Rational Equivariant Forms

Transcription

1 CRM-CICMA-Concordia University Mai 1, 2011 Atkin s Memorial Lecture and Workshop This is joint work with Abdellah Sebbar.

2 Notation Let us fix some notation: H := {z C; I(z) > 0}, H := H P 1 (Q), SL 2 (Z) := the modular group, α z := az + b ( ) a b cz + d, α = SL 2 (Z), z C. c d Γ is a subgroup of SL 2 (Z) of finite index, which we call a modular subgroup. (We would like to mention that all what we present here, in fact, holds for any discrete subgroup of SL 2 (R)).

3 Preliminaries: The Schwarz derivative: For a meromorphic function on a domain (open and connected) of C, the Schwarz derivative, denoted {f, z}, is defined by It satisfies the following rules. ( ) f ( ) f 2 {f, z} = 2 (1) f f = 2f f 3f 2 f 2. Chain rule: If w is a function of z then {f, z} = (dw/dz) 2 {f, w} + {w, z}. ( ) a b Consequently, for α = GL 2 (C), we have c d {w, z} = det α {w, α z} (2) (cz + d) 4

4 Preliminaries: The Schwarz derivative: For a meromorphic function on a domain (open and connected) of C, the Schwarz derivative, denoted {f, z}, is defined by It satisfies the following rules. ( ) f ( ) f 2 {f, z} = 2 (1) f f = 2f f 3f 2 f 2. Chain rule: If w is a function of z then {f, z} = (dw/dz) 2 {f, w} + {w, z}. ( ) a b Consequently, for α = GL 2 (C), we have c d {w, z} = det α {w, α z} (2) (cz + d) 4

5 Preliminaries: The Schwarz derivative: {f, z} = 0 if and only if f is a linear fractional transform of z. {w 1, z} = {w 2, z} if and only if each function (w i ) is a linear fraction of the other. Inversion formula: If w (z 0 ) 0 for some point z 0, then in a neighborhood of z 0, {z, w} = (dz/dw) 2 {w, z}. (3) Lastly, given a meromorphic function f on a domain D of C, then one deduces the following The Schwarz derivative {f, z} of f has a double pole at the critical points of f and is holomorphic elsewhere including at simple poles of f.

6 Modular Schwarzian This is due to J. McKay and A. Sebbar. Suppose now that f is a modular function for some modular subgroup Γ. (M-S) i. If f is a modular function for Γ then {f, z} is a (meromorphic) weight 4 modular form for Γ. ii. If in addition Γ is of genus 0 and f is a Hauptmodul for Γ, then {f, z} is weight 4 (holomorphic if Γ is torsion free) modular form for the normalizer of Γ inside SL 2 (R). What about the converse? In other words, given meromorphic function f on H such that {f, z} is a weight 4 modular form on a modular subgroup Γ, what can be said about the invariance group G f of f.

7 Modular Schwarzian This is due to J. McKay and A. Sebbar. Suppose now that f is a modular function for some modular subgroup Γ. (M-S) i. If f is a modular function for Γ then {f, z} is a (meromorphic) weight 4 modular form for Γ. ii. If in addition Γ is of genus 0 and f is a Hauptmodul for Γ, then {f, z} is weight 4 (holomorphic if Γ is torsion free) modular form for the normalizer of Γ inside SL 2 (R). What about the converse? In other words, given meromorphic function f on H such that {f, z} is a weight 4 modular form on a modular subgroup Γ, what can be said about the invariance group G f of f.

8 The origin In fact, using properties of the Schwarz derivative, we have for α = ( a b c d) Γ {f(α z), α z} = (cz + d) 4 {f(z), z} (modularity) = (cz + d) 4 {f(α z), z} (prpty the Schz. der.) Hence {f(α z), z} = {f(z), z} and therefore f(α z) = Φ α f(z), for some Φ α GL 2 (C).

9 The origin In particular, we have Φ : Γ GL 2 (C) α Φ α is a group homomorphism. Moreover, G f = Ker Φ. Theorem (S) If f is as above and is such that f(z + n) = f(z), n {1, 2, 3, 4, 5}, then f is a modular function for Γ(n) = G f. There are some cases where G f, for instance log(f), f a Hauptmodul of Γ(n) with n as above, is no larger than < T m >, for some m (T = ( ) ). Question: when G f is trivial? This is the case if Φ is injective.

10 The origin In particular, we have Φ : Γ GL 2 (C) α Φ α is a group homomorphism. Moreover, G f = Ker Φ. Theorem (S) If f is as above and is such that f(z + n) = f(z), n {1, 2, 3, 4, 5}, then f is a modular function for Γ(n) = G f. There are some cases where G f, for instance log(f), f a Hauptmodul of Γ(n) with n as above, is no larger than < T m >, for some m (T = ( ) ). Question: when G f is trivial? This is the case if Φ is injective.

11 The origin In particular, we have Φ : Γ GL 2 (C) α Φ α is a group homomorphism. Moreover, G f = Ker Φ. Theorem (S) If f is as above and is such that f(z + n) = f(z), n {1, 2, 3, 4, 5}, then f is a modular function for Γ(n) = G f. There are some cases where G f, for instance log(f), f a Hauptmodul of Γ(n) with n as above, is no larger than < T m >, for some m (T = ( ) ). Question: when G f is trivial? This is the case if Φ is injective.

12 The origin In particular, we have Φ : Γ GL 2 (C) α Φ α is a group homomorphism. Moreover, G f = Ker Φ. Theorem (S) If f is as above and is such that f(z + n) = f(z), n {1, 2, 3, 4, 5}, then f is a modular function for Γ(n) = G f. There are some cases where G f, for instance log(f), f a Hauptmodul of Γ(n) with n as above, is no larger than < T m >, for some m (T = ( ) ). Question: when G f is trivial? This is the case if Φ is injective.

13 Equivariant functions : first examples For the rest of this talk, Φ is the natural injection of Γ into GL 2 (C). Hence, our functions are of the following type. Definition A meromorphic function h on H is called an equivariant function for Γ if it satisfies h(α z) = α h(z), for all α Γ. A first, obvious, example is f(z) = z. A larger class comes from Let f be a weight k, k Z, modular form for Γ. Then the function h f (z) = z + kf(z) f (z) ( ) is an equivariant function.

14 Equivariant functions : first examples For the rest of this talk, Φ is the natural injection of Γ into GL 2 (C). Hence, our functions are of the following type. Definition A meromorphic function h on H is called an equivariant function for Γ if it satisfies h(α z) = α h(z), for all α Γ. A first, obvious, example is f(z) = z. A larger class comes from Let f be a weight k, k Z, modular form for Γ. Then the function h f (z) = z + kf(z) f (z) ( ) is an equivariant function.

15 Equivariant functions : first examples For the rest of this talk, Φ is the natural injection of Γ into GL 2 (C). Hence, our functions are of the following type. Definition A meromorphic function h on H is called an equivariant function for Γ if it satisfies h(α z) = α h(z), for all α Γ. A first, obvious, example is f(z) = z. A larger class comes from Let f be a weight k, k Z, modular form for Γ. Then the function h f (z) = z + kf(z) f (z) ( ) is an equivariant function.

16 The general definition We define a slash operator on equivariant functions via the following action of SL 2 (R) on meromorphic functions on H. For a meromorphic function f and γ = ( a b c d) SL2 (R), we let f [γ](z) = j γ (z) 2 f(γ z) cj γ (z) 1, j γ (z) = cz + d. For a meromorphic function h on H, set H(z) = (h(z) z) 1. Then we have The function h is an equivariant function for Γ if and only if H [γ](z) = H(z) for all γ Γ. Furthermore, H [ γ](z) = H [γ](z) if 1 2 Γ.

17 The general definition We define a slash operator on equivariant functions via the following action of SL 2 (R) on meromorphic functions on H. For a meromorphic function f and γ = ( a b c d) SL2 (R), we let f [γ](z) = j γ (z) 2 f(γ z) cj γ (z) 1, j γ (z) = cz + d. For a meromorphic function h on H, set H(z) = (h(z) z) 1. Then we have The function h is an equivariant function for Γ if and only if H [γ](z) = H(z) for all γ Γ. Furthermore, H [ γ](z) = H [γ](z) if 1 2 Γ.

18 The general definition Definition An equivariant function for Γ is called an equivariant form for Γ if it satisfies 1 h is meromorphic on H; 2 h is meromorphic at the cusps, meaning that the function H [γ](z) is meromorphic at infinity for all γ SL 2 (Z). Example For f M m k (Γ), k 0, the equivariant function h f = z + kf/f (as in ( )) is therefore an equivariant form. Suppose that Γ 1 and Γ 2 are conjugate subgroups of SL 2 (Z), say Γ 1 = αγ 2 α 1, for α GL + 2 (Q). Then if h 1 is an equivariant form on Γ 1, the function h 2 (z) = α 1 h 1 α(z) is an equivariant form on Γ 2.

19 The general definition Definition An equivariant function for Γ is called an equivariant form for Γ if it satisfies 1 h is meromorphic on H; 2 h is meromorphic at the cusps, meaning that the function H [γ](z) is meromorphic at infinity for all γ SL 2 (Z). Example For f M m k (Γ), k 0, the equivariant function h f = z + kf/f (as in ( )) is therefore an equivariant form. Suppose that Γ 1 and Γ 2 are conjugate subgroups of SL 2 (Z), say Γ 1 = αγ 2 α 1, for α GL + 2 (Q). Then if h 1 is an equivariant form on Γ 1, the function h 2 (z) = α 1 h 1 α(z) is an equivariant form on Γ 2.

20 The general definition Definition An equivariant function for Γ is called an equivariant form for Γ if it satisfies 1 h is meromorphic on H; 2 h is meromorphic at the cusps, meaning that the function H [γ](z) is meromorphic at infinity for all γ SL 2 (Z). Example For f M m k (Γ), k 0, the equivariant function h f = z + kf/f (as in ( )) is therefore an equivariant form. Suppose that Γ 1 and Γ 2 are conjugate subgroups of SL 2 (Z), say Γ 1 = αγ 2 α 1, for α GL + 2 (Q). Then if h 1 is an equivariant form on Γ 1, the function h 2 (z) = α 1 h 1 α(z) is an equivariant form on Γ 2.

21 Rational Equivariant forms Definition An equivariant form is called a rational equivariant form if it arises from a modular form (of weight k 0). For c C and n Z, we have h f n = h cf = h f = z + kf/f, f M m k (Γ). For conjugate subgroups Γ 1, Γ 2, if h 1 (z) = z + kf(z)/f (z), f M m k (Γ 1), is a rational equivariant form then so is h 2 (z) = α 1 h 1 α(z), and we have h 2 (z) = z + k(f k[α])(z) (f k [α]) (z).

22 Rational Equivariant forms Definition An equivariant form is called a rational equivariant form if it arises from a modular form (of weight k 0). For c C and n Z, we have h f n = h cf = h f = z + kf/f, f M m k (Γ). For conjugate subgroups Γ 1, Γ 2, if h 1 (z) = z + kf(z)/f (z), f M m k (Γ 1), is a rational equivariant form then so is h 2 (z) = α 1 h 1 α(z), and we have h 2 (z) = z + k(f k[α])(z) (f k [α]) (z).

23 Rational Equivariant forms Definition An equivariant form is called a rational equivariant form if it arises from a modular form (of weight k 0). For c C and n Z, we have h f n = h cf = h f = z + kf/f, f M m k (Γ). For conjugate subgroups Γ 1, Γ 2, if h 1 (z) = z + kf(z)/f (z), f M m k (Γ 1), is a rational equivariant form then so is h 2 (z) = α 1 h 1 α(z), and we have h 2 (z) = z + k(f k[α])(z) (f k [α]) (z).

24 Example A fundamental example corresponds to the modular discriminant h (z) = z + 12 (z)/ (z) = z + 6/πiE 2 (z). Remarks The trivial example h t (z) = z does not fit in the above setting, however we can associate it to modular functions. The zeros and poles of the function H f (z) = (h f (z) z) 1 are all simple and have rational residues. Indeed, if n is the multiplicity of f at z 0 (a pole or a zero), then z 0 is a simple pole of H f with residue n/k. At a cusp s of Γ and γ SL 2 (Z) such that γ s =, one can see that 1 H f [γ 1 ](z) = n z i 2iπ lim Q, kl s where n is the order of infinity in the q s -expansion of f k [γ 1 ](z) and l s is the cusp width at s of Γ. This justifies the use of rational.

25 Example A fundamental example corresponds to the modular discriminant h (z) = z + 12 (z)/ (z) = z + 6/πiE 2 (z). Remarks The trivial example h t (z) = z does not fit in the above setting, however we can associate it to modular functions. The zeros and poles of the function H f (z) = (h f (z) z) 1 are all simple and have rational residues. Indeed, if n is the multiplicity of f at z 0 (a pole or a zero), then z 0 is a simple pole of H f with residue n/k. At a cusp s of Γ and γ SL 2 (Z) such that γ s =, one can see that 1 H f [γ 1 ](z) = n z i 2iπ lim Q, kl s where n is the order of infinity in the q s -expansion of f k [γ 1 ](z) and l s is the cusp width at s of Γ. This justifies the use of rational.

26 : The converse Question: When an equivariant form is a rational equivariant form? The following proves that the conditions of the remark are in fact also sufficient. Theorem Let Γ be a modular subgroup and let h be an equivariant form for Γ. Then h is rational if and only if 1. The poles of H in H are all simple with rational residues. 2. For each cusp s of Γ and γ SL 2 (Z) with γ s =, we have 1 2iπ lim z i H [γ 1 ](z) Q. To prove this one has to establish the following lemmas.

27 : The converse Question: When an equivariant form is a rational equivariant form? The following proves that the conditions of the remark are in fact also sufficient. Theorem Let Γ be a modular subgroup and let h be an equivariant form for Γ. Then h is rational if and only if 1. The poles of H in H are all simple with rational residues. 2. For each cusp s of Γ and γ SL 2 (Z) with γ s =, we have 1 2iπ lim z i H [γ 1 ](z) Q. To prove this one has to establish the following lemmas.

28 : The converse Lemma Let Γ be a modular subgroup, and let h be an equivariant form for Γ. Then H has only a finite number of poles in the closure of a fundamental domain of Γ. Lemma Suppose that h is equivariant for a modular subgroup Γ such that H has only simple poles in H, then the set of the residues at these poles is finite. The theorem then follows by associating to h the function ( z ) f(z) = exp kh(u)d u, z 0 where z 0 H not a zero of H and k is chosen to make disappear the denominators of the (finitely many) rational residues. The function f is in fact a modular form of wght k.

29 : The converse Lemma Let Γ be a modular subgroup, and let h be an equivariant form for Γ. Then H has only a finite number of poles in the closure of a fundamental domain of Γ. Lemma Suppose that h is equivariant for a modular subgroup Γ such that H has only simple poles in H, then the set of the residues at these poles is finite. The theorem then follows by associating to h the function ( z ) f(z) = exp kh(u)d u, z 0 where z 0 H not a zero of H and k is chosen to make disappear the denominators of the (finitely many) rational residues. The function f is in fact a modular form of wght k.

30 A particular case In the case of genus zero subgroups Theorem Let h be an equivariant form on a subgroup Γ of SL 2 (Z) of genus 0 such that h(z) z for all z H. Suppose also that, for every cusp s, if γ SL 2 (Z) is such that γ s =, we have lim z i H [γ 1 ](z) = a s is finite and satisfies (a s /6) πiz. Then 6 h(z) = h (z) = z + πie 2 (z).

31 Are there other examples? Yes! Theorem Let Γ be a modular subgroup and let f and g be modular forms of weights k and k + 2 respectively, then h(z) = z + k is an equivariant form for Γ. f(z) f (z) + g(z) A complete classification will appear in a joint work with Abdellah Sebbar, with a complete structure (an affine space) and geometric correspondences.

32 Are there other examples? Yes! Theorem Let Γ be a modular subgroup and let f and g be modular forms of weights k and k + 2 respectively, then h(z) = z + k is an equivariant form for Γ. f(z) f (z) + g(z) A complete classification will appear in a joint work with Abdellah Sebbar, with a complete structure (an affine space) and geometric correspondences.

33 Are there other examples? Yes! Theorem Let Γ be a modular subgroup and let f and g be modular forms of weights k and k + 2 respectively, then h(z) = z + k is an equivariant form for Γ. f(z) f (z) + g(z) A complete classification will appear in a joint work with Abdellah Sebbar, with a complete structure (an affine space) and geometric correspondences.

34 The Effect of the Schwarz Derivative and the Cohen-Rankin Bracket If h is an equivariant form for Γ, then {h, z} is a weight 4 modular form on Γ. In the case of rational equivariant forms, this is connected to the Cohen-Rankin bracket, which is defined for n 0 with D j = dj by dz j [f, g] n = ( k+n 1 )( l+n 1 ) r+s=n s r D r f D s g, r, s 0. It is known that for f M m k (Γ) and g Mm l (Γ) and for every n 0, the function [f, g] n belongs to M m k+l+2n (Γ). Let f be a modular form of weight k on Γ and h f the corresponding rational equivariant form. Then f 2 h f is a weight 2k + 4 modular form on Γ. Moreover, the poles of {h f, z} are located at the zeros of the second Cohen-Rankin bracket [f, f] 2 of f.

35 The Effect of the Schwarz Derivative and the Cohen-Rankin Bracket If h is an equivariant form for Γ, then {h, z} is a weight 4 modular form on Γ. In the case of rational equivariant forms, this is connected to the Cohen-Rankin bracket, which is defined for n 0 with D j = dj by dz j [f, g] n = ( k+n 1 )( l+n 1 ) r+s=n s r D r f D s g, r, s 0. It is known that for f M m k (Γ) and g Mm l (Γ) and for every n 0, the function [f, g] n belongs to M m k+l+2n (Γ). Let f be a modular form of weight k on Γ and h f the corresponding rational equivariant form. Then f 2 h f is a weight 2k + 4 modular form on Γ. Moreover, the poles of {h f, z} are located at the zeros of the second Cohen-Rankin bracket [f, f] 2 of f.

36 The Effect of the Schwarz Derivative and the Cohen-Rankin Bracket If h is an equivariant form for Γ, then {h, z} is a weight 4 modular form on Γ. In the case of rational equivariant forms, this is connected to the Cohen-Rankin bracket, which is defined for n 0 with D j = dj by dz j [f, g] n = ( k+n 1 )( l+n 1 ) r+s=n s r D r f D s g, r, s 0. It is known that for f M m k (Γ) and g Mm l (Γ) and for every n 0, the function [f, g] n belongs to M m k+l+2n (Γ). Let f be a modular form of weight k on Γ and h f the corresponding rational equivariant form. Then f 2 h f is a weight 2k + 4 modular form on Γ. Moreover, the poles of {h f, z} are located at the zeros of the second Cohen-Rankin bracket [f, f] 2 of f.

37 An Application: The zeros of E 2 Theorem 6 The Eisenstein series E 2 has infinitely many zeros in the half-strip S = {τ H, 1 2 < Re(τ) 1 2 }. The proof is as follows. One first observes that, for each z H such that h (z) Q, the SL 2 (Z)-equivalence class of z contains a zero of E 2. Fix such a point z 0. Then, in any neighborhood U of z 0, the function h takes infinitely many rational values, and so any neighborhood of z 0 contains infinitely many points that are equivalent to a zero of E 2. If U is sufficiently small then, each equivalence class of a point of U meets U in at most 3 points (about the vertices of a fundamental domain of SL 2 (Z)). In fact, as z 0 cannot be an elliptic point, one can choose U such that U is in the interior of some fundamental domain for SL 2 (Z), and then the points of U are two by two inequivalent.

38 An Application: The zeros of E 2 Theorem 6 The Eisenstein series E 2 has infinitely many zeros in the half-strip S = {τ H, 1 2 < Re(τ) 1 2 }. The proof is as follows. One first observes that, for each z H such that h (z) Q, the SL 2 (Z)-equivalence class of z contains a zero of E 2. Fix such a point z 0. Then, in any neighborhood U of z 0, the function h takes infinitely many rational values, and so any neighborhood of z 0 contains infinitely many points that are equivalent to a zero of E 2. If U is sufficiently small then, each equivalence class of a point of U meets U in at most 3 points (about the vertices of a fundamental domain of SL 2 (Z)). In fact, as z 0 cannot be an elliptic point, one can choose U such that U is in the interior of some fundamental domain for SL 2 (Z), and then the points of U are two by two inequivalent.

39 An Application: The zeros of E 2 So, U contains infinitely many points in the equivalence class of distinct zeros of E 2. Note that the strict monotonicity of E 2 on the imaginary axis provides us with the point z 0. Finally, shift all the zeros to the strip S and notice that E 2 is invariant under translation and has integer coefficients.

40 The Cross-ratio of Equivariant Forms Let h 1, h 2, h 3, h 4 be four equivariant forms on a modular subgroup Γ. Define a function f as the cross-ratio of these four elements f = (h 1, h 2 ; h 3, h 4 ) = (h 1 h 3 )(h 2 h 4 ) (h 2 h 3 )(h 1 h 4 ). Then, if h 2 h 3 and h 1 h 4, the function f is a modular function on Γ. This actually gives a parametrization of equivariant forms by modular functions. We have 1728(h t, h E6 ; h E4, h ) = j.

41 The Cross-ratio of Equivariant Forms Let h 1, h 2, h 3, h 4 be four equivariant forms on a modular subgroup Γ. Define a function f as the cross-ratio of these four elements f = (h 1, h 2 ; h 3, h 4 ) = (h 1 h 3 )(h 2 h 4 ) (h 2 h 3 )(h 1 h 4 ). Then, if h 2 h 3 and h 1 h 4, the function f is a modular function on Γ. This actually gives a parametrization of equivariant forms by modular functions. We have 1728(h t, h E6 ; h E4, h ) = j.

42 T hank you f or your attention! & Special thanks to Ramin!