We 55
6.4 know that any elliptic curve is isomorphic to a Hesse cubic curve. Let us
give another cubic equation for an elliptic curve, called a Weierstrass equation. Its
coecients will give us new examples of modular forms. Recall that dim Th(k, )ab =
k. Let use <, > to denote the linear span. We have
Th(1, ) 1 1 =< 1 1 (z ; ) >=< T >;
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22
Th(2, ) =< T 2 , X >,
Th(3, ) 1 1 =< T 3 , T X , Y >,
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for some functions X Th(2, ), Y Th(3, ) 1 1 . Now the following seven func22
tions
T 6, T 4X , T 2X 2, X 3, T 3Y , T X Y , Y 2
all belong to the space Th(6, ). They must be linearly dependent and we have
aT 6 + bT 4 X
2
+ cT 2 X
2
3
+ dX
+ eT 3 Y + f T X Y + gY
2
= 0.
(6.12)
Assume g = 0, d = 0. It is easy to nd
X = X + T 2 ,
Y = Y + XT + T 3
which reduces this expression to the form
Y 2 T X 3 AXT 4 BT 6 = 0,
(6.13)
for some scalars A, B . Let
(z ) = X/T 2 ,
1 (z ) = Y /T 3 .
Dividing (6.13) by T 6 we obtain a relation
1 (z )2 = (z + )3 A(z ) + B.
(6.14)
Since both X and T 2 belong to the same space Th(2, ) the functions (z ), 1 (z ) have
periods Z + Z and meromorphic on C. As we shall see a little later, 1 (z ) = d .
dz
Consider the map
E = C/ P2 ,
z (T (z )3 , T (z )X (z ), Y (z )).
Since T (z )3 , T (z )X (z ), Y (z ) all belong to the same space Th(3, ) 1 , 1 this map is
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well-dened and holomorphic. It diers from the map from Example 3.2 only by a
composition with a translation on E and a linear change of the projective coordinates
coordinates. This is because, for any f Th(k; ) we have
2
ei[a
+2(z +a)b]
f (z +
b + a
) Th(k; )ab
k
(see Lecture 3). So it is an isomorphism onto its image. The relation (6.13) tells us
that the image is the plane projective curve of degree 3 given by the equation
y 2 t x3 Axt2 Bt3 = 0,
(6.15)
Now it is clear why we assumed that the coecients d, g in (6.12) are not equal to
zero. If g = 0, we obtain an equation f (x, y, t) = 0 for the image of E in which y