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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

    Masters of Science



    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


  • Chapter 1


    The following work is a Masters 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


  • that the following holds: For a prime ideal p of O not contained in Sn(K) and a (mod pm) cusp

    form F Mk(n)Op with k > 2n, there exists G Sk(n)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(r)Kf , n > r, a Hecke eigenform. For the

    Klingen-Eisenstein series [f ]nr attached to f , we choose a prime ideal p in OKf with p 6 Sn(Kf )

    such that (n)p ([f ]

    nr) = p([f ]nr) m, m Z1. Then there exists F Sk(n)Op such that

    [f ]nr F (mod pm) for some 0 6= pm.

    Kikuta and Takemori assert that this is a generalization of Ramanujans congruence

    11(n) (n) (mod 691).

    They make this assertion since 11(n) denotes the nth Fourier coefficient of the Eisenstein series

    of weight 12, which we denote in Section 3.4 as (1)12 , and (n) is the nth Fourier coefficient of

    Ramanujans function [9]. More precisely, they are assert that they are generlizing the result that

    691(1)12 is a (mod 691) cusp form of degree 1 that is congruent modulo 691 to the Ramanujans

    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 Matn(R) by t.

    For matrices A,B Matn(R) Matn(C) we define A[B] = tBAB.

    For V GLn(R) we define U(V ) =

    V 0n0n

    tV 1

    .For tS = S Matn(R) we define T (s) =

    In S0n In


  • 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 .


  • Chapter 2

    Siegel Modular Forms

    2.1 Siegel Modular Group

    For n Z1 define J2n =

    0n InIn 0n

    where 0n is the nn zero matrix and In is the nnidentity 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) = { Mat2n(R) : tJ = ()J, () R}.

    We call () the similitude factor of . Therefore, writing =

    A BC D

    , where A,B,C,D Matn(R), yields the following relations;

    tAC = tCA, tDB = tBD, tAD tCB = ()In,

    tDA tBC = ()In, and 1 = ()1

    tD tBtC 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

    n = Sp2n(Z) = ker{ : GSp2n(Z) R}.


  • Note that () = 1 if and only if tJ = 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 Z = (AZ +B)(CZ +D)1

    where =

    A BC D

    GSp+2n(R). For this to be a well defined group action we must check thefollowing:

    1. CZ +D is invertible;

    2. Z Hn;

    3. I2n Z = Z;

    4. () Z = ( Z) for all , 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)

    = ()InZ ()InZ

    = ()(Z Z)

    = 2i()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


  • vector v Cn such that (CZ +D)v = 0. This would then yield

    tvY v =i2()1

    t(v)(t(CZ +D)(Az +B) t(AZ +B)(CZ +D)



    (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 Z Hn. We start by checking for symmetry

    t(CZ +D)( Z t( Z))(CZ +D) = t(CZ +D)(AZ +B) t(AZ +B)(CZ +D)

    = Z tZ

    = 0n.

    Since CZ +D is invertible this implies Z is symmetric. Similarly, we have

    t(CZ +D)Im( Z)(CZ +D) = i

    2t(CZ +D)

    ( Z t( Z)

    )(CZ +D)


    (t(CZ +D)(AZ +B) t(AZ +B)(CZ +D)

    )= ()Y

    > 0n.

    The final inequality holds since Y is positive definite and GSp+2n(R) means () > 0. We thus

    conclude that Z Hn.

    The third condition is clear since

    I2n Z = (InZ + 0n)(0nZ + In)1 = Z.


  • For the last condition let =

    A BC D

    and =A BC D

    be in GSp+2n(R). We then get ( Z) =

    ((AZ +B)(C Z +D)1


    (A(AZ +B)(C Z +D)1 +B

    ) (C(AZ +B)(C Z +D)1 +D

    )1= (A(AZ +B) +B(C Z +D)) (C(AZ +B) +D(C Z +D))


    = ((AA +BC )Z + (AB +BD)) ((CA +DC )Z + (CB +DD))1

    = () Z.

    Thus, our definition for Z defines a group action of GSp+2n(R) of Hn. Since n 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 |k)(Z) := ()nkn(n+1)

    2 j(, Z)kf( Z), where j(, Z) = det(CZ +D)

    for =

    A BC 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 n is a

    holomorphic function f : Hn C such that for all n we have (f |k)(Z) = f(Z).

    The collection of modular forms of weight k and level n is a C-vector space which we denote

    Mk(n) (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.


  • Furthermore, f has Fourier expansion

    f(Z) =


    a(T ; f)qT , qT = exp{2itr(TZ)}, Z Hn,


    n = {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


    where dX =ijdXi,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 n cusp forms we introduce the Siegel


    : Mk(n) Mk(n1)

    f 7 limt



    , Z Hn1, 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 (f) in Mk(n1).

    It is crucial that we acknowledge that 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(n1) Mk(r+1)


    since each above is differs from the rest. To take forms from Mk(n) to Mk(

    r) as in the sequence

    above we use nr.


  • Using this operator, we define Sk(n) := ker 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(1). We then


    (f) = limt


    = limt

    (a0(f) +


    ai(f) e2i(it))

    = a0(f) +



    (ai(f) e2t

    )= a0(f) +





    = 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 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 Mk(n).

    Let f Mk(n), T n1, Z Hn1, and X = Re(Z ). We have

    a(T ; (f)) =

    X mod 1

    ((f))(Z )e2itr(TZ)dX


    X mod 1



    Z it

    e2itr(T Z)dX

    = limt

    X mod 1


    Z it







    = a



    ; f .


  • To get the third equality use Theorem 4.3 of [23] to get



    e2itr(T Z) =



    e2itr(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



    .Thus, f Sk(n) if and only if a



    ; f = 0 for all T n1. We can strengthen

    the result to f Sk(n) if and only if a(T ; f) = 0 for all 0 T n such that 0 6< T . To get this

    we use the identity

    a(tUTU ; f) = det(U)ka(T ; f), for all 0 T n, 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)


    Proof Let f Mk(n) with kn 1 mod 2 and let 0 T n. Observe that t(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 n 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(n) with k odd. Applying the Siegel operator,


  • we see that (f) Mk(n1) but

    k(n 1) 1 mod 2

    and hence f ker , 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 interacts with the fourier coefficients of a form f Mk(n), we now

    consider the more general result for nr. The closed form of the iterative application is given by

    nr : Mk(n) Mk(r)

    f 7 limt


    Z itInr

    , Z Hr, t Rwhere Inr is the n r n r identity matrix. To see the motivation behind this definition,

    consider the Fourier coefficients, a(T ; nr(f)), of a form f Mk(n). The affect of nr on a

    form f Mk(n) is shown by its affect on the Fourier coefficients

    a(T ; nr(f)) = a



    ; f .

    Using this and taking r 1, we have

    a(T ; nr(f)) = a



    ; nr(f)

    = a




    ; f

    = a



    ; f

    = a(T ; nr+1(f)).

    Thus, for any f Mk(n) and r 1 the Fourier coefficients of nr(f) and nr+1(f) coincide.

    Since the coefficients of a power series, and in particular those of a Fourier expansion, uniquely


  • determine the function, we must have nr = nr+1. Furthermore, induction now grants us

    rj nr = nj for 0 j r n.

    By Corollary 5.5 of [23], it is known that is surjective for even k > 2n, so by the first

    isomorphism theorem of vector spaces we have


    n) = Mk(n)/ ker 'Mk(n1).

    Since Mk(n) is a C-vector space we can express it as the direct sum of Sk(n) and its compliment,

    Sk(n)c. We wish to address the structure of Sk(

    n)c. To do so we introduce the Petersson inner


    Definition 2.2.3 Let f, g Mk(n) with at least one a cusp form. We define the Petersson inner

    product as

    f, g :=

    n\Hnf(Z)g(Z) det(Y )kdn(Z)

    where dn(Z) = det(Y )(n+1)

    ij dxijdyij, Z = X + iY , and xij , yij are the i, jth entry of X,Y ,


    Lemma 2.2.4 For n Z1 the measure dn is invariant under the action of GSp+2n(R) on Hn.

    That is, dn(Z) = dn(Z) for all GSp+2n(R).

    Proof Let g = GSp+2n(R) and define X = Re(Z) and Y = Im(Z). The Jacobian for the change

    of variables of Z 7 gZ is given by


    ((X, Y )

    (X,Y )


    (X, Y )

    (X,Y )=

    XX Y XX



    .Since the action of is holomorphic on Hn we have the Cauchy-Riemann equations





    Y= Y


    Using these relations we see that

    12I i2I i2I


    XX Y XX



    I iIiI I

    =XX iXY 0

    0 X

    X + iX


    =(gZ)Z 0

    0 (gZ)Z


  • Consequently, we have


    ((X, Y )

    (X,Y )

    )= det







    We now write dZ = (dzij) and d(gZ) for the matrices of the differentials of Z and Z,

    respectively. Let =

    A BC D

    , then (Z)(CZ +D) = (AZ +D). We haved(Z)(CZ +D) + (Z)CdZ = AdZ.

    Multiplying both sides on the left byt(CZ +D) produces

    t(CZ +D)AdZ = d(Z)[CZ +D] +

    t(CZ +D)(Z)CdZ

    = d(Z)[CZ +D] +t(CZ +D)

    t((AZ +B)(CZ +D)1)CdZ

    = d(Z)[CZ +D] +t(AZ +B)CdZ.

    Rewriting, we have

    d(Z)[CZ +D] =t(CZ +D)AdZ t(AZ +B)CdZ

    = Z(tCA tAC)dZ + (tDA tBC)dZ

    = ()dZ since GSp+2n(R).

    We then have

    d(Z) =t(CZ +D)1()dZ(CZ +D)1 = dZ

    [()(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(Z)

    Zmeasures changes in Z as Z varies and

    we have d(Z) = dZ[

    ()(CZ +D)1], we obtain




    )= det(

    ()(CZ +D)1)n+1 = ()n(n+1)/2 det(CZ +D)(n+1).


  • Similarly, we obtain




    )= ()n(n+1)/2det(CZ +D)


    Combining all the results have

    dn(Z) = det(Y)(n+1) det

    ((X, Y )

    (X,Y )




    (()n det(Y )

    |det(CZ +D)|2









    =|det(CZ +D)|2n+2

    ()n(n+1) det(Y )n+1()n(n+1)|det(CZ +D)|2n2



    = det(Y )(n+1)ij


    = dn(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(n) = Sk(


    we take Sk(n) as our choice for Sk(

    n)c. Since is surjective for k > 2n, we obtain (Sk(



    Mk(n1) = Sk(

    n1)Sk(n1). We then define Mkn,n1 = Sk(n) 1(Sk(

    n1)). Iter-

    ating this process we obtain the following subspaces due to [15] using the notation of [10]

    Mkn,r =

    Sk(n) r = n

    {f Sk(n) : (f) Mkn1,r} 0 r < n.

    More precisely, since Mk(n) is finite-dimensional we are able to use induction to identify these

    spaces as

    Mkn,r = ker n+r+1/ ker n+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


  • following

    Mkn,r1 Mkn,r2 = {f Sk(

    n) : (f) Mkn1,r1 Mkn1,r2}

    = {f Sk(n) : nr2(f) Mkr2,r1 Mkr2,r2}

    = {f Sk(n) : nr2(f) Mkr2,r1 Sk(r2)}.

    However, since r2 > r1 we have Mkr2,r1 = {g Sk(

    r2) : (g) Mkr21,r1}. By orthogonality

    of Mkr2,r1 and Mkr2,r2 = Sk(

    r2) for r1 < r2, we must then have Mkr2,r1 Sk(

    r2) = {0}, showing

    nr2(f) = 0. However, by the definition of Mkn,r2 , anything that vanishes under nr2 must be 0.

    Therefore we conclude Mkn,r1 Mkn,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 n is surjective and maps Mk(n) to Mk(

    0) = Mk0,0 = C we can


    Mk(n) = Mkn,0 + M

    kn,1 + + M

    kn,n1 + M


    This combined with the subspaces being pairwise disjoint we get

    Mk(n) = Mkn,0 M

    kn,1 M

    kn,n1 M


    = Mkn,0 Mkn,1 M

    kn,n1 Sk(n).

    Thus, given f Mk(n) we have a unique fr Mkn,r such that f =nr=1 fr. For each fr

    we have nr(fr) = fgr Mkr,r = Sk(r), noting that nn 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.


  • For 0 r < n, we define the Klingen parabolic subgroup

    n,r :=


    n .

    We construct the Klingen Eisenstein series associated to f Sk(r) by

    Ekn,r(f)(Z) =

    n,r\nj(, Z)kf(( Z)), Z Hn,

    where ( Z) is the upper left r r matrix of Z = (AZ +B)(CZ +D)1.

    As an example, take n = 1, r = 0 and f = 1 Mk(0) . We have

    Ek1,0(1)(z) =

    a bc d

    1,0\1(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 1 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 =


    1 =


    0 1

    1 : Z .

    If =

    a bc d

    SL2(Z), then we have gcd(c, d) = 1 and

    1,0 =

    a+ c b+ d

    c d

    1 : Z .

    Since I2 1,0 we can map the pair (c,d) to (c, d). For simplicity, we focus on the pair (c, d).


  • Now suppose =

    a bc d

    SL2(Z), then (a, b) is a solution to the Diophantine equationxc 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 Z such that a = a+ c and b = b+ d. Consequently we have


    a+ c b+ dc d


    0 1

    a bc d

    1,0.Therefore, we have a bijection of sets given by

    1,0\1 {

    (c, d) Z2 : gcd(c, d) = 1}


    a bc d

    (c, d)and hence

    Ek1,0(1)(z) =

    1,0\1(cz + d)k =


    (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(r) the series Ekn,r(f) converges to a classical Siegel

    modular form of weight k in Mk(n) and satisfies nr(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(r) 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.


  • Thus, up to some constant, we have the majorant Ekn,r(f)(Z) in Hn by

    Gkn,r(Z) =

    n,r\ndet(Im( Z)) k2 |j(, Z)|k.

    We study this series on vertical strips of the Siegel upper half-space given by

    Bn(d) = {Z Hn : tr(X2) d1, 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( Z))|j(, Z)|2 cdet(Im( iIn))|j(, iIn)|2

    for all Z Bn(d), 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


    00 2

    Ir 3

    0 Inr

    where Z = (Z) + i(Z) we can obtain the uniform convergence of Gkn,r(Z) by studying

    Gkn,r(iIn) =

    n,r\ndet(Im( iIn))

    k2 |j(, iIn)|k



    k2 |j(, iIn)|k



    k2 det((iIn))

    k2 |j(, iIn)|k



    k2 det(In)

    k2(|j(, iIn)|2

    ) k2 |j(, iIn)|k=


    det(2(iIn))k2 .


  • Note that the second to last equality arise from the identity

    t(CZ +D)(Z)(CZ +D) =

    t(CZ +D)Im( Z)(CZ +D) = Y

    for n, noting that changes if we change . In [10] we see why this is absolutely convergent

    and consequently get that Ekn,r(f) converges to a form in Mk(n).

    We now consider nr(Ekn,r(f)) and note that the absolute convergence of the series

    Ekn,r(f)(Z) =

    n,r\nj(, Z)kf(( Z)), Z Hn,

    allows us to apply nr term by term in the summation. The first element in the sum we consider

    is any representative


    Ar,r Ar,nr Br,r Br,nr

    Anr,r Anr,nr Bnr,r Bnr,nr

    Cr,r Cr,nr Dr,r Dr,nr

    0nr,r 0nr,nr 0nr,r Dnr,nr

    n,r n

    and since this matrix is in n we have the following relations

    Ar,r Br,rCr,r Dr,r

    r, Anr,nrtDnr,nr = Inr,and Ar,nr

    tDnr,nr = 0r,nr = Cr,nrtDnr,nr.

    Since Dnr,nr is invertible we must have Ar,nr = Cr,nr = 0r,nr. Thus a representative of n,r

    produces the following; Ar,rZ +Br,r Br,nrAnr,rZ

    +Bnr,r itAnr,nr +Bnr,nr

    Cr,rZ +Dr,r Dr,nr

    0nr,r Dnr,nr



    Ar,rZ +Br,r Br,nrAnr,rZ

    +Bnr,r itAnr,nr +Bnr,nr

    (Cr,rZ +Dr,r)1 (Cr,rZ +Dr,r)1Dr,nr(Dnr,nr)1

    0nr,r (Dnr,nr)1


  • =

    (Ar,rZ +Br,r)(Cr,rZ +Dr,r)1

    .Next, we see that


    Cr,rZ +Dr,r Dr,nr0nr,r Dnr,nr

    = det(Cr,rZ +Dr,r) det(Dnr,nr) = det(Cr,rZ +Dr,r)sinceDnr,nr and its inverse are integral matrices. Using these results, writing g(Z) = j(, Z)


    Z)), and recalling that k is even, we have

    nr(g(Z)) = limt







    = limt


    C Z 0r,nr

    0nr,r itInr



    (Ar,rZ +Br,r)(Cr,rZ +Dr,r)1

    = (det(Cr,rZ +Dr,r))kf

    (Ar,rZ +Br,r)(Cr,rZ +Dr,r)1

    , 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,rCr,r Dr,r

    r.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,nr0nr,r itInr

    and Y = Y 0r,nr

    0nr,r tInr

    .Rewriting the identity

    t(CZ +D)(Z)(CZ +D) =

    t(CZ +D)Im( Z)(CZ +D) = Y


  • produces the expression

    1(Z) 00 12 (Z)

    = (Z)1= (CZ +D)(Y 1)

    t(CZ +D)

    = (CZ +D)

    (Y )1 0r,nr0nr,r it1Inr

    t(CZ +D)=

    Y 1[(ZtCnr,r + tDnr,r)] + t1[(ittCnr,nr + tDnr,nr)]

    .Thus we have

    12 (Z) = Y1[(ZtCnr,r +

    tDnr,r)] + t1[(ittCnr,nr +


    allowing us to simplify the following expression

    det(Im( Z))|j(, Z)|2 = det()|j(, Z)|2


    det(2)|j(, Z)|2

    = det(12 ) det(Y )

    = det(12 ) det(Y)tnr

    = det(t12 ) det(Y)

    = det(Y )P (t),

    where P (t) = det(tY 1[ZtCnr,r +tDnr,r] + Inr[it

    tCnr,nr +tDnr,nr]). Thus using this

    equality we have

    Gkn,r(Z) =

    n,r\n(det(Y )P (t))

    k2 .

    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


    P (t)k2 . Clearly, P (t) is a polynomial in t and thus vanishes for deg(P ) > 0. Thus we will


  • assume deg(P ) = 0 and show this only occurs for n,r. To see this we note that for t 0, both

    tY 1[ZtCnr,r +tDnr,r] and Inr[it

    tCnr,nr +tDnr,nr]

    are positive semi-definite since Y 1 and Inr are positive definite. Furthermore their sum must be

    positive definite since the left-hand side of

    det(Im( Z))|j(, Z)|2 = det(Y )P (t)

    is non-zero for all Z Hn and n,r. As a consequence we have

    det(Inr[ittCnr,nr +

    tDnr,nr]) P (t)

    for all t and where we have equality when 0r,r = Y1[ZtCnr,r +

    tDnr,r]. Since Y1 is positive

    definite, this occurs when 0nr,r = ZtCnr,r+

    tDnr,r which holds if and only if 0nr,r = Cnr,r =

    Dnr,r. This then leaves

    P (t) = det(Inr[ittCnr,nr +


    = det(t2Cnr,nrtCnr,nr +Dnr,nr


    which still produces a polynomial in t unless 0 = Cnr,nr. Thus we have limt

    P (t)k2 vanishes

    unless n,r\n has 0 = Cnr,r = Dnr,r and 0 = Cnr,nr. This is precisely when n,r.

    Therefore all terms in the series in Gkn,r(Z), and thus in Ekn,r(Z), with away from n,r vanish for

    Z =

    Z 00 itInr

    as t. Thus for Z Hr, 0 r n, f Sk(n), and k > n+ r + 1 even


  • we have:

    nr(Ekn,r(f))(Z) = lim




    C Z 0r,nr

    0nr,r itInr



    Z 0r,nr0nr,r itInr




    C Z 0r,nr

    0nr,r itInr



    Z 0r,nr0nr,r itInr

    = (det(Cr,rZ +Dr,r))kf

    (Ar,rZ +Br,r)(Cr,rZ +Dr,r)1

    , 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,rCr,r Dr,r

    r.Thus we see that Ekn,r( ) is a right inverse to nr 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(r).

    We do this in order to extend the definition of the operator [ ]nr to all of Mk(r). For f = [fj ]rj

    with fj Sk(j) we define [f ]nr = [fj ]nj . Using this notion as well as our desire for a linear operator,

    we take f Mk(r) and get the r + 1-tuple of cusp forms, (f0, , fr), mentioned at the end of

    Section 2.2 and define

    [f ]nr =rj=0

    [fj ]nj =



    Note that this agrees with our construction in the previous section since we have seen nj([fj ]nj ) =

    fj .

    Using this relation we introduce the notation

    En,j,k =[Sk(


    We then have the decomposition


    nr =




  • 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, gM :=nj=0

    nj([fj ]

    nj ),

    nj([gj ]nj )



    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, gM where

    at least one of f and g is a cusp form. Without loss of generality, let g Sk(n) and write

    f =nj=0[fj ]

    nj . Since f Mk(n) = Sk(n) Sk(n) and thus can also write f = fc + fp

    where fc Sk(n) and fp Sk(n). 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, gM = fn, g+n1j=0

    nj([fj ]nj ), 0 since f Sk(n)

    = fn, g+n1j=0


    = fn, g

    = fc, g

    = fc, g+ fp, g since fp Sk(n)

    = fc + fp, g

    = f, g.

    Now consider f En,`,k and g En,j,k where ` 6= j and, without loss of generality, 0 ` 0 and K is a number field, then

    Mk(n)Q Q K = Mk(

    n)K .

    Proof Let {f1, . . . , fm} be as defined above and consider the following map extended linearly

    : Mk(n)Q Q K Mk(


    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(n)K

    Mk(n). There exists {j}mj=1 C such that f =

    mj=1 jfj . Note that this is equivalent to

    a(T ; f) =


    ja(T ; fj) for all T n.

    Since the fj form a basis, it is possible to take a subset {Ti}mi=1 n 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 =t(1 m) and

    a =t(a(T1; f) a(Tm; f)) produces A = a. Let Aj denote the matrix produced by replacing the

    jth column of A with a. By Cramers rule, for all j, we then have

    j =det(Aj)

    det(A) K since det(A) Q,det(Aj) K.

    Consequentlymj=1(fj j) Mk(n)Q Q K and is thus in the preimage of f , demonstrating

    that is surjective.

    Now supposeri=1(hi i) ker. Combining this with the fact that {f1, . . . , fm} is a


  • basis, we obtain

    0 =








    where ci,j Q=




    )fj .

    By definition of a basis we must then haveri=1 ici,j = 0 for all j. Using this produces


    (hi i) =ri=1


    ci,jfj i




    (ci,jfj i)




    (fj ici,j)









    (fj 0)

    = 0.

    Therefore ker = {0}, showing that the map is injective. Combining this with the surjectivity we

    have that 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(n)Q, there exists c Z>0 such that cf


    Corollary 2.4.3 Let K be a number field and consider a subring K R OK . Then, for every

    f Mk(n)R, there exists c Z>0 such that cf Mk(n)OK .

    Proof Let f Mk(n)R Mk(n)Q. By Lemma 2.4.2 there exists c Z>0 such that cf


  • Mk(n)O. However, we also have that cf Mk(n)R Mk(n)K , and thus

    cf Mk(n)O Mk(n)K = Mk(n)OK = Mk(n)OK .

    We note that c is not unique; in fact, if we define the ideal

    (f)K := { OK : f Mk(n)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(n)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(n)R, we have

    Mk(n)R = Mk(n)OK OK R

    as R-modules.

    Proof We define the following map of R-modules

    R : Mk(n)R Mk(n)OK OK R

    f 7 uf u1,

    where we take u R (f)K , thus u1 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 u1 = uf vv1u1

    = uvf (uv)1

    = vf uu1v1

    = vf v1.


  • 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 (0) = 0 1. For arbitrary f Mk(n)R and R K with , OK and R

    we have,

    R(f) =

    (uf u1)

    = uf u1

    = uf (u)1

    = (u)



    ) (u)1

    = R




    We wish to extend R linearly by defining




    = j=1


    To see this is well defined let f1, f2, f3 Mk(n)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

    R(f1 + f2) = R(f3)

    = u3f3 u13

    = u3f3 (u1u2)(u1u2u3)1

    = (u1u2u3)(f1 + f2) (u1u2u3)1

    =(u2u3(u1f1) (u1u2u3)1

    )+(u1u3(u2f2) (u1u2u3)1


    (u1f1 u11

    )+(u2f2 u12

    )= R(f1) + R(f2).

    Since R 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 = R(f) = uf u1


  • if and only if one of the following three situations occurs:

    1. u = 0,

    2. u1 = 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(n)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 u1 6= 0. Thus we

    must have f = 0, demonstrating kerR = {0}, that is, R is injective.

    Consequently, we know that Mk(n)R is isomorphic to the image of R. Let




    )Mk(n)OK OK R

    with j , j OK and j R. We obtain a preimage through the following process.


    fj jj


    jfj 1j


    jjjfj 1j





    = R



    .Thus R is an isomorphism and we have proven the proposition.

    Taking R = Q, we obtain Mk(n)Q = Mk(n)Z Z Q. Combining this with Lemma 2.4.1


    Mk(n)K = Mk(n)Q Q K

    = Mk(n)Z Z QQ K

    = Mk(n)Z Z K.


  • This is necessary result for Mizumotos proof of the following lemma from [17].

    Lemma 2.4.5 ([17], Lemma A.4) Let {1, . . . , `} OK be an integral basis of K; that is, for

    any K there exists {i}`i=1 Q such that =i=1

    ii. Then

    Mk(n)OK =


    Mk(n)Z i

    for n, k Z>0.

    Proof The first step is to show that

    Mk(n)OK =


    Mk(n)Z i.

    The inclusion is obvious so we turn our attention to . Let f Mk(n)OK . Since we identify

    Mk(n)Z Z K = Mk(

    n)K via

    f =


    jfj mj=1

    (fj j) with m Z>0,

    there exist f1, , fr Mk(n)Z and ai K such that f =ri=1 aifi. Furthermore, since

    {1, . . . , `} OK is an integral basis of K there exist cij Q such that

    ai =j=1

    cijj .

    Thus we have

    f =













    hjj .


  • Since the hj s are Q-linear combinations of f1, , fr Mk(n)Z, we have that all hj Mk(n)Q.

    Now for all T n we have

    a(T ; f) =j=1

    a(T ;hj)j .

    Since {1, . . . , g} is an integral basis of K and a(T ;) OK , we must have a(T ;hj) Z for all

    j since the representation fo a(T ;) is unique with respect to the basis {1, . . . , `} and OK =

    Z[1, . . . , `] as a free Z-module. Consequently, we have hj Mk(n)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 n. Since {1, . . . , `} 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 n 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(n)OK

    = Mk(n)Z Z OK through the linear extension

    of the map

    f =


    hjj 7gj=1

    (hj j).

    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(n)Op with n, k Z>0, there exists a p-unit

    u OK such that uf Mk(n)OK .

    In other words, for all f Mk(n)Op there exists u (OK p) Op such that uf

    Mk(n)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(n)OK OK Op.

    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(n)Op that Op (f)K 6= . Thus we apply Proposition 2.4.4 to get the result.


  • We immediately have the following corollary.

    Corollary 2.4.8 Let p be prime and Z(p) the integers localized at (p). Then


    = Mk(n)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(n)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(n)OK OK Op by Corollary 2.4.7


    n)Z Z OK)OK Op by Lemma 2.4.5

    = Mk(n)Z Z Op

    = Mk(n)Z Z(Z(p) Z(p) Op

    )= Mk(n)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 nr 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

    n(N) :=

    A BC B

    n : A D In (mod N), B C 0n (mod N)

    of n is called the principle subgroup of level N .


  • Observe that n(N) is the kernel of the surjective group homomorphism

    n = Sp2n(Z) Sp2n(Z/NZ)A BC D

    7A BC D

    (mod N).Thus we have n(N) / n as well as

    [n : n(N)] = #Sp2n(Z/NZ)

  • Therefore we get the union is disjoint by taking distinct orbits ngi = ngi. To see that the union

    is finite we use the notation = g1ng n 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 is a congruence subgroup.

    Proof Since g is a rational matrix, there exists N Z>0 such that Ng, Ng1 Mat2n(Z). Note,

    by definition, we have n(N) n. Set N = N3 and observe that

    gn(N)g1 g (I2n +N Mat2n(Z)) g1

    = g(I2n + N

    3 Mat2n(Z))g1

    = I2n + (Ng)(N Mat2n(Z)


    I2n + N Mat2n(Z).

    Furthermore, it is clear that every element of gn(N)g1 has determinant equal to 1. Combining

    results we have gn(N)g1 n(N) and, equivalently,

    n(N) g1n(N)g g1ng.

    Since, by definition, n(N) n we therefore have

    n(N) g1ng n =

    and conclude that is a congruence subgroup.

    Lemma 2.5.4 The set {i} is a set of coset represeentatives of \n if and only if {gi} = {gi} is

    a set of orbit representatives for n\ngn.

    Proof Consider the map of sets

    n n\ngn

    7 ng.


  • This map is surjective since for any n n\ngn there exist 1, 2 n such that

    n = n1g2 = ng2

    which is the image of 2. Now suppose ng1 =

    ng2; then we must have 112 g1ng. We

    also have by definition that 112 n. Consequently 1

    12 g1ng n = , that is, two

    elements of n produce the same orbit in n\ngn if they differ by an element of . Thus the full

    preimage of ng is the coset , demonstrating that there is a bijection between the coset space

    and orbit space. Consequently {i} is a complete set of cosets if and only if {ngi} = {ngi} is

    a complete set of orbits, therefore demonstrating the desired result.

    Since is a congruence subgroup of n, as demonstrated by Lemma 2.5.3, we have the set

    of coset representatives, {i}, of \n is finite. Therefore, by Lemma 2.5.4, we have {gi} = {gi}

    is a finite set of orbit respresentatives for n\ngn. We then conclude that ngn =i

    ngi is a

    finite disjoint union.

    Definition 2.5.5 For g GSp+2n(Q), we define the weight k, ngn operator by

    f [ngn]k :=i

    f |kgi

    where the gi are defined by ngn =



    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 ngn,

    g GSp+2n(Q) with coefficients in C. That is

    Tn =



    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.


  • Given two double cosets

    ngn =


    ngi and ngn =



    where (g) = |n\ngn| = [n : ], we have

    (ngn) (ngn) =

    nhnngngnc(g, g;h)nhn

    where c(g, g;h) is the number of pairs gi, gj such that gig

    j nh. 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 [ngn]k)[ngn]k = f [(

    ngn) (ngn)]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) :=


    aj [ngj

    n]k : m Z0, 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 Cmj=1

    aj [ngj

    n] 7mj=1

    ([ngjn] 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 [ngj



  • with aj R. For T acting on a modular form, f , we write T f where

    T f := fj=1

    aj [ngj

    n]k :=j=1

    ajf [ngj

    n]k .

    Since T is a linear operator, it is natural to ask if it has any eigenvectors; that is, are there any

    f Mk(n) such that T f = (T ; f)f for some (T ; f) C? The short answer is yes, and of

    particular interest are eigen modular forms; modular forms satisfying T f = (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 ngn = 1an(ag)n. This is desirable due to the following proposition.

    Proposition 2.5.7 ([1], Lemma 3.3.6) Let GSp+2n(Q) Mat2n(Z), then the double coset

    nn has a unique representative of the form

    g = diag(a1, . . . , an, d1, . . . , dn)

    with aj , dj Z satisfying aj > 0, ajdj = () 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

    1 = SL2(Z) and GSp+2 (Q) = GL+2 (Q). Let =

    a bc d

    GSp+2 (Q) Mat2(Z). We then have() = ad bc. Using lemma 3.2.2 of [1] we have

    22 = SL2(Z)SL2(Z) = SL2(Z)

    e1 00 e2

    SL2(Z)with e|e2 and note that this is the just the matrix of elementary divisors of over Z. Setting a1 = e1

    and d1 = e2 we get a1|d1 and a1d1 = ad bc = (), thus showing the base case.

    We now assume proposition 2.5.7 holds for all degrees n < m and consider the case n = m.


  • To utilize the induction hypothesis we will show that

    nn = n

    a1 01,n1 0 01,n1

    0n1,1 A4 0n1,1 B4

    0 01,n1 d1 01,n1

    0n1,1 C4 0n1,1 D4



    A4 B4C4 D4

    GSp+2(n1)(Q) Mat2(n1)(Z). To do this, we first assume that is primitivesince if it is not that we set a1 = gcd(), i.e., a1 is the greatest common divisor of the entries of

    and instead would work with where = a1.

    Let = () be the minimum of the greatest common divisors coming from the columns of

    . Let i denote the column corresponding to and note that we can assume i n since we would

    otherwise consider J2n. 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

    V 00 tV 1

    . Now that we have the firstcolumn of produces the values , we assume = 1. In this case, we use Lemma 1.3.9 of [1] to pick

    a representative of n such that the first column is given by t(

    1 0 . . . 0

    ) R2n. Using the

    block form, we may now assume that we have

    1 A2 B1 B2

    0n1,1 A4 B3 B4

    0 C2 D1 D2

    0n1,1 C4 D3 D4

    and observe that for W =

    1 A20n1,1 In1

    we have

    W 0n0n



    a1 01,n1 B1 B2

    0n1,1 A4 B3 B4

    0 C3 D1 D2

    0n1,1 C4 D3 D4



  • 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 to be of the form we just obtained, we

    consider the product

    In S0n In

    where S =B1 B2tB2 0n1

    .This produces a matrix of the form

    1 01,n1 0 01,n1

    0n1,1 A4 B3 B4

    0 C2 D1 D2

    0n1,1 C4 D3 D4


    Using the relation AtB = BtA produces B3 = 0n1,1. Similarly,tAC = tCA produces C2 = 01,n1.

    Combing these facts with the formulas AtD BtC = ()In and tAD tCB = ()In produces

    D1 = (), D2 = 01,n1, D3 = 0n1,1, and

    A4 B4C4 D4

    n1. Thus, our induction hypothesis onthe degree n would complete the proof for = 1. We now consider the case > 1. In this case, we

    can follow the same construction up to a representative of the form

    A2 B1 B2

    0n1,1 A4 B3 B4

    0 C2 D1 D2

    0n1,1 C4 D3 D4


    We then use right multiplication by the matrices U

    1 A20n1,1 In1

    , T 0 B2tB2 0n1

    , and T B1 01,n1

    01,n1 0n1

    where the entries of A2 A2, B2 B2, and B1 B1 are all in the set {1, 2, . . . , }. It should also

    be noted that the B1 reference is the B1 block of


    1 A20n1,1 In1

    T 0 B2tB2 0n1

    ,i.e., it is necessary to reduce A2 and B2 before B1.


  • We denote this new reduced matrix by 0 and observe that since the the first entry of every

    column is less than or equal to that we must have (0) () = . If (0) = () = , then

    we must have divides all the entries of ; however, this is a contradiction since we assumed is

    primitive and > 1. We must then take 1 (0) < and we repeat the process on 0. We then

    iterate the process until (0) = and we apply the earlier case of = 1; that is, we now can pick a

    representative for satisfying

    nn = n

    1 01,n1 0 01,n1

    0n1,1 A4 0n1,1 B4

    0 01,n1 () 01,n1

    0n1,1 C4 0n1,1 D4


    with =

    A4 B4C4 D4

    GSp+2(n1)(Q)Mat2(n1)(Z) with () = (). Our induction hypothesisthen gives the unique representative diag(a2, . . . , an, d2, . . . , dm) of the double coset


    satisfying ai|ai+1, an|dn, and aidi = () = (). Moreover, there exist =

    1 23 4

    and =

    1 23


    such that = diag(a2, . . . , an, d2, . . . , dm). We then have

    1 01,n1 0 01,n1

    0n1,1 1 0n1,1 2

    0 01,n1 () 01,n1

    0n1,1 3 0n1,1 4

    1 01,n1 0 01,n1

    0n1,1 A4 0n1,1 B4

    0 01,n1 () 01,n1

    0n1,1 C4 0n1,1 D4

    1 01,n1 0 01,n1

    0n1,1 1 0n1,1


    0 01,n1 () 01,n1

    0n1,1 3 0n1,1



    1 01,n1 0 01,n1

    0n1,1 diag(a2, . . . , an) 0n1,1 0n1

    0 01,n1 () 01,n1

    0n1,1 0n1 0n1,1 diag(d2, . . . , dn)


    Thus, by induction we have the result for primitive .

    Recall that if is not primitive then we take a1 to be the greatest common divisor of the


  • entries of . We then have = a1 where is primitive and thus nn has a unique representative

    given by diag(1, a2, . . . , an, (

    ), d2, . . . , dn). We observe that () = (a1

    ) = a21(). Combing

    this with nn = a1nn, we have the unique representative

    diag(a1, a1a2, . . . , a1a

    n, a1(

    ), a1d2, . . . , a1d

    n) = diag(a1, . . . , an, d1, . . . , dn).

    We will always choose g in ngn as the unique diagonal matrix described above. Doing

    so shows that ngn = ngn where g = (g)g1. This comes from combining the identity

    tgJg = (g)J with the fact that we pick g to be diagonal and so

    g = tg = (g)Jg1J1 = JgJ1 ngn.

    More generally we have g = J tgJ1 and (g) = g. Using these relations we obtain the


    ngi =


    = ngn

    = (ngn)

    = Jt(ngn)J1

    = JntgnJ1






    gi (n)


    gi n.


    ngi = ngn = ngn =


    n, we have

    f |kgi =

    f |kgj . (2.1)


  • Recall that the summation is finite. It is useful to see the action of the slash operator

    associated with gi . The action is given by

    (f |kgi )(Z) = j(gi , Z)k(gi )nkn(n+1)

    2 f(gi Z)

    = (gi)nkj(g1i , Z)

    k(gi)nkn(n+1)2 f((gi)In (g1i Z))

    = j(g1i , Z)k(gi)

    n(n+1)2 f(g1i 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 i = giZ when necessary and will write Yi = Im(giZ) = Im(i).

    Note that

    f1 [ngn] , f2 =

    f1|kgi, f2

    =f1|kgi, f2



    det(Y )k (f1|kgi) (Z)f2(Z)dn(Z)



    det(Y )k(gi)nkn(n+1)2 j(gi, Z)




    j(gi, Z)k det(Y i )

    k(gi)n(n+1)2 f1(giZ)f2(Z)dn(Z)



    j(gi, g1i i)

    k det(Y i )k(gi)

    n(n+1)2 f1(i)f2(g1i i)dn(g

    1i i)



    det(Y i )kf1(i)(gi)

    n(n+1)2 j(g1i , i)

    kf2(g1i i)dn(i)



    det(Y i )kf1(i)(f2|kgi ) (i)dn(i) by (2.2)

    =f1, f2|kgi



    = f1, f2 [ngn] due to (2.1).


  • To obtain the fifth equality we use the identity

    t(CiZ +Di)Y

    i (CiZ +Di) = (gi)Y.

    Taking the derivative of both sides then produces

    det(Y )k = (gi)nk det(Y )kj(gi, Z)

    kj(gi, Z)k.

    To obtain the seventh equality we use the identity

    j(, Z) = j(, Z)j(, Z)

    which allows us to write the following.

    j(gi, g1i i)

    k = j(gi, g1i i)



    1i , i)


    j(g1i , i)k

    =j(I2n, i)


    j(g1i , i)k

    = j(g1i , i)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(n) 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

    ngn =p



  • Proposition 2.5.8 ([1], Proposition 3.3.9) Let M , M n. Suppose that the quotients





    d1(M )

    )= 1,


    (nMn) (nM n) = nMM n.

    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 nMn from Proposition 2.5.7. It

    is also important to note that Andrianov proves this for arbitrary level, n(N), and in that case the

    double cosets are of the form n0 (N)Mn0 (N) where

    n0 (N) =

    A BC D

    n : C 0 (mod N) n(N).

    We note that for full level we have n0 (1) = n = n(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

    ngn =p


    p prime


    where we gp is defined as the gp in the finite product for p|(g) and I2n for p - (g). Through this

    mapping we obtain the decomposition

    T(Q) =p


    where Tp is the set of formal sums with coefficients in Q of double cosets ngpn satisfying (gp) = pn

    for some n Z. By Theorem 3.3.23 of [1] we know that Tp is generated by elements of the form


  • T (p) = T (n)(p) = n

    In 00 pIn


    Ti(p2) = T

    (n)i =


    Ini 0 0 0

    0 pIi 0 0

    0 0 p2Ini 0

    0 0 0 pIi


    for i = 1, . . . , n, and T(n)n (p2)1 = [npI2n


    which we show is in T by showing it is in T(Q).

    Working over Q we see that

    nI2nn = n(pI2n)(p


    = npI2nnnp1I2n


    = Tn(p2)np1I2n


    We are able to split the double coset by Proposition 2.5.8 since






    )= gcd




    )= gcd(1, 1) = 1.

    This shows that over Q we have

    Tnn (p2)1 = np1I2n

    n = p1nI2nn 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 (n1)1 (p2), . . . , T

    (n1)n1 (p

    2)] given by T 7 T where T is

    Hecke operator satisfying T = T . Taking [ngpn] Tp we then have

    [ngpn] = p` [ngp


    where p` is the largest power of p occuring in the denominator of the entries of gp and gp = p`gp. If


  • gp = I2n then we have

    (f [ngp


    = (p`f [nI2n


    = p` (f [nI2nn]k) = p

    ` ((f))[n1I2(n1)


    since is C-linear, Mk(n) is invariant under [nI2nn]k, and Mk(n1) is invariant under[n1I2(n1)

    n1]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 gives

    the result for single cosets as well as extends to formal sums of cosets, thereby giving the following


    Proposition 2.5.9 There exists a mapping T 7 T of T(n)p to T(n1)p with the following properties.

    (T f) = T (f) for all f Mk(n) 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 ngn =p

    ngpn, we also have the expression

    f [ngn]k = f




    =((f [ngp1

    n]k) )

    [ngpmn]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(n1)p such that

    (f [ngp


    = T p ((f)) .

    Using this we have

    (f [ngn]k) = (((

    f [ngp1n]k) )


    = T pm((((

    f [ngp1n]k) ) [



    = T pm( (T p1 ((f))

    ))= (T pm T

    p1) ((f)) .


  • Since 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 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(n1)(R) such that (T f) = T ((f)). We can extend this result via induction

    to obtain `(T f) = T (`(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(r) be an eigenform for some fixed r 0 and

    even k > n+ r + 1 with n r. Then [f ]nr Mk(n) 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(j).

    Since Sk(j) has a basis of eigenforms. Expressing f as linear combination of the basis shows that

    T f Sk(j) 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(j) and f0 orthogonal to

    En,j,k. We then consider (T f) = [gj ]n1j + (f0) and note that it is equal to

    (T f) = T (f) = [hj ]n1j

    since (f) En1,j,k which is stable under T(n1) by our induction hypothesis. Combining the

    equations yields 0 = [gj ]n1j [hj ]

    n1j + (f0) = [gj hj ]

    n1j + (f0). Since f0 is orthogonal to

    En,j,k, we have (f0) is orthogonal to En1,j,k. Thus we must have gj = hj and (f0) = 0. Since we

    now have f0 Sk(n), it suffices to also show that f0 Sk(n) to obtain En,j,k is stable under

    T(n). To see this, consider any Sk(n) and observe that T f, = f, T = 0 since Hecke

    operators are hermitian, Sk(n) is stable under T(n), and En,j,k is orthogonal to Sk(n). We then

    use this to obtain

    0 = T f, = [gj ]nj , + f0, = f0,

    for all Sk(n) since [gj ]nj , = 0. Thus, f0 Sk(n) 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(r) 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


  • there exists some gj satisfying T [fj ]nj = [gj ]nj , yielding

    T [f ]nr =rj=0

    [gj ]nj and

    nr(T [f ]nr ) =rj=0

    [gj ]rj .

    We also have

    nr(T [f ]nr ) = Tnr([f ]nr ), T


    = T f

    = (T ; f)f

    = (T ; f)rj=0

    [fj ]rj



    [(T ; f)fj ]rj .

    Combining these expressions and using the linearity of []rj yields

    0 =


    [gj (T ; f)fj ]rj .

    Applying r to both sides of the expression yields 0 = gr (T ; f)fr, that is, gr =

    (T ; f)fr and 0 =r1j=0[gj (T

    ; f)fj ]rj . Applying

    r1 then produces gr1 = (T ; f)fr1.

    Continuing this process, we obtain gj = (T ; f)fj and consquently

    T [f ]nr =rj=0

    [gj ]nj =


    [(T ; f)fj ]nj = (T; f)


    [fj ]nj = (T

    ; f)[f ]nr .

    Therefore [f ]nr is an eigenform with (T ; [f ]nr ) = (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(n) be an eigenform. The field Kf = Q({(T ; f) :

    T f = (T ; f)f, T T(n)}) is a totally real number field.

    Proof Let T(n) = HomC(T(n),C). For T(n) we define

    Mk(n;) = {f Mk(n) : T f = (T ; f)f}


  • and

    m() = dimC (Mk(n;)) 0.

    Let (T) = { T : m() 1}. Then we have M = #(T) dimC (Mk(n)). Write Q() =

    Q((T(Q))). Now take Aut(C) and for every g =T0 a(T ; g)q

    T Mk(n) set (g) =T0 (a(T ; g))q

    T . We know from Section 2.4 that Mk(n)Q Q C = Mk(

    n), thus we identify

    g Mk(n) withi gi ci Mk(n)Q Q C. Since Q is fixed by every field automorphism of C

    we have

    (g) = (

    gi ci)

    with ci C, gi Mk(n)Q


    (gi) (ci)


    gi (ci) Mk(n)Q Q C = Mk(n).

    For T we also define ()(T ) = ((T )) for T 1 T(Q)Q C = T. It is also known

    from Equation 2.4 of [27] that Mk(n)Q is T(Q)-stable. Consequently, for g Mk(

    n;) we have

    T (g) = T (

    gi ci)

    with ci C, gi Mk(n)Q

    = T(

    gi (ci))

    =T gi (ci)

    = (

    T gi ci)

    = (T

    gi ci)

    = (T g)

    = ((T ; g)g)

    = ()(T )(g).

    We see that if g Mk(n;) then (g) Mk(n;()). Thus, for any Aut(C)

    and (T) then () (T) and (Q()) = Q(()). Since the automorphisms of C act as

    permutations on the elements of (T), which is finite, we see that Q((T))/Q is Galois, where

    Q((T)) is the composite field. Furthermore, EndQ(Q()) Gal(Q((T))/Q) Sym(m), yielding

    #EndQ(Q()) #Gal(Q((T))/Q) M !


  • showing that Q() is a finite extension.

    To get Q() 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(n) an eigenform we have (T ; f) (T) and thus have Kf :=

    Q({(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(r) is an eigenform then [f ]nr Mk(n)Kf .


  • Chapter 3

    Results of T. Kikuta and S.


    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(n)Op . We say F is a mod pm cusp form if (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(n)Op with k > 2n, there exists G Sk(n)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 p be the normalized additive valuation with respect to p, i.e., p(p) = 1. For

    F Mk(n)K we define the following;

    p(F ) := min{p(a(T ;F )) : T n}

    (n)p (F ) := min{p(a(T ;F )) : T n, rank(T ) = n

    } for 0 n n.


  • Main Corollary Let k > 2n be even and f Sk(r)Kf , n > r, a Hecke eigenform. For the

    Klingen-Eisenstein series [f ]nr attached to f , suppose there exists a prime ideal p in OKf with p 6

    Sn(Kf ) such that (n)p ([f ]

    nr ) = p(([f ]

    nr ))m, for some m Z1. Then there exists F Sk(n)Op

    such that [f ]nr F mod pm for some 0 6= pm.

    3.2 Lemmas and Their Proofs

    By Theorem 2.3 of [6, pp. 150] we know that1k

    Mk(n)Z is finitely generated; that is,

    there exists a set of generators {1, , t} such that


    Mk(n)Z = Z[1, , t]

    as Z-algebras. By Corollary 2.4.7, we have that Mk(n)Z(p)= Mk(n)Z Z Z(p) for any prime p.

    Combining these facts we get


    Mk(n)Z(p) =



    n)Z Z Z(p))




    Z Z(p)= Z[1, , t]Z Z(p)

    = Z(p)[1, , t].

    This allows us to introduce the following lemma.

    Lemma 3.2.1 Let M be a natural number. For any prime p, the algebrak>M

    Mk(n)Z(p) is finitely

    generated over Z(p).

    Proof Let {1, , t} be the set of generators of the Z(p)-algebra

    kMk(n)Z(p) , where i

    Mki(n)Z(p) . For each i, let i Z0 be the smallest integer such that the weight of

    ii , iki, is

    strictly greater than M . We prove thatk>M

    Mk(n)Z(p) is generated over Z(p) by monomials of the


    11 , , tt , (3.1)

    i11 itt (i1k1 + + itkt > M, 0 ij < 2j). (3.2)


  • Note that any F Mk(n)Z(p) can be written as a linear combination of monomials

    a11 att with i1k1 + + itkt = k. Thus it suffices to show that F =


    att can be generated

    by monomials of the form (3.1) and (3.2) for k > M .

    For each ai we can write ai = qii + ri for qi, ri Z with 0 ri < i. Thus we write

    F = a11 att = (


    rtt )


    (ii )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(n)Z(p) . Thus we rewrite our expression of F as

    F = (r11 rj1j1



    rtt )(jj

    )qj1i 6=j

    (ii )qi .

    It is clear that(jj

    )qj1i 6=j (

    ii )

    qi is generated by monomials of the form (3.1), so it remains to

    show that r11 rj1j1



    rtt is of the form (3.2).

    First, each ri 6= rj satisfies 0 ri < i < 2i. For rj we have 0 rj +j < j +j = 2j .

    Thus each individual exponent satisfies its respective bounds. Now observe that the weight of the

    monomial is given by

    (rj + j)kj +i 6=j

    riki = jkj +




    > M

    by the definition of j . Thus r11




    rtt 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 thatk>M

    Mk(n)Z(p) = Z(p)[f1, , fs] with

    fi Mki(n)Z(p) . Then we havek>M

    Mk(n)Op = Op[f1, , fs] for any prime ideal p above p.


  • Proof Let Z(p)[x1, . . . , xs] be a polynomial ring and let I be the kernel of the Z(p)-algebra morphism

    : Z(p)[x1, , xs] Z(p)[f1, , fs]

    defined by (xi) = fi for 1 i s. By Corollary 2.4.8 we have Mk(n)Z(p)Z(p)Op= Mk(n)Op ,

    thus it follows that





    n)Z(p) Z(p) Op)




    )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 -operator K : Mk(n)K Mk(n1)K is


    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

    MKS 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(n)id Mk(n)

    Mk(n1) 0

    = = =

    Sk(n)K K C

    id1 Mk(n)K K CK1 Mk(n1)K K C

    We know the first sequence is short exact and wish to get the second one is as well by showing the


  • two sequences are isomorphic. This amounts to showing



    = =

    Mk(n)K K C

    K1 Mk(n1)K K C

    commutes. To see this, we consider a basis {f1, , fm} of Mk(n)K and with the standard identi-

    fication obtain {f11, , fm1} is a basis of Mk(n)KK C. Thus we see that {f1, , fm} also

    serves as a basis of Mk(n). 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 K(fi) 1.

    Going across and then down we obtain the composition given by

    fi 7 (fi) 7 (fi) 1.

    Since essentially strips away the Fourier coefficients of fi with full rank, we have (fi)

    Mk(n1)K and more importantly (fi) = K(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(n)K K Cid1 Mk(n)K K C

    K1 Mk(n1)K K C 0.

    Since we know C is faithfully flat over K we then have

    0 Sk(n)Kid Mk(n)K

    K Mk(n1)K 0.

    Therefore, we conclude that the restricted Siegel operator K : Mk(n)K Mk(n1)K is surjec-


    In order to define Sn(K), we first make the two following oberservations. First, by Lem-


  • mas 3.2.1 and 3.2.2 we have


  • Since Pj(x1, . . . , xs) is a polynomial in s variables we have

    (Gj) = Pj((F1), . . . ,(Fs)) = P (f1, . . . , fs) = gj .

    We now show that if there exists some i such that p(Gi) < 0 for any choice of Gi 1K (gi) then

    there must be some j such that p(Fj) < 0 for all Fj 1K (fj).

    Since p is an additive valuation and Pi(x1, . . . , xs) is a finite sum of monomials in s vari-

    ables, one of the monomials, say s`=1

    xe`` with 0 6= Op, must satisfy

    p (P (F1, . . . , Fs)) p



    F e``


    Consequently, we have

    0 > p(Gi)

    p (P (F1, . . . , Fs))

    = p



    F e``


    = p +




    e`p(F`) since Op.

    By definition of a polynomial, we must have e` 0 for all `. Hence there must exist some j such

    that p(Fj) < 0. Since Fj 1K (fj), the only way to choose a different form Fj 1K (fj) is if we

    have Fj = Fj + H for some H Sk(n), and hence (Fj) = (Fj) = fj . Using this relation we

    observe that

    (Pi(F1, . . . , Fj1, Fj , Fj+1, . . . , Fs)) = Pi((F1), . . . ,(Fj1),(Fj),(Fj+1), . . . ,(Fs))

    = Pi(f1, . . . , fj1, fj , fj+1, . . . , fs)

    = gi.


  • Thus Pi(F1, . . . , Fj1, Fj , Fj+1, . . . , Fs) 1K (gi) and thus we have


    (Pi(F1, . . . , Fj1, Fj , Fj+1, . . . , Fs)

    )< 0.

    If we are able to choose Fj such that p(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 1K (fj) satisfies p(Fj) < 0. Thus we see that

    p Sn(K) is independent of the set of generators.

    Lemma 3.2.6 For d = [K : Q]

  • It follows that d1Gi 1Q (fi). By our initial assumption, p - d so (p)(d1Gi) 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(n)Op with k > 2n, there exists G Sk(n)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


  • In particular, since Fi Mk(n)Op for all i, we then have P (F1, . . . Fs) Mk(n)Op . Since

    F Mk(n)Op we then obtain G ker K = Sk(n)Op Sk(

    n)K . Therefore we have

    F = P (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(r)Kf , n > r, a Hecke eigenform. For the

    Klingen-Eisenstein series [f ]nr attached to f , suppose there exists a prime ideal p in OKf with

    p 6 Sn(Kf ) such that (n)p ([f ]nr ) = p(([f ]nr )) m, for some m Z1. Then there exists F

    Sk(n)Of,p , where Of,p is the localization of OKf at p, such that [f ]nr F mod pm for some

    0 6= pm.

    Proof Let f Sk(r)Kf , then by [18] we know [f ]nr Mk(n)Kf and [f ]n1r Mk(n1)Kf . We

    further suppose that p / Sn(Kf ) such that (n)p ([f ]nr ) = p(([f ]nr ))m, for some constant m Z1.

    By Corollary 2.4.3, there exists some c Z>0 such that c[f ]n1r Mk(n1)OKf . It is acceptable

    to weaken the choice of c so that we have c[f ]n1r Mk(n1)Of,p

    where Of,p is the localization

    of OKf at p. Either way, we obtain

    p(c[f ]n1r ) = p(c([f ]

    nr )) 0.

    Moreover, we have

    (n)p (c[f ]

    nr ) = p(c([f ]

    nr ))m m.

    We now take some 0 6= pm and observe that

    (n)p (c[f ]

    nr ) = p() +

    (n)p (c[f ]

    nr ) = p() + p(c([f ]

    nr ))m mm = 0.

    We now set = c, noting that we have pm, and observe that [f ]nr Mk(n)Of,p

    . Further-

    more, we have

    p(([f ]nr )) = p(c[f ]

    n1r ) = p() + p(c[f ]

    n1r ) m.


  • This tells us that [f ]n1r vanishes (mod pm), demonstrating that [f ]nr Mk(n)O

    f,pis a

    (mod pm) cusp form. We now apply the Main Theorem to get F Sk(n)Of,p

    such that

    [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 (n)k be the normalized weight k Siegel-Eisenstein series; that is,

    (n)k (Z) = [1]

    n(Z) =

    n,0\nj(, Z)k

    with a(


    )= 1 where 0n is the n n 0 matrix. It is known that (n)k Mk(n)Q and in

    particular, by [7], that

    10 = 43867

    212 35 52 7 53






    12 =131 593

    213 37 53 72 337


    (2)4 )

    3 + 2 53((2)6 )2 691(2)12

    )are normalized cusp forms in M10(

    2)Q and M12(2)Q, respectively. In [8] these forms are further

    normalized to

    X10 = 410 and X12 = 1212

    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(2)Z for k = 10, 12.

    Theorem 3.4.1 ([19], Theorem 4.3) Assume p 5. If F Mk(2)Z(p) , then there exists a

    unique polynomial P Z(p)[x1, x2, x3, x4] such that

    F = P(

    (2)4 ,

    (2)6 , X10, X12


    Note that [19] uses the notation 10 and 12 instead of our X10 and X12, but states earlier


  • 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 k2Z

    Mk(2)Z(p) = Z(p)


    (2)4 ,

    (2)6 , X10, X12


    By [24], we have (3)k



    )for k > 4 since

    (3)k = [1]

    (3) and 1 Sk(0). We define

    F10 =43867

    210 35 52 7 53





    )F12 =

    131 593211 36 53 72 337


    (3)4 )

    3 + 2 53((3)6 )2 691(3)12


    and immediately see that Q(F10) = X10 and Q(F12) = X12. Thus it is clear that S3(Q) is

    contained in the set of primes p satisfying (p)



    )< 0 for k = 4, 6 or (p)(Fk) < 0 for k = 10, 12.

    For the primes satisfying (p)



    )< 0 for k = 4, 6 we refer to Bocherer [2]. For (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 1Q (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 k = (2)k . Furthermore, we let S12(1) be Ramanujans delta function.

    To simplify notation, we adopt the standard notation (m, r, n) to represent

    m r/2r/2 n

    2.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((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


  • p - 2 3. Suppose k is even and F Mk(2)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 Ramanujans -function and set f12 = 7 S12(1). We then have [f12] is a mod

    7 cusp form. By the main corollary, there exists a cusp form F12 S12(2) such that [f12] F12

    mod 7. To confirm this numerically we consider F12 := X12 S12(2) and observe the following


    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(1) = 1 and thus posses a unique newform C16 S16(1)Z with a(1;C16) = 1. Setting

    f16 = 72 11 C16 we obtain a cusp form f16 S16(1) satisfying a(1; f16) = 72 11. Since C16


  • 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

    ([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() where 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(1) and take F16 S16(2) 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(2) such that [f16] F16 = g for some p\p2. We then

    apply the theorem again, this time to 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 1([f16] F16) 0 (mod p). Thus applying the theorem repeatedly yields

    [f16] F16 (mod p2).

    Weight 20

    It is known that dimS20(1) = 1 and thus, via scaling, we have f20 S20(1) 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(2) with explicit


    F20 = 38



    (2)6 X10 +



    )2X12 1785600X210


    We have the following table of Fourier coefficients.


  • 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(1) = 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(2) to

    be the cusp form from [14] scaled by 29 . Thus we have

    F22 =2





    )3X10 5



    )2X10 + 30


    (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


    [f22] F22 (mod 61).


  • 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 4 = (2)4 and 6 =

    (2)6 . Throughout this section it is important to note that

    any form a4b6X


    d12 with c > 0 or d > 0 is a cusp form since



    d12) = (4)

    a (6)

    b (X10)

    c (X12)


    It is important to recall thatk2Z

    Mk(2)Z(p) = Z(p)


    (2)4 ,

    (2)6 , X10, X12


    Weight 12

    We start by highlighting why Takemori and Kikuta chose f12 = 7. It is known by [14]


    []21 =1



    172826 +


    7X12 =


    26 3334


    26 3326 +

    25 32


    This shows that (2)7 ([]

    21) = 1 while 7(([]21)) = 7() = 0 since a(1; ) = 1 and S12(1).

    Hence, we take 7 (7) in order to get

    7[]21 =7



    172826 + 288X12 X12 (mod 7).

    Weight 16

    Let C16 S16(1) 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 =


    26 3344


    26 334

    26 +






    26 3344


    26 334

    26 +

    24 3 5 9772 11

    6X10 27 32

    7 114X12

    It is immediate that we have

    49[C16]21 36X10 (mod 7);


  • however, to match the corollary we must consider 49[C16]21 (mod 49). In that case we have

    49[C16]21 456X10 284X12 (mod 49).

    We need to show that the right-hand side does not vanish, otherwise this is a trivial example. Using

    Takemoris code from GitHub associate with his paper [22] we have a((1, 0, 1);6X10) = 2 and

    a((1, 0, 1);4X12) = 10, demonstratin