10Modular forms and rep theory
III Modular Forms and L-functions
10 Modular forms and rep theory
In this final chapter, we are going to talk about the relation between modular
forms and representation theory. The words “representation theory” are a bit
vague. We are largely going to talk about automorphic representations, and this
is related to Langlands programme.
Recall that f is a modular form on SL
2
(Z) if
(i) f is holomorphic H → C
(ii) f |
k
γ = (cz + d)
−k
f(γ(z)) = f(z) for all
γ =
a b
c d
∈ SL
2
(Z)
(iii) It satisfies suitable growth conditions at the cusp ∞.
Let’s look at the different properties in turn. The second is the modularity
condition, which is what gave us nice properties like Hecke operators. The
growth condition is some “niceness” condition, and for example this gives the
finite-dimensionality of the space of modular forms.
But how about the first condition? It seems like an “obvious” condition to
impose, because we are working on the complex plane. Practically speaking,
it allows us to use the tools of complex analysis. But what if we dropped this
condition?
Example. Recall that we had an Eisenstein series of weight 2,
E
2
(z) = 1 − 24
X
n≥1
σ
1
(n)q
n
.
This is not a modular form. Of course, we have
E
2
(
z
) =
E
2
(
z
+ 1), but we saw
that
E
2
−
1
z
− z
2
E
2
(z) =
12z
2πi
6= 0.
However, we can get rid of this problem at the expense of making a non-
holomorphic modular form. Let’s consider the function
f(z) =
1
y
=
1
Im(z)
= f(z + 1).
We then look at
f
−
1
z
− z
2
f(z) =
|z|
2
y
−
z
2
y
=
z(¯z − z)
y
= −2iz.
Aha! This is the same equation as that for
E
2
apart from a constant factor. So
if we let
˜
E
2
(z) = E
2
(z) −
3
πy
,
then this satisfies
˜
E
2
(z) =
˜
E
2
(z + 1) = z
−2
˜
E
2
−
1
z
.
The term
3
πy
certainly tends to 0 rapidly as
|z| → ∞
, so if we formulate the
growth condition in (iii) without assuming holomorphicity of
f
, then we will find
that
˜
E
2
satisfies (ii) and (iii), but not (i). This is an example of a non-holomorphic
modular form of weight 2.
Perhaps this is a slightly artificial example, but it is one.
Let’s explore what happens when our functions satisfy (ii) and (iii), but not
(i).
Definition
(Non-holomorphic modular forms)
.
We let
W
k
(Γ(1)) be the set of
all C
∞
functions H → C such that
(ii) f |
k
γ = f for all γ ∈ Γ(1)
(iii) f
(
x
+
iy
) =
O
(
y
R
) as
y → ∞
for some
R >
0, and the same holds for all
derivatives.
Note that the notation is not standard.
Before we proceed, we need to introduce some notation from complex analysis.
As usual, we write z = x + iy, and we define the operators
∂
∂z
=
1
2
∂
∂x
+
∂
i∂y
∂
∂¯z
=
1
2
∂
∂x
−
∂
i∂y
.
We can check that these operators satisfy
∂z
∂z
=
∂¯z
∂¯z
= 1,
∂¯z
∂z
=
∂z
∂¯z
= 0.
Moreover, the Cauchy–Riemann equations just says
∂f
∂ ¯z
= 0, and if this holds,
then the complex derivative is just
∂f
∂z
. Thus, if we are working with potentially
non-holomorphic functions on the complex plane, it is often useful to consider
the operators
∂
∂z
and
∂
∂ ¯z
separately.
Using this notation, given f ∈ W
k
, we have
f ∈ M
k
⇐⇒
∂f
∂¯z
= 0.
So suppose
f
is not holomorphic, then
∂f
∂ ¯z
6
= 0. We can define a new operator by
L
∗
k
(f) = −2iy
2
∂f
∂¯z
.
Note that this is slightly strange, because we have a subscript
k
, but the function
doesn’t depend on
k
. Also, we put a star up there for some reason. It turns out
there is a related operator called
L
k
, which does depend on
k
, and this
L
∗
k
is a
slight modification that happens not to depend on k.
This has the following properties:
Proposition.
– We have L
∗
k
f = 0 iff f is holomorphic.
– If f ∈ W
K
(Γ(1)), then g ≡ L
∗
k
f ∈ W
k−2
(Γ(1)).
Thus, L
∗
k
is a “lowering” operator.
Proof. The first part is clear. For the second part, note that we have
f(γ(z)) = (cz + d)
k
f(z).
We now differentiate both sides with respect to
¯z
. Then (after a bit of analysis),
we find that
(c¯z + d)
−2
∂f
∂¯z
(γ(z)) = (cz + d)
k
∂f
∂¯z
.
On the other hand, we have
(Im γ(z))
2
=
y
2
|cz + d|
4
.
So we find
g(γ(z)) = −2i
y
2
|2z + d|
4
(c¯z + d)
2
(cz + d)
k
∂f
∂¯z
= (cz + d)
k−2
g(z).
The growth condition is easy to check.
Example. Consider
˜
E
2
defined previously. Since E
2
is holomorphic, we have
L
∗
k
˜
E
2
=
6i
π
y
2
∂
∂¯z
1
y
= constant,
which is certainly a (holomorphic) modular form of weight 0.
In general, if
L
∗
k
f
is actually holomorphic, then it is in
M
k−2
. Otherwise, we
can just keep going! There are two possibilities:
– For some 0 ≤ ` <
k
2
, we have
0 6= L
∗
k−2`
···L
∗
k−2
L
∗
k
f ∈ M
k−2`
.
–
The function
g
=
L
∗
2
L
∗
4
···L
∗
k
f ∈ W
0
(Γ(1)), and is non-zero. In this case,
g(γ(z)) = g(z) for all γ ∈ SL
2
(Z).
What does
W
0
(Γ(1)) look like? Since it is invariant under Γ(1), it is just a
C
∞
function on the fundamental domain
D
satisfying suitable
C
∞
conditions on the
boundary. This space is huge. For example, it contains any
C
∞
function on
D
vanishing in a neighbourhood of the boundary.
This is too big. We want to impose some “regularity” conditions. Previously,
we imposed a very strong regularity condition of holomorphicity, but this is too
strong, since the only invariant holomorphic functions are constant.
A slightly weaker condition might be to require it is harmonic, i.e.
˜
∆f
=
∂
2
f
∂x
2
+
∂
2
f
∂y
2
= 0. However, the maximum principle also implies f must vanish.
A weaker condition would be to require that
f
is an eigenfunction of
˜
∆
, but
there is a problem that
˜
∆
is not invariant under Γ(1). It turns out we need a
slight modification, and take
∆ = −y
2
∂
2
∂x
2
+
∂
2
∂y
2
.
It is a straightforward verification that this is indeed invariant under
SL
2
(
R
), i.e.
∆(f(γ(z))) = (∆f)(γ(z)).
In fact, this ∆ is just the Laplacian under the hyperbolic metric.
Definition
(Maass form)
.
A Maass form on
SL
2
(
Z
) is an
f ∈ W
0
(Γ(1)) such
that
∆f = λf
for some λ ∈ C.
There are interesting things we can prove about these. Recall that our
first examples of modular forms came from Eisenstein series. There are also
non-holomorphic Eisenstein series.
Example. Let s ∈ C and Re(s) > 0. We define
E(z, s) =
1
2
X
(c,d)=1c,d∈Z
y
s
|cz + d|
2s
=
1
2
X
γ=±
(
∗ ∗
c d
)
∈±
(
1 ∗
0 1
)
\PSL
2
(Z)
(Im γ(z))
s
.
It is easy to see that this converges. From the second definition, we see that
E
(
z, s
) is invariant under Γ(1), and after some analysis, this is
C
∞
and satisfies
the growth condition.
Finally, we check the eigenfunction condition. We can check
∆y
s
= −y
2
∂
2
∂y
2
(y
s
) = s(1 − s)y
s
.
But since ∆ is invariant under SL
2
(R), it follows that we also have
∆E(z, s) = s(1 − s)E(z, s).
In the case of modular forms, we studied the cusp forms in particular. To
study similar phenomena here, we look at the Fourier expansion of
f
. We have
the periodicity condition
f(x + iy + 1) = f (x + iy).
Since this is not holomorphic, we cannot expand it as a function of
e
2πiz
. However,
we can certainly expand it as a function in e
2πix
. Thus, we write
f(x + iy) =
∞
X
n=−∞
F
n
(y)e
2πinx
.
This looks pretty horrible, but now recall that we had the eigenfunction condition.
Then we have
λf = ∆f = −y
2
∞
X
n=−∞
(F
00
n
(y) − 4πn
2
F
n
(y))e
2πinx
.
This tells us F
n
(y) satisfies the differential equation
−y
2
F
00
n
(y) + (λ − 4π
2
n
2
y
2
)F
n
(y) = 0. (∗)
It isn’t terribly important what exactly the details are, but let’s look what
happens in particular when n = 0. Then we have
y
2
F
00
0
+ λF
0
= 0.
This is pretty easy to solve. The general solution is given by
F
0
= Ay
s
+ By
s
0
,
where s and s
0
= 1 − s are the roots of s(1 − s) = λ.
What about the other terms? We see that if
y
is large, then, if we were an
applied mathematician, then we would say the
λF
(
y
) term is negligible, and
then the equation looks like
F
00
n
(y) = 4π
2
n
2
F (y).
This has two independent solutions, and they are
e
±2πny
. It is in fact true that
the true solutions to the equation grow as
e
±2πny
for large
y
. To satisfy the
growth condition, we must only pick those that grow as
e
−2πny
. We call this
κ
|n|,λ
(y). These are known as the Bessel functions.
Thus, we find that we have
f(z) = Ay
s
+ By
1−s
| {z }
“constant term”
+
X
n6=0
a
n
(f)κ
|n|,λ
(y)e
2πinx
.
The exact form isn’t really that important. The point is that we can separate
out these “constant terms”. Then it is now not difficult to define cusp forms.
Definition
(Cusp form)
.
A Maass form is a cusp form if
F
1
= 0, i.e.
A
=
B
= 0.
Similar to modular forms, we have a theorem classifying Maass cusp forms.
Theorem
(Maass)
.
Let
S
Maass
(Γ(1)
, λ
) be the space of Maass cusp forms with
eigenvalue
λ
. This space is finite-dimensional, and is non-zero if and only if
λ ∈ {λ
n
: n ≥ 0}, where {λ
n
} is a sequence satisfying
0 < λ
0
< λ
1
< λ
2
< ··· → ∞.
Given this, we can define Hecke operators just as for holomorphic forms (this
is easier as
k
= 0), and most of the theory we developed for modular forms carry
over.
Even though we proved all these very nice properties of these cusps forms, it
took people a lot of time to actually come up with such a cusp form! Nowadays,
we are able to compute these with the aid of computers, and there exists tables
of λ’s and Hecke eigenforms.
Now recall that we had this mysterious operator
L
∗
k
= −2iy
2
∂
∂¯z
,
which had the property that if f |
k
γ = f, then (L
∗
k
f) |
k−2
γ = (L
∗
k
f).
With a bit of experimentation, we can come up with something that raises
the weight.
Definition (R
∗
k
). Define
R
∗
k
= 2i
∂
∂z
+
1
y
k.
Now this has a property that
Proposition. If f |
k
γ = f, then (R
∗
k
f) |
k+2
γ = R
∗
k
f.
Note that this time, since we are differentiating against
z
, the
cz
+
d
term
will be affected, and this is where the
1
y
k term comes in.
Suppose we have
f
=
f
0
∈ M
k
(Γ(1)). Then we can apply
R
to it to obtain
f
1
= R
∗
k
f
0
. We can now try to apply L
∗
k+2
to it. Then we have
L
∗
k+2
R
∗
k
f = −2iy
2
∂
∂¯z
2if
0
+
k
y
f
= −2iy
2
kf
∂y
−1
∂¯z
= −kf.
So we don’t get anything new.
But of course, we can continue in the other direction. We can recursively
obtain
f
2
= R
∗
k+2
f
1
, f
3
= R
∗
k+4
f
2
, ··· .
Then we can compute L
∗
k+2n
and R
∗
k+2n
of these, and we find that
(R
∗
L
∗
− L
∗
R
∗
)f
n
= (k + 2n)f
n
.
This looks suspiciously like the representation of the Lie algebra of
sl
2
, where we
have operators that raise and lower weights. The only slightly non-trivial part is
that this is an infinite-dimensional representation, as we can keep on raising and
(at least in general) it doesn’t get to 0.
It turns out it is much easier to make sense of this by replacing functions on
H
with functions on
G
=
SL
2
(
R
). By the orbit-stabilizer theorem, we can write
H = G/K, where
K = SO(2) = {g ∈ SL
2
(R) : g(i) = 1} =
r
θ
=
cos θ sin θ
−sin θ cos θ
.
Recall that we defined the function
j
(
γ, z
) =
cz
+
d
, where
γ
=
a b
c d
. This
satisfied the property
j(γδ, z) = j(γ, δ(z))j(δ, z).
The main theorem is the following:
Proposition.
For Γ
⊆
Γ(1), there is a bijection between functions
f
:
H → C
such that
f |
k
γ
=
f
for all
γ ∈
Γ, and functions Φ :
G → C
such that Φ(
γg
) = Φ(
g
)
for all γ ∈ Γ and Φ(gr
θ
) = e
ikθ
Φ(g).
The real reason of this is that such an
f
is a section of a certain line bundle
L
k
on Γ
\H
= Γ
\G/K
. The point is that this line bundle can be made trivial
either by pulling to
H
=
G/K
, or to Γ
\G
. Of course, to actually prove it, we
don’t need such fancy language. We just need to write down the map.
Proof. Given an f , we define
Φ(g) = (ci + d)
−k
f(g(i)) = j(g, i)
−k
f(g(i)).
We can then check that
Φ(γg) = j(γg, i)
−k
f(γ(g(i)))
= j(γg, i)
−k
j(γ, g(i))
k
f(g(i))
= Φ(g).
On the other hand, using the fact that r
θ
is in the stabilizer of i, we obtain
Φ(gr
θ
) = j(gr
θ
, i)
−k
f(gr
θ
(i))
= j(gr
θ
, i)
−k
f(g(i))
= j(g, r
θ
(i))j(r
θ
, 1)f (g(i))
= Φ(g)j(r
θ
, i)
−k
.
But j(r
θ
, i) = −sin θ + cos θ. So we are done.
What we can do with this is that we can cast everything in the language of
these functions on
G
. In particular, what do these lowering and raising operators
do? We have our
C
∞
function Φ : Γ
\G → C
. Now if
X ∈ g
=
sl
2
(
R
), then this
acts on Φ by differentiation, since that’s how Lie algebras and Lie groups are
related. Explicitly, we have
XΦ =
d
dt
t=0
Φ(ge
Xt
).
When we compute these things explicitly, we find that, up to conjugacy,
L
∗
and
R
∗
just correspond to the standard elements
X
−
=
0 0
1 0
, X
+
=
0 1
0 0
∈ sl
2
,
and we have
[X
+
, X
−
] = 2H, H =
1 0
0 −1
.
Then the weight k is just corresponds to H.
What does the Laplacian correspond to? It corresponds to a certain product
of these operators, the Casimir operator , given by
Ω = X
+
X
−
+ X
−
X
+
+
1
2
H
2
.
This leads to the notion of automorphic forms.
Definition
(Automorphic form)
.
An automorphic form on Γ is a
C
∞
function
Φ : Γ\G → C such that Φ(gr
θ
) = e
ikθ
Φ(g) for some k ∈ Z such that
ΩΦ = λΦ
for some λ ∈ C, satisfying a growth condition given by
Φ
a b
c d
≤ polynomial in a, b, c, d.
The condition of this being a cusp function is then
Z
1
0
Φ
1 x
0 1
g
dx = 0.
These things turn out to be exactly what we’ve looked at before.
Proposition. The set of cuspoidal automorphic forms bijects with representa-
tions of
sl
2
generated by holomorphic cusp forms
f
and their conjugates
¯
f
, and
Maass cusp forms.
The holomorphic cusp forms
f
generate a representation of
sl
2
with lowest
weight; The conjugates of holomorphic cusp forms generate those with highest
weight, while the Maass forms generate the rest.
This is now completely susceptible to generalization. We can replace
G,
Γ
with any semi-simple Lie group (e.g.
SL
n
(
R
),
Sp
2n
(
R
)), and Γ by some arithmetic
subgroup. This leads to the general theory of automorphic forms, and is one
half of the Langlands’ program.