WEN-CHING WINNIE LI AND YOTSANAN MEEMARK
Abstract. In this paper, we study the Drinfeld cusp forms for Γ1(T ) and Γ(T ) using
Teitelbaum’s interpretation as harmonic cocycles. We obtain explicit eigenvalues of Hecke operators associated to degree one prime ideals acting on the cusp forms for Γ1(T ) of
small weights and conclude that these Hecke operators are simultaneously diagonalizable. We also show that the Hecke operators are not diagonalizable in general for Γ1(T ) of
large weights, and not for Γ(T ) even of small weights. The Hecke eigenvalues on cusp forms for Γ(T ) with small weights are determined and the eigenspaces characterized.
1. Introduction
Hecke operators played a crucial role in the study of the arithmetic of classical modular forms. Their actions on cusp forms are skew Hermitian with respect to the Petersson inner product, and hence they are diagonalizable. This property is fundamental in un-derstanding the classical cusp forms.
The function field analogue of the Poincare upper half plane is the Drinfeld upper half plane. Parallel to the classical modular forms, there are the Drinfeld modular forms introduced by Goss in [Gos80]. He also defined the Hecke operators in a similar way. While certain arithmetic properties are alike for classical and Drinfeld modular forms, there are also sharp differences. For instance, B¨ockle [B¨oc04] showed that the Eichler-Shimura correspondence over a function field associates a Drinfeld (cuspidal) common
Date: August 31, 2007.
2000 Mathematics Subject Classification. Primary: 11F52; Secondary: 20E08.
Key words and phrases. Drinfeld cusp forms; Harmonic cocycles; Hecke operators.
The research of the first author is supported in part by the NSF grant DMS-0457574. Part of the research was performed while she was visiting the National Center for Theoretical Sciences, Mathematics Division, in Hsinchu, Taiwan. She would like to thank the Center for its support and hospitality. The second author was supported in part by Grants for Development of New Faculty Staff from Chulalongkorn University, Thailand.
eigenform of Hecke operators to a degree one, instead of degree two as in the classical case, Galois representation, reflecting different multiplicative relations on Hecke operators. Moreover, since the domain and image of Drinfeld modular forms have the same positive characteristic, there is no adequate analog of the Petersson inner product. Hence the diagonalizability of the Hecke operators on Drinfeld forms still remains an open question. Using the residue map, Teitelbaum [Tei91] in 1991 gave an interpretation of Drinfeld cusp forms as harmonic cocycles on the directed edges of a regular tree T . The actions of the Hecke operators were carried over to harmonic cocycles by B¨ockle [B¨oc04]. Since the directed edges of T are parametrized by cosets of PGL2 over a local field F modulo its Iwahori subgroup I, the Drinfeld cusp forms for a congruence subgroup Γ can then be regarded as vector-valued left Γ-equivariant functions on PGL2(F )/I, and hence they are determined by the values on Γ\PGL2(F )/I. This viewpoint is quite helpful in com-putation when a fundamental domain is easily described. Another advantage is that, by means of the strong approximation theorem, the Drinfeld cusp forms can also be seen as equivariant functions in adelic setting. This approach appeared in Gekeler and Reversat [GR96] and also in B¨ockle [B¨oc04].
Let K = Fq(T ) be the rational function field. The arithmetic of Drinfeld modular forms
for the full modular group GL2(Fq[T ]) was studied extensively in [Gos80] and [Gek88].
Using geometric methods, B¨ockle and Pink investigated in [B¨oc04] the structure of double cusp forms for Γ1(T ) with weight k ≤ q + 2. They also computed the Hecke eigenvalues for weight 4 double cusp forms.
The purpose of this paper is to study Drinfeld cusp and double cusp forms for the congruence subgroups Γ1(T ) and Γ(T ) of GL2(Fq[T ]), with emphasis on the behavior of
the Hecke operators. Working with harmonic cocycles, we determine the eigenvalues and the corresponding eigenspaces for Hecke operators at degree one places of K. As we shall see, the diagonalizability of the Hecke operators depends on the group and also the weight. More precisely, the Hecke operators on the space of cusp forms of Γ1(T ) are diagonalizable for small weights k ≤ q, but not for large weights k > q in general. Further, as we pass from Γ1(T ) to its subgroup Γ(T ), the distinct eigenvalues for Hecke operators on cusp forms with weights k ≤ q remain the same although the multiplicities may differ. We
also characterize each eigenspace. Explicit computations show that the Hecke actions on the spaces of cusp forms and double cusp forms for Γ(T ) of small weights change from diagonalizable to not diagonalizable as the weight increases.
This paper is organized as follows. The Drinfeld cusp forms and properties of the tree are reviewed in Sections 2 and 3, respectively. Harmonic cocycles are recalled in Section 4. In Section 5 we summarize Teitelbaum’s isomorphism between Drinfeld cusp forms and harmonic cocycles and describe B¨ockle’s criterion of double cusp forms as harmonic cocycles. The actions of the Hecke operators on harmonic cocycles are introduced in Section 6. The body of this paper is Sections 7 and 8, dealing with cusp forms for Γ1(T ) and Γ(T ), respectively. The final section gives examples of the Hecke actions on the cusp forms for Γ(T ) for weights k = 3, 4 and 5, making explicit the main results of the paper. This paper grows out of the second author’s thesis [Mee06], written under the direction of the first author.
2. Drinfeld Cusp Forms
Let K = F(T ) be the rational function field over the finite field F with q elements. Write ∞ for the place of K with 1/T as a uniformizer. Then A = F[T ] is the ring of functions in K regular outside ∞. Denote by K∞ the completion of K at ∞, O∞ its
ring of integers, and P∞ the maximal ideal in O∞. Let C = b¯K∞ be the completion of an
algebraic closure of K∞.
The Drinfeld upper half plane Ω = C r K∞ is endowed with a rigid analytic
struc-ture, on which GL2(K∞) acts by fractional linear transformations. For γ = (a bc d) ∈
GL2(K∞), m, k ∈ Z and f : Ω → C, define
(f |
k,m
γ)(z) := f (γz)(det γ)m(cz + d)−k.
Let Γ be a congruence subgroup of the modular group GL2(A). It has finitely many cusps, represented by Γ\P1(K). A rigid analytic function f : Ω → C is called a Drinfeld
cusp form for Γ of weight k and type m for Γ if it satisfies
(i) f |
k,m
γ = f for all γ ∈ Γ;
(iii) f vanishes at all cusps.
The cusp forms for Γ of weight k and type m form a vector space Sk,m(Γ) over C. It
contains a subspace S2
k,m(Γ) of double cusp forms, which vanish at all cusps at least twice.
Remark. While the weight can be any integer, the possible type is an element in Z/(mΓ), where mΓ is the order of det(Γ), a subgroup of F×q. Thus Sk,m(Γ) 6= 0 implies k ≡ 2m
mod (mΓ). In particular, if mΓ = 1, which is the case to be considered in this paper, then for fixed k, all Sk,m(Γ) are identical, and the same holds for Sk,m2 (Γ).
The following dimension formula for cusp forms was computed by Teitelbaum.
Proposition 1 ([Tei91]). Let gΓ be the genus of Γ\ ¯Ω and hΓ the number of cusps of Γ\Ω.
If Γ is p0-torsion free and m
Γ = 1, then
dimCSk,m(Γ) = (k − 1)(gΓ+ hΓ− 1).
3. The Tree T
The coset space PGL2(K∞)/PGL2(O∞) =: T may be interpreted as a (q + 1)-regular
tree on which the group GL2(K∞) acts by left translations. The vertices of T are
the cosets PGL2(K∞)/PGL2(O∞), while the directed edges of T are parametrized by
PGL2(K∞)/I∞, where
I∞ = {(a bc d ) ∈ GL2(O∞) : c ∈ P∞}/{(a 00 a ) ∈ GL2(O∞)}
is the Iwahori subgroup of PGL2(O∞). The edge represented by g ∈ GL2(K∞) will be
abbreviated as hgi.
As in Serre [Ser80], a vertex or edge of T is called Γ-stable if its stabilizer in Γ is trivial; otherwise it is Γ-unstable. Let T∞ be the subgraph of T consisting of unstable
vertices and edges. Then S0 = Vert(T ) r Vert(T∞) is the set of stable vertices and
S1 = [Edge(T ) r Edge(T∞)]/± is the set of non-oriented stable edges.
Two paths in T are considered equivalent if they differ at only finitely many edges. An
end of T is an equivalence class of infinite paths {e1, e2, . . . }. There is a canonical bijection between the set of ends and P1(K
corresponding to the cusps. The stabilizer of an unstable vertex v fixes a unique rational end, and similarly for an unstable edge e; denote them by b(v) and b(e), respectively. An edge w of T is a source of an unstable edge e if w has the same orientation as e and there exists an unstable boundary vertex v of w such that the path from v to its end b(v) passes through e. If e is stable, then it is its own source. Denote by src(e) the set of all sources of e. There are certain inaccuracies in [Tei91] concerning the sources of an edge. We thank the referee for pointing them out.
4. Harmonic Cocycles
For k ≥ 0 and m ∈ Z, let V (k, m) be the (k − 1)-dimensional vector space over C with a basis {XjYk−2−j : 0 ≤ j ≤ k − 2} endowed with the action of GL
2(K∞) given by
γ = (a b
c d ) : XjYk−2−j 7→ (det γ)m−1(dX − bY )j(−cX + aY )k−2−j
for all 0 ≤ j ≤ k − 2. This then induces the action of γ = (a b
c d ) ∈ GL2(K∞) on the dual
space Hom(V (k, m), C) by sending w ∈ Hom(V (k, m), C) to
(γw)(XjYk−2−j) = (det γ)1−mw((aX + bY )j(cX + dY )k−2−j) for 0 ≤ j ≤ k − 2.
A harmonic cocycle of weight k and type m for Γ is a function c from the set of directed edges of T to Hom(V (k, m), C) satisfying
(a) For all vertices v of T ,
X
e7→v
c(e) = 0,
where e runs through all edges in T with terminal vertex v;
(b) For all edges e of T , c(¯e) = −c(e), where ¯e denotes e with reversed orientation;
(c) It is Γ-equivariant, namely, for all edges e and elements γ ∈ Γ, c(γe) = γ(c(e)).
The last condition means
for all (a b
c d ) ∈ Γ and 0 ≤ j ≤ k − 2. Let Hk,m(Γ) denote the space of harmonic cocycles
of weight k and type m for Γ.
As observed by Teitelbaum [Tei91], the value of a cocycle c ∈ Hk,m(Γ) at a directed
edge e is the sum of c evaluated at the source of e. Consequently, cocycles in Hk,m(Γ) are
determined by their values on Γ\S1.
5. Cusp Forms and Harmonic Cocycles
There is a building map from Ω to T commuting with the action of GL2(K∞) (cf. [Fv04]
and [Tei91]). Using it one can define, for any C-valued holomorphic 1-form f (z)dz on Ω, the residue Resef (z)dz at any directed edge e of T . This in turn gives a way to associate
harmonic cocycles to cusp forms. More precisely, for each cusp form f ∈ Sk,m(Γ), define
the function Res(f ) from the directed edges of T to Hom(V (k, m), C) by assigning, for any directed edge e, the values of Res(f )(e) at the basis elements XjYk−2−j to be
Res(f )(e)(XjYk−2−j) = Resezjf (z)dz
(5.1)
for all 0 ≤ j ≤ k − 2. Then properties (a) and (b) follow from the rigid analytic residue theorem, and (c) from the modularity of f . Therefore Res(f ) lies in Hk,m(Γ).
Theorem 2 (Teitelbaum [Tei91]). The residue map Res : Sk,m(Γ) → Hk,m(Γ) is an
isomorphism.
Thus we identify cusp forms with harmonic cocycles. This allows us to view cusp forms for Γ as vector valued left Γ-equivariant functions on PGL2(K∞)/I∞, or left GL2 (K)-equivariant functions on the adelic group GL2(AK) by applying the strong approximation
theorem (cf. [GR96] and [Rev00]). When k = 2, such functions are C-valued and Γ-equivariance becomes Γ-invariance. Indeed, some harmonic cocycles can be lifted to Z-valued functions on GL2(K)\GL2(AK), as remarked in [GR96], [Rev00] and [B¨oc04].
Denote by H2
k,m(Γ) the image of Sk,m2 (Γ) under the Res map. To describe double cusp
forms as cocycles, we define the source of an end [s] to be
where t(e) denotes the terminal vertex of e. The following result of B¨ockle characterizes the image of double cusp forms under the residue map.
Theorem 3 (B¨ockle [B¨oc04]). Let Γ[s] denote the Γ-stabilizer of an end [s] representing
a cusp of Γ. Then
(a) The subspace of V (k, m) stabilized by Γ[s], denoted V (k, m)Γ[s], is one-dimensional. (b) Γ[s] acts freely on src([s]) with finitely many orbits, represented by edges e[s]1 ,
. . . , e[s]ls.
(c) Let f ∈ Sk,m(Γ) and c = Res(f ). Then f is a double cusp form if and only if for
any cusp [s], Pls
i=1c(e
[s]
i )(gs) = 0 for any generator gs of V (k, m)Γ[s].
Combined with Proposition 1, one obtains the dimension formula for the space of double cusp forms:
Proposition 4 (B¨ockle [B¨oc04]). Let gΓ be the genus of Γ\ ¯Ω and hΓ the number of cusps
of Γ\Ω. If Γ is p0-torsion free and m
Γ = 1, then dimCSk,m2 (Γ) = gΓ if k = 2; (k − 2)(gΓ+ hΓ− 1) + gΓ− 1 if k > 2. 6. Hecke Operators
We shall focus on the congruence groups Γ = Γ1(T ) and Γ(T ) defined as Γ1(T ) = {(a bc d) ∈ GL2(A) : a ≡ d ≡ 1 and c ≡ 0 mod T } and
Γ(T ) = {(a b
c d) ∈ GL2(A) : a ≡ d ≡ 1 and b ≡ c ≡ 0 mod T } .
They are p0-torsion free. Let P 6= (T ) be a maximal ideal of A; choose the generator P
to be the irreducible polynomial in P satisfying P (0) = 1. Suppose deg P = d. Then Γ(T ) (P 0 0 1) Γ(T ) = Γ(T ) (P 00 1) t G b∈A,deg b<d Γ(T )¡1 b(1−P ) 0 P ¢ .
The Hecke operator at P is defined using the coset representatives of this double coset:
TP = Pk−m−1 £ (P 0 0 1) + X b∈A,deg b<d ¡1 b(1−P ) 0 P ¢¤ ,
which acts on a holomorphic function f on Ω via | k,m TP. That is, TPf (z) = ¡ f | k,m TP ¢ (z) = Pk−m−1£f | k,m (P 0 0 1) (z) + X b∈A,deg b<d f | k,m ¡1 b(1−P ) 0 P ¢ (z)¤.
The generator P is chosen in order to avoid the use of characters. Here we have followed the normalization in B¨ockle [B¨oc04], which is a constant multiple of that defined by Goss [Gos80]. It is easy to check that TP sends Sk,m(Γ) to itself and preserves the double cusp
forms. For two prime ideals P and Q not equal to (T ), TP commutes with TQ.
The action of the Hecke operator TP can be transported to harmonic cocycles by means of the residue map. This was carried out in [B¨oc04]. Precisely, TP sends c ∈ Hk,m(Γ) to
a harmonic cocycle whose value at a directed edge e of T is (6.1) TPc(e) = Pk−m−1 ³ (P 0 0 1)−1c ((P 00 1) e) + X b∈A,deg b<d ¡1 b(1−P ) 0 P ¢−1 c¡¡1 b(1−P ) 0 P ¢ e¢´.
This formula will be used to compute the eigenvalues and eigenfunctions of Hecke op-erators. As we shall see from the cases Γ = Γ1(T ) and Γ(T ), the Hecke operators are sometimes diagonalizable and sometimes not, depending on the group and the weight.
7. Cusp Forms for Γ1(T )
In this section we consider cusp forms and double cusp forms for Γ1(T ). We may choose as a fundamental domain of Γ1(T )\T the path connecting the cusp [∞] =
¡1 0 ¢ and cusp [0] =¡01¢, as shown below. [∞] · · · · ¿¡1 0 0 T2 ¢ h(T 0←−0 1)i −→ D³ 1 0 0 T2 ´E( 1 0 0 T) γ0=h(0 11 0)i ←− −→ ¯ γ0=h(1 00 T)i (1 0 0 1) h(0 T 1 0)i ←− −→ h(1 0 0 1)i (T 0 0 1) ¿ · · · · [0] It contains no stable vertices and one stable edge h(0 1
1 0)i, denoted by γ0. Then gΓ1(T ) = 0
so that dimCSk,m(Γ1(T )) = k − 1 by Proposition 1, and dimCS2,m2 (Γ1(T )) = 0 and dimCSk,m2 (Γ1(T )) = k − 3 for k ≥ 3 by Proposition 4. Theorem 3 of [Tei91] implies that any harmonic cocycle c for Γ1(T ) automatically vanishes on all edges of the fundamental domain except γ0 and its two neighboring edges up to orientation. Further, the value of c at γ0 determines its values at the two neighboring edges by harmonicity. Therefore to de-termine a harmonic cocycle for Γ1(T ), it suffices to first know its value in Hom(V (k, m), C)
at γ0, and then extend to other edges by Γ1(T )-equivariancy and harmonicity. This is the strategy we shall use to compute the action of the Hecke operators.
The stabilizers of the cusps [∞] and [0] are (Γ1(T ))[∞] = {(1 c0 1) : c ≡ 0 mod T } and (Γ1(T ))[0] = {(1 0c 1) : c ≡ 0 mod T }, respectively. Thus V (k, m)(Γ1(T ))[∞] and
V (k, m)(Γ1(T ))[0]are generated by Yk−2and Xk−2, respectively. Also, (Γ
1(T ))[∞]\src([∞]) =
{¯γ0 = h(1 00 T)i} and (Γ1(T ))[0]\src([0]) = {γ0}. Recall that ¯γ0 is the opposite of γ0. Hence by Theorem 3, we have
Proposition 5. S2
k,m(Γ1(T )) = {c ∈ Sk,m(Γ1(T )) : c(γ0)(Yk−2) = c(γ0)(Xk−2) = 0}. Now we study the action of the Hecke operators TP on Sk,m(Γ1(T )), where P is gen-erated by P = 1 + αT . Using equation (6.1), harmonicity and Γ1(T )-equivariancy, and noting q is the cardinality of the field F, we get, for 0 ≤ j ≤ k − 2,
(7.1) TPc(γ0)(XjYk−2−j) = c(γ0) ³ Xj(P Y )k−2−j + bq−1j c X m=0 ³j−m(q−1)X l=0 ¡ j l+m(q−1) ¢¡k−2−j l ¢ (1 − P )l− Pk−2−j¡m(q−1)j ¢´Xj−m(q−1)Y(k−2−j)+m(q−1) + bk−2−jq−1 c X n=1 ³ k−2−jX l=n(q−1) ¡ j l−n(q−1) ¢¡k−2−j l ¢ (1 − P )l− Pj¡n(q−1)k−2−j¢Tn(q−1) ´ Xj+n(q−1)Y(k−2−j)−n(q−1) ´ .
For each 0 ≤ j ≤ k − 2, define the harmonic cocycle cj by specifying its value at γ0 by: (7.2) cj(γ0)(XjYk−2−j) = 1 and cj(γ0)(XlYk−2−l) = 0 for l 6= j.
Further, put, for 0 ≤ j ≤ k − 2 and a degree one polynomial Q = 1 + βT , the polynomial
(7.3) λj(Q) = j X l=0 ¡j l ¢¡k−2−j l ¢ (1 − Q)l = min{j,k−2−j}X l=0 ¡j l ¢¡k−2−j l ¢ (−βT )l.
Then λj(Q) has degree at most min{j, k − 2 − j}. Note that λ0(Q) = λk−2(Q) = 1 and
λj(Q) = λk−2−j(Q) for all 0 ≤ j ≤ k − 2.
To see the behavior of the Hecke operators, we distinguish two cases, according to the weight being small or large. First assume q ≥ k ≥ 2. In this case (7.1) is reduced to (7.4) TPc(γ0)(XjYk−2−j) = λj(P )c(γ0)(XjYk−2−j).
Therefore each cj is an eigenfunction of TP with eigenvalue λj(P ). We have shown
Theorem 6. Let P be a prime ideal of A generated by P with P (0) = 1 and deg P = 1.
Suppose q ≥ k ≥ 2. Then
(1) Each cj, 0 ≤ j ≤ k − 2, is an eigenfunction of the Hecke operator TP with
eigenvalue λj(P ); and
(2) The Hecke operators at the ideals of degree one are simultaneously diagonalized on
Hk,m(Γ1(T )) with respect to the basis cj, 0 ≤ j ≤ k − 2.
It is natural to ask if the cj, 0 ≤ j ≤ k − 2, are also common eigenfunctions of the
Hecke operators TP for prime ideals P of degree d > 1; and if so, find the eigenvalues. Our computations lead to the following
Conjecture. Let P be a prime ideal of A generated by P with P (0) = 1 and deg P =
d ≥ 1. Suppose q ≥ k ≥ 2. Let θ be a root of P . Then each cj, 0 ≤ j ≤ k − 2, is
an eigenfunction of the Hecke operator TP with eigenvalue λj(P ) :=
Qd−1
i=0 λj(1 + θq
i
T ).
Consequently, the Hecke operators are simultaneously diagonalized on Sk,m(Γ1(T )). We have verified this conjecture for d ≤ 2. Another evidence is provided by [B¨oc04] for the case k = 4 and all d.
Remark. If we factor the polynomial λj(1+T ) =
Qdeg λj(1+T )
s=1 (1+δsT ), then the eigenvalue
λj(P ) above can also be expressed as
Qdeg λj(1+T )
s=1 P (δsT ).
It is worth pointing out that the degree of λj(P ) above is at most d(k − 2)/2. This may
be regarded as the Ramanujan conjecture on Drinfeld cusp forms.
Notice that for k ≤ q + 2 and 1 ≤ j ≤ q − 2, equation (7.1) is easily reduced to (7.4) as well. Therefore for P of degree 1, k = q + 1 and 1 ≤ j ≤ k − 3, one gets
TPcj = λj(P )cj.
Recall that a double cusp form c for Γ1(T ) satisfies c(γ0)(Yk−2) = c(γ0)(Xk−2) = 0. Therefore c1, . . . , ck−3 form a basis of the subspace of double cusp forms, on which a
Proposition 7. Let P be a prime ideal of A generated by the polynomial P of degree 1
with P (0) = 1. If q + 2 ≥ k ≥ 4, then for all c ∈ S2
k,m(Γ1(T )) and 1 ≤ j ≤ k − 3, one has
TPc(γ0)(XjYk−2−j) = λj(P )c(γ0)(XjYk−2−j).
Proof. It remains to prove the proposition for the case k = q + 2, and j = 1 or k − 3. In
this case, equation (7.1) gives, for c ∈ S2
k,m(Γ1(T )),
TPc(γ0)(XYk−3) = λ1(P )c(γ0)(XYk−3)
and
TPc(γ0)(Xk−3Y ) = λk−3(P )c(γ0)(Xk−3Y )
since c(γ0)(Yk−2) = c(γ0)(Xk−2) = 0. ¤
Corollary 8. Let P be a degree one prime ideal of A generated by the polynomial P
with P (0) = 1. If q + 2 ≥ k ≥ 4, then cj, 1 ≤ j ≤ k − 3, are eigenfunctions of the
Hecke operator TP with eigenvalue λj(P ). Further, the Hecke operators for degree one
prime ideals are simultaneously diagonalized on S2
k,m(Γ1(T )) with respect to the basis cj,
1 ≤ j ≤ k − 3.
Note that there are no nonzero double cusp forms for weight k < 4. The above result for k = 4 is Proposition 15.6 of [B¨oc04], proved by B¨ockle and Pink.
We now consider the case of general weight k. Assume P = (P ), where deg P = 1 and
P (0) = 1. Again, we appeal to (7.1). For i = 0, 1, . . . , q − 2 and mi = 0, 1, . . . , bk−2−iq−1 c,
we have
TPc(γ0)(Xi+mi(q−1)Yk−2−(i+mi(q−1))) = c(γ0) ³
Xi+mi(q−1)(P Y )k−2−(i+mi(q−1))
+ mi X m=0 ³i+(miX−m)(q−1) l=0 ¡i+mi(q−1) l+m(q−1) ¢¡k−2−(i+mi(q−1)) l ¢ (1 − P )l
− Pk−2−(i+mi(q−1))¡i+mi(q−1)
m(q−1) ¢´ Xi+(mi−m)(q−1)Yk−2−(i+(mi−m)(q−1)) + bk−2−i q−1Xc−mi n=1 ³k−2−(i+mXi(q−1)) l=n(q−1) ¡i+mi(q−1) l−n(q−1) ¢¡k−2−(i+mi(q−1)) l ¢ (1 − P )l
− Pi+mi(q−1)¡k−2−(i+mi(q−1))
n(q−1)
¢
Recall the function cj defined by (7.2). For i = 0, 1, · · · , q − 2, denote by Sk,m(Γ1(T ))i the
subspace of Sk,m(Γ1(T )) generated by {ci, ci+(q−1), · · · , ci+bk−2−i
q−1 c(q−1)} so that Sk,m(Γ1(T )) = Lq−2
i=0Sk,m(Γ1(T ))i. The above calculation proves the following
Theorem 9. Let P = (P ), where deg P = 1 and P (0) = 1. Then for each i = 0, 1, . . . ,
q−2, Sk,m(Γ1(T ))i is invariant under TP. The action of TP restricted to Sk,m(Γ1(T ))i with
respect to the basis {ci, ci+(q−1), · · · , ci+bk−2−iq−1 c(q−1)} is represented by the matrix [TP]i =
α(i)0,0+ Pk−2−i β(i)
0,1 β (i) 0,2 . . . β (i) 0,bk−2−i q−1 c α(i)1,1 α (i) 1,0+ Pk−2−(i+(q−1)) β (i) 1,1 . . . β (i) 1,bk−2−iq−1 c−1
α(i)2,2 α(i)2,1 α(i)2,0+ Pk−2−(i+2(q−1)) . . . β(i)
2,bk−2−iq−1 c−2 .. . ... ... . .. ... α(i) bk−2−iq−1 c,bk−2−iq−1 c α (i) bk−2−iq−1 c,bk−2−iq−1 c−1 α (i) bk−2−iq−1 c,bk−2−iq−1 c−2 . . . α (i) bk−2−iq−1 c,0+ P k−2−(i+bk−2−iq−1 c(q−1)) where α(i)mi,m= i+(mi−m)(q−1)X l=0 ¡i+mi(q−1) l+m(q−1) ¢¡k−2−(i+mi(q−1)) l ¢
(1 − P )l− Pk−2−(i+mi(q−1))¡i+mm(q−1)i(q−1)¢ and βm(i)i,n = k−2−(i+mXi(q−1)) l=n(q−1) ¡i+mi(q−1) l−n(q−1) ¢¡k−2−(i+mi(q−1)) l ¢ (1 − P )l − Pi+mi(q−1)¡k−2−(i+mn(q−1)i(q−1))¢Tn(q−1)
for mi = 0, 1, . . . , bk−2−iq−1 c, 0 ≤ m ≤ mi and 1 ≤ n ≤ bk−2−iq−1 c − mi.
Using geometric arguments, B¨ockle and Pink computed the above structures for the space of double cusp forms of k = 5, q = 2 and k = 6, q = 3 in Proposition 15.3 of [B¨oc04]. To illustrate the above theorem, we give two examples of cusp forms with weights k > q; in the first each Hecke action is diagonalizable, while in the second it is not.
Example 10. q = 3, k = 5 and P = 1 + αT, where α = ±1. There are two invariant subspaces under TP, namely, S5,m(Γ1(T ))0 and S5,m(Γ1(T ))1 spanned by {c0, c2} and
{c1, c3}, respectively. With respect to these bases, we have [TP]0 = (αT P1 0) and [TP]1 = ¡ P αT3 0 1 ¢ ,
Example 11. q = 2 and k = 5. There is only one polynomial P = 1 + T to consider. Further there is only one residue class mod q − 1 given by i = 0, so one has
[TP]0 = µ 1 0 0 0 T2 1 T2 T3 T T 1 T3 0 0 0 1 ¶ .
Thus TP has the eigenvalue 1 of multiplicity two with two linearly independent eigenfunc-tions c0 and c3, and the eigenvalue 1 + T3/2 of multiplicity two with only one linearly independent eigenfunction T1/2c
1+ c2. Hence TP is not diagonalizable on S5,m(Γ1(T )). Further, since c1 and c2 span the space of the double cusp forms S5,m2 (Γ1), this shows that the Hecke operator TP is not diagonalizable on S5,m2 (Γ1(T )) either. Observe that, unlike the case k ≤ q + 2, there is a non-polynomial eigenvalue 1 + T3/2.
8. Cusp Forms for Γ(T )
In this section, we work with Γ = Γ(T ). A fundamental domain of Γ(T )\T contains
q + 1 rays, corresponding to the cusps [∞] =¡10¢ and [r] = ¡r1¢, r ∈ F, one stable vertex (1 0
0 1) and q + 1 stable edges γr := h(r 11 0)i, r ∈ F, and γ∞ := h(1 00 1)i. Thus gΓ(T ) = 0 so that dimCSk,m(Γ(T )) = (k − 1)q by Proposition 1, and dimCS2,m2 (Γ(T )) = 0 and dimCSk,m2 (Γ(T )) = (k − 2)q − 1 for k ≥ 3 by Proposition 4. To determine a harmonic
cocycle for Γ(T ), as noted in Sec. 4, one needs to know only its values at γr, r ∈ F,
and its value at γ∞ is determined by the hamornicity condition c(γ∞) +
P
r∈Fc(γr) = 0.
The stabilizer of the cusp [∞] (resp. [r], r ∈ F) is Γ[∞] = {(1 c0 1) : c ≡ 0 mod T } (resp. Γ[r] =
©¡
1+rc −r2c
c 1−rc ¢
: c ≡ 0 mod Tª) so that V (k, m)Γ[∞] (resp. V (k, m)Γ[r]) is spanned
by Yk−2 (resp. (X − rY )k−2). Moreover, Γ
[∞]\src([∞]) = {γ∞} and Γ[r]\src([r]) = {γr},
r ∈ F. Thus by Theorem 3, the double cusp forms can be described as follows.
Proposition 12. A harmonic cocycle c ∈ Hk,m(Γ(T )) lies in Hk,m2 (Γ(T )) if and only if
c(γ∞)(Yk−2) = 0 and c(γr)((X − rY )k−2) = 0 for all r ∈ F.
Next we study the action of the Hecke operator TP at c ∈ Hk,m(Γ(T )). Recall that a
harmonic cocycle takes values in Hom(V (k, m), C). In view of the above proposition, it turns out that the action is best described if, for all r ∈ F, the basis (X − rY )jYk−2−j,
the directed edge γr. Therefore we shall describe the action using such bases. To ease our
notation, for c ∈ Hk,m(Γ(T )), r ∈ F, and 0 ≤ j ≤ k − 2, let
(8.1) Z(c, r, j) = c(γr)((X − rY )jYk−2−j).
Assume that P is generated by P = 1+αT with α ∈ F×. Again, we use (6.1), harmonicity
and Γ(T )-equivariancy to arrive at the main identity of the Hecke action:
(8.2) Z(TPc, r, j) = Pk−2−jZ(c, r, j) − Pj bk−2−jq−1 c X n=1 ¡k−2−j n(q−1) ¢ Tn(q−1)Z(c, r, j + n(q − 1)) +X b6=r hXj u=0 (b − r)j−u ³ Pk−2−j¡uj¢− u X l=0 ¡ j u−l ¢¡k−2−j l ¢ (1 − P )l ´ Z(c, b, u) − k−2 X u=j+1 k−2−jX l=u−j ¡ j u−l ¢¡k−2−j l ¢ (1 − P )l(b − r)j−uZ(c, b, u)i.
Notice that when j = k − 2, (8.2) becomes
(8.3) Z(TPc, r, k − 2) = Z(c, r, k − 2) for all r ∈ F. Moreover, for j = 0 and r ∈ F we have
Z(TPc, r, 0) = Pk−2Z(c, r, 0) − bk−2 q−1c X n=1 ¡ k−2 n(q−1) ¢ Tn(q−1)Z(c, r, n(q − 1)) +X b6=r ³ (Pk−2− 1)Z(c, b, 0) − k−2 X u=1 (1 − P )u(b − r)−uZ(c, b, u) ´ .
Summing over all r ∈ F and using harmonicity, we get
−TPc(γ∞)(Yk−2) = I + II, where I =X r∈F ³ Pk−2Z(c, r, 0) − bk−2 q−1c X n=1 ¡ k−2 n(q−1) ¢ Tn(q−1)Z(c, r, n(q − 1))´ and II =Pb∈F³Pr6=b(Pk−2− 1)Z(c, b, 0) −Pk−2 u=1(1 − P )u P r6=b(b − r)−uZ(c, b, u) ´ =Pb∈F ³ −(Pk−2− 1)Z(c, b, 0) +Pbk−2q−1c n=1 (1 − P )n(q−1)Z(c, b, n(q − 1)) ´ .
Combined, this gives
(8.4) TPc(γ∞)(Yk−2) = c(γ∞)(Yk−2).
The equations (8.3) and (8.4) then imply
Proposition 13. Let c ∈ Sk,m(Γ(T )) be an eigenfunction of TP, where P 6= (T ) has
degree 1. If it is not a double cusp form, then the eigenvalue is 1.
Assume further that q ≥ k ≥ 2. In this case (8.2) is reduced to (8.5) Z(TPc, r, j) = j−1 X u=0 X b∈F (b − r)j−u ³ Pk−2−j¡uj¢− u X l=0 ¡ j u−l ¢¡k−2−j l ¢ (1 − P )l ´ Z(c, b, u) + [Pk−2−j − λ j(P )] X b∈F Z(c, b, j) + λj(P )Z(c, r, j) − k−2 X u=j+1 k−2−jX l=u−j ¡ j u−l ¢¡k−2−j l ¢ (1 − P )lX b6=r (b − r)j−uZ(c, b, u) = j−1 X u=0 αu(j, P ) X b∈F (b − r)j−uZ(c, b, u) + [Pk−2−j − λ j(P )] X b∈F Z(c, b, j) + λj(P )Z(c, r, j) − k−2 X u=j+1 βu(j, P ) X b6=r (b − r)j−uZ(c, b, u), where λj(P ) = Pj l=0 ¡j l ¢¡k−2−j l ¢ (1 − P )l is given by (7.3), αu(j, P ) = Pk−2−j ¡j u ¢ − u X l=0 ¡ j u−l ¢¡k−2−j l ¢ (1 − P )l for 0 ≤ u ≤ j − 1, and βu(j, P ) = k−2−jX l=u−j ¡ j u−l ¢¡k−2−j l ¢ (1 − P )l for j + 1 ≤ u ≤ k − 2.
For r ∈ F and 0 ≤ j ≤ k − 2, denote by c(r)j the function
c(r)j (γr)((X − rY )jYk−2−j) = 1 and c(r)j (γs)((X − rY )lYk−2−l) = 0 if s 6= r or l 6= j.
Let cj =
P
r∈Fc
(r)
j . Then TPcj = λj(P )cj, that is, cj is an eigenfunction of TP with eigenvalue λj(P ). Observe that cj are liftings of the eigenfunctions of Sk,m(Γ1(T )).
Our next goal is to show that λj(P ) are the eigenvalues for the Hecke operator TP on
Sk,m(Γ(T )) when q ≥ k. For this, we need
Lemma 14. Suppose that c is an eigenfunction of the Hecke operator TP on Sk,m(Γ(T ))
with eigenvalue λ 6= λn(P ) for all 0 ≤ n ≤ k − 2. Then for each 0 ≤ n ≤ k − 2 and each
r ∈ F, there are constants A(n)u ∈ F(T ) for n + 1 ≤ u ≤ k − 2 such that
(8.6)n (λ − λn(P ))Z(c, r, n) = k−2 X u=n+1 A(n) u X b6=r (b − r)n−uZ(c, b, u).
Grant this lemma. By applying (8.6)n repeatedly from n = k − 2 down to n = 0, we
deduce that c = 0. This proves
Theorem 15. Let P = (P ) 6= (T ) be a degree one prime ideal of A. For q ≥ k ≥ 2
the distinct eigenvalues for the Hecke operator TP on Sk,m(Γ(T )) are the distinct λj(P ),
0 ≤ j ≤ k − 2.
Let c be an eigenfunction of TP with eigenvalue λ. Then (8.5) gives rise to
(λ − λj(P ))Z(c, r, j) = j−1 X u=0 αu(j, P ) X b∈F (b − r)j−uZ(c, b, u) +X b∈F [Pk−2−j− λj(P )]Z(c, b, j) − k−2 X u=j+1 βu(j, P ) X b6=r (b − r)j−uZ(c, b, u)
for all 0 ≤ j ≤ k − 2 and r ∈ F. Summing over all r ∈ F, we get, for each 0 ≤ j ≤ k − 2,
(8.7) (λ − λj(P )) X r∈F Z(c, r, j) = 0. Hence if λ 6= λj(P ), then P r∈FZ(c, r, j) = 0 so that (8.8)j (λ − λj(P ))Z(c, r, j) = j−1 X u=0 αu(j, P ) X b∈F (b − r)j−uZ(c, b, u) − k−2 X u=j+1 βu(j, P ) X b6=r (b − r)j−uZ(c, b, u)
for all 0 ≤ j ≤ k − 2 and r ∈ F. When j = 0, the first sum on the right side is void and hence (8.6)0 holds with A(0)u = βu(0, P ) for 1 ≤ u ≤ k − 2. We shall prove Lemma 14 by
induction on n. To proceed, we prove an identity which will be used repeatedly in the computations to follow.
Proposition 16. For 1 ≤ l, t ≤ k − 2 ≤ q − 2 and any C-valued function X(s) on F, we have X b∈F X s∈F s6=b (b − r)t (s − b)lX(s) = P s∈F(−1)l+1 ¡t l ¢ (s − r)t−lX(s) if t > l; P s∈F(−1)l+1X(s) if t = l; 0 if t ≤ l.
Proof. Let 1 ≤ l, t ≤ k − 2 ≤ q − 2. Then
X b∈F X s∈F s6=b (b − r)t (s − b)lX(s) = X s∈F X b∈F b6=s ((b − s) + (s − r))t (s − b)l X(s) =X s∈F t X i=0 (−1)l¡ti¢(s − r)t−iX b6=s (b − s)i−lX(s). Since 1 ≤ l, t ≤ k − 2 ≤ q − 2, Pb∈F b6=s(b − s)
i−l vanishes unless i = l in which case it is −1,
so X b∈F X s∈F s6=b (b − r)t (s − b)lX(s) = P s∈F(−1)l+1 ¡t l ¢ (s − r)t−lX(s), if t > l; 0, if t < l. If t = l, then X b∈F X s∈F s6=b ³b − r s − b ´t X(s) =X s∈F X b∈F b6=s ³s − r s − b − 1 ´t X(s) =X s∈F X b6=s ³Xt−1 i=0 ¡t i ¢ (−1)i³s − r s − b ´t−i + (−1)t´X(s) = (−1)t+1X s∈F X(s).
This proves the proposition. ¤
Proof of Lemma 14. We shall apply Proposition 16 to X(s) = Z(c, s, j), in which case
the sum is equal to 0 when t = l because of (8.7) and the assumption λ 6= λj(P ) for all
j. Assume that the statement is valid up to n, where 0 ≤ n < k − 2. That is, for all
0 ≤ j ≤ n and b ∈ F, we have (8.9)j Z(c, b, j) = 1 λ − λj(P ) k−2 X u=j+1 A(j) u X s6=b (s − b)j−uZ(c, s, u).
Substituting (8.9)0 into (8.8)n+1, we get (λ − λn+1(P ))Z(c, r, n + 1) = n X u=0 αu(n + 1, P ) X b∈F (b − r)(n+1)−uZ(c, b, u) − k−2 X u=n+2 βu(n + 1, P ) X b6=r (b − r)(n+1)−uZ(c, b, u) =X b∈F α0(n+1,P ) λ−λ0(P ) k−2 X u=1 A(0)u X s6=b (b−r)n+1 (s−b)u Z(c, s, u) + n X u=1 A(n+1),0u X b∈F (b − r)(n+1)−uZ(c, b, u) + k−2 X u=n+2 A(n+1)u X b6=r (b − r)(n+1)−uZ(c, b, u).
Here A(n+1),0u = αu(n + 1, P ), 1 ≤ u ≤ n + 1, depend only on u and n. By Proposition
16, the first triple sum of the right hand side is equal to
n X u=1 α0(n+1,P ) λ−λ0(P ) X s∈F (−1)u+1¡n+1 u ¢ (s − r)(n+1)−uZ(c, s, u),
which can be combined with the middle double sum of the right hand side to bring the above identity to the following form:
(λ−λn+1(P ))Z(c, r, n + 1) = n X u=1 A(n+1),1u X b∈F (b − r)(n+1)−uZ(c, b, u) + k−2 X u=n+2 A(n+1)u X b6=r (b − r)(n+1)−uZ(c, b, u).
Next we replace Z(c, b, 1) above by (8.9)1 and use Proposition 16 to express (λ−λn+1(P ))
times Z(c, r, n + 1) as a linear combination of Pb∈F(b − r)n+1−uZ(c, b, u) for 2 ≤ u ≤ n
and Pb6=r(b − r)n+1−uZ(c, b, u) for n + 2 ≤ u ≤ k − 2 with coefficients A(n+1),2
u depending
only on n and u. Repeat this procedure. After n − 1 iterations, we arrive at
(λ − λn+1(P ))Z(c, r, n + 1) = A(n+1),n n X b∈F (b − r)Z(c, b, n) + k−2 X u=n+2 A(n+1) u X b6=r (b − r)(n+1)−uZ(c, b, u).
For the final calculation, use (8.9)n to get (λ − λn+1(P ))Z(c, r, n + 1) = A(n+1),nn λ−λn(P ) X b∈F k−2 X u=n+1 A(n) u X s6=b b−r (s−b)u−nZ(c, s, u) + k−2 X u=n+2 A(n+1) u X b6=r (b − r)(n+1)−uZ(c, b, u) = k−2 X u=n+2 A(n+1) u X b6=r (b − r)(n+1)−uZ(c, b, u).
Hence Lemma 14 follows by induction. ¤
The techniques used to prove Lemma 14 can be extended to describe the eigenspaces of TP. Let c be an eigenfunction of TP with eigenvalue λn(P ). The relations among
Z(c, r, j) for r ∈ F and 0 ≤ j ≤ k − 2 are distinguished by two cases, according to λj(P )
equal to λn(P ) or not.
For those l with λl(P ) 6= λn(P ), the equation (8.8)l gives
Z(c, b, l) = 1 λn(P )−λl(P ) hXl−1 u=0 αu(l, P ) X s∈F (s − b)l−uZ(c, s, u) − k−2 X u=l+1 βu(l, P ) X s6=b (s − b)l−uZ(c, s, u) i
for all b ∈ F. Further, for such l we havePb∈FZ(c, b, l) = 0 by (8.7). Let l0 < l1 < · · · < lt
be the distinct l’s such that λl(P ) 6= λn(P ). Then the same inductive procedure as in the
proof of Lemma 14 yields, for each lv, 0 ≤ v ≤ t,
(8.10)lv Z(c, b, lv) = X 0≤u<lv λu(P )=λn(P ) A(lv) u X s∈F (s − b)lv−uZ(c, s, u) + k−2 X u=lv+1 A(lv) u X s6=b (s − b)lv−uZ(c, s, u)
for some explicitly determined elements A(lv)u in F(T ) depending only on u and P .
Let i be an index such that λi(P ) = λn(P ). The Hecke action (8.5) gives rise to
(8.11) 0 = i−1 X u=0 αu(i, P ) X b∈F (b − r)i−uZ(c, b, u) + [Pk−2−i− λi(P )] X b∈F Z(c, b, i) − k−2 X u=i+1 βu(i, P ) X b6=r (b − r)i−uZ(c, b, u).
By successively substituting (8.10)lv into (8.11), starting with v = 0 and ending with
v = t, and simplifying the expression using Proposition 16 at each step, we eliminate all Z(c, b, l)’s in the equation (8.11) with λl(P ) 6= λn(P ) and arrive at an identity of the form
(8.12)i,r 0 = X 0≤u≤k−2 λu(P )=λn(P ) Cu(i, P ) X b6=r (b − r)i−uZ(c, b, u)
for some explicitly determined elements Cu(i, P ) in F(T ) depending only on i, u and P .
We have shown
Theorem 17. Suppose q ≥ k ≥ 2. Let P = (P ), where P ∈ F[T ] has degree one and
P (0) = 1. Then λi(P ), 0 ≤ i ≤ k − 2, with suitable multiplicities are the eigenvalues
of the Hecke operator TP on Sk,m(Γ(T )). For 0 ≤ n ≤ k − 2, set An = {i : 0 ≤ i ≤
k − 2 and λi(P ) = λn(P )} and denote the integers in [0, k − 2] r An by l0 < · · · < lt.
Let c be an eigenfunction in Hk,m(Γ(T )) with eigenvalue λn(P ). Then c is determined by
Z(c, b, u) with u ∈ An and b ∈ F subject to the conditions (8.12)i,r for i ∈ An and r ∈ F.
The remaining Z(c, b, l)’s are determined recursively by (8.10)lv from v = t to v = 0.
9. Examples
To illustrate Theorem 17, we compute the action of TP on Hk,m(Γ) for small weights
k = 3, 4, 5. None of these are diagonalizable with respect to the Hecke operator. Let c be
an eigenfunction.
(i) q ≥ k = 3. Here λ0(P ) = λ1(P ) = 1. It follows from (8.5) that X b∈F Z(c, b, 0) =X b6=r Z(c, b, 1) r − b
for all r ∈ F. We shall solve this linear system. Fix a generator a of F× and arrange the
elements of F in the order 0, a, a2, . . . , aq−1. Express the above system in matrix form
(9.1) 0 −1 a −a12 −a13 . . . −aq−11 1
a 0 a−a1 2 a−a1 3 . . . a−a1q−1 1
a2 a21−a 0 a2−a1 3 . . . a2−a1q−1
... ... ... ... . .. ...
1
aq−1 aq−11−a aq−11−a2 aq−11−a3 . . . 0
Z(c, 0, 1) Z(c, a, 1) Z(c, a2, 1) ... Z(c, aq−1, 1) = c c c ... c ,
where c =Pb∈FZ(c, b, 0). We determine the nullity of the coefficient matrix M. Write
M = 1 0 0 . . . 0 0 1 a 0 . . . 0 0 0 1 a2 . . . 0 ... ... ... ... ... 0 0 0 . . . 1 aq−1 0 −1 a −a12 −a13 . . . −aq−11 1 0 1
1−a 1−a12 . . . 1−a1q−2
1 1
1−aq−2 0 1−a1 . . . 1−a1q−3 ... ... ... ... . .. ... 1 1 1−a 1 1−a2 1−a13 . . . 0 .
Call the second matrix on the right hand side C. Note that Nul(M) = Nul(C). Consider the submatrix obtained from C by deleting the first row and the first column
C0 = 0 1
1−a 1−a12 . . . 1−a1q−2 1
1−aq−2 0 1−a1 . . . 1−a1q−3 ... ... ... . .. ... 1
1−a 1−a12 1−a13 . . . 0
,
which is a (q − 1) × (q − 1) circulant matrix. Then v0 j = 1 aj a2j ... a(q−2)j , j = 1, 2, . . . , q − 1, are q − 1 linearly independent eigenvectors of C0 with eigenvalue
aj 1 − a + a2j 1 − a2 + · · · + a(q−2)j 1 − aq−2 = j,
as a consequence of the following lemma.
Lemma 18. For j = 1, 2, . . . , q − 1 and l ≥ 1, we have
q−2 X n=1 ajn (1 − an)l = (−1) l−1¡j l ¢ .
Proof. We shall prove this lemma by induction on l. For l = 1, we compute q−2 X n=1 ajn 1 − an = q−2 X n=1 ajn− 1 + 1 1 − an = q−2 X n=1 h −(1 + an+ · · · + a(j−1)n) + 1 1 − an i .
Since a has order q −1 and 1 ≤ j ≤ q −1,Pq−2n=1ain = −ai(q−1)= −1 for all i = 1, . . . , j −1.
As Pq−2n=1 1
1−an = −1, the above sum is equal to
q−2
X
n=1
ajn
1 − an = −(q − 2) + (j − 1) − 1 = j.
Next, we assume that Pq−2n=1 ajn
(1−an)l = (−1)l−1 ¡j
l
¢
for all j = 1, . . . , q − 1. Then
q−2 X n=1 ajn (1 − an)l+1 = q−2 X n=1 ajn− 1 + 1 (1 − an)l+1 = q−2 X n=1 h −1 + a + · · · + a (j−1)n (1 − an)l + 1 (1 − an)l+1 i = −[−1 + (−1)l−1¡1 l ¢ + · · · + (−1)l−1¡j−1 l ¢ ] − 1 = (−1)l¡ j l+1 ¢
by the Pascal’s triangle identity Pmi=1¡il¢=¡m+1l+1¢. The lemma follows by induction. ¤
Back to the matrix C. The vectors v0 = 1 1 1 ... 1 and vj = 0 1 aj ... a(q−2)j , j = 1, . . . , q−1, are
q linearly independent eigenvectors of C with the eigenvalues 0 and j, respectively. Since
our field has characteristic p, this shows that the nullity(C) = q/p. If c =Pb∈FZ(c, b, 0) =
0, then we obtain (q − 1) + q/p linearly independent eigenvectors for TP. When c 6= 0, note that v = 0 ca ca2 ... caq−1
is a solution of (9.1). Together with the homogeneous ones, we have q + q/p linearly independent eigenvectors for TP, all with eigenvalue 1. Since 1 is the only eigenvalue of TP, its total multiplicity 2q, thus TP is not diagonalizable. We record this result in
Proposition 19. Suppose F has cardinality q ≥ 3 and characteristic p. For a maximal
degree one ideal P 6= (T ), 1 is the only eigenvalue of the Hecke operator TP on S3,m(Γ(T )).
The eigenspace of TP has dimension q + q/p, while the space S3,m(Γ(T )) has dimension 2q. Consequently, TP is not diagonalizable on S3,m(Γ(T )).
As the dimension of the 1-eigenspace of TPon S3,m2 (Γ(T )) is q−1, which is dimCS3,m2 (Γ(T )), so TP is diagonalizable on S3,m2 (Γ(T )).
(ii) q ≥ k = 4. In this case λ0(P ) = λ2(P ) = 1 and λ1(P ) = −P + 2. A similar computation yields
Proposition 20. Suppose F has cardinality q ≥ 4 and characteristic p. For a maximal
degree one ideal P 6= (T ), 1 and 2 − P are the two distinct eigenvalues of the Hecke operator TP on S4,m(Γ(T )). The 1-eigenspace has dimension q + 2q/p if p > 2 and
dimension q + q/p if p = 2. The (2 − P )-eigenspace has dimension q. Moreover, TP is
not diagonalizable on S4,m(Γ(T )).
One checks that TP on S4,m2 (Γ(T )) is diagonalizable since dimC S4,m2 (Γ(T )) = 2q − 1, the 1-eigenspace is (q − 1)-dimensional, and the (2 − P )-eigenspace has dimension q. (iii) q ≥ k = 5. In this case λ0(P ) = λ3(P ) = 1 and λ1(P ) = λ2(P ) = −2P + 3. First we assume p > 2 so that 1 6= −2P + 3. To determine the 1-eigenspace, consider the equations from (8.5) with j = 0, 1, 2: (9.2) 0 = (P2+ P + 1)X b∈F Z(c, b, 0) + 3X b6=r Z(c, b, 1) b − r − 3(P − 1) X b6=r Z(c, b, 2) (b − r)2 + (P − 1)2 X b6=r Z(c, b, 3) (b − r)3 , 2Z(c, r, 1) = (P + 1)X b∈F (b − r)Z(c, b, 0) + (P + 3)X b∈F Z(c, b, 1) − (P − 3)X b6=r Z(c, b, 2) b − r − (P − 1) X b6=r Z(c, b, 3) (b − r)2 , and 2Z(c, r, 2) =X b∈F (b − r)2Z(c, b, 0) + 3X b∈F (b − r)Z(c, b, 1) + 3X b∈F Z(c, b, 2) +X b6=r Z(c, b, 3) b − r
for all r ∈ F. Summing the second and third equations over all r, we getPr∈FZ(c, r, 1) =
0 and Pr∈FZ(c, r, 2) = 0, which lead to the following simplifications of the second and
third equations: (9.3) Z(c, r, 1) = P + 1 2 X b∈F (b− r)Z(c, b, 0)−P − 3 2 X b6=r Z(c, b, 2) b − r − P − 1 2 X b6=r Z(c, b, 3) (b − r)2 , and (9.4) Z(c, r, 2) = 1 2 X b∈F (b − r)2Z(c, b, 0) +3 2 X b∈F (b − r)Z(c, b, 1) + 1 2 X b6=r Z(c, b, 3) b − r
for all r ∈ F. Plugging (9.3) into (9.4) and using Proposition 16 to simplify, we get (9.5) Z(c, r, 2) = 1 2 X b∈F (b − r)2Z(c, b, 0) + 1 2 X b6=r Z(c, b, 3) b − r
for all r ∈ F. Substituting (9.3) and (9.5) into (9.2) and simplifying the result using Proposition 16, we obtain (9.6) X b∈F Z(c, b, 0) =X b6=r Z(c, b, 3) (r − b)3
for all r ∈ F. To solve the above linear system, we employ the same method as in case (i), that is, computing the nullity of
C = 0 −1 a3 −a16 −a19 . . . −a3(q−1)1 1 0 1
(1−a)3 (1−a12)3 . . . (1−a1q−2)3
1 1
(1−aq−2)3 0 (1−a)1 3 . . . (1−a1q−3)3
... ... ... ... . .. ...
1 1
(1−a)3 (1−a12)3 (1−a13)3 . . . 0
.
By Lemma 18, the vectors v0 = 1 1 1 ... 1 and vj = 0 1 aj ... a(q−2)j , j = 1, . . . , q −1, are q linearly independent eigenvectors of C with the eigenvalues 0 and ¡3j¢, respectively. Therefore the nullity of C is 3q/p when p > 3 and is q/p when p = 3, which yields the number of linearly independent eigenvectors if c := Pb∈FZ(c, b, 0) = 0. When c 6= 0, we note that
v = 0 ca3 ca6 ... ca3(q−1)
is a solution of (9.6). Together with the homogeneous ones, we see that the 1-eigenspace of TP has dimension q + 3q/p if p > 3 and q + q/p if p = 3.
Next we determine the eigenvectors with eigenvalue −2P + 3. Such eigenvectors are double cusp forms by Proposition 13, so Z(c, r, 3) = 0 for all r ∈ F. Thus the equations from (8.5) with j = 0, 1, 2 can be simplified as
Z(c, r, 0) = −3 2 X b6=r Z(c, b, 1) b − r + 3(P − 1) 2 X b6=r Z(c, b, 2) (b − r)2 , 0 = (P + 1)X b∈F (b − r)Z(c, b, 0) + (P + 3)X F Z(c, b, 1) − (P − 3)X b6=r Z(c, b, 2) b − r ,
and 0 = X b∈F (b − r)2Z(c, b, 0) + 3X b∈F (b − r)Z(c, b, 1) + 3X b∈F Z(c, b, 2)
for all r ∈ F. Substituting the first relation into the second and the third, and simplifying the resulting expressions by using Proposition 16, we arrive at
(9.7) X b∈F Z(c, b, 1) = 2X b6=r Z(c, b, 2) r − b and (9.8) X b∈F Z(c, b, 2) = 0
for all r ∈ F. Write c =Pb∈FZ(c, b, 1). Solve the system (9.7) using the same method as
(9.1). When c = 0, we get homogeneous solutions Z(c,0,2) Z(c,a,2) Z(c,a2,2) ... Z(c,aq−1,2) = 1 1 1 ... 1 or 0 1 aj ... a(q−2)j ,
j = 1, . . . , q − 1; when c 6= 0, we get a nonhomogeneous solution
Z(c,0,2) Z(c,a,2) Z(c,a2,2) ... Z(c,aq−1,2) = 1 2 0 ca ca2 ... caq−1
. Note that all solutions satisfy the equation (9.8). Thus the (−2P + 3)-eigenspace of TP has dimension q + q/p. Combined with the dimension of 1-eigenspace, we conclude that TP is not diagonalizable on S5,m(Γ(T )) since the space has dimension 4q. We summarize the above discussion in
Proposition 21. Suppose F has cardinality q ≥ 4 and characteristic p > 2. For a
maximal degree one ideal P 6= (T ), 1 and −2P + 3 are the two distinct eigenvalues of the Hecke operator TP on S5,m(Γ(T )). The 1-eigenspace has dimension q + 3q/p if p > 3 and
dimension q + q/p if p = 3. The (−2P + 3)-eigenspace has dimension q + q/p. Further, TP is not diagonalizable on S5,m(Γ(T )).
Now we turn to the case when p = 2. In this case, we have only one eigenvalue, namely, 1. Then (8.5) for j = 0, 1, 2 become
0 = (P2+ P + 1)X b∈F Z(c, b, 0) +X b6=r Z(c, b, 1) b − r + (P − 1) X b6=r Z(c, b, 2) (b − r)2 + (P − 1)2 X b6=r Z(c, b, 3) (b − r)3 , 0 =X b∈F (b − r)Z(c, b, 0) +X b∈F Z(c, b, 1) +X b6=r Z(c, b, 2) b − r + X b6=r Z(c, b, 3) (b − r)2 , and 0 = X b∈F (b − r)2Z(c, b, 0) +X b∈F (b − r)Z(c, b, 1) +X b∈F Z(c, b, 2) +X b6=r Z(c, b, 3) (b − r)
for all r ∈ F. Observe that we can represent the above system as a homogeneous matrix equation Mx = 0, where M is a 3q × 4q matrix. Moreover, it is clear that rank M > 1. Thus the eigenspace is has dimension less than 4q, so that the Hecke operator TP is not diagonalizable. Therefore we have shown
Proposition 22. Suppose F has cardinality q ≥ 4 and characteristic p = 2. For a
maximal degree one ideal P 6= (T ), 1 is the only eigenvalue of the Hecke operator TP
on S5,m(Γ(T )). The eigenspace of TP has dimension less than 4q, the dimension of
S5,m(Γ(T )). Hence TP is not diagonalizable on S5,m(Γ(T )).
As for the action of TP on S5,m2 (Γ(T )), by the same computation as before, we see that for q odd, the 1-eigenspace is (q−1)-dimensional and the (3−2P )-eigenspace has dimension
q + q/p so that the total dimension is less than 3q − 1, the dimension of S2
5,m(Γ(T )); for
q even, the matrix M is 3q × 3q with rank at least two, thus the eigenspace is at most
(3q − 2)-dimensional. Hence in both cases, TP on S5,m2 (Γ(T )) is not diagonalizable.
Remark. For Drinfeld cusp forms, what happens in case (iii) is representative of the general weights. For example, when the weight k = 6, we have three distinct eigenvalues 1, 4 − 3P and 6 − 6P + P2 if p 6= 3 and two distinct eigenvalues 1 and P2 if p = 3. The computations for Z(c, b, u) are similar.
References
[B¨oc04] G. B¨ockle, An Eichler-Shimura isomorphism over function fields between Drin-feld modular form and cohomology classes of crystals, preprint 2004, available at http://www.exp-math.uni-essen.de/∼boeckle.
[Gek88] E.-U. Gekeler, On the coefficients of Drinfeld modular forms, Invent. Math. 93 (1988), 667-700. [GR96] E.-U. Gekeler and M. Reversat, Jacobians of Drinfeld modular curves, J. Reine Agnew. Math.
476 (1996), 27-93.
[Fv04] J. Fresnel and M. van der Put, Rigid Analytic Geometry and Its Applications, Birkh¨auser, 2004. [Gos80] D. Goss, Modular forms for Fr[T ], J. Reine Agnew. Math. 317 (1980) 16-39.
[Mee06] Y. Meemark, Operators Based on Double Cosets of GL2, Ph.D. Thesis 2006, The Pennsylvania
State University.
[Rev00] M. Reversat, On modular Forms of characteristic p > 0, J. Number Theory 84 (2000) 214-229. [Ser80] J. P. Serre, Trees, Springer, 1980.
[Tei91] J. T. Teitelbaum, The poisson kernel for Drinfeld modular curves, J. Amer. Math. Soc. 4 (1991), 491-511.
Wen-Ching Winnie Li, Department of Mathematics, The Pennsylvania State University, University Park, PA 16802
E-mail address: [email protected]
Yotsanan Meemark, Department of Mathematics, Faculty of Science, Chulalongkorn University, Bangkok, 10330 THAILAND
