Home Triviality of the Hecke action on ordinary Drinfeld cuspforms of level Γ1(tn )
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Triviality of the Hecke action on ordinary Drinfeld cuspforms of level Γ1(tn )

  • Shin Hattori ORCID logo EMAIL logo
Published/Copyright: September 29, 2022

Abstract

Let k 2 and n 1 be any integers. In this paper, we prove that all Hecke operators act trivially on the space of ordinary Drinfeld cuspforms of level Γ 1 ( t n ) and weight k .

1 Introduction

Let p be a rational prime, let q > 1 be a p-power integer, let A = 𝔽 q [ t ] , K = 𝔽 q ( t ) and K = 𝔽 q ( ( 1 / t ) ) . Let be the ( 1 / t ) -adic completion of an algebraic closure of K and put Ω = K , which has a natural structure as a rigid analytic variety over K . For any non-zero element 𝔫 A , we put

Γ 1 ( 𝔫 ) = { γ SL 2 ( A ) | γ ( 1 * 0 1 ) mod 𝔫 } .

For any arithmetic subgroup Γ of SL 2 ( A ) and integer k 2 , a rigid analytic function f : Ω is called a Drinfeld modular form of level Γ and weight k if it satisfies

f ( a z + b c z + d ) = ( c z + d ) k f ( z ) for any  ( a b c d ) Γ

and a certain regularity condition at cusps. A Drinfeld modular form is called a cuspform if it vanishes at all cusps, and a double cuspform if it vanishes twice at all cusps. They form -vector spaces S k ( Γ ) and S k ( 2 ) ( Γ ) , respectively. These spaces admit a natural action of Hecke operators.

Let A be an irreducible polynomial, K the -adic completion of K and the -adic completion of an algebraic closure of K . For the algebraic closure K ¯ of K in , we fix an embedding ι : K ¯ .

Suppose that divides 𝔫 . The Hecke operator at acting on S k ( Γ 1 ( 𝔫 ) ) is denoted by U . Note that any eigenvalue of U is an element of K ¯ . We say f S k ( Γ 1 ( 𝔫 ) ) is ordinary (with respect to ι ) if f is in the generalized eigenspace belonging to an eigenvalue λ K ¯ satisfying ι ( λ ) 𝒪 × . We denote the subspace of ordinary Drinfeld cuspforms by S k ord ( Γ 1 ( 𝔫 ) ) . It is an analogue of the notion of ordinariness for elliptic modular forms studied in [9].

Let us focus on the case 𝔫 = t n and = t with some integer n 1 . In this case, the structure of S k ord ( Γ 1 ( t n ) ) seems quite simple. For n = 1 , it is known that all Hecke operators act trivially on the one-dimensional -vector space S k ord ( Γ 1 ( t ) ) (see [8, Proposition 4.3]). In this paper, we prove that this holds in general, as follows.

Theorem 1.1 (Theorem 4.9).

Let k 2 and n 1 be any integers. Then we have

dim S k ord ( Γ 1 ( t n ) ) = q n - 1

and all Hecke operators act trivially on S k ord ( Γ 1 ( t n ) ) .

This result suggests that Hida families for Drinfeld cuspforms should be trivial for the level Γ 1 ( t n ) .

For Drinfeld modular forms, it is well known that the weak multiplicity one, which states that any Hecke eigenform is determined up to a scalar multiple by the Hecke eigenvalues, is false. Gekeler [3, Section 7] raised a question if the property holds when we fix the weight. Theorem 1.1 gives a negative answer to it (see also [2, Examples 15.4 and 15.7] for a variant ignoring Hecke eigenvalues at places dividing the level).

For the proof of Theorem 1.1, we study a subspace S k of S k = S k ( Γ 1 ( t n ) ) containing S k ( 2 ) = S k ( 2 ) ( Γ 1 ( t n ) ) . It consists of cuspforms which vanish twice at unramified cusps (Section 3.3). We show that all Hecke operators act trivially on S k / S k and U t is nilpotent on S k / S k ( 2 ) (Lemma 3.9 and Proposition 3.10). Then, using the constancy of the dimension of S k ord ( Γ 1 ( t n ) ) with respect to k (see [8, Proposition 3.4]), we reduce Theorem 1.1 to showing that the dimension of S 2 ord ( Γ 1 ( t n ) ) is no more than q n - 1 (Theorem 3.11).

Consider the multiplicative group Θ n = 1 + t A / t n A , which acts on S k ( Γ 1 ( t n ) ) via the diamond operator. To obtain the upper bound of the dimension, the key point is the freeness of S 2 ( Γ 1 ( t n ) ) as a module over the group ring [ Θ n ] (Proposition 4.8): From the fact that S 2 ord ( Γ 1 ( t ) ) is one-dimensional [6, Lemma 2.4] and another constancy result of the dimension of the ordinary subspace [8, Proposition 3.5], we see that the Θ n -fixed part of S 2 ord ( Γ 1 ( t n ) ) is also one-dimensional. Thus the freeness implies that it injects into a single component [ Θ n ] of the free [ Θ n ] -module S 2 ( Γ 1 ( t n ) ) , which gives the desired bound.

The paper is organized as follows. In Section 2, we will recall the definition of Hecke operators and study their effect on Fourier expansions of Drinfeld cuspforms at cusps. In Section 3, we will define the subspace S k and study its properties analytically. In Section 4, using the description of Drinfeld cuspforms via harmonic cocycles on the Bruhat–Tits tree [13, 2], we will give an explicit basis of the -vector space S 2 ( Γ 1 ( t n ) ) and a description of the diamond operator in terms of the basis. These enable us to show the freeness and Theorem 1.1.

2 Drinfeld cuspforms of level Γ 1 ( 𝔫 )

Let k 2 be any integer and 𝔫 any element of A 𝔽 q . In this section, we study Hecke operators acting on S k ( Γ 1 ( 𝔫 ) ) . For any group Γ acting on a set X, we denote the stabilizer of x X in Γ by Stab ( Γ , x ) .

2.1 Cusps and uniformizers

Consider the action of SL 2 ( A ) on 1 ( ) defined by

( a b c d ) ( x y ) = ( a x + b y c x + d y ) .

We refer to any element of 1 ( K ) as a cusp. For any arithmetic subgroup Γ of SL 2 ( A ) , put

Cusps ( Γ ) = Γ \ 1 ( K ) .

We abusively identify an element of Cusps ( Γ ) with a cusp representing it.

Next we recall the definition of the uniformizer at each cusp [5, (2.7)], following the normalization of [3, (4.1)]. Let C be the Carlitz module. It is the Drinfeld module of rank one over A defined by the homomorphism of 𝔽 q -algebras

A = 𝔽 q [ t ] End ( 𝔾 a ) , t ( Z t Z + Z q ) ,

where we put 𝔾 a = Spec ( A [ Z ] ) . For any a A , we denote by Φ a C ( Z ) the element of A [ Z ] such that the image of a by the map above is defined by ( Z Φ a C ( Z ) ) .

For any subgroup 𝔟 of A containing a non-zero ideal of A, we define

e 𝔟 ( z ) = z 0 b 𝔟 ( 1 - z b ) ,

which is an entire function on Ω. Let π ¯ be a Carlitz period, so that

(2.1) Φ t C ( π ¯ e A ( z ) ) = π ¯ e A ( t z ) .

For any integer l 0 , we put

u 𝔟 ( z ) = 1 π ¯ e 𝔟 ( z ) , u ( z ) = u A ( z ) , u l ( z ) = u ( t l ) ( z ) = 1 t l u ( z t l ) .

Since 𝔫 A 𝔽 q , the group Γ 1 ( 𝔫 ) is p -torsion free. For any cusp s 1 ( K ) , choose ν s SL 2 ( A ) satisfying ν s ( ) = s and put

𝔟 s = { b A | ( 1 b 0 1 ) Stab ( ν s - 1 Γ 1 ( 𝔫 ) ν s , ) } ( 𝔫 ) .

Then we refer to the function

u s ( z ) := u 𝔟 s ( z )

as the uniformizer at s for Γ 1 ( 𝔫 ) . Note that 𝔟 s depends only on s up to a multiple of an element of 𝔽 q × . Thus 𝔟 s and u s ( z ) are independent of the choice of ν s if 𝔟 s is an ideal of A for some choice of ν s . For example, we have 𝔟 = A for any choice of ν and the uniformizer at is u ( z ) .

For any function f on Ω, integer k 2 and γ GL 2 ( K ) , we define the slash operator by

( f | k γ ) ( z ) = det ( γ ) k - 1 ( c z + d ) - k f ( a z + b c z + d ) , γ = ( a b c d ) .

Then, for any f S k ( Γ 1 ( 𝔫 ) ) , we can write

( f | k ν s ) ( z ) = i 1 a i u s ( z ) i , a i

when the ( 1 / t ) -adic absolute value | u s ( z ) | of u s ( z ) is sufficiently small. We refer to it as the Fourier expansion of f at the cusp s and put

ord ( s , f ) = min { i 1 a i 0 } .

The latter is independent of the choice of ν s .

Lemma 2.1.

Let m A be any monic irreducible polynomial and i 1 any integer. Then:

  1. deg ( β ) < deg ( 𝔪 ) u ( z + β 𝔪 ) = 𝔪 u ( z ) .

  2. If i 2 , then deg ( β ) < deg ( 𝔪 ) u ( z + β 𝔪 ) i 𝔪 u ( z ) 2 A [ u ( z ) ] .

  3. u ( 𝔪 z ) u ( z ) 2 A [ [ u ( z ) ] ] .

Here the sum deg ( β ) < deg ( m ) runs over the set of β A satisfying deg ( β ) < deg ( m ) .

Proof.

Put r = deg ( 𝔪 ) and

Φ 𝔪 C ( Z ) = 𝔪 Z + c 1 Z q + + c r - 1 Z q r - 1 + Z q r .

We denote by C [ 𝔪 ] the kernel of the multiplication by 𝔪 on C and by C [ 𝔪 ] ( ) the A-module of -valued points of it. Then we have

(2.2) z 0 b C [ 𝔪 ] ( ) ( 1 - z b ) = 𝔪 - 1 Φ 𝔪 C ( z ) .

Let α i be the coefficient of Z q i in 𝔪 - 1 Φ 𝔪 C ( Z ) . By [7, Lemma 3.2], we have α i A for any 0 i r - 1 and α r = 𝔪 - 1 .

Let G i , 𝔪 ( X ) be the i-th Goss polynomial with respect to the 𝔽 q -vector space C [ 𝔪 ] ( ) . Then [3, computation above (7.3)] gives

(2.3) deg ( β ) < deg ( 𝔪 ) ( z + β 𝔪 ) i = G i , 𝔪 ( 𝔪 u ( z ) ) .

For i = 1 , we have G i , 𝔪 ( X ) = X and (i) follows. For i 2 , [3, (3.8)] and (2.2) show that G i , 𝔪 ( X ) has no linear term and G i , 𝔪 ( 𝔪 X ) 𝔪 A [ X ] , which yields (ii). Moreover, we have

u ( 𝔪 z ) = u ( z ) q r 1 + c r - 1 u ( z ) q r - q r - 1 + + 𝔪 u ( z ) q r - 1 ,

which implies (iii). ∎

Put ζ t l = π ¯ e A ( 1 t l ) , so that Φ t l C ( ζ t l ) = 0 by (2.1).

Lemma 2.2.

Let l 1 be any integer. For any β F q , we have

u l - 1 ( z + β t ) t u l ( z ) A [ ζ t l ] [ [ u l ( z ) ] ] .

Here A [ ζ t l ] is the A-subalgebra of C generated by ζ t l .

Proof.

This follows from

u l - 1 ( z + β t ) = t t l π ¯ e A ( z + β t l ) = t t l π ¯ e A ( z t l ) 1 1 + t l π ¯ e A ( β t l ) t l π ¯ e A ( z t l )
= t u l ( z ) 1 1 + t l β ζ t l u l ( z ) .

2.2 Hecke operators

Now we recall the definition of Hecke operators (for example, see [8, Section 3.1]). Let 𝔪 A be any monic irreducible polynomial. Then the Hecke operator T 𝔪 acting on S k ( Γ 1 ( 𝔫 ) ) is defined as

T 𝔪 f = ξ f | k ξ ,

where ξ runs over any complete set of representatives of the coset space

(2.4) Γ 1 ( 𝔫 ) \ Γ 1 ( 𝔫 ) ( 1 0 0 𝔪 ) Γ 1 ( 𝔫 ) .

When 𝔪 𝔫 , we write T 𝔪 also as U 𝔪 .

Let 𝔞 A be any element which is prime to 𝔫 . Take any matrix η 𝔞 , SL 2 ( A ) satisfying

η 𝔞 , ( * * 0 𝔞 ) mod 𝔫

and put

ξ 𝔞 , = η 𝔞 , ( 𝔞 0 0 1 ) .

Note that we have

η 𝔞 , Γ 1 ( 𝔫 ) η 𝔞 , - 1 = Γ 1 ( 𝔫 ) , ξ 𝔞 , Γ 1 ( 𝔞 𝔫 ) ξ 𝔞 , - 1 Γ 1 ( 𝔫 ) .

Hence we obtain

(2.5) f S k ( Γ 1 ( 𝔫 ) ) f | k η 𝔞 , S k ( Γ 1 ( 𝔫 ) ) , f | k ξ 𝔞 , S k ( Γ 1 ( 𝔞 𝔫 ) ) .

For any α ( A / ( 𝔫 ) ) × , we choose a lift 𝔞 A of α and put

α 𝔫 f = f | k η 𝔞 ,

for any f S k ( Γ 1 ( 𝔫 ) ) , which is independent of the choices of 𝔞 and η 𝔞 , . Then α α 𝔫 defines an action of the group ( A / ( 𝔫 ) ) × on S k ( Γ 1 ( 𝔫 ) ) .

Lemma 2.3.

For any α ( A / ( n ) ) × , the diamond operator α n commutes with all Hecke operators.

Proof.

Let 𝔪 A be any monic irreducible polynomial. First suppose 𝔪 𝔫 . Write

η 𝔞 , = ( S S T T )

with some S , S , T , T A satisfying T 0 , T 𝔞 mod 𝔫 and S T - S T = 1 . Since S is prime to 𝔫 , there exists β A satisfying β S S mod 𝔫 . Then we have

η 𝔞 , - 1 ( 1 0 0 𝔪 ) η 𝔞 , Γ 1 ( 𝔫 ) ( 1 β 0 𝔪 ) = Γ 1 ( 𝔫 ) ( 1 0 0 𝔪 ) ( 1 β 0 1 ) ,

which yields

(2.6) Γ 1 ( 𝔫 ) η 𝔞 , - 1 ( 1 0 0 𝔪 ) η 𝔞 , Γ 1 ( 𝔫 ) = Γ 1 ( 𝔫 ) ( 1 0 0 𝔪 ) Γ 1 ( 𝔫 ) .

The lemma in this case follows from this equality.

Next suppose 𝔪 𝔫 . Note that the natural map

SL 2 ( A ) SL 2 ( A / ( 𝔫 ) ) × SL 2 ( A / ( 𝔪 ) )

is surjective. Since α 𝔫 is independent of the choices of 𝔞 and η 𝔞 , , we may assume that η 𝔞 , satisfies

η 𝔞 , ( 1 0 0 1 ) mod 𝔪 .

Then we have

η 𝔞 , - 1 ( 1 0 0 𝔪 ) η 𝔞 , Γ 1 ( 𝔫 ) ( 1 0 0 𝔪 )

and (2.6) holds also in this case, which yields the lemma. ∎

Let us give an explicit description of the Hecke operator T 𝔪 . For any β A satisfying deg ( β ) < deg ( 𝔪 ) , put

ξ 𝔪 , β = ( 1 β 0 𝔪 ) .

When 𝔪 = t , we also write ξ β for ξ t , β . Then the operator U 𝔪 for 𝔪 𝔫 is given by

( U 𝔪 f ) ( z ) = deg ( β ) < deg ( 𝔪 ) ( f | k ξ 𝔪 , β ) ( z ) = 1 𝔪 deg ( β ) < deg ( 𝔪 ) f ( z + β 𝔪 ) .

When 𝔪 𝔫 , the set

{ ξ 𝔪 , β deg ( β ) < deg ( 𝔪 ) } { ξ 𝔪 , }

forms a complete set of representatives of the coset space (2.4) and thus

T 𝔪 f = deg ( β ) < deg ( 𝔪 ) f | k ξ 𝔪 , β + f | k ξ 𝔪 , .

3 U t -operator of level Γ 1 ( t n )

Let k 2 and n 1 be any integers. In the rest of the paper, we assume 𝔫 = t n .

In this section, we study the operator U t acting on S k ( Γ 1 ( t n ) ) , and prove a criterion, in terms of U t , for all Hecke operators to act trivially on S k ord ( Γ 1 ( t n ) ) (Theorem 3.11). We denote S k ( Γ 1 ( t n ) ) and S k ( 2 ) ( Γ 1 ( t n ) ) also by S k and S k ( 2 ) , respectively.

Put A n = A / ( t n ) . Let v t be the t-adic valuation on K normalized as v t ( t ) = 1 . For any c A n - 1 , take any lift c ~ A of c and put

v ¯ t ( c ) = min { v t ( c ~ ) , n - 1 } ,

which is independent of the choice of c ~ .

3.1 Cusps of Γ 1 ( t n )

For any c , d A n - 1 , put

h ¯ ( c , d ) = ( 1 1 + t d 0 t c 1 + t d ) SL 2 ( A n ) .

Since the natural map SL 2 ( A ) SL 2 ( A n ) is surjective, we can take a lift h ( c , d ) Γ 1 ( t ) of h ¯ ( c , d ) by this map.

Lemma 3.1.

Let ( c , d ) be any element of A n - 1 2 . Suppose that an element h Γ 1 ( t ) satisfies

h ( * * t c 1 + t d ) mod t n .

Then h Γ 1 ( t n ) h ( c , d ) .

Proof.

We have det ( h h ( c , d ) - 1 ) = 1 and

h h ( c , d ) - 1 ( * * t c 1 + t d ) ( 1 + t d 0 - t c 1 1 + t d ) = ( * * 0 1 ) mod t n .

Hence the ( 1 , 1 ) -entry of the matrix h h ( c , d ) - 1 is also congruent to one modulo t n and thus we have h h ( c , d ) - 1 Γ 1 ( t n ) . ∎

From Lemma 3.1, we see that the set

{ h ( c , d ) c , d A n - 1 }

forms a complete set of representatives of Γ 1 ( t n ) \ Γ 1 ( t ) .

Note that for

S B ( A ) = { ( * * 0 * ) SL 2 ( A ) }

the map

Γ 1 ( t ) \ SL 2 ( A ) / S B ( A ) Γ 1 ( t ) \ 1 ( K ) , γ γ ( )

is bijective. Hence we obtain

Cusps ( Γ 1 ( t ) ) = { , 0 } .

Consider the natural map

Cusps ( Γ 1 ( t n ) ) Cusps ( Γ 1 ( t ) ) .

For { , 0 } , we denote by Cusps ( Γ 1 ( t n ) ) the inverse image of by this map. Then we have a bijection

Γ 1 ( t n ) \ Γ 1 ( t ) / Stab ( Γ 1 ( t ) , ) Cusps ( Γ 1 ( t n ) ) , γ γ ( ) .

From the equalities

Stab ( Γ 1 ( t ) , ) = { ( 1 b 0 1 ) | b A } ,
Stab ( Γ 1 ( t ) , 0 ) = { ( 1 0 t c 1 ) | c A } ,

we can show the following lemma.

Lemma 3.2.

The following statements hold:

  1. Let Λ be a subset of A n - 1 2 which forms a complete set of representatives for the equivalence relation

    ( c , d ) ( c , d ) c = c and  d - d c A n - 1 .

    Then the set

    { h ( c , d ) ( ) ( c , d ) Λ }

    forms a complete set of representatives of Cusps ( Γ 1 ( t n ) ) .

  2. The set

    { h ( 0 , d ) ( 0 ) d A n - 1 }

    forms a complete set of representatives of Cusps 0 ( Γ 1 ( t n ) ) .

Lemma 3.3.

Let ( c , d ) be any element of A n - 1 2 . Put m = v ¯ t ( c ) [ 0 , n - 1 ] .

  1. For s = h ( c , d ) ( ) , we have

    𝔟 s = ( t n - 1 - m ) , u s ( z ) = u n - 1 - m ( z ) = 1 t n - 1 - m u ( z t n - 1 - m ) .

  2. For s = h ( 0 , d ) ( 0 ) , we have

    𝔟 s = ( t n ) , u s ( z ) = u n ( z ) = 1 t n u ( z t n ) .

Proof.

For any x A , the element

(3.1) h ( c , d ) ( 1 x 0 1 ) h ( c , d ) - 1 SL 2 ( A )

is congruent modulo t n to

( 1 - t c x 1 + t d x ( 1 + t d ) 2 - t 2 c 2 x 1 + t c x 1 + t d )

and thus the element of (3.1) lies in Γ 1 ( t n ) if and only if

v ¯ t ( x ) max { n - 1 - m , n - 2 - 2 m } = n - 1 - m ,

which yields (i).

For (ii), observe

h ( 0 , d ) ( 0 ) = h ( 0 , d ) J ( ) , J = ( 0 - 1 1 0 ) .

Since

h ( 0 , d ) J ( 1 x 0 1 ) J - 1 h ( 0 , d ) - 1 ( 1 0 - x ( 1 + t d ) 2 1 ) mod t n ,

the element of the left-hand side lies in Γ 1 ( t n ) if and only if x ( t n ) . This concludes the proof. ∎

3.2 Hecke operators of level Γ 1 ( t n )

Lemma 3.4.

For any f S k ( Γ 1 ( t n ) ) , any monic irreducible polynomial m A and d A n - 1 , we have

( T 𝔪 f ) | k h ( 0 , d ) = { deg ( β ) < deg ( 𝔪 ) f | k h ( 0 , d ) ξ β ( 𝔪 = t ) , deg ( β ) < deg ( 𝔪 ) f | k h ( 0 , d ) ξ 𝔪 , β + f | k h ( 0 , d ) ξ 𝔪 , ( 𝔪 t ) .

Moreover, when m t , we can write

( f | k h ( 0 , d ) ξ 𝔪 , ) ( z ) = i 2 c i u ( z ) i , c i ,

if | u ( z ) | is sufficiently small.

Proof.

Since f | k h ( 0 , d ) = 1 + t d t n f , Lemma 2.3 shows the former assertion.

Let us show the latter assertion for 𝔪 t . We have

( f | k h ( 0 , d ) ξ 𝔪 , ) ( z ) = 𝔪 k - 1 ( f | k h ( 0 , d ) η 𝔪 , ) ( 𝔪 z ) .

For any x A , observe

h ( 0 , d ) η 𝔪 , ( 1 x 0 1 ) ( h ( 0 , d ) η 𝔪 , ) - 1 Γ 1 ( t n ) ,

which shows that the uniformizer at the cusp h ( 0 , d ) η 𝔪 , ( ) is u ( z ) . Then we can write

( f | k h ( 0 , d ) η 𝔪 , ) ( z ) = i 1 b i u ( z ) i , b i ,

and the assertion follows from Lemma 2.1(iii). ∎

Lemma 3.5.

Let β F q and ( c , d ) A n - 1 2 be any elements. Then:

  1. ξ β h ( c , d ) Γ 1 ( t n ) h ( t c , d - β c ) ξ β .

  2. If β 0 , then

    ξ β h ( c , d ) J Γ 1 ( t n ) h ( β - 1 ( 1 + t d ) , d - β c ) ( 1 0 0 t ) ( β - 1 0 β - 1 ) .

  3. ξ 0 h ( c , d ) J Γ 1 ( t n ) h ( t c , d ) J ( t 0 0 1 ) .

Proof.

Write

h ( c , d ) = ( P t n Q t R S ) , P , Q , R , S A .

Since S P mod t , we have t - 1 ( S - P ) A and the element

ξ β h ( c , d ) ξ β - 1 = ( P + t β R β ( S - P t ) - β 2 R + t n - 1 Q t 2 R S - t β R ) Γ 1 ( t )

satisfies

ξ β h ( c , d ) ξ β - 1 ( * * t 2 c 1 + t ( d - β c ) ) mod t n .

Thus Lemma 3.1 shows (i).

For (ii), the matrix ξ β h ( c , d ) J equals

( t n β - 1 Q + S β ( S - P t ) + t n - 1 Q - β 2 R t β - 1 S S - t β R ) ( 1 0 0 t ) ( β - 1 0 β - 1 ) .

The first matrix lies in Γ 1 ( t ) , and it is congruent modulo t n to

( * * t β - 1 ( 1 + t d ) 1 + t ( d - β c ) ) .

By Lemma 3.1, this matrix is contained in Γ 1 ( t n ) h ( β - 1 ( 1 + t d ) , d - β c ) and (ii) follows.

For (iii), the matrix ξ 0 h ( c , d ) J equals

( 1 0 0 t ) ( P t n Q t R S ) ( 0 - 1 1 0 ) = ( P t n - 1 Q t 2 R S ) J ( t 0 0 1 ) .

The first matrix of the right-hand side lies in Γ 1 ( t ) , and it is congruent modulo t n to

( * * t 2 c 1 + t d ) ,

from which (iii) follows by Lemma 3.1. ∎

Lemma 3.6.

Let a , c , d A n - 1 be any elements. Take any lift a A of 1 + t a A n . Then we have

η 𝔞 , h ( c , d ) Γ 1 ( t n ) h ( ( 1 + t a ) c , a + d + t a d ) .

Proof.

Since 𝔞 1 mod t , the matrix η 𝔞 , lies in Γ 1 ( t ) . Thus the lemma follows from

η 𝔞 , h ( c , d ) ( * * t ( 1 + t a ) c ( 1 + t a ) ( 1 + t d ) ) mod t n .

3.3 Unramified double cuspforms

Put

S k = { f S k ord ( h ( 0 , d ) ( ) , f ) 2  for any  d A n - 1 } .

Lemma 3.7.

The C -subspace S k is stable under all Hecke operators.

Proof.

Let f be any element of S k and 𝔪 A any monic irreducible polynomial. By Lemma 3.3(i) the uniformizer at the cusp h ( 0 , d ) ( ) is u ( z ) and we can write

( f | k h ( 0 , d ) ) ( z ) = i 2 a i u ( z ) i , a i .

Then Lemma 2.1(ii) shows that the term

deg ( β ) < deg ( 𝔪 ) f | k h ( 0 , d ) ξ 𝔪 , β

in the equality of Lemma 3.4 has no linear term of u ( z ) . Thus the lemma follows from the latter assertion of Lemma 3.4. ∎

For any f S k and d A n - 1 , we write

( f | k h ( 0 , d ) ) ( z ) = i 1 a i u ( z ) i , a i ,

and put L d ( f ) = a 1 . Then the -linear map

L : S k / S k d A n - 1 , f ( L d ( f ) ) d

is injective.

Lemma 3.8.

One has

dim S k / S k = q n - 1 .

In particular, the map L is bijective.

Proof.

We denote Cusps ( Γ 1 ( t n ) ) also by Cusps . By Lemma 3.2(i), the points

h ( 0 , d ) ( ) , d A n - 1

form a subset Cusps of cardinality q n - 1 of Cusps . We abusively identify Cusps and Cusps with the reduced divisors they define on the Drinfeld modular curve X 1 ( t n ) over , and put D = Cusps + Cusps . Let g be the genus of X 1 ( t n ) and h the number of cusps. Since 0 Cusps Cusps , we have h > q n - 1 .

Let ω ¯ be the Hodge bundle on X 1 ( t n ) , so that deg ( ω ¯ 2 ) = 2 g - 2 + 2 h and deg ( ω ¯ ) 0 (see for example [7, Corollary 4.2] with Δ = { 1 } ). For k 2 , we have

deg ( ω ¯ k ( - D ) ) = k deg ( ω ¯ ) - deg ( D )
= ( k - 2 ) deg ( ω ¯ ) + 2 g - 2 + h - q n - 1 2 g - 1 .

As S k can be identified with H 0 ( X 1 ( t n ) , ω ¯ k ( - D ) ) , the Riemann–Roch theorem implies

dim S k = deg ( ω ¯ k ( - D ) ) + 1 - g = ( k - 1 ) ( g - 1 + h ) - q n - 1 .

From dim S k = ( k - 1 ) ( g - 1 + h ) (see [2, Proposition 5.4]), we obtain

dim S k / S k = q n - 1 .

Since the both sides of the injection L have the same dimension, it is a bijection. ∎

Lemma 3.9.

All Hecke operators act trivially on S k / S k .

Proof.

Let 𝔪 A be any monic irreducible polynomial. Take any element f S k . By Lemma 2.1 and Lemma 3.4, we obtain L d ( T 𝔪 f ) = L d ( f ) for any d A n - 1 and the injectivity of the map L shows T 𝔪 f f mod S k . This concludes the proof. ∎

3.4 Nilpotency of U t on S k / S k ( 2 )

For any integer i, put

C i = { ( c , d ) A n - 1 2 v ¯ t ( c ) i } .

To study the U t -action on S k , we define

S k , i = { f S k ord ( h ( c , d ) ( ) , f ) 2  for any  ( c , d ) C i }

so that

S k = S k , n - 1 S k , n - 2 S k , 0 = S k , - 1 S k ( 2 ) .

Proposition 3.10.

Let i [ 0 , n - 1 ] be any integer. We have:

  1. U t ( S k , i ) S k , i - 1 .

  2. U t ( S k , 0 ) S k ( 2 ) .

In particular, the operator U t acting on S k / S k ( 2 ) is nilpotent.

Proof.

For assertion (i), take any f S k , i and ( c , d ) C i - 1 . We need to show

(3.2) ord ( h ( c , d ) ( ) , U t f ) 2 .

Since the case of c = 0 follows from Lemma 3.7, we may assume c 0 . Put m = v ¯ t ( c ) . For any β 𝔽 q , we have ( t c , d - β c ) C i and the assumption yields v ¯ t ( t c ) = m + 1 . By Lemma 3.3(i), we can write

( f | k h ( t c , d - β c ) ) ( z ) = j 2 a j ( β ) u n - 2 - m ( z ) j , a j ( β ) ,

and Lemma 3.5(i) yields

( ( U t f ) | k h ( c , d ) ) ( z ) = β 𝔽 q ( f | k ξ β h ( c , d ) ) ( z ) = β 𝔽 q ( f | k h ( t c , d - β c ) ξ β ) ( z )
= 1 t β 𝔽 q j 2 a j ( β ) u n - 2 - m ( z + β t ) j .

Since the uniformizer at h ( c , d ) ( ) is u n - 1 - m ( z ) , Lemma 2.2 gives inequality (3.2).

Let us show assertion (ii). Take any f S k , 0 and d A n - 1 . Since we already know U t f S k , 0 by (i), it is enough to show

(3.3) ord ( h ( 0 , d ) ( 0 ) , U t f ) 2 .

By Lemma 3.3(ii), the uniformizer at h ( 0 , d ) ( 0 ) = h ( 0 , d ) J ( ) is u n ( z ) .

Consider the equality

(3.4) ( U t f ) | k h ( 0 , d ) J = β 𝔽 q × f | k ξ β h ( 0 , d ) J + f | k ξ 0 h ( 0 , d ) J .

For the first term in the right-hand side of (3.4), we have

v ¯ t ( β - 1 ( 1 + t d ) ) = 0

and by Lemma 3.3 (i) we can write

( f | k h ( β - 1 ( 1 + t d ) , d ) ) ( z ) = j 2 a j u n - 1 ( z ) j , a j .

Then Lemma 3.5(ii) gives

( f | k ξ β h ( 0 , d ) J ) ( z ) = t k - 1 ( β - 1 t ) - k j 2 a j u n - 1 ( β z - 1 β - 1 t ) j
= β k t j 2 a j β - 2 j u n - 1 ( z - β - 1 t ) j

and by Lemma 2.2 this term lies in u n ( z ) 2 [ [ u n ( z ) ] ] .

For the second term on the right-hand side of (3.4), write

( f | k h ( 0 , d ) J ) ( z ) = j 1 a j u n ( z ) j , a j .

By Lemma 3.5(iii), we have

( f | k ξ 0 h ( 0 , d ) J ) ( z ) = t k - 1 ( f | k h ( 0 , d ) J ) ( t z ) = t k - 1 j 1 a j u n ( t z ) j .

Since Lemma 2.1(iii) shows

u n ( t z ) u n ( z ) 2 [ [ u n ( z ) ] ] ,

we obtain inequality (3.3). This concludes the proof of the proposition. ∎

Recall that we fixed an embedding ι t : K ¯ t . We say that λ K ¯ is a t-adic unit if ι t ( λ ) 𝒪 t × .

Theorem 3.11.

For any integer k 2 , the following are equivalent:

  1. U t acting on S k ( 2 ) ( Γ 1 ( t n ) ) has no t - adic unit eigenvalue.

  2. U t acting on S k has no t - adic unit eigenvalue.

  3. dim S k ord ( Γ 1 ( t n ) ) q n - 1 .

  4. U t acting on S 2 ( 2 ) ( Γ 1 ( t n ) ) is nilpotent.

  5. U t acting on S 2 is nilpotent.

  6. dim S 2 ord ( Γ 1 ( t n ) ) q n - 1 .

If these equivalent conditions hold, then for any k 2 we have

dim S k ord ( Γ 1 ( t n ) ) = q n - 1

and all Hecke operators act trivially on S k ord ( Γ 1 ( t n ) ) .

Proof.

The equivalence of (i) and (ii) follows from Proposition 3.10. Note that the multiplicity μ of t-adic unit eigenvalues of U t acting on S k is equal to dim S k ord ( Γ 1 ( t n ) ) . By Lemma 3.8 and Lemma 3.9, the only t-adic unit eigenvalue of U t acting on S k / S k is one, with multiplicity q n - 1 . Hence μ q n - 1 , and the equality holds if and only if there is no other t-adic unit eigenvalue on S k . The latter condition means that (ii) holds. This implies that (ii) and (iii) are equivalent, and that (iii) is equivalent to dim S k ord ( Γ 1 ( t n ) ) = q n - 1 .

By [8, (2.6) and Proposition 2.2], any eigenvalue of U t acting on S 2 ( Γ 1 ( t n ) ) is algebraic over 𝔽 q . Thus U t acts on a subspace of S 2 ( Γ 1 ( t n ) ) without t-adic unit eigenvalue if and only if the action is nilpotent. This shows the equivalence of (iv)(vi). The equivalence of (iii) and (vi) follows from [8, Proposition 3.4 (1)].

If these conditions hold, then we have dim S k ord ( Γ 1 ( t n ) ) = q n - 1 and the natural map

S k ord ( Γ 1 ( t n ) ) S k / S k

is an isomorphism compatible with all Hecke operators. Now the last assertion follows from Lemma 3.9. ∎

Since X 1 ( t ) is of genus zero, we have S 2 ( 2 ) ( Γ 1 ( t ) ) = 0 and the nilpotency of U t acting on it holds trivially. Thus Theorem 3.11 yields the following corollary, which reproves [6, Lemma 2.4] and [8, Proposition 4.3] without using the theory of A-expansions [11] or Bandini–Valentino’s formula [1, (4.2)].

Corollary 3.12.

For any k 2 , we have

dim S k ord ( Γ 1 ( t ) ) = 1

and all Hecke operators act trivially on S k ord ( Γ 1 ( t ) ) .

Note that by [4, Corollary 5.7] the genus of X 1 ( t n ) is

(3.5) 1 + q 2 n - 2 - ( n + 1 ) q n - 1 + ( n - 1 ) q n - 2

and for n 2 it is zero only if n = q = 2 . Thus the nilpotency of U t acting on S 2 ( 2 ) ( Γ 1 ( t n ) ) seems non-trivial in general. We will prove it in a rather indirect way (Corollary 4.10).

4 Freeness and triviality

In this section, we prove the triviality of the Hecke action on S k ( Γ 1 ( t n ) ) for any k 2 and n 1 (Theorem 4.9). Put Θ n = 1 + t A n A n × . The key point of the proof is to show that S 2 ( Γ 1 ( t n ) ) , which we consider as a [ Θ n ] -module via the diamond operator, is the direct sum of copies of [ Θ n ] (Proposition 4.8). For this, we need a description of S 2 ( Γ 1 ( t n ) ) using harmonic cocycles on the Bruhat–Tits tree.

4.1 Bruhat–Tits tree and Γ 1 ( t n )

We consider K 2 as the set of row vectors, and define an action of GL 2 ( K ) on K 2 by

γ ( x 1 , x 2 ) = ( x 1 , x 2 ) γ - 1 .

Let 𝒯 be the Bruhat–Tits tree for SL 2 ( K ) (see for example [12, Chapter II, Section 1], [4, Section 1] and [2, Section 3.1]). Recall that the set 𝒯 0 of vertices of 𝒯 is by definition the set of K × -equivalence classes of 𝒪 K -lattices in K 2 , where 𝒪 K is the ring of integers of K . The action induces an action of GL 2 ( K ) on the tree 𝒯 , and also on the oriented tree 𝒯 o associated to 𝒯 . We denote by 𝒯 1 o the set of oriented edges. For any e 𝒯 1 o , the origin, the terminus and the opposite edge of e are denoted by o ( e ) , t ( e ) and - e , respectively. Then the group { ± 1 } acts on 𝒯 1 o by ( - 1 ) e = - e , which commutes with the action of GL 2 ( K ) .

Put π = 1 / t , which is a uniformizer of K . For any integer i, let v i be the class of the lattice 𝒪 K ( π i , 0 ) 𝒪 K ( 0 , 1 ) . Then we have

( π - i 0 0 1 ) v 0 = v i .

We denote by e i the oriented edge with origin v i and terminus v i + 1 .

For any subgroup Γ of SL 2 ( A ) , we say e 𝒯 1 o is Γ-stable if Stab ( Γ , e ) = { 1 } , and Γ-unstable otherwise. We define Γ-stability of a vertex similarly. The set of Γ-stable edges is denoted by 𝒯 1 o , Γ - st . For Γ = Γ 1 ( t ) , we know [10, Section 7] that the set of Γ 1 ( t ) -stable edges is equal to Γ 1 ( t ) J ( ± e 0 ) with

J = ( 0 - 1 1 0 ) .

Lemma 4.1.

A complete set of representatives of Γ 1 ( t n ) \ T 1 o , Γ 1 ( t ) - st / { ± 1 } is given by

Λ 1 , n = { h ( c , d ) J e 0 c , d A n - 1 } .

Proof.

Consider the natural map

(4.1) Λ 1 , n Γ 1 ( t n ) \ 𝒯 1 o , Γ 1 ( t ) - st / { ± 1 } ,

which is surjective since 𝒯 1 o , Γ 1 ( t ) - st = Γ 1 ( t ) J ( ± e 0 ) . Suppose that ( c , d ) , ( c , d ) A n - 1 2 satisfy

γ h ( c , d ) J e 0 = h ( c , d ) J e 0 or γ h ( c , d ) J ( - e 0 ) = h ( c , d ) J e 0

with some γ Γ 1 ( t n ) . For the former case, since J e 0 is Γ 1 ( t ) -stable, we have

γ h ( c , d ) = h ( c , d )

and thus ( c , d ) = ( c , d ) . For the latter case, we have

J - 1 h ( c , d ) - 1 γ h ( c , d ) J v 0 = v 1 .

Since the distance of v 0 and v 1 is one, it contradicts [12, Chapter II, Section 1.2, Corollary]. Hence the map (4.1) is also injective. ∎

4.2 Harmonic cocycles

In this subsection, we recall a description of Drinfeld cuspforms using harmonic cocycles due to Teitelbaum [13], following [2] and [8].

Let k 2 be any integer. We denote by H k - 2 ( ) the -subspace of homogeneous polynomials of degree k - 2 in the polynomial ring [ X , Y ] . We consider the left action of GL 2 ( K ) on it defined by

γ ( X , Y ) = ( X , Y ) γ .

We put

V k ( ) = Hom ( H k - 2 ( ) , ) ,

on which GL 2 ( K ) acts naturally. For ξ GL 2 ( K ) , ω V k ( ) and P ( X , Y ) H k - 2 ( ) , the action is given by

( ξ ω ) ( P ( X , Y ) ) = ω ( ξ - 1 P ( X , Y ) ) = ω ( P ( ( X , Y ) ξ - 1 ) ) .

Definition 4.2.

A map c : 𝒯 1 o V k ( ) is called a harmonic cocycle of weight k over if the following two conditions hold:

  1. For any v 𝒯 0 , we have

    e 𝒯 1 o , t ( e ) = v c ( e ) = 0 .

  2. For any e 𝒯 1 o , we have c ( - e ) = - c ( e ) .

For any arithmetic subgroup Γ of SL 2 ( A ) , we say c is Γ-equivariant if c ( γ e ) = γ c ( e ) for any γ Γ and e 𝒯 1 o . We denote the -vector space of Γ-equivariant harmonic cocycles of weight k over by C k har ( Γ ) .

Let Γ be an arithmetic subgroup of SL 2 ( A ) which is p -torsion free. In this case, for any Γ-unstable vertex v, the group Stab ( Γ , v ) fixes a unique rational end which we denote by b ( v ) .

Definition 4.3.

A Γ-stable edge e 𝒯 1 o is called a Γ-source of an edge e 𝒯 1 o if the following conditions hold:

  1. If e is Γ-stable, then e = e .

  2. If e is Γ-unstable, then a vertex v of e is Γ-unstable, e lies on the unique half line from v to b ( v ) and e has the same orientation as e with respect to this half line.

The set of Γ-sources of e is denoted by src Γ ( e ) .

For any harmonic cocycle c : 𝒯 1 o V k ( ) of weight k over , we have

(4.2) c ( e ) = e src Γ ( e ) c ( e ) .

We denote by S k ( Γ ) the -vector space of Drinfeld cuspforms of level Γ and weight k. Then, for any rigid analytic function f on Ω and e 𝒯 1 o , Teitelbaum defined an element Res ( f ) ( e ) V k ( ) , which gives a natural isomorphism of -vector spaces

(4.3) Res Γ : S k ( Γ ) C k har ( Γ ) , f ( e Res ( f ) ( e ) ) ,

see [13, Theorem 16]. Note that we are following the normalization in [2, Theorem 5.10]. Moreover, by [2, (17)], the slash operator can be read off via the corresponding harmonic cocycle by

(4.4) Res ( f | k γ ) ( e ) = γ - 1 Res ( f ) ( γ e ) .

Lemma 4.4.

One has

dim C 2 har ( Γ 1 ( t n ) ) = q 2 ( n - 1 ) .

Proof.

By [4, Proposition 5.6] (or Lemma 3.2), the number h of cusps of X 1 ( t n ) equals

h = ( n + 1 ) q n - 1 - ( n - 1 ) q n - 2 .

Thus the lemma follows from [2, Proposition 5.4], (3.5) and (4.3). ∎

Lemma 4.5.

Let c be any element of C 2 har ( Γ 1 ( t n ) ) . Then:

  1. For any γ Γ 1 ( t n ) and e 𝒯 1 o , we have c ( γ e ) = c ( e ) .

  2. c is determined by its restriction to the subset Λ 1 , n of Lemma 4.1.

Proof.

Since the group GL 2 ( K ) acts trivially on V 2 ( ) = , we have

c ( γ e ) = γ c ( e ) = c ( e )

and assertion (i) follows.

For assertion (ii), it suffices to show that if the restriction of c to Λ 1 , n is zero, then c ( e ) = 0 for any e 𝒯 1 o . By (4.2), we may assume that e is Γ 1 ( t ) -stable. Then it is written as e = ± γ e with some e Λ 1 , n and γ Γ 1 ( t n ) , which yields c ( e ) = ± γ c ( e ) = 0 . This concludes the proof. ∎

Corollary 4.6.

The C -linear map

C 2 har ( Γ 1 ( t n ) ) e Λ 1 , n , c ( c ( e ) ) e Λ 1 , n

is an isomorphism.

Proof.

By Lemma 4.5(ii), the map is injective. Since Λ 1 , n = q 2 ( n - 1 ) , Lemma 4.4 implies that it is an isomorphism. ∎

By Corollary 4.6, there exists a unique element [ c , d ] C 2 har ( Γ 1 ( t n ) ) satisfying

[ c , d ] ( h ( c , d ) J e 0 ) = { 1 if  ( c , d ) = ( c , d ) , 0 otherwise .

The set { [ c , d ] c , d A n - 1 } forms a basis of the -vector space C 2 har ( Γ 1 ( t n ) ) .

4.3 Proof of the main theorem

Consider the subgroup Θ n = 1 + t A n of A n × . Via the isomorphism Res Γ 1 ( t n ) of (4.3), the diamond operator α t n acting on S 2 ( Γ 1 ( t n ) ) induces an operator on C 2 har ( Γ 1 ( t n ) ) , which we also denote by α t n . In particular, the group Θ n acts on C 2 har ( Γ 1 ( t n ) ) via α α t n .

Lemma 4.7.

For any a , c , d A n - 1 , the action of 1 + t a Θ n on [ c , d ] is given by

1 + t a t n [ c , d ] = [ ( 1 + t a ) - 1 c , ( 1 + t a ) - 1 ( d - a ) ] .

Proof.

By (4.4) and Lemma 3.6, for any c , d A n - 1 we have

( 1 + t a t n [ c , d ] ) ( h ( c , d ) J e 0 ) = [ c , d ] ( h ( ( 1 + t a ) c , a + d + t a d ) J e 0 ) ,

which is equal to one if ( c , d ) = ( ( 1 + t a ) - 1 c , ( 1 + t a ) - 1 ( d - a ) ) and zero otherwise. This concludes the proof. ∎

Proposition 4.8.

The C [ Θ n ] -module S 2 ( Γ 1 ( t n ) ) is isomorphic to the direct sum of q n - 1 copies of C [ Θ n ] .

Proof.

It suffices to show the assertion for C 2 har ( Γ 1 ( t n ) ) . Take any ( c , d ) A n - 1 2 . We claim that the Θ n -orbit

{ 1 + t a t n [ c , d ] a A n - 1 }

of [ c , d ] is of cardinality q n - 1 . Indeed, if one has 1 + t a t n [ c , d ] = 1 + t a t n [ c , d ] for some a , a A n - 1 , then Lemma 4.7 yields

( 1 + t a ) - 1 ( d - a ) = ( 1 + t a ) - 1 ( d - a ) ,

which is equivalent to ( 1 + t d ) ( a - a ) = 0 and we obtain a = a .

Denote by V ( c , d ) the -subspace of C 2 har ( Γ 1 ( t n ) ) spanned by the Θ n -orbit of [ c , d ] . Then V ( c , d ) is stable under the Θ n -action and dim V ( c , d ) = q n - 1 . Consider the map

[ Θ n ] V ( c , d ) , α α t n [ c , d ] .

It is a homomorphism of [ Θ n ] -modules which is surjective. Since the both sides have the same dimension, it is an isomorphism. Since the -vector space C 2 har ( Γ 1 ( t n ) ) is the direct sum of -subspaces V ( c , d ) , the proposition follows from Lemma 4.4. ∎

Theorem 4.9.

We have

dim S 2 ord ( Γ 1 ( t n ) ) = q n - 1

and all Hecke operators act trivially on S k ord ( Γ 1 ( t n ) ) for any k 2 .

Proof.

By Theorem 3.11, it is enough to show dim S 2 ord ( Γ 1 ( t n ) ) q n - 1 . Put

Γ 0 p ( t n ) = { γ SL 2 ( A ) | γ mod t n ( 1 + t A n A n 0 1 + t A n ) } ,

as in [8, Section 3]. Then the Θ n -fixed part of S 2 ( Γ 1 ( t n ) ) is S 2 ( Γ 0 p ( t n ) ) . Since the Hecke operator U t commutes with the action of Θ n and it is defined by the same formula for the levels Γ 1 ( t n ) and Γ 0 p ( t n ) (see [8, Section 3.1]), we see that S 2 ord ( Γ 1 ( t n ) ) is stable under the Θ n -action and

S 2 ord ( Γ 1 ( t n ) ) Θ n = S 2 ord ( Γ 0 p ( t n ) ) ,

where the right-hand side is the ordinary subspace of S 2 ( Γ 0 p ( t n ) ) defined similarly to the case of S 2 ( Γ 1 ( t n ) ) . Then [8, Proposition 3.5] and Corollary 3.12 yield

dim S 2 ord ( Γ 1 ( t n ) ) Θ n = dim S 2 ord ( Γ 0 p ( t n ) ) = dim S 2 ord ( Γ 1 ( t ) ) = 1 .

On the other hand, Proposition 4.8 gives an injection of [ Θ n ] -modules

S 2 ord := S 2 ord ( Γ 1 ( t n ) ) i = 1 q n - 1 V i , V i = [ Θ n ] .

Let I be the set of integers M [ 1 , q n - 1 ] such that there exists an injection of [ Θ n ] -modules S 2 ord i = 1 M V i . Then I is nonempty and let m be its minimal element.

Now we reduce ourselves to showing m = 1 . Suppose m > 1 and consider an injection S 2 ord i = 1 m V i . Since Θ n is an abelian p-group and contains no non-trivial p-power root of unity, Schur’s lemma implies that the only irreducible representation of Θ n over is the trivial representation. Since both of

S 2 ord V 1 , S 2 ord i = 2 m V i

are [ Θ n ] -submodules of S 2 ord , if one of them is non-zero, then it contains the trivial representation. Since the -vector space ( S 2 ord ) Θ n is one-dimensional, we see that either of them is zero. Thus either of the induced maps

S 2 ord ( i = 1 m V i ) / V 1 i = 1 m - 1 V i , S 2 ord ( i = 1 m V i ) / ( i = 2 m V i ) V 1

is injective, which contradicts the minimality of m. This concludes the proof of the theorem. ∎

Theorem 3.11 and Theorem 4.9 yield the following corollary.

Corollary 4.10.

The operator U t acting on S 2 ( 2 ) ( Γ 1 ( t n ) ) is nilpotent.

Remark 4.11.

By Theorem 3.11, if we could prove the nilpotency of U t acting on S 2 ( 2 ) ( Γ 1 ( t n ) ) directly, then Theorem 4.9 would follow. As the proof of Theorem 4.9 indicates, the reason we can bypass it is that we know the dimension of S 2 ord ( Γ 1 ( t ) ) because X 1 ( t ) is of genus zero. The author has no idea of how to show the nilpotency directly.

Award Identifier / Grant number: JP17K05177

Award Identifier / Grant number: JP20K03545

Funding statement: This work was supported by JSPS KAKENHI Grant Numbers JP17K05177 and JP20K03545.

Acknowledgements

The author would like to thank Ernst-Ulrich Gekeler and Federico Pellarin for helpful conversations on this topic, Gebhard Böckle for pointing out an error in a previous manuscript and the anonymous referee for careful reading and valuable comments. A part of this work was carried out during the author’s visit to Université Jean Monnet. He wishes to thank its hospitality.

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Received: 2021-01-09
Revised: 2022-07-27
Published Online: 2022-09-29
Published in Print: 2022-11-01

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