Sharedwww / tables / Notes / refineogg.texOpen in CoCalc
Author: William A. Stein
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9\title{\bf\Huge \mbox{A refinement of Ogg's conjecture}\vspace{5ex}}
10\author{\LARGE William A. Stein}
11\date{\Large \today \vspace{2ex}\\}
13\newcounter{Pagecount}
14\markboth{}{W.\thinspace{}A. Stein, A refinement of Ogg's conjecture}
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16        \S\thePagecount. #1}\vspace{1ex}}
17\theoremstyle{plain}
18\newtheorem{thm}{Theorem}
19\newtheorem{prp}[thm]{Proposition}
20\newtheorem{cor}[thm]{Corollary}
21\newtheorem{conj}[thm]{Conjecture}
22\DeclareMathOperator{\numer}{numer}
23\DeclareMathOperator{\order}{order}
24\newcommand{\eisen}{\numer\left(\frac{p-1}{12}\right)}
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65
66\begin{document}
67
68\Large
69\maketitle
70\pagenumbering{Roman}
71
72\setcounter{page}{0}
73
74\Page{The conjecture}
75\begin{bulletlist}
76\item $p$ prime
77\item $f = q+a_2 q^2 + \cdots \in S_2(\Gamma_0(p);\C)$
78        eigenform for $\T=\Z[\ldots T_n\ldots]$
79\item $I_f := \ker(\T \xrightarrow{T_p \mapsto a_p} K_f=\Q(\ldots a_n\ldots))$
80\item $A_f := J_0(p)/I_f J_0(p)$, abelian variety, dimension $d:=[K_f:\Q]$
81\item $\Adual_f :=$ dual of $A_f$
82\item $\pi: J_0(p)\ra A_f$
83\end{bulletlist}
84
85{\bf Associated finite groups:}
86\begin{bulletlist}
87\item $A_f(\Q)_{\tor}$ rational torsion
88\item $\Phi_{A_f}(\Fpbar)$ component group of closed fiber of N\'{e}ron model
89\end{bulletlist}
90
91{\bf Motivation:} Both quantities are closely related to quantities that
92appear in the conjecture of Birch and Swinnerton-Dyer, so we must
93understand them completely.
94
95\begin{conj}
96\begin{itemize}
97\item[(a)] We have
98$$\#\Phi_{A_f}(\Fpbar) = \#A_f(\Q)_{\tor} 99 % = \#\Adual_f(\Q)_{\tor} 100% = \#\Phi_{A_f}(\Fp) 101 = \order(\pi(0-\infty))$$
102\item[(b)] (More tentative)
103We have
104   $$\prod \#A_f(\Q)_{\tor} = 105 \numer\left(\frac{p-1}{12}\right),$$
106where the product is over the Galois conjugacy classes.
107\end{itemize}
108\end{conj}
109
110{\bf Why?}
111\begin{bulletlist}
112\item It is the nicest possible answer.
113\item All of my computer computations suggest it.
114%\item Part (a) is true for elliptic curves (Mestre-Oesterle, 1989)
115\end{bulletlist}
116{\bf Remark:} Totally false when the level
117is composite! Counterexamples abound, e.g., $E=${\bf 33A},
118$E(\Q)=\Z/2\cross\Z/2$, $\Phi_E(\Fbar_3)=\Z/6\Z$.
119
120
121\Page{Computational evidence}
122\begin{bulletlist}
123\item Conjecture~1 true for all $305$ of the $A_f$ of conductor $p\leq 631$.
124\item For all $666$ of the $A_f$ of conductor $p\leq 1571$ the conjecture
125is true, up to powers of $2$.
126I haven't checked the $2$-power.  Same statement for $880$ of the
127abelian varieties up to level $2411$ (some levels missed).
128\end{bulletlist}
129
130{\bf Example 1:}
131$$\begin{array}{lrrrrr} 132 J_0(389) \sim& A_1 &\cross A_2&\cross A_3 &\cross A_6 &\cross A_{20}\\ 133\#A_f(\Q)_{\tor}& 1 & 1& 1&1&97\\ 134\#\Phi_f(\Fpbar)&1&1&1&1&97\\ 135\frac{L(1)}{\Omega}&0&0&0&0&\frac{2^{11}\cdot 5^2}{97} 136\end{array}$$
137
138$$\numer\left(\frac{389-1}{12}\right) = 97$$
139
140
141
142{\bf Example 2:}
143$$\begin{array}{lrrrr} 144 J_0(113) \sim&A_1 &\cross A_2&\cross A_3 & \cross B_3\\ 145 \#A_f(\Q)_{\tor}&2&2&1&7\\ 146 \#\Phi_f(\Fpbar)&2&2&1&7\\ 147\frac{L(1)}{\Omega}&\frac{1}{2}&\frac{1}{2}&0&\frac{8}{7} 148\end{array}$$
149
150$$\numer\left(\frac{113-1}{12}\right) = 28$$
151
152(Remark: Probably $\text{Sha}(B_3)$ has order $8$,
153and is explained by the Mordell-Weil group of $A_3$.)
154
155
156\Page{Theoretical evidence}
157{\bf Ribet and Mestre-Oesterl\'{e} (J. Reine Angew. Math., 1989):}
158If $A_f$ is an elliptic curve, then Conjecture~1(a) true.
159
160{\bf Mazur:}
161\begin{bulletlist}
162\item (Ogg's conjecture)
163 $$\# J_0(p)(\Q)_{\tor} = \# \Phi_{J_0(p)}(\Fpbar) 164 = \numer\left(\frac{p-1}{12}\right).$$
165   So my conjecture can be viewed as a refinement of Ogg's.
166\item If $\ell \mid\# A_f(\Q)$ then
167$\ell \mid \eisen$\\(Mazur's {\em Modular curves}\dots, page 141).
168
169\item Conjecture 1(b): Mazur's comment on the conjecture is:
170\begin{quote}
171  What you are conjecturing is something I don't think I ever thought
172of---at least in those terms---It has some implications which are interesting
173(to me). For example suppose that a prime q divides the numerator of
174$(p-1)/12$  but $q^2$ does not; let $A$ be the $q$-Eisenstein
175quotient. Then only
176one optimal quotient of $A$ would be allowed to have nontrivial $q$-torsion,
177and the other(s), if there are any, would, I guess, then have to settle for
178a $\mu_q$ subgroup.  I suppose, though, that it is very often the case that
179(under the above hypotheses of $q$) the $q$-Eisenstein quotient is simple.
180\end{quote}
181\end{bulletlist}
182
183{\bf Ribet:}
184If $\ell\mid \#\Phi_{A_f}(\Fpbar)$ then $\rho_{f,\m}$ {\em tends}
185to be finite for some $\m$ containing $\ell$.  If so, and $\rho_{f,\m}$
187
188{\bf Remark:} If $\Phi_J\ra \Phi_{A_f}$ is surjective, then
189the primes  on the left hand side of Conjecture 1b divide the right
190hand side.  So we make the following weaker conjecture.
191
192\begin{conj}
193The natural map $\Phi_J \into \Phi_{A_f}$ is surjective.
194\end{conj}
195
196\Page{Grothendieck's monodromy pairing}
197\vspace{1ex}
198\begin{bulletlist}
199\item $J:=J_0(p)$.
200\item $A:=A_f$.
201\item $\cA:=$ N\'{e}ron model of $A$.
202\item $T_A := \text{(torus of closed fiber of$\cA$)} = \cA_{\Fp}^o$,
203since $A$ is purely toric.
204\item $X_A := \Hom(T_A,\Gm)$; $A\mapsto X_A$ contravariant;
205      $X_A$ free abelian of rank $X_A=\dim A$.
206\end{bulletlist}
207The key diagrams are as follows.
208$$\[email protected]=6pc{\Adual \[email protected]{^(->}[r]^{\pi'}\ar[dr]_{\theta} 209 & J \[email protected]{->>}[d]^{\pi}\\ 210 &A} 211\qquad\qquad\qquad 212 \[email protected]=6pc{X_{A} \[email protected]{^(->}[r]^{\pi^*} \ar[dr]^{\theta^*} 213 & X_{J} \[email protected]{->>}[d]^{\pi_*} \\ 214 & X_{\Adual}\[email protected]/^1.5pc/[ul]^{\theta_*}} 215$$
216
217{\bf Modular degree:} $m_A = \sqrt{\deg \theta}$.
218
219{\bf Monodromy:}
220$$\langle \quad , \quad \rangle : X_J \cross X_J \ra \Z$$
221$$\langle \quad , \quad \rangle : X_A \cross X_{\Adual} \ra \Z$$
222
223{\bf Component groups:}
224$$0\ra X_J \ra \Hom(X_J,\Z) \ra \Phi_J \ra 0$$
225$$0\ra X_{\Adual} \ra \Hom(X_A,\Z) \ra \Phi_A \ra 0$$
226
227\Page{Formulas for $\Phi_A$}
228$S := \text{ saturation of$\pi^* X_A$}$\\
229For any $L\subset S$ of finite index, define
230\begin{eqnarray*}
232 \Phi_L &:=& \coker(X_J \ra \Hom(L,\Z))
233\end{eqnarray*}
234
235\begin{thm}[-]
236Let $L\subset S$ be of finite index; then
237\begin{enumerate}
238\item \vspace{-2ex}$\Phi_A = \Phi_{\pi^* X_A}$, i.e.,
239     $X_J \ra \Hom(\pi^* X_A, \Z) \ra \Phi_A \ra 0$
240 is exact;
241\item $\displaystyle \frac{\#\Phi_A}{m_A} = \frac{\#\Phi_L}{m_L}$;
242\item $\image(\Phi_{J}\ra\Phi_A) \isom \Phi_S$.
243\end{enumerate}
244\end{thm}
245
246\begin{cor}
247$$\#\coker(\Phi_J\ra\Phi_A) = \frac{\#\Phi_A}{\#\Phi_S} = \frac{m_A}{m_S},$$
248in particular, $m_S \mid m_A$.
249\end{cor}
250
251{\bf Conclusion:}
252$$\[email protected]=1.5cm{ 253 & *++++[F-,]{\Phi_J\onto \Phi_A} \[email protected]{<=>}[dl]_{\txt{easy }}\[email protected]{<=>}[dr]^{\txt{ hard}}\\ 254 \mbox{\pi^* X_A \text{ is saturated}} & &{m_A=m_S}}$$
255
256{\bf Observation:}
257\begin{eqnarray*}
259m_A &=& \#(H_1(A,\Z)^+/\pi_*(H_1(J,\Z)^+[I_f]))\cdot (2-\text{power})
260\end{eqnarray*}
261so if $H_1(J,\Z)^+\tensor\Zp \isom X_J\tensor\Zp$ as $\T\tensor\Zp$-modules,
262then $$p\nmid \#\coker(\Phi_J\ra\Phi_A).$$
263
264\Page{Level lowering}
265The component group $\Phi_A$  of $A$ frequently forces'' various
266$A[\m]$ to arise from a finite flat group scheme.
267Ribet's theorem implies such~$\m$ are reducible, and then
268Mazur's theorem implies such $\m$ Eisenstein.
269
270Suppose $\ell\mid\#\Phi_A$. Does $\ell\mid \numer(\frac{p-1}{12})$?\\
271$X:=X_A, \quad Y:=X_{\Adual}$
272$$\xymatrix{ 273 & 0\ar[r]\ar[d]& Y\ar[d]\ar[rr]^{\ell}& & Y\ar[d]\ar[r]& Y/\ell Y\ar[r]\ar[d]& 0\\ 2740\ar[r]& {\Hom(X/\ell X,\mu_\ell)}\ar[d]\ar[r] 275 & T\ar[d]\ar[rr]^{\cdot^\ell}&& T\ar[d]\ar[r] & 0\ar[d]\\ 276 0\ar[r]& {A[\ell]}\ar[r]& A\ar[rr]^{\ell}&& A\ar[r]& 0}$$
277
278Monodromy: $0 \ra Y \ra \Hom(X,\Z) \ra \Phi_A \ra 0$, gives
279$$0 \ra \Phi_A[\ell] \ra Y\tensor \Z/\ell\Z \ra \Hom(X,\Z/\ell\Z) 280 \ra \Phi_A \tensor\Z/\ell\Z\ra 0.$$
281Snake gives: $0 \ra \Hom(X/\ell X,\mu_{\ell}) 282 \ra A[\ell] \xrightarrow{\,f\,} Y/\ell Y \ra 0.$\\
283So:
284$$\xymatrix{ 285 & & f^{-1}(\Phi_A[\ell])\[email protected]{^(->}[d] 286 & {\Phi_A[\ell]} \[email protected]{^(->}[d]\\ 287 0\ar[r]&{\Hom(X/\ell X,\mu_{\ell})} 288 \ar[r]& {A[\ell]}\ar[r] & {Y/\ell Y}\ar[r]&0. 289}$$
290
291The maximal finite part of $A[\ell]$ is $f^{-1}(\Phi_A[\ell])$,
292which is {\em bigger} than $\Hom(X/\ell X,\mu_{\ell})$.
293Is it possible to show that the $f^{-1}(\Phi_A[\ell])$
294in $A[\ell]$ forces some $A[\m]$ to be finite?
295Example: If $\dim A=1$ then $Y/\ell Y$ has dimension one
296and the sequence becomes
297$$0 \ra \Hom(X/\ell X,\mu_{\ell}) 298 \ra A[\ell] \ra \Phi_A[\ell] \ra 0$$
299so $A[\ell]$ is finite, so $\ell\nmid\eisen$.
300
301\end{document}
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