\documentclass{slides} \usepackage{amsmath} \usepackage{amssymb} \usepackage[all]{xy} \DeclareFontEncoding{OT2}{}{} % to enable usage of cyrillic fonts \newcommand{\textcyr}[1]{% {\fontencoding{OT2}\fontfamily{wncyr}\fontseries{m}% \fontshape{n}\selectfont #1}} \newcommand{\Sha}{{\mbox{\textcyr{Sh}}}} \newcommand{\hra}{\hookrightarrow} \newcommand{\C}{\mathbf{C}} \newcommand{\Q}{\mathbf{Q}} \newcommand{\ra}{\rightarrow} \newcommand{\T}{\mathbf{T}} \newcommand{\Z}{\mathbf{Z}} \newcommand{\tor}{\mbox{\tiny\rm tor}} \newcommand{\m}{\mathfrak{m}} \newcommand{\ncisom}{\approx} % noncanonical isomorphism \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 $J_0(N)$ $$\xymatrix{ A_f^{\vee}\ar@{^(->}[r] & J_0(N)\ar@{->>}[d]\\ & A_f }$$ 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's Motif\footnote{``Motif'' in the sense of unifying idea or recurring theme.}: Visualize $\Sha$!} Given $A \hra J$, let $B=J/A$. We have: $$\xymatrix{ 0\ar[r] &{A(\Q)} \ar[r] & {J(\Q)} \ar[r] & {B(\Q)} \ar`r[d]`d[]`[dlll]`d[][dll]\\ &H^1(\Q,A)\ar[r] &H^1(\Q,J)\ar[r] &{\cdots}\\ } $$ %$$0 \ra A(\Q) \ra J(\Q) \ra B(\Q) \ra H^1(\Q,A) \ra H^1(\Q,J) \ra \cdots$$ The {\em visible part} of $H^1(\Q,A)$ is the image of $B(\Q)$. Points in $B(\Q)$ can be ``written down.'' {\bf Observation:} (Greenberg) Any $c\in H^1(\Q,A)$ is visible in $J=\mbox{\rm Res}_{K/\Q}(A_K)$, where~$K$ splits~$c$. {\bf Question:} How much of $\Sha(A_f^{\vee})$ is visible in $J_0(N)$? \end{slide} \begin{slide} \head{Experimental answer ($N$ 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}{lc} \hspace{1em}$A_f$ & $\sqrt{\#\Sha/ 2^?}$\\\hline {\bf 389E} & 5 \\ {\bf 433D} & 7 \\ {\bf 563E} & 13\\ {\bf 571D} & 3\\ {\bf 709C} & 11\\ {\bf 997H} & 9\\ {\bf 1061D} & 151\\ {\bf 1091C} & 7 (inv)\\ \end{tabular}\hspace{1em} \begin{tabular}{lc} \hspace{1em}$A_f$ & $\sqrt{\#\Sha/ 2^?}$\\\hline {\bf 1171D} & 11\\ {\bf 1283C} & 5 (inv)\\ {\bf 1429B} & 5 (inv)\\ {\bf 1481C} & 13 (inv)\\ {\bf 1483D} & 15\\ {\bf 1531D} & 73\\ {\bf 1559B} & 11 (inv)\\ {\bf 1567D} & $7\cdot 41$ \end{tabular} \end{center} Spectacular examples:\vspace{-1ex} \begin{center} \begin{tabular}{lcc} \hspace{1em}$A_f$ & $\mbox{\rm dim}(A_f)$ & $\sqrt{\#\Sha/ 2^?}$\\\hline {\bf 2111B} & 112 & 211 (inv, eisen)\\ {\bf 2333C} & 101 & 83341 \\ {\bf 2593B} & 109 & $67\cdot 2213$\\ \end{tabular} \end{center} \end{slide} \begin{slide} \head{Verifying 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 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 $$B(\Q)/p B(\Q) \subset \mbox{\rm visible part of } \Sha(A).$$} Since $B[p]\subset A\cap B$, $$\xymatrix{ & & B\ar[d] \ar[r]^{p}& B\ar[d]\\ 0\ar[r]& A\ar[r]&J\ar[r]&C\ar[r] &0.}$$ {\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 reduction and comp.\ grp.\ hypothesis to get into $\Sha(A)$. \end{slide} \begin{slide} \head{Positive evidence for BSD} In the 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 $L(B,1)$ 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{First example of visibility at higher level} Cremona-Mazur discovered elliptic curve $E$ at level $5389=17\cdot 317$ s.t.\vspace{-2ex} \begin{center} BSD $\Longrightarrow$ $\#\Sha(A_f)=3^2$.\vspace{-2ex} \end{center} $$E:\quad y^2 + xy + y = x^3 - 35590x - 2587197$$ At level $37723 = 7\cdot 5389 = 7\cdot 17\cdot 317$, I found $$F:\quad y^2 - y = x^3 + x^2 + 34x - 248,$$ with $F[3] \ncisom E[3]$. Theorem 1 can be modified to show that $\Sha(E)$ is nontrivial and visible in $J_0(37723)$. \end{slide} \begin{slide} \head{Can we construct points when $L(E,1)$ vanishes to order $>1$?} {\bf Hard problem:} {\em Suppose $L(E,1)$ and $L'(E,1)$ vanish. Show $E(\Q)$ is infinite.} If one could deduce the existence of nontrivial visible elements of $\Sha(A_f)$ in new way, then perhaps one could deduce that the rank of $E$ is $>0$ without computing points on $E$. 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}