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Ramanujan type congruences for the Klingen-Eisensteinseries
A Project
Presented to
the Graduate School of
Clemson University
In Partial Fulfillment
of the Requirements for the Degree
Master’s of Science
Mathematics
by
Hugh Roberts Geller
December 2016
Accepted by:
Dr. Jim Brown, Committee Chair
Dr. Kevin James
Dr. Hui Xue
Table of Contents
Title Page . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . i
1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
2 Siegel Modular Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42.1 Siegel Modular Group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42.2 Modular Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.3 Klingen Eisenstein Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 152.4 R-modules of modular forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 252.5 The Hecke Algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
3 Results of T. Kikuta and S. Takemori . . . . . . . . . . . . . . . . . . . . . . . . . 523.1 Definitions, Remarks, and Main Results . . . . . . . . . . . . . . . . . . . . . . . . . 523.2 Lemmas and Their Proofs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 533.3 Proof of the main theorem and its corollary . . . . . . . . . . . . . . . . . . . . . . . 603.4 Numerical Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 623.5 New Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67
4 Future Work . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72
Bibliography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74
ii
Chapter 1
Introduction
The following work is a Master’s project meant as introduction to the rich and complex
topic that is Siegel modular forms. In this first chapter we introduce the motivation of the paper
by T. Kikuta and S. Takemori cited as [9] as well as introducing some notation that will be used
consistently throughout this write-up.
In Chapter 2, we introduce the concept of a Siegel modular form. We introduce several
concepts and relate them to the degree one case, which is more commonly referred to as elliptic
modular forms. The reasons for doing this are two-fold. First, we wish to demonstrate how Siegel
modular forms act a generalization of the well-known elliptic modular forms. Second, we do this in
order to understand the work of [9].
In Chapter 3 we will directly address the findings of [9]. We fill in the details of their proofs
in order to fully understand their stated main theorem and its corollary. In short, the paper studies
congruences of the Fourier coefficients of Siegel modular forms. In particular, they define what it
means for a modular form to be a (mod pm) cusp form. We define this precisely in Definition 3.1.1
but for now it suffices to say that a degree n Siegel modular form is a (mod pm) cusp form if every
Fourier coefficient of rank 0 ≤ r ≤ n− 1 vanishes modulo pm.
The main theorem states that given a number field K for nearly every prime ideal p in
the ring of algebraic integers, (mod pm) cusp forms are congruent to true cusp forms of the same
weight. The precise theorem is as follows
Main Theorem ([9]) There exists a finite set Sn(K) of prime ideals in K depending on n such
1
that the following holds: For a prime ideal p of O not contained in Sn(K) and a (mod pm) cusp
form F ∈Mk(Γn)Opwith k > 2n, there exists G ∈ Sk(Γn)Op
such that F ≡ G (mod pm).
The corollary gives conditions under which we can take what is known as a Hecke eigen
cusp form and force a constant multiple of its attached Klingen-Eisenstein series to be congruent to
a true cusp form modulo pm. Precisely, it states
Main Corollary ([9]) Let k > 2n be even and f ∈ Sk(Γr)Kf, n > r, a Hecke eigenform. For the
Klingen-Eisenstein series [f ]n−r attached to f , we choose a prime ideal p in OKfwith p 6∈ Sn(Kf )
such that υ(n)p ([f ]n−r) = υp(Φ[f ]n−r) −m, m ∈ Z≥1. Then there exists F ∈ Sk(Γn)Op
such that
α[f ]n−r ≡ F (mod pm) for some 0 6= α ∈ pm.
Kikuta and Takemori assert that this is a generalization of Ramanujan’s congruence
σ11(n) ≡ τ(n) (mod 691).
They make this assertion since σ11(n) denotes the nth Fourier coefficient of the Eisenstein series
of weight 12, which we denote in Section 3.4 as ψ(1)12 , and τ(n) is the nth Fourier coefficient of
Ramanujan’s ∆ function [9]. More precisely, they are assert that they are generlizing the result that
691ψ(1)12 is a (mod 691) cusp form of degree 1 that is congruent modulo 691 to the Ramanujan’s ∆
function, the unique full level, degree 1, weight 12 cusp form.
1.1 Notation
We fix the following notation.
Let R be a commutative ring with identity.
Let Matn(R) be the set of n× n matrices with entries in R.
We denote the tranpose of a matrix γ ∈ Matn(R) by tγ.
For matrices A,B ∈ Matn(R) ⊂ Matn(C) we define A[B] = tBAB.
For V ∈ GLn(R) we define U(V ) =
V 0n
0ntV −1
.
For tS = S ∈ Matn(R) we define T (s) =
In S
0n In
.
2
Let K be a number field with ring of integers OK .
For a Hecke Eigenform f , let Kf be the number field generated by adjoining the Hecke
eigenvalues of f .
3
Chapter 2
Siegel Modular Forms
2.1 Siegel Modular Group
For n ∈ Z≥1 define J2n =
0n In
−In 0n
where 0n is the n×n zero matrix and In is the n×n
identity matrix. Note, we will write J for J2n when the dimension is clear from the context. Using
this matrix, we define the general symplectic group GSp2n(R) by
GSp2n(R) = {γ ∈ Mat2n(R) : tγJγ = µ(γ)J, µ(γ) ∈ R×}.
We call µ(γ) the similitude factor of γ. Therefore, writing γ =
A B
C D
, where A,B,C,D ∈
Matn(R), yields the following relations; tAC = tCA, tDB = tBD, tAD − tCB = µ(γ)In,
tDA− tBC = µ(γ)In, and γ−1 = µ(γ)−1
tD −tB
−tC tA
.
We will work with various subgroups of GSp2n(R). For R ⊆ R are the interested in the
subgroup matrices with positive similitude factor, denoted GSp+2n(R). We are particularly interested
in the subgroup of symplectic, 2n× 2n matrices with integral entries. We refer to this group as the
Siegel modular group (of degree n) and write
Γn = Sp2n(Z) = ker{µ : GSp2n(Z)→ R×}.
4
Note that µ(γ) = 1 if and only if tγJγ = J.
We will be working over the Siegel upper half-space of degree n and which is given by
Hn = {Z ∈ Matn(C) : tZ = Z, Z = X + iY, X, Y ∈ Matn(R), Y > 0}
where Y > 0 means Y is positive definite. For the degree 1 case, we see that this matches the
definition of the upper half plane. We define an action of GSp+2n(R) on Hn by
Z 7→ γ · Z = (AZ +B)(CZ +D)−1
where γ =
A B
C D
∈ GSp+2n(R). For this to be a well defined group action we must check the
following:
1. CZ +D is invertible;
2. γ · Z ∈ Hn;
3. I2n · Z = Z;
4. (γγ′) · Z = γ · (γ′ · Z) for all γ, γ′ ∈ GSp+2n(R) and all Z ∈ Hn.
To check the first condition we use the following identity
t(CZ +D)(AZ +B)− t
(AZ +B)(CZ +D) = (tDA− tBC)Z +t Z(tCB − tAD)
= µ(γ)InZ − µ(γ)InZ
= µ(γ)(Z − Z)
= 2iµ(γ)Y.
We now recall that a n × n matrix is invertible if and only if it gives a bijective map on an n-
dimensional vector space to itself. Thus, if CZ + D is not invertible then there exists a non-zero
5
vector v ∈ Cn such that (CZ +D)v = 0. This would then yield
tvY v =−i2µ(γ)−1t(v)
(t(CZ +D)(Az +B)− t
(AZ +B)(CZ +D))v
=−i2µ(γ)−1
(t((CZ +D)v)(AZ +B)v − t
(v)t(AZ +B)(CZ +D)v
)= 0,
but this contradicts the fact that Y is assumed to be positive definite. Therefore no such v ∈ Cn
can exist and we conclude CZ +D is invertible.
We now must show that γ · Z ∈ Hn. We start by checking for symmetry
t(CZ +D)(γ · Z − t
(γ · Z))(CZ +D) =t(CZ +D)(AZ +B)− t
(AZ +B)(CZ +D)
= Z − tZ
= 0n.
Since CZ +D is invertible this implies γ · Z is symmetric. Similarly, we have
t(CZ +D)Im(γ · Z)(CZ +D) =
−i2t(CZ +D)
(γ · Z − t
(γ · Z))
(CZ +D)
=−i2
(t(CZ +D)(AZ +B)− t
(AZ +B)(CZ +D))
= µ(γ)Y
> 0n.
The final inequality holds since Y is positive definite and γ ∈ GSp+2n(R) means µ(γ) > 0. We thus
conclude that γ · Z ∈ Hn.
The third condition is clear since
I2n · Z = (InZ + 0n)(0nZ + In)−1 = Z.
6
For the last condition let γ =
A B
C D
and γ′ =
A′ B′
C ′ D′
be in GSp+2n(R). We then get
γ · (γ′ · Z) = γ ·((A′Z +B′)(C ′Z +D′)−1
)=
(A(A′Z +B′)(C ′Z +D′)−1 +B
) (C(A′Z +B′)(C ′Z +D′)−1 +D
)−1
= (A(A′Z +B′) +B(C ′Z +D′)) (C(A′Z +B′) +D(C ′Z +D′))−1
= ((AA′ +BC ′)Z + (AB′ +BD′)) ((CA′ +DC ′)Z + (CB′ +DD′))−1
= (γγ′) · Z.
Thus, our definition for γ · Z defines a group action of GSp+2n(R) of Hn. Since Γn and
GSp+2n(R) are subgroups of GSp+
2n(R) (for R a subring of R), we get both subgroups give a group
action on Hn via the same action.
2.2 Modular Forms
Using the group action defined in Section 2.1, we define the weight k slash operator on
functions f : Hn → C by
(f |kγ)(Z) := µ(γ)nk−n(n+1)
2 j(γ, Z)−kf(γ · Z), where j(γ, Z) = det(CZ +D)
for γ =
A B
C D
∈ GSp+2n(R). We now give the definition of a Siegel modular form.
Definition 2.2.1 A classical Siegel modular form of weight k, degree n > 1, and level Γn is a
holomorphic function f : Hn → C such that for all γ ∈ Γn we have (f |kγ)(Z) = f(Z).
The collection of modular forms of weight k and level Γn is a C-vector space which we denote
Mk(Γn) (when the degree is clear we will simply write Mk). We note that by Theorem 2 of Chapter
2, Section 4 of [11] that this vector space is finite-dimensional.
For degree 1 classical Siegel modular forms we are simply working with elliptic modular
forms and require f to be holomorphic at the cusps, that is the orbit of i∞ under the group action.
Initially this was a requirement for any degree but Max Koecher proved in 1954 that for degree
n > 1 we automatically get the function is holomorphic at the cusps.
7
Furthermore, f has Fourier expansion
f(Z) =∑
0≤T∈Λn
a(T ; f)qT , qT = exp{2πitr(TZ)}, Z ∈ Hn,
where
Λn = {T = (tij) ∈ Matn(Q) : tT = T and tii, 2tij ∈ Z}.
That is, the sum runs over all semi-positive definite, half-integral matrices. This series is uniformly
convergent on compact subsets of Hn ([23], page 9).
It is important to note that the Fourier coefficients are given by
a(T ; f) =
∫X mod 1
f(Z)e−2πitr(TZ)dX
where dX =∏i≤jdXi,j . Also, X mod 1 means Xi,j mod 1 for i ≤ j and Xi,j is the (i, j)-th entry of
X. We note that these coefficients are independent of Y as demonstrated on pages 32 and 33 of
[25]. We will revisit the Fourier expansion once we have a definition for the subspace of cusp forms.
To motivate the idea of weight k, degree n, level Γn cusp forms we introduce the Siegel
operator
Φ : Mk(Γn) → Mk(Γn−1)
f 7→ limt→∞
f
Zit
, Z ∈ Hn−1, t ∈ R.
It is trivial to see this defines a linear map due to basic limit laws. Page 11 of [23] explains
how to identify Φ(f) in Mk(Γn−1).
It is crucial that we acknowledge that Φ depends on the n though the notation does not
reflect this. It should be clear from the context which degree we are working with. The notation
can be difficult to follow when we consider the iterative mapping given by
Mk(Γn)Φ→Mk(Γn−1)
Φ→ · · · Φ→Mk(Γr+1)Φ→Mk(Γr)
since each Φ above is differs from the rest. To take forms from Mk(Γn) to Mk(Γr) as in the sequence
above we use Φn−r.
8
Using this operator, we define Sk(Γn) := ker Φ to be the subspace of cusp forms. To see
why this makes sense we visit the case of elliptic modular forms, that is, take f ∈Mk(Γ1). We then
have
Φ(f) = limt→∞
f(it)
= limt→∞
(a0(f) +
∞∑i=1
ai(f) · e2πi(it)
)
= a0(f) +
∞∑i=1
limt→∞
(ai(f) · e−2πt
)= a0(f) +
∞∑i=1
ai(f)(
limt→∞
e−2πt)
= a0(f).
Note that since the Fourier expansion is uniformly convergent we can use the dominated convergence
theorem to exchange the limit and summation in line 3.
Thus, f ∈ ker Φ if and only if a0(f) = 0, which agrees with the elliptic definition of cusp
forms. Revisiting the formula for the Fourier coefficients allows us to determine when exactly
f ∈ ker Φ ⊂Mk(Γn).
Let f ∈Mk(Γn), T ′ ∈ Λn−1, Z ′ ∈ Hn−1, and X ′ = Re(Z ′). We have
a(T ′; Φ(f)) =
∫X′ mod 1
(Φ(f))(Z ′)e2πitr(T ′Z′)dX ′
=
∫X′ mod 1
limt→∞
f
Z ′it
e2πitr(T ′Z′)dX ′
= limt→∞
∫X mod 1
f
Z ′it
e
2πitr
T′
0
Z′
it
dX
= a
T ′
0
; f
.
9
To get the third equality use Theorem 4.3 of [23] to get
∣∣∣∣∣∣∣fZ ′
it
e2πitr(T ′Z′)
∣∣∣∣∣∣∣ =
∣∣∣∣∣∣∣fZ ′
it
∣∣∣∣∣∣∣∣∣∣e2πitr(T ′Z′)
∣∣∣ ≤Mt · 1 = Mt
where Mt = M is fixed for large enough t. Thus we can apply the dominated convergence theorem
in order to exchange the limit and integral. In the same line, we switch from X ′ to X where
X = Re
Z ′it
=
X ′0
.
Thus, f ∈ Sk(Γn) if and only if a
T ′
0
; f
= 0 for all T ′ ∈ Λn−1. We can strengthen
the result to f ∈ Sk(Γn) if and only if a(T ; f) = 0 for all 0 ≤ T ∈ Λn such that 0 6< T . To get this
we use the identity
a(tUTU ; f) = det(U)−ka(T ; f), for all 0 ≤ T ∈ Λn, and U ∈ GLn(Z).
It is important to note the following consequence of this relation.
Lemma 2.2.2 ([23], Corollary 4.2) A classical Siegel Modular form of weight k with kn ≡ 1(mod 2)
vanishes.
Proof Let f ∈Mk(Γn) with kn ≡ 1 mod 2 and let 0 ≤ T ∈ Λn. Observe thatt(−In)T (−In) = T .
Thus if kn ≡ 1(mod 2), then we have
a(T ; f) = a(t(−In)T (−In); f)
= det(−In)−ka(T ; f)
= (−1)−nka(T ; f)
= −a(T ; f).
Thus we have a(T ; f) = 0 for all T ∈ Λn and therefore f vanishes. �
Using this result, we see that every Siegel modular form of odd weight, even degree, and full level
must be a cusp form. To see this, suppose f ∈ Mk(Γn) with k odd. Applying the Siegel operator,
10
we see that Φ(f) ∈Mk(Γn−1) but
k(n− 1) ≡ 1 mod 2
and hence f ∈ ker Φ, proving it is a cusp form. Consequently the main theorem of [9] is trivial for
k odd. Moving forward, unless otherwise specified, we take k even.
Since we know how Φ interacts with the fourier coefficients of a form f ∈Mk(Γn), we now
consider the more general result for Φn−r. The closed form of the iterative application is given by
Φn−r : Mk(Γn) → Mk(Γr)
f 7→ limt→∞
f
Z ′itIn−r
, Z ′ ∈ Hr, t ∈ R
where In−r is the n − r × n − r identity matrix. To see the motivation behind this definition,
consider the Fourier coefficients, a(T ′; Φn−r(f)), of a form f ∈ Mk(Γn). The affect of Φn−r on a
form f ∈Mk(Γn) is shown by its affect on the Fourier coefficients
a(T ′; Φn−r(f)) = a
T ′
0n−r
; f
.
Using this and taking r ≥ 1, we have
a(T ′; ΦΦn−r(f)) = a
T ′
01
; Φn−r(f)
= a
T ′
01
0n−r
; f
= a
T ′
0n−r+1
; f
= a(T ′; Φn−r+1(f)).
Thus, for any f ∈Mk(Γn) and r ≥ 1 the Fourier coefficients of ΦΦn−r(f) and Φn−r+1(f) coincide.
Since the coefficients of a power series, and in particular those of a Fourier expansion, uniquely
11
determine the function, we must have Φ ◦ Φn−r = Φn−r+1. Furthermore, induction now grants us
Φr−j ◦ Φn−r = Φn−j for 0 ≤ j ≤ r ≤ n.
By Corollary 5.5 of [23], it is known that Φ is surjective for even k > 2n, so by the first
isomorphism theorem of vector spaces we have
Mk(Γn)/Sk(Γn) = Mk(Γn)/ ker Φ 'Mk(Γn−1).
Since Mk(Γn) is a C-vector space we can express it as the direct sum of Sk(Γn) and its compliment,
Sk(Γn)c. We wish to address the structure of Sk(Γn)c. To do so we introduce the Petersson inner
product.
Definition 2.2.3 Let f, g ∈Mk(Γn) with at least one a cusp form. We define the Petersson inner
product as
〈f, g〉 :=
∫Γn\Hn
f(Z)g(Z) det(Y )kdνn(Z)
where dνn(Z) = det(Y )−(n+1)∏i≤j dxijdyij, Z = X + iY , and xij , yij are the i, jth entry of X,Y ,
respectively.
Lemma 2.2.4 For n ∈ Z≥1 the measure dνn is invariant under the action of GSp+2n(R) on Hn.
That is, dνn(γZ) = dνn(Z) for all γ ∈ GSp+2n(R).
Proof Let g = GSp+2n(R) and define X∗ = Re(γZ) and Y ∗ = Im(γZ). The Jacobian for the change
of variables of Z 7→ gZ is given by
det
(∂(X∗, Y ∗)
∂(X,Y )
)where
∂(X∗, Y ∗)
∂(X,Y )=
∂X∗
∂X∂Y ∗
∂X
∂X∗
∂Y∂Y ∗
∂Y
.
Since the action of γ is holomorphic on Hn we have the Cauchy-Riemann equations
∂X∗
∂X=∂Y ∗
∂Yand
∂X∗
∂Y= −∂Y
∗
∂X.
Using these relations we see that
12I − i
2I
− i2I
12I
∂X∗
∂X∂Y ∗
∂X
∂X∗
∂Y∂Y ∗
∂Y
I iI
iI I
=
∂X∗
∂X − i∂X∗
∂Y 0
0 ∂X∗
∂X + i∂X∗
∂Y
=
∂(gZ)∂Z 0
0 ∂(gZ)∂Z
.
12
Consequently, we have
det
(∂(X∗, Y ∗)
∂(X,Y )
)= det
(∂(γZ)
∂Z
)det
(∂(γZ)
∂Z
).
We now write dZ = (dzij) and d(gZ) for the matrices of the differentials of Z and γZ,
respectively. Let γ =
A B
C D
, then (γZ)(CZ +D) = (AZ +D). We have
d(γZ)(CZ +D) + (γZ)CdZ = AdZ.
Multiplying both sides on the left byt(CZ +D) produces
t(CZ +D)AdZ = d(γZ)[CZ +D] +
t(CZ +D)(γZ)CdZ
= d(γZ)[CZ +D] +t(CZ +D)
t((AZ +B)(CZ +D)−1)CdZ
= d(γZ)[CZ +D] +t(AZ +B)CdZ.
Rewriting, we have
d(γZ)[CZ +D] =t(CZ +D)AdZ − t
(AZ +B)CdZ
= Z(tCA− tAC)dZ + (tDA− tBC)dZ
= µ(γ)dZ since γ ∈ GSp+2n(R).
We then have
d(γZ) =t(CZ +D)−1µ(γ)dZ(CZ +D)−1 = dZ
[√µ(γ)(CZ +D)−1
].
From page 9 of [11], it is known that the determinant of the linear map Z 7→ tMZM = Z[M ] on
symmetric matrices, Z, is given by detMn+1. Since∂(γZ)
∂Zmeasures changes in γZ as Z varies and
we have d(γZ) = dZ[√
µ(γ)(CZ +D)−1], we obtain
det
(∂(γZ)
∂Z
)= det(
√µ(γ)(CZ +D)−1)n+1 = µ(γ)n(n+1)/2 det(CZ +D)−(n+1).
13
Similarly, we obtain
det
(∂(γZ)
∂Z
)= µ(γ)n(n+1)/2det(CZ +D)
−(n+1).
Combining all the results have
dνn(γZ) = det(Y ∗)−(n+1) det
(∂(X∗, Y ∗)
∂(X,Y )
)∏i≤j
dxijdyij
=
(µ(γ)n det(Y )
|det(CZ +D)|2
)−(n+1)
det
(∂(γZ)
∂Z
)det
(∂(γZ)
∂Z
)∏i≤j
dxijdyij
=|det(CZ +D)|2n+2
µ(γ)n(n+1) det(Y )n+1µ(γ)n(n+1)|det(CZ +D)|−2n−2
∏i≤j
dxijdyij
= det(Y )−(n+1)∏i≤j
dxijdyij
= dνn(Z).
Therefore the measure is invariant under the action of GSp+2n(R) on Hn.
Furthermore, the integral defined in the Petersson inner product converges when f or g is
a cusp form. Due to the orthogonal decomposition,
Mk(Γn) = Sk(Γn)⊕Sk(Γn)⊥,
we take Sk(Γn)⊥ as our choice for Sk(Γn)c. Since Φ is surjective for k > 2n, we obtain Φ(Sk(Γn)⊥
)=
Mk(Γn−1) = Sk(Γn−1)⊕Sk(Γn−1)⊥. We then define Mkn,n−1 = Sk(Γn)⊥ ∩Φ−1
(Sk(Γn−1)
). Iter-
ating this process we obtain the following subspaces due to [15] using the notation of [10]
Mkn,r =
Sk(Γn) r = n
{f ∈ Sk(Γn)⊥ : Φ(f) ∈Mkn−1,r} 0 ≤ r < n.
More precisely, since Mk(Γn) is finite-dimensional we are able to use induction to identify these
spaces as
Mkn,r = ker Φn+r+1/ ker Φn+r.
Our identification makes it clear that these spaces are pair-wise disjoint but we explicitly show this
for clarity. Consider r1 6= r1 and, without loss of generality, let 0 ≤ r1 < r2 < n. We then have the
14
following
Mkn,r1 ∩Mk
n,r2 = {f ∈ Sk(Γn)⊥ : Φ(f) ∈Mkn−1,r1 ∩Mk
n−1,r2}
= {f ∈ Sk(Γn)⊥ : Φn−r2(f) ∈Mkr2,r1 ∩Mk
r2,r2}
= {f ∈ Sk(Γn)⊥ : Φn−r2(f) ∈Mkr2,r1 ∩Sk(Γr2)}.
However, since r2 > r1 we have Mkr2,r1 = {g ∈ Sk(Γr2)⊥ : Φ(g) ∈ Mk
r2−1,r1}. By orthogonality
of Mkr2,r1 and Mk
r2,r2 = Sk(Γr2) for r1 < r2, we must then have Mkr2,r1 ∩Sk(Γr2) = {0}, showing
Φn−r2(f) = 0. However, by the definition of Mkn,r2 , anything that vanishes under Φn−r2 must be 0.
Therefore we conclude Mkn,r1 ∩Mk
n,r2 = {0} for r1 6= r2. Note that we did not need to check the
cases 0 ≤ r1 < r2 = n since the result would follow trivially from the definition of Mkn,r.
Since k > 2n we have Φn is surjective and maps Mk(Γn) to Mk(Γ0) = Mk0,0 = C we can
write
Mk(Γn) = Mkn,0 + Mk
n,1 + · · ·+ Mkn,n−1 + Mk
n,n.
This combined with the subspaces being pairwise disjoint we get
Mk(Γn) = Mkn,0 ⊕Mk
n,1 ⊕ · · · ⊕Mkn,n−1 ⊕Mk
n,n
= Mkn,0 ⊕Mk
n,1 ⊕ · · · ⊕Mkn,n−1 ⊕Sk(Γn).
Thus, given f ∈Mk(Γn) we have a unique fr ∈Mkn,r such that f =
∑nr=1 fr. For each fr
we have Φn−r(fr) = fgr ∈ Mkr,r = Sk(Γr), noting that Φn−n is just the identity operator yielding
fn = gn. Therefore, we have a (n+ 1)-tuple of cusp forms (g0, g1, . . . , gn) associated to f . The next
section will highlight the significance of this association.
2.3 Klingen Eisenstein Series
In the previous section we saw how to take a Siegel modular form of arbitrary degree and
obtain a tuple of cusp forms. This, however, supposes Siegel modular forms of degree greater than
1 exist. In this section we give a construction of degree n Siegel modular forms from degree r forms.
We will then revisit the association of a modular form with a tuple of cusp forms.
15
For 0 ≤ r < n, we define the Klingen parabolic subgroup
∆n,r :=
∗ ∗
0n−r,n+r ∗
∈ Γn
.
We construct the Klingen Eisenstein series associated to f ∈ Sk(Γr) by
Ekn,r(f)(Z) =∑
γ∈∆n,r\Γn
j(γ, Z)−kf((γ · Z)∗), Z ∈ Hn,
where (γ · Z)∗ is the upper left r × r matrix of γ · Z = (AZ +B)(CZ +D)−1.
As an example, take n = 1, r = 0 and f = 1 ∈Mk(Γ0) . We have
Ek1,0(1)(z) =∑
a b
c d
∈∆1,0\Γ1
(cz + d)−k.
We will show that this is precisely the level 1, weight k Eisenstein series for elliptic modular forms.
To see this we only need to show there is a bijection between cosets of ∆1,0 in Γ1 and relatively
prime pairs (c, d) ∈ Z2 up to sign; that is, (c, d) and (−c,−d) correspond to the same coset but
(c,−d) produces a different coset.
We start by giving a more explicit description of ∆1,0 ⊂ Sp2(Z) = SL2(Z):
∆1,0 =
∗ ∗
01−0,1+0 ∗
∈ Γ1
=
±1 α
0 ±1
∈ Γ1 : α ∈ Z
.
If γ =
a b
c d
∈ SL2(Z), then we have gcd(c, d) = 1 and
∆1,0γ =
±a+ αc ±b+ αd
±c ±d
∈ Γ1 : α ∈ Z
.
Since −I2 ∈ ∆1,0 we can map the pair (−c,−d) to (c, d). For simplicity, we focus on the pair (c, d).
16
Now suppose γ′ =
a′ b′
c d
∈ SL2(Z), then (a′, b′) is a solution to the Diophantine equation
xc− yd = 1.
If we consider (a, b) as our initial solution the equation, then it is an elementary result that any
solution is of the form
(a+ ct, b+ dt) where t ∈ Z,
and thus there exists some α ∈ Z such that a′ = a+ αc and b′ = b+ αd. Consequently we have
γ′ =
a+ αc b+ αd
c d
=
1 α
0 1
a b
c d
∈ ∆1,0γ.
Therefore, we have a bijection of sets given by
∆1,0\Γ1 ←→{
(c, d) ∈ Z2 : gcd(c, d) = 1}
∆1,0
a b
c d
←→ (c, d)
and hence
Ek1,0(1)(z) =∑
γ∈∆1,0\Γ1
(cz + d)−k =∑
gcd(c,d)=1
(cz + d)−k,
which is precisely the level 1, weight k Eisenstein series for elliptic modular forms.
Moreover, we have the following theorem,
Theorem 2.3.1 ([23], Theorem 5.4) Let n ≥ 1 and 0 ≤ r ≤ n and k > n + r + 1 be integers
with k even. Then for every cusp form f ∈ Sk(Γr) the series Ekn,r(f) converges to a classical Siegel
modular form of weight k in Mk(Γn) and satisfies Φn−r(Ekn,r(f)) = f .
Using [10] we first show the convergence of Ekn,r(f) and then address the iterated Siegel
operator. To do this we show the series converges absolutely-uniformly. For f ∈ Sk(Γr) we know by
proposition 4 (p. 45) of [11] that the quantity det(Y )k/2f(Z) is bounded for all X + iY = Z ∈ Hr.
17
Thus, up to some constant, we have the majorant Ekn,r(f)(Z) in Hn by
Gkn,r(Z) =∑
γ∈∆n,r\Γn
det(Im(γ · Z)∗)−k2 |j(γ, Z)|−k.
We study this series on vertical strips of the Siegel upper half-space given by
Bn(d) = {Z ∈ Hn : tr(X2) ≤ d−1, Y ≥ dIn}
for d > 0. On these strips we have the following lemma.
Lemma 2.3.2 ([10], p. 33) For n ≥ 1 and d > 0 there exists c > 0, dependent only on n and d,
so that
det(Im(γ · Z)∗)|j(γ, Z)|2 ≥ cdet(Im(γ · iIn)∗)|j(γ, iIn)|2
for all Z ∈ Bn(d), γ ∈ Sp2n(R), and 0 ≤ r ≤ n.
Note that r in the lemma is in reference to (Z)∗ meaning take the upper-left r × r matrix
of Z. Combining this lemma with the decomposition
η =
η∗ 0
0 η2
Ir η3
0 In−r
where γ · Z = ξ(Z) + iη(Z) we can obtain the uniform convergence of Gkn,r(Z) by studying
Gkn,r(iIn) =∑
γ∈∆n,r\Γn
det(Im(γ · iIn)∗)−k2 |j(γ, iIn)|−k
=∑
γ∈∆n,r\Γn
det(η∗(iIn))−k2 |j(γ, iIn)|−k
=∑
γ∈∆n,r\Γn
det(η2(iIn))k2 det(η(iIn))−
k2 |j(γ, iIn)|−k
=∑
γ∈∆n,r\Γn
det(η2(iIn))k2 det(In)−
k2
(|j(γ, iIn)|−2
)− k2 |j(γ, iIn)|−k
=∑
γ∈∆n,r\Γn
det(η2(iIn))k2 .
18
Note that the second to last equality arise from the identity
t(CZ +D)η(Z)(CZ +D) =
t(CZ +D)Im(γ · Z)(CZ +D) = Y
for γ ∈ Γn, noting that η changes if we change γ. In [10] we see why this is absolutely convergent
and consequently get that Ekn,r(f) converges to a form in Mk(Γn).
We now consider Φn−r(Ekn,r(f)) and note that the absolute convergence of the series
Ekn,r(f)(Z) =∑
γ∈∆n,r\Γn
j(γ, Z)−kf((γ · Z)∗), Z ∈ Hn,
allows us to apply Φn−r term by term in the summation. The first element in the sum we consider
is any representative
γ =
Ar,r Ar,n−r Br,r Br,n−r
An−r,r An−r,n−r Bn−r,r Bn−r,n−r
Cr,r Cr,n−r Dr,r Dr,n−r
0n−r,r 0n−r,n−r 0n−r,r Dn−r,n−r
∈ ∆n,r ⊂ Γn
and since this matrix is in Γn we have the following relations
Ar,r Br,r
Cr,r Dr,r
∈ Γr, An−r,n−rtDn−r,n−r = In−r,
and Ar,n−rtDn−r,n−r = 0r,n−r = Cr,n−r
tDn−r,n−r.
Since Dn−r,n−r is invertible we must have Ar,n−r = Cr,n−r = 0r,n−r. Thus a representative of ∆n,r
produces the following; Ar,rZ∗ +Br,r Br,n−r
An−r,rZ∗ +Bn−r,r itAn−r,n−r +Bn−r,n−r
Cr,rZ∗ +Dr,r Dr,n−r
0n−r,r Dn−r,n−r
−1
=
Ar,rZ∗ +Br,r Br,n−r
An−r,rZ∗ +Bn−r,r itAn−r,n−r +Bn−r,n−r
(Cr,rZ
∗ +Dr,r)−1 −(Cr,rZ
∗ +Dr,r)−1Dr,n−r(Dn−r,n−r)
−1
0n−r,r (Dn−r,n−r)−1
19
=
(Ar,rZ∗ +Br,r)(Cr,rZ
∗ +Dr,r)−1 ∗
∗ ∗
.
Next, we see that
det
Cr,rZ∗ +Dr,r Dr,n−r
0n−r,r Dn−r,n−r
= det(Cr,rZ∗ +Dr,r) det(Dn−r,n−r) = ±det(Cr,rZ
∗ +Dr,r)
sinceDn−r,n−r and its inverse are integral matrices. Using these results, writing g(Z) = j(γ, Z)−kf((γ·
Z)∗), and recalling that k is even, we have
Φn−r(g(Z)) = limt→∞
j
γ,Z∗
itIn−r
−k
f
γ ·
Z∗itIn−r
∗
= limt→∞
det
C Z∗ 0r,n−r
0n−r,r itIn−r
+D
−k
f
(Ar,rZ
∗ +Br,r)(Cr,rZ∗ +Dr,r)
−1 ∗
∗ ∗
∗
= (±det(Cr,rZ∗ +Dr,r))
−kf
(Ar,rZ
∗ +Br,r)(Cr,rZ∗ +Dr,r)
−1 ∗
∗ ∗
∗ , γ ∈ ∆n,r
= det(Cr,rZ∗ +Dr,r)
−kf((Ar,rZ∗ +Br,r)(Cr,rZ
∗ +Dr,r)−1) since k is even.
= f(Z∗) since
Ar,r Br,r
Cr,r Dr,r
∈ Γr.
We will now show that the iterated Siegel operator forces the other terms in Ekn,r(f) to 0.
To get this result, we return to Gkn,r; by showing the summands become zero away from ∆n,r we
show the same applies for Ekn,r(f). Writing Z∗ = X∗ + iY ∗, we have
Z =
Z∗ 0r,n−r
0n−r,r itIn−r
and Y =
Y ∗ 0r,n−r
0n−r,r tIn−r
.
Rewriting the identity
t(CZ +D)η(Z)(CZ +D) =
t(CZ +D)Im(γ · Z)(CZ +D) = Y
20
produces the expression
η∗−1(Z) 0
0 η−12 (Z)
= η(Z)−1
= (CZ +D)(Y −1)t(CZ +D)
= (CZ +D)
(Y ∗)−1 0r,n−r
0n−r,r −it−1In−r
t(CZ +D)
=
∗ ∗
∗ Y ∗−1[(Z∗tCn−r,r + tDn−r,r)] + t−1[(ittCn−r,n−r + tDn−r,n−r)]
.
Thus we have
η−12 (Z) = Y ∗−1[(Z∗tCn−r,r + tDn−r,r)] + t−1[(ittCn−r,n−r + tDn−r,n−r)]
allowing us to simplify the following expression
det(Im(γ · Z)∗)|j(γ, Z)|2 = det(η∗)|j(γ, Z)|2
=det(η(Z))
det(η2)|j(γ, Z)|2
= det(η−12 ) det(Y )
= det(η−12 ) det(Y ∗)tn−r
= det(tη−12 ) det(Y ∗)
= det(Y ∗)P (t),
where P (t) = det(tY ∗−1[Z∗tCn−r,r + tDn−r,r] + In−r[ittCn−r,n−r + tDn−r,n−r]). Thus using this
equality we have
Gkn,r(Z) =∑
γ∈∆n,r\Γn
(det(Y ∗)P (t))− k
2 .
Since the Siegel operator uses the limit as t → ∞ and we know the series is absolutely convergent,
we apply the limit term-wise. Moreover, since det(Y ∗) is independent of t we need only focus on
limt→∞
P (t)−k2 . Clearly, P (t) is a polynomial in t and thus vanishes for deg(P ) > 0. Thus we will
21
assume deg(P ) = 0 and show this only occurs for γ ∈ ∆n,r. To see this we note that for t ≥ 0, both
tY ∗−1[Z∗tCn−r,r + tDn−r,r] and In−r[ittCn−r,n−r + tDn−r,n−r]
are positive semi-definite since Y ∗−1 and In−r are positive definite. Furthermore their sum must be
positive definite since the left-hand side of
det(Im(γ · Z)∗)|j(γ, Z)|2 = det(Y ∗)P (t)
is non-zero for all Z ∈ Hn and γ ∈ ∆n,r. As a consequence we have
det(In−r[ittCn−r,n−r + tDn−r,n−r]) ≤ P (t)
for all t and where we have equality when 0r,r = Y ∗−1[Z∗tCn−r,r + tDn−r,r]. Since Y ∗−1 is positive
definite, this occurs when 0n−r,r = Z∗tCn−r,r+ tDn−r,r which holds if and only if 0n−r,r = Cn−r,r =
Dn−r,r. This then leaves
P (t) = det(In−r[ittCn−r,n−r + tDn−r,n−r])
= det(t2Cn−r,n−rtCn−r,n−r +Dn−r,n−r
tDn−r,n−r),
which still produces a polynomial in t unless 0 = Cn−r,n−r. Thus we have limt→∞
P (t)−k2 vanishes
unless γ ∈ ∆n,r\Γn has 0 = Cn−r,r = Dn−r,r and 0 = Cn−r,n−r. This is precisely when γ ∈ ∆n,r.
Therefore all terms in the series in Gkn,r(Z), and thus in Ekn,r(Z), with γ away from ∆n,r vanish for
Z =
Z∗ 0
0 itIn−r
as t→∞. Thus for Z∗ ∈ Hr, 0 ≤ r ≤ n, f ∈ Sk(Γn), and k > n+ r + 1 even
22
we have:
Φn−r(Ekn,r(f))(Z∗) = limt→∞
∑γ∈∆n,r\Γn
det
C Z∗ 0r,n−r
0n−r,r itIn−r
+D
−k
f
γ ·
Z∗ 0r,n−r
0n−r,r itIn−r
∗
=∑
γ∈∆n,r\Γn
limt→∞
det
C Z∗ 0r,n−r
0n−r,r itIn−r
+D
−k
f
γ ·
Z∗ 0r,n−r
0n−r,r itIn−r
∗
= (±det(Cr,rZ∗ +Dr,r))
−kf
(Ar,rZ
∗ +Br,r)(Cr,rZ∗ +Dr,r)
−1 ∗
∗ ∗
∗ , γ ∈ ∆n,r
= det(Cr,rZ∗ +Dr,r)
−kf((Ar,rZ∗ +Br,r)(Cr,rZ
∗ +Dr,r)−1) since k is even.
= f(Z∗) since
Ar,r Br,r
Cr,r Dr,r
∈ Γr.
Thus we see that Ekn,r( · ) is a right inverse to Φn−r when we restrict to degree r cusp forms. From
here we use the notation of [12] and write
[f ]nr := Ekn,r(f) for f ∈ Sk(Γr).
We do this in order to extend the definition of the operator [ · ]nr to all of Mk(Γr). For f = [fj ]rj
with fj ∈ Sk(Γj) we define [f ]nr = [fj ]nj . Using this notion as well as our desire for a linear operator,
we take f ∈ Mk(Γr) and get the r + 1-tuple of cusp forms, (f0, · · · , fr), mentioned at the end of
Section 2.2 and define
[f ]nr =r∑j=0
[fj ]nj =
r∑j=0
Ekn,j(fj).
Note that this agrees with our construction in the previous section since we have seen Φn−j([fj ]nj ) =
fj .
Using this relation we introduce the notation
En,j,k =[Sk(Γj)
]nj.
We then have the decomposition
[Mk(Γn)]nr =
r⊕j=0
En,j,k.
23
It is important to note that this decomposition is one into orthogonal subspaces with respect to the
Petersson inner product. However, to see this we use Maass’ extension of the inner product [15,
p. 96] that says for f =∑rj=0[fj ]
nj and g =
∑rj=0[gj ]
nj we write
〈f, g〉M :=
n∑j=0
⟨Φn−j([fj ]
nj ),Φn−j([gj ]
nj )⟩
=
n∑j=0
〈fj , gj〉
which is a finite sum of convergent inner products, since both fj , gj are cusp forms for all j.
To see that this agrees with the original definition we consider 〈f, g〉 and 〈f, g〉M where
at least one of f and g is a cusp form. Without loss of generality, let g ∈ Sk(Γn) and write
f =∑nj=0[fj ]
nj . Since f ∈ Mk(Γn) = Sk(Γn) ⊕ Sk(Γn)⊥ and thus can also write f = fc + fp
where fc ∈ Sk(Γn) and fp ∈ Sk(Γn)⊥. Recall that at the end of Section 2.2 that we showed the
(n+ 1)-tuple of cusp forms (f0, f1, . . . , fn) associated with f is unique, hence we must have fn = fc.
We see that
〈f, g〉M = 〈fn, g〉+
n−1∑j=0
〈Φn−j([fj ]nj ), 0〉 since f ∈ Sk(Γn)
= 〈fn, g〉+
n−1∑j=0
0
= 〈fn, g〉
= 〈fc, g〉
= 〈fc, g〉+ 〈fp, g〉 since fp ∈ Sk(Γn)⊥
= 〈fc + fp, g〉
= 〈f, g〉.
Now consider f ∈ En,`,k and g ∈ En,j,k where ` 6= j and, without loss of generality, 0 ≤ ` <
24
j ≤ n. Then both forms have expressions f = [f`]n` and g = [gj ]
nj , and thus we have
〈f, g〉M =⟨Φn−`([f`]
n` ),Φn−`([g`]
n` )⟩
+⟨Φn−j([fj ]
nj ),Φn−j([gj ]
nj )⟩
= 〈f`, g`〉+ 〈fj , gj〉
= 〈f`, 0〉+ 〈0, gj〉 by definition of f, g
= 0.
Thus we see that f ∈ En,`,k and g ∈ En,j,k are orthogonal for 0 ≤ ` < j ≤ n. For the specific case of
n = j we have
En,n,k = [Sk(Γn)]nn = Sk(Γn),
thus showing degree n cusp forms are orthogonal to all Klingen-Eisenstein series.
2.4 R-modules of modular forms
Recall that Mk(Γn) is a finite-dimensional C-vector space.
Given a subring R ⊂ C we wish to consider the R-module
Mk(Γn)R = {f ∈Mk(Γn) : a(T ; f) ∈ R for all T ∈ Λn}
and the corresponding cusp forms
Sk(Γn)R = Sk(Γn) ∩Mk(Γn)R.
It is natural to ask what information on Mk(Γn) can be obtained by studying Mk(Γn)R. In
[21], Theorem 3 implies
Mk(Γn) ∼= Mk(Γn)Q ⊗Q C
for k even and gives a singular condition for the result to hold when k is odd. We do not state the
condition as [20] shows it holds for all odd k ∈ Z and at no other point will we use the condition.
Note, this isomorphism is often exploited and sometimes the two sets are written as being equal
instead of isomorphic. We also note that this isomorphism demonstrates that Mk(Γn)Q is a finite
dimensional Q-vector space since tensoring with C is merely an extension of scalars. Therefore, if
25
{f1, . . . , fm} is Q-basis of Mk(Γn)Q then it also acts as a C-basis of Mk(Γn). Thus, for f ∈Mk(Γn)
we identify it in Mk(Γn)Q ⊗Q C by
f =
m∑j=1
αjfj ←→m∑j=1
(fj ⊗ αj).
Using this identification we prove the following lemma.
Lemma 2.4.1 If k ∈ Z>0 and K is a number field, then
Mk(Γn)Q ⊗Q K ∼= Mk(Γn)K .
Proof Let {f1, . . . , fm} be as defined above and consider the following map extended linearly
ψ : Mk(Γn)Q ⊗Q K −→ Mk(Γn)K
f ⊗ c 7−→ cf.
It is clear that this is a linear map of K-modules. To see this map is surjective, let f ∈Mk(Γn)K ⊂
Mk(Γn). There exists {αj}mj=1 ⊂ C such that f =∑mj=1 αjfj . Note that this is equivalent to
a(T ; f) =
m∑j=1
αja(T ; fj) for all T ∈ Λn.
Since the fj form a basis, it is possible to take a subset {Ti}mi=1 ⊂ Λn such that a(Ti; f) 6= 0 for
some i and det(A) 6= 0 where A = (a(Ti, fj)) ∈ Matm(Q). Supposing otherwise would imply linear
dependence amongst the fj , contradicting the definition of a basis. Setting α =t(α1 · · ·αm) and
a =t(a(T1; f) · · · a(Tm; f)) produces Aα = a. Let Aj denote the matrix produced by replacing the
jth column of A with a. By Cramer’s rule, for all j, we then have
αj =det(Aj)
det(A)∈ K since det(A) ∈ Q,det(Aj) ∈ K.
Consequently∑mj=1(fj ⊗ αj) ∈ Mk(Γn)Q ⊗Q K and is thus in the preimage of f , demonstrating
that ψ is surjective.
Now suppose∑ri=1(hi ⊗ αi) ∈ kerψ. Combining this with the fact that {f1, . . . , fm} is a
26
basis, we obtain
0 =
r∑i=1
αihi
=
r∑i=1
αi
m∑j=1
ci,jfj
where ci,j ∈ Q
=
m∑j=1
(r∑i=1
αici,j
)fj .
By definition of a basis we must then have∑ri=1 αici,j = 0 for all j. Using this produces
r∑i=1
(hi ⊗ αi) =
r∑i=1
m∑j=1
ci,jfj ⊗ αi
=
r∑i=1
m∑j=1
(ci,jfj ⊗ αi)
=
m∑j=1
r∑i=1
(fj ⊗ αici,j)
=
m∑j=1
(fj ⊗
r∑i=1
αici,j
)
=
m∑j=1
(fj ⊗ 0)
= 0.
Therefore kerψ = {0}, showing that the map is injective. Combining this with the surjectivity we
have that ψ is an isomorphism. �
Revisiting [21], we get the following lemma and introduce a corollary to highlight our in-
tended use.
Lemma 2.4.2 ([21], Theorem 1) Let Q denote an algebraic closure of Q, and O the ring of all
algebraic integers in Q. Then, for every f ∈ Mk(Γn)Q, there exists c ∈ Z>0 such that cf ∈
Mk(Γn)O.
Corollary 2.4.3 Let K be a number field and consider a subring K ⊇ R ⊇ OK . Then, for every
f ∈Mk(Γn)R, there exists c ∈ Z>0 such that cf ∈Mk(Γn)OK.
Proof Let f ∈ Mk(Γn)R ⊂ Mk(Γn)Q. By Lemma 2.4.2 there exists c ∈ Z>0 such that cf ∈
27
Mk(Γn)O. However, we also have that cf ∈Mk(Γn)R ⊂Mk(Γn)K , and thus
cf ∈Mk(Γn)O ∩Mk(Γn)K = Mk(Γn)O∩K = Mk(Γn)OK. �
We note that c is not unique; in fact, if we define the ideal
(f)K := {α ∈ OK : αf ∈Mk(Γn)OK}
of OK then it is easy to see that {0} ( cOK ⊆ (f)K for any c satisfying the result of Corollary
2.4.3. Since we know the ideals (f)K are strictly larger than the zero-ideal, it is worth asking if
there exist rings, R, satisfying OK ⊆ R ⊆ K such that R× ∩ (f)K 6= ∅ for all f ∈Mk(Γn)R. Taking
R = K is clearly true and later we will see that for a prime ideal p ⊆ OK , that the localization with
respect to p, Op, also satisfies this property. Knowing that the set of rings satisfying this property
is non-empty warrants the consideration of the following proposition.
Proposition 2.4.4 For all rings R satisfying OK ⊆ R ⊆ K such that R× ∩ (f)K 6= ∅ for all
f ∈Mk(Γn)R, we have
Mk(Γn)R ∼= Mk(Γn)OK⊗OK
R
as R-modules.
Proof We define the following map of R-modules
ϕR : Mk(Γn)R −→ Mk(Γn)OK⊗OK
R
f 7−→ uf ⊗ u−1,
where we take u ∈ R× ∩ (f)K , thus u−1 is defined. Note that the choice of u does not matter since
for any u, v ∈ R× ∩ (f)K we have u, v ∈ OK since (f)K ⊆ OK and hence
uf ⊗ u−1 = uf ⊗ vv−1u−1
= uvf ⊗ (uv)−1
= vf ⊗ uu−1v−1
= vf ⊗ v−1.
28
Thus we see the map is well-defined. Now observe that if we take f to be the 0 function and u = 1
then we get ϕ(0) = 0⊗ 1. For arbitrary f ∈Mk(Γn)R and αβ ∈ R ⊆ K with α, β ∈ OK and β ∈ R×
we have,
α
βϕR(f) =
α
β(uf ⊗ u−1)
= uf ⊗ α
βu−1
= uαf ⊗ (uβ)−1
= (uβ)
(α
βf
)⊗ (uβ)−1
= ϕR
(α
βf
).
We wish to extend ϕR linearly by defining
ϕR
∑j=1
fj
=∑j=1
ϕR(fj).
To see this is well defined let f1, f2, f3 ∈Mk(Γn)R such that f3 = f1 + f2. Let u1, u2, u3 be chosen
so that ui ∈ R× ∩ (fi)K . Note that the units of any ring is a multiplicatively closed set and thus
any product ue11 ue22 u
e33 ∈ R× for any ej ∈ Z. Thus we see that
ϕR(f1 + f2) = ϕR(f3)
= u3f3 ⊗ u−13
= u3f3 ⊗ (u1u2)(u1u2u3)−1
= (u1u2u3)(f1 + f2)⊗ (u1u2u3)−1
=(u2u3(u1f1)⊗ (u1u2u3)−1
)+(u1u3(u2f2)⊗ (u1u2u3)−1
)=
(u1f1 ⊗ u−1
1
)+(u2f2 ⊗ u−1
2
)= ϕR(f1) + ϕR(f2).
Since ϕR is a well-defined linear map, it is natural to ask about its kernel and image. Since R is a
subring of a field it is an integral domain, thus we have
0 = ϕR(f) = uf ⊗ u−1
29
if and only if one of the following three situations occurs:
1. u = 0,
2. u−1 = 0,
3. f = 0.
We note that this is an if and only if relation since for a pure tensor to be zero there must
be an element in the equivalence class such that one of the two elements in the tensor vanishes.
However, since R is an integral domain and f ∈Mk(Γn)R, we have uf vanishes if and only if u = 0
or f = 0. Since u ∈ R× we must have u 6= 0. By the same rational we also have u−1 6= 0. Thus we
must have f = 0, demonstrating kerϕR = {0}, that is, ϕR is injective.
Consequently, we know that Mk(Γn)R is isomorphic to the image of ϕR. Let
∑j=1
(fj ⊗
αjβj
)∈Mk(Γn)OK
⊗OKR
with αj , βj ∈ OK and βj ∈ R×. We obtain a preimage through the following process.
∑j=1
fj ⊗αjβj
=∑j=1
αjfj ⊗ β−1j
=∑j=1
βjαjβjfj ⊗ β−1
j
=∑j=1
ϕR
(αjβjfj
)
= ϕR
∑j=1
αjβjfj
.
Thus ϕR is an isomorphism and we have proven the proposition. �
Taking R = Q, we obtain Mk(Γn)Q ∼= Mk(Γn)Z ⊗Z Q. Combining this with Lemma 2.4.1
yields
Mk(Γn)K ∼= Mk(Γn)Q ⊗Q K
∼= Mk(Γn)Z ⊗Z Q⊗Q K
∼= Mk(Γn)Z ⊗Z K.
30
This is necessary result for Mizumoto’s proof of the following lemma from [17].
Lemma 2.4.5 ([17], Lemma A.4) Let {ω1, . . . , ω`} ⊂ OK be an integral basis of K; that is, for
any α ∈ K there exists {βi}`i=1 ⊂ Q such that α =∑i=1
βiωi. Then
Mk(Γn)OK=⊕i=1
Mk(Γn)Z · ωi
for n, k ∈ Z>0.
Proof The first step is to show that
Mk(Γn)OK=∑i=1
Mk(Γn)Z · ωi.
The inclusion ⊃ is obvious so we turn our attention to ⊂. Let f ∈ Mk(Γn)OK. Since we identify
Mk(Γn)Z ⊗Z K = Mk(Γn)K via
f =
m∑j=1
αjfj ←→m∑j=1
(fj ⊗ αj) with m ∈ Z>0,
there exist f1, · · · , fr ∈ Mk(Γn)Z and ai ∈ K such that f =∑ri=1 aifi. Furthermore, since
{ω1, . . . , ω`} ⊂ OK is an integral basis of K there exist cij ∈ Q such that
ai =∑j=1
cijωj .
Thus we have
f =
r∑i=1
aifi
=
r∑i=1
∑j=1
cijωj
fi
=∑j=1
(r∑i=1
cijfi
)ωj
=∑j=1
hjωj .
31
Since the hj ’s are Q-linear combinations of f1, · · · , fr ∈Mk(Γn)Z, we have that all hj ∈Mk(Γn)Q.
Now for all T ∈ Λn we have
a(T ; f) =∑j=1
a(T ;hj)ωj .
Since {ω1, . . . , ωg} is an integral basis of K and a(T ;ϕ) ∈ OK , we must have a(T ;hj) ∈ Z for all
j since the representation fo a(T ;ϕ) is unique with respect to the basis {ω1, . . . , ω`} and OK =
Z[ω1, . . . , ω`] as a free Z-module. Consequently, we have hj ∈ Mk(Γn)Z for all j; thus giving the
summation we set out to demonstrate.
The next step is to show that this summation is in fact a direct sum. Suppose f = 0,
then a(T ; f) = 0 for all T ∈ Λn. Since {ω1, . . . , ω`} is a basis, the only linear combination of these
elements that produces 0 is the trivial one, and thus a(T ;hj) = 0 for all T ∈ Λn and all j. Therefore
hj = 0 for all j, showing that the sum is actually a direct sum, yielding the desired result. �
We note this is equivalent to Mk(Γn)OK
∼= Mk(Γn)Z ⊗Z OK through the linear extension
of the map
f =
g∑j=1
hjωj 7→g∑j=1
(hj ⊗ ωj).
To understand the work of [9] requires an analogous result localizing both Z and OK at prime ideals.
To get this result we need to show that R = Op satisfies the requirements of Proposition 2.4.4. To
see this we use the following lemma.
Lemma 2.4.6 ([17], Lemma A.3) For any f ∈ Mk(Γn)Op with n, k ∈ Z>0, there exists a p-unit
u ∈ OK such that uf ∈Mk(Γn)OK.
In other words, for all f ∈ Mk(Γn)Op there exists u ∈ (OK − p) ⊂ O×p such that uf ∈
Mk(Γn)OK. Thus we have u ∈ O×p ∩ (f)K giving the following corollary to Proposition 2.4.4.
Corollary 2.4.7 Let K be a number field with ring of integers OK . Let Op be the localization of
OK at the prime ideal p ⊆ OK . Then
Mk(Γn)Op∼= Mk(Γn)OK
⊗OKOp.
Proof Set R = Op, the localiztion of OK at the prime ideal p. By Lemma 2.4.6 we have seen for
all f ∈Mk(Γn)Op that O×p ∩ (f)K 6= ∅. Thus we apply Proposition 2.4.4 to get the result.
32
We immediately have the following corollary.
Corollary 2.4.8 Let p be prime and Z(p) the integers localized at (p). Then
Mk(Γn)Z(p)
∼= Mk(Γn)Z ⊗Z Z(p).
Proof Set K = Q and apply Corollary 2.4.7. �
Corollary 2.4.9 Let Op be the localization of OK at the prime ideal p ⊆ OK . Then
Mk(Γn)Op∼= Mk(Γn)Z(p)
⊗Z(p)Op,
where (p) = Z ∩ p.
Proof Recall for rings R ⊂ S that S ∼= R ⊗R S. Since (p) ⊂ p and Z ⊂ OK , we have Z(p) ⊂ Op.
Using this observe that
Mk(Γn)Op∼= Mk(Γn)OK
⊗OKOp by Corollary 2.4.7
∼=(Mk(Γn)Z ⊗Z OK
)⊗OK
Op by Lemma 2.4.5
∼= Mk(Γn)Z ⊗Z Op
∼= Mk(Γn)Z ⊗Z(Z(p) ⊗Z(p)
Op)
∼= Mk(Γn)Z(p)⊗Z(p)
Op by Corollary 2.4.8. �
2.5 The Hecke Algebra
The natural question to ask now is whether or not the operators Φn−r and [·]nr preserve any
special properties of modular forms. To answer that question we need to introduce the following.
Definition 2.5.1 For N ∈ Z>0, the subgroup
Γn(N) :=
A B
C B
∈ Γn : A ≡ D ≡ In (mod N), B ≡ C ≡ 0n (mod N)
of Γn is called the principle subgroup of level N .
33
Observe that Γn(N) is the kernel of the surjective group homomorphism
Γn = Sp2n(Z) −→ Sp2n(Z/NZ)A B
C D
7−→
A B
C D
(mod N).
Thus we have Γn(N) / Γn as well as
[Γn : Γn(N)] = #Sp2n(Z/NZ) <∞.
Definition 2.5.2 A subgroup Γ ⊆ Γn such that Γn(N) ⊂ Γ ⊂ Γn is called a congruence subgroup of level N .
Technically, if N is a multiple of N then we have Γn(N) ⊂ Γn(N) and we have any congru-
ence subgroup of level N is also a congruence subgroup N . To remove any ambiguity, we will refer
to Γ as a congruence subgroup of level N if N is minimal with respect to Γn(N) ⊂ Γ ⊂ Γn. We
further note that Γn(1) = Γn.
Since the principle congruence subgroups have finite index in Γn, it is easy to see that any
congruence subgroup also has finite index. To see this observe that if Γn(N) ⊆ Γ ⊂ Γn then
[Γn : Γ] ≤ [Γn : Γ][Γ : Γn(N)] = [Γn : Γn(N)] <∞.
These definitions and observations are necessary to show that we can write
ΓngΓn =∐i
Γngi, g, gi ∈ GSp+2n(Q).
We start by demonstrating that ΓngΓn can be expressed as such a union. To do so, observe
that ΓngΓn = {γ1gγ2 : γ1, γ2 ∈ Γn} and consequently we have the orbit space Γn\ΓngΓn = {Γngγ :
γ ∈ Γn}. Now suppose that we have γ1, γ2 such that Γngγ1∩Γngγ2 6= ∅. There must exist h1, h2 ∈ Γn
such that h1gγ1 = h2gγ2. This is equivalent to γ1 = g−1h−11 h2gγ2 and thus
Γngγ1 = Γngg−1h−11 h2gγ2
= Γnh−11 h2gγ2
= Γngγ2.
34
Therefore we get the union is disjoint by taking distinct orbits Γngi = Γngγi. To see that the union
is finite we use the notation Γ = g−1Γng ∩ Γn and utilize analogues to the techniques used to get
Lemmas 5.1.1 and 5.1.2, respectively, of [5].
Lemma 2.5.3 Let g ∈ GSp+2n(Q), then Γ is a congruence subgroup.
Proof Since g is a rational matrix, there exists N ∈ Z>0 such that Ng, Ng−1 ∈ Mat2n(Z). Note,
by definition, we have Γn(N) ⊂ Γn. Set N = N3 and observe that
gΓn(N)g−1 ⊂ g (I2n +N ·Mat2n(Z)) g−1
= g(I2n + N3 ·Mat2n(Z)
)g−1
= I2n + (Ng)(N ·Mat2n(Z)
)(Ng−1)
⊂ I2n + N ·Mat2n(Z).
Furthermore, it is clear that every element of gΓn(N)g−1 has determinant equal to 1. Combining
results we have gΓn(N)g−1 ⊂ Γn(N) and, equivalently,
Γn(N) ⊂ g−1Γn(N)g ⊂ g−1Γng.
Since, by definition, Γn(N) ⊂ Γn we therefore have
Γn(N) ⊂ g−1Γng ∩ Γn = Γ
and conclude that Γ is a congruence subgroup. �
Lemma 2.5.4 The set {γi} is a set of coset represeentatives of Γ\Γn if and only if {gi} = {gγi} is
a set of orbit representatives for Γn\ΓngΓn.
Proof Consider the map of sets
Γn −→ Γn\ΓngΓn
γ 7−→ Γngγ.
35
This map is surjective since for any Γnα ∈ Γn\ΓngΓn there exist γ1, γ2 ∈ Γn such that
Γnα = Γnγ1gγ2 = Γngγ2
which is the image of γ2. Now suppose Γngγ1 = Γngγ2; then we must have γ1γ−12 ∈ g−1Γng. We
also have by definition that γ1γ−12 ∈ Γn. Consequently γ1γ
−12 ∈ g−1Γng ∩ Γn = Γ, that is, two
elements of Γn produce the same orbit in Γn\ΓngΓn if they differ by an element of Γ. Thus the full
preimage of Γngγ is the coset Γγ, demonstrating that there is a bijection between the coset space
and orbit space. Consequently {Γγi} is a complete set of cosets if and only if {Γngγi} = {Γngi} is
a complete set of orbits, therefore demonstrating the desired result. �
Since Γ is a congruence subgroup of Γn, as demonstrated by Lemma 2.5.3, we have the set
of coset representatives, {γi}, of Γ\Γn is finite. Therefore, by Lemma 2.5.4, we have {gi} = {gγi}
is a finite set of orbit respresentatives for Γn\ΓngΓn. We then conclude that ΓngΓn =∐i
Γngi is a
finite disjoint union.
Definition 2.5.5 For g ∈ GSp+2n(Q), we define the weight k, ΓngΓn operator by
f [ΓngΓn]k :=∑i
f |kgi
where the gi are defined by ΓngΓn =∐i
Γngi.
The above is an example of a Hecke operator. More generally, a Hecke operator is any
element of the Hecke algebra.
Definition 2.5.6 The Hecke algebra, Tn, is the set of finite formal sums of double cosets ΓngΓn,
g ∈ GSp+2n(Q) with coefficients in C. That is
Tn =
m∑j=1
cjΓngjΓ
n : cj ∈ C, gj ∈ GSp+2n(Q)
.
We will write T for Tn except for the cases when we are moving from the space of modular
forms with degree n to degree r; in those cases we will use the exponent notation to be clear as to
which space we are working with. Further note that the above definition gives T as a C-module.
36
Given two double cosets
ΓngΓn =
ν(g)∐i=1
Γngi and Γng′Γn =
ν(g′)∐j=1
Γng′j
where ν(g) = |Γn\ΓngΓn| = [Γn : Γ], we have
(ΓngΓn) · (Γng′Γn) =∑
ΓnhΓn⊂ΓngΓng′Γn
c(g, g′;h)ΓnhΓn
where c(g, g′;h) is the number of pairs gi, g′j such that gig
′j ∈ Γnh. In Lemma 3.1.5 of [1], Andri-
anov proves this formula comes from wanting multiplication of double cosets to correspond with
composition of Hecke operators, that is, the desire to have
(f [ΓngΓn]k)[Γng′Γn]k = f [(ΓngΓn) · (Γng′Γn)]k.
Since we have c(g, g′;h) ∈ Z for all g, g′, h it is easy to see that we can form an R-algebra for
Z ⊆ R ⊆ C, a subring, by defining
T(R) :=
m∑j=1
aj [ΓngjΓn]k : m ∈ Z≥0, aj ∈ R, gj ∈ GSp+
2n(Q)
.
From this definition, is easy to see that taking R = Q and considering the map
T −→ T(Q)⊗Q Cm∑j=1
aj [ΓngjΓn] 7−→
m∑j=1
([ΓngjΓn]⊗ aj)
yields an isomorphism as C-algebras. Moving forward, we will use T ∈ T to denote an arbitrary
Hecke operator. That is, for T ∈ T(R) we have
T =∑j=1
aj [ΓngjΓn]k
37
with aj ∈ R. For T acting on a modular form, f , we write T f where
T f := f∑j=1
aj [ΓngjΓn]k :=
∑j=1
ajf [ΓngjΓn]k .
Since T is a linear operator, it is natural to ask if it has any eigenvectors; that is, are there any
f ∈ Mk(Γn) such that T f = λ(T ; f)f for some λ(T ; f) ∈ C? The short answer is yes, and of
particular interest are eigen modular forms; modular forms satisfying T f = λ(T ; f)f for all T ∈ T.
Most of the results on Hecke operators will be shown on a single arbitrary coset with the
general result being obtained by extending it linearly. Since we take g ∈ GSp+2n(Q) there exists
a ∈ Z>0 such that ag ∈ GSp+2n(Q)∩Mat2n(Z), therefore if we are working in T(R) with R ⊃ Q then
we can write ΓngΓn = 1aΓn(ag)Γn. This is desirable due to the following proposition.
Proposition 2.5.7 ([1], Lemma 3.3.6) Let γ ∈ GSp+2n(Q) ∩ Mat2n(Z), then the double coset
ΓnγΓn has a unique representative of the form
g = diag(a1, . . . , an, d1, . . . , dn)
with aj , dj ∈ Z satisfying aj > 0, ajdj = µ(γ) for all j. Furthermore an|dn, aj |aj+1 for j =
1, . . . , n− 1.
Proof Andrianov proves the result by induction on the degree n. For the case n = 1 we have
Γ1 = SL2(Z) and GSp+2 (Q) = GL+
2 (Q). Let γ =
a b
c d
∈ GSp+2 (Q) ∩Mat2(Z). We then have
µ(γ) = ad− bc. Using lemma 3.2.2 of [1] we have
Γ2γΓ2 = SL2(Z)γSL2(Z) = SL2(Z)
e1 0
0 e2
SL2(Z)
with e|e2 and note that this is the just the matrix of elementary divisors of γ over Z. Setting a1 = e1
and d1 = e2 we get a1|d1 and a1d1 = ad− bc = µ(γ), thus showing the base case.
We now assume proposition 2.5.7 holds for all degrees n < m and consider the case n = m.
38
To utilize the induction hypothesis we will show that
ΓnγΓn = Γn
a1 01,n−1 0 01,n−1
0n−1,1 A4 0n−1,1 B4
0 01,n−1 d1 01,n−1
0n−1,1 C4 0n−1,1 D4
Γn
where
A4 B4
C4 D4
∈ GSp+2(n−1)(Q) ∩Mat2(n−1)(Z). To do this, we first assume that γ is primitive
since if it is not that we set a1 = gcd(γ), i.e., a1 is the greatest common divisor of the entries of γ
and instead would work with γ′ where γ = a1γ′.
Let δ = δ(γ) be the minimum of the greatest common divisors coming from the columns of
γ. Let i denote the column corresponding to δ and note that we can assume i ≤ n since we would
otherwise consider γJ2n. We can further assume that i = 1 since we can use a permutation matrix
V ∈ SLn(Z) to interchange the 1st and ith columns via γ
V 0
0 tV −1
. Now that we have the first
column of γ produces the values δ, we assume δ = 1. In this case, we use Lemma 1.3.9 of [1] to pick
a representative of Γnγ such that the first column is given by t
(1 0 . . . 0
)∈ R2n. Using the
block form, we may now assume that we have
γ
1 A2 B1 B2
0n−1,1 A4 B3 B4
0 C2 D1 D2
0n−1,1 C4 D3 D4
and observe that for W =
1 −A2
0n−1,1 In−1
we have
γ
W 0n
0ntW−1
=
a1 01,n−1 B1 B2
0n−1,1 A4 B3 B4
0 C3 D1 D2
0n−1,1 C4 D3 D4
.
39
It is important to note that the subblocks forming the blocks B,C,D are not necessarily the same
as the ones we start with. Now assuming that we chose γ to be of the form we just obtained, we
consider the product
γ
In S
0n In
where S =
−B1 −B2
−tB2 0n−1
.
This produces a matrix of the form
1 01,n−1 0 01,n−1
0n−1,1 A4 B3 B4
0 C2 D1 D2
0n−1,1 C4 D3 D4
.
Using the relation AtB = BtA produces B3 = 0n−1,1. Similarly, tAC = tCA produces C2 = 01,n−1.
Combing these facts with the formulas AtD − BtC = µ(γ)In and tAD − tCB = µ(γ)In produces
D1 = µ(γ), D2 = 01,n−1, D3 = 0n−1,1, and
A4 B4
C4 D4
∈ Γn−1. Thus, our induction hypothesis on
the degree n would complete the proof for δ = 1. We now consider the case δ > 1. In this case, we
can follow the same construction up to a representative of the form
δ A2 B1 B2
0n−1,1 A4 B3 B4
0 C2 D1 D2
0n−1,1 C4 D3 D4
.
We then use right multiplication by the matrices U
1 −A2
0n−1,1 In−1
, T
0 −B2
−tB2 0n−1
, and T
−B1 01,n−1
01,n−1 0n−1
where the entries of A2− δA2, B2− δB2, and B1− δB1 are all in the set {1, 2, . . . , δ}. It should also
be noted that the B1 reference is the B1 block of
γU
1 −A2
0n−1,1 In−1
T
0 −B2
−tB2 0n−1
,
i.e., it is necessary to reduce A2 and B2 before B1.
40
We denote this new reduced matrix by γ0 and observe that since the the first entry of every
column is less than or equal to δ that we must have δ(γ0) ≤ δ(γ) = δ. If δ(γ0) = δ(γ) = δ, then
we must have δ divides all the entries of γ; however, this is a contradiction since we assumed γ is
primitive and δ > 1. We must then take 1 ≤ δ(γ0) < δ and we repeat the process on γ0. We then
iterate the process until δ(γ0) = and we apply the earlier case of δ = 1; that is, we now can pick a
representative for γ satisfying
ΓnγΓn = Γn
1 01,n−1 0 01,n−1
0n−1,1 A4 0n−1,1 B4
0 01,n−1 µ(γ) 01,n−1
0n−1,1 C4 0n−1,1 D4
Γn,
with γ∗ =
A4 B4
C4 D4
∈ GSp+2(n−1)(Q)∩Mat2(n−1)(Z) with µ(γ∗) = µ(γ). Our induction hypothesis
then gives the unique representative diag(a2, . . . , an, d2, . . . , dm) of the double coset Γn−1γ∗Γn−1
satisfying ai|ai+1, an|dn, and aidi = µ(γ∗) = µ(γ). Moreover, there exist ξ =
ξ1 ξ2
ξ3 ξ4
and
ξ′ =
ξ′1 ξ′2
ξ′3 ξ′4
such that ξγ∗ξ′ = diag(a2, . . . , an, d2, . . . , dm). We then have
1 01,n−1 0 01,n−1
0n−1,1 ξ1 0n−1,1 ξ2
0 01,n−1 µ(γ) 01,n−1
0n−1,1 ξ3 0n−1,1 ξ4
1 01,n−1 0 01,n−1
0n−1,1 A4 0n−1,1 B4
0 01,n−1 µ(γ) 01,n−1
0n−1,1 C4 0n−1,1 D4
1 01,n−1 0 01,n−1
0n−1,1 ξ′1 0n−1,1 ξ′2
0 01,n−1 µ(γ) 01,n−1
0n−1,1 ξ′3 0n−1,1 ξ′4
=
1 01,n−1 0 01,n−1
0n−1,1 diag(a2, . . . , an) 0n−1,1 0n−1
0 01,n−1 µ(γ) 01,n−1
0n−1,1 0n−1 0n−1,1 diag(d2, . . . , dn)
.
Thus, by induction we have the result for primitive γ.
Recall that if γ is not primitive then we take a1 to be the greatest common divisor of the
41
entries of γ. We then have γ = a1γ′ where γ′ is primitive and thus Γnγ′Γn has a unique representative
given by diag(1, a′2, . . . , a′n, µ(γ′), d′2, . . . , d
′n). We observe that µ(γ) = µ(a1γ
′) = a21µ(γ′). Combing
this with ΓnγΓn = a1Γnγ′Γn, we have the unique representative
diag(a1, a1a′2, . . . , a1a
′n, a1µ(γ′), a1d
′2, . . . , a1d
′n) = diag(a1, . . . , an, d1, . . . , dn).
�
We will always choose g in ΓngΓn as the unique diagonal matrix described above. Doing
so shows that ΓngΓn = Γng∨Γn where g∨ = µ(g)g−1. This comes from combining the identity
tgJg = µ(g)J with the fact that we pick g to be diagonal and so
g = tg = µ(g)Jg−1J−1 = Jg∨J−1 ∈ Γng∨Γn.
More generally we have g∨ = J tgJ−1 and (gγ)∨ = γ∨g∨. Using these relations we obtain the
following.
∐Γngi = ΓngΓn
= Γng∨Γn
= (ΓngΓn)∨
= Jt(ΓngΓn)J−1
= JΓntgΓnJ−1
=∐
JΓngiΓ−1
=∐
(Γngi)∨
=∐
g∨i (Γn)∨
=∐
g∨i Γn.
Since∐
Γngi = ΓngΓn = Γng∨Γn =∐
g∨j Γn, we have
∑f |kgi =
∑f |kg∨j . (2.1)
42
Recall that the summation is finite. It is useful to see the action of the slash operator
associated with g∨i . The action is given by
(f |kg∨i )(Z) = j(g∨i , Z)−kµ(g∨i )nk−n(n+1)
2 f(g∨i Z)
= µ(gi)−nkj(g−1
i , Z)−kµ(gi)nk−n(n+1)
2 f(µ(gi)In · (g−1i Z))
= j(g−1i , Z)−kµ(gi)
−n(n+1)2 f(g−1
i Z). (2.2)
We could further analyze this function; however, the current formulation is sufficient for showing
that the Hecke operators are hermitian with respect to the Petersson inner product. In showing this,
we will make the change of variable ωi = giZ when necessary and will write Y ∗i = Im(giZ) = Im(ωi).
Note that
〈f1 [ΓngΓn] , f2〉 =⟨∑
f1|kgi, f2
⟩=
∑〈f1|kgi, f2〉
=∑ ∫
Γn\Hn
det(Y )k (f1|kgi) (Z)f2(Z)dνn(Z)
=∑ ∫
Γn\Hn
det(Y )kµ(gi)nk−n(n+1)
2 j(gi, Z)−kf1(giZ)f2(Z)dνn(Z)
=∑ ∫
Γn\Hn
j(gi, Z)k det(Y ∗i )kµ(gi)−n(n+1)
2 f1(giZ)f2(Z)dνn(Z)
=∑ ∫
Γn\giHn
j(gi, g−1i ωi)
k det(Y ∗i )kµ(gi)−n(n+1)
2 f1(ωi)f2(g−1i ωi)dνn(g−1
i ωi)
=∑ ∫
Γn\Hn
det(Y ∗i )kf1(ωi)µ(gi)−n(n+1)
2 j(g−1i , ωi)−kf2(g−1
i ωi)dνn(ωi)
=∑ ∫
Γn\Hn
det(Y ∗i )kf1(ωi)(f2|kg∨i ) (ωi)dνn(ωi) by (2.2)
=∑〈f1, f2|kg∨i 〉
=⟨f1,∑
f2|kg∨i⟩
= 〈f1, f2 [ΓngΓn]〉 due to (2.1).
43
To obtain the fifth equality we use the identity
t(CiZ +Di)Y
∗i (CiZ +Di) = µ(gi)Y.
Taking the derivative of both sides then produces
det(Y )k = µ(gi)−nk det(Y ∗)kj(gi, Z)kj(gi, Z)k.
To obtain the seventh equality we use the identity
j(γδ, Z) = j(γ, δZ)j(δ, Z)
which allows us to write the following.
j(gi, g−1i ωi)
k = j(gi, g−1i ωi)
k
=j(gig
−1i , ωi)
k
j(g−1i , ωi)k
=j(I2n, ωi)
k
j(g−1i , ωi)k
= j(g−1i , ωi)
−k.
Thus we see that any Hecke operator represented by a single coset is hermitian. The result extends
to all Hecke opertors due to linearity.
Since the Hecke operators are Hermitian with respect to the Petersson inner product on
cusp forms, a finite dimensional C-vector space, we know from linear algebra that Sk(Γn) then has
a basis of orthogonal eigenforms.
Another reason to choose g to be diagonal, is we can then write it as a finite product of
diagonal matrices,∏p gp, where each gp has diagonal entries restricted to powers of p. Combining
this with the following proposition allows us to write
ΓngΓn =∏p
ΓngpΓn.
44
Proposition 2.5.8 ([1], Proposition 3.3.9) Let M , M ′ ∈ Γn. Suppose that the quotients
gcd
(a1(M)
d1(M),a1(M ′)
d1(M ′)
)= 1,
then
(ΓnMΓn) (ΓnM ′Γn) = ΓnMM ′Γn.
To clarify, the notation a1(M) and d1(M) reference, respectively, the first and the (n+ 1)st
diagonal entries of the unique representative of the double coset ΓnMΓn from Proposition 2.5.7. It
is also important to note that Andrianov proves this for arbitrary level, Γn(N), and in that case the
double cosets are of the form Γn0 (N)MΓn0 (N) where
Γn0 (N) =
A B
C D
∈ Γn : C ≡ 0 (mod N)
⊇ Γn(N).
We note that for full level we have Γn0 (1) = Γn = Γn(1). We can apply this proposition to our case
since the entries of gp and gp′ are coprime for p 6= p′. We can use this relation to identify
ΓngΓn =∏p
ΓngpΓn ←→
⊗p prime
ΓngpΓn
where we gp is defined as the gp in the finite product for p|µ(g) and I2n for p - µ(g). Through this
mapping we obtain the decomposition
T(Q) =⊗p
Tp,
where Tp is the set of formal sums with coefficients in Q of double cosets ΓngpΓn satisfying µ(gp) = pn
for some n ∈ Z. By Theorem 3.3.23 of [1] we know that Tp is generated by elements of the form
45
T (p) = T (n)(p) = Γn
In 0
0 pIn
Γn,
Ti(p2) = T
(n)i = Γn
In−i 0 0 0
0 pIi 0 0
0 0 p2In−i 0
0 0 0 pIi
Γn
for i = 1, . . . , n, and T(n)n (p2)−1 = [ΓnpI2nΓn]
−1which we show is in T by showing it is in T(Q).
Working over Q we see that
ΓnI2nΓn = Γn(pI2n)(p−1I2n)Γn
= ΓnpI2nΓnΓnp−1I2nΓn
= Tn(p2)Γnp−1I2nΓn.
We are able to split the double coset by Proposition 2.5.8 since
gcd
(a1(pI2n)
d1(pI2n)
a1(p−1I2n)
d1(p−1I2n)
)= gcd
(p
p,p−1
p−1
)= gcd(1, 1) = 1.
This shows that over Q we have
Tnn (p2)−1 = Γnp−1I2nΓn = p−1ΓnI2nΓn ∈ T(Q).
We now consider Theorem 2 of [26] which tells us that there exists an epimorphism from
C[T (p), T(n)1 (p2), . . . , T
(n)n (p2)] → C[T (p), T
(n−1)1 (p2), . . . , T
(n−1)n−1 (p2)] given by T 7→ T ∗ where T is
Hecke operator satisfying ΦT = T ∗Φ. Taking [ΓngpΓn] ∈ Tp we then have
[ΓngpΓn] = p−` [ΓngpΓ
n]
where p` is the largest power of p occuring in the denominator of the entries of gp and gp = p`gp. If
46
gp = I2n then we have
Φ(f [ΓngpΓ
n]k)
= Φ(p−`f [ΓnI2nΓn]k
)= p−`Φ (f [ΓnI2nΓn]k) = p−` (Φ(f))
[Γn−1I2(n−1)Γ
n−1]k
since Φ is C-linear, Mk(Γn) is invariant under [ΓnI2nΓn]k, and Mk(Γn−1) is invariant under[Γn−1I2(n−1)Γ
n−1]k. If gp 6= I2n then it is of a form on which we can apply Theorem 3.3.23 of [1]
to get that it is generated by elements of the form T (p) and T(n)i (p). Thus the linearity of Φ gives
the result for single cosets as well as extends to formal sums of cosets, thereby giving the following
proposition.
Proposition 2.5.9 There exists a mapping T 7→ T ∗ of T(n)p to T(n−1)
p with the following properties.
• Φ(T f) = T ∗Φ(f) for all f ∈Mk(Γn) and T ∈ T(n)p ,
• the mapping is an epimorphism.
Since for a general diagonal matrix g as described in Proposition 2.5.7 we have the the
decomposition ΓngΓn =∏p ΓngpΓ
n, we also have the expression
f [ΓngΓn]k = f
[∏p
ΓngpΓn
]k
=((f [Γngp1Γn]k
)· · ·)
[ΓngpmΓn]k .
We only concern ourselves with the full set of distinct primes, p1, . . . , pm, occuring in the entries of
g since for p not in that list gp = I2n and acts trivially on the modular form. Using Proposition
2.5.9, let T ∗p be the element of T(n−1)p such that
Φ(f [ΓngpΓ
n]k)
= T ∗p (Φ(f)) .
Using this we have
Φ (f [ΓngΓn]k) = Φ(((
f [Γngp1Γn]k)· · ·)
[ΓngpmΓn]k)
= T ∗pm(Φ(((
f [Γngp1Γn]k)· · ·) [
Γngpm−1Γn]k
))...
= T ∗pm(· · ·(T ∗p1 (Φ(f))
))= (T ∗pm · · · T
∗p1) (Φ(f)) .
47
Since Φ is a linear operator this result extends linearly to all of T(n)(Q). Furthermore, since the ring
we are looking at the algebra over only determines the coefficients of the formal sums, the linearity
of Φ shows that the result holds for T(n)(R) for Q ⊆ R ⊆ C. That is to say, for all T ∈ T(n)(R)
there exists T ∗ ∈ T(n−1)(R) such that Φ(T f) = T ∗(Φ(f)). We can extend this result via induction
to obtain Φ`(T f) = T ∗(Φ`(f)) for ` ≥ 0 where T ∈ T(n)(R) and T ∗ ∈ T(`)(R). This is necessary to
prove the following theorem.
Theorem 2.5.10 ([12],Theorem 1) Let f ∈ Mk(Γr) be an eigenform for some fixed r ≥ 0 and
even k > n+ r + 1 with n ≥ r. Then [f ]nr ∈Mk(Γn) is also an eigenform.
Proof We first claim that En,j,k is stable under T(n). Observe that if g ∈ Ej,j,k, then f ∈ Sk(Γj).
Since Sk(Γj) has a basis of eigenforms. Expressing f as linear combination of the basis shows that
T f ∈ Sk(Γj) for all T ∈ T(r). We now suppose for induction that the result hold for f ∈ Er,j,k for
r < n < 12k and consider the case f ∈ En,j,k.
Consider T T(n). We then have T f = [gj ]nj + f0 for some gj ∈ Sk(Γj) and f0 orthogonal to
En,j,k. We then consider Φ(T f) = [gj ]n−1j + Φ(f0) and note that it is equal to
Φ(T f) = T ∗Φ(f) = [hj ]n−1j
since Φ(f) ∈ En−1,j,k which is stable under T(n−1) by our induction hypothesis. Combining the
equations yields 0 = [gj ]n−1j − [hj ]
n−1j + Φ(f0) = [gj − hj ]n−1
j + Φ(f0). Since f0 is orthogonal to
En,j,k, we have Φ(f0) is orthogonal to En−1,j,k. Thus we must have gj = hj and Φ(f0) = 0. Since we
now have f0 ∈ Sk(Γn), it suffices to also show that f0 ∈ Sk(Γn)⊥ to obtain En,j,k is stable under
T(n). To see this, consider any ψ ∈ Sk(Γn) and observe that 〈T f, ψ〉 = 〈f, T ψ〉 = 0 since Hecke
operators are hermitian, Sk(Γn) is stable under T(n), and En,j,k is orthogonal to Sk(Γn). We then
use this to obtain
0 = 〈T f, ψ〉 = 〈[gj ]nj , ψ〉+ 〈f0, ψ〉 = 〈f0, ψ〉
for all ψ ∈ Sk(Γn) since 〈[gj ]nj , ψ〉 = 0. Thus, f0 ∈ Sk(Γn)⊥ but since it is also a cusp form we must
have f0 = 0 and T f = [gj ]nj ∈ En,j,k. Thus by induction, we have En,j,k is stable under T(n) for all
j ≤ n < 12k.
We now consider f ∈ Mk(Γr) satisfying the assumptions of the theorem. We then have
f =∑rj=0[fj ]
rj and consequently [f ]nr =
∑rj=0[fj ]
nj . Since En,j,k is stable under T(n), for each j
48
there exists some gj satisfying T [fj ]nj = [gj ]
nj , yielding
T [f ]nr =
r∑j=0
[gj ]nj and Φn−r(T [f ]nr ) =
r∑j=0
[gj ]rj .
We also have
Φn−r(T [f ]nr ) = T ∗Φn−r([f ]nr ), T ∗ ∈ T(r)
= T ∗f
= λ(T ∗; f)f
= λ(T ∗; f)r∑j=0
[fj ]rj
=
r∑j=0
[λ(T ∗; f)fj ]rj .
Combining these expressions and using the linearity of [·]rj yields
0 =
r∑j=0
[gj − λ(T ∗; f)fj ]rj .
Applying Φr to both sides of the expression yields 0 = gr − λ(T ∗; f)fr, that is, gr =
λ(T ∗; f)fr and 0 =∑r−1j=0[gj − λ(T ∗; f)fj ]
rj . Applying Φr−1 then produces gr−1 = λ(T ∗; f)fr−1.
Continuing this process, we obtain gj = λ(T ∗; f)fj and consquently
T [f ]nr =
r∑j=0
[gj ]nj =
r∑j=0
[λ(T ∗; f)fj ]nj = λ(T ∗; f)
r∑j=0
[fj ]nj = λ(T ∗; f)[f ]nr .
Therefore [f ]nr is an eigenform with λ(T ; [f ]nr ) = λ(T ∗; f) for every T ∈ T(n). �
To make use of this result we need the following theorem.
Theorem 2.5.11 ([13], Theorem 1) Let f ∈ Sk(Γn) be an eigenform. The field Kf = Q({λ(T ; f) :
T f = λ(T ; f)f, T ∈ T(n)}) is a totally real number field.
Proof Let T(n) = HomC(T(n),C). For λ ∈ T(n) we define
Mk(Γn;λ) = {f ∈Mk(Γn) : T f = λ(T ; f)f}
49
and
m(λ) = dimC (Mk(Γn;λ)) ≥ 0.
Let Λ(T) = {λ ∈ T : m(λ) ≥ 1}. Then we have M = #Λ(T) ≤ dimC (Mk(Γn)). Write Q(λ) =
Q(λ(T(Q))). Now take σ ∈ Aut(C) and for every g =∑T≥0 a(T ; g)qT ∈ Mk(Γn) set σ(g) =∑
T≥0 σ(a(T ; g))qT . We know from Section 2.4 that Mk(Γn)Q ⊗Q C ∼= Mk(Γn), thus we identify
g ∈Mk(Γn) with∑i gi ⊗ ci ∈Mk(Γn)Q ⊗Q C. Since Q is fixed by every field automorphism of C
we have
σ(g) = σ(∑
gi ⊗ ci)
with ci ∈ C, gi ∈Mk(Γn)Q
=∑
σ(gi)⊗ σ(ci)
=∑
gi ⊗ σ(ci) ∈Mk(Γn)Q ⊗Q C ∼= Mk(Γn).
For λ ∈ T we also define σ(λ)(T ) = σ(λ(T )) for T ⊗ 1 ∈ T(Q)⊗Q C ∼= T. It is also known
from Equation 2.4 of [27] that Mk(Γn)Q is T(Q)-stable. Consequently, for g ∈Mk(Γn;λ) we have
T σ(g) = T σ(∑
gi ⊗ ci)
with ci ∈ C, gi ∈Mk(Γn)Q
= T(∑
gi ⊗ σ(ci))
=∑T gi ⊗ σ(ci)
= σ(∑
T gi ⊗ ci)
= σ(T∑
gi ⊗ ci)
= σ(T g)
= σ(λ(T ; g)g)
= σ(λ)(T )σ(g).
We see that if g ∈ Mk(Γn;λ) then σ(g) ∈ Mk(Γn;σ(λ)). Thus, for any σ ∈ Aut(C)
and λ ∈ Λ(T) then σ(λ) ∈ Λ(T) and σ(Q(λ)) = Q(σ(λ)). Since the automorphisms of C act as
permutations on the elements of Λ(T), which is finite, we see that Q(Λ(T))/Q is Galois, where
Q(Λ(T)) is the composite field. Furthermore, EndQ(Q(λ)) ⊆ Gal(Q(Λ(T))/Q) ⊂ Sym(m), yielding
#EndQ(Q(λ)) ≤ #Gal(Q(Λ(T))/Q) ≤M !
50
showing that Q(λ) is a finite extension.
To get Q(λ) ⊆ R we recall that all T ∈ T are hermitian with respect to the Petersson inner
produce and thus have only real eigenvalues.
Lastly, for f ∈ Sk(Γn) an eigenform we have λ(T ; f) ∈ Λ(T) and thus have Kf :=
Q({λ(T ; f)}) is a totally real number field. �
We conclude that by stating that it is known by Theorem A(n,r) (found in Appendix A) of
[18] that if f ∈ Sk(Γr) is an eigenform then [f ]nr ∈Mk(Γn)Kf.
51
Chapter 3
Results of T. Kikuta and S.
Takemori
3.1 Definitions, Remarks, and Main Results
Let K be an algebraic number field and O = OK its ring of integers. For a prime ideal
p ⊂ O, let Op be the localization of O at p.
Definition 3.1.1 Let F ∈Mk(Γn)Op . We say F is a mod pm cusp form if Φ(F ) ≡ 0 mod pm.
Main Theorem There exists a finite set Sn(K) of prime ideals in K depending on n such that
the following holds. For a prime ideal p of O not contained in Sn(K) and a mod pm cusp form
F ∈Mk(Γn)Opwith k > 2n, there exists G ∈ Sk(Γn)Op
such that F ≡ G mod pm.
Note: The set Sn(K) is defined at the end of Section 3.2. It is presented along with proofs
of some of its properties. We also note that the theorem is trivial for odd weight k because there
are no non-cusp forms of odd weight.
Letting νp be the normalized additive valuation with respect to p, i.e., νp(p) = 1. For
F ∈Mk(Γn)K we define the following;
νp(F ) := min{νp(a(T ;F )) : T ∈ Λn}
ν(n′)p (F ) := min{νp(a(T ;F )) : T ∈ Λn, rank(T ) = n′} for 0 ≤ n′ ≤ n.
52
Main Corollary Let k > 2n be even and f ∈ Sk(Γr)Kf, n > r, a Hecke eigenform. For the
Klingen-Eisenstein series [f ]nr attached to f , suppose there exists a prime ideal p in OKfwith p 6∈
Sn(Kf ) such that ν(n)p ([f ]nr ) = νp(Φ([f ]nr ))−m, for some m ∈ Z≥1. Then there exists F ∈ Sk(Γn)Op
such that α[f ]nr ≡ F mod pm for some 0 6= α ∈ pm.
3.2 Lemmas and Their Proofs
By Theorem 2.3 of [6, pp. 150] we know that⊕1≤k
Mk(Γn)Z is finitely generated; that is,
there exists a set of generators {ψ1, · · · , ψt} such that
⊕k≥1
Mk(Γn)Z ∼= Z[ψ1, · · · , ψt]
as Z-algebras. By Corollary 2.4.7, we have that Mk(Γn)Z(p)
∼= Mk(Γn)Z ⊗Z Z(p) for any prime p.
Combining these facts we get
⊕k≥1
Mk(Γn)Z(p)=
⊕k≥1
(Mk(Γn)Z ⊗Z Z(p)
)
=
⊕k≥1
Mk(Γn)Z
⊗Z Z(p)
= Z[ψ1, · · · , ψt]⊗Z Z(p)
= Z(p)[ψ1, · · · , ψt].
This allows us to introduce the following lemma.
Lemma 3.2.1 Let M be a natural number. For any prime p, the algebra⊕k>M
Mk(Γn)Z(p)is finitely
generated over Z(p).
Proof Let {ψ1, · · · , ψt} be the set of generators of the Z(p)-algebra⊕
kMk(Γn)Z(p), where ψi ∈
Mki(Γn)Z(p)
. For each i, let αi ∈ Z≥0 be the smallest integer such that the weight of ψαii , αiki, is
strictly greater than M . We prove that⊕k>M
Mk(Γn)Z(p)is generated over Z(p) by monomials of the
form
ψα11 , · · · , ψαt
t , (3.1)
ψi11 · · ·ψitt (i1k1 + · · ·+ itkt > M, 0 ≤ ij < 2αj). (3.2)
53
Note that any F ∈ Mk(Γn)Z(p)can be written as a linear combination of monomials
ψa11 · · ·ψatt with i1k1 + · · ·+ itkt = k. Thus it suffices to show that F = ψa11 · · ·ψ
att can be generated
by monomials of the form (3.1) and (3.2) for k > M .
For each ai we can write ai = qiαi + ri for qi, ri ∈ Z with 0 ≤ ri < αi. Thus we write
F = ψa11 · · ·ψatt = (ψr11 · · ·ψ
rtt )
t∏i=1
(ψαii )
qi .
If r1k1 + · · · rtkt > M then we are done, so we suppose the contrary. Then there must be some
qj ≥ 1, otherwise we would have
a1k1 + · · · atkt = r1k1 + · · · rtkt ≤M
contradicting F ∈⊕k>M
Mk(Γn)Z(p). Thus we rewrite our expression of F as
F = (ψr11 · · ·ψrj−1
j−1 ψrj+αj
j ψrj+1
j+1 · · ·ψrtt )(ψαj
j
)qj−1∏i 6=j
(ψαii )
qi .
It is clear that(ψαj
j
)qj−1∏i 6=j (ψαi
i )qi is generated by monomials of the form (3.1), so it remains to
show that ψr11 · · ·ψrj−1
j−1 ψrj+αj
j ψrj+1
j+1 · · ·ψrtt is of the form (3.2).
First, each ri 6= rj satisfies 0 ≤ ri < αi < 2αi. For rj we have 0 ≤ rj +αj < αj +αj = 2αj .
Thus each individual exponent satisfies its respective bounds. Now observe that the weight of the
monomial is given by
(rj + αj)kj +∑i 6=j
riki = αjkj +
t∑i=0
riki
≥ αjkj
> M
by the definition of αj . Thus ψr11 · · ·ψrj−1
j−1 ψrj+αj
j ψrj+1
j+1 · · ·ψrtt is of the form (3.2) and consequently
F is generated by a finite product of monomials of the form of (3.1) and (3.2). �
Lemma 3.2.2 Let M be a natural number. Assume that⊕k>M
Mk(Γn)Z(p)= Z(p)[f1, · · · , fs] with
fi ∈Mki(Γn)Z(p)
. Then we have⊕k>M
Mk(Γn)Op = Op[f1, · · · , fs] for any prime ideal p above p.
54
Proof Let Z(p)[x1, . . . , xs] be a polynomial ring and let I be the kernel of the Z(p)-algebra morphism
ϕ : Z(p)[x1, · · · , xs]→ Z(p)[f1, · · · , fs]
defined by ϕ(xi) = fi for 1 ≤ i ≤ s. By Corollary 2.4.8 we have Mk(Γn)Z(p)⊗Z(p)
Op ∼= Mk(Γn)Op ,
thus it follows that
⊕k>M
Mk(Γn)Op∼=
⊕k>M
(Mk(Γn)Z(p)
⊗Z(p)Op)
=
(⊕k>M
Mk(Γn)Z(p)
)⊗Z(p)
Op
∼= (Z(p)[x1, · · · , xs]/I)⊗Z(p)Op
∼= Op[x1, · · · , xs]/IOp
∼= Op[f1, · · · , fs].�
Lemma 3.2.3 For k > 2n, the restricted Siegel Φ-operator ΦK : Mk(Γn)K → Mk(Γn−1)K is
surjective.
Proof To get this result, we utilize the fact that C is faithfully flat over any number field K. By
Theorem 7.2 of [16] any ring S ⊃ K is said to be faithfully flat if given any K-module, M , we have
M⊗KS 6= 0. Since K is a number field, we have C ⊃ K. Furthermore, K is a field so any K-module,
M , is actually a K-vector space. Therefore M ⊗K C is an extension of scalars and thus a non-trivial
C vector space. Since C is faithfully flat over K, we know that any sequence of K-modules is exact
if and only if the corresponding sequence obtained by tensoring with C is exact. We now consider
the following diagram.
0 −→ Sk(Γn)id−→ Mk(Γn)
Φ−→ Mk(Γn−1) −→ 0
∼=↓ ∼=↓ ∼=↓
Sk(Γn)K ⊗K C id⊗1−→ Mk(Γn)K ⊗K C ΦK⊗1−→ Mk(Γn−1)K ⊗K C
We know the first sequence is short exact and wish to get the second one is as well by showing the
55
two sequences are isomorphic. This amounts to showing
Mk(Γn)Φ−→ Mk(Γn−1)
∼=↓ ∼=↓
Mk(Γn)K ⊗K C ΦK⊗1−→ Mk(Γn−1)K ⊗K C
commutes. To see this, we consider a basis {f1, · · · , fm} of Mk(Γn)K and with the standard identi-
fication obtain {f1⊗1, · · · , fm⊗1} is a basis of Mk(Γn)K⊗K C. Thus we see that {f1, · · · , fm} also
serves as a basis of Mk(Γn). Since linear maps of vector spaces are completely determined by how
they act on a basis we will show that diagram commutes on an arbitrary basis element fi. Going
down and across we obtain the following composition of mappings
fi 7−→ fi ⊗ 1 7−→ ΦK(fi)⊗ 1.
Going across and then down we obtain the composition given by
fi 7−→ Φ(fi) 7−→ Φ(fi)⊗ 1.
Since Φ essentially strips away the Fourier coefficients of fi with full rank, we have Φ(fi) ∈
Mk(Γn−1)K and more importantly Φ(fi) = ΦK(fi). Hence we have the diagram commutes on
a basis and hence every element. Moreover, we now have that the two sequences are isomorphic,
proving that the following sequence is exact
0 −→ Sk(Γn)K ⊗K C id⊗1−→ Mk(Γn)K ⊗K C ΦK⊗1−→ Mk(Γn−1)K ⊗K C −→ 0.
Since we know C is faithfully flat over K we then have
0 −→ Sk(Γn)Kid−→ Mk(Γn)K
ΦK−→ Mk(Γn−1)K −→ 0.
Therefore, we conclude that the restricted Siegel operator ΦK : Mk(Γn)K →Mk(Γn−1)K is surjec-
tive. �
In order to define Sn(K), we first make the two following oberservations. First, by Lem-
56
mas 3.2.1 and 3.2.2 we have⊕
2n<k
Mk(Γn−1)Op is a finitely generated Op-algebra. Restricting
to⊕
2n<k∈2ZMk(Γn−1)Op , we find that we still have a finitely generated Op-algebra though the
number of generators may increase, i.e., if f1 and f2 are both generators with odd weights k1
and k2 then they may need to be replaced with f21 , f1f2, and f2
2 having respective weights 2k1,
k1 + k2, and 2k2. The other observation is that given a set of generators {f1, . . . , fs} of the algebra⊕2n<k∈2ZMk(Γn−1)Op , Lemma 3.2.3 tells us that Φ−1
K (fi) 6= ∅ for 1 ≤ i ≤ s.
Definition 3.2.4 Let Sn(K) be the set of all primes ideals of p ⊂ OK such that given a set of
generators {f1, . . . , fs} of⊕
2n<k∈2ZMk(Γn−1)Op , there exists some i such that for all Fi ∈ Φ−1
K (fi)
we have νp(Fi) < 0.
Lemma 3.2.5 Whether or not p ∈ Sn(K) depends on n but not on the set of generators of⊕2n<k∈2Z
Mk(Γn−1)Op .
Proof Suppose for some p ⊂ O we have two sets of generators, {f1, . . . , fs} and {g1, . . . , gr}, of⊕2n<k∈2Z
Mk(Γn−1)Op . It suffices to show that if for 1 ≤ i ≤ s we take Fi ∈ Mk(Γn)Op with
Φ(Fi) = fi, then for every 1 ≤ j ≤ r we can get Gj ∈ Mk(Γn)Op such that Φ(Gj) = gj . To see
this observe that gj ∈⊕
2n<k∈2ZMk(Γn−1)Op = Op [f1, . . . , fs] means there exists a polynomial
Pj ∈ Op[x1, . . . , xs] such that
gj = Pj(f1, . . . , fs).
Taking Fi ∈Mk(Γn)Op with Φ(Fi) = fi, set
Gj = Pj(F1, . . . , Fs).
Recall the limit of a product of functions if the product of the limits, provided the limit of each
function exists. Keeping this in mind, we observe that
Φ(f · g) = limt→∞
fZ
it
g
Zit
=
limt→∞
f
Zit
limt→∞
g
Zit
= (Φ(f))(Φ(g)).
57
Since Pj(x1, . . . , xs) is a polynomial in s variables we have
Φ(Gj) = Pj(Φ(F1), . . . ,Φ(Fs)) = P (f1, . . . , fs) = gj .
We now show that if there exists some i such that νp(Gi) < 0 for any choice of Gi ∈ Φ−1K (gi) then
there must be some j such that νp(Fj) < 0 for all Fj ∈ Φ−1K (fj).
Since νp is an additive valuation and Pi(x1, . . . , xs) is a finite sum of monomials in s vari-
ables, one of the monomials, say αs∏`=1
xe`` with 0 6= α ∈ Op, must satisfy
νp (P (F1, . . . , Fs)) ≥ νp
(α
s∏`=1
F e``
).
Consequently, we have
0 > νp(Gi)
≥ νp (P (F1, . . . , Fs))
= νp
(α
s∏`=1
F e``
)
= νp +
s∑`=1
e`νp(F`)
≥s∑`=1
e`νp(F`) since α ∈ Op.
By definition of a polynomial, we must have e` ≥ 0 for all `. Hence there must exist some j such
that νp(Fj) < 0. Since Fj ∈ Φ−1K (fj), the only way to choose a different form Fj ∈ Φ−1
K (fj) is if we
have Fj = Fj + H for some H ∈ Sk(Γn), and hence Φ(Fj) = Φ(Fj) = fj . Using this relation we
observe that
Φ(Pi(F1, . . . , Fj−1, Fj , Fj+1, . . . , Fs)) = Pi(Φ(F1), . . . ,Φ(Fj−1),Φ(Fj),Φ(Fj+1), . . . ,Φ(Fs))
= Pi(f1, . . . , fj−1, fj , fj+1, . . . , fs)
= gi.
58
Thus Pi(F1, . . . , Fj−1, Fj , Fj+1, . . . , Fs) ∈ Φ−1K (gi) and thus we have
νp
(Pi(F1, . . . , Fj−1, Fj , Fj+1, . . . , Fs)
)< 0.
If we are able to choose Fj such that νp(Fj) ≥ 0 then we do so and there exists another form in
the set {F1, . . . , Fs} with negative valuation. In that case we would repeat this process. In a worst
case scenario, we would repeat this process no more than s− 1 times in order to isolate a form with
negative valuation which cannot be replaced with another that has non-negative valuation. Thus
without loss of generality we can assume that our first choice Fj cannot be replaced by a form with
non-negative valuation. That is, any choice of Fj ∈ Φ−1K (fj) satisfies νp(Fj) < 0. Thus we see that
p ∈ Sn(K) is independent of the set of generators. �
Lemma 3.2.6 For d = [K : Q] <∞ we have
{p : p ∩ Z = (p) ∈ Sn(Q), p - d} ⊆ Sn(K) ⊆ {p : p ∩ Z = (p) ∈ Sn(Q)}.
Proof We start by proving the second containment. Let p ∈ Sn(K) and {f1, . . . , fs} be a set
of generators of⊕
2n<k∈2ZMk(Γn−1)Z(p). By Lemma 3.2.2 we have
⊕2n<k∈2Z
Mk(Γn−1)Op =
Op[f1, . . . , fs]. By definition, since p ∈ Sn(K), there exists an i with 1 ≤ i ≤ s such that for
all Fi ∈ Φ−1K (fi) we have νp(Fi) < 0. In particular, for all Fi ∈ Φ−1
Q (fi) ⊆ Φ−1K (fi) we have
ν(p)(Fi) ≤ νp(Fi) < 0. Thus (p) ∈ Sn(Q) and p ∈ {p : p ∩ Z = (p) ∈ Sn(Q)}.
We now prove the first containment. Let {f1, . . . , fs} be as before and further suppose that
p ∩ Z = (p) ∈ Sn(Q) with p - d = [K : Q]. Now suppose for contradiction that for every 1 ≤ i ≤ s
there exists Fi ∈ Φ−1K (fi) such that νp(Fi) ≥ 0. Consider the forms given by
Gi =∑
σ∈Emb(K,C)
Fσi ∈Mk(Γn)Q.
Since νp(Fi) ≥ 0 we have ν(p)(Gi) ≥ 0 and thus Gi ∈Mk(Γn)Z(p). Furthermore, since any embed-
ding of K into C fixes Q and since we have fi ∈ Mk(Γn)Z(p)⊂ Mk(Γn)Q we obtain Φ(Fσi ) = fi.
Thus
Φ(Gi) = Φ
∑σ∈Emb(K,C)
Fσi
=∑
σ∈Emb(K,C)
Φ(Fσi ) =∑
σ∈Emb(K,C)
fi = dfi.
59
It follows that d−1Gi ∈ Φ−1
Q (fi). By our initial assumption, p - d so ν(p)(d−1Gi) ≥ 0, contradicting
that p ∩ Z = (p) ∈ Sn(Q). �
3.3 Proof of the main theorem and its corollary
We now have the necessary information to prove Theorem 3.1.2, the main theorem of [9].
We recall the statement is as follows.
Main Theorem There exists a finite set Sn(K) of prime ideals in K depending on n such that
the following holds. For a prime ideal p of O not contained in Sn(K) and a mod pm cusp form
F ∈Mk(Γn)Opwith k > 2n, there exists G ∈ Sk(Γn)Op
such that F ≡ G mod pm.
Proof Recall that the case of odd weight is trivial; thus we focus on even weight. Let {f1, . . . , fs}
be a set of generators of⊕
2n<k∈2ZMk(Γn−1)Op . Suppose that F ∈Mk(Γn)Op is a mod pm cusp
form, that is,
Φ(F ) ≡ 0 mod pm.
It follows by definition of νp that there exists some γ ∈ pm such that γ−1Φ(F ) ∈ Mk(Γn−1)Op .
Note that there exists some T0 ∈ Λn−1 such that a(T0; Φ(F )) = νp(Φ(F )), in which case we can
take γ = a(T0; Φ(F )). Since
γ−1Φ(F ) ∈Mk(Γn−1)Op ⊂⊕
2n<k∈2ZMk(Γn−1)Op
there exists a polynomial P ∈ Op[x1, . . . , xs] such that
γ−1Φ(F ) = P (f1, . . . , fs).
Since p ∈ Sn(K), for every 1 ≤ i ≤ s there exists some Fi ∈ Φ−1K (fi) such that νp(Fi) ≥ 0. Thus
F ∈ γP (F1, . . . , Fs) + ker ΦK
and specifically, there exists some G ∈ ker ΦK = Sk(Γn)K such that
F = γP (F1, . . . , Fs) +G.
60
In particular, since Fi ∈ Mk(Γn)Op for all i, we then have P (F1, . . . Fs) ∈ Mk(Γn)Op . Since
F ∈Mk(Γn)Op we then obtain G ∈ ker ΦK = Sk(Γn)Op ⊂ Sk(Γn)K . Therefore we have
F = γP (F1, . . . , Fs) +G
≡ 0 +G mod pm
≡ G mod pm. �
With the main theorem of [9] proved, we move onto their main corollary.
Main Corollary Let k > 2n be even and f ∈ Sk(Γr)Kf, n > r, a Hecke eigenform. For the
Klingen-Eisenstein series [f ]nr attached to f , suppose there exists a prime ideal p in OKfwith
p 6∈ Sn(Kf ) such that ν(n)p ([f ]nr ) = νp(Φ([f ]nr )) − m, for some m ∈ Z≥1. Then there exists F ∈
Sk(Γn)Of,p, where Of,p is the localization of OKf
at p, such that α[f ]nr ≡ F mod pm for some
0 6= α ∈ pm.
Proof Let f ∈ Sk(Γr)Kf, then by [18] we know [f ]nr ∈Mk(Γn)Kf
and [f ]n−1r ∈Mk(Γn−1)Kf
. We
further suppose that p /∈ Sn(Kf ) such that ν(n)p ([f ]nr ) = νp(Φ([f ]nr ))−m, for some constant m ∈ Z≥1.
By Corollary 2.4.3, there exists some c ∈ Z>0 such that c[f ]n−1r ∈ Mk(Γn−1)OKf
. It is acceptable
to weaken the choice of c so that we have c[f ]n−1r ∈ Mk(Γn−1)O
f,pwhere Of,p is the localization
of OKfat p. Either way, we obtain
νp(c[f ]n−1r ) = νp(cΦ([f ]nr )) ≥ 0.
Moreover, we have
ν(n)p (c[f ]nr ) = νp(cΦ([f ]nr ))−m ≥ −m.
We now take some 0 6= β ∈ pm and observe that
ν(n)p (βc[f ]nr ) = νp(β) + ν
(n)p (c[f ]nr ) = νp(β) + νp(cΦ([f ]nr ))−m ≥ m−m = 0.
We now set α = βc, noting that we have α ∈ pm, and observe that α[f ]nr ∈Mk(Γn)Of,p
. Further-
more, we have
νp(Φ(α[f ]nr )) = νp(βc[f ]n−1r ) = νp(β) + νp(c[f ]n−1
r ) ≥ m.
61
This tells us that α[f ]n−1r vanishes (mod pm), demonstrating that α[f ]nr ∈ Mk(Γn)O
f,pis a
(mod pm) cusp form. We now apply the Main Theorem to get F ∈ Sk(Γn)Of,p
such that
α[f ]nr ≡ F (mod pm).
3.4 Numerical Examples
We now consider the numerical examples of [9]. To start, we review their work giving a
superset of S3(Q). Let ψ(n)k be the normalized weight k Siegel-Eisenstein series; that is,
ψ(n)k (Z) = [1]n(Z) =
∑γ∈∆n,0\Γn
j(γ, Z)−k
with a(
0n;ψ(n)k
)= 1 where 0n is the n × n 0 matrix. It is known that ψ
(n)k ∈ Mk(Γn)Q and in
particular, by [7], that
χ10 = − 43867
212 · 35 · 52 · 7 · 53
(ψ
(2)4 ψ
(2)6 − ψ(2)
10
)and
χ12 =131 · 593
213 · 37 · 53 · 72 · 337
(3272(ψ
(2)4 )3 + 2 · 53(ψ
(2)6 )2 − 691ψ
(2)12
)are normalized cusp forms in M10(Γ2)Q and M12(Γ2)Q, respectively. In [8] these forms are further
normalized to
X10 = −4χ10 and X12 = 12χ12
so that a
1 1/2
1/2 1
;Xk
= 1 for k = 10, 12. Furthermore, by Theorem 1 of [8] we have
Xk ∈ Sk(Γ2)Z for k = 10, 12.
Theorem 3.4.1 ([19], Theorem 4.3) Assume p ≥ 5. If F ∈ Mk(Γ2)Z(p), then there exists a
unique polynomial P ∈ Z(p)[x1, x2, x3, x4] such that
F = P(ψ
(2)4 , ψ
(2)6 , X10, X12
).
Note that [19] uses the notation χ10 and χ12 instead of our X10 and X12, but states earlier
62
in the paper that Xk is normalized at a
1 1/2
1/2 1
;Xk
= 1 for k = 10, 12. We note that this
is equivalent to ⊕k∈2Z
Mk(Γ2)Z(p)= Z(p)
[ψ
(2)4 , ψ
(2)6 , X10, X12
].
By [24], we have ψ(3)k ∈ Φ−1
Q(ψ
(2)k
)for k > 4 since ψ
(3)k = [1](3) and 1 ∈ Sk(Γ0). We define
F10 =43867
210 · 35 · 52 · 7 · 53
(ψ
(3)4 ψ
(3)6 − ψ(3)
10
)F12 =
131 · 593
211 · 36 · 53 · 72 · 337
(3272(ψ
(3)4 )3 + 2 · 53(ψ
(3)6 )2 − 691ψ
(3)12
)
and immediately see that ΦQ(F10) = X10 and ΦQ(F12) = X12. Thus it is clear that S3(Q) is
contained in the set of primes p satisfying ν(p)
(ψ
(3)k
)< 0 for k = 4, 6 or ν(p)(Fk) < 0 for k = 10, 12.
For the primes satisfying ν(p)
(ψ
(3)k
)< 0 for k = 4, 6 we refer to Bocherer [2]. For ν(p)(Fk) < 0 for
k = 10, 12 it is clear that the primes we wish to avoid are contained in {2, 3, 5, 7, 53, 337} since we
are not guaranteed that there is a choice in Φ−1
Q (Xk) where any of those primes no longer occur in
the denominater of a Fourier coefficient.
We now review some of the numerical examples of Corollary 3.1.3 as given by [9] for the
case of degree 2. For simplicty, we will drop the exponent notation for the Siegel-Eisenstein sereies;
that is, we will write ψk = ψ(2)k . Furthermore, we let ∆ ∈ S12(Γ1) be Ramanujan’s delta function.
To simplify notation, we adopt the standard notation (m, r, n) to represent
m r/2
r/2 n
∈ Λ2.
Combining Theorem 1.1 and Proposition 2.2 of [4] we get the following theorem from [9].
Theorem 3.4.2 ([9], Theorem 4.1) Let p ≥ 5 be a prime, k even, and F ∈Mk(Γ(2))Z(p). If
a((m, r, n);F ) ≡ 0 (mod p)
for every 0 ≤ m,n ≤ k10 and r with 4mn− r2 ≥ 0, then we have F ≡ 0 (mod p).
The authors of [9] then state the theorem is easily modified to obtain the following.
Theorem 3.4.3 Let K be a number field with ring of integers O. Let p be a prime ideal such that
63
p - 2 · 3. Suppose k is even and F ∈Mk(Γ2)Op with
a((m, r, n);F ) ≡ 0 (mod p)
for every 0 ≤ m,n ≤ k10 and r with 4mn− r2 ≥ 0, then we have F ≡ 0 (mod p).
From these two theorems it becomes apparent that it suffices to check the following Fourier
coefficents based off of the even weight k of a degree two Siegel modular form.
T = (1, 0, 1), (1, 1, 1) if 10 ≤ k ≤ 18,
T = (1, 0, 1), (1, 0, 2), (1, 1, 1), (1, 1, 2), (2, 0, 2), (2, 1, 2), (2, 2, 2) if 20 ≤ k ≤ 28.
Though the theorems above call for more Fourier coefficients to be checked, in the case of even
weight k, we have the coefficients for (m, r, n), (n, r,m), (m,−r, n), and (n,−r,m) are all the same.
Weight 12 Example
Consider Ramanujan’s ∆-function and set f12 = 7∆ ∈ S12(Γ1). We then have [f12] is a mod
7 cusp form. By the main corollary, there exists a cusp form F12 ∈ S12(Γ2) such that [f12] ≡ F12
mod 7. To confirm this numerically we consider F12 := X12 ∈ S12(Γ2) and observe the following
congruences.
T = (m, r, n) a(T ; [f12]) a(T ;F12) Modulo 7
(1,0,1) 1242 10 3
(1,1,1) 92 1 1
Using Theorem 3.4.2 we see that [f12]− F12 vanishes modulo 7 and thus
[f12] ≡ F12 mod 7.
Weight 16
Consider the quadratic x2 − x − 12837 and one of its roots a; set K = Q(a). By [14] that
dimS16(Γ1) = 1 and thus posses a unique newform C16 ∈ S16(Γ1)Z with a(1;C16) = 1. Setting
f16 = 72 · 11 · C16 we obtain a cusp form f16 ∈ S16(Γ1) satisfying a(1; f16) = 72 · 11. Since C16
64
has integer Fourier coefficients, we must than have 72 · 11 divides all the coefficients of f16, i.e., f16
vanishes (mod 72). We take p = (7, a+4) and immediately see that [f16] is a mod p2 cusp form since
Φ([f16]) = f16. Moreover, by [14], we know that the space of degree 1, weight 30, newforms is spanned
by two Hecke eigenforms with coefficients in Q(α) where α satisfies 0 = x2 − 8640x − 454569984.
We note that −192a + 4416 also satisfies this quadratic. From [14], we take the newform that has
−192a + 4416 as its eigenvalue for the T (2) Hecke operator, for elliptic modular forms. We denote
this eigenform as g30 ∈ S30(Γ1) and take F16 ∈ S16(Γ2) to be its Saito-Kurokawa lift when we
normalize it as below.
T = (m, r, n) a(T ; [f16]) a(T ;F16) Modulo p2
(1,0,1) 5394 80a+3600 4
(1,1,1) 124 8a+1248 26
We apply Theorem 3.4.3 repeatebly where our first application of yields [f16] − F16 ≡ 0 (mod p),
i.e., there exists some form g ∈ M16(Γ2) such that [f16] − F16 = αg for some α ∈ p\p2. We then
apply the theorem again, this time to α−1([f16] − F16). We note that this is well defined since
[f16]− F16 vanishes (mod p), and since the required coefficients are congruent (mod p2) we must
have α−1([f16]− F16) ≡ 0 (mod p). Thus applying the theorem repeatedly yields
[f16] ≡ F16 (mod p2).
Weight 20
It is known that dimS20(Γ1) = 1 and thus, via scaling, we have f20 ∈ S20(Γ1) such that
a(1; f20) = 11 · 712. This shows that [f20] is a mod 712 cusp form. From [14], we take the weight 20,
degree 2 cusp form listed as “interesting” and scale it by -38 to obtain F20 ∈ S20(Γ2) with explicit
formula
F20 = 38
(ψ
(2)4 ψ
(2)6 X10 +
(ψ
(2)4
)2
X12 − 1785600X210
).
We have the following table of Fourier coefficients.
65
T = (m, r, n) a(T ; [f20]) a(T ;F20) Modulo 712
(1,0,1) 10386 304 304
(1,0,2) 1925356716 198816 2217
(1,1,1) 76 76 76
(1,1,2) 162929376 4256 4257
(2,0,2) 1238800286736 -335343616 3868
(2,1,2) 385264596000 278989920 816
(2,2,2) 9084897120 -63912960 1879
Thus we see from Theorem 3.4.3 that [f20]− F20 vanishes (mod 712) and hence
[f20] ≡ F20 (mod 712).
Weight 22
We consider weight 22 since dimS22(Γ1) = 1 and thus can obtain a cusp form f22 satisfying
a(1; f22) = 7 · 13 · 17 · 61 · 103. Thus we have [f22] is a mod 61 cusp form. We set F22 ∈ S22(Γ2) to
be the cusp form from [14] scaled by 29 . Thus we have
F22 =2
9
(−61
(ψ
(2)4
)3
X10 − 5(ψ
(2)4
)2
X10 + 30ψ(2)4 ψ
(2)6 X12 + 80870400X10X12
).
We have the following table of Fourier coefficients.
T = (m, r, n) a(T ; [f22]) a(T ;F22) Modulo 61
(1,0,1) -179610 96 35
(1,0,2) -133169475780 -1728 41
(1,1,1) -740 -8 53
(1,1,2) -8620265280 -10752 45
(2,0,2) 54428790246720 -313368576 14
(2,1,2) 15093047985984 142287360 41
(2,2,2) 223472730240 17725440 60
Thus, by Theorem 3.4.2, we see that [f22]−F22 vanishes (mod 61), allowing us to conclude
that
[f22] ≡ F22 (mod 61).
66
3.5 New Examples
In order to understand the nature of the congruences given by the main corollary of [9] we
provide some more examples. These examples are restricted to degree 2 forms, so we simplify our
notation by writing ψ4 = ψ(2)4 and ψ6 = ψ
(2)6 . Throughout this section it is important to note that
any form ψa4ψb6X
c10X
d12 with c > 0 or d > 0 is a cusp form since
Φ(ψa4ψb6X
c10X
d12) = Φ (ψ4)
aΦ (ψ6)
bΦ (X10)
cΦ (X12)
d.
It is important to recall that⊕k∈2Z
Mk(Γ2)Z(p)= Z(p)
[ψ
(2)4 , ψ
(2)6 , X10, X12
].
Weight 12
We start by highlighting why Takemori and Kikuta chose f12 = 7∆. It is known by [14]
that
[∆]21 =1
1728ψ3
4 −1
1728ψ2
6 +288
7X12 =
1
26 · 33ψ3
4 −1
26 · 33ψ2
6 +25 · 32
7X12.
This shows that ν(2)7 ([∆]21) = −1 while ν7(Φ([∆]21)) = ν7(∆) = 0 since a(1; ∆) = 1 and ∆ ∈ S12(Γ1).
Hence, we take 7 ∈ (7) in order to get
7[∆]21 =7
1728ψ3
4 −7
1728ψ2
6 + 288X12 ≡ X12 (mod 7).
Weight 16
Let C16 ∈ S16(Γ1) be the unique normalized newform of weight 16 and degree 1. Note that
this corresponds to the choice, f16, in [9] through the equation f16 = 72 · 11C16. By [14] we have the
following Klingen-Eisenstein series scaled so that a((1, 0, 0); [C16]21) = 1.
[C16]21 =1
26 · 33ψ4
4 −1
26 · 33ψ4ψ
26 +
23280
539ψ6X10 −
1152
77ψ4X12
=1
26 · 33ψ4
4 −1
26 · 33ψ4ψ
26 +
24 · 3 · 5 · 97
72 · 11ψ6X10 −
27 · 32
7 · 11ψ4X12
It is immediate that we have
49[C16]21 ≡ 3ψ6X10 (mod 7);
67
however, to match the corollary we must consider 49[C16]21 (mod 49). In that case we have
49[C16]21 ≡ 45ψ6X10 − 28ψ4X12 (mod 49).
We need to show that the right-hand side does not vanish, otherwise this is a trivial example. Using
Takemori’s code from GitHub associate with his paper [22] we have a((1, 0, 1);ψ6X10) = −2 and
a((1, 0, 1);ψ4X12) = 10, demonstrating that
a((1, 0, 1); 45ψ6X10 − 28ψ4X12) ≡ 45a((1, 0, 1);ψ6X10)− 28a((1, 0, 1);ψ4X12) (mod 49)
≡ 22 (mod 49).
Thus 49[C16]21 ≡ 45ψ6X10 − 28ψ4X12 (mod 49) is a non-vanishing example.
We can perform the saem analysis using the prime 11. In that case we obtain
11[C16]21 ≡ 3ψ6X10 + 2ψ4X12.
As before, we wish to show that the right-hand side does not vanish. To do so, we consider the
following.
a((1, 0, 1); 3ψ6X10 + 2ψ4X12) ≡ 3a((1, 0, 1);ψ6X10) + 2a((1, 0, 1);ψ4X12) (mod 11)
≡ 3 (mod 11)
This demonstrates that any choice of prime p such that ν(2)p (f) = νp(Φ(f)) −m for some
m ∈ Z>0 can be used to construct an example.
Weight 18
Let C18 ∈ S18(Γ1) be the unique normalized newform of weight 18 and degree 1. We then
have
[C18]21 =1
1728ψ3
4ψ6 −1
1728ψ3
6 −11448
143ψ2
4X10 +108144
1001ψ6X12
=1
26 · 33ψ3
4ψ6 −1
26 · 33ψ3
6 −23 · 33 · 53
11 · 13ψ2
4X10 +24 · 32 · 571
7 · 11 · 13ψ6X12
68
We then have [C18]21 ∈ M18(Γ2)Q and f18 := 1729728[C18]21 ∈ M18(Γ2)Z. We can then identify
these forms respectively in M18(Γ2)K and M18(Γ2)OKwhere K = Q(α) ' Q[x]/(x2 − x− 589050).
Let F18 ∈ S18(Γ2) be the unique Maass form given by [14] with explciit formula
F18 = 1295ψ24X10 + (α− 575)ψ6X12 ∈ S18(Γ2)OK
.
Before checking for congruences we note that Q(α) = Q(√d) where d = 2356201 is the discriminant
of x2 − x − 589050. This equality allows us to utilize Theorem 8.3 of [3] to see that the ideals
(7), (11), (13) ⊂ OK all split. Using Corollary 8.4 of [3] we then have (7, α), (11, α+10), (13, α+4) ⊂
OK are prime ideals lying above (7), (11), and (13), respectively.
The first case we consider is congruences with respect to p1 = (7, α). Observe that we have
the following table of Fourier coefficients.
T = (m, r, n) a(T ; f18) a(T ;F18) Modulo p1
(1,0,1) -2643840 10α− 8340 -3
(1,1,1) -34560 α+ 720 -1
Using Theorem 3.4.3 we then have f18 − F18 ≡ 0 (mod p1) which is equivalent to
1729728[C18]21 ≡ F18 (mod p1).
The second case we consider is p2 = (11, α+10). For this case, we use 3f18 = 5189184[C18]21.
T = (m, r, n) a(T ; 3f18) a(T ;F18) Modulo p2
(1,0,1) -7931520 10α− 8340 -5
(1,1,1) -103680 α+ 720 -3
Using Theorem 3.4.3 we then have 3f18 − F18 ≡ 0 (mod p2) which is equivalent to
5189184[C18]21 ≡ F18 (mod p2).
The last case we consider for weight 18 is p3 = (13, α+ 4) and −11f18 = −19022058[C18]21.
T = (m, r, n) a(T ;−11f18) a(T ;F18) Modulo p3
(1,0,1) 29082240 10α− 8340 5
(1,1,1) 380160 α+ 720 1
69
By Theorem 3.4.3 we then have F18 + 11f18 ≡ 0 (mod p3) which is equivalent to
−19022058[C18]21 ≡ F18 (mod p3).
Weight 20
Let C20 ∈ S20(Γ1) be the unique normalized newform of weight 20 and degree 1. We then
have
[C20]21 =1
26 · 33ψ5
4 −1
26 · 33ψ2
4ψ26 +
25 · 3 · 5 · 23 · 421
11 · 712ψ4ψ6X10 −
25 · 32 · 977
712ψ2
4X12 −214 · 37 · 53 · 17
11 · 712X2
10.
We consider the case p = 11 we have
11[C20]21 ≡ 7psi4ψ6X10 + 3X210 (mod 11).
Using [22] we see that a((1, 0, 1);ψ4ψ6X10) = −2 and a((1, 0, 1);X210) = 0. We see that
a((1, 0, 1); 7ψ4ψ6X10 + 3X210) = −14 ≡ 8 (mod 11)
showing that the right-hand side does not vanish. We go even further by considering the field
Q(β) ∼= Q[x]/(x2− x− 15934380). Since x2− x− 15934380 ≡ x(x− 1) (mod 11) we know 11 splits.
We consider the prime ideal p = (11, β) and the form g20 := −95819328[C20]21. We set G20S20(Γ2)
to be the unique Maass form given by [14] with explicit formula
G20 = (10β + 40100)ψ4ψ6X10 + (−7β − 21560)ψ24X12 + 6749568000X2
10.
Using [14] we obtain the following table of Fourier coefficients.
70
T = (m, r, n) a(T ; 5g20) a(T ;G20) Modulo p
(1,0,1) -89735040 −90β − 295800 1
(1,0,2) -16635082026240 −27036β − 78025680 4
(1,1,1) -656640 3β + 18540 5
(1,1,2) -1407709808640 −5544β − 19679520 -3
(2,0,2) -10703234477399040 −10036800β + 11326214400 -5
(2,1,2) -3328686109440000 −3997080β − 40732178400 -1
(2,2,2) -78493511116800 287712β + 8190149760 1
By Theorem 3.4.3 we have 5g20 −G20 ≡ 0 (mod p) which is equivalent to
−479096640[C20]21 ≡ G20 (mod p).
71
Chapter 4
Future Work
One possible direction for future work is given in [9]. The authors pose the following
question.
Question: Given degree n, does there exist an explicit bound Cn such that
maxSn(Q) < Cn?
They pose this question since we know S2(Q) ⊂ {2, 3} and S3(Q) ⊂ {2, 3, 5, 7, 53, 337}. We the
significance of such a bound is seen in Lemma 3.2.6, where Sn(K) is related back to Sn(Q). Hence,
knowing a bound on on primes in Sn(Q) would provide more information about the primes occuring
in Sn(K) for arbitrary number field K.
Another direction would be to pursue the same result but for degree n, level N , weight k
modular forms given by Mk(Γn(N)). For the results used in [9] to hold we have two main sticking
points:
1. Does Theorem 2.3 of [6] hold for arbitrary level? That is; do we have
⊕k≥1
Mk(Γn(N))Z = Z[φ1, . . . , φm]
for some m ∈ Z≥1 and generators φ1, . . . , φm?
2. For which k do we have
Mk(Γn(N)) = Mk(Γn(N))Q ⊗Q C?
72
To answer the first question requires developing a background in algebraic geometry. We can tackle
the second question by combining Theorem 1 of [21] with the result from the same paper stating
Mk(Γn(N)) = Mk(Γn(N))Z ⊗Z C
for all even k as well as for any odd k satisfying
0 6= A(Γn(N),Q) =
{f
g| f ∈Mk+m(Γn(N))Q, g ∈Mm(Γn(N))Q,m ∈ Z≥0
}.
This should follow from [20] giving A(Γn,Q) 6= 0 due to theta functions.
A final direction worth considering would be the case of Hilbert modular forms as they are
another multivariables generalization of elliptic modular forms. Many of Shimura’s results on Siegel
modular forms hold for Hilbert modular forms as demonstrated by [20]. The approach may have to
differ as the main theorem of [9] depended on using the Siegel operator, in particular its surjectivity,
to transition between degree n− 1 and degree n Siegel modular forms.
73
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