Results on the standard twist of L -functions Jerzy Kaczorowski - PowerPoint PPT Presentation

Notation. The standard twist. The case of half-integral weight cusp forms L -functions. The general case. Results on the standard twist of L -functions Jerzy Kaczorowski (joint work with Alberto Perelli) Adam Mickiewicz University, Poznań, Poland and Institute of Mathematics of the Polish Academy of Sciences Cetraro, 2019

Notation. The standard twist. The case of half-integral weight cusp forms L -functions. The general case. General notation. S — the Selberg class. S ♯ — the extended Selberg class. For F ∈ S ♯ , σ > 1 ∞ ∞ a F ( n ) a ( n ) � � F ( s ) = = n s n s n = 1 n = 1 Functional equation Φ( s ) = ω Φ( 1 − s ) , where Φ( s ) = Q s � r j = 1 Γ( λ j s + µ j ) F ( s ) , and r � 0, Q > 0, λ j > 0, ℜ µ j � 0, | ω | = 1.

Notation. The standard twist. The case of half-integral weight cusp forms L -functions. The general case. Invariants Degree: r � d F := 2 λ j j = 1

Notation. The standard twist. The case of half-integral weight cusp forms L -functions. The general case. Invariants Degree: r � d F := 2 λ j j = 1 Conductor: r λ 2 λ j q F := ( 2 π ) d F Q 2 � j j = 1

Notation. The standard twist. The case of half-integral weight cusp forms L -functions. The general case. Invariants Degree: r � d F := 2 λ j j = 1 Conductor: r λ 2 λ j q F := ( 2 π ) d F Q 2 � j j = 1 ξ - and θ -invariants: r ( µ j − 1 � θ F := ℑ ξ F ξ F := 2 2 ) , j = 1

Notation. The standard twist. The case of half-integral weight cusp forms L -functions. The general case. Invariants Degree: r � d F := 2 λ j j = 1 Conductor: r λ 2 λ j q F := ( 2 π ) d F Q 2 � j j = 1 ξ - and θ -invariants: r ( µ j − 1 � θ F := ℑ ξ F ξ F := 2 2 ) , j = 1 The root number: r λ − 2 i ℑ ( µ j ) � ω F := ω j j = 1

Notation. The standard twist. The case of half-integral weight cusp forms L -functions. The general case. The standard twist d := { F ∈ S ♯ : d F = d } S ♯ Definition Let F ∈ S # d , d > 0. For a real α > 0 and σ > 1 we define the standard twist by the formula ∞ a ( n ) n s e ( − n 1 / d α ) . � F ( s , α ) = n = 1 ( e ( θ ) := exp ( 2 π i θ ))

Notation. The standard twist. The case of half-integral weight cusp forms L -functions. The general case. The standard twist 1. We write n α = q F d − d α d and � a ( n ) if n α = n ∈ N a ( n α ) = 0 otherwise .

Notation. The standard twist. The case of half-integral weight cusp forms L -functions. The general case. The standard twist 1. We write n α = q F d − d α d and � a ( n ) n α = n ∈ N if a ( n α ) = 0 otherwise . 2. Spec ( F ) := { α > 0 : a ( n α ) � = 0 }

Notation. The standard twist. The case of half-integral weight cusp forms L -functions. The general case. Theorem (J.K.-A.P. – 2005) F ( s , α ) has meromorphic continuation to C . Moreover, F ( s , α ) is entire if α �∈ Spec ( F ) . Otherwise, F ( s , α ) has at most simple poles at the points s k = d + 1 − k d − i θ F k � 0 d , 2 d with a ( n α ) res s = s 0 F ( s , α ) = c F ( c F � = 0 ) . n 1 − s 0 α

Notation. The standard twist. The case of half-integral weight cusp forms L -functions. The general case. Some applications of the standard twist The standard twist proved to be the central object in the Selberg class theory.

Notation. The standard twist. The case of half-integral weight cusp forms L -functions. The general case. Some applications of the standard twist I. S ♯ d = ∅ if 0 < d < 1. [E. Richert, S. Bochner, B. Conrey-A. Ghosh, G. Molteni] Proof: Suppose there exists F ∈ S ♯ d with 0 < d < 1. Take α ∈ Spec ( F ) . Then F ( s , α ) has a pole at s = s 0 . But ℜ ( s 0 ) = 1 2 + 1 2 d > 1 – a contradiction.

Notation. The standard twist. The case of half-integral weight cusp forms L -functions. The general case. Some applications of the standard twist II. Description of the structure of S ♯ 1 (J.K.-A.P. – 1999). If d = 1, the standard twist is linear ∞ a ( n ) e ( − α n ) � F ( s , α ) = n s n = 1 Thus F ( s , α + 1 ) = F ( s , α ) Comparing residues at s = s 0 we see that coefficients a ( n ) are q -periodic ( q - the conductor of F ). = ⇒ F ( s ) is a linear combination of Dirichlet L -functions ( mod q ) .

Notation. The standard twist. The case of half-integral weight cusp forms L -functions. The general case. Some applications of the standard twist III. S ♯ d = ∅ for 1 < d < 2 (J.K.-A.P. – 2011). Main idea of the proof. Suppose that F ∈ S ♯ has degree 1 < d < 2, and consider ∞ a ( n ) � F ( s , f ) = n s e ( − f ( n , α )) n = 1 where N α j ξ κ j � f ( ξ, α ) = , α = ( α 0 , . . . , α N ) , α 0 > 0 j = 0 κ 0 > κ 1 > . . . > κ N > 0 κ 0 > 1 / d

Notation. The standard twist. The case of half-integral weight cusp forms L -functions. The general case. Some applications of the standard twist Two basic operations T : F ( s , f ) �→ F ( s ∗ , f ∗ ) where f ∗ denotes (suitably defined) ‘conjugated’ exponent of a similar form as f but possibly with different exponents and coefficients. S : F ( s , f ) �→ F ( s , ξ + f ) Consider the group S = < S , T > which acts on twists F ( s , f ) .

Notation. The standard twist. The case of half-integral weight cusp forms L -functions. The general case. Some applications of the standard twist We take α 0 ∈ Spec ( F ) and the exponent f 0 ( ξ ) = α 0 ξ 1 / d , so that F ( s , f 0 ) is the standard twist of F . Then it is proved that there exists g = STS m N S . . . S m 1 TS m 1 TS ∈ S such that g ( F ( s , f 0 )) = F ( s ∗ , f ∗ ) with ℜ ( s ∗ 0 ) > 1 ⇒ F ( s , f ∗ ) has Now, F ( s , f 0 ) has a pole at s = s 0 = a pole at s = s ∗ 0 - a contradiction.

Notation. The standard twist. The case of half-integral weight cusp forms L -functions. The general case. Some applications of the standard twist n = 1 a ( n ) n − s has meromorphic IV. Let F ( s ) = � ∞ continuation to C with at most one singularity, a pole at s = 1. Moreover, let Φ( s ) = ω Φ( 1 − s ) ( | ω | = 1 ) � √ � s 5 Φ( s ) = Γ( s + µ ) F ( s ) ( ℜ ( µ ) � 0 ) 2 π a ( n ) ≪ n ε ∞ b ( n )Λ( n ) ( b ( n ) ≪ n θ , θ < 1 / 2 ) . � log F ( s ) = n s n = 2

Notation. The standard twist. The case of half-integral weight cusp forms L -functions. The general case. Some applications of the standard twist Theorem (J.K.-A.P. – 2018) There exists k ∈ N , χ ( mod5 ) such that ℜ ( µ ) = k − 1 χ ( − 1 ) = ( − 1 ) k 2 and either F ( s ) = ζ ( s ) L ( s , χ ) (if F ( s ) is polar) or F ( s ) = L ( s + µ, f ) for certain newform f ∈ S k (Γ 0 ( 5 ) , χ ) .

Notation. The standard twist. The case of half-integral weight cusp forms L -functions. The general case. The general problem: Problem: Describe finer properties of the standard twist. In particular: (1) does it satisfy functional equation relating s to 1 − s ? (2) What is the polar structure of F ( s , α ) when α ∈ Spec ( F ) ? (3) Give precise convexity bounds for the Lindelöff µ -function µ ( s , α ) = inf { λ | F ( σ + it ) = O ( | t | λ ) as t → ∞} . (4) Determine location of the zeros (trivial, nontrivial). (5) Other.

Notation. The standard twist. The case of half-integral weight cusp forms L -functions. The general case. The case of half-integral weight cusp forms L -functions. Let f be a cusp form of half-integral weight κ = k / 2 and level N , where k > 0 is an odd integer and 4 | N , and L f ( s ) be the associated Hecke L -function. Then L f ( s ) is entire and satisfies the functional equation Λ f ( s ) = ω Λ f ∗ ( κ − s ) where � √ � s N Λ f ( s ) = Γ( s ) L f ( s ) 2 π | ω | = 1 and f ∗ is related to f by the slash operator. Note that L f ∗ ( s ) is also entire and has properties similar to L f ( s ) .

Notation. The standard twist. The case of half-integral weight cusp forms L -functions. The general case. The case of half-integral weight cusp forms L -functions. Extra notation  − e i πµ a ∗ ( ν 2 ) if ν � 1    e i π ( 1 c ∗ l ( ν 2 ) = 2 + l − µ ) a ∗ ( ν 2 ) if − ν α < ν < − 1  e − i πµ a ∗ ( ν 2 ) if ν < − ν α   √ ν = √ n ν α = √ n α = 1 N α , ( n � 1 ) 2 c ∗ ( ν 2 ) F + � l ( s , ν ) = 1 2 + l | ν + ν α | 2 s − 1 2 − l | ν | ν> − ν α c ∗ ( ν 2 ) F − � l ( s , ν ) = 1 2 + l | ν + ν α | 2 s − 1 2 − l | ν | ν< − ν α F ∗ l ( s , α ) = e − i π s F + l ( s ) + e i π s F − l ( s )

Notation. The standard twist. The case of half-integral weight cusp forms L -functions. The general case. The case of half-integral weight cusp forms L -functions. Theorem ((J.K.-A.P. – 2018)) (1) The functions F ∗ l ( s , α ) are entire. (2) We have � √ � 1 − 2 s h ∗ ω N a l Γ( 2 ( 1 − s ) − 1 � 2 − l ) F ∗ √ F ( s , α ) = l ( 1 − s , α ) 4 π i 2 π l = 0 (( h ∗ = max ( 0 , [ κ ] − 1 ))

Notation. The standard twist. The case of half-integral weight cusp forms L -functions. The general case. The case of half-integral weight cusp forms L -functions. Corollary For α ∈ Spec ( F ) , the standard twist F ( s , α ) has a finite number of poles. They could be at the points s = s l = 3 4 − l ( l = 0 , . . . , h ∗ ) 2