Author: William A. Stein
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23\newcommand{\ncisom}{\approx}   % noncanonical isomorphism
24
25
26\title{Visible Shafarevich-Tate groups
27       of modular abelian varieties\footnote{Thanks
28       for inviting me!}\vspace{1em}\\
29{\small (Utrecht Arithmetic Geometry Workshop)}}
30\author{William A. Stein}
31\date{Tuesday, 27 June 2000}
33\begin{document}
34\maketitle
35
36\begin{slide}
38Let $f=\sum a_n q^n\in S_2(\Gamma_0(N);\C)$ be a newform.\\
39\mbox{}$\qquad f \leadsto$ optimal quotient $A_f$ of $J_0(N)$
40$$\xymatrix{ 41 A_f^{\vee}\[email protected]{^(->}[r] & J_0(N)\[email protected]{->>}[d]\\ 42 & A_f 43}$$
44Shafarevich-Tate group of $A$:
45$$\Sha(A) := \ker\left(H^1(\Q,A) \ra \prod_v H^1(\Q_v,A)\right)$$
46{\bf Conjecture:} (Birch, Swinnerton-Dyer, Tate)\vspace{-3ex}
47\begin{enumerate}
48\item {\em Formula:} $\Sha(A_f)$ is finite and
49$$L(A_f,1) = 50\frac{\Omega_{A_f}\cdot 51 \prod_{p\mid N} c_p}{\#A_f(\Q)\cdot \#A_f^{\vee}(\Q)}\cdot 52\#\Sha(A_f).$$
53\vspace{-5ex}
54\item {\em Rank:} Order of vanishing of $L(A_f,s)$ at $1$ equals
55rank of $A_f(\Q)$.  (Regulators, etc.)
56
57\end{enumerate}
58
59
60\end{slide}
61
62\begin{slide}
63\head{Mazur: Visualize $\Sha$!}
64
65Given $A \hra J$, let $B=J/A$.  We have:
66$$\xymatrix{ 670\ar[r] 68&{A(\Q)} 69\ar[r] 70& {J(\Q)} 71\ar[r] 72& {B(\Q)} 73\arr[d]d[][dlll]d[][dll]\\ 74&H^1(\Q,A)\ar[r] 75&H^1(\Q,J)\ar[r] 76&{\cdots}\\ 77} 78$$
79%$$0 \ra A(\Q) \ra J(\Q) \ra B(\Q) \ra H^1(\Q,A) \ra H^1(\Q,J) \ra \cdots$$
80The {\em visible or effaceable part} of $H^1(\Q,A)$ is the image
81of $B(\Q)$; equivalently, the kernel of
82$H^1(\Q,A) \ra H^1(\Q,J)$.
83Visible, in the sense that torsor is fiber over point in $J(\Q)$.
84
85{\bf Observation:} (Greenberg, ---)  Any $c\in H^1(\Q,A)$,
86is visible in $J=\mbox{\rm Res}_{K/\Q}(A_K)$, where~$K$ splits~$c$.\\
87{P{\tiny ROOF}:}
88Natural map $\iota:A\hookrightarrow J$ induces
89$$H^1(\Q,A) \xrightarrow{\iota_*} H^1(\Q,J) 90 = H^1(K,A)$$
91(by Shapiro's Lemma), so $\iota_*(c) = 0.$
92
93Conrad: Abelian varieties over infinite fields embed in Jacobians.
94
95{\bf Mazur's query:} Is $\Sha(A_f^{\vee})$ visible in $J_0(N)$?
96
97\end{slide}
98
99\begin{slide}
100\head{The data ($N$ prime)}
101Assume BSD.  There are $38$ rank zero $A_f^{\vee}$
102of prime level $\leq 2161$ with nontrivial odd part of $\Sha$.
103Of these, $22$ have all this $\Sha$ visible. E.g.,
104\begin{center}
105\begin{tabular}{lcll}
106\hspace{1em}$A_f$ & dim & $\sqrt{\#\Sha/ 2^?}$ & notes\\\hline
107{\bf 389} & 20 & 5 &\\
108{\bf 433} & 16 & 7 &\\
109{\bf 563} & 31& 13 &\\
110{\bf 571} & 2 & 3 &\\
111{\bf 709} & 30 & 11 &\\
112{\bf 997} & 42 & 9 & \\
113{\bf 1061} & 46 & 151 & \\
114{\bf 1091} & 62 & 7  & inv\\
115{\bf 1171} & 53  & 11 & \\
116{\bf 1283} & 62 & 5 & inv\\
117{\bf 1429} & 53 & 5 & inv\\
118{\bf 1481} & 62 & 14 & inv\\
119  $\cdots$  & $\cdots$ & $\cdots$ \\
120{\bf 2111} & 112 & 211  & inv, not eisen\\
121{\bf 2333} & 101 & 83341 & \\
122\end{tabular}
123\end{center}
124
125\end{slide}
126
127\begin{slide}
129{\bf Theorem 1. (---)}
130{\em Let $A, B \subset J$ s.t. $(A \cap B)(\C)$ and $A(\Q)$ finite.
131Assume $B$ has toric reduction at each $\ell \mid N_J$.
132Let $p$ be an odd prime (or principle prime ideal) not dividing $N_J$,
133the orders of comp.\ grps.\ of $A$ and $B$,
134$(J/B)(\Q)_{\tor}$, $B(\Q)_{\tor}$, and $B[p]\subset A$.  Then
135$$B(\Q)/p B(\Q) \subset \Sha(A)^{\vis}.\vspace{-5ex}$$}
136
137{P{\tiny ROOF SKETCH}:} Since $B[p]\subset A\cap B$,
138$B\ra C$ factors through $p$:
139$$\xymatrix{ 140& & B\ar[d] \ar[r]^{p}& B\ar[d]\\ 141 0\ar[r]& A\ar[r]&J\ar[r]&C\ar[r] &0.}$$
142{\bf Key diagram:}
143$$\xymatrix{ 1440 \ar[r] & B(\Q) \ar[r]^{p}\ar[d] & B(\Q)\ar[dr]^{\pi} \ar[r]\ar[d] 145 & B(\Q)/pB(\Q)\ar[r]\ar[d] & 0\\ 1460 \ar[r] & J(\Q)/A(\Q)\ar[r] & C(\Q) \ar[r] & H^1(\Q,A)^{\text{vis}} \ar[r] & 0\\ 147}$$
148Snake lemma implies $B(\Q)/p B(\Q)$ injects into $H^1(\Q,A)^{\text{vis}}$.
149Use subtle exactness properties of N\'eron models
150to get into $\Sha(A)^{\text{vis}}$.
151\end{slide}
152
153\begin{slide}
155Theorem 1 can be refined: condition that $p\nmid$ geometric component
156group of $B$ replaced by
157$$p \nmid \prod_{\ell} 158 \#\Phi_{B,\ell}(\F_\ell)\cdot \#H^1(\F_\ell,\Phi_{B,\ell}).$$
159
160A while ago, Cremona and Mazur discovered three elliptic curves, {\bf
1612849A}, {\bf 4343B}, {\bf 5389A}, in which BSD predicts odd invisible
162$\Sha$.\vspace{-3ex}
163
164\begin{center}
165\begin{tabular}{lcl}
166\hspace{1em}$A_f$ & $\sqrt{\#\Sha/ 2^?}$ & becomes visible\\\hline
167{\bf 2849A} & $3$ & $8547 = 3\cdot 2849 = 3\cdot 7\cdot 11\cdot 37$\\
168{\bf 5389A} & $3$ & $37723 = 7 \cdot 5389 = 7\cdot 17\cdot 317$\\\hline
169{\bf 1429B} & $5$ & $2858 = 2\cdot 1429$ (dim $B=2$) \\
170\end{tabular}\hspace{1em}
171\end{center}
172Now we know {\em unconditionally} that there is odd visible $\Sha$!
173
174{\bf Conjecture:} Let $c\in \Sha(A_f^{\vee})[p]$.  Then there exists an integer
175$M$ such that $\iota_{*}(c)=0$ for $\iota:A_f^{\vee}\ra J_0(NM)$ one of the
176natural maps.   (Require $p\nmid \deg(\iota)$.)
177
178\end{slide}
179
180
181\begin{slide}
183In the eventually
184visible examples above, we have {\em unconditionally proved} that
185up to a $2$-power $\Sha(A_f)$ is as big as BSD predicts.
186
187Vanishing of $L(B,1)$ usually forces vanishing of $L(A_f,1)$ modulo $p$,
188so we are very unlikely to observe that $\Sha(A_f)$ is bigger than
189BSD predicts.
190\end{slide}
191
192
193\begin{slide}
195
196{\bf Hard open problem:} {\em Suppose $L(E,1)$ and $L'(E,1)$ vanish.
197Show $E(\Q)$ is infinite.}
198
199Formulate visible analogue of BSD conjecture.
200Deduce the existence of nontrivial visible
201elements of $\Sha(A_f)$ in new way, and
202conclude that the rank of $E$ is $>0$ without computing
203points on~$E$.
204
205{\em Test question:} Prove that the first elliptic curve of rank $>1$
206has rank $>1$ {\em without} explicitely finding any points.
207This is the curve $E$ labeled {\bf 389A}.
208
209Strategy: Let~$A$ be {\bf 389E}, so $E \subset A\cap E$.
210Use Euler system of Heegner points'' to deduce that
211$5$ divides cardinality of {\em visible part} of $\Sha(A)$,
212then use the key diagram
213to deduce that $E(\Q)/5 E(\Q)$ is nontrivial
214(assume $E(\Q)=0$ and use the snake lemma to derive a