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\title{Lecture 35: The Birch and Swinnerton-Dyer Conjecture, Part 2}
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\section{The BSD Conjecture}
Let $E$ be an elliptic curve over~$\Q$ given by an equation
$$
  y^2 = x^3 + ax + b
$$
with $a,b\in\Z$.  For $p\nmid \Delta = -16(4a^3 + 27b^2)$,
let 
$a_p = p+1 - \# E(\Z/p\Z)$.
Let
$$
  L(E,s) = \prod_{p\nmid\Delta} \frac{1}{1-a_p p^{-s} + p^{1-2s}}.
$$

\begin{theorem}[Breuil, Conrad, Diamond, Taylor, Wiles]\mbox{}\\
\mbox{}\hspace{5em}$L(E,s)$ extends to an analytic function on all of $\C$.
\end{theorem}

\begin{conjecture}[Birch and Swinnerton-Dyer]
The Taylor expansion of $L(E,s)$ at $s=1$ has the form
$$
  L(E,s) = c(s-1)^r + \text{\rm  higher order terms}
$$
with $c\neq 0$ and $E(\Q)\ncisom \Z^r \cross E(\Q)_{\tor}$.
\end{conjecture}

A special case of the conjecture is the assertion that $L(E,1)=0$ if
and only if $E(\Q)$ is infinite.  The assertion ``$L(E,1)=0$ implies
that $E(\Q)$ is infinite'' is the part of the conjecture that secretely
motives much of my own research.

\section{What is Known}
On page 5 of Wiles's paper, he discusses the history
of the following theorem.
\begin{theorem}[Gross,  Kolyvagin, Zagier, et al.]
Suppose that 
$$
  L(E,s) = c(s-1)^r + \text{\rm  higher order terms}
$$
with $r\leq 1$.  Then the Birch and Swinnerton-Dyer conjecture
is true for~$E$, that is, 
$E(\Q)\ncisom \Z^r\oplus E(\Q)_{\tor}.$
\end{theorem}

I suspect that most elliptic curves satisfy the hypothesis of the
above theorem, i.e., they have rank $0$ or $1$.  For example, almost
96\% of the ``first $78198$'' elliptic curves have $r\leq 1$.  I
suspect that the curves with $r>1$ have ``density'' $0$ amongst all
elliptic curves.  This doesn't mean that we are done.  In practice it
is often the curves with $r>1$ that are interesting and useful, and
experts can still be observed saying ``almost nothing is known about the
Birch and Swinnerton-Dyer conjecture''.

%The Birch and Swinnerton-Dyer conjecture has inspired the creation of
%a vast web of conjectures by the likes of Beilinson, Deligne, Mazur,
%Tate, Bloch, Kato, Perrin-Riou, and others.

\section{How to Compute $L(E,s)$ with a Computer}
\subsection{Best Models}
Let~$E$ be an elliptic curve over~$\Q$, defined by
a Weierstrass equation
$$
  y^2 + a_1 xy + a_3 y = x^3 + a_2 x^2 + a_4 x + a_6.
$$
There are many choices of Weierstrass equations that
define an elliptic curve that is ``essentially the same''
as~$E$.  E.g., you found others by completing the square.
Among all of these, there is a best possible model, which
is the one with smallest discriminant.  It can be computed
in PARI as follows:
\begin{verbatim}
? E = ellinit([0,0,0,-43,166]);
? E.disc
%61 = -6815744
? E = ellchangecurve(E,ellglobalred(E)[2])
%62 = [1, -1, 1, -3, 3, ...]
? E.disc
%63 = -1664
\end{verbatim}
Thus $y^2 + xy +y = x^3 -x^2 -3x +3$ is a ``better''
model than $y^2 = x^3 - 43x + 166$.  

\hd{WARNING:} Some of the elliptic curves functions in PARI will {\em
LIE} if you give as input an elliptic curve that is defined by a model
that isn't the best possible.  These devious liars include 
{\tt elltors, ellap, ellak}, and {\tt elllseries}.

\subsection{Formula for $L(E,s)$}
As mentioned before, the PARI function {\tt elllseries} 
can compute $L(E,s)$.   I figured out how this function works, and
explain it below.

Because~$E$ is modular, one can show that we have the following
rapidly-converging series expression for $L(E,s)$, for $s>0$:
$$
L(E,s) = N^{-s/2}\cdot (2\pi)^s\cdot  \Gamma(s)^{-1}\cdot
\sum_{n=1}^{\infty} a_n \cdot \left(F_n(s-1) - \eps F_n(1-s)\right)
$$
where
$$
  F_n(t) = 
           \Gamma\left(t+1, \frac{2\pi n}{\sqrt{N}}\right)
\cdot \left(\frac{\sqrt{N}}{2\pi n}\right)^{t+1}.
$$
Here 
$$
  \Gamma(z) = \int_{0}^{\infty} t^{z-1} e^{-t} dt
$$
is the {\em $\Gamma$-function} (e.g., $\Gamma(n) = (n-1)!$),
and 
$$
  \Gamma(z,\alpha) = \int_{\alpha}^{\infty} t^{z-1} e^{-t}dt
$$
is the {\em incomplete $\Gamma$-function}.
The number~$N$ is called the {\em conductor} of~$E$ and is
very similar to the discriminant of~$E$; it is only divisible
by primes that divide the best possible discriminant of~$E$.  
You can compute~$N$ using the PARI command {\tt ellglobalred(E)[1]}.

As usual, for $p\nmid \Delta$, we have
$$
  a_p = p+1 - \# E(\Z/p\Z),
$$
and for $r\geq 2$,
$$
  a_{p^r} = a_{p^{r-1}}a_p - p a_{p^{r-2}},
$$
and
$a_{nm} = a_n a_m$ if $\gcd(n,m)=1$
(I won't define the $a_p$ when $p\mid \Delta$, but it's
not difficult.)
Finally, $\eps$ depends only on~$E$ and is either $+1$ or $-1$.
I won't define $\eps$ either, but you can compute it in PARI
using {\tt ellrootno(E)}.

At $s=1$, the formula can be massively simplified, and we have
$$
  L(E,1) = (1+\eps) \cdot \sum_{n=1}^{\infty} 
              \frac{a_n}{n} e^{-2\pi n/\sqrt{N}}.
$$
This sum converges rapidly, because
$e^{-2\pi n/\sqrt{N}}\ra 0$ quickly as $n\ra \infty$.


Monday's lecture will be filled with numerical examples and numerical
evidence for the Birch and Swinnerton-Dyer conjecture.   Wednesday's
lecture will be a review for the take-home {\bf FINAL EXAM}.




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