CoCalc Public Fileswww / tables / Notes / compgp.tex
Author: William A. Stein
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9\title{\bf\Huge \mbox{Component groups of optimal quotients}\vspace{5ex}}
10\author{\LARGE William A. Stein}
11\date{\Large \today \vspace{2ex}\\}
13\newcounter{Pagecount}
14\markboth{}{W.\thinspace{}A. Stein, Component groups of optimal quotients}
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25\newtheorem{conj}[thm]{Conjecture}
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71
72\begin{document}
73
74\Large
75\maketitle
76\tableofcontents
77\pagenumbering{Roman}
78\setcounter{page}{0}
79
80\Page{Ribet's letter}
81{\bf November, 1987:} Ribet wrote a letter to Mestre describing
82component groups of elliptic curves at primes of multiplicative reduction.
83\begin{bulletlist}
84\item $N=pM$ with $(M,p)=1$.
85\item $E :=$ an elliptic curve of conductor $N$.
86\item $\pi: J_0(N)\ra E$ with $\ker(\pi)$ connected.
87\item $m_E :=$ modular degree of $E$.
88\item $T_J :=$ torus of $\cJ_0(N)/\Fp$.
89\item $X_J := \Hom(T_J,\Gm)\isom\Div^0(X_0(M)(\Fpbar)^{\ss})$
90\item $X_E := \Hom(T_E,\Gm) =: X_{\Edual}$.
91\end{bulletlist}
92{\bf Monodromy:}\\
93$\displaystyle\langle \quad , \quad \rangle_J : X_J \cross X_J \ra \Z$,
94
95$\displaystyle \langle C, C' \rangle := \begin{cases} 960&\text{if$C\neq C'$},\\ 97w_C = \frac{\#\Aut(C)}{2} & \text{otherwise.} 98\end{cases}$
99
100$\displaystyle\langle \quad , \quad \rangle : X_E \cross X_{\Edual} \ra \Z$
101
102{\bf Component groups:}
103$$\xymatrix{ 104 0\ar[r]&X_J\ar[r]& {\Hom(X_J,\Z)}\ar[r] &{\Phi_J} \ar[r]& 0\\ 105 0\ar[r]&X_{\Edual}\ar[r]&{\Hom(X_E,\Z)}\ar[r]&{\Phi_E}\ar[r]& 0 106}$$
107
108Let $\pi^*:X_E\ra X_J$ be the natural map and write
109$$\pi^* X_E = \Z\cdot \sum n_C[C].$$
110\begin{thm}[Ribet]
111\begin{eqnarray*}
112\#\coker(\Phi_J\ra \Phi_E) &=& \gcd(\ldots n_C \ldots),\\
113\#\Phi_E &=& \gcd_{C,C'}(w_C n_C - w_{C'} n_{C'}),\\
114m_E\cdot \#\Phi_E &=& \sum w_C n_C^2 = \langle \sum n_C [C], \sum n_C[C] \rangle.
115\end{eqnarray*}
116\end{thm}
117
118\Page{Computing $\Phi_E$ without finding Tate's~$q$}
119Let $I\subset\T$ be the kernel of $T_n \mapsto a_n \in \Z$.
120$$\xymatrix{*++{X_E} \[email protected]{^(->}[rrr]^{\pi^*}&&&*+{X_J[I]\subset X_J}}$$
121Can compute $X_J[I]$, but the index is often not~$1$:
122$$[X_J[I] : \pi^* X_E] = 1 \iff \#\coker(X_J\ra X_E)=1.$$
123{\bf Idea:}
124$$m_E = \frac{m_E\cdot\#\Phi_E}{\#\Phi_E} 125 = \frac{\sum w_C n_C^2}{\gcd(w_C n_C - w_{C'} n_{C'})}.$$
126We {\em can} compute $m_E$ using modular symbols.\\
127We can compute $\lambda\sum n_C[C]$ for some unknown $\lambda$
128(secretly, $\lambda = 1/\#\coker(X_J\ra X_E)$).
129Then
130$$\frac{\sum w_C \lambda^2 n_C^2} 131 {\gcd(w_C \lambda n_C - w_{C'} \lambda n_{C'})} 132 = \lambda \cdot m_E.$$
133Thus we can compute~$\lambda$.
134
135{\bf Example:}
136\begin{bulletlist}
137\item $p=3$, $M=11$
138\item $f= q + q^2 - q^3 - q^4 - 2q^5 + \cdots \in S_2(33)^{\new}$
139\item $X_0(11)(\Fbar_3)^{\ss} = \Z C_1 \oplus \Z C_2$, $w_1=w_2=1$
140\item $X=X[I]=\Z(C_1-C_2)$
141\item $m_E=3$
142\end{bulletlist}
143Thus $$\lambda\cdot 3 = \frac{1^2+(-1)^2}{2}=1;$$
144  so $\lambda=\frac{1}{3}$, and
145     $$\pi^*X_E = \Z(3 C_1 - 3 C_2).$$
146Conclusion: $\#\Phi_{E} = 6$ and $\#\coker(\Phi_J\ra \Phi_E)=3$.\\
147Remark: $f$ is congruent mod~$3$ to a form of level $11$.
148
149
150\Page{Component groups of abelian varieties}
151Let $f=\sum a_n q^n \in S_2(\Gamma_0(N))$ be a newform.\\
152Set $I := \ker(\T \ra \Q(\ldots a_n \ldots)).$
153
154$$\xymatrix 155{ &IJ_0(N)\ar[d]\\ 156*++{\Adual} \[email protected]{^(->}[r]^{\pi^{\vee}}\ar[dr]_{\theta} 157 & J \[email protected]{->>}[d]^{\pi}\\ 158 &A} 159\qquad\qquad\qquad 160 \[email protected]=6pc{ 161 *++{X_{A}} \ar[r]^{\pi^*} \ar[dr]^{\theta^*} 162 & X_{J} \ar[d]^{\pi_*} \\ 163 & X_{\Adual}\[email protected]/^1.5pc/[ul]^{\theta_*}} 164$$
165
166{\bf Modular degree:}
167$m_A := \sqrt{\deg(\theta)}$
168
169Consider $L\subset X_J[I]$ of finite index.
170\begin{eqnarray*}
172 \Phi_L &:=& \coker(X_J \ra \Hom(L,\Z))
173\end{eqnarray*}
174
175\begin{thm}[---]
176For any $L$ as above,
177\begin{eqnarray*}
178\frac{\#\Phi_A}{m_A} &=& \frac{\#\Phi_L}{m_L},\\
179\image(\Phi_J\ra\Phi_A) &=& \Phi_{X[I]},
180\#\coker(\Phi_J\ra\Phi_A) = \frac{m_A}{m_{X[I]}},\\
181m_A &=& \sqrt{\#(H_1(A)/\pi_* H_1(J)[I])}.
182\end{eqnarray*}
183\end{thm}
184
185{\bf Example:} $p=3$, $N=3\cdot 17$:
186\begin{bulletlist}
187\item $f = q + \alp q^2 + \cdots \in S_2(\Gamma_0(3\cdot 17))$,
188$\alp^2 + \alp-4=0$.
189\item $\dim J_0(51) =13$ and $\dim A = 2$.
190\item $m_A=8$.
191\item Compute $L=X[I]$; find $\#\Phi_L = 4$ and $m_L = 2$.
192\end{bulletlist}
193$$\frac{\#\Phi_{A}}{8} = \frac{4}{2} 194\implies \#\Phi_{A} = 16 195\text{ and } \#\coker(\Phi_J\ra\Phi_A) =\frac{8}{2} = 4.$$
196Remark: $f$ is congruent to a newform of level $17$ mod $3$.
197
198
199\Page{Lemmas}
200The remainder of the lecture is devoted to proving
201that $\frac{\#\Phi_A}{m_A} = \frac{\#\Phi_L}{m_L}$.
202$$\[email protected]=3pc{ 203 *++{X_{A}} \[email protected]{^(->}[r]^{\pi^*} \ar[dr]^{\theta^*} 204 & X_{J} \[email protected]{->>}[d]^{\pi_*} \\ 205 & X_{\Adual}\[email protected]/^1.5pc/[ul]^{\theta_*}} 206$$
207{\em injectivity}:
208$\theta_*\pi_*\pi^*=\theta_*\theta^*=[\deg(\theta)]\neq 0$
209and $X_A$ is free.\\
210{\em surjectivity}: proved later.
211
212Restrict $\langle \, , \, \rangle_J$ to obtain
213$$\alp : X_J \ra \Hom(\pi^* X_A, \Z).$$
214\begin{lem}\label{lem:1}
215   $\ker(\pi_*)=\ker(\alp)$
216\end{lem}
217\begin{proof}
218$x \in \ker(\pi_*)$, $z\in{}X_A$\\
219$\alp(x)(\pi^* z) = \langle x, \pi^* z\rangle 220 = \langle \pi_* x, z\rangle 221 = \langle 0, z \rangle = 0 222\implies x \in \ker(\alp)$\\
223Thus $\ker(\pi_*)\subset \ker(\alp)$.\\
224$\dim \coker(\pi_*) = \dim A = \dim\coker(\alp) 225 \implies \ker(\pi_*) = \ker(\alp).$
226\end{proof}
227
228\begin{lem}\label{lem:2}
229$\Phi_{\pi^* X_A} \isom \Phi_A$.
230\end{lem}
231\begin{proof}
232$$\xymatrix{0\ar[r] 233 & X_J/\ker(\alp)\ar[d]^{\isom} \ar[r] 234 & {\Hom(\pi^* X_A,\Z)}\ar[r] \ar[d]^{\isom} & \coker(\alp)\ar[r]\ar[d] & 0\\ 235 0\ar[r] & X_{A'}\ar[r] & {\Hom(X_A,\Z)}\ar[r] & {\Phi_A}\ar[r] & 0}$$
236The middle isomorphism is induced by the isomorphism $X_A \ra \pi^* X_A$.\\
237Commutes: $x \in X_J$, $z\in X_A$, then
238$\langle x,\pi^* z\rangle_J = \langle \pi_* x, z\rangle_A$.\\
239Thus $\coker(\alp)\isom \Phi_A$, as claimed.
240\end{proof}
241
242\Page{Proof of theorem}
243\begin{lem}\label{lem:3}
244$\frac{\#\Phi_L}{m_L}$ does not depend on~$L$.
245\end{lem}
246\begin{proof}
247Consider $L'\subset X_J[I]$ finite index; set $a = [L:L'] \in \Q$.\\
248Then
249 $$m_{L'} = [X_{A^{\vee}}:\pi_* L'] 250 = [X_{A^{\vee}}:\pi_* L] 251 \cdot [ \pi_* L : \pi_* L'] 252 = m_L \cdot a$$
253since $\pi_*|_{X_J[I]}$ is injective. Also,
254\begin{eqnarray*}
255 \#\Phi_{L'} &=& \#\coker(X_J\ra\Hom(L',\Z))\\
256             &=& \#\coker(X_J\ra\Hom(L,\Z))\cdot[L:L']=\#\Phi_L\cdot a.
257\end{eqnarray*}
258For this, think of lattices inside $\Hom(X_J[I],\Z)$.
259\end{proof}
260
261{\em Proof of theorem.}
262By Lemma~\ref{lem:3}, we may assume that $L=\pi^* X_A$.\\
263By Lemma~\ref{lem:2}, $\Phi_L \isom \Phi_A$.
264\begin{eqnarray*}
266     &=& [X_{\Adual} : \pi_*\pi^* X_A]\\
267     &=& [X_{\Adual} : \theta^* X_A]\\
268     &=& \#\coker(\theta^*)\\
269     &=& m_A.
270\end{eqnarray*}
271The final equality is proved below.
272
273\Page{Analysis of the modular polarization}
274{\bf Uniformization:}
275$A$ is purely toric, so over $\Qp^{\ur}$:
276$$\xymatrix{ 277 0\ar[r]\ar[d] & X_{A}\ar[r]\ar[d] & X_{\Adual}\ar[r]\ar[d] 278 & \coker(\theta_a)\ar[d]\\ 279 \ker(\theta_t)\ar[r]\ar[d]& T_{\Adual}\ar[d]\ar[r] 280 & T_A\ar[r]\ar[d] & 0\\ 281 \ker(\theta)\ar[r] & A^{\vee}\ar[r]^{\theta} & A}$$
282The snake lemma gives
283$$0 \ra \ker(\theta_t) \ra \ker(\theta) \ra \coker(\theta^*)\ra 0.$$
284The dual of
285$$\xymatrix{ 286 & X_{A}\ar[r]\ar[d] & X_{A^{\vee}}\ar[d]\ar[r] & \coker(\theta_a) \\ 287 \ker(\theta_t)\ar[r] & T_{\Adual} \ar[r]^{\theta} & T_{A}}$$
288is
289$$\xymatrix{ 290 & T_A & T_{\Adual}\ar[l]_{\theta^{\vee}} 291 & \coker(\theta_a)^{\vee}\ar[l]\\ 292 \ker(\theta_t)^{\vee}& X_{\Adual}\ar[l]\ar[u] &X_A\ar[l]\ar[u] 293}$$
294But $\theta^{\vee}=\theta$, since $\theta$ arises from the $\theta$-divisor, so
295     $$\coker(\theta^*)^\vee \isom \ker(\theta_t).$$
296In particular,
297 $$\#\ker(\theta) = (\#\coker(\theta^*))^2,$$
298so
299$$m_A = \#\coker(\theta^*).$$
300
301
302\Page{Surjectivity of $\pi_*$}
303There are easier ways, but in the interest of suggesting
304a more general theorem we use the Raynaud uniformization.
305
306\begin{lem}\label{lem:surj}
307The map $\pi_*:X_J\ra X_{\Adual}$ is surjective.
308\end{lem}
309\begin{proof}
310Because $G_J$ is simply connected, $\pi$ induces a map
311$G_J\ra T_A$.
312Because $\pi$ is surjective and $T_A$ is a
313torus, the map $G_J\ra T_A$ is surjective.
314The snake lemma applied to the following diagram gives
315a surjective map from $B=\ker(\pi)$ to
316$M=\coker(X_J\ra X_{\Adual})$.
317$$\xymatrix{ 318 & X_J\ar[r]\ar[d] & X_{\Adual}\ar[r]\ar[d] & M\ar[r]\ar[d]& 0 \\ 319 & G_J\ar[r]\ar[d] & T_A\ar[d]\ar[r] & 0 \\ 320B\ar[r] & J\ar[r]^{\pi}& A 321}$$
322Because $\pi$ is optimal, $B$ is connected so $M$ must also be connected.
323Since $M$ is discrete it follows that $M=0$.
324
325Remark: Abbes pointed out that the condition that
326$A$ be purely toric can be relaxed.
327\end{proof}
328
329{\bf Remarks:}
330\begin{bulletlist}
331\item In the above theorem, $J_0(N)$ can be replaced by any
332semistable Jacobian and $A$ by any purely toric optimal quotient.
333\item Mazur, Raynaud, and Edixhoven have given an explicit formula
334for $\Phi_{J_0(Mp)}$ for $p\geq 5$.
335\end{bulletlist}
336
337\end{document}
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