39
This gives the Euler identity
X
Y
(1)r q r(3r+1)/2 =
(1 q m ).
r Z
(4.23)
m=1
In particular, we get the following Fourier expansion for the Dedekinds function ( ):
1
( ) = q 24
X
(1)r q r(3r+1)/2 .
r Z
The positive integers of the form n + (k 2) n(n2
2
LECTURE 1. BINARY QUADRATIC FORMS
Let x = m1 v + m2 w . The length of x is given by the formula
x|2 = |m1 v + m2 w|2 = (m1 , m2 )
vv
vw
vw
ww
m1
m2
=
am2 + 2bm1 m2 + cm2 ,
1
2
where
a = v v,
b = v w,
c = w w.
(1.1)
Let us consider the (binary) quadratic
56
LECTURE 6. MODULAR FORMS
enters only in the rst degree. Thus we can express y in terms of x, t and obtain that
E is isomorphic to P1 (C). If d = 0 we obtain that f could be chosen of degree 2.
Again this is impossible. Note that we also have in (6.13)
57
After all of these normalizations, the elliptic function (z ) with respect to
is uniquely determined by the conditions (6.18) and (6.19). It is called called the
Weierstrass function with respect to the lattice .
One can nd explicitly the function (z
58
LECTURE 6. MODULAR FORMS
as a holomorphic map from C \ cfw_e1 , e2 , e3 to (C/ )/(z z ). It can be shown that
it extends to a holomorphic isomorphism from the Weiertrass cubic y 2 = 4x3 g2 x g3
onto (C/ ) \ cfw_0. This is the inverse of the map given
59
Setting t = e2iz , we rewrite the left-hand side as follows:
cot(z ) =
Xm
eiz + eiz
t+1
cos z
= i iz
) = i
= i (1 2
t ).
sin z
e
eiz
t1
m=0
Dierentiating k 1 2 times in z , we get
X
(k 1)!
X
(z + m)k = (2i)k
mZ
mk1 tm .
m=1
This gives us the needed
61
Example 6.4. We apply the previous theorem to (z ; ) = (z ) and z =
case
= 0 (2) = cfw_
(1) : 2| .
1
.
2
In this
Now, replacing z with z/( + ) in (6.22), we get
(
+
z
;
) = ( + )2 [z 2 +
+ +
X
(m,n)=(0,0)
1
].
z m( + ) + n( + )
Since Z + Z = Z( +
62
LECTURE 6. MODULAR FORMS
6.5 Show that
0
(z1 )
det @ (z2 )
r(z3 )
(z1 )
(z2 )
(z3 )
1
1
1A = 0
1
whenever z1 + z2 + z3 = 0. Deduce from this an explicit formula for the group law on
the projective cubic curve y 2 t = 4x3 g2 xt2 g3 t3 .
6.6 (Weierstr
63
(iii) Show that L ( ) is almost modular form for the group
0 (N ) = cfw_
SL(2, Z) : N |c,
i.e.
L (
+
) = ( + )k/2 (d)L ( ),
+
where (d) = ( (1)
d
k
2
D
0 (N ),
) is the quadratic residue symbol.
(iv) Prove that L ( ) is a modular form for 0 (2) w
Lecture 7
The Algebra of Modular
Forms
7.1 Let be a subgroup of nite index of (1). We set
Mk () = cfw_modular forms of weight k with respect to ,
We also denote by Mk ()0 the subspace of cuspidal modular forms. It is clear that
Mk () is a vector space ove
3
and hence
a
c
b
d
=
ab
cd
.
This can be also expressed by saying that the form f is obtained from the form
f by using the change of variables
x x + y,
y x + y.
We write this in the form
f = M f.
According to Lagrange two binary quadratic forms f and g a
66
LECTURE 7. THE ALGEBRA OF MODULAR FORMS
Note that when f is a modular form with respect to a group we have
g (f ) = (f ),
g .
For each H let
8
>2
<
m = 3
>
:
1
if (1) i,
if (1) e2i/3 ,
otherwise.
(7.3)
Lemma 7.1. Let f ( ) be a modular form of weight k
67
Use the function q = e2i to map the segment cfw_ : |Re | 1 , Im = h onto the
2
1
circle C : |q | = e2h . When we move along the segment from the point 2 + ih to the
1
point 2 + ih the image point moves along the circle in the clockwise way. We have
1
2
68
LECTURE 7. THE ALGEBRA OF MODULAR FORMS
In particular, this is true for g2 . For any other f M2 (1) we have f /g2 is (1)
invariant and also holomorphic at (since g2 is not a cusp form). This shows that
f /g2 is constant and
M2 (1) = Cg2 .
Similar argum
69
7.3 Let us give some examples.
Example 7.1. We know that the Eisenstein series E2k is a modular form of weight
2
2
k with respect to (1). Since M4 (1) = Cg2 = CE4 , comparing the constant
coecients in the Fourier expansions we obtain
E8 =
(8) 2
E4 .
2
55
6.4 We 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.
54
LECTURE 6. MODULAR FORMS
Here we used that (T S ) = (ST )1 since (T 2 S 2 ) = 1 and similarly
(T ST ) = (ST S )1 , (T ST S ) = (ST ST )1 ,
(T ST ST ) = (ST ST S )1 .
Also (ST ) = (ST ST )1 . Thus it is enough to verify that the elements S, ST, ST S,
T
Lecture 5
Transformations of Theta
Functions
5.1 Let us see now that the theta constants ab and their derivatives ab satisfy
the functional equation similar to (4.2). This will imply that certain powers of theta
constants are modular forms. For brevity we
42
LECTURE 5. TRANSFORMATIONS OF THETA FUNCTIONS
Thus for any Th(k; )ab , we have
(z ( + ) Th(e ; Z + Z),
where
=
+
,
+
em+n (z ) = e(m+n
)( + ) (z (
+ ).
We have, using (5.1),
e1 (z ) = e + (z ( + ) = e2i(ab ) eik(2z( +)+
eik ( +)(z+1)
2
2
( + )z ) i
43
where
k
k
, a + b
).
2
2
Summarizing we obtain that, for any (z, ) Th(k; )ab ,
(a , b ) = (a b +
(5.4)
2
eik ( +)z ( + )z ; ) Th(k, )a b .
(5.5)
`
Now let us replace with its inverse . We rewrite (5.13) and (5.14) as
2
eik ( +)z ( + )z ; ) Th(k, )a
w
44
LECTURE 5. TRANSFORMATIONS OF THETA FUNCTIONS
Now take M = ( 0 1 ) . We have
10
eiz
2
/
00 (z/ ; 1/ ) = B00 (z ; )
for some B depending only on . Plugging in z = 0 and applying (4.2), we get
B = i ,
(5.10)
where the square root takes positive values on
45
Proof. We shall prove in the next lecture that it is enough to check this for generators
of the group SL(2, Z). Also we shall show that the group SL(2, Z) is generated by the
matrices M1 = ( 1 1 ) , M2 = ( 0 1 ) , I. We have from (4.14) and (4.15)
01
1
Lecture 1
Binary Quadratic Forms
1.1 The theory of modular form originates from the work of C.F. Gauss of
1831 in which he gave a geometrical interpretation of some basic notions of
number theory.
Let us start with choosing two non-proportional vectors in
46
LECTURE 5. TRANSFORMATIONS OF THETA FUNCTIONS
Exercises
5.1 Show that the constant (M ) in (5.16) is equal to i
odd. If is odd and is even, it is equal to
0
symbol, where we also set ( 1 ) = 1.
ei/4 ( ).
1
2
( | ) when is even and is
Here ( x ) is the
48
LECTURE 5. TRANSFORMATIONS OF THETA FUNCTIONS
5.7 Dene the Weierstrass -function by
2
z
2
z 2 ( 1 1 /61 1 1 ) 1 1 ( ; )
1
1
22
(z ; 1 , 2 ) = 1 e
22
22
1 1 (0)
.
22
Show that
(i) (z ; 1 , 2 ) does not depend on the basis 1 , 2 of the lattice ;
(ii) (
Lecture 6
Modular Forms
6.1 We have seen already in Lecture 5 (5.2) and Corollary 5.3 that the functions
( )4k = 00 (0; )4k (resp. ( )24 ) satisfy the functional equation
f ( + 2) = f ( ),
f (1/ ) = 2 f ( ),
(resp.
f ( + 1) = f ( ),
f (1/ ) = 12 f ( ).
In
50
LECTURE 6. MODULAR FORMS
of in the space of holomorphic functions on H dened by
(g )(z ) = |k g 1 .
Note that we switched here to g
1
(6.6)
in order to get
(gg ) = (g ) (g ).
It follows from the above that to check (6.1) for some subgroup it is enough