Sharedwww / tables / Notes / compgp.texOpen in CoCalc
Author: William A. Stein
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% compgroup.tex
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\title{\bf\Huge \mbox{Component groups of optimal quotients}\vspace{5ex}}
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\author{\LARGE William A. Stein}
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\date{\Large \today \vspace{2ex}\\}
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\pagestyle{myheadings}
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\markboth{}{W.\thinspace{}A. Stein, Component groups of optimal quotients}
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\theoremstyle{plain}
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\newtheorem{thm}{Theorem}
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\newtheorem{lem}[thm]{Lemma}
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\newtheorem{conj}[thm]{Conjecture}
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\begin{document}
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\Large
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\maketitle
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\tableofcontents
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\pagenumbering{Roman}
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\setcounter{page}{0}
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\Page{Ribet's letter}
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{\bf November, 1987:} Ribet wrote a letter to Mestre describing
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component groups of elliptic curves at primes of multiplicative reduction.
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\begin{bulletlist}
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\item $N=pM$ with $(M,p)=1$.
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\item $E := $ an elliptic curve of conductor $N$.
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\item $\pi: J_0(N)\ra E$ with $\ker(\pi)$ connected.
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\item $m_E := $ modular degree of $E$.
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\item $T_J := $ torus of $\cJ_0(N)/\Fp$.
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\item $X_J := \Hom(T_J,\Gm)\isom\Div^0(X_0(M)(\Fpbar)^{\ss})$
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\item $X_E := \Hom(T_E,\Gm) =: X_{\Edual}$.
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\end{bulletlist}
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{\bf Monodromy:}\\
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$\displaystyle\langle \quad , \quad \rangle_J : X_J \cross X_J \ra \Z$,
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$\displaystyle \langle C, C' \rangle := \begin{cases}
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0&\text{if $C\neq C'$},\\
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w_C = \frac{\#\Aut(C)}{2} & \text{otherwise.}
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\end{cases}$
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$\displaystyle\langle \quad , \quad \rangle : X_E \cross X_{\Edual} \ra \Z$
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{\bf Component groups:}
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$$\xymatrix{
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0\ar[r]&X_J\ar[r]& {\Hom(X_J,\Z)}\ar[r] &{\Phi_J} \ar[r]& 0\\
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0\ar[r]&X_{\Edual}\ar[r]&{\Hom(X_E,\Z)}\ar[r]&{\Phi_E}\ar[r]& 0
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}$$
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Let $\pi^*:X_E\ra X_J$ be the natural map and write
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$$\pi^* X_E = \Z\cdot \sum n_C[C].$$
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\begin{thm}[Ribet]
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\begin{eqnarray*}
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\#\coker(\Phi_J\ra \Phi_E) &=& \gcd(\ldots n_C \ldots),\\
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\#\Phi_E &=& \gcd_{C,C'}(w_C n_C - w_{C'} n_{C'}),\\
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m_E\cdot \#\Phi_E &=& \sum w_C n_C^2 = \langle \sum n_C [C], \sum n_C[C] \rangle.
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\end{eqnarray*}
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\end{thm}
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\Page{Computing $\Phi_E$ without finding Tate's~$q$}
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Let $I\subset\T$ be the kernel of $T_n \mapsto a_n \in \Z$.
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$$\xymatrix{*++{X_E} \[email protected]{^(->}[rrr]^{\pi^*}&&&*+{X_J[I]\subset X_J}}$$
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Can compute $X_J[I]$, but the index is often not~$1$:
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$$[X_J[I] : \pi^* X_E] = 1 \iff \#\coker(X_J\ra X_E)=1.$$
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{\bf Idea:}
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$$m_E = \frac{m_E\cdot\#\Phi_E}{\#\Phi_E}
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= \frac{\sum w_C n_C^2}{\gcd(w_C n_C - w_{C'} n_{C'})}.$$
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We {\em can} compute $m_E$ using modular symbols.\\
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We can compute $\lambda\sum n_C[C]$ for some unknown $\lambda$
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(secretly, $\lambda = 1/\#\coker(X_J\ra X_E)$).
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Then
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$$\frac{\sum w_C \lambda^2 n_C^2}
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{\gcd(w_C \lambda n_C - w_{C'} \lambda n_{C'})}
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= \lambda \cdot m_E.$$
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Thus we can compute~$\lambda$.
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{\bf Example:}
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\begin{bulletlist}
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\item $p=3$, $M=11$
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\item $f= q + q^2 - q^3 - q^4 - 2q^5 + \cdots \in S_2(33)^{\new}$
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\item $X_0(11)(\Fbar_3)^{\ss} = \Z C_1 \oplus \Z C_2$, $w_1=w_2=1$
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\item $X=X[I]=\Z(C_1-C_2)$
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\item $m_E=3$
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\end{bulletlist}
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Thus $$\lambda\cdot 3 = \frac{1^2+(-1)^2}{2}=1;$$
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so $\lambda=\frac{1}{3}$, and
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$$\pi^*X_E = \Z(3 C_1 - 3 C_2).$$
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Conclusion: $\#\Phi_{E} = 6$ and $\#\coker(\Phi_J\ra \Phi_E)=3$.\\
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Remark: $f$ is congruent mod~$3$ to a form of level $11$.
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\Page{Component groups of abelian varieties}
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Let $f=\sum a_n q^n \in S_2(\Gamma_0(N))$ be a newform.\\
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Set $I := \ker(\T \ra \Q(\ldots a_n \ldots)).$
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$$\xymatrix
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{ &IJ_0(N)\ar[d]\\
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*++{\Adual} \[email protected]{^(->}[r]^{\pi^{\vee}}\ar[dr]_{\theta}
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& J \[email protected]{->>}[d]^{\pi}\\
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&A}
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\qquad\qquad\qquad
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\[email protected]=6pc{
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*++{X_{A}} \ar[r]^{\pi^*} \ar[dr]^{\theta^*}
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& X_{J} \ar[d]^{\pi_*} \\
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& X_{\Adual}\[email protected]/^1.5pc/[ul]^{\theta_*}}
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$$
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{\bf Modular degree:}
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$m_A := \sqrt{\deg(\theta)}$
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Consider $L\subset X_J[I]$ of finite index.
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\begin{eqnarray*}
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m_L &:=& \#(X_{\Adual}/\pi_* L)\\
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\Phi_L &:=& \coker(X_J \ra \Hom(L,\Z))
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\end{eqnarray*}
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\begin{thm}[---]
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For any $L$ as above,
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\begin{eqnarray*}
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\frac{\#\Phi_A}{m_A} &=& \frac{\#\Phi_L}{m_L},\\
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\image(\Phi_J\ra\Phi_A) &=& \Phi_{X[I]},
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\#\coker(\Phi_J\ra\Phi_A) = \frac{m_A}{m_{X[I]}},\\
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m_A &=& \sqrt{\#(H_1(A)/\pi_* H_1(J)[I])}.
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\end{eqnarray*}
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\end{thm}
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{\bf Example:} $p=3$, $N=3\cdot 17$:
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\begin{bulletlist}
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\item $f = q + \alp q^2 + \cdots \in S_2(\Gamma_0(3\cdot 17))$,
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$\alp^2 + \alp-4=0$.
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\item $\dim J_0(51) =13$ and $\dim A = 2$.
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\item $m_A=8$.
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\item Compute $L=X[I]$; find $\#\Phi_L = 4$ and $m_L = 2$.
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\end{bulletlist}
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$$\frac{\#\Phi_{A}}{8} = \frac{4}{2}
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\implies \#\Phi_{A} = 16
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\text{ and } \#\coker(\Phi_J\ra\Phi_A) =\frac{8}{2} = 4.$$
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Remark: $f$ is congruent to a newform of level $17$ mod $3$.
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\Page{Lemmas}
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The remainder of the lecture is devoted to proving
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that $\frac{\#\Phi_A}{m_A} = \frac{\#\Phi_L}{m_L}$.
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$$\[email protected]=3pc{
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*++{X_{A}} \[email protected]{^(->}[r]^{\pi^*} \ar[dr]^{\theta^*}
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& X_{J} \[email protected]{->>}[d]^{\pi_*} \\
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& X_{\Adual}\[email protected]/^1.5pc/[ul]^{\theta_*}}
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$$
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{\em injectivity}:
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$\theta_*\pi_*\pi^*=\theta_*\theta^*=[\deg(\theta)]\neq 0$
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and $X_A$ is free.\\
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{\em surjectivity}: proved later.
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Restrict $\langle \, , \, \rangle_J$ to obtain
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$$\alp : X_J \ra \Hom(\pi^* X_A, \Z).$$
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\begin{lem}\label{lem:1}
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$\ker(\pi_*)=\ker(\alp)$
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\end{lem}
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\begin{proof}
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$x \in \ker(\pi_*)$, $z\in{}X_A$\\
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$\alp(x)(\pi^* z) = \langle x, \pi^* z\rangle
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= \langle \pi_* x, z\rangle
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= \langle 0, z \rangle = 0
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\implies x \in \ker(\alp)$\\
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Thus $\ker(\pi_*)\subset \ker(\alp)$.\\
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$\dim \coker(\pi_*) = \dim A = \dim\coker(\alp)
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\implies \ker(\pi_*) = \ker(\alp).$
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\end{proof}
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\begin{lem}\label{lem:2}
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$\Phi_{\pi^* X_A} \isom \Phi_A$.
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\end{lem}
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\begin{proof}
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$$\xymatrix{0\ar[r]
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& X_J/\ker(\alp)\ar[d]^{\isom} \ar[r]
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& {\Hom(\pi^* X_A,\Z)}\ar[r] \ar[d]^{\isom} & \coker(\alp)\ar[r]\ar[d] & 0\\
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0\ar[r] & X_{A'}\ar[r] & {\Hom(X_A,\Z)}\ar[r] & {\Phi_A}\ar[r] & 0}$$
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The middle isomorphism is induced by the isomorphism $X_A \ra \pi^* X_A$.\\
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Commutes: $x \in X_J$, $z\in X_A$, then
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$\langle x,\pi^* z\rangle_J = \langle \pi_* x, z\rangle_A$.\\
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Thus $\coker(\alp)\isom \Phi_A$, as claimed.
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\end{proof}
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\Page{Proof of theorem}
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\begin{lem}\label{lem:3}
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$\frac{\#\Phi_L}{m_L}$ does not depend on~$L$.
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\end{lem}
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\begin{proof}
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Consider $L'\subset X_J[I]$ finite index; set $a = [L:L'] \in \Q$.\\
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Then
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$$ m_{L'} = [X_{A^{\vee}}:\pi_* L']
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= [X_{A^{\vee}}:\pi_* L]
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\cdot [ \pi_* L : \pi_* L']
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= m_L \cdot a$$
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since $\pi_*|_{X_J[I]}$ is injective. Also,
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\begin{eqnarray*}
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\#\Phi_{L'} &=& \#\coker(X_J\ra\Hom(L',\Z))\\
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&=& \#\coker(X_J\ra\Hom(L,\Z))\cdot[L:L']=\#\Phi_L\cdot a.
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\end{eqnarray*}
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For this, think of lattices inside $\Hom(X_J[I],\Z)$.
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\end{proof}
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{\em Proof of theorem.}
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By Lemma~\ref{lem:3}, we may assume that $L=\pi^* X_A$.\\
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By Lemma~\ref{lem:2}, $\Phi_L \isom \Phi_A$.
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\begin{eqnarray*}
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m_L &=& [X_{\Adual}:\pi_* L]\\
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&=& [X_{\Adual} : \pi_*\pi^* X_A]\\
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&=& [X_{\Adual} : \theta^* X_A]\\
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&=& \#\coker(\theta^*)\\
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&=& m_A.
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\end{eqnarray*}
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The final equality is proved below.
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\Page{Analysis of the modular polarization}
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{\bf Uniformization:}
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$A$ is purely toric, so over $\Qp^{\ur}$:
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$$\xymatrix{
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0\ar[r]\ar[d] & X_{A}\ar[r]\ar[d] & X_{\Adual}\ar[r]\ar[d]
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& \coker(\theta_a)\ar[d]\\
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\ker(\theta_t)\ar[r]\ar[d]& T_{\Adual}\ar[d]\ar[r]
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& T_A\ar[r]\ar[d] & 0\\
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\ker(\theta)\ar[r] & A^{\vee}\ar[r]^{\theta} & A}$$
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The snake lemma gives
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$$ 0 \ra \ker(\theta_t) \ra \ker(\theta) \ra \coker(\theta^*)\ra 0.$$
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The dual of
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$$\xymatrix{
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& X_{A}\ar[r]\ar[d] & X_{A^{\vee}}\ar[d]\ar[r] & \coker(\theta_a) \\
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\ker(\theta_t)\ar[r] & T_{\Adual} \ar[r]^{\theta} & T_{A}}$$
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is
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$$\xymatrix{
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& T_A & T_{\Adual}\ar[l]_{\theta^{\vee}}
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& \coker(\theta_a)^{\vee}\ar[l]\\
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\ker(\theta_t)^{\vee}& X_{\Adual}\ar[l]\ar[u] &X_A\ar[l]\ar[u]
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}$$
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But $\theta^{\vee}=\theta$, since $\theta$ arises from the $\theta$-divisor, so
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$$\coker(\theta^*)^\vee \isom \ker(\theta_t).$$
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In particular,
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$$\#\ker(\theta) = (\#\coker(\theta^*))^2,$$
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so
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$$m_A = \#\coker(\theta^*).$$
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\Page{Surjectivity of $\pi_*$}
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There are easier ways, but in the interest of suggesting
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a more general theorem we use the Raynaud uniformization.
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\begin{lem}\label{lem:surj}
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The map $\pi_*:X_J\ra X_{\Adual}$ is surjective.
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\end{lem}
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\begin{proof}
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Because $G_J$ is simply connected, $\pi$ induces a map
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$G_J\ra T_A$.
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Because $\pi$ is surjective and $T_A$ is a
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torus, the map $G_J\ra T_A$ is surjective.
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The snake lemma applied to the following diagram gives
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a surjective map from $B=\ker(\pi)$ to
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$M=\coker(X_J\ra X_{\Adual})$.
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$$\xymatrix{
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& X_J\ar[r]\ar[d] & X_{\Adual}\ar[r]\ar[d] & M\ar[r]\ar[d]& 0 \\
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& G_J\ar[r]\ar[d] & T_A\ar[d]\ar[r] & 0 \\
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B\ar[r] & J\ar[r]^{\pi}& A
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}$$
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Because $\pi$ is optimal, $B$ is connected so $M$ must also be connected.
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Since $M$ is discrete it follows that $M=0$.
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Remark: Abbes pointed out that the condition that
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$A$ be purely toric can be relaxed.
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\end{proof}
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{\bf Remarks:}
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\begin{bulletlist}
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\item In the above theorem, $J_0(N)$ can be replaced by any
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semistable Jacobian and $A$ by any purely toric optimal quotient.
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\item Mazur, Raynaud, and Edixhoven have given an explicit formula
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for $\Phi_{J_0(Mp)}$ for $p\geq 5$.
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\end{bulletlist}
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\end{document}
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