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\begin{document}
\title{Lecture VI\\ Curves of genus 0 and 1}
\author{Kapil Hari Paranjape}
\maketitle
We study the ``simplest'' possible curves. Since we have declared
them to be simple we can ask more difficult questions about them!
The same algebraic variety may come in many different guises so an
interesting problem is to ``recognise'' algebraic varieties. Thus the
curve $\bbP^1$ also appears as the curve in $\bbP^2$ defined by the
equation $XY-Z^2=0$; for that matter there is also the rational
normal curve of degree $d$ in $\bbP^d$. It is an interesting exercise
to show that a non-degenerate (doesn't lie in a hyperplane)
irreducible curve of degree $d$ in $\bbP^d$ is isomorphic to $\bbP^1$
\emph{once} one is given a point on it. Thus the existence of rational
points is quite important in recognising varieties. For example the
curve in $\bbP^2$ defined by $X^2+Y^2+Z^2=0$ (in characteristic different
from 2) is isomorphic to the curve defined by $XY=Z^2$ after a linear
change of co-ordinates in a field where the first curve has a point.
As a consequence of the Riemann-Roch theorem for curves \emph{any}
smooth irreducible curve of genus 0 can be realised as a conic in
$\bbP^2$; thus, a very interesting number-theoretic question is to try
to understand which non-algebraically closed fields $k$ have the property
that a conic over this field always has a point with co-ordinates in
this field. As an exercise you can examine the case when $k$ is
finite.
Let us move on to curves of genus 1. Such curves come in a number of
different guises. The most classical is the Tate-Weierstrass form
\[
y^2 + a_1 x y + a_3 y = x^3 + a_2 x^2 + a_4 x + a_6
\]
As usual, we will ``complete'' this curve to make it projective by
taking the equation
\[
Y^2 Z + a_1 X Y Z + a_3 Y Z^2 =
X^3 + a_2 X^2 Z + a_4 X Z^2 + a_6 Z^3
\]
which defines a curve in $\bbP^2$; we note that $(0:1:0)$ is a point
of this curve which is sometimes called the origin and sometimes
called the point at infinity.
However, a curve of genus 1 can come in a number of other guises. For
example, we can increase the degree of the right-hand side in the
above equation to obtain
\[
y^2 + a_1 x y + a_2 y = x^4 + b_1 x^3 + b_2 x^2 + b_3 x + b_4
\]
We again need to ``projectivise'' which we do by considering the
equation
\[
Y^2 + a_1 W X Y + a_2 W^2 Y =
X^4 + b_1 X^3 W + b_2 X^2 W^2 + b_3 X W^3 + b_4 W^4
\]
in the homogeneous polynomial ring $k[W,X,Y]$ where $W$ and $X$ have
degree 1 and $Y$ has degree $2$. Equivalently, we can make the
substitutions $X^2=T$, $WX=U$ and $W^2=V$ to obtain the equation
\[
Y^2 + a_1 U Y + a_2 V Y =
T^2 + b_1 T U + b_2 T V + b_3 U V + b_4 V^2
\]
in the homogeneous co-ordinate ring $k[T,U,V,Y]/(U^2-TV)$.
We note that the latter curve can be thought of the the curve defined
by two quadratic equations $Q_i(T,U,V,Y)=0$ for $i=1,2$. So that is
yet another guise in which curves of genus 1 appear.
In fact, unlike that case of curves of genus 0, curves of genus 1
can appear in infinitely many different guises, one for each integer
$n>1$. The situation becomes much simpler however, if we assume that $k$
is algebraically closed or more simply that there is at least one point
(in other words if the object of our study is not pointless!). If a
smooth irreducible curve of genus 1 \emph{has} as a point with
co-ordinates in the field $k$ then all these different disguises collapse
and only the Weierstrass form remains. The reason is the Riemann-Roch theorem
once again. An elliptic curve is defined to be a curve $E$ of genus 1
with a chosen point $o$.
The Tate-Weierstrass form of the elliptic curve with the specially
chosen point $(0:1:0)$ has another special property. We will show
that there is a natural commutative group law on this curve which
thus becomes a projective group variety. Consider a pair of points
$(p,q)$ and $(r,s)$ on the curve and consider the parametric line
given by $t(p,q)+(1-t)(r,s)$ that passes through these points.
Substituting this in the equation
\[
y^2 + a_1 x y + a_3 y = x^3 + a_2 x^2 + a_4 x + a_6
\]
we obtain a cubic polynomial equation $f(t)=0$
with coefficients in the ring $k[a_1,a_2,a_3,a_4,a_6]$. We further
note that $t=0$ and $t=1$ are solutions of $f(t)=0$ so that
$f(t)/(t^2-t)$ is a linear polynomial. This gives us a solution
for $t$ which we substitute in order to obtain the point $(u,v)$
which is the third point on the line. It is clear that the line $x=u$
also contains the point $(u,a_1 u + a_3 - v)$. The group law on the
curve is defined by
\[
(p,q)\oplus (r,s) = (u, a_1 u + a_3 - v)
\]
One has to also take the special cases where $p=r$ into account.
The appropriate formulae can be written down in those cases as well.
It is probably easier to understand the group law more geometrically
via Riemann-Roch. Let $D$ be any divisor on the elliptic curve $E$
and consider the divisor $D-(\deg(D)-1)o$. The latter is a divisor of
degree 1. From Riemann-Roch it follows that this divisor is linear
equivalent to an effective divisor of degree 1; in other words, there
is a point $p(D)$ such that $D$ is linearly equivalent to the divisor
$(\deg(D)-1)o+p(D)$. Thus the group structure on divisors modulo
linear equivalence becomes a group structure on points of $E$ with
$o$ playing the role of identity.
Since we have an abelian group, it is natural to ask whether it has
torsion subgroups for various orders. We will approach this problem
from a slightly different viewpoint. Let $O$ and $I$ be conics. Pick a
point $A_0$ of $O$ and pick one line $L_0$ out of the two lines that
contain $A_0$ and are tangent to $I$. Let $A_1$ be the second point of
intersection of $A_0$ with $O$; so $L_0$ is one of the two lines
which contain $A_1$ and are tangent to $I$. Let $L_1$ by the second
tangent to $I$ that contains $A_1$. Clearly we can iterate this
construction which leads to the following interesting property:
\emph{If this construction is cyclic, then there is an $n$ such that this
construction gives a cycle of length $n$ for \emph{all} starting
positions}.
What is the relation of this construction with elliptic curves? Let
$G(X,Y,Z)$ be the equation for $I$ and $F(X,Y,Z)$ be the equation for
$O$. Consider the curve $C$ in $\bbP^3$ defined by the equations
\begin{eqnarray*}
W^2 & = & G(X,Y,Z)\\
0 & = & F(X,Y,Z)
\end{eqnarray*}
As seen before, $C$ is a curve of genus 1. Since our construction
depends on picking a point $A_0$ of $O$ and a tangent line $L_0$ of
$I$, each of these conics has a point; as seen above, this means that
we can parametrically solve them. Specifically, by a linear change of
co-ordinates we can assume that $G(X,Y,Z)=Y^2-XZ$.
Now, the surface $Q$ defined by the equation $W^2=Y^2-XZ$ has a
parametric solution
\[ A_{pq}=(W:X:Y:Z) = (p-q:2pq:p+q:2) \]
This parametrisation is used to express $Q$ in the form
$\bbP^1\times\bbP^1$ where $p$ and $q$ represent the co-ordinates of
the the point in each factor. In particular, each point
$A_{p,q}$ of $Q$
determines a pair of lines $H_q$ and $V_p$ in $Q$ given respectively
in parametric form by
\[
t \mapsto (t-q:2tq:t+q:2) \text{~and~}
t \mapsto (p-t:2pt:p+t:2)
\]
The image of these lines in the $(X:Y:Z)$ plane under projection from
$(W:X:Y:Z)=(1:0:0:0)$ are the lines $T_q$ and $T_p$ where, for
each $r$, the line $T_r$
\[
t \mapsto (2rt:t+r:2)
\]
is tangent to the conic $I$ at the point $(r^2:r:1)$. In fact
$T_p$ and $T_q$ are precisely the two lines that contain the point
$(2pq:p+q:2)$ and are tangent to the conic $I$. (Exercise: Carry out
the above construction without the use of co-ordinates).
Now, the point $A_{p,q}$ lies on the curve $C$ if and only if
its image lies on $O$. Conversely, given a point $A_0$ on $O$ and a
line $L_0$ that passes through $A_0$ and is tangent to $I$, let $r$
be such that $L_0=T_r$. This means that $A_0=(x:y:z)=(2tr:t+r:2)$, so
that if we take $p_0=r$ and $q_0=(y-rz)/z$, then $(x:y:z)=(2p_0q_0:p_0+q_0:2)$.
In particular, the point $A_{p_0,q_0}$ maps to $A_0$ under the above
projection; note again that $A_{p_0,q_0}$ lies on $C$. We could equally
well have taken the point $A_{q_0,p_0}$; as
a convention we will pick $A_{p_0,q_0}$. In this case $V_{p_0}$ is the line in
$Q$ that maps to $L_0=T_r=T_{p_0}$. Let us now follow the
construction.
The line $L_1$ is the second tangent to $I$ from the point $A_0$.
Thus, it is the image of the line $H_{q_0}$. The point $A_1$ is the
second point of intersection of $L_1$ with the conic $O$; this is
the image of a (unique) point $A_{p_1,q_0}$ that is the second
intersection of $H_{q_0}$ with $C$. Continuing this approach, we see
that $L_2$ is the image of $V_{p_1}$ which intersects $C$ in the
point $A_{p_1,q_1}$ that lies over $A_2$. Now, we are in a position
similar to the starting point and we can iterate.
It may appear that the ``lifted'' construction corresponds to two
steps of the original construction, but note that we will consider an
iteration to be complete even if we reach $A_{q_0,p_0}$ instead of
$A_{p_0,q_0}$.
To see what this has to do with elliptic curves, we note that the
divisors $H_q\cap C$ are all linearly equivalent to each other
(similar story with the divisors $V_p\cap C$). Now consider the
divisor
\[ D=D_1=V_{p_1}\cap C - H_{q_0}\cap C=A_{p_1,q_1}-A_{p_0,q_0} \]
Treating $C$ as an elliptic curve with the origin $A_{p_0,q_0}$, we
can identify the divisor $D$ with the point $A_{p_1,q_1}$. Note that
the divisor
\[ D_2 = V_{p_2}\cap C - H_{q_1}\cap C=A_{p_2,q_2}-A_{p_1,q_1} \]
is linearly equivalent to $D$ as remarked above. Thus
\[
A_{p_2,q_2}-A_{p_0,q_0} = D_1+D_2 = 2D =
(A_{p_1,q_1}\oplus A_{p_1,q_1}) - A_{p_0,q_0}
\]
Thus the above iterated construction is identified with the sequence of
positive integer multiples of $A_{p_1,q_1}$ in the group law on the
elliptic curve. It is clear that the construction cycles if and only
if this point is of finite order. Finally, we use the linear
equivalence to again note that regardless of which point $P$ of $C$
we take to be the origin, the order of the point corresponding to $D$
is the same.
The above study of a pair of quadratic forms has been taken up in
depth in the work of U. Bhosle of the TIFR along with others like
S. Ramanan of TIFR and P. E. Newstead of Liverpool.
\end{document}