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% computing_cp.tex
\documentclass[11pt]{article}
\include{macros}
\title{Component Groups of Optimal Quotients of $J_0(N)$}
\begin{document}
\maketitle
\begin{abstract}
\end{abstract}
\section{Introduction}
\comment{
What are the simplest naturally occuring abelian varieties over $\Q$? 
In my opinion they are the ``optimal'' quotients of the Jacobian $J_0(N)$
of the modular curve $X_0(N)=\overline{\sltwoz\backslash\h}$.
Let $f\in S_2(\Gamma_0(N))$ be a newform.  Then Shimura associated to 
$f$ a quotient $A_f$ of $J_0(N)$ having all of the right naturality
properties.  These quotients are optimal in the sense that the kernel
inside $J_0(N)$ is itself an abelian variety. 
The Shimura-Taniyama conjecture, mostly proved by recent work
of Wiles and Taylor-Wiles, asserts that every elliptic curve over
$\Q$ is isogeneous to some $A_f$.  Thus understanding the abelian varieties
$A_f$ is paramount importance!

Neron associated to any abelian variety a model of the integer ring
which is an abelian scheme and whose points are just what one guesses
by looking at the points over the base field.  Let $\cA_f$ be the
Neron model of $A_f$.  Suppose $p$ is a prime which divides the level
$N$, so that $\cA_f/\F_p$ isn't an abelian variety, and hence has no 
right to be connected. The group of components 
$\Phi_{A_f,p}$ is of important in many ways.  For example,
according to the Birch and Swinnerton-Dyer conjecture,
the order of its rational subgroup appears in the numerator 
of a certain rational number associated to the $L$-function 
of $A_f$.
}

Let $f\in S_2(\Gamma_0(N))$ be a newform and $A_f$ the corresponding
optimal quotient of $J_0(N)$.  
Let $\Phi_{A_f,p}$ be the special fiber of the Neron model of $A_f$ at $p$.
The integer $\#\Phi_{A_f,p}$ is of great arithmetic interest. For example,
it appears in the Birch and Swinnerton-Dyer conjecture (I lie: actually the
cardinality of the rational component group appears there).  

\begin{center}
How can we compute $\#\Phi_{A_f,p}$?
\end{center}

When $A_f=E$ is an elliptic curve the Riemann-Roch theorem gives us
a powerful tool: we have a simple explicit Weierstrass equation which defines $E$.  
Using this Tate provided an efficient algorithm for computing the order
of $\Phi_{E,p}$.  I doubt that there is an analogue of the Weierstrass 
equation when $\dim A_f > 1$!

Ribet, in a letter to Mestre, considered the case in which $A_f=E$ 
is an elliptic curve which has multiplicative reduction at $p$, 
i.e., $\ord_p(N)=1$.  Using Grothendieck's monodromy pairing 
on character groups of certain tori he was able to provide a 
``Weierstrass-free'' formula for $\Phi_{E,p}$.
In this paper we generalize Ribet's method to $A_f$ of arbitrary
dimension and thus obtain an explicit formula for $\#\Phi_{A_f,p}$.

The formula is:
$$\#\Phi_{A_f,p} =
   \sqrt{\deg(\delta)}
     \cdot \frac{\# \coker (X \ra \Hom(X[\p],\Z))^2}
                {\disc(X[\p] \cross X[\p] \ra \Z)}$$
where 
\begin{itemize}
\item $\delta:A_f^{\vee}\ra A_f$ is the modular polarization.
\item $X$ is the character group of the torus associated to $J_0(N)/\Fp$.
\item $\p_f=\Ann_{\T}(f)$, where $\T$ is the Hecke algebra.
\item $X\cross X\ra \Z$ is the Monodromy pairing.
\end{itemize}
Each of these quantities can be explicitely computed.  See
\cite{kohel}, \cite{mestre}, and \cite{stein}.

\section{The Monodromy Pairing}
[to be written -- will contain the basic facts]

\section{Ribet's Formulas}
[to be written -- will contain Ribet's formulas as motivation.]

\section{General Formula}
In this section we prove our main theorem.

Let $f\in S_2(\Gamma_0(N))$ be a newform and suppose $p|N$ but $p\nmid M=N/p$. 
The Hecke algebra $\T$ acts on the fundamental group $X_J$ of $J_0(N)$.
The prime ideal 
           $$\p_f=\Ann_\T(f) \subset \T$$
cuts out a submodule 
        $$V_f=X_J[\p_f] \subset X_J$$
of $X_J$.
\begin{theorem}
$$\#\Phi_{A_f,p} =
   \sqrt{\deg(\delta)}
     \cdot \frac{\# \coker (X_J \ra \Hom(V_f,\Z))^2}
                {\disc(V_f, V_f)}$$
\end{theorem}

\vspace{.3in}
We now begin the proof.
\vspace{.3in}

{\noindent \bf Fundamental diagrams:}
\begin{center}
\begin{picture}(310,150)
\put(0,110){$A^{\vee}$}
\put(70,110){$J$}
\put(70,40){$A$}
\put(75,100){\vector(0,-1){45}}
\put(75,63){\vector(0,-1){3}}
\put(13,105){\vector(1,-1){55}}
\qbezier(65,45)(10,60)(5,103)\put(5,103){\vector(0,1){1}}
\put(18,113){\vector(1,0){48}}
\qbezier(18,113)(14,115)(18,117)

\put(40,117){$\pi^{\vee}$}
\put(40,80){$\delta$}
\put(20,50){$\hat{\delta}$}
\put(80,80){$\pi$}


\put(230,110){$X_A$}
\put(300,110){$X_J$}
\put(300,40){$X_{A^{\vee}}$}
\put(305,100){\vector(0,-1){45}}
\put(305,63){\vector(0,-1){3}}
\put(243,105){\vector(1,-1){55}}
\qbezier(290,45)(240,60)(237,103)\put(237,103){\vector(0,1){1}}
\put(248,113){\vector(1,0){48}}
\qbezier(248,113)(244,115)(248,117)

\put(270,117){$\pi^*$}
\put(270,80){$\delta^*$}
\put(250,50){$\hat{\delta}^*$}
\put(310,80){$\pi_*$}
\end{picture}
\end{center}

{\noindent \bf Monodromy pairings:}

$$\langle \quad , \quad \rangle_J : X_J \cross X_J \ra \Z$$
$$\langle \quad , \quad \rangle_A : X_A^{\vee} \cross X_A \ra \Z$$
\vspace{.4in}

{\noindent \bf Component groups:}
$$0\ra X_J \ra \Hom(X_J,\Z)\ra \Phi_{J,p} \ra 0$$
$$0\ra X_{A^{\vee}} \ra \Hom(X_A,\Z)\ra \Phi_{A,p} \ra 0$$
\vspace{.4in}

{\noindent \bf Step 1. Formulas:} 
\begin{lemma}
Let $x\in X_{A^{\vee}}$, $y\in X_A$.  Then
$$\langle \pi^*\hat{\delta}^* x, \pi^* y\rangle_J = 
      \deg(\delta) \langle x, y \rangle_A.$$
\end{lemma}
\begin{proof}
This is a direct computation, using the functorial properties of
the objects involved.
\begin{eqnarray*}
\langle \pi^*\hat{\delta}^* x, \pi^* y\rangle_J &=& 
            \langle \pi_*\pi^*\hat{\delta}^* x, y\rangle_A \\
        &=& \langle \delta^* \hat{\delta}^* x, y\rangle_A \\
        &=& \langle (\hat{\delta} \circ \delta)^* x, y\rangle_A\\
        &=& \deg(\delta) \langle x, y \rangle_A
\end{eqnarray*}
\end{proof}

\begin{proposition}
$$\#\Phi_{A,p} = \frac{\disc(\pi^*\hat{\delta}^* X_{A^{\vee}}, \pi^* X_A)}{\deg(\delta)^d}.$$
\end{proposition}
\begin{proof}
By the lemma we have
$$\disc(\pi^*\hat{\delta}^* X_{A^{\vee}}, \pi^* X_A) = \deg(\delta)^d
        \cdot \disc(X_{A^{\vee}},X_A).$$
Now use that 
$$\#\Phi_{A,p}=\#\coker(X_{A^{\vee}}\ra X_A)= \disc(X_{A^{\vee}},X_A).$$
\end{proof}

\begin{proposition}\label{discformula}
$$\#\Phi_{A,p} = \frac{[X_A:\hat{\delta}^* X_{A^\vee}]\cdot
                            \disc(\pi^* X_A,\pi^* X_A)}
                  {\deg(\delta)^d}.$$
\end{proposition}
\begin{proof}
Using the formula from the previous proposition we have
\begin{eqnarray*}
\#\Phi_{A,p} &=& \frac{\disc(\pi^*\hat{\delta}^* X_{A^{\vee}}, \pi^* X_A)}
                      {\deg(\delta)^d} \\
            &=& \frac{[\pi^* X_A : \pi^*\hat{\delta}^*X_{A^{\vee}}]\cdot
                            \disc (\pi^* X_A, \pi^* X_A)}
                     {\deg(\delta)^d}
\end{eqnarray*}
Note that $\pi^*$ is an injection so it does not change
the index:
$$[\pi^*X_A : \pi^*\hat{\delta}^* X_{A^{\vee}}]
     = [X_A:\hat{\delta}^* X_{A^\vee}].$$
\end{proof}

{\noindent \bf Step 2. Identify $\Phi_{A,p}$ inside of $X_J$:}

\begin{proposition}\label{snakecokernel}
$$\Phi_{A,p} \isom \coker(X_J \ra \Hom(\pi^* X_A, \Z))$$
\end{proposition}
\begin{proof}
Let $\tilde{X}_J$ denote the image of $X_J$ in $\Hom(\pi^*X_A,\Z)$. 
We have the following diagram with exact rows and columns:
$$
\begin{matrix}
   &    &            &      &                  &      & 0 & \\
   &    &            &      &                  &      & \da & \\
   &    &            &      &     0             & \lra & K & \\
   &    &            &      &     \da             &      & \da & \\
  0&\lra& \tilde{X}_J& \lra & \Hom(\pi^*X_A,\Z) & \lra & W & \lra 0 \\
   &    &  \da \isom     &      &   \da \isom            &   &  \da & \\
  0&\lra& X_{A^{\vee}}& \lra & \Hom(X_A,\Z) & \lra & \Phi_{A,p} & \lra 0 \\
   &    &   \da         &      &     \da             &      & \da & \\
   &    &    0        &  \lra   &     0             & \lra & L &  \\
   &    &            &      &                  &      & \da & \\
   &    &            &      &                  &      & 0 & \\
\end{matrix}
$$
The snake lemma implies that the sequence
$$0 \ra K \ra 0 \ra 0 \ra L \ra 0$$
is exact, 
hence $K=L=0$ and so $W= \Phi_{A,p}$.
\end{proof}

{\noindent \bf Step 3. Relate $\pi^*X_A$ to $V_f$:}
Let 
$$V_f = X_J[\p_f] = \bigcap_{t\in\Ann_{\T}(f)} \ker(t) \subset X_J.$$
By multiplicity one and $\T$-invariance of $\pi^*$,
$$\pi^* X_A \subset V_f,$$
and the index is finite.  Let 
   $$ m = [V_f : \pi^* X_A].$$

The discriminant expression of $\Phi_{A,p}$ is ``homogeneous of degree
$2$ in $V_f$'':
\begin{proposition}\label{homsquare}
$$\#\Phi_{A,p} = \frac{[X_A:\hat{\delta}^* X_{A^\vee}]\cdot m^2 \cdot 
                            \disc(V_f, V_f)}
                  {\deg(\delta)^d}.$$
\end{proposition}
\begin{proof}
Note that
$$\disc(\pi^* X_A, \pi^* X_A) = m^2 \disc(V_f,V_f).$$
Now use (\ref{discformula}).
\end{proof}

The cokernel expression for $\Phi_{A,p}$ is ``homogeneous of degree
$1$ in $V_f$'':
\begin{proposition}\label{homunit}
$$\#\Phi_{A,p} = m\cdot \#\coker( X_J\ra \Hom(V_f,\Z)).$$
\end{proposition}
\begin{proof}
Use (\ref{snakecokernel}) and properties of cokernels.
\end{proof}

{\noindent \bf Step 4. Cancel $m$:}
\begin{proposition}
$$\#\Phi_{A,p} 
   = \frac{\deg{\delta}^d}{[X_A:\hat{\delta}^* X_{A^{\vee}}]}
    \cdot \frac{\#\coker(X_J\ra \Hom(V_f,\Z))^2}{\disc(V_f,V_f)}$$
\end{proposition}
\begin{proof}
Let $c=\#\Phi_{A,p}$.  Using (\ref{homsquare}) and (\ref{homunit}) we obtain
$$
 c = \frac{c^2}{c} 
   = \frac{m^2\cdot \#\coker(X_J \ra \Hom(V_f,\Z))^2}
          {\frac{[X_A:\hat{\delta}^* X_{A^{\vee}}]\cdot m^2\cdot \disc(V_f,V_f)}
                {\deg(\delta)^d}}
$$
Cancelling the $m^2$'s gives the formula.
\end{proof}

{\noindent \bf Step 5. Analyze $\deg(\delta)$:}
\begin{proposition}
$$\frac{\deg(\delta)^d}{[X_A:\hat{\delta}^* X_{A^{\vee}}]}
   = [X_{A^{\vee}} : \delta^* X_{A}]$$
\end{proposition}
\begin{proof}
We have
\begin{eqnarray*}
\deg(\delta)^d &=& [X_A:\deg(\delta) X_A] \\
        &=& [X_A : \hat{\delta}^* X_{A^{\vee}}]
              \cdot [\hat{\delta}^* X_{A^{\vee}} : 
                      \hat{\delta}^* \delta^* X_{A^{\vee}}] \\
        &=& [X_A : \hat{\delta}^* X_{A^{\vee}}]
               \cdot [X_{A^{\vee}} : 
                      \delta^* X_{A^{\vee}}]
\end{eqnarray*}
The last step uses that $\hat{\delta}^*$ is injective.
Now divide both sides by $ [X_A : \hat{\delta}^* X_{A^{\vee}}]$.
\end{proof}

Putting everything together we obtain
$$\#\Phi_{A,p} 
 = \frac{\#\coker(X_A\ra X_{A^{\vee}}) \cdot \#\coker(X_J\ra \Hom(V,\Z))^2}
       {\disc(V_f, V_f)}.$$
Thus to prove the theorem it remains only to show that 
$$\#\coker(X_A\ra X_{A^{\vee}}) = \sqrt{\deg(\delta)}.$$
{\noindent \bf Step 6. Analyze cokernel of $\delta^*$:}
\begin{theorem}
$$\#\coker(X_A\ra X_{A^{\vee}}) = \sqrt{\deg(\delta)}.$$
\end{theorem}
\begin{proof}
The construction of Neron models commutes with unramified base
extension.  Since $\ord_p(N)=1$ we can pass to an unramified
extension of $\Q_p$ so that $A$ has universal cover $\Gm^d$. 
Applying the Raynaud-van der Put theorem we obtain
an exact commutative diagram of rigid analytic spaces:
$$\begin{matrix}
        &         &     &   0     &           &     0     &          &      &          \\
        &         &     &  \da    &           &    \da    &          &      &          \\
0  \lra &     0   &\lra &  X_A    &    \lra   &X_{A^\vee} & \lra     &  L   &  \lra 0  \\
        &   \da   &     &   \da   &           &   \da     &          &  \da &          \\
0  \lra & \Lambda & \lra&  \Gm^d  & \lra      & \Gm^d     & \lra     &   0  & \lra 0   \\
        & \da     &     &   \da   &           &  \da      &          &  \da &          \\
0  \lra &   K     & \lra&  A^{\vee}& \lra     & A         & \lra     &   0  & \lra 0   \\
        &         &     &   \da   &           & \da       &          &      &          \\
        &         &     &    0    &           &     0     &          &      &          \\

\end{matrix}$$
where all maps are induced by the map $\delta:A^{\vee}\ra A$, so
$K=\Ker(\delta)$ and $L=\coker(X_A\ra X_{A^{\vee}})$.
Applying the snake lemma, with the connecting map going from $K$ to $L$, 
we obtain the exact sequence
$$0 \ra \Lambda \ra K \ra L \ra 0.$$

Now compute the $1$-motive dual, i.e., $M'=\Hom(M,\Gm)$, of the above exact diagram  to obtain:
$$\begin{matrix}
0  \lla &    0    & \lla&  \Gm^d  & \lla      & \Gm^d     & \lla     &   L'  & \lla 0   \\
        & \ua     &     &   \ua   &           &  \ua      &          &  \ua &          \\
0  \lla &\Lambda' &\lla & X_{A^\vee}&    \lla   &X_{A}    & \lla     &  0   &  \lla 0  \\
        &  \ua    &     &   \ua   &           & \ua       &          &  \ua &          \\   
       &  K'     & \lla  &    0    &  \lla   &     0      & \lla       &   0   &          \\
\end{matrix}$$
Here $A'=A=0$, $(\Gm^d)'=X_A$ or $X_{A^{\vee}}$.
Note that dualizing is not in general exact, though it is exact on the $X_A$ and $\Gm^d$ parts.

Adding in the cokernels at the top gives:
$$\begin{matrix}
        &         &     &   0     &           &     0     &          &      &          \\
        &         &     &  \ua    &           &    \ua    &          &      &          \\
        &   0    & \lla&  A      & \lla    &   A^{\vee} & \lla     &   K'  & \lla 0   \\
        &   \ua   &     &   \ua   &           &   \ua     &          &  \ua &          \\
0  \lla &    0    & \lla&  \Gm^d  & \lla      & \Gm^d     & \lla     &   L'  & \lla 0   \\
        & \ua     &     &   \ua   &           &  \ua      &          &  \ua &          \\
0  \lla &\Lambda' &\lla & X_{A^\vee}&    \lla   &X_{A}    & \lla     &  0   &  \lla 0  \\
        &  \ua    &     &   \ua   &           & \ua       &          &    &          \\   
       &  K'     & \lla  &    0    &  \lla   &     0      &          &      &          \\
\end{matrix}$$
All of the maps in this diagram are induced by 
$$\delta^{\vee}:A^{\vee} \ra A.$$
But $\delta$ is induced by the canonical polarization of the Jacobian $J_0(N)$,
so $\delta^{\vee}=\delta$.  This is the same diagram as before, but viewed
from a different point of view!  Thus:
\begin{eqnarray*}
  K' &=& K\\
  L' &=& \Lambda\\
  \Lambda' &=& L\\
\end{eqnarray*}
If $G$ is a finite group scheme then $\#G = \#G'$. 
Thus the exact sequence $0\ra \Lambda \ra K \ra L \ra 0$ together with
the above equalities yields
$$\#K = \# L \cdot \# \Lambda = \# L \cdot \# L' = (\#L)^2.$$
This completes the proof.
\end{proof}

\begin{question}
In the proof we determined the structure of 
$\coker(X_A\ra X_{A^{\vee}})$
in terms of the abelian group $\Ker(\delta)$.  It is possible
to compute $\Ker(\delta)$ explicitely hence it also possible
to compute $\coker(X_A\ra X_{A^{\vee}})$.  Is
it possible to somehow use this to obtain the {\em structure}
of $\Phi_{A,p}$, instead of just the order?
\end{question}

\section{Examples}
[a couple of explicit worked examples illustrating various principles]


\section{Tables}
[table of all $c_p$ for level $\leq 100$.]

[table of supringsly LARGE $c_p$.]

[table of analytic orders of $\Sha$ for all $A_f$ of rank
$0$ and prime level $\leq 1500$.]
The table will essentially look like:
\begin{center}
\begin{tabular}{|l|c|c|c|c|}\hline
  $f$ & $\odd(L(A_f,1)/\Omega(A_f))$ & $\numer((N-1)/12)$ & $\deg(\delta)$ & $c_p$ \\\hline\hline
{\bf 389E20} & 	$5^2/97$ & $97$     & $5^2\cdot 2^?$   &    $97$\\
{\bf 433D16}&	$7^2/3^2$ & $2^2\cdot3^2$  &  $3^2\cdot 7^2\cdot 37^2\cdot 2^?$ &   \\
{\bf 563E31}&	$13^2/281$& $281$ &  $13^2\cdot 2^?$  &                          \\
{\bf 571D2}&	$3^2/1$  & $5\cdot 19$   & $2^4\cdot 3^4\cdot 127^2$ &          $1$\\
{\bf 709C30}&	$11^2/59$ & $59$       & $11^2\cdot 2^?$ &  \\
{\bf 997H42}&		$3^4/83$  & $83$ & & \\               
{\bf 1061D46}&		$151^2/(5\cdot53)$ 	&	$5\cdot53$ &  $61^2\cdot 151^2\cdot 179^2\cdot 2^?$ & \\
{\bf 1091C62}&		$7^2/(5\cdot109)$	&	$5\cdot109$&            $2^?$ & \\
{\bf 1171D53}&		$11^2/(3\cdot5\cdot13)$	&	$3\cdot5\cdot13$& & \\
{\bf 1283C62}&		$5^2/641$	&	$641$& & \\
{\bf 1429B64}&		$5^2/(7\cdot17)$	&	$7\cdot17$& & \\
{\bf 1481C71}&		$13^2/(5\cdot37)$	&	$2\cdot5\cdot37$& & \\
{\bf 1483D67}&		$3^2\cdot5^2/(13\cdot19)$	&	$13\cdot19$& &\\\hline
\end{tabular}
\end{center}
\comment{
{\bf 389E20} & 	$\frac{5^2}{97}$ & $97$     & $5^2\cdot 2^?$   &    $97$\\
{\bf 433D16}&	$\frac{7^2}{3^2}$ & $2^2\cdot3^2$  &  $3^2\cdot 7^2\cdot 37^2\cdot 2^?$ &   \\
{\bf 563E31}&	$\frac{13^2}{281}$& $281$ &  $13^2\cdot 2^?$  &                          \\
{\bf 571D2}&	$\frac{3^2}{1}$  & $5\cdot 19$   & $2^4\cdot 3^4\cdot 127^2$ &          $1$\\
{\bf 709C30}&	$\frac{11^2}{59}$ & $59$       & $11^2\cdot 2^?$ &  \\
{\bf 997H42}&		$\frac{3^4}{83}$  & $83$ & & \\               
{\bf 1061D46}&		$\frac{151^2}{5\cdot53}$ 	&	$5\cdot53$ &  $61^2\cdot 151^2\cdot 179^2\cdot 2^?$ & \\
{\bf 1091C62}&		$\frac{7^2}{5\cdot109}$	&	$5\cdot109$&            $2^?$ & \\
{\bf 1171D53}&		$\frac{11^2}{3\cdot5\cdot13}$	&	$3\cdot5\cdot13$& & \\
{\bf 1283C62}&		$\frac{5^2}{641}$	&	$641$& & \\
{\bf 1429B64}&		$\frac{5^2}{7\cdot17}$	&	$7\cdot17$& & \\
{\bf 1481C71}&		$\frac{13^2}{5\cdot37}$	&	$2\cdot5\cdot37$& & \\
{\bf 1483D67}&		$\frac{3^2\cdot5^2}{13\cdot19}$	&	$13\cdot19$& &\\\hline
}


\begin{thebibliography}{HHHHHHH}         
\bibitem[K]{kohel} D. Kohel, {\em Hecke module structure of quaternions},
preprint, (1998).

\bibitem[M]{mestre} J.F. Mestre,
   {\em La m\'{e}thode des graphs. Exemples et applications}, 
Taniguchi Symp., Proceedings of the international 
conference on class numbers and fundamental units of 
algebraic number fields (Katata, 1986), 217--242, Nagoya Univ., Nagoya, (1986).

\bibitem[S]{stein} W.A. Stein,
{\em The Modular Polarization of a Quotient of $J_0(N)$},
in preparation, (1999).

\end{thebibliography} \normalsize\vspace*{1 cm}

\end{document}