CoCalc Shared Fileswww / tables / computing_cp.tex
Author: William A. Stein
1% computing_cp.tex
2\documentclass[11pt]{article}
3\include{macros}
4\title{Component Groups of Optimal Quotients of $J_0(N)$}
5\begin{document}
6\maketitle
7\begin{abstract}
8\end{abstract}
9\section{Introduction}
10\comment{
11What are the simplest naturally occuring abelian varieties over $\Q$?
12In my opinion they are the optimal'' quotients of the Jacobian $J_0(N)$
13of the modular curve $X_0(N)=\overline{\sltwoz\backslash\h}$.
14Let $f\in S_2(\Gamma_0(N))$ be a newform.  Then Shimura associated to
15$f$ a quotient $A_f$ of $J_0(N)$ having all of the right naturality
16properties.  These quotients are optimal in the sense that the kernel
17inside $J_0(N)$ is itself an abelian variety.
18The Shimura-Taniyama conjecture, mostly proved by recent work
19of Wiles and Taylor-Wiles, asserts that every elliptic curve over
20$\Q$ is isogeneous to some $A_f$.  Thus understanding the abelian varieties
21$A_f$ is paramount importance!
22
23Neron associated to any abelian variety a model of the integer ring
24which is an abelian scheme and whose points are just what one guesses
25by looking at the points over the base field.  Let $\cA_f$ be the
26Neron model of $A_f$.  Suppose $p$ is a prime which divides the level
27$N$, so that $\cA_f/\F_p$ isn't an abelian variety, and hence has no
28right to be connected. The group of components
29$\Phi_{A_f,p}$ is of important in many ways.  For example,
30according to the Birch and Swinnerton-Dyer conjecture,
31the order of its rational subgroup appears in the numerator
32of a certain rational number associated to the $L$-function
33of $A_f$.
34}
35
36Let $f\in S_2(\Gamma_0(N))$ be a newform and $A_f$ the corresponding
37optimal quotient of $J_0(N)$.
38Let $\Phi_{A_f,p}$ be the special fiber of the Neron model of $A_f$ at $p$.
39The integer $\#\Phi_{A_f,p}$ is of great arithmetic interest. For example,
40it appears in the Birch and Swinnerton-Dyer conjecture (I lie: actually the
41cardinality of the rational component group appears there).
42
43\begin{center}
44How can we compute $\#\Phi_{A_f,p}$?
45\end{center}
46
47When $A_f=E$ is an elliptic curve the Riemann-Roch theorem gives us
48a powerful tool: we have a simple explicit Weierstrass equation which defines $E$.
49Using this Tate provided an efficient algorithm for computing the order
50of $\Phi_{E,p}$.  I doubt that there is an analogue of the Weierstrass
51equation when $\dim A_f > 1$!
52
53Ribet, in a letter to Mestre, considered the case in which $A_f=E$
54is an elliptic curve which has multiplicative reduction at $p$,
55i.e., $\ord_p(N)=1$.  Using Grothendieck's monodromy pairing
56on character groups of certain tori he was able to provide a
57Weierstrass-free'' formula for $\Phi_{E,p}$.
58In this paper we generalize Ribet's method to $A_f$ of arbitrary
59dimension and thus obtain an explicit formula for $\#\Phi_{A_f,p}$.
60
61The formula is:
62$$\#\Phi_{A_f,p} = 63 \sqrt{\deg(\delta)} 64 \cdot \frac{\# \coker (X \ra \Hom(X[\p],\Z))^2} 65 {\disc(X[\p] \cross X[\p] \ra \Z)}$$
66where
67\begin{itemize}
68\item $\delta:A_f^{\vee}\ra A_f$ is the modular polarization.
69\item $X$ is the character group of the torus associated to $J_0(N)/\Fp$.
70\item $\p_f=\Ann_{\T}(f)$, where $\T$ is the Hecke algebra.
71\item $X\cross X\ra \Z$ is the Monodromy pairing.
72\end{itemize}
73Each of these quantities can be explicitely computed.  See
74\cite{kohel}, \cite{mestre}, and \cite{stein}.
75
76\section{The Monodromy Pairing}
77[to be written -- will contain the basic facts]
78
79\section{Ribet's Formulas}
80[to be written -- will contain Ribet's formulas as motivation.]
81
82\section{General Formula}
83In this section we prove our main theorem.
84
85Let $f\in S_2(\Gamma_0(N))$ be a newform and suppose $p|N$ but $p\nmid M=N/p$.
86The Hecke algebra $\T$ acts on the fundamental group $X_J$ of $J_0(N)$.
87The prime ideal
88           $$\p_f=\Ann_\T(f) \subset \T$$
89cuts out a submodule
90        $$V_f=X_J[\p_f] \subset X_J$$
91of $X_J$.
92\begin{theorem}
93$$\#\Phi_{A_f,p} = 94 \sqrt{\deg(\delta)} 95 \cdot \frac{\# \coker (X_J \ra \Hom(V_f,\Z))^2} 96 {\disc(V_f, V_f)}$$
97\end{theorem}
98
99\vspace{.3in}
100We now begin the proof.
101\vspace{.3in}
102
103{\noindent \bf Fundamental diagrams:}
104\begin{center}
105\begin{picture}(310,150)
106\put(0,110){$A^{\vee}$}
107\put(70,110){$J$}
108\put(70,40){$A$}
109\put(75,100){\vector(0,-1){45}}
110\put(75,63){\vector(0,-1){3}}
111\put(13,105){\vector(1,-1){55}}
112\qbezier(65,45)(10,60)(5,103)\put(5,103){\vector(0,1){1}}
113\put(18,113){\vector(1,0){48}}
114\qbezier(18,113)(14,115)(18,117)
115
116\put(40,117){$\pi^{\vee}$}
117\put(40,80){$\delta$}
118\put(20,50){$\hat{\delta}$}
119\put(80,80){$\pi$}
120
121
122\put(230,110){$X_A$}
123\put(300,110){$X_J$}
124\put(300,40){$X_{A^{\vee}}$}
125\put(305,100){\vector(0,-1){45}}
126\put(305,63){\vector(0,-1){3}}
127\put(243,105){\vector(1,-1){55}}
128\qbezier(290,45)(240,60)(237,103)\put(237,103){\vector(0,1){1}}
129\put(248,113){\vector(1,0){48}}
130\qbezier(248,113)(244,115)(248,117)
131
132\put(270,117){$\pi^*$}
133\put(270,80){$\delta^*$}
134\put(250,50){$\hat{\delta}^*$}
135\put(310,80){$\pi_*$}
136\end{picture}
137\end{center}
138
139{\noindent \bf Monodromy pairings:}
140
141$$\langle \quad , \quad \rangle_J : X_J \cross X_J \ra \Z$$
142$$\langle \quad , \quad \rangle_A : X_A^{\vee} \cross X_A \ra \Z$$
143\vspace{.4in}
144
145{\noindent \bf Component groups:}
146$$0\ra X_J \ra \Hom(X_J,\Z)\ra \Phi_{J,p} \ra 0$$
147$$0\ra X_{A^{\vee}} \ra \Hom(X_A,\Z)\ra \Phi_{A,p} \ra 0$$
148\vspace{.4in}
149
150{\noindent \bf Step 1. Formulas:}
151\begin{lemma}
152Let $x\in X_{A^{\vee}}$, $y\in X_A$.  Then
153$$\langle \pi^*\hat{\delta}^* x, \pi^* y\rangle_J = 154 \deg(\delta) \langle x, y \rangle_A.$$
155\end{lemma}
156\begin{proof}
157This is a direct computation, using the functorial properties of
158the objects involved.
159\begin{eqnarray*}
160\langle \pi^*\hat{\delta}^* x, \pi^* y\rangle_J &=&
161            \langle \pi_*\pi^*\hat{\delta}^* x, y\rangle_A \\
162        &=& \langle \delta^* \hat{\delta}^* x, y\rangle_A \\
163        &=& \langle (\hat{\delta} \circ \delta)^* x, y\rangle_A\\
164        &=& \deg(\delta) \langle x, y \rangle_A
165\end{eqnarray*}
166\end{proof}
167
168\begin{proposition}
169$$\#\Phi_{A,p} = \frac{\disc(\pi^*\hat{\delta}^* X_{A^{\vee}}, \pi^* X_A)}{\deg(\delta)^d}.$$
170\end{proposition}
171\begin{proof}
172By the lemma we have
173$$\disc(\pi^*\hat{\delta}^* X_{A^{\vee}}, \pi^* X_A) = \deg(\delta)^d 174 \cdot \disc(X_{A^{\vee}},X_A).$$
175Now use that
176$$\#\Phi_{A,p}=\#\coker(X_{A^{\vee}}\ra X_A)= \disc(X_{A^{\vee}},X_A).$$
177\end{proof}
178
179\begin{proposition}\label{discformula}
180$$\#\Phi_{A,p} = \frac{[X_A:\hat{\delta}^* X_{A^\vee}]\cdot 181 \disc(\pi^* X_A,\pi^* X_A)} 182 {\deg(\delta)^d}.$$
183\end{proposition}
184\begin{proof}
185Using the formula from the previous proposition we have
186\begin{eqnarray*}
187\#\Phi_{A,p} &=& \frac{\disc(\pi^*\hat{\delta}^* X_{A^{\vee}}, \pi^* X_A)}
188                      {\deg(\delta)^d} \\
189            &=& \frac{[\pi^* X_A : \pi^*\hat{\delta}^*X_{A^{\vee}}]\cdot
190                            \disc (\pi^* X_A, \pi^* X_A)}
191                     {\deg(\delta)^d}
192\end{eqnarray*}
193Note that $\pi^*$ is an injection so it does not change
194the index:
195$$[\pi^*X_A : \pi^*\hat{\delta}^* X_{A^{\vee}}] 196 = [X_A:\hat{\delta}^* X_{A^\vee}].$$
197\end{proof}
198
199{\noindent \bf Step 2. Identify $\Phi_{A,p}$ inside of $X_J$:}
200
201\begin{proposition}\label{snakecokernel}
202$$\Phi_{A,p} \isom \coker(X_J \ra \Hom(\pi^* X_A, \Z))$$
203\end{proposition}
204\begin{proof}
205Let $\tilde{X}_J$ denote the image of $X_J$ in $\Hom(\pi^*X_A,\Z)$.
206We have the following diagram with exact rows and columns:
207$$208\begin{matrix} 209 & & & & & & 0 & \\ 210 & & & & & & \da & \\ 211 & & & & 0 & \lra & K & \\ 212 & & & & \da & & \da & \\ 213 0&\lra& \tilde{X}_J& \lra & \Hom(\pi^*X_A,\Z) & \lra & W & \lra 0 \\ 214 & & \da \isom & & \da \isom & & \da & \\ 215 0&\lra& X_{A^{\vee}}& \lra & \Hom(X_A,\Z) & \lra & \Phi_{A,p} & \lra 0 \\ 216 & & \da & & \da & & \da & \\ 217 & & 0 & \lra & 0 & \lra & L & \\ 218 & & & & & & \da & \\ 219 & & & & & & 0 & \\ 220\end{matrix} 221$$
222The snake lemma implies that the sequence
223$$0 \ra K \ra 0 \ra 0 \ra L \ra 0$$
224is exact,
225hence $K=L=0$ and so $W= \Phi_{A,p}$.
226\end{proof}
227
228{\noindent \bf Step 3. Relate $\pi^*X_A$ to $V_f$:}
229Let
230$$V_f = X_J[\p_f] = \bigcap_{t\in\Ann_{\T}(f)} \ker(t) \subset X_J.$$
231By multiplicity one and $\T$-invariance of $\pi^*$,
232$$\pi^* X_A \subset V_f,$$
233and the index is finite.  Let
234   $$m = [V_f : \pi^* X_A].$$
235
236The discriminant expression of $\Phi_{A,p}$ is homogeneous of degree
237$2$ in $V_f$'':
238\begin{proposition}\label{homsquare}
239$$\#\Phi_{A,p} = \frac{[X_A:\hat{\delta}^* X_{A^\vee}]\cdot m^2 \cdot 240 \disc(V_f, V_f)} 241 {\deg(\delta)^d}.$$
242\end{proposition}
243\begin{proof}
244Note that
245$$\disc(\pi^* X_A, \pi^* X_A) = m^2 \disc(V_f,V_f).$$
246Now use (\ref{discformula}).
247\end{proof}
248
249The cokernel expression for $\Phi_{A,p}$ is homogeneous of degree
250$1$ in $V_f$'':
251\begin{proposition}\label{homunit}
252$$\#\Phi_{A,p} = m\cdot \#\coker( X_J\ra \Hom(V_f,\Z)).$$
253\end{proposition}
254\begin{proof}
255Use (\ref{snakecokernel}) and properties of cokernels.
256\end{proof}
257
258{\noindent \bf Step 4. Cancel $m$:}
259\begin{proposition}
260$$\#\Phi_{A,p} 261 = \frac{\deg{\delta}^d}{[X_A:\hat{\delta}^* X_{A^{\vee}}]} 262 \cdot \frac{\#\coker(X_J\ra \Hom(V_f,\Z))^2}{\disc(V_f,V_f)}$$
263\end{proposition}
264\begin{proof}
265Let $c=\#\Phi_{A,p}$.  Using (\ref{homsquare}) and (\ref{homunit}) we obtain
266$$267 c = \frac{c^2}{c} 268 = \frac{m^2\cdot \#\coker(X_J \ra \Hom(V_f,\Z))^2} 269 {\frac{[X_A:\hat{\delta}^* X_{A^{\vee}}]\cdot m^2\cdot \disc(V_f,V_f)} 270 {\deg(\delta)^d}} 271$$
272Cancelling the $m^2$'s gives the formula.
273\end{proof}
274
275{\noindent \bf Step 5. Analyze $\deg(\delta)$:}
276\begin{proposition}
277$$\frac{\deg(\delta)^d}{[X_A:\hat{\delta}^* X_{A^{\vee}}]} 278 = [X_{A^{\vee}} : \delta^* X_{A}]$$
279\end{proposition}
280\begin{proof}
281We have
282\begin{eqnarray*}
283\deg(\delta)^d &=& [X_A:\deg(\delta) X_A] \\
284        &=& [X_A : \hat{\delta}^* X_{A^{\vee}}]
285              \cdot [\hat{\delta}^* X_{A^{\vee}} :
286                      \hat{\delta}^* \delta^* X_{A^{\vee}}] \\
287        &=& [X_A : \hat{\delta}^* X_{A^{\vee}}]
288               \cdot [X_{A^{\vee}} :
289                      \delta^* X_{A^{\vee}}]
290\end{eqnarray*}
291The last step uses that $\hat{\delta}^*$ is injective.
292Now divide both sides by $[X_A : \hat{\delta}^* X_{A^{\vee}}]$.
293\end{proof}
294
295Putting everything together we obtain
296$$\#\Phi_{A,p} 297 = \frac{\#\coker(X_A\ra X_{A^{\vee}}) \cdot \#\coker(X_J\ra \Hom(V,\Z))^2} 298 {\disc(V_f, V_f)}.$$
299Thus to prove the theorem it remains only to show that
300$$\#\coker(X_A\ra X_{A^{\vee}}) = \sqrt{\deg(\delta)}.$$
301{\noindent \bf Step 6. Analyze cokernel of $\delta^*$:}
302\begin{theorem}
303$$\#\coker(X_A\ra X_{A^{\vee}}) = \sqrt{\deg(\delta)}.$$
304\end{theorem}
305\begin{proof}
306The construction of Neron models commutes with unramified base
307extension.  Since $\ord_p(N)=1$ we can pass to an unramified
308extension of $\Q_p$ so that $A$ has universal cover $\Gm^d$.
309Applying the Raynaud-van der Put theorem we obtain
310an exact commutative diagram of rigid analytic spaces:
311$$\begin{matrix} 312 & & & 0 & & 0 & & & \\ 313 & & & \da & & \da & & & \\ 3140 \lra & 0 &\lra & X_A & \lra &X_{A^\vee} & \lra & L & \lra 0 \\ 315 & \da & & \da & & \da & & \da & \\ 3160 \lra & \Lambda & \lra& \Gm^d & \lra & \Gm^d & \lra & 0 & \lra 0 \\ 317 & \da & & \da & & \da & & \da & \\ 3180 \lra & K & \lra& A^{\vee}& \lra & A & \lra & 0 & \lra 0 \\ 319 & & & \da & & \da & & & \\ 320 & & & 0 & & 0 & & & \\ 321 322\end{matrix}$$
323where all maps are induced by the map $\delta:A^{\vee}\ra A$, so
324$K=\Ker(\delta)$ and $L=\coker(X_A\ra X_{A^{\vee}})$.
325Applying the snake lemma, with the connecting map going from $K$ to $L$,
326we obtain the exact sequence
327$$0 \ra \Lambda \ra K \ra L \ra 0.$$
328
329Now compute the $1$-motive dual, i.e., $M'=\Hom(M,\Gm)$, of the above exact diagram  to obtain:
330$$\begin{matrix} 3310 \lla & 0 & \lla& \Gm^d & \lla & \Gm^d & \lla & L' & \lla 0 \\ 332 & \ua & & \ua & & \ua & & \ua & \\ 3330 \lla &\Lambda' &\lla & X_{A^\vee}& \lla &X_{A} & \lla & 0 & \lla 0 \\ 334 & \ua & & \ua & & \ua & & \ua & \\ 335 & K' & \lla & 0 & \lla & 0 & \lla & 0 & \\ 336\end{matrix}$$
337Here $A'=A=0$, $(\Gm^d)'=X_A$ or $X_{A^{\vee}}$.
338Note that dualizing is not in general exact, though it is exact on the $X_A$ and $\Gm^d$ parts.
339
340Adding in the cokernels at the top gives:
341$$\begin{matrix} 342 & & & 0 & & 0 & & & \\ 343 & & & \ua & & \ua & & & \\ 344 & 0 & \lla& A & \lla & A^{\vee} & \lla & K' & \lla 0 \\ 345 & \ua & & \ua & & \ua & & \ua & \\ 3460 \lla & 0 & \lla& \Gm^d & \lla & \Gm^d & \lla & L' & \lla 0 \\ 347 & \ua & & \ua & & \ua & & \ua & \\ 3480 \lla &\Lambda' &\lla & X_{A^\vee}& \lla &X_{A} & \lla & 0 & \lla 0 \\ 349 & \ua & & \ua & & \ua & & & \\ 350 & K' & \lla & 0 & \lla & 0 & & & \\ 351\end{matrix}$$
352All of the maps in this diagram are induced by
353$$\delta^{\vee}:A^{\vee} \ra A.$$
354But $\delta$ is induced by the canonical polarization of the Jacobian $J_0(N)$,
355so $\delta^{\vee}=\delta$.  This is the same diagram as before, but viewed
356from a different point of view!  Thus:
357\begin{eqnarray*}
358  K' &=& K\\
359  L' &=& \Lambda\\
360  \Lambda' &=& L\\
361\end{eqnarray*}
362If $G$ is a finite group scheme then $\#G = \#G'$.
363Thus the exact sequence $0\ra \Lambda \ra K \ra L \ra 0$ together with
364the above equalities yields
365$$\#K = \# L \cdot \# \Lambda = \# L \cdot \# L' = (\#L)^2.$$
366This completes the proof.
367\end{proof}
368
369\begin{question}
370In the proof we determined the structure of
371$\coker(X_A\ra X_{A^{\vee}})$
372in terms of the abelian group $\Ker(\delta)$.  It is possible
373to compute $\Ker(\delta)$ explicitely hence it also possible
374to compute $\coker(X_A\ra X_{A^{\vee}})$.  Is
375it possible to somehow use this to obtain the {\em structure}
376of $\Phi_{A,p}$, instead of just the order?
377\end{question}
378
379\section{Examples}
380[a couple of explicit worked examples illustrating various principles]
381
382
383\section{Tables}
384[table of all $c_p$ for level $\leq 100$.]
385
386[table of supringsly LARGE $c_p$.]
387
388[table of analytic orders of $\Sha$ for all $A_f$ of rank
389$0$ and prime level $\leq 1500$.]
390The table will essentially look like:
391\begin{center}
392\begin{tabular}{|l|c|c|c|c|}\hline
393  $f$ & $\odd(L(A_f,1)/\Omega(A_f))$ & $\numer((N-1)/12)$ & $\deg(\delta)$ & $c_p$ \\\hline\hline
394{\bf 389E20} & 	$5^2/97$ & $97$     & $5^2\cdot 2^?$   &    $97$\\
395{\bf 433D16}&	$7^2/3^2$ & $2^2\cdot3^2$  &  $3^2\cdot 7^2\cdot 37^2\cdot 2^?$ &   \\
396{\bf 563E31}&	$13^2/281$& $281$ &  $13^2\cdot 2^?$  &                          \\
397{\bf 571D2}&	$3^2/1$  & $5\cdot 19$   & $2^4\cdot 3^4\cdot 127^2$ &          $1$\\
398{\bf 709C30}&	$11^2/59$ & $59$       & $11^2\cdot 2^?$ &  \\
399{\bf 997H42}&		$3^4/83$  & $83$ & & \\
400{\bf 1061D46}&		$151^2/(5\cdot53)$ 	&	$5\cdot53$ &  $61^2\cdot 151^2\cdot 179^2\cdot 2^?$ & \\
401{\bf 1091C62}&		$7^2/(5\cdot109)$	&	$5\cdot109$&            $2^?$ & \\
402{\bf 1171D53}&		$11^2/(3\cdot5\cdot13)$	&	$3\cdot5\cdot13$& & \\
403{\bf 1283C62}&		$5^2/641$	&	$641$& & \\
404{\bf 1429B64}&		$5^2/(7\cdot17)$	&	$7\cdot17$& & \\
405{\bf 1481C71}&		$13^2/(5\cdot37)$	&	$2\cdot5\cdot37$& & \\
406{\bf 1483D67}&		$3^2\cdot5^2/(13\cdot19)$	&	$13\cdot19$& &\\\hline
407\end{tabular}
408\end{center}
409\comment{
410{\bf 389E20} & 	$\frac{5^2}{97}$ & $97$     & $5^2\cdot 2^?$   &    $97$\\
411{\bf 433D16}&	$\frac{7^2}{3^2}$ & $2^2\cdot3^2$  &  $3^2\cdot 7^2\cdot 37^2\cdot 2^?$ &   \\
412{\bf 563E31}&	$\frac{13^2}{281}$& $281$ &  $13^2\cdot 2^?$  &                          \\
413{\bf 571D2}&	$\frac{3^2}{1}$  & $5\cdot 19$   & $2^4\cdot 3^4\cdot 127^2$ &          $1$\\
414{\bf 709C30}&	$\frac{11^2}{59}$ & $59$       & $11^2\cdot 2^?$ &  \\
415{\bf 997H42}&		$\frac{3^4}{83}$  & $83$ & & \\
416{\bf 1061D46}&		$\frac{151^2}{5\cdot53}$ 	&	$5\cdot53$ &  $61^2\cdot 151^2\cdot 179^2\cdot 2^?$ & \\
417{\bf 1091C62}&		$\frac{7^2}{5\cdot109}$	&	$5\cdot109$&            $2^?$ & \\
418{\bf 1171D53}&		$\frac{11^2}{3\cdot5\cdot13}$	&	$3\cdot5\cdot13$& & \\
419{\bf 1283C62}&		$\frac{5^2}{641}$	&	$641$& & \\
420{\bf 1429B64}&		$\frac{5^2}{7\cdot17}$	&	$7\cdot17$& & \\
421{\bf 1481C71}&		$\frac{13^2}{5\cdot37}$	&	$2\cdot5\cdot37$& & \\
422{\bf 1483D67}&		$\frac{3^2\cdot5^2}{13\cdot19}$	&	$13\cdot19$& &\\\hline
423}
424
425
426\begin{thebibliography}{HHHHHHH}
427\bibitem[K]{kohel} D. Kohel, {\em Hecke module structure of quaternions},
428preprint, (1998).
429
430\bibitem[M]{mestre} J.F. Mestre,
431   {\em La m\'{e}thode des graphs. Exemples et applications},
432Taniguchi Symp., Proceedings of the international
433conference on class numbers and fundamental units of
434algebraic number fields (Katata, 1986), 217--242, Nagoya Univ., Nagoya, (1986).
435
436\bibitem[S]{stein} W.A. Stein,
437{\em The Modular Polarization of a Quotient of $J_0(N)$},
438in preparation, (1999).
439
440\end{thebibliography} \normalsize\vspace*{1 cm}
441
442\end{document}
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