(\Q)} \arr[d]d[][dlll]d[][dll]\\ &H^1(\Q,A)\ar[r] &H^1(\Q,J)\ar[r] &{\cdots}\\ }$$
%$$0 \ra A(\Q) \ra J(\Q) \ra B(\Q) \ra H^1(\Q,A) \ra H^1(\Q,J) \ra \cdots$$
The {\em visible or effaceable part} of $H^1(\Q,A)$ is the image
of $B(\Q)$; equivalently, the kernel of
$H^1(\Q,A) \ra H^1(\Q,J)$.
Visible, in the sense that torsor is fiber over point in $J(\Q)$.

{\bf Observation:} (Greenberg, ---)  Any $c\in H^1(\Q,A)$,
is visible in $J=\mbox{\rm Res}_{K/\Q}(A_K)$, where~$K$ splits~$c$.\\
{P{\tiny ROOF}:}
Natural map $\iota:A\hookrightarrow J$ induces
$$H^1(\Q,A) \xrightarrow{\iota_*} H^1(\Q,J) = H^1(K,A)$$
(by Shapiro's Lemma), so $\iota_*(c) = 0.$

Conrad: Abelian varieties over infinite fields embed in Jacobians.

{\bf Mazur's query:} Is $\Sha(A_f^{\vee})$ visible in $J_0(N)$?

\end{slide}

\begin{slide}
\head{The data ($N$ prime)}
Assume BSD.  There are $38$ rank zero $A_f^{\vee}$
of prime level $\leq 2161$ with nontrivial odd part of $\Sha$.
Of these, $22$ have all this $\Sha$ visible. E.g.,
\begin{center}
\begin{tabular}{lcll}
\hspace{1em}$A_f$ & dim & $\sqrt{\#\Sha/ 2^?}$ & notes\\\hline
{\bf 389} & 20 & 5 &\\
{\bf 433} & 16 & 7 &\\
{\bf 563} & 31& 13 &\\
{\bf 571} & 2 & 3 &\\
{\bf 709} & 30 & 11 &\\
{\bf 997} & 42 & 9 & \\
{\bf 1061} & 46 & 151 & \\
{\bf 1091} & 62 & 7  & inv\\
{\bf 1171} & 53  & 11 & \\
{\bf 1283} & 62 & 5 & inv\\
{\bf 1429} & 53 & 5 & inv\\
{\bf 1481} & 62 & 14 & inv\\
$\cdots$  & $\cdots$ & $\cdots$ \\
{\bf 2111} & 112 & 211  & inv, not eisen\\
{\bf 2333} & 101 & 83341 & \\
\end{tabular}
\end{center}

\end{slide}

\begin{slide}
\head{Criterion for visibility}
{\bf Theorem 1. (---)}
{\em Let $A, B \subset J$ s.t. $(A \cap B)(\C)$ and $A(\Q)$ finite.
Assume $B$ has toric reduction at each $\ell \mid N_J$.
Let $p$ be an odd prime (or principle prime ideal) not dividing $N_J$,
the orders of comp.\ grps.\ of $A$ and $B$,
$(J/B)(\Q)_{\tor}$, $B(\Q)_{\tor}$, and $B[p]\subset A$.  Then
$$B(\Q)/p B(\Q) \subset \Sha(A)^{\vis}.\vspace{-5ex}$$}

{P{\tiny ROOF SKETCH}:} Since $B[p]\subset A\cap B$,
$B\ra C$ factors through $p$:
$$\xymatrix{ & & B\ar[d] \ar[r]^{p}& B\ar[d]\\ 0\ar[r]& A\ar[r]&J\ar[r]&C\ar[r] &0.}$$
{\bf Key diagram:}
$$\xymatrix{ 0 \ar[r] & B(\Q) \ar[r]^{p}\ar[d] & B(\Q)\ar[dr]^{\pi} \ar[r]\ar[d] & B(\Q)/pB(\Q)\ar[r]\ar[d] & 0\\ 0 \ar[r] & J(\Q)/A(\Q)\ar[r] & C(\Q) \ar[r] & H^1(\Q,A)^{\text{vis}} \ar[r] & 0\\ }$$
Snake lemma implies $B(\Q)/p B(\Q)$ injects into $H^1(\Q,A)^{\text{vis}}$.
Use subtle exactness properties of N\'eron models
to get into $\Sha(A)^{\text{vis}}$.
\end{slide}

\begin{slide}
\head{Visibility at higher level}
Theorem 1 can be refined: condition that $p\nmid$ geometric component
group of $B$ replaced by
$$p \nmid \prod_{\ell} \#\Phi_{B,\ell}(\F_\ell)\cdot \#H^1(\F_\ell,\Phi_{B,\ell}).$$

A while ago, Cremona and Mazur discovered three elliptic curves, {\bf
2849A}, {\bf 4343B}, {\bf 5389A}, in which BSD predicts odd invisible
$\Sha$.\vspace{-3ex}

\begin{center}
\begin{tabular}{lcl}
\hspace{1em}$A_f$ & $\sqrt{\#\Sha/ 2^?}$ & becomes visible\\\hline
{\bf 2849A} & $3$ & $8547 = 3\cdot 2849 = 3\cdot 7\cdot 11\cdot 37$\\
{\bf 5389A} & $3$ & $37723 = 7 \cdot 5389 = 7\cdot 17\cdot 317$\\\hline
{\bf 1429B} & $5$ & $2858 = 2\cdot 1429$ (dim $B=2$) \\
\end{tabular}\hspace{1em}
\end{center}
Now we know {\em unconditionally} that there is odd visible $\Sha$!

{\bf Conjecture:} Let $c\in \Sha(A_f^{\vee})[p]$.  Then there exists an integer
$M$ such that $\iota_{*}(c)=0$ for $\iota:A_f^{\vee}\ra J_0(NM)$ one of the
natural maps.   (Require $p\nmid \deg(\iota)$.)

\end{slide}

\begin{slide}
\head{Evidence for the BSD conjecture}
In the eventually
visible examples above, we have {\em unconditionally proved} that
up to a $2$-power $\Sha(A_f)$ is as big as BSD predicts.

Vanishing of $L(B,1)$ usually forces vanishing of $L(A_f,1)$ modulo $p$,
so we are very unlikely to observe that $\Sha(A_f)$ is bigger than
BSD predicts.
\end{slide}

\begin{slide}
\head{Query: Constructing points?}

{\bf Hard open problem:} {\em Suppose $L(E,1)$ and $L'(E,1)$ vanish.
Show $E(\Q)$ is infinite.}

Formulate visible analogue of BSD conjecture.
Deduce the existence of nontrivial visible
elements of $\Sha(A_f)$ in new way, and
conclude that the rank of $E$ is $>0$ without computing
points on~$E$.

{\em Test question:} Prove that the first elliptic curve of rank $>1$
has rank $>1$ {\em without} explicitely finding any points.
This is the curve $E$ labeled {\bf 389A}.

Strategy: Let~$A$ be {\bf 389E}, so $E[5] \subset A\cap E$.
Use Euler system of Heegner points'' to deduce that
$5$ divides cardinality of {\em visible part} of $\Sha(A)$,
then use the key diagram
to deduce that $E(\Q)/5 E(\Q)$ is nontrivial
(assume $E(\Q)=0$ and use the snake lemma to derive a
contradiction).

\end{slide}

\end{document}