Sharedwww / Tables / 65.texOpen in CoCalc
Author: William A. Stein
1
% 65.tex
2
% GOALS:
3
4
\documentclass{article}
5
\textwidth=1.08\textwidth
6
\hoffset=-.17in
7
\bibliographystyle{amsalpha}
8
\include{macros}
9
%\usepackage[all]{xy}
10
11
\begin{document}
12
%\CompileMatrices % xymatrix
13
%\UseTips % xymatrix option, use computer modern arrow tips (look better)
14
%\OnlyOutlines
15
16
\par\noindent
17
\myhead{The BSD Conjecture for $\mathbf{J_0(65)}$}{0.1}
18
19
20
\section*{Introduction}
21
The Jacobian $J=J_0(65)$ of the modular curve $X_0(65)$ is
22
a $5$ dimensional abelian variety whose Mordell-Weil group
23
has 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
$curve$.
28
It is the modular Jacobian of lowest level for which
29
the Birch and Swinnerton-Dyer conjecture (BSD) predicts
30
that the Shafarevich-Tate group is nontrivial [Verify this!
31
It might VERY WELL be false!!]
32
33
Let $L(J,s)$ be the canonical $L$-series of $J$. The Birch and
34
Swinnerton-Dyer conjecture asserts that
35
$$L'(J,1) \qe
36
\frac{|\Sha|\cdot c_5\cdot c_{13}}
37
{T^2}
38
\cdot R \cdot \Omega_J$$
39
where
40
$\Sha$ is the Shafarevich-Tate group, $T$ is the torsion
41
subgroup of $J(\Q)$, $c_5$ and $c_{13}$ are the number
42
of 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 $J(\R)$
44
with respect to a basis of N\'{e}ron differentials.
45
In this paper we compute each of the quantities
46
$c_5$, $c_{13}$, $T$, $R$, $L'(J,1)$ and a quantity
47
which differs from $\Omega_J$ by a Manin constant
48
which is probably equal to $1$.
49
50
51
The goal of this short note is to compute $|\Sha|$, assuming
52
the truth of the BSD conjecture.
53
\tableofcontents
54
55
\section{Some equations}
56
To be concrete, we give equations for the abelian varieties
57
under consideration, though we will not make use of these
58
equations.
59
According to \cite{galbraith} the curve $X_0(65)$ is cut out
60
by the following equations:
61
\begin{eqnarray*}
62
v^2-y^2+2x^2+2z^2&=&0\\
63
v^2-2y^2+z^2+w^2-2wx&=&0\\
64
y^2+2yz+2z^2-w^2-x^2&=&0
65
\end{eqnarray*}
66
According to \cite{cremona:algs, empirical},
67
the $3$ optimal new factors $A$, $B$, and $C$ are all Jacobians
68
of 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}
77
A lucky fact is that $J_0(65)$ is {\em new} in the sense that
78
$X_0(5)$ and $X_0(13)$ have genus $0$.
79
This makes it reasonable to expect that the following
80
assumption holds:
81
\begin{quote}
82
{\bf Assumption:} The Manin constant is $1$, so that the real
83
volume $\Omega_J$ is equal to the volume computed using
84
a basis of N\'{e}ron differentials.
85
\end{quote}
86
87
\subsection{Leading coefficient at $\mathbf{s=1}$}
88
The canonical $L$-function associated to an abelian variety
89
is invariant under isogeny. In particular,
90
since $J\sim A\cross B\cross C$ we have
91
$L(J,s) = L(A\cross B\cross C,s) = L(A,s)L(B,s)L(C,s)$.
92
Since $B$ and $C$ have rank $0$ and $A$ has rank $1$, the
93
leading coefficient of $L(J,s)$ as $s=1$ is
94
the derivative $L'(J,s)/1!=L'(J,s)$.
95
By 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*}
101
Using the that $L(A,1)=0$, taking
102
$L'(A,1)$ from \cite{cremona:algs}, and
103
computing $L(B,1)$, $L(C,1)$ using \hecke{} \cite{stein:hecke} (500
104
terms of $q$-expansion), we obtain
105
\begin{eqnarray*}
106
L'(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}
113
I guess that $|J(\Q)_{\tor}|=168=2^3\cdot3\cdot7$.
114
The upper bound coming from counting points on $J(\F_p)$
115
using Hecke operators is $2^3\cdot 3\cdot 7$. Each
116
of $A$, $B$, and $C$ possess exactly one nontrivial
117
rational $2$-torsion point.
118
We also have $|A(\Q)_{\tor}|=2$,
119
$|B(\Q)_{\tor}|=6$, and $|C(\Q)_{\tor}=14$,
120
so $2\cdot 3\cdot 7$ is a lower bound.
121
122
After computing all other invariants (see below)
123
we find that the only possible value for $J(\Q)_{\tor}$ so
124
that the order of $\Sha$ predicted by BSD is an
125
integer is $|J(\Q)_{\tor}|=168$.
126
127
\subsection{Regulator}
128
There is an injective map $A\hookrightarrow J$
129
and the image generates the $J(\Q)$, up to torsion.
130
{\em I GUESS} that
131
$\Reg(J) = \Reg(A).$
132
We find that
133
$$\Reg(A)\approx 0.3755140986612663218044728765$$
134
(using \pari{}'s {\tt ellheightmatrix(e,[[1,0]])}, then
135
double checking in Cremona's tables.)
136
137
{\bf WARNING:} I have not proved that $\Reg(J)=\Reg(A)$.
138
139
140
\subsection{Real volume}
141
The volume with respect to an integral basis at infinity is
142
$$\Omega_J \approx 93.23516483763867.$$
143
This was computed using \hecke{}.
144
145
146
147
\subsection{Tamagawa numbers}
148
We have
149
$$\overline{c}_5 = 42,\qquad \overline{c}_{13} = 6.$$
150
These values were computed both by the character group method
151
(see below) and via the Edixhoven-Mazur-Raynaud formula.
152
Using a more careful analysis of character groups,
153
we find that $c_5=14$ and $c_{13}=6$. The details are
154
given below.
155
156
\subsubsection{Detailed computation of $\mathbf{c_5}$}
157
Let $\cD$ be the free abelian group generated by
158
the supseringular points on the modular curve
159
$X_0(13)/\F_5$. The character group $\cX_{5}(J)$
160
of the toric part of $X_0(65)/\F_5$ is the subgroup
161
of $\cD$ of elements of degree $0$ (the
162
degree is the sum of the coefficients on the
163
canonical basis of superingular points).
164
165
\begin{proposition} $c_5 = 14$\end{proposition}
166
Some 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
171
of 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 $\mathbf{\Phi_5(J)}$:}
182
Choose the following basis for the group
183
$\cX=\cX_{5}(J)$ of divisors of degree $0$:
184
$$b_1=e_1-e_2, \,b_2=e_1-e_3,\, b_3=e_1-e_4, \, b_4=e_1-e_5,\,
185
b_5 = e_1-e_6.$$
186
The component group $\Phi$ fits into the sequence
187
$$0 \lra \cX \lra \Hom(\cX,\Z) \lra \Phi \lra 0.$$
188
Thus $\Phi$ is the quotient of the group $b_1^*,\ldots, b_5^*$
189
by the subgroup generated by
190
$$\left\{\begin{array}{l}
191
6b_1^*+3b_2^*+3b_3^*+3b_4^*+3b_5^*\\
192
3b_1^*+4b_2^*+3b_3^*+3b_4^*+3b_5^*\\
193
3b_1^*+3b_2^*+4b_3^*+3b_4^*+3b_5^*\\
194
3b_3^*+3b_2^*+3b_3^*+4b_4^*+3b_5^*\\
195
3b_3^*+3b_2^*+3b_3^*+3b_4^*+4b_5^*
196
\end{array}\right\}$$
197
Letting $[x]$ denote a class in the quotient group, we find
198
$$[b_2^*]=[b_3^*]=[b_4^*]=[b_5^*]=3[b_1^*]$$
199
and that $42[b_1^*]=0$.
200
201
To compute the subgroup fixed by $\Frob=T_p$
202
we must find the action of $\Frob$ on $\Hom(\cX,\Z)$.
203
The 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
206
set of elements $x\in\cX\tensor\Q$ such that
207
$\langle x, y \rangle \in\Z$ for all $y\in\cX$.
208
The dual basis $b_1^*,b_2^*,b_3^*,b_4^*,b_5^*\in\cX\tensor\Q$
209
is a set of elements so that $b_i^*(b_j)=\delta_{ij}$
210
(Kronecker's $\delta$).
211
For convenience, we will work instead in $\cD\tensor\Q$.
212
If $v\in \cD\tensor\Q$ corresponds to $b_1^*$ then we
213
have the matrix equation
214
$$
215
\left(\begin{matrix}
216
1&-1&0&0&0&0\\
217
1&0&-1&0&0&0\\
218
1&0&0&-1&0&0\\
219
1&0&0&0&-1&0\\
220
1&0&0&0&0&-1\\
221
1&1&3&3&3&3
222
\end{matrix}\right)
223
\cdot
224
\left(\begin{matrix}
225
3&0&0&0&0&0\\
226
0&3&0&0&0&0\\
227
0&0&1&0&0&0\\
228
0&0&0&1&0&0\\
229
0&0&0&0&1&0\\
230
0&0&0&0&0&1
231
\end{matrix}\right)
232
\cdot v =
233
\left(\begin{matrix}
234
1\\0\\0\\0\\0\\0\end{matrix}
235
\right).
236
$$
237
Letting $a$ denote the first matrix, and $b$ the second diagonal
238
one, we find that the action of
239
$\Frob$ on $\Hom(X,\Z)$ with respect to the
240
basis $b_1^*, b_2^*, b_3^*, b_4^*, b_5^*, \Eis^*$
241
is given by
242
243
$$a\cdot b\cdot \Frob\cdot (a\cdot b)^{-1}
244
=\left(\begin{matrix}
245
-1&0&0&0&0&0\\
246
-1&0&1&0&0&0\\
247
-1&1&0&0&0&0\\
248
-1&0&0&0&1&0\\
249
-1&0&0&1&0&0\\
250
0&0&0&0&0&1
251
\end{matrix}\right).$$
252
Therefore
253
$$\Frob(b_1^*)=-(b_1^*+b_2^*+b_3^*+b_4^*+b_5^*)=-(13b_1^*).$$
254
Multiplication 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}
260
Some 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
266
of 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:}
277
Choose the following basis for the group $\cX=\cX_{13}(J)$ of divisors
278
of degree $0$:
279
$$b_1=e_1-e_2, \,b_2=e_1-e_3,\, b_3=e_1-e_4, \, b_4=e_1-e_5,\,
280
b_5 = e_1-e_6.$$
281
The component group $\Phi$ fits into the sequence
282
$$0 \lra \cX \lra \Hom(\cX,\Z) \lra \Phi \lra 0.$$
283
Since the monodromy weights are trivial,
284
$\Phi$ is the quotient of the group $b_1,\ldots, b_5$
285
by the subgroup generated by
286
$$\left\{\begin{array}{l}
287
2b_1^*+b_2^*+b_3^*+b_4^*+b_5^*,\\
288
b_1^*+2b_2^*+b_3^*+b_4^*+b_5^*,\\
289
b_1^*+b_2^*+2b_3^*+b_4^*+b_5^*,\\
290
b_1^*+b_2^*+b_3^*+2b_4^*+b_5^*,\\
291
b_1^*+b_2^*+b_3^*+b_4^*+2b_5^*.
292
\end{array}\right\}$$
293
Backsubstituting, we find
294
$$[b_1^*]=[b_2^*]n=[b_3^*]=[b_4^*]=[b_5^*]$$
295
and that $6[b_1^*]=0$.
296
297
Just as before we compute
298
$$
299
a\cdot b \cdot \Frob \cdot (a\cdot b)^{-1}
300
=
301
=\left(\begin{matrix}
302
-1&0&0&0&0&0\\
303
-1&0&1&0&0&0\\
304
-1&1&0&0&0&0\\
305
-1&0&0&0&1&0\\
306
-1&0&0&1&0&0\\
307
0&0&0&0&0&1
308
\end{matrix}\right).
309
$$
310
Therefore
311
$$\Frob(b_1^*)=-(b_1^*+b_2^*+b_3^*+b_4^*+b_5^*)=-(5b_1^*).$$
312
Since $-5\con 1\pmod{6}$ the action of Frobenius is trivial
313
so $c_{13}=6$.
314
315
316
317
318
319
320
\section{Analytic order of the Shafarevich-Tate group}
321
The BSD conjecture predicts that
322
$$L'(J,1)\qe \frac{|\Sha|\cdot c_5\cdot c_{13}}
323
{|J(\Q)_{\tor}|^2}
324
\cdot \Reg \cdot \Omega_J$$
325
Assuming 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}
341
342
343