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\par\noindent
\myhead{The BSD Conjecture for $\mathbf{J_0(65)}$}{0.1}

\section*{Introduction}
The Jacobian $J=J_0(65)$ of the modular curve $X_0(65)$ is
a $5$ dimensional abelian variety whose Mordell-Weil group
has rank $1$.   It is isogenous over $\Q$ to a product
$A\cross B \cross C$ where $A$ is the elliptic curve
$y^2 =$, and $B$ is the Jacobian of the curve
$curve$, and $C$ is the Jacobian of
$curve$.
It is the modular Jacobian of lowest level for which
the Birch and Swinnerton-Dyer conjecture (BSD) predicts
that the Shafarevich-Tate group is nontrivial [Verify this!
It might VERY WELL be false!!]

Let $L(J,s)$ be the canonical $L$-series of $J$.  The Birch and
Swinnerton-Dyer conjecture asserts that
$$L'(J,1) \qe \frac{|\Sha|\cdot c_5\cdot c_{13}} {T^2} \cdot R \cdot \Omega_J$$
where
$\Sha$ is the Shafarevich-Tate group, $T$ is the torsion
subgroup of $J(\Q)$, $c_5$ and $c_{13}$ are the number
of components of the special fiber of the N\'{e}ron model of $J$,
$R$ is the regulator and $\Omega_J$ is the measure of $J(\R)$
with respect to a basis of N\'{e}ron differentials.
In this paper we compute each of the quantities
$c_5$, $c_{13}$, $T$, $R$, $L'(J,1)$ and a quantity
which differs from $\Omega_J$ by a Manin constant
which is probably equal to $1$.

The goal of this short note is to compute $|\Sha|$, assuming
the truth of the BSD conjecture.
\tableofcontents

\section{Some equations}
To be concrete, we give equations for the abelian varieties
under consideration, though we will not make use of these
equations.
According to \cite{galbraith} the curve $X_0(65)$ is cut out
by the following equations:
\begin{eqnarray*}
v^2-y^2+2x^2+2z^2&=&0\\
v^2-2y^2+z^2+w^2-2wx&=&0\\
y^2+2yz+2z^2-w^2-x^2&=&0
\end{eqnarray*}
According to \cite{cremona:algs, empirical},
the $3$ optimal new factors $A$, $B$, and $C$ are all Jacobians
of curves:
\begin{eqnarray*}
A &=& \Jac(y^2+xy=x^3-x)\\
B &=& \Jac(y^2=-x^6+10x^5-32x^4+20x^3+40x^2+6x-1\\
C &=& \Jac(y^2+(x^3+1)y = -4x^6+9x^4+7x^3+18x^2-10)
\end{eqnarray*}

\section{Computing the invariants}
A lucky fact is that $J_0(65)$ is {\em new} in the sense that
$X_0(5)$ and $X_0(13)$ have genus $0$.
This makes it reasonable to expect that the following
assumption holds:
\begin{quote}
{\bf Assumption:} The Manin constant is $1$, so that the real
volume $\Omega_J$ is equal to the volume computed using
a basis of N\'{e}ron differentials.
\end{quote}

\subsection{Leading coefficient at $\mathbf{s=1}$}
The canonical $L$-function associated to an abelian variety
is invariant under isogeny.  In particular,
since $J\sim A\cross B\cross C$  we have
$L(J,s) = L(A\cross B\cross C,s) = L(A,s)L(B,s)L(C,s)$.
Since $B$ and $C$ have rank $0$ and $A$ has rank $1$, the
leading coefficient of $L(J,s)$ as $s=1$ is
the derivative $L'(J,s)/1!=L'(J,s)$.
By the product rule we have
\begin{eqnarray*}
L'(J,s) &=& L'(A,s)L(B,s)L(C,s)\\
&&+ L(A,s)L'(B,s)L(C,s)
+ L(A,s)L(B,s)L'(C,s).
\end{eqnarray*}
Using the that $L(A,1)=0$, taking
$L'(A,1)$ from \cite{cremona:algs}, and
computing $L(B,1)$, $L(C,1)$ using \hecke{} \cite{stein:hecke} (500
terms of $q$-expansion), we obtain
\begin{eqnarray*}
L'(J,1) &=& L'(A,1)L(B,1)L(C,1)\\
&\approx & 0.50533434230686\cdot 0.91225087803795579498
\cdot 0.45206783187309768436\\
&\approx& 0.2083995171877410261569391894
\end{eqnarray*}

\subsection{Torsion}
I guess that $|J(\Q)_{\tor}|=168=2^3\cdot3\cdot7$.
The upper bound coming from counting points on $J(\F_p)$
using Hecke operators is $2^3\cdot 3\cdot 7$.  Each
of $A$, $B$, and $C$ possess exactly one nontrivial
rational $2$-torsion point.
We also have $|A(\Q)_{\tor}|=2$,
$|B(\Q)_{\tor}|=6$, and $|C(\Q)_{\tor}=14$,
so $2\cdot 3\cdot 7$ is a lower bound.

After computing all other invariants (see below)
we find that the only possible value for $J(\Q)_{\tor}$ so
that the order of $\Sha$ predicted by BSD is an
integer is $|J(\Q)_{\tor}|=168$.

\subsection{Regulator}
There is an injective map $A\hookrightarrow J$
and the image generates the $J(\Q)$, up to torsion.
{\em I GUESS} that
$\Reg(J) = \Reg(A).$
We find that
$$\Reg(A)\approx 0.3755140986612663218044728765$$
(using \pari{}'s {\tt ellheightmatrix(e,[[1,0]])}, then
double checking in Cremona's tables.)

{\bf WARNING:} I have not proved that $\Reg(J)=\Reg(A)$.

\subsection{Real volume}
The volume with respect to an integral basis at infinity is
$$\Omega_J \approx 93.23516483763867.$$
This was computed using \hecke{}.

\subsection{Tamagawa numbers}
We have
$$\overline{c}_5 = 42,\qquad \overline{c}_{13} = 6.$$
These values were computed both by the character group method
(see below) and via the Edixhoven-Mazur-Raynaud formula.
Using a more careful analysis of character groups,
we find that $c_5=14$ and $c_{13}=6$.   The details are
given below.

\subsubsection{Detailed computation of $\mathbf{c_5}$}
Let $\cD$ be the free abelian group generated by
the supseringular points on the modular curve
$X_0(13)/\F_5$.  The character group $\cX_{5}(J)$
of the toric part of $X_0(65)/\F_5$ is the subgroup
of $\cD$ of elements of degree $0$ (the
degree is the sum of the coefficients on the
canonical basis of superingular points).

\begin{proposition} $c_5 = 14$\end{proposition}
Some data about the character group $\cX_{5}(J)$.

{\bf Monodromy weights:} $w_1=w_2=3, w_3=w_4=w_5=w_6=1$.

{\bf Operators:} The matrix of $T_{5}=\Frob$ on the group
of divisors on the supersingular points:
$$\Frob_5 = \left(\begin{matrix} 0&1&0&0&0&0\\ 1&0&0&0&0&0\\ 0&0&0&1&0&0\\ 0&0&1&0&0&0\\ 0&0&0&0&0&1\\ 0&0&0&0&1&0 \end{matrix}\right)$$

{\bf Presentation of the component group $\mathbf{\Phi_5(J)}$:}
Choose the following basis for the group
$\cX=\cX_{5}(J)$ of divisors of degree $0$:
$$b_1=e_1-e_2, \,b_2=e_1-e_3,\, b_3=e_1-e_4, \, b_4=e_1-e_5,\, b_5 = e_1-e_6.$$
The component group $\Phi$ fits into the sequence
$$0 \lra \cX \lra \Hom(\cX,\Z) \lra \Phi \lra 0.$$
Thus $\Phi$ is the quotient of the group $b_1^*,\ldots, b_5^*$
by the subgroup generated by
$$\left\{\begin{array}{l} 6b_1^*+3b_2^*+3b_3^*+3b_4^*+3b_5^*\\ 3b_1^*+4b_2^*+3b_3^*+3b_4^*+3b_5^*\\ 3b_1^*+3b_2^*+4b_3^*+3b_4^*+3b_5^*\\ 3b_3^*+3b_2^*+3b_3^*+4b_4^*+3b_5^*\\ 3b_3^*+3b_2^*+3b_3^*+3b_4^*+4b_5^* \end{array}\right\}$$
Letting $[x]$ denote a class in the quotient group, we find
$$[b_2^*]=[b_3^*]=[b_4^*]=[b_5^*]=3[b_1^*]$$
and that $42[b_1^*]=0$.

To compute the subgroup fixed by $\Frob=T_p$
we must find the action of $\Frob$ on $\Hom(\cX,\Z)$.
The monodromy pairing $\langle\,,\,\rangle : \cX\cross\cX\ra \Z$ is
$\T$-invariant, and thus we can find the $\T$-module
$\Hom(\cX,\Z)$ inside of $\cX\tensor\Q$.  It is the
set of elements $x\in\cX\tensor\Q$ such that
$\langle x, y \rangle \in\Z$ for all $y\in\cX$.
The dual basis $b_1^*,b_2^*,b_3^*,b_4^*,b_5^*\in\cX\tensor\Q$
is a set of elements so that $b_i^*(b_j)=\delta_{ij}$
(Kronecker's $\delta$).
For convenience, we will work instead in $\cD\tensor\Q$.
If $v\in \cD\tensor\Q$ corresponds to $b_1^*$ then we
have the matrix equation
$$\left(\begin{matrix} 1&-1&0&0&0&0\\ 1&0&-1&0&0&0\\ 1&0&0&-1&0&0\\ 1&0&0&0&-1&0\\ 1&0&0&0&0&-1\\ 1&1&3&3&3&3 \end{matrix}\right) \cdot \left(\begin{matrix} 3&0&0&0&0&0\\ 0&3&0&0&0&0\\ 0&0&1&0&0&0\\ 0&0&0&1&0&0\\ 0&0&0&0&1&0\\ 0&0&0&0&0&1 \end{matrix}\right) \cdot v = \left(\begin{matrix} 1\\0\\0\\0\\0\\0\end{matrix} \right).$$
Letting $a$ denote the first matrix, and $b$ the second diagonal
one, we find that the action of
$\Frob$ on $\Hom(X,\Z)$ with respect to the
basis $b_1^*, b_2^*, b_3^*, b_4^*, b_5^*, \Eis^*$
is given by

$$a\cdot b\cdot \Frob\cdot (a\cdot b)^{-1} =\left(\begin{matrix} -1&0&0&0&0&0\\ -1&0&1&0&0&0\\ -1&1&0&0&0&0\\ -1&0&0&0&1&0\\ -1&0&0&1&0&0\\ 0&0&0&0&0&1 \end{matrix}\right).$$
Therefore
$$\Frob(b_1^*)=-(b_1^*+b_2^*+b_3^*+b_4^*+b_5^*)=-(13b_1^*).$$
Multiplication by $-13$ on a cyclic group of order
$42$ has exactly $14$ fixed points, so $c_5=14$.

\subsubsection{Detailed computation of $\mathbf{c_{13}}$}

\begin{proposition} $c_{13} = 6$\end{proposition}
Some data about the character group
$\cX_{13}(J)$.

{\bf Monodromy weights:} $w_1=w_2=w_3=w_4=w_5=w_6=1$.

{\bf Operators:} The matrix of $T_{13}=\Frob$ on the group
of divisors on the supersingular points:
$$\Frob_{13} = \left(\begin{matrix} 0&1&0&0&0&0\\ 1&0&0&0&0&0\\ 0&0&0&1&0&0\\ 0&0&1&0&0&0\\ 0&0&0&0&0&1\\ 0&0&0&0&1&0 \end{matrix}\right)$$

{\bf Presentation of the group:}
Choose the following basis for the group $\cX=\cX_{13}(J)$ of divisors
of degree $0$:
$$b_1=e_1-e_2, \,b_2=e_1-e_3,\, b_3=e_1-e_4, \, b_4=e_1-e_5,\, b_5 = e_1-e_6.$$
The component group $\Phi$ fits into the sequence
$$0 \lra \cX \lra \Hom(\cX,\Z) \lra \Phi \lra 0.$$
Since the monodromy weights are trivial,
$\Phi$ is the quotient of the group $b_1,\ldots, b_5$
by the subgroup generated by
$$\left\{\begin{array}{l} 2b_1^*+b_2^*+b_3^*+b_4^*+b_5^*,\\ b_1^*+2b_2^*+b_3^*+b_4^*+b_5^*,\\ b_1^*+b_2^*+2b_3^*+b_4^*+b_5^*,\\ b_1^*+b_2^*+b_3^*+2b_4^*+b_5^*,\\ b_1^*+b_2^*+b_3^*+b_4^*+2b_5^*. \end{array}\right\}$$
Backsubstituting, we find
$$[b_1^*]=[b_2^*]n=[b_3^*]=[b_4^*]=[b_5^*]$$
and that $6[b_1^*]=0$.

Just as before we compute
$$a\cdot b \cdot \Frob \cdot (a\cdot b)^{-1} = =\left(\begin{matrix} -1&0&0&0&0&0\\ -1&0&1&0&0&0\\ -1&1&0&0&0&0\\ -1&0&0&0&1&0\\ -1&0&0&1&0&0\\ 0&0&0&0&0&1 \end{matrix}\right).$$
Therefore
$$\Frob(b_1^*)=-(b_1^*+b_2^*+b_3^*+b_4^*+b_5^*)=-(5b_1^*).$$
Since $-5\con 1\pmod{6}$ the action of Frobenius is trivial
so $c_{13}=6$.

\section{Analytic order of the Shafarevich-Tate group}
The BSD conjecture predicts that
$$L'(J,1)\qe \frac{|\Sha|\cdot c_5\cdot c_{13}} {|J(\Q)_{\tor}|^2} \cdot \Reg \cdot \Omega_J$$
Assuming this and solving for $|\Sha|$ we obtain
\begin{eqnarray*}
|\Sha|
&=& \frac{L'(J,1)}{\Reg \cdot \Omega_J}
\cdot \frac{J(\Q)_{\tor}^2}{c_5\cdot c_{13}}\\
&=& \frac{0.2083995171877410261569391894}
{0.3755140986612663218044728765\cdot 93.23516483763867}
\cdot \frac{168^2}{14\cdot 6}\\
&\approx& \frac{1}{168}
\cdot \frac{168^2}{14\cdot 6}\\
&\approx& 2.
\end{eqnarray*}

\bibliography{biblio}

\end{document}