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Author: William A. Stein
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\begin{document}
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\title{Constructing elements in Shafarevich-Tate groups of modular motives}
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\author{Neil Dummigan}
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\author{William A. Stein}
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\date{July 10th, 2001}
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\subjclass{11F33, 11F67, 11G40.}
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\keywords{modular form,$L$-function, Bloch-Kato conjecture,
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Shafarevich-Tate group.}
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\address{University of Sheffield\\ Department of Pure
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Mathematics\\
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Hicks Building\\ Hounsfield Road\\ Sheffield, S3 7RH\\
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U.K.}
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\address{Harvard University\\ Mathematics Department\\
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1 Oxford Street\\ Cambridge, MA 02138\\ U.S.A.}
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\email{n.p.dummigan@shef.ac.uk} \email{was@math.harvard.edu}
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\begin{abstract}
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\end{abstract}
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\maketitle
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\section{Introduction}
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\section{Motives and Galois representations}
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Let $f=\sum a_nq^n$ be a newform of weight $k\geq 2$ for
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$\Gamma_0(N)$, with coefficients in an algebraic number field $E$.
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A theorem of Deligne \cite{De1} implies the existence, for each
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(finite) prime $\lambda$ of $E$, of a continuous representation
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$$\rho_{\lambda}:\Gal(\Qbar/\QQ)\rightarrow \Aut(V_{\lambda})$$
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($V_{\lambda}$ is a two-dimensional vector space over
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$E_{\lambda}$), such that
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\begin{enumerate}
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\item $\rho_{\lambda}$ is unramified at $p$ for all primes $p$ not dividing
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$lN$ (where $\lambda \mid l$);
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\item if $\Frob_p$ is an arithmetic Frobenius element at such a $p$ then the
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characteristic polynomial of $\Frob_p^{-1}$ acting on
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$V_{\lambda}$ is $x^2-a_px+p^{k-1}$.
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\end{enumerate}
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Following Scholl \cite{Sc}, $V_{\lambda}$ may be constructed as
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the $\lambda$-adic realisation of a Grothendieck motive $M_f$.
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There are also Betti and de Rham realisations $V_B$ and $V_{\dR}$,
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both $2$-dimensional $E$-vector spaces. For details of the
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construction see \cite{Sc}. The de Rham realisation has a Hodge
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filtration $V_{\dR}=F^0\supset F^1=\ldots =F^{k-1}\supset
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F^k=\{0\}$. The Betti realisation $V_B$ comes from singular
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cohomology, while $V_{\lambda}$ comes from \'etale $l$-adic
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cohomology. There are natural isomorphisms $V_B\otimes
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E_{\lambda}\simeq V_{\lambda}$. Using a basis for singular
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cohomology with $\ZZ$-coefficients, we get
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$\Gal(\Qbar/\QQ)$-stable $O_{\lambda}$-modules $T_{\lambda}$
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inside each $V_{\lambda}$. Define
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$A_{\lambda}=V_{\lambda}/T_{\lambda}$. There are two kinds of
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twist we shall have to consider. There is the Tate twist
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$V_{\lambda}(j)$ (for an integer $j$), which amounts to
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multiplying the action of $\Frob_p$ by $p^j$. For $D$ the
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discriminant of a quadratic field, there is the quadratic twist
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$V_{\lambda}(\chi_D)$, which is the tensor product of
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$V_{\lambda}$ with a one-dimensional space on which
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$\Gal(\Qbar/\QQ)$ acts via the quadratic character $\chi_D$.
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Following \cite{BK} (Section 3), for $p\neq l$ (including
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$p=\infty$) let
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$$H^1_f(\QQ_p,V_{\lambda}(\chi_D,j))=\ker (H^1(D_p,V_{\lambda}(\chi_D,j))\rightarrow
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H^1(I_p,V_{\lambda}(\chi_D,j))).$$ The subscript $f$ stands for
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``finite part''. $D_p$ is a decomposition subgroup at a prime
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above $p$, $I_p$ is the inertia subgroup, and the cohomology is
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for continuous cocycles and coboundaries. For $p=l$ let
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$$H^1_f(\QQ_l,V_{\lambda}(\chi_D,j))=\ker (H^1(D_l,V_{\lambda}(\chi_D,j))\rightarrow
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H^1(D_l,V_{\lambda}(\chi_D,j)\otimes B_{\cris}))$$ (see Section 1
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of \cite{BK} for definitions of Fontaine's rings $B_{\cris}$ and
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$B_{dR}$). Let $H^1_f(\QQ,V_{\lambda}(\chi_D,j))$ be the subspace
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of elements of $H^1(\QQ,V_{\lambda}(\chi_D,j))$ whose local
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restrictions lie in $H^1_f(\QQ_p,V_{\lambda}(\chi_D,j))$ for all
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primes $p$.
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There is a natural exact sequence
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$$\begin{CD}0@>>>T_{\lambda}(\chi_D,j)@>>>V_{\lambda}(\chi_D,j)@>\pi>>A_{\lambda}(\chi_D,j)@>>>0\end{CD}.$$
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Let
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$H^1_f(\QQ_p,A_{\lambda}(\chi_D,j))=\pi_*H^1_f(\QQ_p,V_{\lambda}(\chi_D,j))$.
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Define the $\lambda$-Selmer group \newline
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$H^1_f(\QQ,A_{\lambda}(\chi_D,j))$ to be the subgroup of elements
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of $H^1(\QQ,A_{\lambda}(\chi_D,j))$ whose local restrictions lie
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in $H^1_f(\QQ_p,A_{\lambda}(\chi_D,j))$ for all primes $p$. Note
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that the condition at $p=\infty$ is superfluous unless $l=2$.
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Define the Shafarevich-Tate group
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$$\Sha(\chi_D,j)=\oplus_{\lambda}H^1_f(\QQ,A_l(\chi_D,j))/\pi_*H^1_f(\QQ,V_l(\chi_D,j)).$$
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The length of its $\lambda$-component may be taken for the
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exponent of $\lambda$ in an ideal of $O_E$, which we call
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$\#\Sha(\chi_D,j)$. We shall only concern ourselves with the case
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$j=k/2$, and write $\Sha(\chi_D)$ for $\Sha(\chi_D,j)$.
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Define the set of global points
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$$\Gamma_{\QQ}(\chi_D)=\oplus_{\lambda}H^0(\QQ,A_{\lambda}(\chi_D,k/2)).$$
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This is analogous to the group of rational torsion points on an
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elliptic curve. The length of its $\lambda$-component may be taken
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for the exponent of $\lambda$ in an ideal of $O_E$, which we call
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$\#\Gamma_{\QQ}(\chi_D)$.
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\section{The Bloch-Kato conjecture}
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The Bloch-Kato conjecture for the motive $M_f(k/2)$ predicts that
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$$\frac{L(f,k/2)}{\vol_{\infty}}=\frac{\left(\prod_pc_p(k/2)\right)\cdot\#\Sha}{(\#\Gamma_{\QQ})^2}.$$
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The local factors $\vol_{\infty}$ and $c_p(k/2)$ depend on the
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choice of a basis for $V_{\dR}$, but their product does not. For
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$l\neq p$, $\ord_{\lambda}(c_p(\chi_D,j))$ is defined to be
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$$\length H^1_f(\QQ_p,T_{\lambda}(\chi_D,j))_{\tors.}-\ord_{\lambda}((1-\chi_D(p)a_pp^{-j}+p^{k-1-2j})).$$
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We omit the definition of $\ord_{\lambda}(c_p(\chi_D,j))$ for
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$\lambda\mid p$, which requires one to assume Fontaine's de Rham
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conjecture (\cite{Fo}, Appendix A6). The above formula is to be
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interpreted as an equality of fractional ideals of $E$. (Strictly
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speaking, the conjecture in \cite{BK} is only given for $E=\QQ$.)
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Similarly, the Bloch-Kato conjecture for $M_f(\chi_D,k/2)$
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predicts that
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$$\frac{L(f,\chi_D,k/2)}{\vol_{\infty}(\chi_D)}=\frac{\left(\prod_pc_p(k/2,\chi_D)\right)\#\Sha(\chi_D)}{(\#\Gamma_{\QQ}(\chi_D))^2}.$$
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A careful treatment of real periods
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such as $\vol_{\infty}(\chi_D)$ is given in \cite{De2}. The
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calculation of the periods of Artin motives in terms of
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generalized Gauss sums, in Section 6 of \cite{De2}, yields the
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following lemma.
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\begin{lem}
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If $\chi_D(-1)=1$ (i.e. if $D>0$) then
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$\vol_{\infty}(\chi_D)=\vol_{\infty}/\sqrt{D}.$
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\end{lem}
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\begin{lem} Let $p\nmid N$ be a prime, $j$ an integer
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and $D$ a quadratic discriminant. Then the fractional ideal
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$c_p(\chi_D,j)$ is supported at most on divisors of $2$ and
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divisors of $p$, while $c_p(j)$ is supported at most on divisors
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of $p$.
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\end{lem}
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\begin{proof}
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As on p. 30 of \cite{Fl1}, for odd $l\neq p$,
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$\ord_{\lambda}(c_p(\chi_D,j))$ is the length of the finite
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$O_{\lambda}$-module
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$H^0(\QQ_p,H^1(I_p,T_{\lambda}(\chi_D,j))_{\tors}),$ where $I_p$
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is an inertia group at $p$. If $p$ does not divide $D$ then
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$T_{\lambda}(\chi_D,j)$ is a trivial $I_p$-module, so
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$H^1(I_p,T_{\lambda}(\chi_D,j))$ is torsion-free. In general,
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since $l\neq p$,$$H^1(I_p,T_{\lambda}(\chi_{D},j))\simeq
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T_{\lambda}(\chi_{D},j)/(1-\tau)T_{\lambda}(\chi_{D},j),$$ where
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$\tau$ is a topological generator of the tame quotient of $I_p$.
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If $p\mid D$ then thanks to the quadratic twist ramified at $p$,
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$\,\tau$ acts as multiplication by $-1$ on
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$T_{\lambda}(\chi_{D},j)$, hence
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$H^1(I_p,T_{\lambda}(\chi_{D},j))$ is trivial (remember $l\neq
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2$).
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The assertion about $c_p(j)$ may be proved similarly to the case
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$p\nmid D$ above.
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\end{proof}
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\begin{lem} Let $D$ be a quadratic discriminant, and $p$ a prime such that
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$\chi_D(p)=1$. Then $c_p(\chi_D,j)=c_p(j)$.
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\end{lem}
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\begin{proof}
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This is simply a consequence of the fact that the
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definition of $\ord_{\lambda}(c_p(j))$ depends only on
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$T_{\lambda}$ as a $\Gal(\Qbar_p/\QQ_p)$-module.
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\end{proof}
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\begin{lem} $\ord_{\lambda}(\#\Gamma_{\QQ}(\chi_D,j))=0$ unless
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the coefficients of $f=\sum a_nq^n$ satisfy the congruence
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$a_p\equiv \chi_D(p)(p^j+p^{k-1-j})\pmod{\lambda}$ for all primes
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$p\nmid lN$.
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\end{lem}
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This follows from the interpretation of $a_p$ as a trace of
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Frobenius. See \cite{SwD}.
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Putting together the above lemmas we arrive at the following:
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\begin{prop}\label{sha} Let $q\nmid N$ be an odd prime. Let $D>0$ be a
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quadratic discriminant such that $\chi_D(q)=1$ and $\chi_D(p)=1$
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for all primes $p\mid N$. Let $\qq\mid q$ be a prime of $E$ such
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that neither
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$$a_p\equiv \chi_D(p)(p^{k/2}+p^{(k/2)-1})\pmod{\qq}$$
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nor
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$$a_p\equiv p^{k/2}+p^{(k/2)-1}\pmod{\qq}$$
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holds for all $p\nmid qN$. If
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$$\ord_{\qq}(L(f,k/2)/\sqrt{D}L(f,\chi_D,k/2))>0$$
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then the Bloch-Kato conjecture predicts that
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$\ord_{\qq}(\#\Sha)>0$.
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\end{prop}
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\begin{prop}\label{triv}
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Suppose that $p\nmid DN$, $p>k$ and that $f$ is not
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congruent to another cusp form for $\Gamma_0(N)$ modulo any
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divisor of $p$. Then, for $(k/2)\leq j\leq k-1$, $c_p(j)$ is
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trivial and $c_p(\chi_D,j)$ is at worst supported on divisors of
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$2$.
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\end{prop}
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This may be proved in a similar manner to Theorem 7.6 of
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\cite{Du1}. The third condition in the statement of the
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proposition ensures that the denominator of the projector used to
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cut out the motive $M$ is coprime to $p$. Bearing in mind Flach's
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generalisation of the Cassels-Tate pairing \cite{Fl2}, we see the
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following corollary.
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\begin{cor} The Bloch-Kato conjecture predicts that, for $D$ as in
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Proposition \ref{sha}, the fractional ideal
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$(L(f,k/2)/\sqrt{D}L(f,\chi_D,k/2))$ is a square, up to divisors
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of $2$ and of primes not satisfying the conditions of Proposition
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\ref{triv}.
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\end{cor}
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\section{Constructing elements of the Shafarevich-Tate group}
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Let $f=\sum a_nq^n$ and $g=\sum b_nq^n$ be newforms of equal
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weight $k\geq 2$ for $\Gamma_0(N)$. Let $E$ be a number field
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large enough to contain all the coefficients $a_n$ and $b_n$.
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Suppose that $\qq\mid q$ is a prime of $E$ such that $f\equiv
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g\pmod{\qq}$, i.e. $a_n\equiv b_n\pmod{\qq}$ for all $n$. For $f$
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we have defined $V_{\lambda}$, $T_{\lambda}$ and $A_{\lambda}$.
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Let $V'_{\lambda}$, $T'_{\lambda}$ and $A'_{\lambda}$ be the
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corresponding objects for $g$. Let $A[\lambda]$ denote the
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$\lambda$-torsion in $A_{\lambda}$. Since $a_p$ is the trace of
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$\Frob_p^{-1}$ on $V_{\lambda}$, it follows from the Cebotarev
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Density Theorem that $A[\qq]$ and $A'[\qq]$ are isomorphic as
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$\Gal(\Qbar/\QQ)$-modules.
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Suppose that $L(g,k/2)=0$. If the sign in the functional equation
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is positive, this implies that the order of vanishing at $s=k/2$
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is at least $2$. According to the Beilinson-Bloch conjecture
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\cite{B}, the order of vanishing of $L(g,s)$ at $s=k/2$ is the
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rank of the group $\CH_0^{k/2}(M_g)(\QQ)$ of $\QQ$-rational,
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null-homologous algebraic cycles on the motive $M_g$, modulo
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rational equivalence.
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Via the $\qq$-adic Abel-Jacobi map, $\CH_0^{k/2}(M_g)(\QQ)$ maps
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to $H^1(\QQ,V'_{\qq}(k/2))$, and Nekov\'ar shows in \cite{Ne2}
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that its image is contained in the subspace
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$H^1_f(\QQ,V'_{\qq}(k/2))$. See also 0.13 of \cite{Ne1}. If, as
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expected, the $\qq$-adic Abel-Jacobi map is injective, we get
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(assuming also the Beilinson-Bloch conjecture) a subspace of
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$H^1_f(\QQ,V'_{\qq}(k/2))$ of dimension equal to the order of
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vanishing of $L(g,s)$ at $s=k/2$.
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If $L(f,k/2)\neq 0$ then it is expected that
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$H^1_f(\QQ,V_{\qq}(k/2))=0$, so that
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$\Sha=H^1_f(\QQ,A_{\qq}(k/2))$.
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\begin{thm}\label{local}
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Let $q\nmid N$ be a prime satisfying $q>k$. Suppose that
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$H^1_f(\QQ,V'_{\qq}(k/2))$ has dimension $r>0$. Suppose that
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$A[\qq]$ and $A'[\qq]$ are irreducible representations of
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$\Gal(\Qbar/\QQ)$ and that, for all primes $p\mid N$, $\,p\neq \pm
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1\pmod{q}$. Suppose also that neither $f$ nor $g$ is congruent
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modulo $\qq$ to any newform of trivial character and level
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dividing $N/p$, with $p$ a prime exactly dividing $N$. Then the
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$\qq$-torsion subgroup of $H^1_f(\QQ,A_{\qq}(k/2))$ has
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$\FF_{\qq}$-rank at least $r-1$.
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\end{thm}\edit{There's something wrong with the statement
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of this theorem, as none of the objects depend on~$g$.
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I think $H^1_f(\QQ,A_{\qq}(k/2))$
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should be replaced by $H^1_g(\QQ,A_{\qq}(k/2))$.}
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\begin{proof}Take a non-zero class $d\in
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H^1_f(\QQ,V'_{\qq}(k/2))$, by continuity and rescaling we may
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assume that it lies in $H^1_f(\QQ,T'_{\qq}(k/2))$ but not in $\qq
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H^1_f(\QQ,T'_{\qq}(k/2))$. Then by reduction we get a non-zero
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class $c\in H^1(\QQ,A'[\qq](k/2))\simeq H^1(\QQ,A[\qq](k/2))$.
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Since $a_p\equiv p^{k/2}+p^{(k/2)-1}\pmod{\qq}$ (for all $p\nmid
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qN$) does not hold (by irreducibility), $H^0(\QQ,A[\qq](k/2))$ is
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trivial, so $H^1(\QQ,A[\qq](k/2))$ injects into
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$H^1(\QQ,A_{\qq}(k/2))$, and we get a non-zero class $\gamma\in
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H^1(\QQ,A_{\qq}(k/2))$.
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First we will show that $\res_p(\gamma)\in
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H^1_f(\QQ_p,A_{\qq}(k/2))$, for all (finite) primes $p\nmid qN$.
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Since in this case $A'_{\qq}(k/2)$ is unramified at $p$, $H^0(I_p,
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A'_{\qq}(k/2))=A'_{\qq}(k/2)$, which is $\qq$-divisible. Therefore
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$H^1(I_p,A'[\qq](k/2))$ injects into $H^1(I_p,A'_{\qq}(k/2))$, and
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it follows from $d\in H^1_f(\QQ,V'_{\qq}(k/2))$ that the image of
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$\gamma$ in $H^1(I_p,A_{\qq}(k/2))$ is zero. By line 3 of p. 125
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of \cite{Fl2}, $H^1_f(\QQ_p,A_{\qq}(k/2))$ is equal to (not just
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contained in) the kernel of the map from $H^1(\QQ_p,A_{\qq}(k/2))$
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to $H^1(I_p,A_{\qq}(k/2))$, so we have shown that
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$\res_p(\gamma)\in H^1_f(\QQ_p,A_{\qq}(k/2))$.
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Next we will show that $\res_p(\gamma)\in
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H^1_f(\QQ_p,A_{\qq}(k/2))$ for $p\mid N$. Our first task is to
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show that $H^0(I_p, A'_{\qq}(k/2))$ is $\qq$-divisible. It
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suffices to show that $$\dim H^0(I_p,A'[\qq](k/2))=\dim
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H^0(I_p,V'_{\qq}(k/2)).$$ If the dimensions differ then, given
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that $g$ is not congruent modulo $\qq$ to a newform of level
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strictly dividing $N$, and since $p\neq \pm 1\pmod{q}$,
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Propositions 3.1 and 2.3 of \cite{L} tell us that $A'[\qq](k/2)$
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is unramified at $p$ and that $\ord_p(N)=1$. (Here we are using
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Carayol's result that $N$ is the prime-to-$q$ part of the
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conductor of $V_{\qq}$ \cite{Ca1}.) But then Theorem 1 of
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\cite{JL} implies the existence of a newform of trivial character
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and level dividing $N/p$, congruent to $g$ modulo $\qq$. This
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contradicts our hypotheses, and we have established that $H^0(I_p,
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A'_{\qq}(k/2))$ is $\qq$-divisible.
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It follows that the image of $c\in H^1(\QQ,A[\qq](k/2))$ in
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$H^1(I_p,A[\qq](k/2))$ is zero. By inflation-restriction,
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$\res_p(c)$ then comes from $H^1(D_p/I_p,H^0(I_p,A[\qq](k/2)))$.
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The order of this group is the same as the order of the group
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$H^0(\QQ_p,A[\qq](k/2))$, which we claim is trivial. By the work
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of Carayol \cite{Ca1}, $p\mid N$ implies that $V_{\qq}(k/2)$ is
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ramified at $p$, so $\dim H^0(I_p,V_{\qq}(k/2))=0$ or $1$. As
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above, we see that $\dim H^0(I_p,V_{\qq}(k/2))=\dim
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H^0(I_p,A[\qq](k/2))$, so we need only consider the case where
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this common dimension is $1$. The (motivic) Euler factor at $p$
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for $M_f$ is $(1-\alpha p^{-s})^{-1}$, where $\Frob_p^{-1}$ acts
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as multiplication by $\alpha$ on the one-dimensional space
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$H^0(I_p,V_{\qq}(k/2))$. It follows from Theor\'eme A of
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\cite{Ca1} that this is the same as the Euler factor at $p$ of
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$L(f,s)$. By the work of Atkin and Lehner \cite{AL}, it then
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follows that $p^2\nmid N$ and $\alpha=-w_pp^{(k/2)-1}$, where
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$w_p=\pm 1$ is such that $W_pf=w_pf$. Twisting by $k/2$,
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$\Frob_p^{-1}$ acts on $H^0(I_p,V_{\qq}(k/2))$ (hence also on
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$H^0(I_p,A[\qq](k/2))$ as $\pm p^{-1}$. Since $p\neq \pm
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1\pmod{q}$, we see that $H^0(\QQ_p,A[\qq](k/2))$ is trivial. Hence
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$\res_p(c)=0$ so $\res_p(\gamma)=0$ and certainly lies in
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$H^1_f(\QQ_p,A_{\qq}(k/2))$.
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It remains to deal with the local condition at $p=q$. Since
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$q\nmid N$ is a prime of good reduction for the motive $M_f$,
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$\,V_{\qq}$ is a crystalline representation of
392
$\Gal(\Qbar_q/\QQ_q)$, meaning $D_{\cris}(V_{\qq})$ and $V_{\qq}$
393
have the same dimension, where
394
$D_{\cris}(V_{\qq}):=H^0(\QQ_q,V_{\qq}\otimes_{\QQ_q} B_{\cris})$.
395
(This is a consequence of Theorem 5.6 of \cite{Fa1}.) It follows
396
from Theorem 4.1(ii) of \cite{BK} that
397
$$D_{\dR}(V_{\qq})/F^{k/2}D_{\dR}(V_{\qq})\simeq H^1_f(\QQ_q,V_{\qq}(k/2)),$$
398
via their exponential map.
399
400
Since $p$ is a prime of good reduction, the de Rham conjecture is
401
a consequence of the crystalline conjecture, which follows from
402
Theorem 5.6 of \cite{Fa1}. Hence
403
$$V_{\dR}\otimes_E E_{\qq}/F^{k/2}V_{\dR}\otimes_E E_{\qq}\simeq H^1_f(\QQ_q,V_{\qq}(k/2)).$$
404
Observe that the dimension of the left-hand-side is $1$.
405
Meanwhile, by 3.8 of \cite{BK}, $H^1_f(\QQ_q,V_{\qq}(k/2))$ is its
406
own annihilator in $H^1(\QQ_q,V_{\qq}(k/2))$ with respect to the
407
Tate duality pairing. Hence $H^1_f(\QQ_q,V_{\qq}(k/2))$ has
408
codimension one in $H^1(\QQ_q,V_{\qq}(k/2))$. It follows that our
409
$r$-dimensional $\FF_{\qq}$-subspace of the $\qq$-torsion in
410
$H^1(\QQ,A_{\qq}(k/2))$ has at least an $(r-1)$-dimensional
411
subspace landing in $H^1_f((\QQ_{\qq},A_{\qq}(k/2))$.
412
413
To justify this argument carefully, we should check that the
414
natural map from $H^1(\QQ_q,V_{\qq}(k/2))$ to
415
$H^1((\QQ_{\qq},A_{\qq}(k/2))$ is surjective. Its cokernel injects
416
into \newline $H^2((\QQ_{\qq},T_{\qq}(k/2))$, which is dual to
417
$H^0((\QQ_{\qq},A_{\qq}(k/2))$ (via local Tate duality). It
418
suffices to check that $H^0((\QQ_{\qq},A[\qq](k/2))$ is trivial.
419
This follows from the description of $A[\qq](k/2)$ as a
420
$\Gal(\Qbar_q/\QQ_q)$-module provided by theorems of Deligne and
421
Fontaine (Theorems 2.5 and 2.6 of \cite{Ed}).
422
\end{proof}
423
424
Theorem 2.7 of \cite{AS} is concerned with verifying local
425
conditions in the case $k=2$, where $f$ and $g$ are associated
426
with abelian varieties $A$ and $B$. (Their theorem also applies to
427
abelian varieties over number fields.) Our restriction outlawing
428
congruences modulo $\qq$ with cusp forms of lower level is
429
analogous to theirs forbidding $q$ from dividing Tamagawa factors
430
$c_{A,l}$ and $c_{B,l}$. (In the case where $A$ is an elliptic
431
curve with $\ord_l(j(A))<0$, consideration of a Tate
432
parametrisation shows that if $q\mid c_{A,l}$, i.e. if
433
$q\mid\ord_l(j(A))$, then it is possible that $A[q]$ is unramified
434
at $l$.)
435
436
\section{Some examples}
437
\subsection{An example of prime level $127$}
438
439
Using modular symbols, we find that there is a newform
440
$$f = q - q^2 - 8q^3 - 7q^4 - 15q^5 + 8q^6 - 25q^7 + \cdots
441
\in S_4(\Gamma_0(127))$$
442
with $L(f,2)=0$.
443
(In fact, $L'(f,2)=0$, as well.)
444
%The sign of the Atkin-Lehner involution $W_{127}$ on~$f$
445
%is $+1$, so $L'(f,2)=0$ as well (look up reference).
446
There is also a newform $g \in S_4(\Gamma_0(127))$ whose
447
Fourier coefficients generate a number field~$K$ of degree~$17$,
448
and $L(g,2)\neq 0$.
449
The newforms~$f$ and~$g$ are congruent
450
modulo a prime $\qq$ of~$K$ of residue characteristic~$43$.
451
The mod~$\qq$ reductions of~$f$ and~$g$ are both equal to
452
$$\fbar = q + 42q^2 + 35q^3 + 36q^4 + 28q^5 + 8q^6 + 18q^7 + \cdots.$$
453
The residue characteristics of congruence primes
454
between the cuspidal and Eisenstein subspaces are~$2$,~$3$,~$7$,
455
and $1613$, so (reference) $\rho_{f,\qq}\approx\rho_{g,\qq}$
456
is irreducible.
457
Since $S_4(\SL_2(\ZZ))=0$, we see that~$\fbar$ does not arise from a
458
level~$1$ form of weight~$4$. Thus, if we assume the Beilinson-Bloch
459
conjecture, Theorem~\ref{local} implies that $\Sha(M_g)$ has
460
order divisible by~$43$.
461
462
463
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\end{document}
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