Part III Modular Forms
Lecturer: Dr. James Newton
Lent term 20141
1
Transcribed by Stiofáin Fordham.
Contents
Lecture 1
1.1. §1: Introduction
1.2. Motivating examples
Lecture 2
2.1. Eisenstein series
Lecture 3
3.1. Fundamental domains
Lecture 4
4.1. §2: Zeroes and poles of modular forms
Lecture 5
5.1. Zero-counting formula
Lecture 6
6.1. Dimension formula
6.2. §3: Congruence subgroups
6.3. §4: Modular forms with weight a congruence subgroup
Lecture 7
7.1. §5: Theta functions
Lecture 8
8.1. §6: Old Forms
8.2. §7: Weight-2 Eisenstein series
Lecture 9
9.1. Structure of Mk (Γ)
Lecture 10
10.1. Riemann surfaces
10.2. §8: Groups acting on Riemann surfaces
Lecture 11
11.1. Groups acting on Riemann surfaces
Lecture 12
12.1. §9: Applications to modular forms
12.2. Cusps and compact modular curves.
Lecture 13
13.1. Cusps and compactifying modular curves
Lecture 14
14.1. Differentials and divisors
14.2. Meromorphic differentials on Riemann surfaces
Lecture 15
15.1. Orders of vanishing for meromorphic differentials
15.2. Divisors
Lecture 16
16.1. The Riemann-Roch theorem
16.2. The genus of modular curves
Lecture 17
3
Page
5
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26
27
27
27
30
31
31
31
32
33
33
34
35
36
37
38
38
38
40
17.1. Modular forms and functions on lattices
17.2. The Hecke operators
Lecture 18
18.1. Matrix version of Hecke operators
Lecture 19
19.1. Properties of the Hecke operators
Lecture 20
20.1. The Petersson inner product
Lecture 21
21.1. The inner-product and the eigenforms
Lecture 22
22.1. L-functions
22.2. The functional equation
Lecture 23
23.1. Hecke eigenforms & L-functions
23.2. Converse theorems of Hecke and Weil
Lecture 24
24.1. §10: Weil’s converse theorem (1967)
24.2. Twists of modular forms
24.3. CM forms
Bibliography
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4
Lecture 1
Lecture 1
17th January 09.00
The web-page for the course is www.dpmms.cam.ac.uk/~jjmn2/MF2014.html. My
email for questions/comments is jjmn2@cam.ac.uk. The books/references
● Diamond, Shurman ‘A first course in modular forms’. Chapters 1-5 contain most of the material for this course [DS05].
● The chapter of Zagier in ‘The 1-2-3 of modular forms’. Very concise
introduction and lots of examples/applications [BvdGHZ08].
● Milne’s lecture notes at www.jmilne.org/math/CourseNotes/mf.html
1.1. §1: Introduction. Some notation
a
Γ(1) = SL2 (Z) = { (
c
b
) ∶ ad − bc = 1; a, b, c, d ∈ Z}
d
called the modular group. We write H = {τ ∈ C∶ Im (τ ) > 0}. The action of Γ(1) on
H is given by
aτ + b
a b
(
)⋅τ =
c d
cτ + d
Exercise: check that this is a group action.
Definition 1.1. Let k be an integer, Γ ⊂ Γ(1) a subgroup of finite index (or
a congruence subgroup), then we say that a meromorphic function f ∶ H → C is
weakly modular of weight k and level Γ if
aτ + b
f(
) = (cτ + d)k f (τ )
cτ + d
a
for all τ ∈ H and (
c
b
) ∈ Γ.
d
E.g. for k = 0: meromorphic functions on Γ/H a Riemann surface, and for k = 2
we have f (τ ) dt meromorphic differential on Γ/H.
For now, a weakly modular function (of weight k, level Γ) is a modular form
if it is holomorphic on H, together with some other condition (*) which I will give
next time. If Γ = Γ(1) “level 1”, then (*) is equivalent to: there exists constants
Y, C ∈ R>0 such that ∣f (τ )∣ ≤ C for Im (τ ) > Y .
1.2. Motivating examples. The first motivating example is the representation numbers for quadratic forms.
Definition 1.2. For k ∈ Z≥1 and n ∈ Z≥0 we define rk (n) := the number of distinct
ways to write n as a sum of k squares.
Remark. We must clarify what we mean by distinct: we count different orderings/signs as distinct e.g. r2 (1) = 4 because
1 = 12 + 02 = 02 + 12 = (−1)2 + (0)2 = 02 + (−1)2
We are thinking of the quadratic form x2 + y 2 here.
Classically this problem was studied by Gauss and Jacobi. Jacobi defined the
θ-function to do this - a function on H defined by
∞
θ(τ ) = ∑ exp(2πin2 τ )
n=−∞
By a formal manipulation one can show
θ(τ )k = ∑ rk (n) exp(2πinτ )
n≥0
5
Lecture 1
For even k, θ(τ )k is a modular form of weight k (level Γ ⊂ Γ(1)). This allows one
to give a proof of
r4 (n) = 8 ∑ d
d>0,d∣n
4∤d
More generally one has
r2k (n) = (nice expression) + (error term)
where the error term is bounded by Deligne’s work on the Weil conjectures.
The next application is convex uniformisation of elliptic curves. If we have a
lattice Λ ⊂ C (a discrete subgroup ≃ Z2 ),we define the Weierstraß ℘-function to be
℘(z, Λ)∶ C/Λ → C
1
1
1
− 2)
℘(z, Λ) = 2 + ∑ (
2
z
ω
ω∈Λ/{0} (z − ω)
and it satisfies ℘(z + ω, Λ) = ℘(z, Λ) for ω ∈ Λ. If we define an elliptic curve
EΛ ∶ y 2 = 4x3 − g4 (Λ)x − g6 (Λ)
where g4 (Λ) = 60 ∑ω∈Λ/{0}
1
ω4
and g6 (Λ) = 140 ∑ω∈Λ/{0}
1
ω6
then the map
′
z ↦ (℘(z, Λ), ℘ (z, Λ))
gives an isomorphism
∼
C/Λ → EΛ (C)
If τ ∈ H, define Λτ = Z ⋅ τ ⊕ Z ⊂ C then the map τ ↦ g4 (Λτ ) gives a modular form
of weight 4 and level 1 and the map τ ↦ g6 (Λτ ) gives a modular form of weight 6
and level 1. Homogeneous polynomials in these are also modular forms e.g.
g4 (Λτ )3 − 27(g6 (Λτ ))2 = disc(EΛτ ) ∼ ∆(τ ) Ramanujan’s ∆-function
is a modular form of weight 12.
Remark. These ideas are what leads to Mazur’s result determining the possibilities
for E(Q)tors with E/Q an elliptic curve.
The next application is about Dirichlet series and L-functions. Define
1
, Re(s) > 1
ns
This has a meromorphic continuation to all of C and a functional equation relating
ζ(s) and ζ(1 − s). There is the Euler product formula
ζ(s) = ∑
ζ(s) = ∏(1 − p−s )−1
p
where the product is taken over p prime. We’ll study how to associate an L-function
L(f, s) to a modular form f with similar properties to ζ(s). A result that we will
prove towards the end of the course is Hecke’s converse theorem which says the
following: suppose that we have a function
an
Z(s) = ∑ s , an ∈ C
n
converging absolutely for Re(s) ≫ 0. Suppose Z(s) has meromorphic continuation
to C and a functional equation. Then
f (τ ) = ∑ an exp(2πinτ )
n≥1
is a modular form.
6
Lecture 2
The final application is Hasse-Weil L-functions. Suppose that E/Q is an elliptic
curve, then one can define a function
an
L(E, s) = ∏ Lp (E, s) = ∑ s
n≥1 n
p prime
for p a prime of good reduction for E, with Ẽp /Fp the reduced elliptic curve, where
Lp (E, s) = (1 − ap p−s + p1−2s )−1
and ap = p + 1 − ∣Ẽp (Fp )∣ and it is absolutely convergent for Re(s) > 2. A famous
theorem is the following
Theorem 1.1 (Wiles, Breuil, Diamond, Taylor). One has that
f (τ ) = ∑ an exp(2πinτ )
n≥1
is a weight 2 modular form.
Remark. This is the “modularity of elliptic curves”. A slightly weaker version was
used to prove Fermat’s last theorem. This theorem is the only way to show that
L(E, s) has analytic continuation to C and the functional equation.
Lecture 2
20th January 09:00
Today we will look at modular forms of level 1 (Γ(1) = SL2 (Z)), their definition
and the Eisenstein series.
Recall that a weakly modular function of weight k (and level 1) is a meromorphic function f ∶ H → C satisfying
f(
aτ + b
) = (cτ + d)k f (τ )
cτ + d
a b
) ∈ Γ(1) and τ ∈ H. The functions f ∶ H → C, weakly modular of weight 0
c d
are precisely the holomorphic functions C → C.
Suppose that f is as above, then
for (
(
1
0
0
) ∈ SL2 (Z)∶ f (τ + 1) = f (τ )
0
Suppose moreover that f is holomorphic on the region
{Im (τ ) > Y } ⊂ H
for some Y ∈ R≥0 . Denote by D∗ the punctured unit disc {0 < ∣q∣ < 1, q ∈ C}.
The map τ ↦ exp(2πiτ ) gives a holomorphic surjection H → D∗ . We can define a
function F ∶ D∗ → C by F (exp(2πiτ )) = f (τ ). The function F is holomorphic on
1
{0 < ∣q∣ < exp(−2πY )} (e.g. write F (q) = f ( 2πi
log(q)) using a suitable branch of
the logarithm). The upshot of this is the following result/definition.
Lemma 2.2. Suppose that f is weakly modular of weight k and holomorphic for
Im (τ ) ≫ 0. Then we have
f (τ ) = F (exp(2πiτ )) = ∑ an (f )q n
n∈Z
where q = exp(2πiτ ) (and this is valid where f is holomorphic). The series is the
Fourier expansion or the q-expansion of f .
Definition 2.3. Suppose that f is weakly modular of weight k.
7
Lecture 2
(1) We say that f is meromorphic at ∞ if f is holomorphic for Im (τ ) ≫ 0,
and in the q-expansion f (τ ) = ∑n∈Z an (f )q n , there exists N ∈ Z such
that an (f ) = 0 for all n < N (i.e. extends to a meromorphic function at
0 ∈ D = {0 ≤ ∣q∣ < 1}),
(2) Similarly, we say that f is holomorphic at ∞ if an (f ) = 0 for all n < 0.
(3) If f is meromorphic at ∞, we say that f is a meromorphic form (or weight
k) (this terminology may be non-standard but it will be useful later),
(4) If f is holomorphic on H and at ∞, then we say that f is a modular form
(of weight k and level 1).
(5) If f is a modular form and a0 (f ) = 0, then we say that f is a cusp form
(or a cuspidal modular form
Lemma 2.3. Suppose that f is weakly modular. Then f is holomorphic at ∞ if
there exists C, Y ∈ R>0 such that ∣f (τ )∣ ≤ C for all τ with Im (τ ) > Y .
Proof. Exercise.
Notation: the set of modular forms of weight k and level 1 will be denoted
Mk (Γ(1)) or Mk for short. The set of cusp forms of weight k and level 1 will be
denoted Sk (Γ(1)) or Sk for short (the German for cusp form is Spitzenform).
The sets Mk and Sk are C-vector spaces, under the obvious scalar multiplication
and addition. Exercise: if f ∈ Mk and g ∈ Ml , then f ⋅ g ∈ Mk+l and if f ∈ Mk and
f ∈ Sl then f ⋅ g ∈ Sk+l , i.e. ⊕k∈Z Mk is a graded C-algebra and ⊕k∈Z Sk is a graded
ideal.
−1 0
= τ ), so f weakly
Remark. Consider the action of (
) ∈ Γ(1) on H (τ ↦ −τ
−1
0 −1
modular of weight k satisfies f (τ ) = (−1)k f (τ ), so for k odd, then f = 0. If Γ ⊂ Γ(1)
−1 0
with (
) ∉ Γ then we could have odd weight things.
0 −1
2.1. Eisenstein series.
Definition 2.4. Let k > 2 be an even integer. The Eisenstein series of weight k is
the function on H given by
1
Gk (τ ) =
∑
(cτ
+
d)k
(c,d)∈Z
(c,d)≠(0,0)
Lemma 2.4. The series for Gk (τ ) is absolutely convergent for τ ∈ H, and it
converges uniformly for τ in a compact subset of H.
Proof. Our picture is
Im
m=2
τ
−1 + τ
1+τ
Re
−1 − τ
1−τ
−τ
8
Lecture 3
and we continue on the lattice - at the mth parallelogram, I get (2m+1)2 −(2m−1)2 =
8m lattice points. If τ is contained in the compact subset of H, we have some
r independent of z such that ∣z∣ ≥ r as z varies over the boundary of the first
parallelogram (z varies over the boundary not just the lattice points). If cτ + d is
in the mth layer, then ∣cτ + d∣ ≥ mr, We have
∑
(c,d)∈Z
(c,d)≠(0,0)
∞
8 ∞
8m
1
1
≤
=
∑
∑
∣cτ + d∣k m=1 (mr)k rk m=1 mk−1
which converges for k > 2.
Remark. Summing in this way its clear that the sum converges for k = 2.
We will show next time that Gk is weakly modular of weight k. It’s clear that
Gk (τ + 1) = Gk (τ ).
Proposition 2.5. The q-expansion of Gk is
Gk (τ ) = 2ζ(k) +
where ζ(k) = ∑n≥1
1
nk
2(2πi)k ∞
n
∑ σk−1 (n)q
(k − 1)! n=1
and σk−1 (n) = ∑ d∣n dk−1 .
d>0
One proof of this will be in the first example sheet. The other proof uses the
Poisson summation formula.
∞
Let h∶ R → C be a continuous function such that h is L1 i.e. ∫−∞ ∣h(x)∣ dx < ∞
and S(x) = ∑d∈Z h(x + d) converges absolutely and uniformly on compact subsets of
∞
R, to an infinitely differentiable function. Then if ĥ(t) = ∫−∞ h(x) exp(−2πixt) dx
we have
S(x) = ∑ h(x + d) = ∑ ĥ(m) exp(2πimx)
m∈Z
d∈Z
Lecture 3
22
nd
January 09:00
Proposition 3.6. Gk is weakly modular of weight k
Proof. We have
Gk (
aτ + b
(cτ + d)k
)= ∑ ′
k
cτ + d
(m,n) (m(aτ + b) + n(cτ + d))
′
1
k
(m,n) (m(aτ + b) + n(cτ + d))
= (cτ + d)k ∑
The expression (am + cn)τ + bm + dn can be written (m, n) (
a
dn). Then right multiplication by (
c
Z2 /{(0, 0)} and itself. So we have
a
c
b
) = (am + cn, bm +
d
b
) ∈ SL2 (Z) gives a bijection between
d
′
′
1
1
=
= Gk (τ )
∑
k
k
(m,n) (mτ + n)
(m,n) (m(aτ + b) + n(cτ + d))
∑
9
Lecture 3
Proposition 3.7. The q-expansion of Gk is
Gk (τ ) = 2ζ(k) +
where ζ(k) = ∑n≥1
1
nk
2(2πi)k ∞
n
∑ σk−1 (n)q
(k − 1)! n=1
and σk−1 (n) = ∑ d∣n dk−1 .
d>0
Proof. We will prove (using Poisson summation) the following claim: fix C ≥
1, τ ∈ H then one has
1
(−2πi)k mk−1 cm
=
q
∑
k
(k − 1)!
m>0
d∈Z (cτ + d)
∑
Assume that this claim holds, then
′
1
1
1
=2∑ k +2∑ ∑
k
k
c≥1 d∈Z (cτ + d)
d≥1 d
(c,d) (cτ + d)
Gk (τ ) = ∑
(−2πi)k mk−1 cm
q
(k − 1)!
c≥1 m≥1
= 2ζ(k) + 2 ∑ ∑
Which gives us the result. So we must prove the claim.
Recall the Poisson summation formula: let h∶ R → C be a continuous function
∞
such that h is L1 i.e. ∫−∞ ∣h(x)∣ dx < ∞ and S(x) = ∑d∈Z h(x+d) converges absolutely
and uniformly on compact subsets of R, to an infinitely differentiable function. Then
∞
if ĥ(t) = ∫−∞ h(x) exp(−2πixt) dx we have
S(x) = ∑ h(x + d) = ∑ ĥ(m) exp(2πimx)
d∈Z
We want to compute ∑d∈Z
have
1
.
(cτ +d)k
m∈Z
We’ll take h = hc where hc (x) =
1
,
k
d∈Z (cτ + x + d)
Sc (x) = ∑
1
(cτ +x)k
then we
Sc (0) = ∑ ĥc (m)
m∈Z
We have
ĥc (m) = ∫
∞
−∞
∞ exp(−2πicmu)
exp(−2πimx)
1
dx = k−1 ∫
du
k
(cτ + x)
c
(τ + u)k
−∞
Suppose we have n ∈ Z: set fn (z) = exp(−2πinz)
(a meromorphic function on C).
zk
One has
z=+∞+τ
1
ĥc (m) = k−1 exp(2πicmτ ) ∫
fcm (z) dz
c
z=−∞+τ
+∞+τ
Define In = ∫−∞+τ fn (z) dz, then to prove the claim, we need to show that In = 0
n
1
for n ≤ 0 and = (−2πi)
for n > 0. Do this using Resz=0 fn (z) = (k−1)!
(−2πin)k−1
(k−1)!
and computing the integrals of fn (z) over contours: for n ≤ 0 we take larger and
larger rectangles whose lower edge is the horizontal line through τ and the upper
edge is far up in the positive imaginary direction (the pole at z = 0 won’t be in
this here - τ ∈ H) (anti-clockwise direction) then for n > 0 we take larger and larger
rectangles whose upper edge is the horizontal line that passes through τ and the
bottom edge is far down in the negative imaginary direction (contains the pole)
(clockwise direction).
k
k−1
10
Lecture 3
n ≤ 0 case
τ
z=0
n > 0 case
Remark. For k positive and even, we have
ζ(k) =
−(2π)k Bk
2(k!)
Where Bk are the Bernoulli numbers defined by the generating function
∞
t
tn
=
B
∑
n
et − 1 n=0
n!
1
and Bn ∈ Q, e.g. B2 = 61 , B4 = − 30
and B6 =
1
42
and Bodd = 0 except for B1 .
Remark. The proposition above shows that Gk is holomorphic at ∞ (no negative
powers), so we have that Gk ∈ Mk (Γ(1)).
Definition 3.5. We define the normalised Eisenstein series to be
Gk (τ )
Ek (τ ) :=
= 1 + ...
2ζ(k)
Remark. Consequence of this is that the q-expansion of Ek (τ ) has rational coefficients.
One upshot of this definition is that it allows us to right down an example of a
cusp form called the Ramanujan ∆-function
∆(τ ) :=
E4 (τ )3 − E6 (τ )2
= q + ⋅ ⋅ ⋅ = ∑ τ (n)q n
1728
n≥1
and the τ (n) is called Ramanujan’s τ -function - we will see later the the τ (n) are
all integers.
Our goal for the next few lectures is to show that the spaces Mk (Γ(1)) are
finite dimensional and compute their dimensions. We will study the action of Γ(1)
on H.
3.1. Fundamental domains.
Definition 3.6. Suppose that a group G acts continuously on a topological space
X. Then a fundamental domain for this action is an open subset F ⊂ X such that
(1) no two distinct points of F are equivalent (in the same G-orbit) under the
action of G,
(2) every point x ∈ X is equivalent to a point in the closure F of F ⊂ X,
Proposition 3.8. The set F = {τ ∈ H∶ ∣τ ∣ > 1, ∣Re(τ )∣ < 12 } ⊂ H is a fundamental
domain for Γ(1) ↻ H.
11
Lecture 4
−1
− 21
1
2
1
The proof will tell us that Γ(1) is generated by the set {S, T } where S =
0 1
1 1
(
) and T = (
). Another consequence of the proof will be that the set
−1 0
0 1
F̃ = F ∪ {τ ∈ H∶ ∣τ ∣ ≥ 1, Re(τ ) = − 12 } ∪ {τ ∈ H∶ ∣τ ∣ = 1, − 12 ≤ Re(τ ) ≤ 0} contains a
unique representative for each point - this is called a strict fundamental domain.
Lecture 4
24th January 09:00
Proposition 4.9. The set F̃ ⊂ H contains a unique element of every Γ(1)-orbit.
Also, Γ(1) is generated by S and T .
Remark. F̃ is a strict fundamental domain for the Γ(1)-action on H implies that
F̃ is a fundamental domain.
0 −1
1 1
Proof. Recall that S = (
) acts by τ ↦ − τ1 and T = (
) acts by
1 0
0 1
τ ↦ τ + 1. Let G be the subgroup of Γ(1) generated by S and T . Fix τ ∈ H. We’ll
a b
show first that F̃ contains an element of G ⋅ τ : for g = (
) ∈ G, then (exercise)
c d
Im (gτ ) =
Im (τ )
∣cτ + d∣2
As (c, d) ∈ Z2 varies over the possible bottom rows for g ∈ G, ∣cτ + d∣ will attain
Im (τ )
a minimum, so ∣cτ
attains a maximum. So fix g0 ∈ G such that Im (g0 τ ) is
+d∣2
maximal. In particular we have
Im (g0 τ )
Im (Sg0 τ ) = Im (−1/g0 τ ) = Im (−g0 τ̄ /∣g0 τ ∣2 ) =
≤ Im (g0 τ )
∣g0 τ ∣2
thus ∣g0 τ ∣ ≥ 1. Since T doesn’t change imaginary parts, Im (T n g0 τ ) is maximal, so
∣T n g0 τ ∣ ≥ 1 for all n ∈ Z. Now it’s clear that we can find g ∈ G such that gτ ∈ F̃: first
find n such that Re(T n g0 τ ) ∈ [−1/2, 1/2). Then if ∣T n g0 τ ∣ = 1 and Re(T n g0 τ ) > 0,
then ST n g0 τ ∈ F̃. Otherwise, T n g0 τ ∈ F̃. This shows that F̃ contains a point of
every G-orbit (hence every Γ(1)-orbit).
It remains to show uniqueness. Suppose we have τ1 , τ2 ∈ F̃ with τ1 ≠ τ2 , with
τ2 = γτ1 for some γ ∈ Γ(1). We have −1/2 ≤ Re(τi ) < 1/2 so γ ≠ ±T n for any n ∈ Z.
√
a b
So if γ = (
), then c ≠ 0. Also, Im (τi ) ≥ 3/2. So we have
c d
√
3
Im (τ1 )
Im (τ1 )
2
≤ Im (τ2 ) =
≤ 2
≤ √
2
2
2
2
∣cτ1 + d∣
c (Im (τ1 ))
c 3
12
Lecture 5
thus c2 ≤ 4/3 so c = ±1. So we have
Im (τ2 ) =
Im (τ1 )
Im (τ1 )
≤
≤ Im (τ1 )
2
∣ ± τ1 + d∣
∣τ1 ∣2
The whole argument is symmetric in τ1 , τ2 thus Im (τ1 ) = Im (τ2 ) and ∣τ1 ∣ = ∣τ2 ∣ = 1.
Hence τ1 = τ2 a contradiction. So F̃ contains a unique element of every γ(1)-orbit.
Last thing is to show that G = Γ(1). Take γ ∈ Γ(1). For any τ ∈ H, there
exists g ∈ G such that gγτ ∈ F̃. If we take τ = 2i, then there exists g ∈ G such that
√
√
a b
gγ(2i) ∈ F̃. Write gγ = (
), then Im (gγ(2i)) = 4c22+d2 ≥ 23 thus 4c2 + d2 ≤ 4/ 3
c d
which implies that c = 0, d = ±1. So gγ = ±T n , and γ ∈ G.
Remark. This proposition is an example of “reduction theory”. It is essentially a
reinterpretation of Gauss’s theory of reduced forms for binary quadratic forms (c.f.
example sheet 1).
Exercise: (1.) Suppose that τ ∈ F̃, then one has
⎧
⎪
{±I}
if τ ≠ ω, i
⎪
⎪
⎪
⎪
⎪
⎪
⟨S⟩
if τ = i
⎪
StabΓ(1) (τ ) = ⎨
⎪
⎛0 −1⎞
⎪
⎪
⎪
⟩ if τ = ω
⟨
⎪
⎪
⎪
⎝1 1 ⎠
⎪
⎩
(2) Compute StabΓ(1) (τ ) for τ ∈ H.
Remark. There is some terminology that comes out of this exercise: we say that
τ ∈ H is elliptic if it is in the Γ(1)-orbit of i or ω. A point is elliptic iff its’ stabiliser
has order 4 or 6. If τ is an elliptic point, its’ order is defined to be 21 #StabΓ(1) (τ ).
Note that Γ(1) ↻ H factors through Γ(1)/{±I}. So we have
order(τ ) = #StabΓ(1)/{±1} (τ )
Remark. We will be interested in describing Γ(1)/H: consider gluing F̃ to itself
by identifying the sides (ω → 1+ω) and the two pieces of the arc. We get C (at least
topologically). Something ‘funny’ happens at the elliptic points (the definition of
‘funny’ shall be given later).
4.1. §2: Zeroes and poles of modular forms.
Definition 4.7. For f a meromorphic form of weight k, τ ∈ H, write ordτ (f )
for the order of vanishing of f at τ . Write f (τ ) = ∑n∈Z an (f )q n , then we define
ord∞ (f ) to be the smallest n such that an (f ) ≠ 0.
The quantity ordτ (f ) depends only on Γ(1) ⋅ τ .
Proposition 4.10. Let f be a non-zero meromorphic form of weight k. Then
ord∞ (f ) +
ordτ (f )
k
=
12
Γ(1)⋅τ ∈Γ(1)/H order(τ )
∑
Lecture 5
27th January 09:00
13
Lecture 5
5.1. Zero-counting formula. Recall that a meromorphic form of weight k
is a weakly modular function of weight k which is meromorphic at ∞. If we write
f (τ ) = F (q) = ∑ an q n
n∈Z
then an = 0 for m ≪ 0, where q = exp(2πiτ ). We defined ord∞ (f ) = min{n∶ an ≠ 0}.
For τ ∈ H, then
⎧
⎪
0
if f is holomorphic at τ , non-vanishing
⎪
⎪
⎪
⎪
ordτ (f ) = ⎨order of vanishing if f (τ ) = 0
⎪
⎪
⎪
⎪
−(order of pole)
if f has a pole at τ
⎪
⎩
and we define
⎧
⎪
1
⎪
⎪
⎪
⎪
nτ = ⎨2
⎪
⎪
⎪
⎪
3
⎪
⎩
τ ∉ Γ(1) ⋅ i, Γ(1) ⋅ ω
τ ∈ Γ(1) ⋅ i
τ ∈ Γ(1) ⋅ ω
Proposition 5.11. Suppose that f is a meromorphic form of weight k. Then one
has
ord∞ (f ) +
ordτ (f )
k
=
nτ
12
Γ(1)τ ∈Γ(1)/H
∑
Remark. Since f (γτ ) = (cτ + d)k f (τ ) and (cτ + d)k is non-vanishing and holomorphic on H, so ordτ (f ) only depends on the orbit Γ(1) ⋅ τ .
Remark. The sum has finitely many non-zero terms: if we consider the fundamental domain for Im (τ ) < C, then f extends to a meromorphic function on the closure
so it has only finitely many zeroes and poles here, then for the part Im (τ ) ≥ C, we
know that it maps to the inside of the disc about q = 0 and F is meromorphic here.
Proof. We will see a proof in the second half of the course using the computation of the degree of a differential on a Riemann surface. Today the proof will
be the same as the one in Serre’s course of arithmetic. The idea is to integrate
f ′ (τ )/(2πif (τ )) around the boundary of F̃ to obtain
1
f ′ (τ )
dτ = ∑ ordτ (f )
∫
2πi ∂ F̃ f (τ )
τ ∈F̃
14
Lecture 5
A
E
C
i
C C′
B
ω
B
′
D′
D
− 12
−1
1+ω
1
2
1
Assume for simplicity that f has no zeroes/poles on ∂ F̃ apart from (possibly) at
i, ω, 1 + ω. Choose points A = − 21 + iY , E = 12 + iY with Y large enough such that
the interior of X contains all zeroes and poles in the interior of F̃. BB ′ , CC ′ and
DD′ are the arcs of circles with small radius r centred on the elliptic points. We
have
1
f ′ (τ )
dτ =
ordτ (f )
∑
∫
2πi C f (τ )
Γ(1)τ ∈Γ(1)/H
nτ =1
′
First, the integral from B to B : as r → 0, then it is the same as integrating around
a circle centred at ω about the arc of π/3 so f (τ ) ∼ a(τ − ω)ordω (f ) and
′
B f ′ (τ )
1
1 1
f ′ (τ )
−ordω (f )
dτ → (
dτ ) =
∫
∫
2πi B f (τ )
6 2πi ○ f (τ )
6
where ∫○ means the integral around the small circle about ω. Similarly, we have
that
D ′ f ′ (τ )
C ′ f ′ (τ )
−ordω (f )
−ordi (f )
dτ
→
,
dτ →
∫
∫
f (τ )
6
f (τ )
2
D
C
E
A
B
Since f (τ ) = f (1 + τ ), we have that ∫D′ = ∫B = ∫A . Since f (−1/τ ) = τ k f (τ ) (using
S from before), then τ ↦ −1/τ maps B ′ C to DC ′ so1
C
∫
B′
f ′ (τ )
dτ
f (τ )
C′
u=−1/τ
=
∫
(
D
f ′ (u) k
+ ) du
f (u) u
1We have f (−1/u) = uk f (u) so f ′ (−1/u) = kuk−1 f (u) + uk f (u) then
f ′ (−1/u) kuk−1 f (u) + uk f ′ (u) k f ′ (u)
=
= +
.
f (−1/u)
uk f (u)
u
f (u)
15
Lecture 5
So that2
′
C f ′ (τ )
D f ′ (τ )
C k
1
1
k
(∫
dτ + ∫
dτ ) =
du →
∫
′
′
2πi B f (τ )
f (τ )
2πi D u
12
C
Finally, we have
A f ′ (τ )
1
dτ
∫
2πi E f (τ )
q=e2πiτ
=
1
F ′ (q)
dq = −ord0 (F ) = −ord∞ (f )
∫′
2πi ○ F (q)
Where ∫○′ means the integral over the small circle about q = 0. Now if we collect
terms the result follows.
Finally, we deal with the case of: if f has zeros/poles on ∂ F̃/{i, ω, 1 + ω} then
we modify the contour: on the left and right edges, if there are zeros/poles then
they will be opposite each other so modify it like
λ
1+λ
and the opposing parts cancel out here. If there are zeros/poles on the edge of the
circle then they will also be in pairs, reflected about the imaginary axis, we modify
it like
− λ1
λ
and again, these cancel each other out.
Some consequences of this: recall that
E43 − E62
, ∆ ∈ S12 (Γ(1))
1728
We have that ord∞ (∆) = 1 (because ∆ = q + . . . ) so we obtain from the above
1 = k/12 − (. . . ) where the ‘. . . ’ are terms that are ≥ 0 (because the q-expansion
starts at q +1 so it cannot have any poles, and hence no negative terms). So the
proposition implies that ∆ has no zeros on H (recall from the introduction that
∆(τ ) = c ⋅ discriminant of the elliptic curve C/Z ⊕ Zτ ).
∆=
Definition 5.8. Define
E4 (τ )3
∆(τ )
It is weakly modular of weight 0, is holomorphic on H and ord∞ (j) = −1.
j(τ ) =
Corollary 5.12. j(τ ) gives a bijection Γ(1)/H → C (the fact that j is weakly
modular of weight 0 implies that j(γτ ) = j(τ ) for all γ ∈ Γ(1)).
Proof. Let z ∈ C, and define fz ∶ H → C by
fz (τ ) = E4 (τ )3 − z∆(τ ) ∈ M12 (Γ(1))
2Let z(t) = exp(2πit) then dz = (2πi)z(t) dt so
∫
C′
D
1/4
f ′ (τ )
dτ = k ∫
(2πi) dt = kπi/6
1/6
f (τ )
16
Lecture 6
We have that fz (τ ) = 0 iff j(τ ) = z. So we need to show that fz has zero set equal
to a Γ(1)-orbit, Γ(1) ⋅ τ . We have that ord∞ (fz ) = 0 (the q-expansion of E4 starts
at 1 and the q-expansion of ∆ starts with q), so the proposition above implies that
ordτ (fz )
ordi (fz ) ordω (fz )
+
+ ∑ ordτ (fz ) =
=1
∑
2
3
nτ
Γ(1)τ ∈Γ(1)/H
All orders are ≥ 0. Now we can see that the only possibilities are that fz has a3
⎧
⎪
simple zero at Γ(1) ⋅ τ , τ non-elliptic,
⎪
⎪
⎪
⎪
⎨double zero at Γ(1) ⋅ i
⎪
⎪
⎪
⎪
triple
zero
at
Γ(1)
⋅
ω
⎪
⎩
Remark. This implies that every elliptic curve over C is isomorphic to C/Λ for
some lattice Λ.
Lecture 6
29
th
January 09:00
6.1. Dimension formula. We take Γ ⫋ Γ(1) = SL2 (Z). Recall we have the
formula
ordτ (f )/nτ = k/12
ord∞ (f ) +
∑
Γ(1)⋅τ ∈Γ(1)/H
for f a meromorphic form of weight k, non-zero.
Lemma 6.13.
(1) If k < 0, then Mk (Γ(1)) = 0.
(2) M0 (Γ(1)) = {constant functions on H} ≅ C.
Proof. (1) is immediate from the previous proposition on the zero counting
formula (if it is holomorphic then it does not have a pole so the terms are all positive
on the left).
(2): if f ∈ M0 (Γ(1)) then write f (τ ) = a0 (f ) − ∑n≥1 an (f )q n so f − a0 (f ) ∈
S0 (Γ(1)) and the zero counting formula implies that f − a0 (f ) = 0 i.e. f is constant
(because f − a0 (f ) must have a zero because every term has a positive power of q
in it).
Lemma 6.14. For every k ≥ 0, dimC Mk (Γ(1)) ≤ ⌊k/12⌋ + 1. Moreover if k ≡
2 ( mod 12) then dimC Mk (Γ(1)) ≤ ⌊k/12⌋.
Proof. Set m = ⌊k/12⌋ + 1 > k/12. Fix P1 , . . . , Pm ∈ Γ(1)/H distinct, nonelliptic orbits. Let f1 , . . . , fm+1 ∈ Mk (Γ(1)) then (using linear algebra) we can
construct f = ∑m+1
i=1 λi fi with zeroes at P1 , . . . , Pm . Now the left-hand side of the
zero counting formula is ≥ m, and the right-hand side is < m, thus f = 0. Hence
dim(Mk ) ≤ m.
Now suppose that k = 12l + 2 for l ≥ 0. We do the same thing with P1 , . . . , Pl :
the left-hand-side of the zero counting formula is ≥ l, the right-hand-side is equal
to l + 16 , so there must be at least a simple zero at i and a double zero at ω to give
7/6 + . . . on the left then in the counting formula by the same argument - choosing
l points this time - we get the left side is ≥ l and the right side is l − 1 < l and since
l = ⌊k/12⌋, the result follows.
Remark. Define M = ⊕k≥0 Mk (Γ(1)) then this is (exercise on example sheet)
contained in the set of holomorphic functions on H.
3One needs solutions to n + n′ /2 + n′′ /3 = 1 and evidently the only ones are (n, n′ , n′′ ) =
(1, 0, 0), (0, 2, 0), (0, 0, 3).
17
Lecture 6
Theorem 6.15. The map
R∶ C[X, Y ] → M
X ↦ E4
Y ↦ E6
is an isomorphism.
Corollary 6.16. The set {E4a E6b ∶ a, b ∈ Z≥0 , 4a + 6b = k} is a basis for Mk (Γ(1))
and
⎧
⎪
⎪⌊k/12⌋ + 1 k ≡/ 2 ( mod 12)
dim Mk (Γ(1)) = ⎨
⎪
⌊k/12⌋
k ≡ 2 ( mod 12)
⎪
⎩
Proof. The first part is immediate and the second part is an exercise.
Proof:(of thm. 6.15). The second part of the above corollary implies that
dimC Mk (Γ(1)) ≤ dimC C[X, Y ]deg k where we have given X degree 4 and Y degree
6. So it is sufficient to prove that R is injective, i.e. that E4 and E6 are algebraically
independent (think of E4 , E6 ⊂ Frac(M ) ⊂ meromorphic functions on H).
Suppose that E4 , E6 are algebraically dependent4. In particular, C(E4 , E6 )/C(E4 )
is a finite field extension. So one can see by drawing out a graph of field extensions that C(E43 , E62 )/C(E43 ) is finite. So E44 , E63 are algebraically dependent, i.e.
2j
∑ λi,j E43i E6 = 0 (where we are summing over some finite set of tuples of nonk/6
negative integers). If we take the weight k component, divide by E6 , . . . so we
3
2
3
2
get a polynomial in E4 /E6 , which is equal to 0. Thus E4 /E6 = λ ∈ C is a constant,
thus (E6 /E4 )2 = E4 /λ is holomorphic, thus E6 /E4 ∈ M2 (Γ(1)) = 0, a contradiction
(the zero counting formula(?)). Hence E4 , E6 are algebraically independent.
Remark. See the proof in Serre’s book [Ser73, ch. vii].
If we have a modular form f = ∑ an (f )q n ∈ Mk (Γ(1)), then it is just an exercise
in linear algebra to find λa,b so that f = ∑a,b λa,b E4a E6b . I will give out the example
sheet on Monday, and we can discuss then when to have the first examples class.
6.2. §3: Congruence subgroups.
Definition 6.9. Let N ∈ Z≥1 . The principal congruence subgroup Γ(N ) ⊂ SL2 (Z)
is
1 0
Γ(N ) = {γ ∈ SL2 (Z), γ ≡ (
) mod N }.
0 1
Lemma 6.17. Γ(N ) is a finite index normal subgroup of SL2 (Z).
Proof. Follows from Γ(N ) = ker (SL2 (Z) → SL2 (Z/N Z)) (because then SL2 (Z)/Γ(N )
injects into SL2 (Z/nZ), a finite group, so it must be of finite index).
Remark. The map SL2 (Z) → SL2 (Z/N Z) is surjective. So [SL2 (Z)∶ Γ(N )] =
#SL2 (Z/N Z).
Definition 6.10. A subgroup Γ ⊂ SL2 (Z) is a congruence subgroup if Γ(N ) ⊂ Γ
for some N (so the congruence subgroups have finite index in SL2 (Z)).
Remark. It is a fact that not every finite index subgroup of SL2 (Z) is a congruence subgroup, but for SLn (Z), n ≥ 3, every finite index subgroup is a congruence
subgroup - the “congruence subgroup property”.
4See proof of proposition 4, §2.1, pg. 15 in [BvdGHZ08].
18
Lecture 7
Definition 6.11. Let N ∈ Z≥1 . Define
a
Γ0 (N ) = {γ = (
c
b
) ∈ SL2 (Z)∶ c ≡ 0 ( mod N )} ⊃ Γ(N )
d
a
Γ1 (N ) = {γ = (
c
b
) ∈ SL2 (Z)∶ a ≡ b ≡ 1 ( mod N ), c ≡ 0 ( mod N )} ⊃ Γ(N )
d
Exercise: compute the index [SL2 (Z)∶ Γ0 (N )] and [SL2 (Z)∶ Γ1 (N )].
6.3. §4: Modular forms with weight a congruence subgroup. Recall
that f ∶ H → C is weakly modular of weight k and level Γ, if
f(
aτ + b
) = (cτ + d)k f (τ )
cτ + d
a b
for all (
) ∈ Γ.
c d
Let Γ ⊂ Γ(1). If α1 . . . , αd are coset representatives of Γ/SL2 (Z), then ⋃i αi F
is a fundamental domain for Γ, where F is the fundamental domain for Γ(1). For
example, if Γ = Γ0 (p) for a prime p, then coset representatives for Γ/SL2 (Z) will
0 −1
0 −1 1 k
be S = (
) and ST k = (
)(
) and the identity, for k = 1, . . . , p − 1.
1 0
1 0
0 1
But then we must deal with the fact that 0 is in here now and so worry about the
condition on f (τ ) as τ → 0.
Lecture 7
st
31 January 09:00
1
We have Γ ⊃ Γ(N ) ∋ (
0
N
) so f (τ + N ) = f (τ ). Let Γ be a congruence
1
1 h
subgroup, then there exists a minimal h ∈ Z≥1 such that (
) ∈ Γ and h is called
0 1
the period of the cusp at infinity. We have f (τ + h) = f (τ ) for any weakly modular
function of level Γ.
Definition 7.12 (pg. 16,[DS05]). Let f ∶ H → C be a weakly modular function of
weight k and level Γ and holomorphic for Im τ ≫ 0. Let qh = exp(2πiτ /h). Define
F(h) (qh ) = f (τ ), then F(h) is a holomorphic function on a punctured disc with
centre 0. So it has a Laurent expansion
F(h) (qh ) = ∑ an (f )qhn
n∈Z
We say that f is meromorphic (resp. holomorphic) at infinity if F(h) extends to a
meromorphic (resp. holomorphic) function at 0.
19
Lecture 7
Next some notation: the “slash operator”. Let γ ∈ SL2 (Z), k ∈ Z and f ∶ H → C.
Define f ∣γ,k ∶ H → C, by
f ∣γ,k (τ ) = (cτ + d)−k f (γτ ),
for γ = (
a
c
b
)
d
Then f is weakly modular of weight k and level Γ iff f is meromorphic and f ∣γ,k = f
for all γ ∈ Γ.
Suppose that f is weakly modular of weight k and level Γ, and α ∈ SL2 (Z).
Then f ∣α,k is weakly modular of weight k and level α−1 Γα.
Definition 7.13. Let f ∶ H → C be weakly modular of weight k and level Γ.
(1) We say that f is a meromorphic form of weight k and level Γ if f ∣α,k is
meromorphic at ∞ for all α ∈ SL2 (Z).
(2) We say that f is a modular form of weight k and level Γ if f is holomorphic
on H and f ∣α,k is holomorphic at ∞ for all α ∈ SL2 (Z).
Moreover f is a cusp form if a0 (f ∣α,k ) is 0 for all α ∈ SL2 (Z).
More notation: the C-vector space of modular forms is denoted Mk (Γ). The
C-vector space of cusp forms is denoted Sk (Γ).
Lemma 7.18. Let f ∶ H → C and α, β ∈ SL2 (Z) and k ∈ Z. Then one has
(f ∣α,k )β,k = f ∣αβ,k
Remark. This says that −∣( ),k defines a right action of SL2 (Z) on the space of
functions from H → C.
Proof. Let
a
α=( 1
c1
b1
),
d1
β=(
a2
c2
b2
)
d2
then f ∣α,k (τ ) = (c1 τ + d1 )−k f (ατ ) so
(f ∣α,k )∣β,k (τ ) = (c2 τ + d2 )−k (c1 (βτ ) + d1 )−k f (αβτ )
−k
a2 τ + b2
a2 τ + b2
) + c2 d1 τ + c1 d2 (
) + d1 d2 ) f (αβτ )
c2 τ + d2
c2 τ + d 2
and the premultiplier can be grouped as
= (c1 c2 τ (
(c1 a2 τ (
−k
c2 τ
d2
c2 τ
d2
+
) + d1 c2 τ + c1 b2 (
+
) + d1 d2 )
c2 τ + d2 c2 τ + d2
c2 τ + d2 c2 τ + d2
−k
= (c1 a2 τ + d1 c2 τ + c1 b2 + d1 d2 )
as required since
f ∣αβ,k = (c1 a2 τ + d1 c2 τ + c1 b2 + d1 d2 )−k f (αβτ )
Some consequences of the above lemma
(1) to show that f is weakly modular of weight k and level Γ then it suffices to
show that f ∣γi ,k = f for γ1 , . . . , γn generators of Γ (a finite index subgroup
of a finitely generated group is finitely generated - I forget to put this as
an exercise on the example sheet).
(2) If f is weakly modular of weight k and level Γ and α ∈ SL2 (Z), then f ∣α,k
only depends on Γα (f ∣γα,k = (f ∣γ,k )∣α,k = f ∣α,k ).
Proposition 7.19 (1). Let Γ(N ) ⊂ Γ and suppose we have f ∶ H → C holomorphic,
n
and weakly modular of weight k and level Γ. We write f (τ ) = ∑n≥0 an (f )qN
.
r
Suppose that there exists C ∈ R≥0 such that ∣an (f )∣ ≤ Cn , for some r ∈ R>0 (where
C, r independent of n). Then f is a modular form of weight k and level Γ.
20
Lecture 8
Remark. We will see later that the converse also holds.
Proof. Exercise. See exercise 1.2.6 in the textbook of Diamond & Shurman
for lots of hints.
7.1. §5: Theta functions. The Jacobi θ-function is defined
∞
θ(τ ) = ∑ exp(2πin2 τ ) = 1 + 2 ∑ q n
2
n≥1
n∈Z
It is easy to check that this series is absolutely and uniformly convergent on compact
subsets of H. The function is holomorphic and θ(τ + 1) = θ(τ ).
Definition 7.14. We define
∞
θ(τ, k) = θ(τ )k = ∑ r(n, k)q n
n=0
for k ∈ Z≥1 and r(n, k) is the number of ways of representing n as a sum of k squares
of integers (same as first lecture).
Proposition 7.20 (2). We have
√
2τ
θ(τ )
i
(the argument of the square root here has real part > 0 so we take the branch of the
square root extending the square root on R>0 ).
θ(−1/4τ ) =
Corollary 7.21. One has
θ(τ /(4τ + 1))2 = (4τ + 1)θ(τ )2 .
Proof. Just manipulating the previous result gives
√
θ(−1/4τ ) = 2τ /iθ(τ ),
√
θ(1/τ ) = −τ /2iθ(−τ /4),
√
θ(1/(τ + 4)) = −(τ + 4)/2iθ(−(τ + 4)/4),
√
θ(τ /(4τ + 1)) = (−1 − 4τ )/2τ iθ((−1 − 4τ )/4τ ).
√
Then we use θ((−1 − 4τ )/4τ ) = θ((−1/4τ ) − 1) = θ(−1/4τ ) = 2τ /iθ(τ ).
Suppose that k ≥ 2 is even. Then
θ(τ, k)∣γ,k/2 = θ(τ, k)
for γ ∈ ⟨ (
1
0
1
1
),(
1
4
0
)⟩
1
If k ≡ 0 (mod 4), this holds moreover for
1
γ ∈⟨±(
0
1
1
),(
1
4
0
)⟩ = Γ
1
On the example sheet Γ = Γ0 (4). So for k ≡ 0 (mod 4) then θ(τ, k) is weakly modular
of weight k/2 and level Γ0 (4). If k ≡ 2 (mod 4), then θ(τ, k) is weakly modular of
weight k/2 and level Γ1 (4).
Proposition 1 implies that θ(τ, 4k) ∈ M2k (Γ0 (4)) (it is enough to show that
θ(τ, 4) ∈ M2 (Γ0 (4)) then take the powers of k(?)). Recall that
r(n, 4) = 8 ∑ d
0<d∣n
4∤d
Next time we will deduce this from the fact that dimC M2 (Γ0 (4)) = 2 and this space
has a basis of Eisenstein series.
21
Lecture 8
Lecture 8
3rd February 09:00
2
We defined θ(τ ) = ∑n∈Z q n then for k ≥ 0
θ(τ, 4k) = (θ(τ ))4k ∈ M2k (Γ0 (4))
Proposition 8.22. One has
√
2τ /iθ(τ )
θ(−1/4τ ) =
Proof. Let h(x) = exp(−πtx) for t ∈ R>0 and x ∈ R then ĥ(y) =
Then Poisson summation implies that
1
√
t
exp(−πy 2 /t).
1
2
2
∑ exp(−πtd ) = √ ∑ exp(−πm /t)
t m∈Z
d∈Z
So for t ∈ R>0 then
1
θ(it/2) = √ θ(i/2t)
t
so we get the result on setting τ = it/2 for t ∈ R>0 (i.e. for τ = s ⋅ i and s > 0). By
the uniqueness of analytic continuation, this implies that we have equality for all
τ ∈ H.
Last time there was a proposition that said that if f is holomorphic at ∞, and
the coefficients of the q-expansion satisfy an (f ) ≤ Cnr then f ∣α,k is holomorphic at
infinity for α ∈ SL2 (Z). If we set
θ(τ, k) = ∑ r(n, k)q n
n≥0
where r(n, k) is the number of way of writing n as a sum of squares - so we deduce
that it is ≤ 2k nk . So we apply proposition 1 to deduce that θ(τ, 4k) ∈ M2k (Γ0 (4)).
8.1. §6: Old Forms.
Lemma 8.23. Let f ∈ Mk (Γ0 (N )) (e.g. N = 1). Let M ≥ 1, then fM ∶ H → C
defined by τ ↦ f (M τ ) is in Mk (Γ0 (M N )).
+b
Remark. The action SL2 (Z) ↻ H is given by τ ↦ aτ
then we can get an action
cτ +d
from
GL+2 (R) = {γ ∈ GL2 (R), det γ > 0}
acting by the same rule.
Proof. We have
fM (τ ) = f ( (
a
Suppose that γ ∈ Γ0 (M N ) and γ = (
c
M
0
0
) ⋅ τ)
1
b
), then
d
fM ∣γ,k (τ ) = (cτ + d)−k f ( (
M
0
0
) γτ )
1
= (cτ + d)−k f ( (
M
0
0
M
)γ (
1
0
= (cτ + d)−k f ( (
1
c/M
22
0
)
1
bM M
)(
d
0
−1
M
(
0
0
) τ)
1
0
) τ)
1
Lecture 8
1
and (
c/M
bM
) ∈ Γ0 (N ) and f has level Γ0 (N ) so
d
= f (M τ ) = fM (τ )
It is clear that fM is holomorphic at infinity: if f (τ ) = ∑ an q n then fM = ∑ an q nM ,
and we have
cM τ + d k
) f ∣α,k (M τ )
fM ∣α,k (τ ) = (
cτ + d
a b
for α = (
) ∈ SL2 (Z). So f ∣α,k (τ ) is bounded for Im τ ≫ 0 (by holomorphy at
c d
infinity). Thus fM ∣α,k (τ ) is bounded for Im τ ≫ 0 thus fM ∣α,k is holomorphic at
infinity.
Now we want to see how to get that formula for the number of squares. Recall
that
r(n, 4) = 8 ∑ d
d∣n
d>0,4/∣d
and the thing on the left comes from θ(τ, 4) ∈ M2 (Γ0 (4)) so we just have to show
that the left hand side is a weight-2 Eisenstein series.
8.2. §7: Weight-2 Eisenstein series. For k > 2 and even we defined
′
Gk (τ ) = ∑
1
(cτ + d)k
This sum does not converge absolutely for k = 2. But we can define
1
1
− 8π ∑ ( ∑
)
2
2
c≠0 d∈Z (cτ + d)
d≠0 d
G2 (τ ) = 2ζ(2) + 2(2πi)2 ∑ σ1 (n)q n = 2 ∑
n≥1
Warning: G2 (τ ) is not weakly modular (we showed that Gk (τ ) (k > 2) was weakly
∼
modular by considering for SL2 (Z) that γ∶ Z2 /{(0, 0)} → Z2 /{(0, 0)} and reordering
the sum ∑(c,d) ). We could get around this by defining
G∗2 (τ ) = G2 (τ ) −
π
Im τ
which then satisfies
G∗2 (
a
for (
c
aτ + b
) = (cτ + d)2 G∗2 (τ )
cτ + d
b
) ∈ SL2 (Z). But this is not holomorphic. An exercise is to show that
d
G2 (τ ) − 2G2 (2τ ) ∈ M2 (Γ0 (2))
G2 (τ ) − 4G2 (4τ ) ∈ M2 (Γ0 (4))
A fact is that dim M2 (Γ0 (4)) = 2 and G2 − 2G2 (2τ ), G2 − 4G2 (4τ ) are a basis
(we have Γ0 (4) ⊂ Γ0 (2) so M2 (Γ0 (2)) ↪ M2 (Γ0 (4)) by f ↦ f ). By comparing
q-expansions, we get the following.
Corollary 8.24. We have
θ(τ, 4) = −
1
(G2 (τ ) − 4G2 (4τ ))
π2
Remark. This implies that r(n, 4) = 8 ∑d∣n,4∤d d.
23
Lecture 9
Theorem 8.25. Let Γ be a congruence subgroup then
Mk (Γ) = {0} (for k < 0)
M0 (Γ) = C
Mk (Γ) is a finite dimensional vector space (for k > 0)
Moreover we have (the slightly stronger statement) that ⊕k≥0 Mk (Γ) is a finitely
generated C-algebra.
Suppose that Γ′ ◁ Γ are congruence subgroups and let G = Γ′ /Γ (described by
elements Γ′ γ). Then G acts on Mk (Γ′ ): f ∈ Mk (Γ′ ) and g ∈ G with g = Γ′ γ, then
define f g = f ∣γ,k . One can check that Mk (Γ′ )G = Mk (Γ) where Mk (Γ′ )G is the
group of invariants. Let f ∈ Mk (Γ′ ), then consider ∏g∈G (f − f g ) = 0 in ⊕k Mk (Γ),
where
g
n
n−1
+ ⋅ ⋅ ⋅ + hn−1 f + hn
∏ (f − f ) = f + hi f
g∈G
then the hi are symmetric so
hi ∈ M○ (Γ′ )G = M○ (Γ)
(they won’t necessarily be of weight k).
Lemma 8.26. One has f ∈ Mk (Γ′ ) (which implies that f satisfies a polynomial
with coefficients in M○ (Γ)).
Lecture 9
7th February 09:00
9.1. Structure of Mk (Γ).
Lemma 9.27. Let Γ′ ◁ Γ be a congruence subgroup. Let f ∈ Mk (Γ′ ) then there
exists hi ∈ Mik (Γ) for i = 1, . . . , n = [Γ ∶ Γ′ ], such that
f n + h1 f n−1 + ⋅ ⋅ ⋅ + hn−1 f + hn = 0
and this equality holds in Mnk (Γ′ ).
Proof. Let G = Γ′ /Γ. then there is an action of G on Mk (Γ′ ) and Mk (Γ′ )G =
Mk (Γ). Then consider ∏g∈G (f − f g ) = 0: in the expansion in powers of f the
coefficients are symmetric polynomials in the f g , so they are G-invariant, thus they
are elements of Mk (Γ).
One has
M (Γ) = ⊕k Mk (Γ) ⊂ ⊕k Mk (Γ′ ) = M (Γ′ )
with M (Γ) ⊂ M (Γ′ ) integral: this follows from the above and the following exercise:
show that for A a subring of B with b ∈ B then b is integral over A iff there exists
a ring C with A ⊂ C ⊂ B with b ∈ C and C is finitely-generated A-module. Then
deduce that à is a subring of B. Then show that if we have an extension of rings
A ⊂ B with B integral over A and B a finitely generated A-algebra, then B is a
finitely generated A-module.
Corollary 9.28. Let Γ be a congruence subgroup and k < 0. Then Mk (Γ) = {0}
and M0 (Γ) is the set of constant functions.
Proof. Assume that Γ ◁ SL2 (Z), and k < 0. Apply lemma 9.27 to f ∈ Mk (Γ)
for Γ ◁ Γ(1) then there are hi ∈ Mik (Γ(1)) = {0} as in the lemma but Mik (Γ(1)) =
{0} since k < 0, so we have f n = 0, thus f = 0. For k = 0, apply the lemma again:
we take elements hi ∈ M0 (Γ(1)) the constants. Hence f is a root of a polynomial
in C[X] so f is a constant function. For general Γ, Γ(N ) ⊂ Γ ⊂ SL2 (Z), we have
24
Lecture 10
Γ(N ) ◁ SL2 (Z). Since Mk (Γ) ⊂ Mk (Γ(N )) so the Γ case follows from the Γ(N )
case.
Theorem 9.29. Let Γ be a congruence subgroup, then
M (Γ) = ⊕k≥0 Mk (Γ)
is a finitely generated C-algebra.
We need the following lemma first.
Lemma 9.30. Let F be a field. Let A ⊂ B be integral domains that are also F algebras (B a commutative F -algebra). Assume that B is integral over A i.e. for all
b ∈ B, there exists elements ai ∈ A such that bn + a1 bn−1 + ⋅ ⋅ ⋅ + an = 0. Assume that
Frac(B)/Frac(A) is a field extension of finite degree. Then A is a finitely generated
F -algebra iff B is a finitely generated F -algebra.
Proof. Let A be a finitely generated F -algebra. Let à be the integral closure
of A in Frac(B) (i.e. Ã is the set of elements of Frac(B) satisfying monic polynomials
with coefficients in A). We have A ⊂ B ⊂ Ã (since B ⊂ Frac(B) and B is integral
over A so we must have B ⊂ Ã). By a result of Emmy Noether: Ã is a finitelygenerated A-module. So if A is Noetherian then B is a finitely-generated A-module,
so B is a finitely generated F -algebra.
Conversely, suppose that b1 , . . . , bn generate B as an F -algebra. Let C ⊂ A ⊂ B
be the F -algebra generated by the coefficients of the polynomials satisfied by the
bi (B is integral over A so take the polynomials that have these as roots). B is a
finitely generated and integral C-algebra, thus B is a finitely-generated C-module,
hence A is a finitely-generated C-module so A is a finitely-generated F -algebra. Proof: (of thm. 9.29). Let Γ(N ) ⊂ Γ with Γ(N )◁SL2 (Z). Let A = M (Γ(1))
(generated by E4 , E6 over C), B = M (Γ(N )) and C = M (Γ). We have A ⊂
C ⊂ B. Our claim is that Frac(B)/Frac(A) is finite (which would imply that
Frac(B)/Frac(C) is finite). Then A is a finitely-generated C-algebra and the previous lemma would imply that B is a finitely-generated C-algebra so C is a finitelygenerated C-algebra.
Proof of claim: let G = Γ(N )/Γ(1) (≅ SL2 (Z/N Z)). G acts on Mk (Γ(N )) for
each k (f ↦ f g = f ∣γ,k where Γ(N )γ = g). So G acts on M (Γ(N )) = B, B G = A
hence G acts on Frac(B). Let x = p/q ∈ Frac(B) then
x ∏ qg ∈ B G = A
g∈G
so x ∈ Frac(A) so Frac(A) = (Frac(B))G (it is clear that Frac(A) ⊂ (Frac(B))G
since B G = A). Then using Artin’s lemma5 we have that Frac(B)/Frac(A) is a
finite extension of fields.
Corollary 9.31. For k ≥ 0, Mk (Γ) is a finite dimensional C-vector space.
Proof. We have shown that ⊕k≥0 Mk (Γ) is a finitely generated C-algebra. By
decomposing a generator into its components of fixed weight, we obtain a generating
set whose elements are modular forms f1 , . . . , fn of weight k1 , . . . , kn . This implies
that Mk (Γ) is spanned by the monomials
{ ∏ fil ∶ li ≥ 0, ∑ ki li = k}
i
i
5Artin’s lemma: let G be a finite group of automorphisms of a field E, with F = E G , then
[E ∶ F ] < ∞ and [E ∶ F ] ≤ ∣G∣.
25
Lecture 10
Lecture 10
7th February 09:00
The examples class is at 3pm on the 17th of February in MR3. For the next section,
c.f. Miranda, Farkas, Kra.
10.1. Riemann surfaces.
Definition 10.15. Let X be a topological space, a presheaf (of Abelian groups)
on X is a pair (F , ρ) comprising
● for U ⊂ X open, an Abelian group F (U ),
● if V ⊂ U ⊂ X open then
ρU
V ∶ F (U ) → F (V )
group homomorphisms with ρU
U the identity and if W ⊂ V ⊂ U then
U
ρVW ○ ρU
V = ρW
Notation if f ∈ F (U ) with V ⊂ U then we write f ∣V for ρU
V (f )
Example.
cts
(1) Let X be any topological space, then (OX
, ρ) is a pre-sheaf, where
cts
OX
= {continuous functions from U → C}
cts
cts
and for V ⊂ U , ρU
V ∶ OX (U ) → OX (V ) given by f ↦ f ∣V .
(2) Let X ⊂ C be open then take (OX , ρ) where
OX (U ) = {holomorphic functions on X}
and
ρU
V
cts
is restriction. One has OX ⊂ OX
.
Definition 10.16. A presheaf F on a topological space X is a sheaf if for any
U ⊂ X open and {Ui }i∈I an open cover of U , then if
● f, g ∈ F (U ) with f ∣Ui = g∣Ui for all i, then f = g,
● if fi ∈ F (Ui ), i ∈ I with fi ∣Ui ∩Uj = fj ∣Ui ∩Uj for all i, j ∈ I, then there exists
(a unique (it is unique by the previous axiom)) f ∈ F (U ) with f ∣Ui = fi
for each i ∈ I.
Example. Examples 1 and 2 above are both actually sheaves.
Definition 10.17. A Riemann surface is a connected Hausdorff topological space
cts
cts
X equipped with a sheaf OX , with OX ⊂ OX
(i.e. OX (U ) ⊂ OX
(U ) for each U )
such that there exists an open cover (atlas) ⋃i∈I Ui = X and homeomorphisms
∼
φi ∶ Ui Ð→ Vi ⊂ C
where the Vi are open, with for U ⊂ Ui open
holomorphic functions
(
)
φi (U ) → C
f ↦f ○φi
∼
OX (U )
an isomorphism.
Definition 10.18. A holomorphic map between Riemann surfaces is a continuous
map f ∶ X → Y such that for U ⊂ Y open, pre-composing with f gives a homomorphism
OY (U ) → OX (f −1 (U ))
φ↦φ○f
This map is denoted
sheaf on Y .
f ♯ ∶ OY
→ f∗ OX where f∗ OX is given by U ↦ OX (f −1 (U )) a
26
Lecture 11
The aim for this and the next lecture: suppose that X is a Riemann surface
with a nice action of a group G (e.g. a congruence subgroup Γ ↻ H). We get a
π
topological space X ↠ G/X and we define OG/X (U ) = OX (π −1 (U ))G .
Theorem 10.32. (G/X, OG/X ) is a Riemann surface and the map π∶ X ↠ G/X
is holomorphic, and in local coordinates at a point x ∈ X, the map π looks like
x ↦ z nx .
Remark. The case of G finite: see the book of Miranda III.3.
10.2. §8: Groups acting on Riemann surfaces. Today, we will just use
the fact that a Riemann suface is Hausdorff and locally compact. Locally compact:
for every point x ∈ X, there exists an open set U ⊂ X, and a compact set K ⊂ X,
with x ∈ U ⊂ K, where K is a compact neighbourhood of x. If every point has an
open neighbourhood homeomorphic to an open disc in C then X is locally compact.
Let G be a group, X a Riemann surface. A holomorphic group action of G on X
is a group action g∗ ∶ X → X with all maps g∗ holomorphic (if e is the identity then
e∗ ∶ X → X is the identity and for g, h ∈ G then (gh)∗ = g∗ ○h∗ ) i.e. a homomorphism
G → Authol (X), where Authol (X) is the group of holomorphic bijections X → X
with holomorphic inverse.
Definition 10.19. Let G ↻ X be a holomorphic group action. We say that G
acts properly if for any pair A, B ⊂ X of compact subsets, the set SA,B = {g ∈
G∶ g∗ (A) ∩ B ≠ ∅} is finite.
Exercise: give G the discrete topology. Consider f ∶ G × X → X × X given by
(g, x) ↦ (x, g∗ x). Show that G acts properly iff f is proper (f is proper if the
inverse image of a compact set is a compact set, so in this case if K ⊂ X × X then
f −1 (K) is compact).
Remark. If G ↻ X is a proper action then for x ∈ X, the stabiliser Gx is finite
(apply the exercise to the compact set {x} × {x}).
Lemma 10.33. If G ↻ X is proper, then for x ∈ X, there is an open neighbourhood
x ∈ Ux (connected with compact closure) satisfying
g∗ (Ux ) ∩ Ux ≠ ∅ ⇔ g∗ x = x
Proof. First we find U ∋ x an open neighbourhood such that gx (U ) ∩ U ≠ ∅
for only finitely many g ∈ G (apply the definition of proper action to a compact
neighbourhood K of x, take U = interior(K)). We have g1 , . . . , gn with g∗ U ∩U ≠ ∅.
If gi,∗ ≠ x, take Vi , Vi′ distinct open neighbourhoods of x, gi,∗ x in U . Let Wi be an
open neighbourhood of x in X with gi,∗ (Wi ) ⊂ Vi′ . Set Ui = Vi ∩ Wi . Check that
Ui ∩ gi Ui = ∅. Let Ux be the intersection of the connected components of ⋂i Ui . Lecture 11
10th February 09:00
11.1. Groups acting on Riemann surfaces. Last time we had lemma 10.33
that if G ↻ X is proper with X a Riemann surface, then there exists an open
neighbourhood Ux ∋ x (connected with compact closure) such that: (gUx ∩ Ux ≠
∅) ⇔ gx = x (i.e. g ∈ Gx ). There is a refined version of this:
Lemma 11.34. Let g ∈ Gx and x ∈ gUx ∩ Ux and G ↻ X as above, then there
exists a Ux as in lemma 1 with gUx = Ux for g ∈ Gx .
27
Lecture 11
Remark. This lemma implies that the G-orbit of Ux in X is ⋃g∈G gUx is
gUx
∐
gGx ∈G/Gx
and gUx depends only on gGx .
Proof. Let U ∋ x be as in lemma 10.33. Set Ux = the connected component
of ⋂g∈Gx gU .
Regarding the quotient topology: let G/X be the set of G-orbits and we have
a map
π∶ X → G/X
x ↦ Gx,
then the topology is defined: a set V ⊂ G/X is open iff π−1 (V ) is open in X.
Exercise: this is the coarsest topology such that every continuous map f ∶ X → Y
where Y is a topological space, with f (gx) = f (x) for all g ∈ G and x ∈ X, factors
through as a continuous map
X
f
π
G/X
Y
From now on we regard G/X as a topological space (with the quotient topology).
Lemma 11.35. The map π∶ X → G/X is continuous and open and G/X is Hausdorff.
Proof. The map π is continuous by definition of quotient topology. Let U ⊂ X
be open. Then we need to show that π −1 (πU ) is open. Indeed, π −1 (πU ) = ⋃g gU a
union of open sets.
It remains to show that G/X is Hausdorff. Suppose that Gx ≠ Gy are distinct
G-orbits. Take Kx ⊃ Ux and Ky ⊃ Uy distinct compact neighbourhoods of x, y ∈ X.
Properness implies that gKx ∩ Ky ≠ ∅ iff g = g1 , . . . , gn . We can shrink Kx , Ky to
ensure that y ∉ gi Kx for i = 1, . . . , n. Now set Vy = Uy ∩(X/ ⋃ni=1 gi Kx ). Observe that
gUx ∩ Vy = ∅ for all g ∈ G. So π(Ux ) and π(Vy ) are disjoint open neighbourhoods
of Gx and Gy.
Corollary 11.36. Let x ∈ X, Ux ∋ x a Gx -stable open neighbourhood as in lemma
11.34. Then we have a diagram
⋃g∈G gUx
π −1 (πUx )
∐g∈G/Gx gUx
gy
Gx /Ux
Gx y
π
πUx
with the bottom row a homeomorphism and Gx /Ux has the quotient topology.
Proof. We need to check that the map gy ↦ Gx y is well-defined and continπ
uous here: the map Ux → πUx is continuous and π(gx y) = π(y) for all y ∈ Gx . So it
factors through
Ux
π
Gx /Ux
28
πUx
Lecture 12
Likewise, the right hand map is constant on G-orbits, hence it factors through πUx .
so
π −1 (πUx )
Gx /Ux
πUx
It follows that the quotient of ⋃g∈G gUx = ∐g∈G/Gx gUx by G is given by
−1
∐ gGx g /(gUx )
q∈G/Gx
and Gx ↻ Ux is the same as gGx g −1 ↻ gUx by y ↦ gy.
Definition 11.20. Let X be a Riemann surface, G ↻ X a proper group action.
Let V ⊂ G/X be open. The set π −1 (V ) has an action of G. So OX (π −1 (V )) has
a right action of G: if f ∶ π −1 (V ) → C, define f g by f g (x) = f (gx). Denote the
G-invariants in OX (π −1 (V )) by OX (π −1 (V ))G .
Definition 11.21. Denote by OG/X the presheaf
V ↦ OX (π −1 (V ))G
(the restriction maps are the obvious ones: if W ⊂ V then
OX (π −1 (V ))G Ð→ OX (π −1 (W ))G ).
restrict
It is easy to check that OG/X is a sheaf. Also if V ⊂ G/X is open then
cts
cts
OG/X (V ) ⊂ OX
(π −1 (V ))G = OG/X
(V ).
Theorem 11.37. (G/X, OG/X ) is a Riemann surface.
Remark. If X is connected then G/X is connected (I stated that all my Riemann
surfaces are assumed connected).
Proof. We need to find charts on open neighbourhoods of every point of G/X.
By corollary 11.36, we can reduce to 0 ∈ X ⊂ C where X is connected and open,
G ↪ Authol (X), G a finite group, stabilising 0 (and for x ≠ 0, Gx = {id}). I will
sketch a proof, the full proof will be in the notes. Claim: there exists an open
neighbourhood U of 0, with gU = U for all g ∈ G and a biholomorphism U ≃ D
(where D is the open unit disc) such that for all g ∈ G, the diagram
U
g
≃
D
U
≃
xζ(g)
D
for a root of unity ζ(g).
The claim implies the theorem: the claim implies that G ≃ Z/nZ, generator
acting by xζ on D, where ζ is a primitive nth root of unity.
G/U ≃ (Z/nZ)/D → D
where the map (Z/nZ)/D is given by z ↦ z n (a holomorphic function on z ∈ D is
invariant under multiplication by ζ iff it is a holomorphic function in z n ).
29
Lecture 12
Lecture 12
12th February 09:00
Theorem 12.38. (G/X, OG/X ) is a Riemann surface. Moreover, π∶ X → G/X
is holomorphic, and for x ∈ X, π has the local form z ↦ z nx in a neighbourhood
of x (i.e. there exists charts around x (with x ∈ Ux ≃ Vx ⊂ C) and π(x) (with
π(x) ∈ Uπ(x) ≃ Vπ(x) ⊂ C) and
x
π
Ux
Uπ(x)
≃
0
Vx
π(x)
≃
z↦z nx
Vπ(x)
0
where nx is the order of Stabr(G) (x), where r(G) ⊂ Aut(X)). Moreover, Stabr(G) (x)
is cyclic of order nx , where r∶ G → Aut(X).
The idea of the proof is to reduce to the case of a finite group acting on D =
{z ∈ C∶ ∣z∣ < 1}, with g0 = 0 for all g ∈ G.
∼
Recall Schwarz’ lemma: if f ∶ R → D is a biholomorphism which fixes 0, then
f (z) = ζz for ζ ∈ C with ∣ζ∣ = 1. So if G ⊂ Authol (D) with g0 = 0 for all g ∈ G with
G finite, then G ≃ Z/nZ, a generator acts by z ↦ ζz with ζ a primitive nth root of
unity.
Now consider the map
α∶ G/D → D
Gz ↦ z n
and we have
D
π
G/D
α
∼
D
z↦z n
Claim. As above, for U ⊂ G/D open, this identifies OG/D (U ) with OD (α(U )), i.e.
(G/D, OG/D )
is a Riemann surface and in fact it is isomorphic to the unit disc.
Proof. We have
OG/D (U ) = {f ∶ π −1 (U ) → C holomorphic, with f (ζz) = f (z) for z ∈ π −1 (U )},
so that f ∈ OG/D (U ) is a holomorphic function in z n , and f (z) = g(z n ) then
g ∈ OD (α(U )).
Proof: (of thm. 12.38). We can assume that G ⊂ Aut(X). We saw that for
x ∈ X, we had a Gx -invarient open neighbourhood Ux ∋ x with Gx /Ux ≃ a neighbourhood of π(x) ∈ G/X (a homeomorphism). This reduces the theorem to the
case where G is finite, and fixes the a point x ∈ X. By a convexity argument and
the Riemann mapping theorem (see the online notes) it follows that we can find a
neighbourhood x ∈ Ux which is G-invariant and Ux ≃ D by x ↦ 0 (biholomorphically) and it follows that this reduces to the case studied previously (c.f. Miranda
III.§3 (for the case that G is finite), and the book of Farkas and Kra).
30
Lecture 13
12.1. §9: Applications to modular forms. First we must obviously show
that we have a proper action, if we want to apply this stuff to modular forms.
Proposition 12.39. Γ(1) ↻ H is a proper group action (so every congruence
subgroup acts properly).
Proof. We want to show that if A, B ⊂ H is compact, then
#{γ ∈ Γ(1)∶ γA ∩ B ≠ ∅} < ∞
m
The idea is that A ⊂ ⋃m
i=1 γi F̃ and B ⊂ ⋃j=1 δj F̃ (you can convince yourself of this
but it is fiddly). Then if γA ∩ B ≠ ∅ then γγi = ±δj for some i, j so γ ∈ a finite set.
An easier proof: we can assume that A, B are rectangles, with Im (A) ⊂ [a1 , a2 ]
and Im (B) ⊂ [b1 , b2 ] and same thing with the real part. Suppose that τ ∈ A, and
a b
γτ ∈ B with γ = (
). Then you can show that if
c d
Im (τ )
≤ b2
∣cτ + d∣2
then the real part of τ is bounded then there are only finitely many possibilities for
c, d. Therefore
b1 ≤
1
γ = ±(
0
n
) {finitely many elements of Γ(1)}
1
And if Re(γτ ) is bounded, then there are only finitely many possibilities for n.
Definition 12.22. If Γ is a congruence subgroup then define
Y (Γ) ∶= Γ/H,
(Riemann surface)
called the open modular curve of level Γ.
Corollary 12.40. The map j∶ H → C (a meromorphic form of weight 0) induces a
biholomorphism j∶ Y (Γ(1)) → C.
Proof. We saw that j is a bijection Γ(1)/H → C. The map j∶ H → C is in
OH (H)Γ(1) = OY (Γ(1)) (Y (Γ(1))). So j induces a holomorphic bijection Y (Γ(1)) →
C, hence a biholomorphism.
π
We have a Riemann surface but we also have this associated map H → Y (Γ(1)) ≃
C, then π has the local form
⎧
⎪
z ↦ z at τ ∈ H non-elliptic,
⎪
⎪
⎪
⎪
⎨z ↦ z 2 at τ ∈ Γ(1) ⋅ i,
⎪
⎪
⎪
⎪
z ↦ z 3 at τ ∈ Γ(1) ⋅ ω.
⎪
⎩
Weakly modular functions of weight 2k and level Γ correspond to meromorphic
k-fold differentials on Y (Γ).
12.2. Cusps and compact modular curves. Weight 0 weakly modular
functions of level Γ(1) (e.g. the j-invariant) are the same thing as meromorphic
functions on Y (Γ(1)). Our strategy of checking that they behave well at infinity was mapping to the unit disc and taking a Laurent expansion about q = 0.
The idea here is that a weight 0 modular form is a holomorphic function on
X(Γ(1)) ⊃ Y (Γ(1)) ≃ C and X(Γ(1)) is compact and ≃ to C ∪ {∞} ≃ PC the
Riemann sphere and the point at infinity will correspond to q = 0.
Lecture 13
14th February 09:00
31
Lecture 13
13.1. Cusps and compactifying modular curves.
Definition 13.23. Let Γ be a congruence subgroup, the cusps of level Γ are denoted
CΓ := Γ/P1 (Q),
where
a
(
c
b x
ax + by
)[ ] = [
].
d y
cx + dy
The cusp at ∞ corresponds to Γ ⋅ [1 ∶ 0].
Remark. N.b. the action of SL2 (Z) on P1 (Q) is transitive.
Definition 13.24. Suppose that s = Γx ∈ CΓ . The width hs of s is defined to be
the index of {±1} ⋅ StabΓ (x) in StabSL2 (Z) (x) (or StabΓ (x) in StabPSL2 (Z) (x) where
SL2 (Z)/{±1} = PSL2 (Z) and Γ is the image of Γ in PSL2 (Z)).
Exercise: hs is independent of the choice of x ∈ S. If γ ∈ SL2 (Z) with γx = ∞
then hs is the width of the cusp ∞ at level γΓγ −1 (which is the same as showing
1 hs
that ± (
) generates StabγΓγ −1 (∞)).
0 1
Example. For Γ = Γ(1) then CΓ = {Γ[1 ∶ 0]} = {∞} we have h∞ = 1. For p prime,
and Γ = Γ0 (p), an exercise is to show that
1
0
CΓ = {Γ [ ] , Γ [ ] },
0
1
1
0
that the other
and since (
1
) ∈ Γ0 (p) it follows that Γ[1 ∶ 0] has width 1, and you can show
1
has width p.
We want to construct compact Riemann surfaces with the underlying set
Y (Γ) ∪ CΓ = X(Γ)
we regard X(Γ) as a topological space.
Definition 13.25. We denote H∗ := H ∪ P1 (Q) called the extended upper halfplane. There is an action SL2 (Z) ↻ H∗ .
We define a topology on H∗ by specifying some open sets and taking the topology generated by them. We have the following open sets
● U ⊂ H open,
● {∞} ∪ UA where UA = {τ ∶ Im (τ ) > A} for all A ∈ R>0 ,
● α({∞} ∪ UA ) for α ∈ SL2 (Z) for A ∈ R>0 (where α({∞} ∪ UA ) is the open
neighbourhood of α∞ ∈ P1 (Q))
An exercise is to show that α({∞} ∪ UA ) looks like a circle in the first quadrant
tangent to real axis at α∞.
H∗ is connected and Hausdorff.
Definition 13.26. We define X(Γ) as a topological space to be Γ/H∗ with the
quotient topology.
Remark. Warning: Γ ↻ H∗ is not proper e.g. StabΓ (∞) is infinite (it is a consequence of the definitions that a proper group action must have finite stabilisers).
Lemma 13.41. If x ∈ P1 (Q), then there exists an open neighbourhood U of x ∈ H∗ ,
such that for γ ∈ Γ we have γU ∩ U ≠ ∅ iff γ ∈ StabΓ (x) = Γx .
32
Lecture 14
Proof. We have αx = ∞ for some α ∈ SL2 (Z). Translating by α: it is enough
to prove the lemma for x = ∞.
Exercise: there exists A large enough such that γUA ∩ UA ≠ ∅ implies that
γ ∈ Γ∞ (i.e. γ is upper triangular).
The point of this lemma is that if we have x ∈ P+ (Q), then there’s an open
neighbourhood U ∋ x in H∗ , such that Y (Γ) ∩ π(U ) (where π∶ H∗ → X(Γ)) is
biholomorphic to
1 hs
⟨(
) ⟩/UA ,
0 1
which is isomorphic to the punctured open disc via τ ↦ exp(2πiτ /hs ) = qhs (where
hs is the width of the cusp S = Γx).
Proposition 13.42. X(Γ) is connected, Hausdorff and compact.
Proof. H∗ is connected so X(Γ) is connected.
We have a continuous map X(Γ) → X(Γ(1)) ≅ P1C , a homeomorphism. To show
that X(Γ) is Hausdorff: let x, y ∈ Y (Γ) then we can separate x, y. If s, s′ ∈ CΓ then
the previous lemma implies that we can separate s, s′ . If s ∈ G and x ∈ Y (Γ) then
we can separate the image of s and the image of x in X(Γ(1)). Taking the inverse
image implies that we can separate in X(Γ).
We have F̃ ⊂ H and define F̃∗ = F̃ ∪ {∞} ⊂ H∗ . The set F̃∗ is compact, and
X(Γ) is the image of
αF̃∗
⋃
Γα∈Γ/Γ(1)
∗
under (H → X(Γ)). So X(Γ) is compact.
Definition 13.27. Let U ⊂ X(Γ) be open. We write
cts
OX(Γ) (U ) = {f ∈ OX(Γ)
(U ), such that f ∣U ∩Y (Γ) ∈ OY (Γ) (U ∩ Y (Γ))}.
Proposition 13.43. (X(Γ), OX(Γ) ) is a Riemann surface.
Proof. OX(Γ) is a sheaf. Every point of Y (Γ) has an open neighbourhood U
with a chart. For the cusp at infinity there is a homeomorphism from π(UA ∪ {∞})
to an open disc, which restricts to a biholomorphism from π(UA ) ⊂ Y (Γ) to a
punctured unit disc. So you can see that this must be a chart on π(UA ∪ {∞})
(a continuous function on an open disc, holomorphic on the punctured disc is a
holomorphic function on the open disc).
Remark. If we think about the map X(Γ) → X(Γ(1)) then in local coordinates
at a cusp s ∈ CΓ , it has the form z ↦ z hs (local coordinates qhs = exp(2πiτ /hs ),
q = qhhss ).
Lecture 14
17th February 9:00
Reminder: examples class is today at 3pm in MR3.
14.1. Differentials and divisors. We have our compactified modular curves
X(Γ), these are compact Riemann surfaces. We want to relate modular forms to
meromorphic differentials on X(Γ) - we use the Rieman-Roch theorem to compute
the dimensions of M2k (Γ) for any congruence subgroup Γ.
a b
Let f ∶ H → C be weakly modular of weight 2 and level Γ. Let γ = (
) then
c d
1
d(γτ )
=
,
dτ
(cτ + d)2
33
Lecture 14
so f (τ ) dτ satisfies f (γτ ) d(γτ ) = f (τ ) dτ . Similarly if f has weight 2k,
f (γτ )(d(γτ ))k = f (τ )(dτ )k .
14.2. Meromorphic differentials on Riemann surfaces.
Definition 14.28. Let U ⊂ C be open and n ∈ Z>0 , we define the C-vector space
of meromorphic differentials of degree n on U to be
Ω⊗n (U ) = {f (z)(dz)n ∶ f meromorphic on U }.
If φ∶ U1 → U2 is holomorphic with U1 , U2 ⊂ C open, then define
φ∗ ∶ Ω⊗n (U2 ) → Ω⊗n (U1 ),
f (z2 )(dz2 )n ↦ f (φ(z1 ))(φ′ (z1 ))n (dz1 )n .
Definition 14.29. Let X be a Riemann surface. If U1 , U2 are two open subsets of
∼
X, with charts φi ∶ Ui → Vi ⊂ C for i = 1, 2, we write
∼
τij := φj ○ φ−1
i ∶ φi (Ui ∩ Uj ) Ð→ φj (Ui ∩ Uj ) .
´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹¸ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¶
´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¸ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¶
⊂Vi
⊂Vj
Then a meromorphic differential of degree n on X is a rule sending charts φ∶ X ⊃
U → V ⊂ C to meromorphic differentials ω(φ) on V , such that for two charts φ1 , φ2
∗
τij
(ω(φj )∣φj (Ui ∩Uj ) ) = ω(φi )∣φi (Ui ∩Uj ) .
Exercise: check that a meromorphic differential on X is uniquely determined by
∼
ω(φi ), with φ∶ Ui → Vi and X = ⋃i Ui , and vice versa.
A consequence of this is that for the sheaf given by
X ⊃ U ↦ {meromorphic differentials of degree n on U },
with restriction maps ρU
V given by pullbacks, then if f ∶ X → Y is holomorphic, then
we can define
f ∗ ∶ {meromorphic differentials on Y } → {meromorphic differentials on X},
using the diagram
X
f
∪
UX
∪
f
∼ φx
C ⊃ VX
Y
UY
∼ φY
f
VY ⊂ C
and ω(φY ) ↦ f ∗ ω(φY ) where f ∗ ω(φX ) is a differential on VX .
Notation: let Ω⊗n (X) be the set of meromorphic differentials of degree n on a
Riemann surface X, and for U ⊂ X let Ω⊗n
X (U ) be the set of meromorphic differentials on U . This makes Ω⊗n
into
a
sheaf
on X.
X
We have
π
H → Y (Γ) ↪ X(Γ).
Definition 14.30. Suppose that ω ∈ Ω⊗n (X(Γ)), then we get π ∗ ω ∈ Ω⊗n (H) and
π ∗ ω = fω (τ )(dτ )n .
Theorem 14.44. Let ω ∈ Ω⊗n (X(Γ)) and π ∗ ω = fω (τ )(dτ )n . Then fω is a meromorphic form of weight 2k and level Γ (weakly modular and meromorphic at ∞).
Moreover, ω ↦ fω gives a C-vector space isomorphism
∼
Ω⊗k (X(Γ)) → Mero2k (Γ)
where Mero2k (Γ) is the set of meromorphic forms or weight 2k and level Γ.
34
Lecture 15
Proof. (1) fω is weakly modular: if γ ∈ Γ we have
γ∶ H
∼
τ ↦γ⋅τ
H
π
π
Y (Γ)id
Y (Γ)
Since π ○ γ = π so π ∗ ω = (π ○ γ)∗ ω = γ ∗ (π ∗ ω). So we have
γ ∗ (fω (τ )(dτ )k ) =
1
fω (γτ )(dτ )k = fω (τ )(dτ )k
(cτ + d)2k
so fω (γτ ) = (cτ + d)2k fω (τ ).
(2) We want to show that for any α ∈ SL2 (Z), then fω ∣α,2k is meromorphic at
∞. If α∶ X(α−1 Γα) ≃ X(Γ) then fα∗ ω = fω ∣α,2k . So it suffices to show that fω is
meromorphic at ∞. For a chart in a neighbourhood of ∞ ∈ X(Γ) we have
∼
{∞} ∪ (Γ∞ /UA ) → open disc
τ ↦ exp(2πiτ /h∞ ) = qh ,
where UA = {Im (τ ) > A}. ω maps to a meromorphic differential g(qh )(dqh )k on the
open disc and this pulls back to fω (τ )(dτ )k on UA , via the map τ ↦ exp(2πiτ /h).
We can see that
2πiqh k
fω (τ ) = (
) g(qh ),
h
hence fω is meromorphic at ∞.
(3) We’ll write down an inverse to the map ω ↦ fω . Let f ∈ Mero2k (Γ). We’ll
construct ω(f ). Suppose that Γ ⋅ τ = x ∈ Y (Γ). We have charts around τ, x such
that π has the form z ↦ z nx , and Γτ acts on z by multiplication by a nx root of
unity ζ. f = g(z) in these coordinates. Since f is a meromorphic form, this implies
that f (τ )(dτ )k is Γ-invariant so g(z)(dz)k is Γτ -invariant. So this means that
g(ζz)ζ k (dz)k = g(ζz)(dζz)k = g(z)(dz)k ,
i.e. g(ζz) = ζ −k g(z) so z k g(z) is Γτ -invariant and = h(z nx ). Now define ω(f ) on
this chart of x by
ω(f ) = (nx z)−k h(z)(dz)k .
Exercise: shows that this pulls back to g(z)(dz)k under the map z ↦ z nx .
Lecture 15
19
th
February 09:00
Last time we had a theorem
Theorem 15.45. Let ω ∈ Ω⊗k (X(Γ)) (the meromorphic differentials of degree k
on X(Γ)) and π∶ H → X(Γ) and π ∗ ω = fω (τ )(dτ )k . The map ω ↦ fω gives an
isomorphism
∼
Ω⊗k (X(Γ)) → Mero2k (Γ)
where Mero2k (Γ) is the set of meromorphic forms of weight 2k and level Γ.
The end of the proof: to show ω ↦ fω is an isomorphism, we will write down
an inverse f ↦ ω(f ). To write down ω(f ) we need to write down compatible
∼
differentials ω(f )(φ) on V , where φ∶ X ⊃ U → V ⊂ C, for charts φ in an atlas. Last
∼
time we wrote down ω(f )(φ) for φ∶ Ux → V a chart on some open neighbourhood
of x for every x ∈ Y (Γ).
35
Lecture 15
In a neighbourhood of ∞: we have a chart φ to an open disc, induced by
τ ↦ exp(2πiτ /h) = qh where h is the width of ∞. Write f (τ ) = F (qh ), then F is a
meromorphic function on the disc. Now set
ω(f )(φ∞ ) = (
2πiqh −k
) F (qh )(dqh )k
h
This differential pulls back to f (τ )(dτ )k under the amp τ ↦ exp(2πiτ /h).
Compatibility of the various ω(f )(φ): by construction, everything pulls back
to f (τ )(dτ )k on π −1 (U ) ⊂ H. Compatibility follows from the subsequence lemma
(if U1 , U2 are two differentials on U1 ∩ U2 , with pull-back to π −1 (U1 ∩ U2 ) ⊂ H the
same, so the lemma implies that they are equal).
π
Lemma 15.46. Let X → Y be a non-constant, holomorphic map of Riemann
surfaces, with Y connected, then π ∗ ∶ Ω⊗n (Y ) → Ω⊗n (X) is injective.
Proof. Exercise.
Recall that for f ∈ Mero2k (Γ), we defined ordx (f ) for x ∈ Y (Γ) with x = Γτ, τ ∈
H, then define
⎧
⎪
⎪order of the zero of f at τ f is holomorphic at τ ,
ordx (f ) = ⎨
⎪
−(order of the pole at τ )
f is non-holomorphic at τ ,
⎪
⎩
for s ∈ C(Γ) and s = Γx with x ∈ P1 (Q), then if α∞ = x, for all α ∈ SL2 (Z) then
ords (f ) = ord∞ (f ∣α,2k ) = {
smallest n such that the coefficient of qhn
}
is non-zero in the qh -expansion of f ∣α,2k .
If h is the width of ∞ for α−1 Γα then ± (
1
0
h
) ∈ α−1 Γα. 2k is even, thus
1
f ∣α,2k (τ + h) = f ∣α,2k (τ )
so we have a Fourier expansion in the variable qh
f ∣α,2k (τ ) = ∑ an qhn
n∈Z
for qh = exp(2πiτ /h).
We have that f ∈ M2k (Γ) iff ordx (f ) ≥ 0 for all x ∈ X(Γ). Also, we have
f ∈ S2k (Γ) iff
⎧
⎪ordx (f ) ≥ 0 for all x ∈ Y (Γ) and,
⎪
⎨
⎪ords (f ) ≥ 1 for all s ∈ C(Γ).
⎪
⎩
15.1. Orders of vanishing for meromorphic differentials. Let X be a
Riemann surface, and ω ∈ Ω⊗n (X).
Definition 15.31. For x ∈ X, we set vx (ω) to be the order of vanishing of f (z) at
z0 , where ω(φ) = f (z)(dz)n for a chart φ∶ U → V where U ⊃ x ↦ z0 ∈ V .
Exercise: for f ∈ Mero2k (Γ), then ω(f ) ∈ Ω⊗k (X(Γ)). Then for x ∈ Y (Γ), we
have
ordx (f ) − k(nx − 1)
,
vx (ω(f )) =
nx
where nx is the order of the stabiliser of τ in Γ/{±I} (π∶ H → X(Γ) has local form
z ↦ z nx from a neighbourhood of τ to a neighbourhood of x). And for s ∈ C(Γ)
then vs (ω(f )) = ords (f ) − k.
Corollary 15.47.
36
Lecture 16
(1) For ω ∈ Ω⊗k (X(Γ)), then fω ∈ M2k (Γ) iff
vx (ω) ≥
−k(nx − 1)
,
nx
for all x ∈ Y (Γ) and vs (ω) ≥ −k for all s ∈ C(Γ).
(2) We have fω ∈ S2k (Γ) iff
vx (ω) ≥
−k(nx − 1)
,
nx
and vs (ω) ≥ 1 − k for all s ∈ C(Γ).
Now we will use the Riemann-Roch formula to compute the dimensions of these
vector spaces.
15.2. Divisors.
Definition 15.32. Let X be a Riemann surface. Denote by Div(X) the free
Abelian group generated by points of x. We write D ∈ Div(X) as
D = ∑ ax [x],
x∈X
where ax ∈ Z and is 0 for all but finitely many x. Define deg(D) = ∑x∈X ax . We
say that D is effective, written D ≥ 0 if ax ≥ 0 for all x ∈ X.
If f is a meromorphic function on X and X is compact then define
div(f ) = ∑ vx (f )[s],
x∈X
where vx (f ) is the order of vanishing of f at x. For ω ∈ Ω⊗n (X) then
div(ω) = ∑ vx (ω)[x].
x∈X
Recall the deg(div(f )) = 0 for f a meromorphic function on a compact Riemann
surface.
Definition 15.33. Let D be a divisor on X (with X compact) then define
L(D) := {f non-zero, meromorphic function ∶ div(f ) + D ≥ 0} ∪ {0},
which is a C-vector space.
Remark. We have L(D) = OX (D)(X) where if U ⊂ X is open, then
OX (D)(U ) = {f non-zero meromorphic on U ∶ div(f ) + D∣U ≥ 0} ∪ {0},
where D∣U is ∑x∈U ax [x]. Since U is just an open set now and not necessarily
compact, the divisor may not be a finite sum, but we can still make sense of this
because div(f ) + D∣U is a divisor.
Example. Let Ωhol
X be the sheaf of holomorphic differentials on X. Let ω0 be a
non-zero meromorphic differential on X (degree 1). Then we have
Ωhol
X ≃ OX (div(ω0 ))
ω ↦ ω/ω0
37
Lecture 16
Lecture 16
21st February 09:00
Definition 16.34. Let ω0 be a non-zero differential of degree 1 on X(Γ) and denote
K = div(ω0 ) = ∑ vx (ω0 )[x].
x∈x(Γ)
For k ∈ Z≥0 define
D(k) := k ⋅ div(ω0 ) + k ∑ [s] + ∑ ⌊
s∈C(Γ)
x∈Y (Γ)
k(nx − 1)
⌋[x],
nx
Dc (k) := D(k) − ∑ [s].
s∈C(Γ)
Define
L(n) := {0} ∪ {meromorphic functions g on X(Γ) such that n + div(g) ≥ 0}.
Theorem 16.48. The map f ↦ ω(f )/ω0k (where ω0 = g(x) dz) induces isomorphisms
∼
M2k (Γ) → L(D(k)),
∼
S2k (Γ) → L(Dc (k)).
Proof. This is immediate by considering the definition of D(k), Dc (k) and
the orders of vanishing of ω(f ). The inverse map is g ↦ fg⋅ω0k ∈ Mero2k (Γ) where
g ⋅ ω0k ∈ Ω⊗k (X(Γ)).
So to compute dim M2k and dim S2k we just need to compute dim L(D(k)) and
dim L(Dc (k)).
Remark. There does exist a non-zero differential ω0 : this follows from a general
result on compact Riemann surfaces, here we have π∶ X(Γ) → X(Γ(1)) ≃ P1C so we
can just pull back a non-zero differential on P1 .
16.1. The Riemann-Roch theorem. Let X be a compact Riemann surface
(connected), and ω0 a non-zero meromorphic differential on X, K = div(ω0 ), D ∈
Div(X).
Theorem 16.49. One has
dim L(D) − dim L(K − D) = deg(D) + 1 − g(X),
where g is the genus.
The Euler characteristic for OX (D):
dim H 0 (X, OX (D)) − dim H 1 (X, OX (D)) = deg(D) + 1 − g(X).
Example. We obtain from the above deg(K) = 2(g − 1) and dim L(K) = g.
16.2. The genus of modular curves.
Theorem 16.50 (Riemann-Hurwitz). Let π∶ X → Y be a non-constant holomorphic
map between compact connected Riemann surfaces. Then we have
2 − 2g(X) = (deg π)(2 − 2g(Y )) − ∑ (ex − 1),
x∈X
where ex is the ramification index - π has the local form z ↦ z ex in a neighbourhood
of x.
We apply this result to π∶ X(Γ) → X(Γ(1)) ≃ P1 , using that X(Γ(1)) has genus
0.
38
Lecture 16
Definition 16.35. Define d = deg(π) = [PSL2 (Z)∶ Γ̄] where Γ̄ is the image of Γ in
PSL2 (Z) = SL2 (Z)/{±I}. Also define
r2 = #{x ∈ Y (Γ)∶ nx = 2}.
If nx = 2 then π(x) = Γ(1) ⋅ i - but the other implication is not true. Also define
r3 = #{x ∈ Y (Γ)∶ nx = 3},
and finally define r∞ = #CΓ = #π −1 (∞).
Theorem 16.51. One has
g(X(Γ)) = 1 +
d r2 r3 r∞
−
−
−
.
12 4
3
2
Remark. This theorem implies that
r2 2r3
d
(16.1)
2g(X) − 2 +
+
+ r∞ = > 0.
2
3
6
Proof. Let deg π = d. For x ∈ Y (Γ), we have ex = nπ(x) /nx which is = 1 if
π(x) ≠ Γ(1) ⋅ i or Γ(1) ⋅ ω, and = nπ(x) otherwise (for s ∈ CΓ , we have es = hs , the
width of the cusp). We let g = g(X(Γ)) then we have
2 − 2g = 2d −
∑
(ex − 1) − ∑ (ex − 1) − ∑ (es − 1).
π(x)=Γ(1)⋅i
π(x)=ω
s∈C(Γ)
Suppose y = Γ(1) ⋅ i or Γ(1) ⋅ ω, then
−1
∑ (ex − 1) = (ny − 1)(#π (y) − rny ),
π(x)=y
For the degree of the morphism
d = ∑ ex = ny (#π −1 (y) − rny ) + rny ,
π(x)=y
then we obtain
∑ (ex − 1) =
π(x)=y
ny − 1
(d − rny ),
ny
Also ∑s∈CΓ (es − 1) = d − r∞ .
Corollary 16.52. Let g be the genus of X(Γ), and let r2 , r3 , r∞ be as above.
(1) For k ≥ 1 we have
k
2k
dim M2k (Γ) = kr∞ + r2 ⌊ ⌋ + r3 ⌊ ⌋ + (2k − 1)(g − 1).
2
3
(2) For k ≥ 2, we have
dim S2k (Γ) = dim M2k − r∞ .
(3) We have
dim S2 (Γ) = g.
Proof. (1) We have dim M2k (Γ) = dim L(D(k)) and
deg(D(k)) = 2k(g − 1)kr∞ + r2 ⌊k/2⌋ + r3 ⌊2k/3⌋
k−1
2k − 2
r2 +
r3
2
3
= (k − 1)(d/6) + 2g − 2 + r∞ ,
≥ 2k(g − 1) + kr∞ +
where we have used (16.1) in the last part. For k ≥ 1, this quantity is > 2g − 2 =
deg(K). So we have deg(K − D(k)) < 0 so dim L(K − D(k)) = 0. Riemann-Roch
then gives
dim L(D(k)) = deg(D(k)) + 1 − g.
39
Lecture 17
Corollary 16.53. We have
dim M2k (Γ0 (4)) = k + 1.
Proof. We have [Γ(1)∶ Γ0 (4)] = 6 = d. We use the fact that we constructed
two linearly independent elements of M2 (Γ0 (4)): E2,2 , E2,4 . This implies that
dim M2 (Γ) − dim S2 (Γ) ≥ 2,
and by the dimensions formula dim S2 (Γ) = r∞ − 1, so r∞ ≥ 3. Since we have
g = 3/2 − r2 /4 − r3 /3 − r∞ /2,
this implies that g = 0 and r∞ = 3 and r2 = r3 = 0 so dim M2k (Γ0 (4)) = k + 1. We
can get formulæ for all Γ(N ), Γ1 (N ), Γ0 (N ).
Lecture 17
24th February 09:00
The idea now, is to construct a collection of commuting endomorphisms of Mk (Γ1 (N )).
The simultaneous eigenvectors for these operators will give a nice basis for Mk (Γ1 (N )).
Later we will see that for f ∈ Sk (Γ1 (N )) with f = ∑n≥1 an q n then we can define
L(f, s) = ∑n≥1 an /ns (which converges for s ∈ C with Re(s) ≫ 0). Then L(f, s)
has a functional equation and analytic continuation to C and f is a Hecke operator
iff L(f, s) = ∏p (local L-factor at p)−1 . Hecke eigenvectors give representations of
Gal(Q̄/Q) and weight-2 Hecke eigenvectors correspond to elliptic curves over Q (c.f.
the later chapters of Diamond and Shurman).
17.1. Modular forms and functions on lattices.
Definition 17.36. A lattice in C is a Z-module L ⊂ C free of rank 2 with generators
ω1 , ω2 ∈ C linearly independent over R (i.e. L ⊂ C a discrete subgroup with L ⊗Z R =
C).
Definition 17.37. For N ∈ Z≥1 , then a Γ1 (N )-level structure on a lattice L is a
point t ∈ C/L of exact order N . The notation we use is
LN = {(L, t)∶ L a lattice, t a Γ1 (N )-level structure}.
For k ∈ Z, then a function F ∶ LN → C has weight k if
F (λL, λt) = λ−k F (L, t),
for all λ ∈ C× .
It is easy to write down examples of such functions: consider
Gk (L) = ∑ `−k ,
0≠`∈L
for k > 2 even, is a weight k function on L1 . Observe that if τ ∈ H, and we write
Lτ = Zτ ⊕ Z, then
Gk (Lτ ) = Gk (τ ),
the Eisenstein series.
Definition 17.38. We define a set
M = {(ω1 , ω2 )∶ ω1 , ω2 ∈ C× , ω1 /ω2 ∈ H}.
40
Lecture 17
Observe that there is a surjective map M → LN : to a pair (ω1 , ω2 ) ∈ M associate
the lattice L(ω1 , ω2 ) := Zω1 ⊕ Zω2 and the Γ1 (N )-level structure ω2 /N + L(ω1 , ω2 ).
The set L has an action of C× on it given by (ω1 , ω2 ) ↦ (Zω1 ⊕ Zω2 , ω2 /N ) and for
λ ∈ C× we have λ(ω1 , ω2 ) = (λω1 , λω2 ). The set M has an action of GL+2 (R) on it:
a b
let γ = (
) ∈ GL+2 (R), then define
c d
γ(ω1 , ω2 ) = (aω1 + bω2 , cω1 + dω2 ).
Lemma 17.54. The map M ↠ LN identifies LN with the quotient Γ1 (N )/M. The
map M ↠ H given by (ω1 , ω2 ) ↦ ω1 /ω2 identifies H with the quotient C× /M.
Proof. Exercise.
The picture for the above is
Γ1 (N )
M
LN
C×
H
C×
Γ1 (N )
Y (Γ1 (N ))
Remark. We can identify the set M with the set of elliptic curves E/C with an
oriented basis of the first homology group H1 (E, Z) and an invariant differential
ω ≠ 0 on E up to isomorphism. We have C/L ≃ E via dz ↔ ω and L ↔ H1 (E, Z).
Similarly
H = {elliptic curves E/C with an oriented basis for H1 (E, Z)}/ ≅,
LN = {elliptic curves E/C with a point of exact order N
and an invariant differential ω0 ≠ 0 on E}/ ≅,
Y (Γ1 (N )) = {elliptic curves E/C with a point of exact order N }/ ≅ .
Definition 17.39. Let F be a function LN → C. Define a function F̃ ∶ M → C by
F̃ (ω1 , ω2 ) = F (Zω1 ⊕ Zω2 , ω2 /N ).
Define f ∶ H → C by
f (τ ) = F̃ (τ, 1) = F (Zτ ⊕ Z, 1/N ).
Lemma 17.55. The above definition identifies
(1) functions F ∶ LN → C of weight k.
(2) Γ1 (N )-invariant functions F̃ ∶ M → C of weight k - i.e.
F̃ (λω1 , λω2 ) = λ−k F̃ (ω1 , ω2 )
for γ ∈ Γ1 (N ), and F̃ (γ(ω1 , ω2 )) = F̃ (ω1 , ω2 ).
(3) (set theoretically) functions f ∶ H → C invariant under the slash operator
∣γ,k for γ ∈ Γ1 (N ).
Proof. Exercise.
Now we say that a weight k function on LN is weakly modular/a meromorphic
form/modular form/cusp form if the associated f ∶ H → C is.
Remark. We can interpret holomorphic weakly modular functions of weight k and
level Γ1 (N ) as holomorphic sections of a line bundle on Y (Γ1 (N )) (so long as there
are no elliptic points; you can check that this is okay for N ≥ 4).
41
Lecture 18
17.2. The Hecke operators.
Definition 17.40. Let F be a function LN → C, and n ∈ Z≥1 . The we define a
new function Tn F , a function LN → C by
1
Tn F (L, t) = ∑ F (L′ , t mod L′ ),
n L′
where the sum is over lattices L′ with L′ ⊃ L and #L′ /L = n, and t mod L′ has
exact order N in C/L′ (for L′ ⊃ L and π∶ C/L ↠ C/L′ an isogeny of degree n such
that π(t) has exact order N , i.e. ker π ∩ ⟨t⟩ = {0}).
Definition 17.41. For n coprime to N , define Tn,n F by
1
1
1
F ( L, t mod L),
n2
n
n
(if n is coprime to N then t mod L/n has exact order N ).
Tn,n F (L, t) =
A proposition that we will prove next time is the following.
Proposition 17.56. We have identities (as endomorphisms of {functions LN → C})
(1) m, n coprime Tm ○ Tn = Tm ,
(2) p∣N prime, r ≥ 1 then Tpr = Tp ○ ⋯ ○ Tp = (Tp )r .
´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¸¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹¶
r times
(3) p ∤ N prime, r ≥ 1 then Tpr ○ Tp = Tpr+1 + pTpr−1 ○ Tp,p ,
(4) Tn ○ Tm,m = Tm,m ○ Tn ,
(5) Tm,m ○ Tn,n = Tmn,mn .
Lecture 18
th
26
February 09:00
Proof. (4),(5) are easy to prove.
(1) Let t ∈ C/Λ have exact order N . We have
1 ′ ′′
′′
′′
∑ ∑ F (L , t mod L )
mn
where ∑′ means summation over L′ ⊃ L of index m and t mod L′ with exact order
N , and ∑′′ means summation over L′′ ⊃ L′ of index n and t mod L′′ of exact order
N . Also we have
1 ′′′
′′
′′
Tmn F (L, t) =
∑ F (L , t mod L )
mn
where ∑′′′ means summation over L′′ ⊃ L of index mn and t mod L′′ of order N .
So we just need to show that if L′′ ⊃ L with index mn, there is a unique lattice L′
with L′′ ⊃ L′ ⊃ L with [L′ ∶ L] = m. If m, n are coprime then L′′ /L has a unique
subgroup of order [L′′ ∶ L] = n. L′ ⊂ L′′ is the preimage of this subgroup.
(2) It is enough to show that Tpr−1 ○ Tp = Tpr for r ≥ 2. Tpr is given by a sum
over L′′ ⊃ L with index pr , such that (t mod L′′ has order N ) iff (t′ = N /p ⋅ t, L′′ /L
does not contain t′ ).
Claim: t′ ∉ L′′ /L implies that L′′ /L is cyclic of order pr . Proof of claim: if
′′
L /L is not cyclic, it must contain p1 L/L ∋ t′ . If L′′ /L is cyclic then there exists a
unique L′ with L ⊂ L′ ⊂ L′′ with [L′ ∶ L] = pr−1 and [L′′ ∶ L′ ] = p (3) Tpr ○ Tp , Tpr+1 , Tpr−1 ○ Tp,p are all given by 1/pr+1 times a sum over L′′ ⊃ L
with [L′′ ∶ L] = pr+1 . Write
(Tm ○ Tn )F (L, T ) =
(∗)F (L, t) =
1
′′
∑ aL′′ F (L , t)
pr+1 L′′
42
Lecture 18
Denote aL′′ in each of the three operators by a, b, c respectively. We want to show
that a = b + pc. It is clear that b = 1. We have
1
1
Tpr−1 ○ Tp,p F (L, t) = Tpr−1 ((L, t) ↦ 2 F ( L, t))
p
p
1 ⋆
= r+1 ∑ F (L′′ , t)
p
where ∑⋆ means summation over L′′ ⊃ p1 L with index pr−1 . So either c = 0 if
L′′ ⊃/ p1 L or = 1 if L′′ ⊃ p1 L.
If c = 0 then L′′ ⊃/ p1 L. Now if L′ ⊂ L′′ with [L′ ∶ L] = p. Therefore L′ = L′′ ∩ p1 L ⫋
1
L. So a = 1 and a = b + pc.
p
If c = 1, then L′′ ⊃ p1 L. Now L′ ⊂ L′′ with [L′ ∶ L] = p correspond to lines in
the 2-dimensional Fp -vector space ( p1 L)/L, so there are p + 1 possibilities for L′ .
Therefore a = p + 1 and a = b + pc.
A consequence of this is that Tp , Tp,p generate a commutative sub-C-algebra of
EndC (functions LN → C),
which contain all of the Tn and Tn,n ’s.
Remark. One has
⎧
1
⎪
⎪
⎪
⎪
⎪
⎪ 1 − Tp X
∑ Tpn X = ⎨
1
⎪
⎪
n=0
⎪
⎪
2
⎪
⎪
⎩ 1 − Tp X + pTp,p X
∞
p ∣ N,
p ∤ N.
Definition 18.42. For d ∈ Z coprime to N , define the diamond operator
⟨d⟩F (L, t) = F (L, dt),
¯
and this only depends on d ∈ (Z/N Z)× .
Lemma 18.57. If F has weight k (i.e. F (λL, λt) = λ−k F (L, t) for λ ∈ C× ) then
Tn F, Tn,n F and ⟨d⟩F have weight k.
Proof. Exercise.
Exercise: if F has weight k then Tn,n F = nk−2 ⟨n⟩F .
We saw last time that weight k functions on LN correspond to functions on H,
invariant under the slash operator ∣γ,k for γ ∈ Γ1 (N ). We need to explicate how Tn
etc. act on the set
{functions on H invariant under ∣γ,k },
in order to show they stabilise the subspace Mk (Γ1 (N )).
18.1. Matrix version of Hecke operators.
Definition 18.43. If n ∈ Z≥1 then define a set
SnN = { (
a
0
b
) ∶ ad = n, a ≥ 1, (a, n) = 1 and 0 ≤ b < d} ⊂ GL+2 (R).
d
Lemma 18.58. For (L, t) ∈ LN , write L = Zω1 + Zω2 , ω1 /ω2 ∈ H, and t = ω2 /N
a b
and for σ = (
) ∈ SnN define Lσ to be the lattices spanned by na ω1 + nb ω2 and
0 d
d
ω (Lσ = n1 σL). Then σ ↦ Lσ is a bijection
n 2
∼
lattices L′ ⊃ L with index n
SnN → {
}.
such that t mod L′ has order N
43
Lecture 19
Proof. Exercise.
Proposition 18.59. Tn , Tn,n and ⟨d⟩ preserve Mk (Γ1 (N )), Sk (Γ1 (N )) (i.e. they
preserve holomorphy and holomorphic at the cusps functions H → C).
Proof. We just need to check Tn ⟨n⟩. If n is coprime to N , let σn ∈ SL2 (Z) be
n̄−1 0
a lift of (
) ∈ SL2 (Z/N Z). Suppose that f ∈ Mk (Γ1 (N )) ↔ F ∶ LN → C. We
0
n̄
have
1
f ∣σn ,k (τ ) = (cτ + d)−k F (Z(σn τ ) + Z, )
N
= F (Z(aτ + b) + Z(cτ + d), n/N )
1
= F (Zτ + Z, n/N ) = (⟨n⟩F )(Zτ + Z, ) = (⟨n⟩f )(τ )
N
Lecture 19
28
th
February 09:00
19.1. Properties of the Hecke operators.
Proposition 19.60. The operators Tn , ⟨n⟩ preserve
Mk (Γ1 (N )), Sk (Γ1 (N )).
Proof. Last time we did ⟨n⟩: ⟨n⟩f = f ∣σn ,k . For τ ∈ H, let Lτ = Zτ ⊕ Z. For
f ∈ Mk (Γ1 (N )) we have a correspondence f ↔ F ∶ LN → C. By definition, we have
Tn f ∶ H → C, given by
1
′
Tn f =
∑ F (L , 1/N )
n L′ ⊃Lτ
index n
1
=
∑ F (Lτ,σ , 1/N )
n σ∈S N
n
a
where σ = (
0
b
), and Lτ,σ = Z aτn+b ⊕ Z nd . We have
d
(Lτ,σ , 1/N ) =
1 aτ + b
(Z
⊕ Z, a/N )
a
d
so we have
1
1
F (Lτ,σ , 1/N ) = ak F (L aτ +b , a/N )
d
n
n
1 k
aτ + b
= a (⟨a⟩f )(
)
n
d
1
= ak dk (⟨a⟩f )∣σ,k
n
= nk−1 (⟨a⟩f )∣σ,k
and for g∶ H → C, g∣σ,k (τ ) = d−k g( aτd+b ). So we have
Tn f = ∑ nk−1 (⟨a⟩f )∣σ,k
N
σ∈Sn
From here it is easy to check that f ∈ Mk (Γ1 (N )) or Sk (Γ1 (N )) thus Tn f ∈
Mk (Γ1 (N )) or Sk (Γ1 (N )).
44
Lecture 19
Remark. For d ∈ (Z/N Z)× we have ⟨d⟩ acting on Mk (Γ1 (N )), which corresponds
to a representation of (Z/N Z)× . Thus
Mk (Γ1 (N )) =
⊕
χ∶(Z/N Z)× →C×
Mk (Γ1 (N ), χ)
where
Mk (Γ1 (N ), χ) = {f ∈ Mk (Γ1 (N ))∶ ⟨d⟩f = χ(d)f for d ∈ (Z/N Z)× }
we say that f has character χ or nebentypus χ.
b
) ↦ d. Γ0 (N ) acts on
d
a b
Mk (Γ1 (N )) by f ↦ f ∣γ,k . This action is given by ⟨d⟩ where γ = (
). In
c d
particular, Mk (Γ1 (N ), triv) = Mk (Γ0 (N )).
Remark. We have Γ1 (N )/Γ0 (N ) ≃ (Z/N Z)× by (
a
c
The notation we use is: Mk (N, χ) = Mk (Γ1 (N ), χ).
Now we would like to understand how the q-expansion changes when we apply
a Hecke operator: how does the q-expansion of Tn f relate to that of f ?
Proposition 19.61. Let f ∈ Mk (M, χ) with f (τ ) = ∑ an q n and for p prime Tp f =
∑n≥0 bn q n then
bn = anp + χ(p)pk−1 an/p
where χ∶ (Z/N Z)× → C× is extended to χ∶ Z → C by setting
⎧
⎪
if (m, N ) ≠ 1,
⎪0
χ(m) = ⎨
⎪
χ(m mod N ) otherwise,
⎪
⎩
and an/p = 0 unless p∣n.
Proof. If p∣N then
SpN = { (
1
0
b
) , 0 ≤ b ≤ p − 1},
p
and if p ∤ N then
SpN = { (
1
0
b
p
) , 0 ≤ b ≤ p − 1} ∪ { (
p
0
0
) }.
1
If p∣N then
Tp (f ) =
so bn =
1
p
1 p−1 τ + b
),
∑ f(
p b=0
p
p−1
∑b=0 anp exp(2πibn) = anp . If p ∤ N then we get
Tp (f ) = ( ∑ anp q n ) + χ(p)pk−1 ⋅
f (pτ )
²
.
=∑n≥0 an/p q n
Now we know how they act in the case of n a prime, so we would like the
general case now. One way of doing this is as follows. Let’s consider the operators
on C[[q]]. Define (for m ≥ 1) an operator
Um ∶ ∑ an q n ↦ ∑ an q n/m ,
n≥0
where q
n/m
n≥0
= 0 unless m∣n, and the operator
Vm ∶ ∑ an q n ↦ ∑ an q mn .
n≥0
n≥0
45
Lecture 20
The above proposition says that Tp = Up + χ(p)pk−1 Vp . We can check that Um ○ Vm
is the identity where
Vm ○ Um ∶ ∑ an q n ↦ ∑ an q n .
n≥0
n≥0
m∣n
We have the identity
(1 − Tp X + χ(p)pk−2 X 2 ) = (1 − Up X)(1 − χ(p)pk−1 Vp X),
in EndC (C[[q]])[X].
Proposition 19.62. We have Tn = T̃n where
T̃n := ∑ χ(d)dk−1 Vd ○ Un/d ,
0<d∣n
where the equality is in EndC (C[[q]]).
Remark. We saw Tn is in the algebra generated by Tp , ⟨p⟩ for p prime dividing n,
so it makes sense to think of Tn as an operator on power series.
Proof. Let a, b be coprime and a, b ≥ 1. We know that Tab = Ta ○ Tb . We can
check that T̃ab = T̃a ○ T̃b (and it follows from Un ○ Vb = Vb ○ Ua for coprime a, b). So
it suffices to prove that Tpr = T̃pr (exercise).
The alternative route is to use a proof similar to the idea used when proving
the Euler-product formula for the zeta-function: we have a formal identity
∞
∑ TN n
n=1
−s
=∏
p∣N
1
1
,
∏
1 − Tp p−s p∤N 1 − Tp p−s + Tp,p p1−2s
= ∏(1 − χ(p)pk−1 Vp p−s )−1 (1 − Up p−s )−1 .
p
Exercise: verify these formal identities. Show that it implies the proposition and
make sense of these identities in some ring.
Corollary 19.63. f ∈ Mk (N, χ) with f (τ ) = ∑n≥0 an q n and Tm f (τ ) = ∑n≥0 bn q n
then
bn =
χ(d)dk−1 amn/d2
∑
0<d∣gcd(m,n)
Proof. This is immediate from the previous proposition.
Definition 19.44. We call f ∈ Mk (N, χ) an eigenform if Tn f = λn f for some
λn ∈ C and all n ≥ 1 and we say that f is normalised if a1 (f ) = 1.
Next time we will prove the following.
Lemma 19.64. If f is a non-constant eigenform, then a1 (f ) ≠ 0, and
λn =
an (f )
a1 (f )
and moreover if a0 (f ) ≠ 0, then
an (f )
= ∑ χ(d)dk−1
a0 (f ) 0<d∣n
for all n ≥ 1.
46
Lecture 20
Lecture 20
3rd March 09:00
Remark. Suppose that we have f ∈ Mk (Γ1 (N ), χ) with f ≠ 0. Then χ(−1) =
(−1)k (exercise). In particular, if χ is trivial, Mk (N, triv) = Mk (Γ0 (N )) ∋ −I, so
Mk (Γ0 (N )) ≠ 0 then k is even.
Lemma 20.65. Let χ be a character of (Z/N Z)× . Suppose that f ∈ Mk (N, χ) =
Mk (Γ1 (N ), χ) is a non-constant eigenform: for n ≥ 1, Tn f = λn f . Let f =
∑n≥0 an q n . Then a1 ≠ 0 and λn = an /a1 . If a0 ≠ 0, then
λn = ∑ χ(d)dk−1 .
d∣n
0<d
Proof. Look at the formula for the q-expansion of Tn f :
λn an = a1 (Tn f ) = an .
So if a1 = 0 then an = 0 for all n ≥ 1 so f is constant. Otherwise λn = an /a1 . The
final statement follows from
a0 (Tn f ) = ∑ χ(d)dk−1 a0 .
d∣n
0<d
Example. Examples of eigenforms.
(1) Ek (τ ) for k ≥ 4 even, λn = ∑ d∣n dk−1 . More generally we can write
0<d
down Eisenstein series Ekχ ∈ Mk (Γ1 (N ), χ) such that λn = ∑ d∣n χ(d)dk−1
0<d
(χ(−1) = (−1)k ) if χ(−1) = (−1)k and k ≥ 2, or k > 2 if χ = triv.
(2) ∆(τ ) ∈ S12 (Γ(1)) is Hecke eigenform (proof: dim S12 (Γ(1)) = 1).
(3) θ(τ )2 = ∑n≥0 r2 (n)q n ∈ M1 (Γ1 (4), χ) and our character χ∶ (Z/4Z)× ≃ {±1}
and χ(−1) = (−1)k , so
M1 (Γ1 (4)) = M1 (Γ1 (4), χ)
with χ the non-trivial character of (Z/4Z)× . dim M1 (Γ1 (4), χ) = 1 (see
the second examples sheet). So f = θ(τ )2 is a Hecke eigenform. a0 (f ) = 1
and a1 (f ) = 4. So the Hecke eigenvalues are
−1
λn = ∑ χ(d) = ∑ ( )
d
d∣n
d∣n
odd
0<d
So we have
an (f ) = 4 ∑ (
d∣n
odd
0<d
−1 k−1
)d
d
)), so r2 (p) = 0 if p ≡ 3
So for p an odd prime then ap (f ) = 4(1 + ( −1
p
(mod 4), and = 8 if p ≡ 1 (mod 4).
) for odd primes p. Then
Let λp = 1 + ( −1
p
Gal(Q(i)/Q)
class field theory → ≃
χ̃
C×
χ
(Z/4Z)×
and λp is the trace of Frobρ or Gal(Q(i)/Q) ↻ (triv) ⊕ χ̃.
47
Lecture 21
Theorem 20.66 (Delinge, Serre). Suppose that f ∈ M1 (N, χ) is an eigenform with
Tn f = λn f . Then there exists a number field K/Q and a representation
ρ∶ Gal(K/Q) ↪ GL2 (C),
such that for p ∤ N , p is unramified in K and for p ∤ N , ρ(Frob) has characteristic
polynomials
X 2 − λp X + χ(p).
Remark. The fact that χ(−1) = (−1)k implies that (det ρ)(complex conj.) = −1 we say that ρ is odd.
Proposition 20.67. If f ∈ Mk (N, χ), then f is a normalised eigenform (ai (f ) = 1)
if and only if
(1) a1 (f ) = 1,
(2) amn (f ) = am (f )an (f ) if m, n are coprime (Tmn = Tm ○ Tn )
(3) p prime, r ≥ 1 then
apr (f )ap (f ) = apr+1 (f ) + χ(p)pk−1 apr−1 (f )
(and the same for Tpr ○ Tp . . . ).
Proof. If f is a normalised eigenform, then (2),(3) follows from the corresponding relations between Hecke operators. Let us assume that (1),(2),(3) hold.
It is enough to show that f is an eigenvector for each p prime (the Tp ’s generate
the Tn ’s). If p ∤ n then
(2)
an (Tp f ) = anp (f ) = ap (f )an (f ).
If p ∣ n then n = pr m with p, m coprime, and you can check that
an (Tp f ) = anp (f ) + χ(p)pk−1 an/p (f ),
which follows from the formula for Tp on q-expansions. Then factoring out using
(2), we get
= am (f )(apr+1 (f ) + χ(p)pk−1 apr−1 (f )),
and then using (3)
= am (f )(apr (f )ap (f )) = ap (f )an (f ),
so an (Tp f ) = ap (f )an (f ) for all n ≥ 1 and all primes p . . . this does not work for the
case n = 0? So let’s assume that a0 (f ) = 0, then Tp (f ) = ap (f )f (will fix later). We want to show that Mk (Γ(1)) has a basis of Hecke eigenforms. We’ll define
an inner product on Sk (N, χ) with respect to which Tn is essentially self-adjoint
(at least for n coprime to N ).
20.1. The Petersson inner product. Define a measure on H: for τ = x + iy
then
dx dy
dµ(τ ) =
y2
This is SL2 (Z)-invariant. For f, g ∈ Mk (Γ), define
⟨f, g⟩ = (⋆) ∫ f (τ )g(τ )(Im τ )k dµ(τ )
F
which is integral over Y (Γ) or X(Γ).
Lecture 21
5
th
March 09:00
48
Lecture 21
21.1. The inner-product and the eigenforms. The proposition from monday: f is a normalised Hecke eigenform iff (multiplicative relations between an (f )).
For the end of the proof: showed that an (Tp f ) = ap (f )an (f ) for all n ≥ 1, so
Tp f − ap (f )f ∈ Mk (N, χ) is constant hence it is equal to zero (for k > 0).
Recall that we had an SL2 (Z)-invariant measure dx dy/y 2 = dµ(τ ) on H (GL+2 (R)invariant). So we can integrate continuous functions on Y (Γ) or X(Γ) with respect
to this measure.
Definition 21.45. Let f be a continuous function on Y (Γ), then
∫
Y (Γ)
f dµ = ∫
F(Γ)
f (Γτ ) dµ(τ )
with F(Γ) a fundamental domain for Γ, for simplicity we restrict to
F(Γ) =
αF(1)
∐
α∈Γ̄/PSL2 (Z)
where F(1) is the standard fundamental domain for SL2 (Z). Remark: this is welldefined i.e. it is independent of choice of F(Γ).
Lemma 21.68. One has
µ(Γ) := ∫
and
∞.
Y (Γ)
1 dµ < ∞
µ(Γ)
= [PSL2 (Z) ∶ Γ̄]
µ(Γ(1))
Proof. Exercise: first check the second part then show explicitly that µ(Γ(1)) <
Definition 21.46. For f, g ∈ Mk (Γ), with one of f, g a cuspform, define
⟨f, g⟩Γ :=
µ(Γ(1))
f (τ )g(τ )(Im τ )k dµ
∫
µ(Γ)
Y (Γ)
(and the integrand here is Γ-invariant).
We’ll show that ⟨f, g⟩Γ converges absolutely, which needs the assumption that
one of f, g is a cuspform.
Lemma 21.69. For f ∈ Sk (Γ), then ∣f (x + iy)∣ ≤ C/y k/2 for x, y ∈ R where C does
not depend on x, y.
Proof. Exercise: define φ(x + iy) = ∣f (x + iy)∣y k/2 . Check that φ descends to
a continuous function on Y (Γ). Then check this extends to a continuous function
on Y (Γ). X(Γ) compact implies that φ is bounded.
Corollary 21.70. For f ∈ Sk (Γ), with f (τ ) = ∑n≥1 an qhn . Then ∣an ∣ ≤ Cnk/2 , called
the ‘trivial bound’.
Proof. Exercise.
Remark. There is a theorem of Deligne: if f ∈ Sk (N, χ) then for n coprime to N, k
k−1
then for all ε > 0 one has ∣an ∣ ≤ Cε n 2 +ε (which follows from the Weil conjectures).
Corollary 21.71. ⟨f, g⟩Γ converges absolutely (if one of f, g ∈ Sk (Γ)).
Proof. We have f g ∈ S2k (Γ) and ∣f (τ )g(τ )∣ = ∣f g(τ )∣ ≤ C/(Im τ )k .
′
′
Exercise: if Γ ⊂ Γ, f, g ∈ Mk (Γ) then we can also think of f, g ∈ Mk (Γ ). Show
that ⟨f, g⟩Γ = ⟨f, g⟩Γ′ (this is the reason for µ(Γ(1))/µ(Γ) factor in the definition).
So we can just write ⟨f, g⟩ for ⟨f, g⟩Γ .
49
Lecture 21
Proposition 21.72. Let f, g ∈ Mk (N, χ) with one of f, g a cusp form and suppose
that n and N are coprime. Then
⟨Tn f, g⟩ = χ(n)⟨f, Tn g⟩
First we need two lemmas
Lemma 21.73. If α ∈ GL+2 (Q), Γ ⊂ SL2 (Z) a congruence subgroup, then SL2 (Z) ∩
αΓα−1 is a congruence subgroup.
Proof. Exercise.
Remark. Let α ∈ GL+2 (Q) and f ∈ Mk (Γ), define
f ∣α,k (τ ) =
a
where α = (
c
Then
aτ + b
1
f(
)
k
(cτ + d)
cτ + d
b
) (warning this is not always the convention used in the literature).
d
f ∣α,k ∈ Mk (α−1 Γα ∩ SL2 (Z))
Lemma 21.74. For α ∈ GL+2 (Q) and f, g ∈ Mk (Γ) then
⟨f, g⟩ = (det α)k ⟨f ∣α,k , g∣α,k ⟩
Proof. Let Γ′ = Γ ∩ αΓα−1 and f, g ∈ Mk (Γ′ ) and f ∣α,k , g∣α,k ∈ Mk (αΓα−1 ). If
F(Γ ) is a fundamental domain for Γ′ , then α−1 F(Γ′ ) is a fundamental domain for
α−1 Γ′ α. Compare the right-hand side as
′
(det α)k
1
[PSL2 (Z) ∶ α−1 Γ′ α]
∫
α−1 F(Γ′ )
f ∣α,k (τ )g∣α,k (τ )(Im τ )k dµ(τ )
(note that [PSL2 (Z) ∶ α−1 Γ′ α] = [PSL2 (Z) ∶ Γ′ ]). Substitute τ by α−1 τ then the
integral becomes
(Im τ )k
dµ(τ )
∫ ′ f (τ )g(τ )
(det α)k
F(Γ )
so we get that the right-hand side is ⟨f, g⟩.
a
Proof:(of prop. 21.72). Recall the set of matrices SnN ∋ σ = (
0
k−1
Tn f = n ∑σ∈SnN χ(a)f ∣σ,k . So we have
b
) and
d
⟨Tn f, g⟩ = nk−1 ∑ χ(a)⟨f ∣σ,k , g⟩
N
σ∈Sn
=
1
∑
n σ∈S N
n
χ(a)
±
⟨f, g∣σ−1 ,k ⟩
root of unity
χ(n)
∑⟨f, χ(d)g∣σ−1 ,k ⟩
n
= χ(n)⟨f, nk−1 ∑ χ(d)g∣A,k ⟩
=
N
σ∈Sn
´¹¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¸ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¹ ¶
Tn g
(where for ⟨f ∣σ,k , g⟩ you choose some subgroup that makes them of the same level,
d −b
you could choose the one with the conjugate of σ, and A = (
).)
0 a
50
Lecture 22
If we restrict the pairing ⟨, ⟩ to Sk (N, χ), we get a Hermitian inner product.
Let’s fix square roots cn of χ(n)−1 for (n, N ) = 1. Then the proposition implies
that ⟨cn Tn f, g⟩ = ⟨f, cn Tn g⟩. So Sk (N, χ) has an orthonormal basis consisting of
eigenvectors for the Tn with (n, N ) = 1.
Lecture 22
7th March 09:00
Last time we had that Sk (N, χ) has a basis consisting of eigenvectors for Tn
with (n, N ) = 1. Recall that Sk (Γ1 (N ))old is the span of f (τ ), f (pτ ) for p∣N prime
and f ∈ Sk (Γ1 (N /p)). Define Sk (Γ1 (N ))new to be the orthogonal complement of
Sk (N )old under the Petersson inner product.
Definition 22.47. Define Sk (N, χ)new to be Sk (Γ1 (N ))new ∩ Sk (N, χ).
Theorem 22.75. Sk (N, χ)new is stable under Tn for all n ≥ 1. If f ∈ Sk (N, χ)new
is an eigenvector for Tn with (n, N ) = 1 then f is an eigenform (i.e. eigenvector
for Tn for all n).
Proof. See Diamond and Shurman, chapter 4, or Miyake’s book “Modular
forms” - see ‘primitive forms’.
In fact if there exists an integer L such that f ∈ Sk (N, χ)new is an eigenvector
for all Tn with (n, L) = 1, then f is an eigenform.
If f ∈ Sk (N, χ)new is an eigenform, it is sometimes referred to as a “newform”.
The space Sk (⋯)new is also referred to as the space of primitive forms.
This theorem has a more natural interpretation in the theory of automorphic
representations - called ‘strong multiplicity one’.
22.1. L-functions. Let f ∈ Mk (Γ) and write f = ∑n≥0 an qhn where qh =
exp(2πiτ /h).
Definition 22.48. Define the L-function
∞
an
s
n=1 n
L(f, s) = ∑
for s ∈ C.
Recall that for f ∈ Sk (Γ), then ∣an ∣ ≤ Cnk/2 . This implies that L(f, s) is
absolutely and uniformly convergent on compact subsets of {s ∈ C∶ Re(s) > k2 + 1}.
Remark. We will probably only look at the case of cusp forms, but for f ∈ Mk (Γ)
then ∣an ∣ ≤ Cnk−1 (k ≥ 2). To see this write Mk (Γ) = Sk (Γ) ⊕ Ek (Γ) where Ek (Γ) is
the space with elements given by explicit Eisenstein series (as in exercise sheet 2).
So in general L(f, s) converges for Re(s) > k.
22.2. The functional equation. This will be the analogue of the equation
relating ζ(s) and ζ(1 − s). We restrict to Γ = Γ1 (N ).
Remark. If Γ is a congruence subgroup, then there exists N and α ∈ GL+2 (Q) such
that αΓα−1 ⊃ Γ1 (N ), so if f ∈ Mk (Γ) then f ∣α−1 ,k ∈ Mk (Γ1 (N )).
Define a matrix
ωN
⎛ 0
:= √
⎝ N
√
−1/ N ⎞
0 ⎠
−1
then you can check that ωN
Γ1 (N )ωN = Γ1 (N ), so if f ∈ Mk (Γ1 (N )) then f ∣ωN ,k ∈
Mk (Γ1 (N )). We have
f ∣ωN ,k (τ ) = N −k/2 τ −k f (−1/N τ )
51
Lecture 23
so if N = 1 then f ∣ωN ,k = f .
Theorem 22.76. Let f ∈ Sk (Γ1 (N )), set g = ik f ∣ωN ,k . Then L(f, s) extends to a
holomorphic function on s ∈ C. Set
Λ(f, s) := N s/2 (2π)−s Γ(s)L(f, s)
Then Λ(f, s) = Λ(g, k − s), the functional equation.
∞
Remark. Recall that Γ(s) = ∫0 ts−1 /et dt, defined for Re(s) > 0, has meromorphic
continuation to s ∈ C with poles at 0, −1, −2, . . . . Γ(s) has no zeroes for s ∈ C.
Proof. Set φ(y) = f (iy) for y ∈ R>0 . The Mellin transform is
F (s) := ∫
∞
φ(y)y s−1 dy
0
Recall that ∣φ(y)∣ ≤ Cy −k/2 so this integral converges at the lower limit for Re(s) >
k
+ 1. We have φ(y) ∼ e−y as y → ∞, so the integral converges at the upper limit
2
for all s ∈ C. We have
φ(1/N u) = N k/2 uk g(iu)
Let f = ∑ an q n , and φ(y) = ∑n≥1 an exp(−2πny), so that
F (s) = ∑ an ∫
n≥1
for Re(s) >
k
2
∞
0
exp(−2πny)y s−1 dy = ∑ an (2πn)−s Γ(s)
n≥1
+ 1. Then
F (s) = N −s/2 Λ(f, s)
Choose A > 0:
A
F (s) = ∫
0
φ(y)y s−1 dy + ∫
∞
φ(y)y s−1 dy
A
and using the substitution u = 1/N y in the first integral, it becomes
∞
∫
g(iu)N k/2−s uk−1−s du
1/N A
which converges for all s, so we get a holomorphic function extending F (s) to s ∈ C.
Thus Λ(f, s) extends to s ∈ C. Since Γ(s) has no zeroes, it follows that L(f, s)
extends to s ∈ C. One can check that N −s/2 Λ(f, s) = N k/2−s N −(k−s)/2 Λ(g, k − s)
(note that N −(k−s)/2 Λ(g, k − s) is just the Mellin transform of g(iu)). Then we can
deduce the functional equation.
Remark. If N = 1 then Λ(f, s) = ik Λ(f, k − s). Next time we will see Hecke’s
converse theorem, which says that this functional equation completely characterises
cusp forms of level 1.
Lecture 23
10th March 09:00
Last time we had for f ∈ Sk (Γ1 (N )) and φ(y) = f (iy) and
F (s) = ∫
∞
φ(y)y s−1 dy,
0
then as y → ∞ then φ(y) = O(e−2πy ) and as y → 0 then φ(y) = O(y −k/2 ) so
the integral converges for Re(s) > 1 + k2 . In fact f (−1/τ ) = τ k f ∣A,k (τ ) where
0 −1
A=(
) and then φ(1/y) = O(y k e−2πiy ) as y → ∞, so the integral converges
1 0
for all s (note that f ∣A,k (τ ) is a cusp form). Then we substitute u = 1/N y to get
the functional equation.
52
Lecture 23
23.1. Hecke eigenforms & L-functions.
Theorem 23.77. Let f ∈ Sk (N, χ) with f = ∑n≥1 an q n , then f is a normalised
eigenform if and only if
L(f, s) =
−s
k−1−2s −1
)
∏ (1 − ap p + χ(p)p
p prime
where χ(p) = 0 if p∣N and = χ(p (mod N )) if p ∤ N .
Lemma 23.78. Suppose we have two sequences {an ∈ C}n≥1 and {bn ∈ C}n≥1 such
that
bn
an
G(s) = ∑ s ,
F (x) = ∑ s ,
n
n
n≥1
n≥1
both converge absolutely and F (s) = G(s) for Re(s) > σ, and uniformly on compact
subsets of {Re(s) > σ}.. Then an = bn for all n ≥ 1.
Proof. See example sheet 3.
Proof:(of thm. 23.77). Suppose that
L(f, s) =
−s
k−1−2s −1
)
∏ (1 − ap p + χ(p)p
p prime
Then the lemma and formal computations shows that the an satisfy a1 = 1, amn =
am an for (m, n) = 1 and apr ap = apr+1 + pk−1 χ(p)apr−1 , which implies that f is a
normalised Hecke eigenform. Conversely, if f is a normalised Hecke eigenform then
an
−s
k−1−2s −1
) = ∑ s = L(f, s)
∏ (1 − ap p + χ(p)p
n≥1 n
p prime
Remark. For k = 2, compare the L-function with the Hasse-Weil L-function of an
elliptic curve.
Theorem 23.79 (Shimura). Let f ∈ S2 (Γ0 (N )), Hecke eigenform (normalised),
with all an ∈ Q, then there is an elliptic curve Ef /Q with L(Ef , s) = L(f, s).
The f gets associated to a differential on X(Γ0 (N )), so we can find Ef as a
quotient of Jac(X0 (N )).
You will prove on example sheet 3 that for f ∈ Sk (Γ0 (N )) a normalised Hecke
eigenform, then there exists a totally real number field F /Q with an ∈ F for all
n ≥ 1.
Shimura also proved that for f ∈ S2 (Γ0 (N )), a Hecke eigenform (normalised),
with all an ∈ Q, then there exists an abelian variety Af /Q, with F ↪ End(AC )⊗Z Q,
and L(Af , s) = L(f, s) × . . . .
A much more difficult thing to prove is the following.
Theorem 23.80 (Wiles, Taylor,. . . ). If E/Q an elliptic curve, then there exists
f ∈ S2 (Γ0 (N )) such that L(f, s) = L(E, s)
A deep result is that if you have an equality of L-functions, then E is isogenous
to Ef (Falting’s isogeny theorem).
23.2. Converse theorems of Hecke and Weil. Hecke proved for the case
N = 1 and Weil proved the general case, we will only sketch Weil’s results though.
A reference for this section: for Hecke results, see Deitmar’s “Automorphic forms”
and for the otther part, just see Miyake’s book.
Proposition 23.81 (Mellin transform). Suppose that g ∈ R>0 → C is twice differentiable, and suppose that there is c ∈ R>0 such that xc−1 g(x), xc g ′ (x) and xc+1 g ′′ (x)
are absolutely integrable on R>0 . Then
G(s) = ∫
∞
xs−1 g(x) dx
0
53
Lecture 24
exists for Re(s) = c and
G(c + it) = O(
1
).
(1 + ∣t∣)2
Moreover,
c+i∞
1
x−s G(s) ds = g(x).
∫
2πi c−i∞
Proof:(sketch). Substitute x = et , and we get
G(s) = ∫ est g(et ) dt.
R
Set s = c + 2πiy, and we get
ect g(et )
´¹¹ ¹ ¹ ¹ ¹ ¹ ¸¹ ¹ ¹ ¹ ¹ ¹ ¶
G(s) = ∫
R
exp(−2πiyt) dt.
Fourier transform
Everything then follows from Fourier inversion for ect g(et ).
Theorem 23.82 (Hecke’s converse theorem). Let {an ∈ C}n≥1 be a sequence satisfys
−s
ing ∣an ∣ ≤ Cnσ for some σ ∈ R>0 . Set Z(s) = ∑∞
n=1 an /n and Λ(s) := (2π) Γ(s)Z(s).
Suppose that Λ(s) extends to a holomorphic function in s ∈ C, with Λ(s) = ik Λ(k−s)
for some integer k. Then f (τ ) = ∑n≥1 an q n is in Sk (Γ(1)).
Proof. Let f = ∑ an q n , gives a holomorphic function on H with f (τ + 1) =
f (τ ). So it suffices to show that f (−1/τ ) = τ k f (τ ) (because
SL2 (Z) = ⟨ (
1
0
1
0
),(
1
1
−1
) ⟩).
0
It suffices to show that f (i/y) = (iy)k f (iy) for y ∈ R>0 (since if they are equal along
the imaginary axis then they must be equal everywhere). Set φ(y) = f (iy), and
check that
Λ(s) = F (s) := ∫ ∫
∞
y s−1 φ(y) dy
0
for Re(s) ≫ 0. Mellin inversion:
φ(y) =
c+i∞
1
Λ(s)y −s ds
∫
2πi c−i∞
we can show that
=
k−c+i∞
1
Λ(s)y −s ds.
∫
2πi k−c−i∞
substitute s = k − s and use fundamental equation to get φ(y) = ik y −k φ(1/y), since
if we do τ ↦ −1/τ twice we get the identity so k is even if f is non-zero, which gives
the result.
Lecture 24
12th March 09:00
24.1. §10: Weil’s converse theorem (1967). This will be like the converse
theorem from last time but for level Γ0 (N ) or Γ1 (N ).
54
Lecture 24
24.2. Twists of modular forms. Let m ∈ Z≥1 and χ∶ (Z/mZ)× → C× primχ′
itive (i.e. not induced by (Z/mZ)× → (Z/dZ)× → C× where d∣m and d ≠ m). Let
f (τ ) = ∑ an q n ∈ Sk (Γ0 (N )), with m, N coprime.
Definition 24.49. Define
fχ (τ ) := ∑ an χ(n)q n .
n≥1
Proposition 24.83. With notation as above, we have fχ ∈ Sk (Γ1 (N m2 ), χ2 ) where
χ2
χ2 ↝ (Z/N m2 Z)× → (Z/mZ)× → C× .
Proof. Let
m
g(χ) = ∑ χ(n) exp(−2πin/m)
n=1
and extend χ to Z in the usual way. A property of Gauss sums is
χ(n) =
1
g(χ) ∑ χ(a) exp(2πian/m).
m
a
So we have
m
fχ = m−1 g(χ) ∑ χ(a)f (τ + a/m)
a=1
m
= m−1 g(χ) ∑ χ̄(a)f ∣A (τ )
a=1
where A = (
1
0
a/m
) and you can check carefully that this implies the proposition.
1
Suppose that f ∈ Sk (Γ0 (N )) satisfies
Λ(f, s) = εΛ(f, k − s)
with ε = ±1 (i.e. f ∣B,k
⎛ 0
= εi f where B = √
⎝ N
−k
√
−1/ N ⎞
).
0 ⎠
Lemma 24.84. Assume that Λ(f, s) = εΛ(f, k − s) and χ∶ (Z/mZ)× → C× . Then
fχ ∣C = εi−k
where C =
⎛ 0
√
⎝m N
g(χ)
χ(−N )fχ
g(χ)
√
−1/m N ⎞
, so
0
⎠
Λ(fχ , s) = εχ Λ(fχ , k − s).
Definition 24.50. Let M be a set of primes. Then we say that M is big if M ∩
{a + nb∶ n ∈ Z} ≠ ∅ for all coprime (a, b).
Example. By Dirichlet’s theorem on primes in an arithmetic progression, if N ∈
Z≥1 then the set {odd primes, p ∤ N } is big.
Theorem 24.85 (Weil). Let {an ∶ n ≥ 1}, with ∣an ∣ = O(nσ ) for some σ ∈ R>0 . Set
f = ∑n≥1 an q n , Z(s) = ∑n≥1 an /ns . Fix N ∈ Z≥0 , and set
Λ(s) = (N )s/2 Γ(s)(2π)−s Z(s).
Suppose that
55
Lecture 24
(1) Λ(s) has holomorphic continuation to s ∈ C and bounded on {Re(s) ∈
[a, b]} for any a, b ∈ R (we say that it is “entire and bounded on vertical
strips”).
(2) there exists a big set M, such that for any primitive char χ of (Z/mZ)× →
C× , with m ∈ M, fχ satisfies part (1), with Λ(fχ , s) = εΛ(fχ̄ , k − s) where
εχ = ε
g(χ)
χ(−N ).
g(χ̄)
Then f ∈ Mk (Γ0 (N )). Moreover, if Z(s) converges absolutely for Re(s) > k − δ, for
some δ > 0, then f ∈ Sk (Γ0 (N )).
Remark. Mk (Γ0 (N )) = Sk (Γ0 (N )) ⊕ Ek (Γ0 (N )), and in the case Ek (Γ0 (N )),
the explicit Eisenstein series expansions have coefficients like σk−1 (n) so ∑ anns will
not converge absolutely for Re(s) < k. I won’t prove this, it has a somewhat
elementary proof but it is not very illuminating. But I will give you an idea of
the proof (see Miyake §4.3 for the details. That Λ(fχ , s) = εχ Λ(fχ̄ , k − s) implies
√
⎛ 0
N⎞
−1/m
−k
√
that fχ ∣A,k = εk i fχ̄ where A =
. Assume this for all χ of
0
⎝m N
⎠
a v
conductors a, b: then for γ = (
) with f ∣γ,k = f . If this holds for enough a, b
uN b
then we get f ∣γ,k = f for all γ ∈ Γ0 (N ).
Now we will see some application of this.
24.3. CM forms. Let K/Q be imaginary and quadratic with discriminant
−d, OK denote the ring of integers. We have algebraic Hecke characters of K:
fractional ideals of K prime to c,
ψ∶ {
} → C×
where c is some ideal in OK
Then ψ((α)) = αt for some α ∈ K × , with α ≡ 1 (mod c) (vP (α − 1) ≥ vP (c) for all
prime P ∣ c). Assume that c is minimal for ψ, i.e. it is maximal as an ideal, but
minimal under the divisibility relation (ψ primitive). Let ωψ ∶ Z → C be given by
a ↦ ψ((a))/at (for a coprime to c, and 0 otherwise). Let ψ be as above and define
fψ =
ψ(a)q N m(a)
∑
a⊂OK ideals
prime to c
Theorem 24.86 (Hecke). We have fψ ∈ St+1 (dN m(c), ωψ ⋅ χK ). χK is the quadratic character of (Z/dZ)× corresponding to K ⊂ Q(µd ) (in fact fψ is a newform
in practice, a Hecke eigenform).
Proof. Use θ-series, or use Weil’s converse theorem
L(fψ , s) = L(ψ, s)
and L(ψ, s) has property (1) in Weil’s theorem.
L(fψ , χ, s) = L(ψ ⋅ (χ ○ N mK/Q ), s)
so fψ is module. If we have a CM elliptic curve over Q and OK ↪ End(EC ), then
there exists ψ with L(E, s) = L(ψ, s) ↝ fψ .
56
Bibliography
[BvdGHZ08] Jan Hendrik Bruinier, Gerard van der Geer, Günter Harder, and Don Zagier. The
1-2-3 of modular forms. Universitext. Springer-Verlag, Berlin, 2008. Lectures from
the Summer School on Modular Forms and their Applications held in Nordfjordeid,
June 2004, Edited by Kristian Ranestad.
[DS05]
Fred Diamond and Jerry Shurman. A first course in modular forms, volume 228 of
Graduate Texts in Mathematics. Springer-Verlag, New York, 2005.
[Ser73]
J.-P. Serre. A course in arithmetic. Springer-Verlag, New York, 1973. Translated
from the French, Graduate Texts in Mathematics, No. 7.
57