Classical Modular Forms
T.N. Venkataramana∗
School of Mathematics, Tata Institute of Fundamental Research,
Colaba, Mumbai, India
Lectures given at the
School on Automorphic Forms on GL(n)
Trieste, 31 July - 18 August 2000
LNS0821002
∗
venky@math.tifr.res.in
Contents
1 Introduction
43
2 A Fundamental Domain for SL(2, Z)
43
3 Modular Forms; Definition and Examples
49
4 Modular Forms and Representation Theory
54
5 Modular Forms and Hecke Operators
62
6 L-functions of Modular Forms
70
Classical Modular Forms
1
43
Introduction
The simplest kind of automorphic forms (apart from Grössencharacters,
which will also be discussed in this conference) are the “elliptic modular
forms”. We will study modular forms and their connection with automorphic forms on GL(2), in the sense of representation theory.
Modular forms arise in many contexts in number theory, e.g. in questions
involving representations of integers by quadratic forms, and in expressing
elliptic curves over Q as quotients Jacobians of modular curves (this was the
crucial step in the proof of Fermat’s Last Theorem by Andrew Wiles), etc.
The simplest modular forms are those on the modular group SL(2, Z)
and we will first define modular forms on SL(2, Z).
To begin with, in section 2, we will describe a fundamental domain for
the action of SL(2, Z) on the upper half plane h. The fundamental domain
will be seen to parametrise isomorphism classes of elliptic curves.
In section 3, we will define modular forms for SL(2, Z) and construct
some modular forms, by using the functions E4 and E6 which we encounter
already in the section on elliptic curves.
In section 4, a representation theoretic interpretation of modular forms
will be given, which will enable us to think of them as automorphic forms
on GL(2, R).
In section 5, we will give an adelic interpretation of modular forms. This
will enable us to think of Hecke operators as convolution operators in the
Hecke algebra; using this, we show the commutativity of the Hecke operators.
We will also prove a special case of the Multiplicity One theorem.
2
A Fundamental Domain for SL(2, Z)
Notation 2.1 Denote by h the “Poincaré upper half-plane” i.e. the space
of complex numbers whose imaginary part is positive:
h = {z ∈ C; z = x + iy, x, y ∈ R, y > 0}.
If z ∈ C, denote respectively by Re(z) and Im(z) the real and imaginary
parts of z.
On the upper half plane h, the group GL(2, R)+ of
real2 × 2 matrices
a b
with positive determinant operates as follows: let g =
∈ GL(2, R)+ ,
c d
and let z ∈ h. Set g(z) = (az + b)/(cz + d). Notice that if cz + d = 0
44
T.N. Venkataramana
and c 6= 0 then, z = −d/c is real, which is impossible since z has positive
imaginary part. Thus, the formula for g(z) makes sense. Observe that
Im(g(z)) = Im(z)(det(g))/ | cz + d |2 .
(1)
The equation (1) shows that the map (g, z) 7→ g(z) takes GL(2, R)+ × h into
h. One checks immediately that this map gives an action of GL(2, R)+ on
the upper half plane h. Note also that
| cz + d |2 = c2 y 2 + (cx + d)2 .
(2)
Therefore, | cz + d |2 ≥ y 2 or 1 according as | c |= 0 or nonzero. Therefore,
Im(γ(z)) ≤ y/ min{1, y 2 } ∀γ ∈ Γ0 ⊂ SL(2, Z),
(3)
where min{1, y 2 } denotes the minimum of 1 and y 2 and Γ0 ⊂ SL(2, Z) is
the group generated by the elements T = ( 10 11 ) and S = ( −10 01 ). Later we
will see that Γ0 is actually SL(2, Z). The element T acts on the upper half
plane h by translation by 1:
1 1
T (z) =
(z) = z + 1 ∀z ∈ h.
(4)
0 1
Similarly, the element S acts by inversion:
0 1
S(z) =
(z) = −1/z ∀z ∈ h.
−1 0
(5)
Consider the set
F = {z ∈ h; −1/2 < Re(z) ≤ 1/2,
| z |≥ 1, and 0 ≤ Re(x)} if | z |= 1.
Theorem 2.2 Given z ∈ h there is a unique point z0 ∈ F and an element
γ ∈ SL(2, Z) such that γ(z) = z0 . Moreover, given γ ∈ SL(2, Z), we have
γ(F ) ∩ F = φ unless γ lies in a finite set ( of elements of SL(2, Z) which
fix the point ω = 1/2 + i31/2 /2 ∈ h or i ∈ h). [one then says that F is a
fundamental domain for the action of SL(2, Z) on the upper half plane
h].
Proof We will first show that any point z on the upper half plane can
be translated by an element of the subgroup Γ0 of SL(2, Z) (generated by
0 1 )) into a point in the “fundamental domain” F .
T ( 10 11 ) and S = ( −1
0
Classical Modular Forms
45
Now, given a real number x, there exists an integer k such that −1/2 <
x + k ≤ 1/2. Therefore, equation (4) shows that given z ∈ h there exists
an integer k such that the real part x′ of T k (z) satisfies the inequalities
−1/2 < x′ ≤ 1/2.
Let y denote the imaginary part of z and denote by Sz the set
Sz = {γ(z); γ ∈ Γ0 , Im(γ(z)) ≥ y , −1/2 < Re(γ(z)) ≤ 1/2} .
We will first show that Sz is nonempty and finite. Let k be as in the previous paragraph. Then −1/2 < Re(T k (z)) ≤ 1/2 and Im(T k (z) = Im(z);
therefore, T k (z) lies in Sz and Sz is nonempty.
Now, equation (3) shows that the imaginary parts of elements of the set
Sz are all bounded from above by y/ min 1, y 2 . By definition, the imaginary
parts of points on Sz are bounded from below by y. The definition of Sz
shows that Sz is a relatively compact subset of h. We get from (3) that
| cz + d |2 ≤ 1; now (2) shows that | c |≤ 1/y 2 . Suppose γ ∈ γ0 = ( ac db ) is
such that γ(z) ∈ Sz then, c is bounded by 1/y 2 and is in a finite set. The
fact that cz + d is bounded now shows that d also lies in a finite set. Since Sz
is relatively compact in h, it follows that γ(z) = (az + b)/(cz + d) is bounded
for all γ(z) ∈ Sz ; therefore, az + b is bounded as well, and hence a and b run
through a finite set. We have therefore proved that Sz is finite.
Let y0 be the supremum of the imaginary parts of the elements of the
finite set Sz ; let S1 = {z ′ ∈ Sz ; Im(z ′ ) = y0 } and let z0 ∈ S1 be an element
whose real part is maximal among elements of S1 . We claim that z0 ∈ F .
First observe that if z ′ ∈ Sz then S(z ′ ) = −1/z ′ has imaginary part y0 / |
z |2 = Im(z ′ )/ | z ′ |2 ≤ y0 whence | z ′ |2 ≥ 1. If | z0 |> 1, then it is immediate
from the definitions of F and Sz that z0 ∈ F . Suppose that | z0 |= 1. Then,
S(z0 ) = −1/z0 also has absolute value 1, its imaginary part is y0 and its real
part is the negative of Re(z0 ); hence S(z0 ) ∈ S1 . The maximality of the real
part of z0 among elements of S1 now implies that Re(z0 ) ≥ 0. Therefore,
z0 ∈ F . We have proved that every element z0 may be translated by an
element of Γ0 into a point in the fundamental domain F .
Suppose now that z ∈ γ −1 (F )∩F for some γ ∈ SL(2, Z). Write γ = ( ac db )
with a, b, c, d ∈ Z and ad − bc = 1. Suppose that Im(γ(z)) ≥ Im(z) = y
(otherwise, replace z by γ(z)). Then, by (3) one gets
(cx + d)2 + c2 y 2 ≤ 1.
(6)
Since z ∈ F , we have x2 + y 2 ≥ 1 and 0 ≤ x ≤ 1/2. Therefore y 2 ≥ 3/4 and
(1 ≥)c2 y 2 ≥ c2 4/3.
(7)
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T.N. Venkataramana
This shows that c2 ≤ 1 since c is an integer.
Suppose c = 0. Then, ad = 1, a, d ∈ Z and we may assume (by
multiplying by the matrix −Id [minus identity] if necessary) that d = 1.
Hence γ = ( 10 1b ). Then, γ(z) = z + b ∈ D which means that 0 ≤ x + b ≤ 1/2
and 0 ≤ x ≤ 1/2. Thus, −1/2 ≤ b ≤ 1/2, i.e. b = 0 and γ is the identity
matrix.
The other possibility is c2 = 1, and by multiplying by the matrix −Id
(minus identity) we may assume that c = 1. Suppose first that d = 0. Then,
bc = −1 whence b = −1. Now, (7) shows that x2 + y 2 ≤ 1. Moreover,
γ(z) = az + b/z = a + bz/ | z |2 = a − z
whence its real part is a − x which lies between 0 and 1/2. Since 0 ≤ x ≤ 1,
it follows that 0 ≤ a ≤ 1. If a = 0 then γ = ( 01 −10 ) and lies in the isotropy
of the point i ∈ h. If a = 1 then, γ = ( 11 −10 ) which lies in the isotropy of the
point ω = 1/2 + i31/2 /2.
We now examine the remaining case of c = 1 and d 6= 0. From (6) we
get (x + d)2 + y 2 ≤ 1. If d ≥ 1 then the inequality 0 ≤ x ≤ 1/2 shows that
1 ≤ d ≤ x + d which contradicts the inequality (x + d)2 + y 2 ≤ 1, which
is impossible. Thus, d ≤ −1; then the inequality 0 ≤ x ≤ 1/2 implies that
x + d ≤ 1/2 + (−1) = −1/2 whence (x + d)2 ≥ 1/4. Since y 2 ≥ 3/4 the
inequality (x+d)2 +y 2 ≤ 1 implies that equalities hold everywhere: y 2 = 3/4,
x = 1/2 and d = −1. Thus, z = ω and z −1 = z 2 . Since 1 = ad−bc = −a−b
(d = −1 and c = 1), and
γ(z) = (az+b)/(z−1) = (az+b)/z 2 = −(az+b)z = a+(−a−b)z = a+z ∈ D,
the real part of γ(z) is a + x = a + 1/2 and is between 0 and 1/2, i.e.
−1
−1/2 ≤ a ≤ 0 i.e. a = 0 and b = −1. Therefore, γ = ( 01 −1
) lies in the
isotropy of ω. This completes the proof of Theorem (2.2).
Corollary 2.3 The group SL(2, Z) is generated by the matrices T = ( 10 11 )
0 1 ).
and S = ( −1
0
Proof In the proof of Theorem (2.2), a point on the upper half plane is
brought into the fundamental domain F by applying only the transformations generated by S and T . The fact that the points on the fundamental domain are inequivalent under the action of SL(2, Z) now implies that
SL(2, Z) is generated by S and T .
Classical Modular Forms
47
(The Corollary can also be proved directly by observing that ST −1 S −1 =
( 11 01 ). Now, the usual row-column reduction of matrices with integral entries
implies that T and ST S −1 generate SL(2, Z)).
Notation 2.4 Elliptic Functions. We recall briefly some facts on elliptic
functions (for a reference to this subsection, see Ahlfors’ book on Complex
Analysis). Given a point τ on the upper half plane h, the space Γτ = Z ⊕ Zτ
of integral linear combinations of 1 and τ forms a discrete subgroup of C
with compact quotient. The quotient
Eτ = C/Γτ
may be realised as the curve in P2 (C) whose intersection with the complement of the plane at infinity is given by
y 2 = 4x3 − g2 x − g3
(8)
The curve Eτ = C/Γτ is called an “elliptic curve”.
The map of C/Γτ to P2 is given by z 7→ (℘′ (z), ℘(z), 1) for z ∈ C. Recall
the definition of ℘: if z ∈ C and does not lie in the lattice Γτ , then write
′
X
℘(z) = 1/z +
(1/(z + w)2 − 1/w2 ),
2
P′
is the sum over all the non-zero points w in the lattice Γτ . The
where
derivative ℘′ (z) of ℘(z) is then given by
X
℘′ (z) =
1/(z + w)3 ,
where the sum is over all the points of the lattice Γτ . One has the equation
(cf. equation (8))
℘′ (z)2 = 4℘(z)3 − g2 (τ )℘(z) − g3 (τ ).
(9)
a b
∈ SL(2, Z) and τ ∈ h, then the elliptic curve Eγ(τ ) is
c d
isomorphic as an algebraic group (which is also a projective variety) to the
elliptic curve Eτ . The explicit isomorphism on C is given by z 7→ z/(cτ + d).
It is also possible to show that if Eτ and Eτ ′ are isomorphic elliptic curves,
then τ ′ is a translate of τ by an element of SL(2, Z).
Thus the fundamental domain F which was constructed in Theorem (2.2) parametrises isomorphism classes of elliptic curves.
If γ =
48
T.N. Venkataramana
In equation (9), recall that the coefficients g2 and g3 are given by
′
g2 (τ ) = 60G4 (τ ) = 60
and
X
(mτ + n)−4
′
X
g3 (τ ) = 140G6 (τ ) = 140
(mτ + n)−6
P′
is the sum over all the pairs of integers (m, n) such that not both
where
m and n are zero. The discriminant of the cubic equation in (9) is given by
1/(16)∆(τ ) where
∆(τ ) = g23 − 27g32 .
(10)
It is well known and easily proved that ℘′ (z) has a simple zero at all the
2-division points 1/2,τ /2 and (1+τ )/2 and that ℘(1/2),℘(τ /2) and ℘((1+
τ )/2) are all distinct. Thus equation (9) transforms to
℘′ (z)2 = 4(℘(z) − ℘(1/2))(℘(z) − ℘(τ /2))(℘(z) − ℘((1 + τ )/2))
(11)
Thus the discriminant of the (nonsingular) cubic in equation (9) is non-zero
and so we obtain that
∆(τ ) 6= 0
(12)
for all τ ∈ h.
Notation 2.5 On the upper half plane h, there is a measure denoted y −2 dxdy,
as z = x + iy varies in h. This measure is easily seen to be invariant under
the action of elements of the group GL(2, R)+ of nonsingular matrices with
positive determinant.
Lemma 2.6 With respect to this measure, the fundamental domain F has
finite volume.
Proof We compute the volume of F . Note that if z = x + iy lies in F ,
then, −1/2 ≤ x ≤ 1/2 and 1/(1 − x2 )1/2 ≤ y < ∞. Thus the volume of F is
the integral
Z ∞
Z 1/2
Z
2
dy/y 2 )
dx(
dxdy/y =
F
−1/2
(1−x2 )1/2
which is easily seen to be π/3. In particular, F has finite volume.
Classical Modular Forms
49
Notation 2.7 Let S denote the inverse image of the fundamental domain
F ⊂ h under the quotient map GL(2, R) → GL(2, R)/O(2)Z = h. Then,
we have proved that GL(2, Z)S = GL(2, R). The set S is called a Siegel
Fundamental Domain.
3
Modular Forms; Definition and Examples
Notation 3.1 Given z ∈ h (h is the upper half plane) and an element
g = ( ac db ), write
j(g, z) = cz + d.
Note that if j(g, z) = 0, then by comparing the real parts and imaginary
parts we get c = 0 and d = 0 which is impossible since ad − bc 6= 0. Thus,
j(g, z) is never zero.
Definition 3.2 A function f : h → C is weakly modular of weight w if
the following two conditions hold.
(1) f is holomorphic on the upper half plane.
(2) for all γ ∈ SL(2, Z), with γ = ( ac db ), we have the equation
f ((az + b)/(cz + d)) = (cz + d)w f (z).
(13)
Given g = ( ac db ) and a function f on the upper half plane h, define
g−1 ∗ f (z) = (cz + d)−w f (g(z)) ∀z ∈ h.
Then, it is easily checked that the map (g, f ) → g−1 ∗ f defines an action
of GL(2, R) on the space of functions on h. Thus, the condition (2) above
is that the function f there is invariant under this action by SL(2, Z). Now
0 1 ) and
by Corollary (2.3), SL(2, Z) is generated by the matrices S = ( −1
0
−1
T = ( 10 11 ). Thus condition (2) is equivalent to saying that γ ∗ f = f for
γ = S, T . This amounts to saying that
f (−1/z) = z w f (z)
(14)
f (z + 1) = f (z).
(15)
and
Note that the invariance of f under the action of −1 where 1 is the
identity matrix in SL(2, Z) implies that f is zero of w is odd: f (z) =
(−1)w f (z). Therefore, we assume from now on (while considering modular
forms for the group SL(2, Z) ) that w = 2k where k is an integer.
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T.N. Venkataramana
Definition 3.3 The map exp : h → D ∗ given by z 7→ e2πiz = q is easily seen
to be a covering map of the upper half plane h onto the set D ∗ of non-zero
complex numbers of modulus less than one. The covering transformations
are generated by T (z) = z + 1. A weakly modular function f is invariant
under T and therefore yields a holomorphic map f ∗ : D ∗ → C given by
f ∗ (q) = f (z) for all z ∈ h. We say that a weakly modular function of weight
w is a modular function of weight w if f ∗ extends to a holomorphic
function of D (the set of complex numbers of modulus less than one) i.e. f ∗
extends to 0 ∈ D.
Let f be a weakly modular function on h. Then, f is a modular function if and only if the function f ∗ has the “Fourier expansion” (or the “qexpansion”)
X
f ∗ (q) =
an q n ,
(16)
n≥0
where an are complex numbers and the summation is over all non-negative
integers n. Observe that a weakly modular function is modular if and only
if it is bounded in the fundamental domain F .
We will say that a modular form is a cusp form if the constant term of
its q-expansion is zero: i.e. a0 = 0 in the notation of equation (15).
Notation 3.4 Examples of modular forms.
First we note that if f and g are modular forms of weights w and w′
then, the product function f g is a modular form of weight ww′ .
We will first prove that for the modular group SL(2, Z), there are no
non-constant “weight zero” modular forms. First note that if f is a weight
zero modular form, then the function f ∗ extends to 0 and hence is bounded
in a disc of radius r < 1. Its inverse image under exp : F → D ∗ is precisely
the set A = {z = x + iy ∈ F ; y > − log r} and f is bounded on the set A.
The complement of the set A in the fundamental domain F is compact, and
f is bounded there as well, whence f is bounded on all of the fundamental
domain F as well as at “infinity”. By the maximum principle, f is constant.
We will now show that there are no modular forms of weight two on
SL(2, Z). Suppose f is one and let F (z) be its integral from z0 to z for
some fixed z0 ∈ h. The modularity of f shows that γ 7→ F (γ(z0 )) gives a
homomorphism from SL(2, Z) to C. But, SL(2, Z) is generated by the finite
order elements S and ST whence, this homomorphism is identically zero.
This and the modularity of f shows that the integral F is invariant under
SL(2, Z). It is easy to show that F ∗ is holomorphic at 0 (integrate both
Classical Modular Forms
51
sides of equation (15)), and use the invariance of F under T ). Hence F is a
modular form of weight zero. By the foregoing paragraph, f is a constant,
i.e. f = 0.
Fix an even positive integer 2k, with k ≥ 2. We will construct a modular
form of degree k as follows. Let τ ∈ h and write (compare the definition of
g2 and g4 in section (2.4))
X ′
G2k (τ ) =
(mτ + n)−2k ,
(17)
P′
is the sum over all the pairs of integers (m, n) not both of which are
where
zero. Then, G2k is easily shown to be a weakly modular function of weight
2k on the upper-half plane. If τ is varying in the fundamental domain and
its imaginary part tends to infinity, then it is clear from the formula for G2k
P′ −2k
n
= 2ζ(2k) where
that G2k (τ ) tends to
X
ζ(s) =
n−s
is the Riemann zeta function (the sum is over all the positive integers n and
in the sum, the real part of s exceeds 1). Consequently, G2k is a modular
form of weight 2k. We will now outline a derivation of the q-expansion of
G2k . Start with the partial fraction expansion
X
πcot(πz) = z −1 +
(z + n)−1 + (z − n)−1
(18)
where the sum is over all positive integers n. This series converges uniformly
on compact subsets of the complement of Z in C.
Write q = e2πiz (where i ∈ h and i2 = −1). Then one has the q-expansion
X
πcot(πz) = πi(q + 1)/(q − 1) = −πi − 2πi
qn
(19)
n≥1
Differentiate 2k- times, the right-hand sides of equations (17) and (18)
with respect to z. We then get the equality
X
X
n2k−1 q n
(20)
(z + n)−2k = ((2k − 1)!)−1 (2πi)2k
n∈Z
n≥1
Fix m and in equation (19) take for z the complex number mτ . Then sum
over all m. We obtain by equations (16) and (18), the q-expansion
X
σ2k−1 q n
(21)
G2k (τ ) = 2ζ(2k) + ((2k − 1)!)−1 (2πi)2k
n≥1
52
T.N. Venkataramana
where for an integer r and n ≥ 1, σr (n) is defined to be the sum
d runs over the positive divisors of n.
By using the power series expansion
X
(1 + x)−2 =
nxn−1
P
dr where
n≥1
and equation (17) one has the power series identity
XX
X
πcot(πz) = z −1 + 2
n2j z 2j−1 = z −1 + 2
ζ(2j)z 2j−1
n≥1 j≥1
(22)
j≥1
P
By comparing the power series expansions cos(x) = m≥0 ((2m)!)−1 x2m
P
and sin(x) = m≥0 ((2m + 1)!)−1 x2m+1 with the right-hand side of equation
(21) one obtains
ζ(2) = π 2 /6, ζ(4) = π 4 /90 andζ(6) = π 6 /(33 .5.7).
(23)
Using (20) and (22) we get
g2 = 60G4 = (4/3)π 4 + 160π 4 (q + · · · )
(24)
where the expression q + · · · is a power series in q with integral coefficients
with the coefficient of q being 1. Similarly, we get (again from (20) and (22))
g3 = 140G6 = (8/27)π 6 − 25 .7π 6 /3(q + · · · )
Therefore, we get, after some calculation, that for all z ∈ h,
X
∆(z) = g2 (z)3 − 27g3 (z)2 = 211 π 12 (q +
τ (n)q n )
(25)
(26)
n≥2
where the τ (n) are integers. We recall that ∆(z) is never zero on the upperhalf plane (section (2.4)). The equation (26) shows that the coefficient of q
in q-expansion of ∆ is non-zero, (and that its constant term is zero).
Lemma 3.5 There are no modular forms of negative weight.
Proof Suppose that f is a modular form of weight −l with l > 0. Form
the product g = f 12 ∆l . Since f and ∆ are modular forms, so is the product.
Since its weight is zero, g is a constant (see the beginning of this subsection).
But, (26) shows that the q-expansion of g has no constant term. Hence g = 0
whence, f = 0.
Classical Modular Forms
53
Lemma 3.6 Suppose that f is a cusp form. Then ∆ divides f i.e., there
is a modular form g such that f = ∆g. In particular, the weight of f is at
least 12.
Proof Consider the quotient g = f /∆. Since ∆ has no zero in h, it follows
that g is holomorphic in h. Clearly, g is weakly modular of weight =weight
of f -12. Now the q-expansion of f (and also ∆), has no constant term; and
the coefficient of q in the q-expansion of ∆ is non-zero. Therefore, g∗ extends
to a holomorphic function in a neighbourhood of 0. That is, g is a modular
function. Since the weight of g is non-negative (by Lemma (3.5)), it follows
that the weight of f must be at least that of ∆, namely, 12.
Corollary 3.7 The space of cusp forms of weight 12 (for SL(2, Z)), is one
dimensional.
Proof If f is a cusp form of weight 12, then f /∆ is a modular form of
weight zero, hence is a constant. That is, the space of cusp forms of weight
12 is spanned by ∆.
Theorem 3.8 The space of modular forms of weight 2k with k ≥ 0 is
n
spanned by the modular forms Gm
4 G6 with 4m + 6n = 2k.
Proof Argue by induction on k. We have already excluded the possibilities
k < 0 and k = 0 and k = 1.
Suppose that k ≥ 2 and that f is modular of weight 2k. First observe
that any integer k ≥ 2 may be written as 2m + 3n for non-negative integers
m and n. Now, the q-expansion of G4 and G6 have non-zero constant term.
n
Hence h = f − λGm
4 G6 ) for a suitable constant λ, has no constant term in its
q-expansion, and is a cusp form. Now, Lemma (3.6) shows that g = h/∆ is a
modular form of weight 2k−12. By induction, g is a linear combination of the
modular forms Ga4 Gb6 with k − 6 = 2a + 3b whence, h is a sum of monomials
of the form Gp4 Gq6 with 2p + 3q = k (recall that ∆ is (60G4 )3 − 27(140G6 )2 ).
Therefore, so is f .
Notation 3.9 Define E2k (z) = G2k /2ζ(2k). Then, it follows from the
Fourier expansion of G2 and G6 that the modular forms E4 and E6 have inP
tegral Fourier coefficients. One sometimes writes ∆(z) = q + n≥2 τ (n)q n .
Then, ∆ has integral Fourier coefficients as well. We now consider the Zmodule spanned by E4m E6n with 4m + 6n = 2k. We get an integral lattice
54
T.N. Venkataramana
of modular forms of weight 2k. We will see later that this integral lattice is
stable under the Hecke operators.
4
Modular Forms and Representation Theory
Notation 4.1 We will begin with some calculations on the Lie
the group GL(2, R). Write,
0 1
1 0
1 0
0
X=
, Y =
, Z=
, and H =
0 0
0 0
0 1
−1
algebra g of
1
.
0
(27)
The complexified Lie algebra of GL(2, R) is M2 (C) the space of 2 × 2
matrices with complex entries; the Lie algebra structure is given by (a, b) 7→
[a, b] = ab − ba; M2 (C) is spanned by X, Y, Z and A. Write A = −iH (where
i ∈ h is the unique element whose square is -1). Then, A acts semisimply
(under the adjoint action) on g with real eigenvalues. Write
g = CE + ⊕ CE − ⊕ CZ ⊕ CA
(28)
where
E − = X + iY − (i/2)A − (i/2)Z and E + = X − iY − (i/2)A + (i/2)Z. (29)
Then E − and E + are eigenvectors for A with eigenvalues −2 and 2 respectively. Of course, on A and Z, A acts by 0. Thus, the complex Lie algebra
spanned by E + , E − and A is isomorphic to sl2 (C).
Definition 4.2 Fix the subgroup K∞ = O(2) of GL(2, R). This is the
group generated by
cosθ sinθ
SO(2) = Rθ =
:θ∈R
(30)
−sinθ cosθ
and
ι=
−1 0
.
0 1
(31)
Then, O(2) is a maximal compact subgroup of GL(2, R). Suppose that (π, V )
is a module for g as well as for O(2) such that the module structures are
compatible. That is, suppose that v ∈ V and ξ ∈ g, and σ ∈ O(2). Then,
π(σ)π(ξ)(v) = π(σ(ξ))(v)
Classical Modular Forms
55
where σ(ξ) is the inner conjugation action of O(2) on the Lie algebra g. One
then says that (π, V ) is a (g, K∞ )-module. If, as a K∞ -module, (π, V ) is a
direct sum of irreducible representations of K∞ with each irreducible representation occurring only finitely many times, then one says that the (g, K∞ )module is admissible. One then sees at once that a (g, K∞ )-submodule (or
a quotient module) of an admissible module is also admissible. One says
that a vector v ∈ V generates (π, V ) as a (g, O(2))-module, if the smallest
submodule of V containing v is all of V .
Notation 4.3 The Tensor Algebra. Given a (complex) vector space
V , denote by T n (V ) = V ⊗n the n-th tensor power of V . This is of
course, spanned by the pure tensors, i.e. vectors of the form v1 ⊗ · · · ⊗ vn
with vi ∈ V . By definition, T 0 (V ) = C and T 1 (V ) = V . Denote by
T (V ) = ⊕T n (V ) where the direct sum is over all the non-negative integers n.
Given non-negative integers m and n, there exists a linear map
T m (V ) ⊗ T n (V ) → T m+n (V )
which on pure tensors is the map
(v1 ⊗ · · · ⊗ vm ) ⊗ (w1 ⊗ · · · ⊗ wn ) 7→ (v1 ⊗ · · · ⊗ vm ⊗ w1 ⊗ · · · wn ).
This extends by linearity to all of T m (V ) ⊗ T n (V ) and thence to all of the
direct sum T (V ). Under this “multiplication”, T (V ) becomes an associative algebra, and is called the tensor algebra of the vector space V . The
subspace T 0 (V ) = C acts simply by scalar multiplication.
Notation 4.4 The Universal Enveloping Algebra. Given now a Lie
algebra g, let u(g) denote the quotient of the tensor algebra T (g) of g, by the
two sided ideal generated by the elements x⊗y −y ⊗x−[x, y], as x and y vary
over the elements of the Lie algebra g. Here, ⊗ denotes the multiplication
in the tensor algebra T (g) and the bracket [x, y] denotes the Lie bracket in g.
The algebra u(g) is called the universal enveloping algebra of g. Note
that g is a subspace of u(g), with [x, y] = xy − yx. Here, x, y are the images
of x, y ∈ g = T 1 (g) under the quotient map T (g) → u(g).
Suppose that u is some algebra over C and f : g → u a linear map such
that f ([x, y]) = f (x)f (y) − f (y)f (x) for all elements x, y ∈ g. Here f (x)f (y)
refers to the product of the two elements in the algebra u. Then, there exists
56
T.N. Venkataramana
a unique algebra map F : u(g) → u which extends f . This is why u(g) is
called the universal enveloping algebra of g.
In particular, if V is a module over g, we have a map f : g → End(V ) of
Lie algebras, and we have f ([x, y]) = [f (x), f (y)] where the bracket structure
on End(V ) is simply the commutator in the algebra End(V ). Therefore, by
the last paragraph, we get a unique extension F : u(g) → End(V ) , with F
an algebra map. In other words, the g module V is naturally a module over
u(g) as well.
Theorem 4.5 The Poincaré-Birkhoff-Witt Theorem: Let a, b and c
be subalgebras of a Lie algebra g such that g = a ⊕ b ⊕ c. Then, one has the
decomposition
u(g) = u(a) ⊗ u(b) ⊗ u(c)
Proof We must prove that every element of u(g) lies in the subspace of
the right-hand side of the above equation. Argue by an induction on the
degree of an element ξ ∈ T n (g). If we have an element yx for example, with
y ∈ c and x ∈ c, then, we may write it as xy − [x, y]. Now xy is in the
above subspace, and since g is by assumption a direct sum of a, b and c, the
element [x, y] also lies in the relevant subspace. We omit the details, since
this would be rather technical, and the reader can easily supply the details.
We now return to the group G = GL(2, R) and its Lie algebra g.
We will now prove the basic fact from representation theory which we
will use.
Theorem 4.6 Let (π, V ) be a (g, K∞ )-module. Suppose that v ∈ V has the
following properties:
(1) v generates V .
(2) The connected component SO(2) of O(2) acts on v by the character
determined by Rθ (v) = e2πiθm v, for some positive integer m (i.e. v is an
eigenvector for A with eigenvalue m).
(3) E − (v) = 0 and Z(v) = 0.
Then the (π, V ) is admissible and irreducible.
Proof Let u(g) denote the universal enveloping algebra of the Lie-algebra
g. One has the decomposition (the Poincaré-Birkhoff-Witt Theorem)
u(g) = u(g).[E − ] + u(g).[Z] ⊕ C[E + ] ⊗ C[A]
(32)
Classical Modular Forms
57
where C[ξ] denotes the algebra generated by the operator ξ. Therefore, if
(as in (31))
−1 0
ι=
0 1
then by assumptions (1) and (2) of the Theorem,
V = C[E + ](v) ⊕ ιC[E + ](v).
(33)
On E + the element A acts by the eigenvalue 2. Therefore, for an integer
p ≥ 0, the element (E + )p (v) is an eigenvector for A with eigenvalue (2p +
m), and ι(E + )p (v) is an eigenvector with eigenvalue (−2p − m) (note that
under the conjugation action of ι, the element A goes to −A, hence ι takes
an r-eigenspace for A into the −r-eigenspace). Since all these weights are
different, equation (33) shows that V is admissible as an SO(2) module (A
generates the complexified Lie algebra of SO(2)). In fact, equation (33)
shows that the multiplicity of an irreducible representation of SO(2) in V is
at most one, i.e. V is admissible.
Suppose that W ⊂ V is a submodule. In the last paragraph, we saw that
the action of A on V is completely reducible; hence the same holds for W .
Suppose that w is a weight vector in W of weight j, say. By replacing w
by ι(w) if necessary, we may assume that j > 0. The last paragraph shows
that j = 2p + m for some p ≥ 0 and also that (E + )p (v) = w (up to scalar
multiples). We may assume that p is the smallest non-negative integer such
that W contains the eigenvector (E + )p (v) = w with eigenvalue 2p + m. The
minimality of p implies that E − (w) = 0. Let W ′ be the submodule of W
generated by the vector w. To prove the irreducibility, it is enough to show
that W ′ = V . We may assume then that W = W ′ .
Since v generates V and Z annihilates v, it follows that Z acts by zero
on all of V . Therefore, the vector w satisfies all the properties that v does
in the assumptions of the Theorem (except that in (2) the eigencharacter is
2p + m). Therefore, cf. equation (33), we have
W = C[E + ](w) ⊕ ιC[E + ](w) = C[E + ](E + )p (v) ⊕ ιC[E + ](E + )p (v).
(34)
Now the equations (33) and (34) show that the codimension of W in V
is finite: dim(V /W ) < ∞. Hence V /W also satisfies the assumptions of
the Theorem (with v ∈ V replaced by its image v ∈ V /W ), but is finite
dimensional. This is impossible by the finite dimensional representation
theory of sl(2, C): a lowest weight vector (i.e. one killed by E − of sl(2))
58
T.N. Venkataramana
cannot have positive weight for A. But v has exactly this property in
V /W . This shows that V /W = 0 i.e. W = V .
Proposition 4.7 Given m > 0, there is a unique irreducible (g, K∞ )module ρm which satisfies the properties of 4.6.
Proof The uniqueness follows easily from the above proof of Theorem (4.6).
Let χm denote the one dimensional complex vector space on which the
group SO(2) acts by the character Rθ 7→ e2πimθ (where Rθ is, as in (30), the
rotation by θ in 2-space). Consider the space
u(g) ⊗ χm .
This is a representation for SO(2) (as well as for the universal enveloping
algebra u(g)). Let ρm be the O(2)-module induced from this SO(2)-module.
Then, ρm satisfies the properties of Theorem (4.6) and is therefore irreducible. Moreover, it is clear that any module V of the type considered in
4.6 is a quotient of ρm . By irreducibility, V = ρm .
Remark 4.8 The modules ρ2k are called the discrete series representations of weight 2k of (g, O(2)). This means the following. Suppose there
exists an irreducible unitary representation of the group GL(2, R), call it ρ.
Suppose that this occurs discretely (i.e. is a closed subspace of ) in a space
of functions L2 (G, ω) which transform according to the unitary character ω
of the centre Z of GL(2, R) and which are square summable with respect to
the Haar measure on the quotient GL(2, R)/Z. Given such a unitary module
ρ, consider the space of vectors whose translates under the compact group
O(2) form a finite dimensional vector space. This is the Harish-Chandra
module of the unitary representation ρ and is a (g, O(2))-module. The
representations ρ2k are the Harish-Chandra modules of discrete series representations of even weight. We will show in the next section, that these are
closely related to modular forms of weight 2k.
There are also the discrete series representations of odd weight, which
we will not discuss, since we are dealing with the group SL(2, Z) and it has
no modular forms of odd weight.
We are now in a position to state the precise relationship of modular
forms with representation theory.
Classical Modular Forms
59
Notation 4.9 Let f be a modular form of weight 2k with k > 0. We will
now construct a function on the group G+ = GL(2, R)+ as follows. Set
Ff (g) = j(g, i)−2k f (g(i))det(g)k
where i ∈ h is the point whose isotropy is the group SO(2) as in equation
(30). As before, j(g, i) = cz + d, where
g = ( ac db ).
By using the modularity of f and the equation j(gh, z) = j(g, h(z))j(h, z)
for the “automorphy factor” j(g, z), it is easy to see that Ff is invariant
under left translation by elements of SL(2, Z) and also under the centre Z∞
of GL(2, R).
We will now check that the (g, O(2))-module generated by Ff is isomorphic to ρ2k , with ρ2k as in 4.7. Note that Ff is contained in the space
C ∞ (Z∞ GL(2, Z)\GL(2, R),
the space of smooth functions on the relevant space and that the latter is a
(g, O(2))-module under right translation by elements of GL(2, R). Moreover,
for all y > 0 and x ∈ R we have
y x
= y k f (x + iy)
(35)
Ff
0 1
The function g 7→ f (g(i)) is right invariant under the action of SO(2)
since i is the isotropy of SO(2). Using the fact that j(Rθ , i) = e−iθ (where
Rθ is as in (30)) one checks that j(gRθ , i) = j(g, i)(e−2iθ ). Therefore, it
follows that
Ff (gRθ ) = Ff (g)e2ikθ .
(36)
0 1 ) (A
This equation implies that under the action of the element A = i( −1
0
generates the Lie algebra of SO(2)), Ff is an eigenvector with eigenvalue 2k.
Compute the action of E − (E − as in (29)) on Ff . Using the invariance
of of Ff under Z∞ and that it is an eigenvector of A with eigenvalue 4k, one
sees that
E − (Ff ) = (X + iY )Ff − ikFf .
Now use equation (35) to conclude that E − Ff = y 2k (∂f /∂z). Since f is
holomorphic, one obtains that E − Ff = 0.
The (g, O(2)) module generated by Ff satisfies the conditions of 4.6.
60
T.N. Venkataramana
Notation 4.10 Growth Properties of Ff . Consider now the growth
properties of Ff . We have the quotient map
GL(2, R)+ → h(⊃ F )
where F is the fundamental domain constructed in section 2. Let S be the
pre-image of F under this quotient map. Then, we have the inclusion
y x
2
: y > 3/4 and − 1/2 < x < 1/2
S ⊂ Z∞ O(2)
0 1
and the latter is a “Siegel set”. Now,
y x
Ff
= y k f (x + iy).
0 1
From the modularity property of f , it follows that f is “bounded at infinity”,
which means that there exists a constant C > 0 such that on the fundamental
domain F of SL(2, Z), the function z 7→ f (z) = f (x + iy) is bounded by C:
| f (z) |≤ C ∀z ∈ F.
Therefore, on the Siegel set S, we have
y x
Ff
≤ Cy k ,
0 1
i.e. Ff has moderate growth on the Siegel Set.
Suppose now that f is a cusp form. Then, the Fourier expansion at
infinity of f is of the form
f (z) =
n=∞
X
an exp(2πinz)
n=1
where a(n) are the Fourier coefficients. The function
X
an q n−1
is clearly bounded in a neighbourhood of infinity in the fundamental domain
F and the complement of a neighbourhood of infinity being compact, it is
bounded on all of F too, by a constant C.
Classical Modular Forms
61
This shows the existence of a constant C > 0 such that for all z ∈ F ,
one has
f (x + iy) ≤ Cexp(−y).
The Haar measure on the group GL(2, R) is the product of the Haar
measure on the group O(2) and the invariant measure dxdy/y 2 on the upper
half plane h = GL(2, R)/O(2) constructed at the end of section 2. This is
an easy exercise.
Thus, the square of the absolute value of Ff integrated on S = π −1 (F )
(where π is the quotient map GL(2, R) → h ) is simply the integral of the
square of the absolute value of f (z) on the domain F . The above estimate
for f shows that this integral over F is finite:
!
Z
Z
∞
1/2
dx
−1/2
(dy/y 2 )exp(−y)
<∞
(1−x2 )1/2
Definition 4.11 Automorphic Forms. Recall the definition of automorphic forms on GL(2, Z). These are smooth functions φ on the quotient
GL(2, Z)\GL(2, R), which are
(1) K-finite. That is, the space of right translates of φ under the compact
group K = O(2) forms a finite dimensional vector space.
(2) The function φ has moderate growth on the Siegel set St,1/2 , i.e.
St,1/2 is a set of the form N1/2 At KZ where, Z is the centre of GL(2, R),
N1/2 is the set of matrices of the form n = ( 10 x1 ) with −1/2 ≤ x ≤ 1/2, and
At is the set of diagonal matrices of the form a = ( y0 10 ) with (0 <)t < y, and
there exists a constant C > 0 such that in the above notation,
φ(nakz) ≤ Cy N
for some integer N and for all elements nakz ∈ St,1/2 in the Siegel set.
(3) There is an ideal I of finite codimension in the centre of the universal
enveloping algebra U (g) which annihilates the smooth function φ.
The last few paragraphs imply the following
Theorem 4.12 Let f be a modular form of weight 2k. Let Ff be the associated function on GL(2, Z)\GL(2, R). Then, Ff is an automorphic form.
Moreover, the (g, O(2)) module generated by Ff is isomorphic to ρ2k with
ρ2k as in 4.7
62
T.N. Venkataramana
Moreover, if f is a cusp form, then, the associated function Ff is rapidly
decreasing on the Siegel domain, and is therefore square summable on the
quotient space Z∞ GL(2, Z)\GL(2, R).
Proof We need only check that an ideal I of finite codimension in the
centre z of the universal enveloping algebra of g annihilates Ff . But, the
module generated by Ff is ρ2k by 4.6 (and 4.7). Now, the 2k eigenspace of
the operator A in the representation ρ2k is one dimensional (and is generated
by Ff ), and z commutes with the action of A (and in fact with all of u(g) as
well). Therefore, the annihilator of Ff in z is an ideal I of codimension one.
Theorem 4.13 The space M2k of modular forms of weight 2k for the group
SL(2, Z) may be identified with the isotypical subspace of the irreducible
(g, O(2)) module ρ2k in the space
C ∞ (Z∞ GL(2, Z)\GL(2, R)).
The isomorphism is obtained by sending a modular form f to the span
of the function Ff under the action of (g, O(2)) (the latter (g, O(2))-module
is isomorphic to ρ2k ).
5
Modular Forms and Hecke Operators
Notation 5.1 Let Af be the ring of finite adeles over Q. Recall that this is
the direct limit (the maps are inclusion maps) as the finite set S of primes
varies, of the product
Y
Y
Qp ×
Zp .
AS =
p∈S
p∈S
/
A∗f
The group of units of Af is the group
of ideles and is the direct limit as
S varies, of
Y
Y
A∗S =
Q∗p ×
Z∗p ,
p∈S
p∈S
/
(where ∗ denotes the group of units of the ring under consideration).
There is a natural inclusion of Q in Af (and hence of Q∗ in A∗f and
of GL(2, Q) in GL(2, Af )). Denote by P the set of primes. The Strong
Approximation Theorem (Chinese Remainder Theorem) implies that
Y
Af = Q +
Zp .
(37)
p∈P
Classical Modular Forms
63
This, and the fact that Z is a principal ideal domain imply that
Y
A∗f = Q∗ .
Z∗p .
(38)
p∈P
From this it is not difficult to deduce that
GL(2, Af ) = GL(2, Q).
Y
GL(2, Zp )
(39)
p∈P
Q
Note that the intersection of GL(2, Q) with Kf = GL(2, Zp ) is precisely
GL(2, Z).
Let A = R × Af be the ring of adeles over Q. Then, Q is diagonally imbedded in A. Hence there is a diagonal imbedding of GL(2, Q)
in GL(2, A) = GL(2, R) × GL(2, Af ). Then, GL(2, Q) is a discrete subgroup
of GL(2, A). Now equation (39) (a consequence of strong approximation)
implies that
Y
GL(2, A) = GL(2, Q)(GL(2, R) ×
GL(2, Zp )).
(40)
p∈P
Now, equation (40) and the last sentence of the previous paragraph imply
that the quotient
Y
GL(2, Zp )).
(41)
GL(2, Q)\GL(2, A) = GL(2, Z)\(GL(2, R) ×
p∈P
Note that GL(2, A) acts by right translations on the left-hand side of the
equation (41).
Notation 5.2 A representation (π, W ) of GL(2, Af ) is said to be smooth
if the isotropy of any vector in W is an open subgroup of GL(2, Af ). Define
the “Hecke algebra” H of GL(2, Af ) as the space of compactly supported
locally constant functions on GL(2, Af ). If W is a smooth representation of
GL(2, Af ), then the Hecke Algebra H also operates on W by “convolutions”:
if µ is a Haar measure on GL(2, Af ), φ ∈ H, and w ∈ W is a vector, then the
W valued function g 7→ φ(g)π(g)w is a locally constant compactly supported
function and hence can be integrated with respect to the Haar measure µ.
Define
Z
φ ∗ w = π(φ)(w) = φ(g)π(g)(w)dµ(g)
(42)
64
T.N. Venkataramana
This gives the GL(2, Af )-module π, the structure of an H-module. As is well
known, the category of smooth representations of GL(2, Af ) is isomorphic
to the category of representations of the Hecke algebra H, the isomorphism
arising from the foregoing action of the Hecke algebra on the smooth module
π.
b is an open compact subgroup of
Notation 5.3 The group K0 = GL(2, Z)
GL(2, Af ) and is the product over all primes p of the groups GL(2, Zp ).
Given g ∈ GL(2, Af ), consider the characteristic function χg of the double
coset set K0 gK0 . Then χg is an element of the Hecke algebra and elements
of H which are bi-invariant under H are finite linear combinations of the
functions χg as g varies. We will refer to the subalgebra generated by these
elements as the ‘unramified Hecke algebra and denote it by H0 .
Under convolution, H is an algebra and H0 is a commutative subalgebra.
Fix a prime p. Let H0 (p) be the subalgebra generated by the elements χMp
and χNp where Mp = ( p0 10 ) and Np = ( p0 0p ). It is easily proved that for
varying p, the algebras H0 (p) generate the unramified Hecke algebra H0 .
Notation 5.4 The equation (41) implies that the space of smooth functions
on Z∞ SL(2, Z)\GL(2, R)+ is isomorphic to the space V0 of K0 -invariant
smooth functions on the quotient GL(2, Q)Z(A)\GL(2, A). On V0 the unramified Hecke algebra operates. Suppose S denotes the image of F × K0 in
Z(A)\GL(2, A). Then, S is contained in a Siegel set S0 whose elements are
of the form
y x
z∞
× k0
0 1
where z∞ ∈ Z∞ , k0 ∈ K0 , | x |< 1/2 and y 2 > 3/4. Suppose that f is a
cusp form for SL(2, Z) and Ff be as in section (4.5). Given g ∈ GL(2, A) =
GL(2, R) × GL(2, Af ), write g = (g∞ , gf ) accordingly. Define the function
Φf on GL(2, Q)\GL(2, A) as follows. Set Φf (g∞ , gf ) = Ff (g∞ ) if gf ∈ K0
and extend to G(A) by demanding that Φf be GL(2, Q)-invariant. The
SL(2, Z)-invariance of Ff implies that Φf is well defined. Now, 4.12 shows
that Φf is an automorphic form on GL(2, A).
By 4.13, Ff is rapidly decreasing on S0 ; moreover, Ff is a cuspidal
automorphic form in the sense that for all g ∈ G(A), the following holds.
Z
Φf (ng)dn = 0
(43)
U (Q)\U (A)
Classical Modular Forms
65
where U is the group of unipotent upper triangular matrices in GL(2) with
ones on the diagonal and dn is the Haar measure on U (A). To prove this,
we note that the vanishing of the integral is unaltered by changing g on the
(1) right by an element of Z∞ O(2) × K0 , since Φf is an eigenvector for
the right translation action by K∞ Z∞ × K0 . We may hence assume that
b an element of G(Af )
g = (g∞ , gf ). Now, up to elements of K0 = GL(2, Z)
is upper triangular;
(2) left by an element of B(Q) since G(Q) normalises U (A) and U (Q),
and preserves the Haar measure on U (A). Note that (the Iwasawa decomposition) the double coset B(Q)\G(Af )/K0 Z(A) is a singleton. Hence, we
may assume that g = g∞ = ( y0 x1 ). Then the above integral is the same as
Z
f (x + n + iy)dn
U (Z)\U (R)
which is nothing but the zero-th Fourier coefficient of f , and by the cuspidality of f , this is zero.
On the Siegel domain, the modular function f satisfies an estimate of
the form
|f (x + iy)| < Cexp(−y)
where C is some constant. This implies that on the Siegel set S, the function
Φf satisfies an estimate of the form
Φf (g) = O(|g|−N )
for some positive integer N . This can be shown to imply that the function
Φf is square summable on the quotient Z(A)GL(2, Q)\GL(2, A) with respect
to the Haar measure. Further, one has the L2 -metric < , > on the space
of cuspidal automorphic forms which translates to the “Petersson” metric
Z
f (z)g(z)y 2k (y −2 dxdy)
< f, g >=
F
for cusp forms f and g of weight k. As before, F is the fundamental domain
for SL(2, Z).
Notation 5.5 From now on, we will fix our attention on cusp forms. We
have the natural inclusion of GL(2, Q) in GL(2, Af ). Let p > 0 be a prime
and let gp = ( p0 10 ) be thought of as an element in GL(2, Af ) under the
66
T.N. Venkataramana
foregoing inclusion. Let Xp denote the characteristic function of the double
coset of K0 through the element gp .
If χp denotes the characteristic function Xp , and f is a cuspidal modular form of weight 2k, then Φ′ = Φf ∗ R(χp ) (where R(φ) denotes the
right convolution by the function φ) is a smooth function on the quotient
GL(2, Q)Z(A)\GL(2, A) whose “infinite” component is still ρ2k (since χp
commutes with the right action of GL(2, R) on the above quotient). Since
χp is K0 invariant, it follows that Φ′ is also right K0 invariant. Therefore,
it corresponds to a modular form g, i.e. Φ′ = Φg . It is easy to show that
Φ′ is cuspidal (the space of cusp forms is stable under right convolutions).
Therefore, g is a cusp form of weight 2k as well. Denote g = T (p)(f ). Then,
T (p) is called the Hecke operator corresponding to the prime p. By noting
that convolution by χp is self-adjoint for the L2 metric on cuspidal automorphic functions on GL(2) one immediately sees that the operators T (p) are
self-adjoint for the Petersson metric on the space of cusp forms of weight
2k. The commutativity of the unramified Hecke algebra implies that the
operators T (p) (as p varies) commute as well.
Definition 5.6 Now a commuting family of self-adjoint operators on a finite dimensional complex vector space can be simultaneously diagonalised.
Consequently, there exists a basis of cusp forms of weight 2k which are simultaneous eigenfunctions for all the Hecke operators T (p); these are called
Hecke eigenforms. If f is a Hecke eigenform for SL(2, Z) and has constant
term 1, then it is called a normalised Hecke eigenform.
Theorem 5.7 The Iwasawa Decomposition: Any matrix in GL(2, Af )
may be written as a product bk with b ∈ B(Af ) (the group of upper triangular
b
matrices), and k ∈ K0 = GL(2, Z).
Proof This is an easy application of the elementary divisors theorem. By
identifying B\G with the projective line P1 , we see that the Iwasawa decomb on P1 (Af ).
position amounts to the transitivity of the action of GL(2, Z)
But, any element of P1 (Af ) may be written as a vector (x, y) ∈ A2f where
for every prime p, the p-th components (xp , yp ) are not both zero.
By changing (x, y) by an element of A∗f if necessary, (x, y) may be asb 2 . Further, x, y may be assumed to be coprime, in the sense
sumed to be in Z
that for every prime p, the p-adic components xp , yp of x, y are coprime.
Now, by writing everything in the notation of row vectors, we want to solve
Classical Modular Forms
b the equation
for g ∈ GL(2, Z)
67
(x, y) = (0, 1)g
(note that the isotropy at (0, 1) is precisely B(Af )). This amounts to finding
b 2 to a basis of Z
b 2 , which can be done precisely because x, y are
(z, t) ∈ Z
coprime.
Notation 5.8 This implies that for a prime p, we have
Xp = ∪( p0 x1 )K0 ∪ ( 10 0p )K0
b
where the union is a disjoint union, and 0 ≤ x ≤ p − 1. Here K0 = GL(2, Z),
as before.
Notation 5.9 We will now state without proof the computation of T (p) for
a prime p. Note that by strong approximation, the K0 invariant function
Φf on the quotient Z(A)GL(2, Q)\GL(2, A) is completely determined by
its values on elements of the form ( y0 x1 ) with y > 0, in the quotient. We
compute (using the description of Xp in the previous section)
Φf ∗ R(χp )( y0 x1 )
and find that this is equal, to
p2k−1 Φg ( y0 x1 )
where
g(z) = (1/p)
X
f ((z + m)/p) + p2k−1 f (pz) = T (p)(f )(z).
0≤m≤p−1
The Fourier coefficients of g at infinity are given by
g(m) = a(mp)
if m is coprime to p and
g(m) = a(mp) + p2k−1 a(m/p)
if p divides m.
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T.N. Venkataramana
Notation 5.10 In particular, if f is an eigenfunction for all the T (p) with
eigenvalue λ(p) say, the equation T (p)f = λ(p)f implies, by comparing the
p-th Fourier coefficients, that a(p) = λ(p)a(1) for each p.
Remark 5.11 In particular, we get from the last two paragraphs, that if f
is an eigenform whose first Fourier coefficient a(1) is zero, then, a(m) = 0 for
all positive m. This easily follows from induction and the formula a(mp) =
λ(p)a(p) if m and p are coprime, and a(m)λ(p) = a(mp) + p2k−1 a(m/p) if p
divides m. Hence f = 0. Thus, we have proved that every Hecke eigenform
is a nonzero multiple of a normalised Hecke eigenform.
Theorem 5.12 (The Multiplicity 1 Theorem): Let f1 and f2 be two
normalised Hecke eigenforms for the action of the Hecke operators T (p) with
the same eigenvalues λ(p) for every prime p. Then, f1 = f2 .
Proof We will prove this by showing that the Fourier coefficients of f1
and f2 are the same. This will imply, by the Fourier expansion for modular
forms, that f1 = f2 . Write f = f1 − f2 .
Now, the first Fourier coefficient of f is zero, since f1 and f2 are normalised. Further, f is also a Hecke eigenform, since f1 and f2 are so, and
with the same eigenvalues. Therefore, by the previous remark, f = 0.
Recall that we have identified [representations π of GL(2, A) whose infinite component π∞ is ρ2k and whose finite component πf contains a non-zero
GL(2, Af ) invariant vector], with [normalised eigenforms f of weight 2k for
the group GL(2, Z)].
Therefore, we have proved that the multiplicity of such a π in the
space of cusp forms on GL(2, A) is one.
Remark 5.13 Later, Cogdell will prove that the multiplicity of a cuspidal
automorphic representation of GL(n) is always 1. This is the famous multiplicity 1 theorem due to Jacquet-Langlands for GL(2) and to PiatetskiiShapiro and Shalika in general.
What we have proved is therefore a very special case when n = 2, the
infinite component is ρ2k , and the representation is unramified at all the
local places.
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69
Definition 5.14 We now define inductively, the operators T (n) as follows.
If m and n are coprime, we define T (mn) = T (m)T (n). This reduces us to
defining T (pm ) where p is a prime. Define recursively the operators T (pm )
by the formula T (p)T (pm ) = T (pm+1 ) + pT (pm−1 ).
This now implies (by the similarity of the recursive formulae for T (n)
P
and the Fourier coefficients a(n)), that if f =
a(n)q n is a normalised
Hecke eigenform, then T (n)f = a(n)f for all n. These T (n)’s are the classical Hecke operators. By construction, they commute, one has T (mn) =
T (m)T (n) if m, n are coprime, and they are self-adjoint for the Petersson
inner product on modular forms of weight 2k.
Theorem 5.15 If f is a normalised Hecke eigenform for SL(2, Z), then,
all its Fourier coefficients are algebraic integers.
Proof We consider the action of the Hecke operators T (n) on the space
0 of cups forms. Note that the space of cusp forms contains the (adM2k
ditive) subgroup L of those cusp forms whose Fourier coefficients are rational integers. This subgroup is stable under the action of the operators
T (n). To see this, first suppose that n = p is a prime. By the formula
aT (p)f (m) = a(pm) + p2k−1 a(m/p) if p divides m and aT (p)f (m) = a(pm)
otherwise, we see that T (p) stabilises the subgroup L. Since the T (p) generate T (n), the operator T (n) also stabilises L.
By induction on k, we see that the space M2k of modular forms of
weight 2k has a basis whose Fourier coefficients are integral: M2k = CE2k ⊕
M2k−12 ∆, and ∆ and E2k have integral Fourier coefficients. This shows that
the subgroup L of the last paragraph contains a basis of the space of cusp
0 .
forms M2k
Since the operator T (n) is self adjoint with respect to a suitable metric
0 , it follows that the eigenvalues of T (n) are all real and are
on the space M2k
all algebraic integers.
Now the Fourier coefficients a(n) of the eigenform f are nothing but
the eigenvalue λ(n) of T (n) corresponding to the eigenvector f , by the last
paragraph of the previous section. Consequently, the Fourier coefficients of
f are all real algebraic integers.
70
T.N. Venkataramana
6
L-functions of Modular Forms
In this section, we define the L-function of a cusp form for SL(2, Z) and
prove that it has analytic continuation to the entire plane and has a nice
functional equation. Later, Cogdell will prove analogous statements for cuspidal automorphic representations for GL(n).
To begin with, we prove an estimate –due to Hecke– for the n-th Fourier
coefficient of a cusp form of weight 2k.
Pn=∞
Lemma 6.1 (Hecke) Let f = n=1
a(n)q n be a cusp form of weight 2k.
Then, there exists a constant C > 0 such that
a(n) ≤ Cnk/2
∀n ≥ 1.
Proof Consider the function φ defined and continuous on the upper half
plane h, given by
φ(z) = y k/2 |f (z)|.
If γ = ( ac db ) ∈ SL(2, Z) and z ′ = γ(z) = x′ + iy ′ , then recall that y ′ =
y/(|cz + d|2 ). Therefore, we obtain from the modularity property of f , that
φ(γ(z)) = φ(z), i.e. φ is invariant under SL(2, Z). Hence, φ is determined
by its restriction to the fundamental domain F . As z tends to infinity in F ,
the cuspidality condition of f shows that f (z) = O(exp(−2πy)). Therefore,
φ is bounded on F and hence on all of the upper half plane h. Thus, there
exists a constant C1 such that
|f (z)| ≤ C1 y −k/2
∀z ∈ h.
Now consider the n-th Fourier coefficient a(n) of f . Clearly,
2πny
a(n) = e
Z
1
f (x + iy)e−2iπnx dx.
0
By applying the foregoing estimate for f to this equation, we obtain
|a(n)| ≤ C1 e2πny y k/2
for all y > 0. Take y = 1/n. We then get
a(n) ≤ (C1 e2π )nk/2 .
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71
Remark 6.2 Deligne has proved that for all primes p, |a(p)| = O(p((k−1)/2)) ,
by linking these estimates with the “Weil Conjectures” for the number of
rational points of algebraic varieties over finite fields.
P
Definition 6.3 If f (z) =
a(n)q n is a cusp form of weight 2k, define for
a complex variable s, the Dirichlet Series L(f, s) by the formula
L(f, s) =
∞
X
a(n)/ns .
n=1
From Lemma (6.1), it follows that the series converges and is holomorphic in
the region Re(s) > 1 + (k/2). The function L(f, s) is called the L-function
of the cusp form f .
Notation 6.4 an integral expression for L(f, s)). We will now write an
integral formula for the L-function of f . First consider the integral
Z ∞
f (iy)y s (dy/y).
0
Since f (iy) = O(exp(−2πy)) for all y > 0 (cf. the proof of Lemma (6.1)), it
follows that if Re(s) > 0, then the integral converges. Let σ be the real part
of s. Then, for each n ≥ 1 the integral
Z ∞
|a(n)|e−2πny y σ (dy/y)
0
converges, and is equal to (|a(n)|/nσ )(2π)−σ Γ(σ) where Γ is the classical
Γ-function:
Z ∞
e−t tz (dt/t).
Γ(z) =
0
From the Hecke estimate a(n) = O(nk/2 ) of Lemma (6.1), it follows that the
infinite sum of these integrals also converges, provided σ > k/2 + 1. Thus,
by the Dominated Convergence Theorem (to justify the interchange of sum
and integral), we obtain the equation
Z ∞
∞
X
a(n)n−s .
f (iy)y s (dy/y) = (2π)−s Γ(s)
0
n=1
We finally obtain the integral expression:
Z ∞
f (iy)y s (dy/y) = (2π)−s Γ(s)L(f, s).
0
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T.N. Venkataramana
Remark 6.5 Notice that the integral expression says that the function
L(f, s) can be analytically continued to the region Re(s) > 0, since the
left-hand side is analytic there, and the function (2π)−s Γ(s) has no zero’s
on the complex plane.
Notation 6.6 The functional Equation. The L-Function and the integral expression could have been defined for any holomorphic function
P
f (z) =
a(n)q n (q = e2πiz ), provided a(n) satisfy a Hecke estimate. We
will now prove a functional equation for L(f, s), by using the modularity
property, especially that f (−1/z) = (−1)k z 2k f (z).
R∞
Consider now the integral I(s) = 0 f (iy)y s (dy/y), which converges for
Re(s) > 0. We write the integral as a sum of the integral from 1 to ∞ and
the integral from from 0 to 1. By making a change of variable y 7→ 1/y we
get,
Z ∞
Z 1
f (−1/(iy))y −s (dy/y).
f (iy)y s (dy/y) =
1
0
Note that as f (1/(iy)) is bounded on the interval [1, ∞], and Re(s) > 0, the
integral on the right side converges. Therefore, we get, using the functional
equation f (−1/iy) = (−1)k y 2k f (iy), that
Z
1
s
k
f (iy)y (dy/y) = (−1)
0
We then get
I(s) =
Z
∞
Z
∞
f (iy)y 2k−s (dy/y).
1
f (iy)(y s + (−1)k y 2k−s )(dy/y).
1
This holds for all s, with Re(s) > 0. We now make the change of variable
s 7→ 2k − s, for s in the region 0 < Re(s) < 2k. Then the above expression
for I(s) shows that I(2k − s) = (−1)k I(s). This is the functional equation
for L(f, s):
(2π)−s Γ(s)L(f, s) = (−1)k (2π)2k−s Γ(2k − s)L(f, 2k − s)
for all s in the region 0 < Re(s) < 2k. The left side of this equation is
analytic in the region Re(s) > 0 and the right side is analytic in the region
Re(s) < 2k. Using the functional equation (and the fact the (2π)−s Γ(s)
never vanishes on the complex plane), we now see that L(f, s) has an analytic
continuation over the entire complex plane.
Classical Modular Forms
73
Notation 6.7 Euler Factors. In the last few sections, we derived the
functional equation and analytic continuation of a cusp form for SL(2, Z).
We now derive an Euler product, for the L-function of a normalised Hecke
eigenform f .
P
n
Let then f = ∞
n=1 a(n)q be a normalised Hecke eigenform of weight
2k. Fix a prime p and consider the infinite sum
Lp (f, s) =
∞
X
a(pm )/pms
m=0
where Re(s) > k/2 + 1. It converges, by the Hecke estimate a(n) = O(nk/2 ).
We now use the relations a(pm )a(p) = a(pm+1 ) + p2k−1 a(pm−1 ). Multiplying
these equations by 1/p(m+1)s and then summing over all m ≥ 0 we get
a(p)Lp (f, s) = (Lp (f, s) − 1)ps + (p2k−1 /p2s )Lp (f, s).
That is,
Lp (f, s)−1 = 1 − a(p)/ps + p2k−1 /p2s .
We now use the fact that a(mn) = a(m)a(n) if m, n are coprime, since
f is a normalised Hecke eigenform. Form the product over all primes p of
these Lp (f, s). We then get (by using the Dominated Convergence Theorem
Q
P
to justify interchanges) the equation
Lp (f, s) =
a(n)/ns = L(f, s).
Thus we have the infinite product expansion (the product being over all
primes p)
Y
1/(1 − a(p)/ps + p2k−1 /p2s ).
L(f, s) =
p
From now on we consider the function L∗ (f, s) = (2π)−s Γ(s)L(f, s) and
refer to this as the L-function of f .
We have thus proved the following Theorem.
P
n
Theorem 6.8 Let f = ∞
of
n=1 a(n)q be a normalised Hecke eigenform
P
weight 2k for SL(2, Z). Then, the L-function L∗ (f, s) = (2π)−s Γ(s) a(n)/ns
converges for Re(s) > k/2, has an analytic continuation to the entire complex plane and satisfies the functional equation
L∗ (f, s) = (−1)k L∗ (f, 2k − s).
Moreover, in the region Re(s) > k/2, one has the Euler product
Y
L∗ (f, s) = (2π)−s Γ(s)
1/(1 − a(p)/ps + p2k−1 /p2s ).
p
74
T.N. Venkataramana
Remark 6.9 Recall that to each Hecke eigenform f of weight 2k, there
corresponds an irreducible cuspidal automorphic representation π(f ) = π =
π∞ ⊗p πp of the (restricted direct product) group GL(2, A) = GL(2, R) ×
Q
p GL(2, Qp ) such that π∞ is the discrete series representation ρ2k and each
πp is an irreducible unramified representation of GL(2, Qp ).
In the lectures of Cogdell, you will see that each cuspidal automorphic
representation π of GL(n, A) has an L-function attached to it –denoted
L(π, s)– which satisfies a functional equation, has an analytic continuation
to the entire plane, and has an Euler product comprising of terms which are
monic polynomials in p−s of degree n.
It turns out that for the representation π(f ) = π attached to the Hecke
eigenform f of weight 2k, the L-function is nothing but L(π, s) = L∗ (f, s +
(k − 1/2)), which can easily be seen to satisfy the equation
L(π, s) = (−1)k L(π, 1 − s).
Moreover, the local factors are of the form
L(πp , s)−1 = (1 − (a(p)/pk−1/2 )/ps + 1/p2s ,
a monic polynomial in in p−s of degree two.