Hecke s Converse Theorem

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1 Hecke s Converse Theorem Christoph Rösch January 7, 2007 Introduction In Andrea s Talk we saw how to get a Dirichlet series from a modular form. Moreover we saw that this Dirichlet series can be analytically continued has a functional equation. The natural question is now: Can we go the other way? Given a Dirichlet series with a certain analytic property, can we get a corresponding modular form? - Hecke s Converse Theorem, which is the topic of this paper, states, that the answer is basically yes. This section is a summary of what we know already about Dirichlet series modular forms. Definition.. (Dirichlet series) A Dirichlet series is a series of the form L(s) n a n n s (s C) for a sequence {a n } n of complex numbers. Theorem.2. (Growth of Fourier Coefficients) If f(z) n a ne nz is a cusp form of weight 2k, then a n O(n k ). Corollary.3. If f(z) n0 a ne nz is a modular form of weight 2k but not a cusp form, then a n O(n 2k ). Definition.4. (L(s, f)) For a holomorphic function on H with Fourier expansion f(z) n0 a ne nz we put L(s; f) : a n n s. n We call L(s; f) the Dirichlet series associated with f. If f is a modular form, we have a n O(n ν ) for a ν > 0. So L(s; f) n a nn s converges absolutely uniformly on any compact subset of Re(s) > + ν.

2 2 ANALYTIC TOOLS 2 Definition.5. (Λ(s; f) Λ N (s; f)) For N > 0 we put Λ N (s; f) : ( 2π N ) s Γ(s)L(s; f). Moreover we define Λ(s; f) : Λ (s; f). Theorem.6. Let f(z) n0 a ne nz g(z) n0 b ne nz be holomorphic functions on H satisfying a n O(n ν ) b n O(n ν ) for a ν > 0. Let k N be positive numbers. If g(z) ( i Nz) k f( /Nz) then both Λ N (s; f) Λ N (s; g) can be analytically continued to the whole s-plane, satisfying the functional equation Λ N (s; f) Λ N (; g), s + b 0 So the question is, can we somehow go the converse direction in Theorem.6? This will be the implication of Theorem 3.. But first we need some analytic tools: 2 Analytic Tools Lemma 2.. (Stirling s estimate for Γ(s)) s C with Arg(s) π, s 0 : with Γ(s) 2πs s /2 e s e H(s) lim H(s) 0 s uniformly on any vertical strip ν σ ν 2. From that we get that ɛ : 0 < ɛ < π/2 z σ + iτ with τ > : Γ(s) Ce π 2 ɛ τ on any vertical strip a σ b. Proof This is a well known estimate for Γ(s). See for example in [3]. Lemma 2.2. (Inverse Mellin Transform of Γ(s)) For t C with Re(t) > 0: e t Γ(s)t s ds (σ > 0). Re(s)σ

3 2 ANALYTIC TOOLS 3 Proof Consider the function F (z) : e ez e σz as a function of z C for a σ > 0. Obviously, F is a Schwarz function so we can compute the Fourier transform of this function don t have to worry about convergency: F (x) So we know: ye z 0 Γ(σ + ix) e ez e σz e ixz dz e ez e (σ+ix)z dz e y y (σ+ix) dy ˇΓ(z) e ez e σz () where ˇΓ(z) is the inverse Fourier transform of Γ. But we can compute ˇΓ(z): ˇΓ(z) Re(s)σ Γ(σ + iy)e izy dy Γ(s)e zs e zσ ds With equation () we get z C: e ez e σz Γ(s)e zs e zσ ds Re(s)σ Since this holds z C, we can choose a z such that e z t for t C with Re(t) > 0: e t t σ Γ(s)t s t σ ds e t Re(s)σ Re(s)σ Γ(s)t s ds This is just a special case of the inverse Mellin transform. For a function f(t) on R 0, the Mellin transform is defined as g(s) : 0 f(t)t s dt if the integral is convergent. There is an inverse transform, which, for σ > 0, has the form f(t) g(s)t s ds Re(s)σ

4 φ(s)e ɛs m O(e τ δ ɛτ m ) ( τ ) 2 ANALYTIC TOOLS 4 Lemma 2.3. (Phragmen-Lindelöf-Principle) Let ν < ν 2 be two real numbers. Define F : {s C ν Re(s) ν 2 } Let φ be a holomorphic function on a domain containing F satisfying φ(s) O(e τ δ ) ( τ ), s σ + iτ uniformly on F with δ > 0. Let b R. If then φ(s) O( τ b ) ( τ ) on Re(s) ν Re(s) ν 2, φ(s) O( τ b ) ( τ ) uniformly on F. Proof By assumption, there is L > 0 such that φ(s) Le τ δ. First consider the case b0. Then there exists M > 0 such that φ(s) M on the lines Re(s) ν Re(s) ν 2. Let m be a positive integer such that m 2 mod 4. Let s σ + iτ C. Re(s m ) Re((σ + iτ) m ) is a polynomial in σ τ. The highest term of τ is τ m, so we have Re(s m ) τ m + O( τ m ) ( τ ), uniformly on F. Therefore Re(s m ) has an upper bound on F. Now take m N so that m > δ Re(s m ) N s F. For all ɛ > 0 we have φ(s)e ɛsm Me ɛn on Re(s) ν Re(s) ν 2, uniformly on F. But O(e τ δ ɛτ m ) 0 uniformly on F as τ. So we can use the maximum principle to see that φ(s)e ɛsm Me ɛn (s F ). Now we let ɛ go to 0. So we get φ(s) M on F, i.e. φ(s) O( τ 0 ). Now let b 0. We define a holomorphic function ψ(s) by ψ(s) (s ν + ) b e b log(s ν+) with the principal branch of log. Since we have Re(log(s ν + )) log s ν + ψ(s) s ν + b τ b ( τ ) uniformly on F. Define φ (s) : φ(s) ψ(s). Then φ (s) satisfies the same assumptions as φ with b 0. So by the case above, φ (s) is bounded on F. So we obtain φ(s) O( τ b ) ( τ ). Now we have the tools we need to prove the main result on the relation between modular forms Dirichlet series:

5 3 HECKE S CONVERSE THEOREM 5 3 Hecke s Converse Theorem We now want to prove, that we can go the other direction in Theorem.6, i.e. for every Dirichlet series with analytic continuation the right type of functional equation, we can find a modular form. Theorem 3.. (Hecke, 936) Let f(z) n0 a ne nz g(z) n0 b ne nz be holomorphic functions on H satisfying a n O(n ν ) b n O(n ν ) for a ν > 0. Let k N be positive numbers. Then the following statements are equivalent: i) g(z) ( i Nz) k f( Nz ). ii) Both Λ N (s; f) Λ N (s; g) can be analytically continued to the whole s-plane, satisfying the functional equation Λ N (s; f) Λ N (; g), s + b 0 Proof i) ii): this is just Theorem.6. ii) i): Let f be such a function fulfilling ii). We have for Re(y) > 0: f(iy) n0 a n e ny Lemma2.2 a n n Re(s)α (2πny) s Γ(s)ds + a 0 (2) for any α > 0. Let ν be such that a n O(n ν ). Choose α > ν +. So L(s; f) : n a nn s is uniformly convergent bounded on Re(s) α. By Stirling s estimate (Lemma 2.), Λ N (s; f) (2π/ N) s Γ(s)L(s; f) is absolutely integrable on Re(s) α therefore we can exchange the order of summation integration: f(iy) (2πy) s Γ(s) a n n s ds + a 0 (3) Re(s)α n ( Ny) s Λ N (s; f)ds + a 0 (4) Re(s)α L(s; f) is bounded on Re(s) α, so by Stirling s estimate, for any µ > 0 Λ N (s; f) O( Im(s) µ ) ( Im(s) ) (5) on Re(s) α. Now take β R such that k β > ν +. With the same argument we get for any µ > 0: Λ N (s; g) Λ N (; f) O( Im(s) µ ) ( Im(s) )

6 3 HECKE S CONVERSE THEOREM 6 on Re(s) β. By assumption s + b 0 is bounded on the strip β Re(s) α. So Λ N (s; f) has only poles at 0 k. By Lemma 2.2, (5) holds uniformly on the domain β Re(s) α. Without loss of generality α > k β < 0. ( Ny) s Λ N (s; f) has simple poles at s 0 s k, with residues a 0 resp. ( Ny) k b 0. By the residue theorem using that (5) holds uniformly on β Re(s) α, we get f(iy) ( Ny) s Λ N (s; f)ds + a 0 Re(s)α Re(s)β ( Ny) s Λ N (s; f)ds + ( Ny) k b 0. Now we can use the functional equation for Λ N (s; f): f(iy) ( Ny) s Λ(; g)ds + ( Ny) k b 0 Re(s)β ( Ny) s k Λ(s; g)ds + ( Ny) k b 0 Re(s)k β ( Ny) k ( ( Ny) s Λ(s; g)ds + b 0 ) Re(s)k β Using the same calculations that led to (4), we can bring this into the form f(iy) ( Ny) k g( iny ). f(z) g(z) are holomorphic on H so we get with z y/i g(z) ( Nz/i) k f( Nz ) g(z) ( Nzi) k f( Nz ) that is exactly i). Remark. Replacing g(z) by i k g(z) we can reformulate Theorem 3.: Let f(z) n0 a ne nz g(z) n0 b ne nz be holomorphic functions on H satisfying a n O(n ν ) b n O(n ν ) for a ν > 0. Let k N be positive numbers. Then the following conditions are equivalent: i) g(z) ( Nz) k f( Nz ). ii) Both Λ N (s; f) Λ N (s; g) can be analytically continued to the whole s-plane, satisfying the functional equation Λ N (s; f) i k Λ N (; g),

7 4 REFERENCES 7 s + ik b 0 Corollary 3.2. Let k 2 be an even number. Assume f(z) n0 a ne nz is holomorphic on H ν > 0 : a n O(n ν ). Then the following statements are equivalent: i) f(z) is a modular form of weight k ii) Λ(s; f) can be analytically continued to the whole s-plane, satisfying the functional equation Λ(s; f) ( ) k/2 Λ(; f), Λ(s; f) + a 0 s + ( )k/2 a 0 Proof Set f g N in Remark. We have seen that for f a holomorphic function on H f is a modular form if only if f(z + ) f(z) f(z) z k f( /z). Since f(z) n0 a ne nz, we have that f(z + ) f(z). So i) in Remark is equivalent to f modular form of weight k. In ii) we need that i k ( ) k/2. There are some more general connections between modular forms Dirichlet series / L-functions; some of them are proved recently, e.g. the Taniyama- Shimura-Weil conjecture; some of them are conjectured but still not proved, e.g. the Langls program. 4 References [] T.Miyake, Modular Forms, Springer-Verlag [2] N.Koblitz, Introduction to Elliptic Curves Modular Forms, GTM 97, Springer-Verlag [3] E.Freitag, R.Busam, Funktionentheorie, Springer-Lehrbuch