\documentclass[12pt]{article}
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\title{Lecture 36: The Birch and Swinnerton-Dyer Conjecture, Part 3}
\begin{document}
\maketitle


\section{A Rationality Theorem}
In the last lecture, I mentioned that it can be surprisingly difficult to
say anything precise about $L(E,s)$, even with the above formulas.
For example, it is a very deep theorem of Gross and Zagier that for
the elliptic curve $y^2 + y = x^3 - 7x + 6$ we have 
$$
  L(E,s) = c(s-1)^3 + \text{ higher order terms},
$$
and nobody has any idea how to prove that there is an elliptic curve
with 
$$
  L(E,s) = c(s-1)^4 + \text{ higher order terms}.
$$  

Fortunately, it is possible to decide whether or not $L(E,1)=0$.
\begin{theorem}
Let $y^2 = x^3 + ax + b$ be an elliptic curve. 
Let 
$$
  \Omega_E = 2^n\int_{\gamma}^{\infty} \frac{dx}{\sqrt{x^3+ax+b}},
$$
where $\gamma$ is the largest real root of $x^3 +ax+b$, and
$n=0$ if $\Delta(E)<0$, $n=1$ if $\Delta(E)>0$.  
Then 
$$
  \frac{L(E,1)}{\Omega_E} \in \Q,
$$
and the denominator is $\leq 24$.
\end{theorem}
In practice, one computes $\Omega_E$ using 
the ``Arithmetic-Geometric Mean'', {\em NOT} numerical
integration.  In PARI, $\Omega_E$ 
is approximated by {\tt E.omega[1]*2\^{}(E.disc>0)}.

\begin{remark}
I don't know if the denominator is ever really as big as~$24$.
It would be a fun student project to either find an example,
or to understand the proof that the quotient is rational and
prove that $24$ can be replaced by something smaller.
\end{remark}


\begin{example}
Let $E$ be the elliptic curve $y^2 = x^3 - 43x + 166$. 
We compute $L(E,1)$ using the above formula and observe
that $L(E,1)/\Omega_E$ appears to be a rational number, as predicted by
the theorem. 
\begin{verbatim}
? E = ellinit([0,0,0,-43,166]);
? E = ellchangecurve(E, ellglobalred(E)[2]);
? eps = ellrootno(E)
%77 = 1
? N = ellglobalred(E)[1]
%78 = 26
? L = (1+eps) * sum(n=1,100, ellak(E,n)/n * exp(-2*Pi*n/sqrt(N)))
%79 = 0.6209653495490554663758626727
? Om = E.omega[1]*2^(E.disc>0)
%80 = 4.346757446843388264631038710
? L/Om
%81 = 0.1428571428571428571428571427
? contfrac(L/Om)
%84 = [0, 7]
? 1/7.0
%85 = 0.1428571428571428571428571428
? elltors(E)
%86 = [7, [7], [[1, 0]]]
\end{verbatim}
Notice that in this example, $L(E,1)/\Omega_E = 1/7 = 1/\#E(\Q)$.
This is shadow of a more refined conjecture of Birch and Swinnerton-Dyer.
\end{example}

\begin{example}
In this example, we verify that $L(E,1)=0$ computationally.
\begin{verbatim}
? E=ellinit([0, 1, 1, -2, 0]);
? L1 = elllseries(E,1)
%4 = -6.881235151133426545894712438 E-29
? Omega = E.omega[1]*2^(E.disc>0)
%5 = 4.980425121710110150642715583
? L1/Omega
%6 = 1.795732353252503036074927634 E-20
\end{verbatim}
\end{example}

\section{Approximating the Rank}
Fix an elliptic curve $E$ over~$\Q$.

The usual method to {\em approximate} the rank is to find a series
that rapidly converges to $L^{(r)}(E,1)$ for $r=0,1,2,3,\ldots$,
then compute $L(E,1)$, $L'(E,1)$, $L^{(2)}(E,1)$, etc., until
one appears to be nonzero.   You can read about this method in \S2.13
of Cremona's book {\em Algorithms for Elliptic Curves}.   For variety,
I will describe a slightly different method that I've played with
recently, which uses the formula for $L(E,s)$ from the last lecture,
the definition of the derivative, and a little calculus.

\begin{proposition}
Suppose that
$$
  L(E,s) = c(s-1)^r + \text{ higher terms}.
$$
Then 
$$ 
  \lim_{s\ra 1}\,
   (s-1)\cdot \frac{L'(E,s)}{L(E,s)} = r.
$$
\end{proposition}
\begin{proof}
Write 
$$
  L(s) = L(E,s) = c_r(s-1)^r + c_{r+1}(s-1)^{r+1} + \cdots.
$$
Then
\begin{align*}
 \lim_{s\ra 1} \, 
  (s-1)\cdot \frac{L'(s)}{L(s)}
   &=  \lim_{s\ra 1} \,
         (s-1)\cdot \frac{r c_r (s-1)^{r-1} + (r+1) c_{r+1}(s-1)^r + \cdots}
          {c_r(s-1)^r + c_{r+1}(s-1)^{r+1} + \cdots}\\
   &=  r\cdot \lim_{s\ra 1} \,
          \frac{c_r (s-1)^{r} + \frac{(r+1)}{r} c_{r+1}(s-1)^{r+1} + \cdots}
          {c_r(s-1)^r + c_{r+1}(s-1)^{r+1} + \cdots}\\
   &= r.
\end{align*}
\end{proof}

Thus the rank~$r$ is ``just''
the limit as $s\ra 1$ of a certain (smooth) function.  We know this
limit is an integer.  But, for example, for the curve 
$$
  y^2 +xy = x^3 - x^2 - 79x + 289
$$
nobody has succeeded in proving that this integer limit is $4$.  (One
can prove that the limit is either $2$ or $4$.)

Using the definition of derivative, we {\em heuristically} approximate 
$(s-1)\frac{L'(s)}{L(s)}$ as follows.  For $|s-1|$ small, we have
\begin{align*}
(s-1)\frac{L'(s)}{L(s)} &= 
    \frac{s-1}{L(s)}\cdot \lim_{h\ra 0}  \frac{L(s+h) - L(s)}{h}\\
    &\approx \frac{s-1}{L(s)}\cdot
             \frac{L(s+(s-1)^2) - L(s)}{(s-1)^2}\\
    &= \frac{L(s^2 - s+1) - L(s)}{(s-1)L(s)}
\end{align*}

\begin{question}
Does
$$
  \lim_{s\ra 1}\, (s-1)\cdot \frac{L'(s)}{L(s)} 
        = \lim_{s\ra 1} \frac{L(s^2 - s+1) - L(s)}{(s-1)L(s)}?
$$
\end{question}

\noindent In any case, we can use this formula in PARI to ``approximate''~$r$.
\begin{verbatim}
? E = ellinit([ 0, 1, 1, -2, 0 ]);
? r(E,s) = L1=elllseries(E,s); L2=elllseries(E,s^2-s+1); (L2-L1)/((s-1)*L1);
? r(E,1.01)
%8 = 2.004135342473941928617680057
? r(E,1.001)
%9 = 2.000431337547225544819319104
\\ One can prove that 2 is the correct limit.
\end{verbatim}

Now let's try the mysterious curve $y^2 +xy = x^3 - x^2 - 79x + 289$
of rank~$4$:
\begin{verbatim}
? E=ellinit([ 1,-1,0,-79,289]);
? r(E,1.001)         \\ takes 6 seconds on PIII 1Ghz
%1 = 4.002222374519085610896440642
? r(E,1.00001) 
%2 = 4.000016181256911064613006133
\end{verbatim}
It certainly looks like
$\lim_{s\ra 1} r(s) = 4$.
We know for a fact that 
$\lim_{s\ra 1} r(s)\in\Z$,
and if only there were a good way to bound the
error we could conclude that
the limit is~$4$.   But this has stumped people for
years, and maybe it is impossible without a very deep
result that somehow interprets this limit in a different
way.  This problem has totally stumped the experts for years.
We desperately need a new idea!!

If one of you wants to do a reading or research project on this
problem in the next year or two, let me know.  One could draw
pictures of $L^{(3)}(E,s)$ or investigate the analogous problem
for other more accessible $L$-series.


\end{document}


? E=ellinit([0,0,1,-7,6]);
? r(E,s) = L1=elllseries(E,s); L2=elllseries(E,s^2-s+1); (L2-L1)/((s-1)*L1);
? r(E,1.001)
%2 = 3.001144104985619206504448552