Number of non-zero coefficients of modular forms modulo p Analytic Aspects of Number Theory, Z¨ urich, 2015 Jo¨ el Bella¨ ıche (first part is joint work with Kannan Soundararajan) Brandeis University May 18, 2015
Objectives n =0 a n q n ∈ F [[ q ]], define Let F be a finite field. For f = � ∞ ψ ( f , x ) = # { n ; 0 ≤ n ≤ x , a n � = 0 } . π ( f , x ) = # { ℓ prime ; ℓ ≤ x , a ℓ � = 0 } . The objective is to give estimates of ψ ( f , x ), π ( f , x ) for x → ∞ , when f is a modular form modulo p , (a) for f fixed; (b) uniformly in f (much harder).
Spaces of modular forms modulo p Fix: a prime p an integer N ≥ 1 (the level ), an element k ∈ Z / ( p − 1) Z (the weight ). Define M k ( N , F p ) ⊂ F p [[ q ]] as the span of all reductions mod p of holomorphic modular forms with coefficients in Z , level Γ 0 ( N ), some weight k ′ such that k ′ (mod p − 1) = k . For F a finite extension of F p , set M k ( N , F ) = M k ( N , F p ) ⊗ F p F
Spaces of modular forms modulo p (continued) Primes dividing Np , and their products, are annoying. Define F k ( N , F ) as the subspace of M k ( N , F ) of f such that a n � = 0 = ⇒ gcd( n , Np ) = 1 . Think of F k ( N , F ) as the “new” part in M k ( N , F ). If f = � a n q n ∈ M k ( N , F ), then f ′ := a n q n ∈ F k ( N 2 , F ) . � ( n , Np )=1 It is no real restriction to consider only F k ( N , F ).
Equivalent for ψ ( f , x ): previous results and statement Assume p ≥ 3. Let 0 � = f ∈ F k ( N , F ). x Serre (1976) : ψ ( f , x ) ≪ (log x ) α for some α > 0. In particular, f is lacunary . x Ahlgren (1999) : ψ ( f , x ) ≫ log x . Chen (2012): ψ ( f , x ) ≫ x (log log x ) N for all N ≥ 1. log x Theorem (B.-Soundararajan, 2014) ψ ( f , x ) ∼ c ( f ) x (log log x ) h ( f ) (log x ) α ( f ) where c ( f ) > 0 , h ( f ) ∈ N , α ( f ) ∈ Q with 0 < α ( f ) ≤ 3 / 4 are effectively computable constants.
Rough skecth of the proof : Hecke operators and Hecke algebras Hecke operators: for ℓ ∤ Np , set � a n q n ) = � ( a n ℓ + ℓ k − 1 a n /ℓ ) q n . T ℓ ( The T ℓ ’s stabilize F k ( N , F ) and commute. But (Serre, Tate, Jochnowitz): only finitely many normalized eigenvectors in F k ( N , F ) (though it is infinite-dimensional.) Use generalized eigenvectors instead. For λ = ( λ ℓ ) ℓ ∤ Np , define F k ( N , F ) λ = { f ∈ F k ( N , F ); ∀ ℓ ∃ n ( T ℓ − λ ℓ ) n f = 0 } . � Then F k ( N , F ) = F k ( N , F ) λ (finite sum) . λ Define the Hecke algebra A λ as the closed subalgebra of End ( F k ( N , F ) λ ) generated by the T ℓ . It is local and complete.
Rough skecth of the proof : the case of a generalized eigenform Fix λ = ( λ ℓ ). Suppose f ∈ F k ( N , F ) λ . Two kinds of ℓ : (i) λ ℓ = 0 or T ℓ locally nilpotent on F k ( N , F ) λ . (ii) λ ℓ � = 0 or T ℓ invertible or T ℓ ∈ A ∗ λ (density α ( f )). Define h ( f ) ( strict order of nilpotence ) as the largest h for which there exist ℓ 1 , . . . , ℓ h of type (i) such that T ℓ 1 . . . T ℓ h f � = 0. Counting n square-free such that a n � = 0. Write n = ℓ 1 . . . ℓ h ℓ ′ 1 . . . ℓ ′ s with ℓ i of type (i), ℓ ′ j of type (ii). a n ( f )= a 1 ( T ℓ ′ 1 ... T ℓ ′ s T ℓ 1 ... T ℓ h f ) � =0 ⇐ ⇒ h ≤ h ( f ) and a 1 ( T ℓ ′ 1 ... T ℓ ′ s g ) � =0 , where g = T ℓ 1 . . . T ℓ h f (finitely many possible g ’s for a given f ) The T ℓ ′ ’s generate a subgroup H of A ∗ λ . T ℓ ′ 1 . . . T ℓ ′ s is a random element of H (random walk). = ⇒ count as if the ℓ i were arbitrary of type (ii) and multiply by the proportion of T ∈ H such that a 1 ( Tg ) � = 0. Then apply Landau’s method.
Uniformity for ψ ( f , x )? Why do we want uniformity? “To take the limit in f .” Let g be a modular form of weight − 1 / 2 (say), e.g. g ( q ) = η ( q ) − 1 = q 1 / 24 � p ( n ) q n modulo p . It is conjectured that ψ ( g , x ) ≍ x . Very little is known about it: not known that ψ ( g , x ) ≫ x 1 / 2+ ǫ for any ǫ > 0 . Define f k as g 1 − p k . The f k are true modular forms mod p . x (log log x ) h k (1) ψ ( f k , x ) ∼ c k . (log x ) α k If we could have x log x for x ≫ A k , (?) ψ ( f k , x ) ≫ we could break the bound x 1 / 2 for ψ ( g , x ). Unfortunately, there is no way to prove (?) with the method we proved (1) : c k goes to 0 very fast, error term is hopelessly large. To compute the equivalent for ψ ( f k , x ) we counted integers n with at least h k := h ( f k ) prime factors, so n ≥ 2 h k and h k tends to grow exponentially in k , so 2 h k grows double-exponentially.
Counting non-zero coefficients at primes: f fixed Counting just prime numbers ℓ such that a ℓ � = 0 is more promising. For f = � a n q n ∈ F k ( N , F ), that set is Frobenian (Serre 1976). So it has a density: x π ( f , x ) ∼ δ ( f ) log x with 0 ≤ δ ( f ) ≤ 1, δ ( f ) ∈ Q . Theorem If f � = 0 , then 0 < δ ( f ) < 1 . Corollary Let f = � a n q n , g = � b n q n ∈ F k ( N , F ) . If a ℓ = b ℓ for a set of density one of ℓ , then f = g.
Counting non-zero coefficients at primes: uniformity? Remember: x (2) π ( f , x ) ∼ δ ( f ) log x , with 0 < δ ( f ) < 1 . To get some uniformity result, like (?), from (2), we need (a) control of δ ( f ), namely that δ ( f ) does not approach 0 (at least for f = f k = g 1 − p k as above). (b) control of the error term (probably using GRH). Both are work in progress. We will discuss a result concerning δ ( f ).
Counting non-zero coefficients at primes: control of δ ( f ) Remember F k ( N , F ) = � λ F k ( N , F ) λ (finite sum) . Theorem (Deligne) To λ is attached a unique semi-simple Galois representation ρ : G Q , Np → GL 2 ( F ) such that tr ¯ ¯ ρ ( Frob ℓ ) = λ ℓ . Let ad 0 ¯ ρ be the adjoint representation, of dimension 3. Theorem Assume that λ is such that ad 0 ¯ ρ is irreducible. Then there exist c , c ′ such that 0 < c < c ′ < 1 and for all f ∈ F ( N , F ) λ , c < δ ( f ) < c ′ .
Counting non-zero coefficients at primes: control of δ ( f ) Theorem Assume that λ is such that ad 0 ¯ ρ is irreducible. Then there exist c , c ′ such that 0 < c < c ′ < 1 and for all f ∈ F ( N , F ) λ , c < δ ( f ) < c ′ . The projective image of ¯ ρ can be abelian (in a torus), dihedral (in the normalizer of a torus), large ( PSL 2 ( F q ) or PGL 2 ( F q )) or exceptional ( A 4 , S 4 or A 5 ). The condition ad 0 ¯ ρ irreducible means that it is exceptional or large. The theorem should also be true for linear combinations of generalized eigenforms.
Counting non-zero coefficients at primes: reducible case What happens when ad 0 ¯ ρ is reducible? The theorem is false, but Say f = � a n q n ∈ F k ( N , F ) λ is abelian if a ℓ depends only on ℓ (mod M ) for some M , say f is dihedral if a ℓ depends only on Frob ℓ in a dihedral extension, special if it is linear combination of abelian and dihedral forms. Conjecture Assume that λ is such that ad 0 ¯ ρ is reducible . (1) There exists c , c ′ such that 0 < c < c ′ < 1 and for all non-special f ∈ F ( N , F ) λ , c < δ ( f ) < c ′ . (2) special forms are rare, i.e. killed by an ideal I in A λ of codim ≥ 1 . (2) is known for N = 1 , p = 2 (Bella¨ ıche-Nicolas-Serre) and N = 1 , p = 3 (Medvedowski). If f ∈ F (1 , F 2 ) is special, h = h ( f ), then δ ( f ) = 2 − u ( h ) − v 2 ( h ) − 1 .
Interlude: pseudo-representations (of dim 2) Let G be a group, A a ring. Representations ρ : G → GL 2 ( A ) don’t glue well. Replace then with their trace and determinant. Definition (Chenevier) A pseudo-rep. of dim 2 of G on A is a pair of maps t , d : G → A , s.t. t (1) = 2. d is a morphism from G to A ∗ ∀ g , h ∈ G , t ( gh ) = t ( hg ). ∀ g , h ∈ G , t ( gh ) + d ( h ) t ( gh − 1 ) = t ( g ) t ( h ). If ρ is a representation, the pair t = tr ρ , d = det ρ is a pseudo-representation. When A is an algebraically closed field, the converse is true.
Theorems on δ ( f ): what about proofs? Theorem 1 If 0 � = f ∈ F k ( F ) , then 0 < δ ( f ) < 1 . Theorem 2 Assume that λ is such that ad 0 ¯ ρ is irreducible. Then there exist c , c ′ such that 0 < c < c ′ < 1 and for all f ∈ F k ( N , F ) λ , c < δ ( f ) < c ′ . Remember the Hecke algebra A λ acting on F k ( N , F ) λ . Similarly, define A k acting on F k ( N , F ). One has A k = � λ A λ . Proposition 1 There exists a pseudo-representation ( t , d ) of ρ : G Q , Np over A k such that t ( Frob ℓ ) = T ℓ and d = ω k − 1 p Both Theorems 1 and 2 can be reduced to results about the image of t .
A result on t ( G Q , Np ) Theorem 3 As a topological vector space, A k is generated by t ( G Q , Np ) . Indeed, t ( G Q , Np ) contains t ( Frob ℓ ) = T ℓ for every ℓ ∤ Np . As a topological algebra, A k is generated by the T ℓ ’s. But the closed subspace generated by t ( G Q , Np ) is an algebra , because t ( g ) t ( h ) = t ( gh ) + d ( h ) t ( gh − 1 ) , and d ( h ) ∈ F p ; and 2 = t (1) ∈ t ( G Q , Np ). QED
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