\documentclass[handout]{beamer}
\usepackage{amssymb,amsmath, amscd}
\usepackage{times, verbatim}
\usepackage{graphicx}
\usepackage{graphics}
\DeclareFontEncoding{OT2}{}{} % to enable usage of cyrillic fonts
  \newcommand{\textcyr}[1]{%
    {\fontencoding{OT2}\fontfamily{wncyr}\fontseries{m}\fontshape{n}%
     \selectfont #1}}
\newcommand{\Sha}{{\mbox{\textcyr{Sh}}}}
\usepackage{fancybox}

\DeclareMathOperator{\vol}{vol}
\DeclareMathOperator{\ord}{ord}
\DeclareMathOperator{\tor}{tor}
\DeclareMathOperator{\rank}{rank}
\DeclareMathOperator{\Reg}{Reg}
\newcommand{\Q}{\mathbb{Q}}

\usepackage{color}
\usepackage{listings} 
\lstdefinelanguage{Sage}[]{Python}
{morekeywords={True,False,sage,singular},
sensitive=true}
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  basicstyle={\small \ttfamily},
  aboveskip=1em,
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\newcommand{\gray}{\color{graycolor}}

\title{The Birch and Swinnerton-Dyer Conjecture}
\author{Benedict Gross and William Stein} 

\date{January 6, 2012 in Boston, MA} 
\begin{document} 
\maketitle 

\begin{frame}[fragile]

\frametitle{Algebraic equations}

\begin{center}
{\includegraphics[height=.4\textheight]{triangle3.pdf}}

\pause
\medskip

{\includegraphics[height=.2\textheight]{pythag.png}}
\end{center}

Pythagoras (600 BCE) \pause ~~~~ Baudh\=ayana (800 BCE)



\end{frame}


\begin{frame}[fragile]
\frametitle{Pythagorean triples}

\medskip

$a^2 + b^2 = c^2$ has solutions $(3,4,5), (5,12,13), (7,24,25),\ldots$

\medskip
\pause
There are more solutions on a Babylonian tablet (1800 BCE):
\medskip

\pause
\begin{tabular}{lr}
\raisebox{-.45\height}{\includegraphics[width=0.7\textwidth]{plimpton.png}}&
\footnotesize 
$\begin{array}{|c|}\hline
\vspace{-2ex}\\
( 3, 4, 5 )\\
( 5, 12, 13 )\\
( 7, 24, 25 )\\
( 9, 40, 41 )\\
( 11, 60, 61 )\\
( 13, 84, 85 )\\
( 15, 8, 17 )\\
( 21, 20, 29 )\\
( 33, 56, 65 )\\
( 35, 12, 37 )\\
( 39, 80, 89 )\\
( 45, 28, 53 )\\
( 55, 48, 73 )\\
( 63, 16, 65 )\\
( 65, 72, 97 )\\
\hline
\end{array}$
\end{tabular}

\end{frame}



\begin{frame}
\frametitle{The general solution of $a^2+b^2=c^2$}



$x = a/c$ and $y = b/c$ satisfy the equation $x^2 + y^2 = 1$


\pause



{\includegraphics[width=.5\textwidth]{param3.pdf}}

\pause


$$ t = \frac{y}{1+x} \pause \qquad\qquad x = \frac{1-t^2}{1+t^2}\qquad y = \frac{2t}{1+t^2}$$
\end{frame}

\begin{frame}



Write $ t = p/q$. Then

$$x = \frac{q^2 - p^2}{q^2 + p^2}\qquad\qquad y = \frac{2qp}{q^2 + p^2}$$

\pause

\medskip

$$a = q^2 - p^2 \qquad\qquad b = 2qp \qquad\qquad c = q^2 + p^2$$

\pause 
$$t = 1/2 \longrightarrow (a,b,c) = (3,4,5)$$


$$t = 2/3 \longrightarrow (a,b,c) = (5,12,13)$$


$$t = 3/4 \longrightarrow (a,b,c) = (7,24,25)$$


\end{frame}

\begin{frame}
\frametitle{Cubic equations}

After linear and quadratic equations come cubic equations, like

$$x^3 + y^3 = 1 \qquad\qquad y^2 + y = x^3 - x$$

\pause

Here there may be either a finite or an infinite number of rational solutions.

\pause

\begin{center}
\includegraphics[width=.4\textwidth]{fermat.jpg}
\end{center}
\end{frame}

\begin{frame}
\frametitle{The graph}

$$y^2 + y = x^3 - x$$

\begin{center}
%\only<1>{\includegraphics[width=.6\textwidth]{secant1.pdf}}%
%\only<2>{\includegraphics[width=.6\textwidth]{secant2.pdf}}%
%\only<3>{
\includegraphics[width=.6\textwidth]{secant3.pdf}%}
\end{center}




\end{frame}

\begin{frame}
\frametitle{The limit of a secant line is a tangent}

$$y^2 + y = x^3 - x$$

\begin{center}
\includegraphics[width=.6\textwidth]{tangent.pdf}
\end{center}

\end{frame}

\begin{frame}[fragile]
\frametitle{Large solutions}

If the number of solutions is infinite, they quickly become large.
$$y^2 + y = x^3 - x$$
\vspace{-10ex}


\scriptsize
\begin{verbatim}
(0, 0)
(1, 0)
(-1, -1)
(2, -3)
(1/4, -5/8)
(6, 14)
(-5/9, 8/27)
(21/25, -69/125)
(-20/49, -435/343)
(161/16, -2065/64)
(116/529, -3612/12167)
(1357/841, 28888/24389)
(-3741/3481, -43355/205379)
(18526/16641, -2616119/2146689)
(8385/98596, -28076979/30959144)
(480106/4225, 332513754/274625)
(-239785/2337841, 331948240/3574558889)
(12551561/13608721, -8280062505/50202571769)
(-59997896/67387681, -641260644409/553185473329)
(683916417/264517696, -18784454671297/4302115807744)
(1849037896/6941055969, -318128427505160/578280195945297)
(51678803961/12925188721, 10663732503571536/1469451780501769)
(-270896443865/384768368209, 66316334575107447/238670664494938073)
\end{verbatim}
% (4881674119706/5677664356225, -8938035295591025771/13528653463047586625)
% (-16683000076735/61935294530404, -588310630753491921045/487424450554237378792)
% (997454379905326/49020596163841, -31636113722016288336230/343216282443844010111)
% (2786836257692691/16063784753682169, -435912379274109872312968/2035972062206737347698803)
% (213822353304561757/158432514799144041, 41974401721854929811774227/63061816101171948456692661)
% (-3148929681285740316/2846153597907293521, -2181616293371330311419201915/4801616835579099275862827431)
% (79799551268268089761/62586636021357187216, -754388827236735824355996347601/495133617181351428873673516736)


\end{frame}

\begin{frame}
\frametitle{Even the simplest solution can be large (Watkins)}
\begin{center}
$\displaystyle{}y^2 + y = x^3 - 5115523309x - 140826120488927$
\end{center}

Numerator of $x$-coordinate of smallest solution (5454 digits):
\begin{center}
\includegraphics[width=\textwidth, height=.3\textheight]{numer.png}
\end{center}
Denominator:

\begin{center}
\includegraphics[width=\textwidth, height=.3\textheight]{denom.png}
\end{center}
\end{frame}



\begin{frame}
\frametitle{The rank}

The rank of $E$ is essentially the number of independent solutions.
\pause
\medskip
\begin{itemize}

\item rank $(E) = 0$ means there are finitely many solutions.

\medskip
\pause

\item rank $(E) > 0$ means there are infinitely many solutions.

\medskip
\pause

\item The curve $E(a)$ with equation  

$$y(y+1) = x(x-1)(x+a)$$

has rank $= 0,1,2,3,4$ for $a=0,1,2,4,16$. % (N = 11, 37, 389, 5077, ?).  


\end{itemize}

%\begin{center}
%\includegraphics[height=.6\textheight]{pics/Ea}
%\end{center}


% Records:
%
% 0 0
% 1 1
% 2 2
% 3 4
% 4 16
% 5 79
% 6 298
%

\end{frame}

\begin{frame}
\frametitle{The rank is finite}

\begin{center}
\includegraphics[height=.45\textheight]{mordell}
\hspace{3em}\includegraphics[height=.45\textheight]{weil}
\end{center}

\pause
\medskip

Can it be arbitrarily large?


\end{frame}

\begin{frame}
\frametitle{The current record is rank($E$) $\geq 28$}

\vspace{1ex}

{\tiny
$
y^2 + xy + y = x^3 - x^2 -
20067762415575526585033208209338542750930230312178956502x +$\vspace{-1.1ex}\\
\hfill$344816117950305564670329856903907203748559443593191803612660082962919394\
48732243429$}

\vspace{-2ex}

\begin{center}
\includegraphics[height=0.7\textheight]{egens.png}
\includegraphics[width=.2\textwidth]{elkies.png}
\end{center}




\end{frame}

\begin{frame}

  Peter Swinnerton-Dyer and Bryan Birch made a prediction for the
  rank, based on the average number of solutions at prime numbers
  $p$.

\begin{center}
\includegraphics[height=.8\textheight]{weddingphoto.png}
\end{center}

\end{frame}

\begin{frame}

\frametitle{Primes}
\medskip

A prime $p$ is a number greater than $1$ that is not divisible by any smaller number.
\pause
\medskip

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67,
71, 73, 79, 83, 89, 97, 101, 103, 107, 109, $\ldots$

\medskip
\pause
There are infinitely many primes. The largest explicit prime known is $2^{43112609} - 1$ with 12,978,189 digits.

\pause 
\medskip

\begin{center}
\includegraphics[width=.8\textwidth]{primepi.pdf}
\end{center}

\end{frame}

% \begin{frame}[fragile]
% \frametitle{The Prime Number $2^{9689}-1=$}
% {\tiny
% \begin{verbatim}
% 478220278805461202952839298660005909741497172402236500851334510991837895094266297027892768
% 611270789458682472098152425631930658505267683408748083442943326479742589324762368833102163
% 320895484735480579994334130982598901374380618710958104314868081377832153049671560156328262
% 441404039814320762203627219040859079053720347525610556407157926386787524098557335652265610
% 854212857732105787905232886503535587361567936365588992571157442015383209175242284304691881
% 142740066213555930351685370397681268638575037622778794958058208183126172570100349820651232
% 987267723348951095346937568303703837399969677158578890563911552261340549570718452415821920
% 822376644205901459333065700972215396237685342377048613857808977562130116781129916640736174
% 660669780818675796691467124607371290420058840892318638773788767529288695379706698096740605
% 353012285353903696549022478492464900795489867850331465554647550450168618735486696437455261
% 412064078294962245202778896213860266593314768769632208950427879162465151931232783175655377
% 937719452467339581928148666857638401959072017941334958297031939388438881049454604034208753
% 656362833215207318161430072176937142623851754052084521466531330118355196259184955893849902
% 534878037671647707393063443684008446825593744345169031599934913766463896897261419901530490
% 654781905622717122494707073971630095377574344130792050186353223446654564569577433188504497
% 825014866346737213039209989485214519099823287877248665051301081676990289251871925006694721
% 570653621624869624056925686555429622155221156042777866254593699880107018616260147647429345
% 983018365127336346273267588306070141035925482914977433929717368076561095959991130918978823
% 835013163567266143596921823997719693387439540399662367558052821120713639637085805605116078
% 177098545257698803233381293927275210194462952749031383555198519709592888523641530178921867
% 514101454120309619127093436903952209828031766894206132557234964363840305648734929088422378
% 629288747223121903238528103409182430661894774072726552428489330447486145494207679904173944
% 716583828167141043583120679050191452732628737033997470720601688256282740427017032260672798
% 034347932642573009183981307771932245539476396060658821432660315614149074055769805516626304
% 444758375671151649018119344223685942415184379538933576543212994405485534515585927342456182
% 514681371472060628778102124092370802149229834963517952727030296297015692768651163505008040
% 728267425236264469571076976886613730278931360967438271901738550848466337347612084356798306
% 505955807293511063754424080735066708298723377976887493898358452309563899612061631863439196
% 711208646438464947096323007272920091258614726799976249670985276950353573392441620265772074
% 124868359220282898331114083392330243391779797699031142584361935093675448381119440881276338
% 808420445180491245438388418080094527562666805762895476338464130510775377324708249580453335
% 571748196502507081973046642282610569751056428979895118219288597635222905389894873761464213
% 9910911535864505818992696826225754111
% \end{verbatim}
% }

%\end{frame}

% \begin{frame}
% \frametitle{Primality testing}

% Determining that $n > 1$ is a prime can be done quickly.

% \pause
% \medskip
% \begin{center}
% "PRIMES is in P" 
% \end{center}
% \medskip
% AKS: Manindra Agrawal, Neeraj Kayal, and Nitin Saxena (2002) 

% \pause
% \medskip

% If $n$ fails the primality test, it is more difficult to factor it.

% \pause
% \medskip

% 123018668453011775513049495838496272077285356959533
% 479219732245215172640050726365751874520219978646938
% 995647494277406384592519255732630345373154826850791
% 702612214291346167042921431160222124047927473779408
% 0665351419597459856902143413 $=$ RSA-$768$ $=$

% \pause
% \medskip

% 334780716989568987860441698482126908177047949837137
% 685689124313889828837938780022876147116525317430877
% 37814467999489\\
% $\times$
% 367460436667995904282446337996279526322791581643430
% 876426760322838157396665112792333734171433968102700
% 92798736308917


% \end{frame}



\begin{frame}

What do we mean by a solution of the cubic equation at the prime number $p$?

\pause
$$y^2 + y = x^3 - x$$

$(x,y) \equiv (3,1)$  is a solution at $p = 11$

\pause
\medskip

There are finitely many solutions $A(p)$ at each prime $p$.

\pause

\begin{center}
\includegraphics[width=.47\textwidth]{A23.pdf}
\hfill
\includegraphics[width=.47\textwidth]{A71.pdf}



\end{center}


\end{frame}

\begin{frame}


It is common to write

$$\#\{\text{solutions mod $p$}\} = A(p) = p + 1 - a(p)$$

\pause

We define the $L$-function of $E$ by the infinite product

$$L(E,s) = \prod_{\text{all }p} (1 - a(p)p^{-s} + p^{1-2s})^{-1} = \sum a(n)n^{-s}$$
\pause

This definition only works in the region $s > 3/2$, where the infinite product converges.

\pause

\begin{center}
\includegraphics[width=.85\textwidth]{lplot32.pdf}
\end{center}

\end{frame}

\begin{frame}

If we formally set $s=1$ in the product, we get

$$\prod_{\text{all }p}(1 - a(p)p^{-1} + p^{-1})^{-1} = \prod_{\text{all }p} \frac{p}{A(p)}$$

\pause
\medskip

If $A(p)$ is large on average compared with $p$, this will approach
0. The larger $A(p)$ is on average, the faster it will tend to 0.

\pause
\medskip

%\begin{center}
%{\scriptsize
%\begin{tabular}{|r|r|r|r|}\hline
 %$p$& $A(p)$ & $\frac{p}{A(p)}$  & $\prod \frac{p}{A_p}$\\\hline
  %23 & 33 &  0.697 &   0.697\\
  %29 & 40 &  0.725 &  0.505\\
  %31 & 40 &  0.775 &  0.392\\
  %37 & 49 &  0.755 &  0.296\\
  %41 & 52 &  0.788 &  0.233\\
  %43 & 56 &  0.768 &  0.179\\
  %47 & 60 &  0.783 &  0.140\\
  %53 & 63 &  0.841 &  0.118\\\hline
%  59 & 72 &  0.819 &  0.097\\
%  61 & 77 &  0.792 &  0.077\\
%  67 & 84 &  0.798 &  0.061\\\hline
%\end{tabular}}
%\end{center}

\includegraphics[width=\textwidth]{Approd.pdf}
\end{frame}


\begin{frame}

\frametitle{The conjecture of Birch and Swinnerton-Dyer}



\begin{enumerate}
\item The function $L(E,s)$ has a natural (analytic) continuation to a neighborhood of $s = 1$.

\pause
\medskip

\item The order of vanishing of $L(E,s)$ at $s =1$ is equal to the rank of $E$.


\pause
\medskip


\item The leading term in the Taylor expansion of $L(E,s)$ at $s=1$ is given by certain arithmetic invariants of $E$.
\end {enumerate}
\medskip
$$L(E,s) = c(E)(s-1)^{\rank(E)} + \cdots $$

\end{frame}

\begin{frame}

The most mysterious arithmetic invariant was studied by John Tate and Igor Shafarevich, who conjectured that it is finite. Tate called this invariant $\Sha$.

\pause

\begin{center}
\includegraphics[height=.45\textheight]{tate.png}
\hspace{3em}\includegraphics[height=.45\textheight]{shafarevich.png}
\end{center}

\

%\begin{center}
%\includegraphics[width=.3\textwidth]{tate.png}
%\end{center}

\end{frame}


  

%\pause
%\begin{center}
%\includegraphics[width=.35\textwidth]{shafarevich.png}
%\end{center}


\begin{frame}
\frametitle{The Birch and Swinnerton-Dyer Conjecture}
\Large
\vspace{-1em}
\begin{align*}
L(E,s) &= c(E) (s-1)^{\rank(E)}+ \cdots \\
c(E) &=
\frac{\Omega_E \cdot \Reg_E \cdot \#\Sha_E \cdot \prod c_p}{\#E(\Q)_{\tor}^2}
\end{align*}


\large
Each quantity on the right measures the size of 
an abelian group attached to $E$.
\vfill

\begin{center}
\includegraphics[width=.6\textwidth]{bsd}
\end{center}

\end{frame}


\begin{frame}
\frametitle{Natural (analytic) continuation}
\mbox{}\vspace{-5em}

The infinite sum $\sum_{n=0}^{\infty}x^n$ converges when $-1 < x < 1$.
\pause

\begin{center}
\includegraphics[width=.7\textwidth]{continue}
\vspace{-10em}

$\displaystyle \frac{1}{1-x} = \sum_{n=0}^{\infty} x^n\qquad\qquad\qquad$
\end{center}


\end{frame}

\begin{frame}
\frametitle{Fermat's Last Theorem}

The natural (analytic) continuation of $L(E,s) = \sum a(n)n^{-s}$ was obtained (in most cases) by Andrew Wiles and Richard Taylor (1995).\pause~ They proved that the function defined by the infinite series

$$F(\tau) = \sum a(n) e^{2\pi i n \tau}$$

is a modular form.

\medskip

\pause

\begin{center}
\hfill
\includegraphics[height=.45\textwidth]{wiles.png}
\hfill
\includegraphics[height=.45\textwidth]{taylor.png}
\hfill
\mbox{}
\end{center}



\end{frame}

\begin{frame}

  Combining a limit formula of Benedict Gross and Don Zagier (1983)
  with work of Victor Kolyvagin (1986) we can now show the following.

\medskip
\pause

If $L(E,1) \neq 0$ the rank is zero, so there are finitely many solutions.

\medskip
\pause

If $L(E,1) = 0$ and $L'(E,1) \neq 0$ the rank is one, so there are infinitely many solutions.

\medskip
\pause

In both cases, we can also show that $\Sha$ is finite.

\pause

\begin{center}
\includegraphics[height=.4\textwidth]{india.png}
\hfill
\includegraphics[height=.4\textwidth]{zagier.png}
\hfill
\includegraphics[height=.4\textwidth]{kolyvagin.png}
\hfill
\end{center}

\end{frame}



\begin{frame}[fragile]
\frametitle{Example: A Rank 1 Curve}
\begin{lstlisting}
sage: E = EllipticCurve([0,0,1,-1,0])
sage: L = E.lseries()
sage: Lser = L.taylor_series(); Lser
0.305999773834052*z + 0.186547797268162*z^2 + ...
sage: c = Lser[1]; c
0.305999773834052
sage: Omega_E = E.period_lattice().omega(); Omega_E
5.98691729246392
sage: Reg_E = E.regulator(); Reg_E
0.0511114082399688
sage: prod_cp = E.tamagawa_product_bsd(); prod_cp 
1
sage: T = E.torsion_order()^2; T 
1
sage: c / (Omega_E * Reg_E * prod_cp / T^2)
1.00000000000000
\end{lstlisting}

\end{frame}




\begin{frame}[fragile]
\frametitle{Example: A Rank 2 Curve}
\begin{lstlisting}
sage: E = EllipticCurve([0,1,1,-2,0])
sage: L = E.lseries()
sage: Lser = L.taylor_series(); Lser
-2.69e-23 + (1.52e-23)*z + 0.75931650028*z^2 + ...
sage: E.rank()
2
sage: c = Lser[2]
sage: Omega_E = E.period_lattice().omega()
sage: Reg_E = E.regulator()
sage: prod_cp = E.tamagawa_product_bsd()
sage: T = E.torsion_order()^2
sage: c / (Omega_E * Reg_E * prod_cp / T^2)
1.00000000000000
sage: S = E.sha(); S
Tate-Shafarevich group ...
sage: S.p_primary_bound(5)  
0
\end{lstlisting}

\vfill

{\bf Open Problem:} {\em Prove that $\Sha(E)$ is finite.}

\end{frame}


\begin{frame}[fragile]
\frametitle{Example: A Rank 4 Curve}
\begin{lstlisting}
sage: E = EllipticCurve([0,15,1,-16,0])
sage: L = E.lseries()
sage: Lser = L.taylor_series(); Lser
4.32e-24 + (-1.96e-23)*z + (2.05e-22)*z^2 
         + (-7.97e-22)*z^3 + 10.84*z^4 - ...
sage: E.rank()
4
sage: c = Lser[4]
sage: Omega_E = E.period_lattice().omega()
sage: Reg_E = E.regulator()
sage: prod_cp = E.tamagawa_product_bsd()
sage: T = E.torsion_order()^2
sage: c / (Omega_E * Reg_E * prod_cp / T^2)
0.99999999999999
\end{lstlisting}
\vfill

{\bf Open Problem:} {\em Prove that $\ord_{s=1} L(E,s)=4$.}
\end{frame}



\begin{frame}
\frametitle{Thank you}

\begin{center}
\includegraphics[height=.9\textheight]{stein_invert}
\end{center}

\end{frame}



% \begin{frame}

% When the order of $L(E,s)$ at $s = 1$ is greater than one we cannot prove anything in general...

% \pause
% \medskip
% The computer has been a great guide.

% \pause

% \medskip


% Here is a summary of the evidence for the simplest rank 2 curve

% $$y(y+1) = x(x-1)(x+2)$$

% \pause
% \begin{itemize}
% \item the order of vanishing is equal to 2 
% \item most primes up to 50,000 do not divide the order of $\Sha$ 
% % no ordinary prime up to 48,859 divides the order of group appearing in leading term $L"(E,1)$(excluding possibly $p=16231$).
% (See the new paper [Stein-Wuthrich].)
% \end{itemize}

% %\pause

% %\begin{center}
% %\includegraphics[width=.5\textwidth]{wstein\.png}
% %\end{center}

% \end{frame}




% \begin{frame}
% \frametitle{The average rank}

%  Manjul Bhargava has recently made progress on the study of the average rank, for ALL cubic
% curves with rational coefficients.

% \pause
% \begin{center}
% \includegraphics[height=.7\textheight]{manjul.png}
% \end{center}

% \end{frame}

% \begin{frame}

% \frametitle{Enumerating the curves}

% \medskip
% \begin{itemize}
% \item Every such curve has a unique equation of the form $y^2 = x^3 + Ax + B$ where $A$ and $B$ are integers (not divisible by $p^4$ and $p^6$, for any prime $p$).
% \pause
% \medskip
% \item Define the height $H(E)$ as the maximum of the positive integers $|A|^3$ and $|B|^2$.
% \pause
% \medskip
% \item For any positive real number $X$, there are only finitely many curves with $H(E) \leq X$. 
% \pause
% \medskip
% \item Call this number $N(X)$. It grows at the same rate as $(X)^{1/2}(X)^{1/3} = X^{5/6}$.
% \end{itemize}

% \end{frame}


% \begin{frame}

% \begin{itemize}
% \item Define the average rank by the limit as $X \rightarrow \infty$ of $$\frac{1}{N(X)} \sum_{H(E)\leq X} rank(E)$$
% \pause
% \medskip
% \item We suspect that this limit exists, and is equal to $1/2$.
% \pause
% \medskip  
% \item In fact, we think that on average half the curves have rank zero and half have rank one.
% \pause
% \medskip
% \item Bhargava and Shankar have shown why there is an upper bound on the limit, and have obtained a specific upper bound which is less than $1$.
% \end{itemize}

% \end{frame}


\end{document}