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\title{Questions About Finiteness of Shafarevich-Tate Groups of Higher Rank Elliptic Curves}
\author{Wei Ho \and William Stein}

\begin{document}
\maketitle
\begin{abstract}

\end{abstract}

\section{Introduction}

\begin{conjecture}\label{conj:finite}
  There are infinitely many pairs $(E,p)$ with $E$ an elliptic curve
  over $\QQ$ such that $E(\QQ)$ has rank $\geq 2$, and the $p$-primary
  part $\Sha(E/\QQ)(p)$ of $\Sha(E/\Q)$ finite.
\end{conjecture}

We will make use of the following two unproved conjectures, which are
consequences of the Birch and Swinnerton-Dyer conjecture:

\begin{conjecture}[Parity Conjecture]\label{conj:parity}
  The parity of the rank of $E(\Q)$ is even if and only if the sign in
  the functional equation for $L(E,s)$ is $+1$.
\end{conjecture}

\begin{conjecture}[Squareness]\label{conj:square}
  For any prime $p$, the order of $\Sha(E/\Q)[p]$ is a perfect square.
\end{conjecture}

\begin{remark}
  Conjecture~\ref{conj:square} is implied by finiteness of
  $\Sha(E/\Q)(p)$, because of the Cassels-Tate pairing.  However, it
  is an a priori weaker statement, since we could have
  $\Sha(E/\Q)\ncisom (\Q_p/\Z_p)^2$, in which case
  $\#\Sha(E/\Q)[p]=p^2$, but $\Sha(E/\Q)$ is not finite.
  Conjecture~\ref{conj:square} is still very difficult; we only know
  it holds for curves with $\ord_{s=1} L(E,s)\leq 1$ and for specific
  examples of curves of rank $\geq 2$.
\end{remark}

We prove Conjecture~\ref{conj:finite}, assuming either
Conjecture~\ref{conj:parity} or Conjecture~\ref{conj:square}.
More specifically, we can prove the following three statements:
\begin{theorem}\label{thm}
\mbox{}\vspace{-1.5em}\\
\begin{enumerate}
\item \label{thm:a} Assuming Conjecture~\ref{conj:parity} or Conjecture~\ref{conj:square} for $p=2$.
Then there are infinitely many elliptic curves of rank exactly $2$ with $\Sha(E/\Q)(2)=0$.
\item Same as \ref{thm:a}, with $p = 2$ replaced by $p = 3$ (in both the hypothesis and conclusion).
\item Same as \ref{thm:a}, with rank exactly 2 replaced by rank exactly 3.
\end{enumerate}
\end{theorem}

We may also relax the assumptions somewhat; we only need the
assumptions for curves in the families we use for these average Selmer
results ($\cF_1$, elliptic curves with one marked point, and
$\cF_2$, elliptic curves with two marked points).



\section{Proofs}

\begin{proof}[Proof of part \label{thm:a} of Theorem~\ref{thm}]
We prove \ref{thm:a}.  The arguments for the other two are identical, using the appropriate Selmer averages.

In the family $\cF_1$ of elliptic curves with one marked point, we use
congruence conditions to construct a positive density family $\cF$ such
that half have root number $+1$ and half have root number $-1$.  
By \cite{}, the average cardinality of the $2$-Selmer group in this family is $6$.
(This average cardinality result is a key new input that makes it possible
to prove this theorem.)

Dokchitsers' result \cite{dokchitser:parity} states that the root
number gives the parity of the $p$-Selmer rank, i.e., root number +1
means even $p$-Selmer rank.  We can ensure that our curves satisfy
the hypothesis of Dokchiters (explain how). 
So in this case, half of the family has even 2-Selmer rank and half has odd.
(This is another key nontrivial new input that makes this result possible.)

Let $p_i$ denote the proportion of curves in $\cF$ with 2-Selmer rank
$i$.  We know (shown elsewhere) that $p_0 = 0$ (this is a statement about specialization
of rank for curves in a family).  We also have 
$$1/2 = p_1 + p_3 + p_5 + \cdots$$
and
$$1/2 = p_2 + p_4 + p_6 + \cdots.$$
Thus
$$6 = 2 p_1 + 4 p_2 + 8 p_3 + 16 p_4 + \cdots.$$
If $p_2 = 0$, then we would have
$$
  6 = 2 p_1 + 8 p_3 + 16 p_4 + \cdots \geq 2 \cdot 1/2 + 16 \cdot 1/2 = 9,
$$
which is a contradiction.  So $p_2 > 0$, i.e., a positive proportion of curves 
in $\F$ have 2-Selmer rank $2$ and root number $+1$.  (In fact, we have that $p_2 \geq 1/4$.)

Since 100\% of curves in $\F$ have rank $\geq $ 1 (as mentioned above), 
Conjecture~\ref{conj:parity} implies that 100\% of curves with 
2-Selmer rank 2 and root number $+1$ (so $p_2$ proportion of $\F$) 
must have algebraic rank $2$, hence for these curves $\Sha(E/\Q)[2]=0$.

Alternatively, assuming instead that $\Sha(E/\Q)[2]$ is a square also implies that 100\% 
of curves with 2-Selmer rank 2 in $\cF$ must have algebraic rank 2 and trivial $\Sha(E/\Q)[2]$.

Clearly (and this time it really is "clearly"!) trivial $\Sha(E/\Q)[2]$ implies
trivial $\Sha(E/\Q)(2)$, and a group of order 1 is finite!  So in fact, we have
produced a fairly large proportion of curves (out of all curves of
rank at least 1) with rank 2 and finite $\Sha(E/\Q)(2)$.
\end{proof}


\section{Other related thoughts and questions}

\begin{enumerate}
\item It seems unlikely that we could remove the conditional assumptions from Theorem~\ref{thm}, 
since then we would have produced a lot of rank 2 curves without knowing where the second point is coming from.

\item Can we vary $p$ instead of the elliptic curve?  Probably not, with these sorts of techniques.

\item Can we parametrize smaller families, e.g., the one $y(y+1) = x(x-1)(x+a)$, in order to get Selmer average results for them?

\item Can we parametrize families of elliptic curves where the root number is always $+1$ or $-1$?

\item Question: are we also (conditionally) showing that for a positive
proportion of all elliptic curves with rank $\geq 2$ that
$\Sha(E/\Q)(6)=0$?  That would be {\bf extremely surprising} (at
least, to me), even conditionally.

\end{enumerate}

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