CoCalc Public Fileswww / Tables / 65.tex
Author: William A. Stein
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16\par\noindent
17\myhead{The BSD Conjecture for $\mathbf{J_0(65)}$}{0.1}
18
19
20\section*{Introduction}
21The Jacobian $J=J_0(65)$ of the modular curve $X_0(65)$ is
22a $5$ dimensional abelian variety whose Mordell-Weil group
23has rank $1$.   It is isogenous over $\Q$ to a product
24$A\cross B \cross C$ where $A$ is the elliptic curve
25$y^2 =$, and $B$ is the Jacobian of the curve
26$curve$, and $C$ is the Jacobian of
27$curve$.
28It is the modular Jacobian of lowest level for which
29the Birch and Swinnerton-Dyer conjecture (BSD) predicts
30that the Shafarevich-Tate group is nontrivial [Verify this!
31It might VERY WELL be false!!]
32
33Let $L(J,s)$ be the canonical $L$-series of $J$.  The Birch and
34Swinnerton-Dyer conjecture asserts that
35$$L'(J,1) \qe 36 \frac{|\Sha|\cdot c_5\cdot c_{13}} 37 {T^2} 38 \cdot R \cdot \Omega_J$$
39where
40$\Sha$ is the Shafarevich-Tate group, $T$ is the torsion
41subgroup of $J(\Q)$, $c_5$ and $c_{13}$ are the number
42of components of the special fiber of the N\'{e}ron model of $J$,
43$R$ is the regulator and $\Omega_J$ is the measure of $J(\R)$
44with respect to a basis of N\'{e}ron differentials.
45In this paper we compute each of the quantities
46$c_5$, $c_{13}$, $T$, $R$, $L'(J,1)$ and a quantity
47which differs from $\Omega_J$ by a Manin constant
48which is probably equal to $1$.
49
50
51The goal of this short note is to compute $|\Sha|$, assuming
52the truth of the BSD conjecture.
53\tableofcontents
54
55\section{Some equations}
56To be concrete, we give equations for the abelian varieties
57under consideration, though we will not make use of these
58equations.
59According to \cite{galbraith} the curve $X_0(65)$ is cut out
60by the following equations:
61\begin{eqnarray*}
62v^2-y^2+2x^2+2z^2&=&0\\
63v^2-2y^2+z^2+w^2-2wx&=&0\\
64y^2+2yz+2z^2-w^2-x^2&=&0
65\end{eqnarray*}
66According to \cite{cremona:algs, empirical},
67the $3$ optimal new factors $A$, $B$, and $C$ are all Jacobians
68of curves:
69\begin{eqnarray*}
70 A &=& \Jac(y^2+xy=x^3-x)\\
71 B &=& \Jac(y^2=-x^6+10x^5-32x^4+20x^3+40x^2+6x-1\\
72 C &=& \Jac(y^2+(x^3+1)y = -4x^6+9x^4+7x^3+18x^2-10)
73\end{eqnarray*}
74
75
76\section{Computing the invariants}
77A lucky fact is that $J_0(65)$ is {\em new} in the sense that
78$X_0(5)$ and $X_0(13)$ have genus $0$.
79This makes it reasonable to expect that the following
80assumption holds:
81\begin{quote}
82{\bf Assumption:} The Manin constant is $1$, so that the real
83volume $\Omega_J$ is equal to the volume computed using
84a basis of N\'{e}ron differentials.
85\end{quote}
86
87\subsection{Leading coefficient at $\mathbf{s=1}$}
88The canonical $L$-function associated to an abelian variety
89is invariant under isogeny.  In particular,
90since $J\sim A\cross B\cross C$  we have
91$L(J,s) = L(A\cross B\cross C,s) = L(A,s)L(B,s)L(C,s)$.
92Since $B$ and $C$ have rank $0$ and $A$ has rank $1$, the
93leading coefficient of $L(J,s)$ as $s=1$ is
94the derivative $L'(J,s)/1!=L'(J,s)$.
95By the product rule we have
96\begin{eqnarray*}
97 L'(J,s) &=& L'(A,s)L(B,s)L(C,s)\\
98         &&+ L(A,s)L'(B,s)L(C,s)
99         + L(A,s)L(B,s)L'(C,s).
100\end{eqnarray*}
101Using the that $L(A,1)=0$, taking
102$L'(A,1)$ from \cite{cremona:algs}, and
103computing $L(B,1)$, $L(C,1)$ using \hecke{} \cite{stein:hecke} (500
104terms of $q$-expansion), we obtain
105\begin{eqnarray*}
106L'(J,1) &=& L'(A,1)L(B,1)L(C,1)\\
107        &\approx & 0.50533434230686\cdot 0.91225087803795579498
108             \cdot 0.45206783187309768436\\
109        &\approx& 0.2083995171877410261569391894
110\end{eqnarray*}
111
112\subsection{Torsion}
113I guess that $|J(\Q)_{\tor}|=168=2^3\cdot3\cdot7$.
114The upper bound coming from counting points on $J(\F_p)$
115using Hecke operators is $2^3\cdot 3\cdot 7$.  Each
116of $A$, $B$, and $C$ possess exactly one nontrivial
117rational $2$-torsion point.
118We also have $|A(\Q)_{\tor}|=2$,
119$|B(\Q)_{\tor}|=6$, and $|C(\Q)_{\tor}=14$,
120so $2\cdot 3\cdot 7$ is a lower bound.
121
122After computing all other invariants (see below)
123we find that the only possible value for $J(\Q)_{\tor}$ so
124that the order of $\Sha$ predicted by BSD is an
125integer is $|J(\Q)_{\tor}|=168$.
126
127\subsection{Regulator}
128There is an injective map $A\hookrightarrow J$
129and the image generates the $J(\Q)$, up to torsion.
130{\em I GUESS} that
131$\Reg(J) = \Reg(A).$
132We find that
133  $$\Reg(A)\approx 0.3755140986612663218044728765$$
134(using \pari{}'s {\tt ellheightmatrix(e,[[1,0]])}, then
135double checking in Cremona's tables.)
136
137{\bf WARNING:} I have not proved that $\Reg(J)=\Reg(A)$.
138
139
140\subsection{Real volume}
141The volume with respect to an integral basis at infinity is
142$$\Omega_J \approx 93.23516483763867.$$
143This was computed using \hecke{}.
144
145
146
147\subsection{Tamagawa numbers}
148We have
149$$\overline{c}_5 = 42,\qquad \overline{c}_{13} = 6.$$
150These values were computed both by the character group method
151(see below) and via the Edixhoven-Mazur-Raynaud formula.
152Using a more careful analysis of character groups,
153we find that $c_5=14$ and $c_{13}=6$.   The details are
154given below.
155
156\subsubsection{Detailed computation of $\mathbf{c_5}$}
157Let $\cD$ be the free abelian group generated by
158the supseringular points on the modular curve
159$X_0(13)/\F_5$.  The character group $\cX_{5}(J)$
160of the toric part of $X_0(65)/\F_5$ is the subgroup
161of $\cD$ of elements of degree $0$ (the
162degree is the sum of the coefficients on the
163canonical basis of superingular points).
164
165\begin{proposition} $c_5 = 14$\end{proposition}
166Some data about the character group $\cX_{5}(J)$.
167
168{\bf Monodromy weights:} $w_1=w_2=3, w_3=w_4=w_5=w_6=1$.
169
170{\bf Operators:} The matrix of $T_{5}=\Frob$ on the group
171of divisors on the supersingular points:
172$$\Frob_5 = \left(\begin{matrix} 173 0&1&0&0&0&0\\ 174 1&0&0&0&0&0\\ 175 0&0&0&1&0&0\\ 176 0&0&1&0&0&0\\ 177 0&0&0&0&0&1\\ 178 0&0&0&0&1&0 179\end{matrix}\right)$$
180
181{\bf Presentation of the component group $\mathbf{\Phi_5(J)}$:}
182Choose the following basis for the group
183$\cX=\cX_{5}(J)$ of divisors of degree $0$:
184 $$b_1=e_1-e_2, \,b_2=e_1-e_3,\, b_3=e_1-e_4, \, b_4=e_1-e_5,\, 185 b_5 = e_1-e_6.$$
186The component group $\Phi$ fits into the sequence
187 $$0 \lra \cX \lra \Hom(\cX,\Z) \lra \Phi \lra 0.$$
188Thus $\Phi$ is the quotient of the group $b_1^*,\ldots, b_5^*$
189by the subgroup generated by
190$$\left\{\begin{array}{l} 1916b_1^*+3b_2^*+3b_3^*+3b_4^*+3b_5^*\\ 1923b_1^*+4b_2^*+3b_3^*+3b_4^*+3b_5^*\\ 1933b_1^*+3b_2^*+4b_3^*+3b_4^*+3b_5^*\\ 1943b_3^*+3b_2^*+3b_3^*+4b_4^*+3b_5^*\\ 1953b_3^*+3b_2^*+3b_3^*+3b_4^*+4b_5^* 196\end{array}\right\}$$
197Letting $[x]$ denote a class in the quotient group, we find
198$$[b_2^*]=[b_3^*]=[b_4^*]=[b_5^*]=3[b_1^*]$$
199and that $42[b_1^*]=0$.
200
201To compute the subgroup fixed by $\Frob=T_p$
202we must find the action of $\Frob$ on $\Hom(\cX,\Z)$.
203The monodromy pairing $\langle\,,\,\rangle : \cX\cross\cX\ra \Z$ is
204$\T$-invariant, and thus we can find the $\T$-module
205$\Hom(\cX,\Z)$ inside of $\cX\tensor\Q$.  It is the
206set of elements $x\in\cX\tensor\Q$ such that
207$\langle x, y \rangle \in\Z$ for all $y\in\cX$.
208The dual basis $b_1^*,b_2^*,b_3^*,b_4^*,b_5^*\in\cX\tensor\Q$
209is a set of elements so that $b_i^*(b_j)=\delta_{ij}$
210(Kronecker's $\delta$).
211For convenience, we will work instead in $\cD\tensor\Q$.
212If $v\in \cD\tensor\Q$ corresponds to $b_1^*$ then we
213have the matrix equation
214$$215\left(\begin{matrix} 216 1&-1&0&0&0&0\\ 217 1&0&-1&0&0&0\\ 218 1&0&0&-1&0&0\\ 219 1&0&0&0&-1&0\\ 220 1&0&0&0&0&-1\\ 221 1&1&3&3&3&3 222\end{matrix}\right) 223\cdot 224\left(\begin{matrix} 225 3&0&0&0&0&0\\ 226 0&3&0&0&0&0\\ 227 0&0&1&0&0&0\\ 228 0&0&0&1&0&0\\ 229 0&0&0&0&1&0\\ 230 0&0&0&0&0&1 231\end{matrix}\right) 232\cdot v = 233\left(\begin{matrix} 234 1\\0\\0\\0\\0\\0\end{matrix} 235 \right). 236$$
237Letting $a$ denote the first matrix, and $b$ the second diagonal
238one, we find that the action of
239$\Frob$ on $\Hom(X,\Z)$ with respect to the
240basis $b_1^*, b_2^*, b_3^*, b_4^*, b_5^*, \Eis^*$
241is given by
242
243$$a\cdot b\cdot \Frob\cdot (a\cdot b)^{-1} 244 =\left(\begin{matrix} 245-1&0&0&0&0&0\\ 246-1&0&1&0&0&0\\ 247-1&1&0&0&0&0\\ 248-1&0&0&0&1&0\\ 249-1&0&0&1&0&0\\ 2500&0&0&0&0&1 251\end{matrix}\right).$$
252Therefore
253$$\Frob(b_1^*)=-(b_1^*+b_2^*+b_3^*+b_4^*+b_5^*)=-(13b_1^*).$$
254Multiplication by $-13$ on a cyclic group of order
255$42$ has exactly $14$ fixed points, so $c_5=14$.
256
257\subsubsection{Detailed computation of $\mathbf{c_{13}}$}
258
259\begin{proposition} $c_{13} = 6$\end{proposition}
260Some data about the character group
261$\cX_{13}(J)$.
262
263{\bf Monodromy weights:} $w_1=w_2=w_3=w_4=w_5=w_6=1$.
264
265{\bf Operators:} The matrix of $T_{13}=\Frob$ on the group
266of divisors on the supersingular points:
267$$\Frob_{13} = \left(\begin{matrix} 268 0&1&0&0&0&0\\ 269 1&0&0&0&0&0\\ 270 0&0&0&1&0&0\\ 271 0&0&1&0&0&0\\ 272 0&0&0&0&0&1\\ 273 0&0&0&0&1&0 274\end{matrix}\right)$$
275
276{\bf Presentation of the group:}
277Choose the following basis for the group $\cX=\cX_{13}(J)$ of divisors
278of degree $0$:
279 $$b_1=e_1-e_2, \,b_2=e_1-e_3,\, b_3=e_1-e_4, \, b_4=e_1-e_5,\, 280 b_5 = e_1-e_6.$$
281The component group $\Phi$ fits into the sequence
282 $$0 \lra \cX \lra \Hom(\cX,\Z) \lra \Phi \lra 0.$$
283Since the monodromy weights are trivial,
284$\Phi$ is the quotient of the group $b_1,\ldots, b_5$
285by the subgroup generated by
286$$\left\{\begin{array}{l} 287 2b_1^*+b_2^*+b_3^*+b_4^*+b_5^*,\\ 288 b_1^*+2b_2^*+b_3^*+b_4^*+b_5^*,\\ 289 b_1^*+b_2^*+2b_3^*+b_4^*+b_5^*,\\ 290 b_1^*+b_2^*+b_3^*+2b_4^*+b_5^*,\\ 291 b_1^*+b_2^*+b_3^*+b_4^*+2b_5^*. 292\end{array}\right\}$$
293Backsubstituting, we find
294$$[b_1^*]=[b_2^*]n=[b_3^*]=[b_4^*]=[b_5^*]$$
295and that $6[b_1^*]=0$.
296
297Just as before we compute
298$$299a\cdot b \cdot \Frob \cdot (a\cdot b)^{-1} 300 = 301 =\left(\begin{matrix} 302-1&0&0&0&0&0\\ 303-1&0&1&0&0&0\\ 304-1&1&0&0&0&0\\ 305-1&0&0&0&1&0\\ 306-1&0&0&1&0&0\\ 3070&0&0&0&0&1 308\end{matrix}\right). 309$$
310Therefore
311$$\Frob(b_1^*)=-(b_1^*+b_2^*+b_3^*+b_4^*+b_5^*)=-(5b_1^*).$$
312Since $-5\con 1\pmod{6}$ the action of Frobenius is trivial
313so $c_{13}=6$.
314
315
316
317
318
319
320\section{Analytic order of the Shafarevich-Tate group}
321The BSD conjecture predicts that
322$$L'(J,1)\qe \frac{|\Sha|\cdot c_5\cdot c_{13}} 323 {|J(\Q)_{\tor}|^2} 324 \cdot \Reg \cdot \Omega_J$$
325Assuming this and solving for $|\Sha|$ we obtain
326\begin{eqnarray*}
327  |\Sha|
328    &=& \frac{L'(J,1)}{\Reg \cdot \Omega_J}
329        \cdot \frac{J(\Q)_{\tor}^2}{c_5\cdot c_{13}}\\
330    &=& \frac{0.2083995171877410261569391894}
331             {0.3755140986612663218044728765\cdot 93.23516483763867}
332        \cdot \frac{168^2}{14\cdot 6}\\
333    &\approx& \frac{1}{168}
334        \cdot \frac{168^2}{14\cdot 6}\\
335    &\approx& 2.
336\end{eqnarray*}
337
338\bibliography{biblio}
339
340\end{document}
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