Invariants of Weil representations

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1 Invariants of Weil representations (joint work with Nils P. Skoruppa) March 17, 2017 Winter seminar March 12-18, 2017 Chalet Fleurs des Neiges, La Plagne, France 1 / 19 Mathematisch-Naturwissenschaftliche Fakulta t Universita t zu Ko ln

2 Finite quadratic modules 2 / 19

3 Finite quadratic modules (A, Q) a finite quadratic module, 2 / 19

4 Finite quadratic modules (A, Q) a finite quadratic module, A is a finite abelian group 2 / 19

5 Finite quadratic modules (A, Q) a finite quadratic module, A is a finite abelian group and Q : A Q/Z a non-degenerate quadratic form 2 / 19

6 Finite quadratic modules (A, Q) a finite quadratic module, A is a finite abelian group and Q : A Q/Z a non-degenerate quadratic form with bilinear form (x, y) := Q(x + y) Q(x) Q(y). 2 / 19

7 Finite quadratic modules (A, Q) a finite quadratic module, A is a finite abelian group and Q : A Q/Z a non-degenerate quadratic form with bilinear form (x, y) := Q(x + y) Q(x) Q(y). The level N of A is the smallest n N, such that nq(x) Z for all x A. 2 / 19

8 Finite quadratic modules (A, Q) a finite quadratic module, A is a finite abelian group and Q : A Q/Z a non-degenerate quadratic form with bilinear form (x, y) := Q(x + y) Q(x) Q(y). The level N of A is the smallest n N, such that nq(x) Z for all x A. Assume that N is odd (for this talk). 2 / 19

9 The Weil representation 3 / 19

10 The Weil representation Let G := SL 2 (Z). 3 / 19

11 The Weil representation Let G := SL 2 (Z). G is generated by S = [ ] and T = [ ]. 3 / 19

12 The Weil representation Let G := SL 2 (Z). G is generated by S = [ ] and T = [ ]. Define a representation ρ = ρ A of G on C[A] = x A Ce x via ρ(t )e x = e(q(x))e x, ρ(s)e x = e( sig(a)/8) A e ( (x, y)) e y. y A 3 / 19

13 The Weil representation Let G := SL 2 (Z). G is generated by S = [ ] and T = [ ]. Define a representation ρ = ρ A of G on C[A] = x A Ce x via ρ(t )e x = e(q(x))e x, ρ(s)e x = e( sig(a)/8) A e ( (x, y)) e y. y A sig(a) Z/8Z can be defined via Milgram s formula e(q(x)) = A e(sig(a)/8). x A 3 / 19

14 The Weil representation Let G := SL 2 (Z). G is generated by S = [ ] and T = [ ]. Define a representation ρ = ρ A of G on C[A] = x A Ce x via ρ(t )e x = e(q(x))e x, ρ(s)e x = e( sig(a)/8) A e ( (x, y)) e y. y A sig(a) Z/8Z can be defined via Milgram s formula e(q(x)) = A e(sig(a)/8). x A This representation is called the Weil representation of G associated with A. 3 / 19

15 Theta functions 4 / 19

16 Theta functions Significance: Theta functions transform with the Weil representation 4 / 19

17 Theta functions Significance: Theta functions transform with the Weil representation If (L, Q) is an even positive definite lattice and L its dual, then A = L /L with Q mod Z is a finite quadratic module. 4 / 19

18 Theta functions Significance: Theta functions transform with the Weil representation If (L, Q) is an even positive definite lattice and L its dual, then A = L /L with Q mod Z is a finite quadratic module. The theta function Θ L (τ) := λ L exp(2πiq(λ)τ)e λ is a vector valued modular form of weight n 2 for ρ A ( λ = λ mod L). 4 / 19

19 Theta functions Significance: Theta functions transform with the Weil representation If (L, Q) is an even positive definite lattice and L its dual, then A = L /L with Q mod Z is a finite quadratic module. The theta function Θ L (τ) := λ L exp(2πiq(λ)τ)e λ is a vector valued modular form of weight n 2 for ρ A ( λ = λ mod L). Modular forms for the Weil representation occur in many places, for instance as input to regularized theta lifts. 4 / 19

20 Invariants 5 / 19

21 Invariants We are interested in the space of G-invariants C[A] G. 5 / 19

22 Invariants We are interested in the space of G-invariants C[A] G. For various reasons: 5 / 19

23 Invariants We are interested in the space of G-invariants C[A] G. For various reasons: dim C C[A] G is one of the terms in the dimension formula for holomorphic cusp forms of weight 2. 5 / 19

24 Invariants We are interested in the space of G-invariants C[A] G. For various reasons: dim C C[A] G is one of the terms in the dimension formula for holomorphic cusp forms of weight 2. Also (the dimension formulas for) modular forms of weight 1 2 and 3 2 involve (the dimension of) C[A]G. 5 / 19

25 Invariants We are interested in the space of G-invariants C[A] G. For various reasons: dim C C[A] G is one of the terms in the dimension formula for holomorphic cusp forms of weight 2. Also (the dimension formulas for) modular forms of weight 1 2 and 3 2 involve (the dimension of) C[A]G. No general, explicit formula known. 5 / 19

26 Observations 6 / 19

27 Observations The representation ρ is in fact defined over Q(ζ N ). 6 / 19

28 Observations The representation ρ is in fact defined over Q(ζ N ). This is clear for ρ(t ) and for ρ(s) use Milgram s formula: e(q(x)) = A e(sig(a)/8). x A 6 / 19

29 Observations The representation ρ is in fact defined over Q(ζ N ). This is clear for ρ(t ) and for ρ(s) use Milgram s formula: e(q(x)) = A e(sig(a)/8). x A The representation ρ factors through a representation of the finite group G N := SL 2 (Z/NZ). 6 / 19

30 Explicit formula Lemma Let g = [ a b 0 d ] GN and x in A. Then where χ(d) = σ d (w)/w with ρ(g)e x = χ(d) e (bdq(x)) e dx, w = x A e (Q(x)) and σ d Gal(K N /Q) with σ d (ζ N ) = ζ d N. 7 / 19

31 Proof 8 / 19

32 Proof 1. Exercise 1: Write [ ] a b 0 d as a product of a diagonal matrix and a power of T. 8 / 19

33 Proof 1. Exercise 1: Write [ ] a b 0 d as a product of a diagonal matrix and a power of T. 2. Answer: [ ] [ a b 0 d = a 0 ] [ ] 1 bd 0 d / 19

34 Proof 1. Exercise 1: Write [ ] a b 0 d as a product of a diagonal matrix and a power of T. 2. Answer: [ ] [ a b 0 d = a 0 ] [ ] 1 bd 0 d This is good because [ ] 1 bd 0 1 ex = e(bdq(x))e x. 8 / 19

35 Proof 1. Exercise 1: Write [ ] a b 0 d as a product of a diagonal matrix and a power of T. 2. Answer: [ ] [ a b 0 d = a 0 ] [ ] 1 bd 0 d This is good because [ ] 1 bd 0 1 ex = e(bdq(x))e x. 4. Only need to consider the action of [ ] a 0 0 d. 8 / 19

36 Proof 1. Exercise 1: Write [ ] a b 0 d as a product of a diagonal matrix and a power of T. 2. Answer: [ ] [ a b 0 d = a 0 ] [ ] 1 bd 0 d This is good because [ ] 1 bd 0 1 ex = e(bdq(x))e x. 4. Only need to consider the action of [ ] a 0 0 d. 5. Exercise 2: Write [ ] a 0 0 d as a word in S and T (in GN )! 8 / 19

37 Proof 1. Exercise 1: Write [ ] a b 0 d as a product of a diagonal matrix and a power of T. 2. Answer: [ ] [ a b 0 d = a 0 ] [ ] 1 bd 0 d This is good because [ ] 1 bd 0 1 ex = e(bdq(x))e x. 4. Only need to consider the action of [ ] a 0 0 d. 5. Exercise 2: Write [ ] a 0 0 d as a word in S and T (in GN )! 6. Answer: [ a 0 ] 0 d = S 1 [ ] 1 d 0 1 S [ 1 a 0 1 ] S [ ] 1 d / 19

38 Proof 1. Exercise 1: Write [ ] a b 0 d as a product of a diagonal matrix and a power of T. 2. Answer: [ ] [ a b 0 d = a 0 ] [ ] 1 bd 0 d This is good because [ ] 1 bd 0 1 ex = e(bdq(x))e x. 4. Only need to consider the action of [ ] a 0 0 d. 5. Exercise 2: Write [ ] a 0 0 d as a word in S and T (in GN )! 6. Answer: [ a 0 ] 0 d = S 1 [ ] 1 d 0 1 S [ 1 a 0 1 ] S [ ] 1 d Exercise 3 (due tomorrow): Show that: ] ex = γ e dx, where [ a 0 0 d γ = σ d (w)/w, w = x A e (Q(x)). 8 / 19

39 Integrality Theorem The space C[A] G is defined over Z, i.e. it has a basis in Z[A]. 9 / 19

40 Integrality Theorem The space C[A] G is defined over Z, i.e. it has a basis in Z[A]. Proof. Use the projection to the invariants and show that it has coefficients in Z (wrt to the standard basis), which follows from: For any [ ] a b c d in G and s (Z/NZ), we have σ s (ρ ([ a b c d ]) ([ ) = ρ a sb ]) s 1 c d 9 / 19

41 Computing the invariants 10 / 19

42 Computing the invariants Our goal is just to compute the dimension of C[A] G 10 / 19

43 Computing the invariants Our goal is just to compute the dimension of C[A] G and optionally also a basis (contained in Z[A] if possible) 10 / 19

44 Computing the invariants Our goal is just to compute the dimension of C[A] G and optionally also a basis (contained in Z[A] if possible) which works in all cases. 10 / 19

45 Identifying invariants 11 / 19

46 Identifying invariants v C[A] G is of course equivalent to: 11 / 19

47 Identifying invariants v C[A] G is of course equivalent to: ρ(t )v = v and ρ(s)v = v. 11 / 19

48 Identifying invariants v C[A] G is of course equivalent to: ρ(t )v = v and ρ(s)v = v. ρ(t )v = v means that supp(v) Iso(A), 11 / 19

49 Identifying invariants v C[A] G is of course equivalent to: ρ(t )v = v and ρ(s)v = v. ρ(t )v = v means that supp(v) Iso(A), where Iso(A) = {x A Q(x) = 0}. 11 / 19

50 Identifying invariants Proposition Let M be a G-submodule of C[A]. Then M G = ( 1 + ρ(s) + ρ(s) 2 + ρ(s) 3) (C[Iso(A)]) C[Iso(A)] Proof. Exercise. 12 / 19

51 The algorithm 13 / 19

52 The algorithm 1. Find the isotropic elements a 1,..., a m and the non-isotropic elements b 1,..., b n in A. 13 / 19

53 The algorithm 1. Find the isotropic elements a 1,..., a m and the non-isotropic elements b 1,..., b n in A. 2. Compute the (m + n) m matrix H such that (Le a1,..., Le am ) = (e a1,..., e am, e b1,..., e bn )H, where L = 1 + ρ(s) + ρ(s) 2 + ρ(s) / 19

54 The algorithm 1. Find the isotropic elements a 1,..., a m and the non-isotropic elements b 1,..., b n in A. 2. Compute the (m + n) m matrix H such that (Le a1,..., Le am ) = (e a1,..., e am, e b1,..., e bn )H, where L = 1 + ρ(s) + ρ(s) 2 + ρ(s) Let U and V be the matrices obtained by extraction the first m and the last n rows of H, respectively. 13 / 19

55 The algorithm 1. Find the isotropic elements a 1,..., a m and the non-isotropic elements b 1,..., b n in A. 2. Compute the (m + n) m matrix H such that (Le a1,..., Le am ) = (e a1,..., e am, e b1,..., e bn )H, where L = 1 + ρ(s) + ρ(s) 2 + ρ(s) Let U and V be the matrices obtained by extraction the first m and the last n rows of H, respectively. 4. Compute a basis V for ker V. 13 / 19

56 The algorithm 1. Find the isotropic elements a 1,..., a m and the non-isotropic elements b 1,..., b n in A. 2. Compute the (m + n) m matrix H such that (Le a1,..., Le am ) = (e a1,..., e am, e b1,..., e bn )H, where L = 1 + ρ(s) + ρ(s) 2 + ρ(s) Let U and V be the matrices obtained by extraction the first m and the last n rows of H, respectively. 4. Compute a basis V for ker V. 5. Return a basis of U(ker V ) = {Ux x V}. 13 / 19

57 The algorithm 1. Find the isotropic elements a 1,..., a m and the non-isotropic elements b 1,..., b n in A. 2. Compute the (m + n) m matrix H such that (Le a1,..., Le am ) = (e a1,..., e am, e b1,..., e bn )H, where L = 1 + ρ(s) + ρ(s) 2 + ρ(s) Let U and V be the matrices obtained by extraction the first m and the last n rows of H, respectively. 4. Compute a basis V for ker V. 5. Return a basis of U(ker V ) = {Ux x V}. 6. Exercise: U seems to be invertible in practice. Tell me why. 13 / 19

58 Reduction mod l 14 / 19

59 Reduction mod l Let l be a prime with l 1 mod N. 14 / 19

60 Reduction mod l Let l be a prime with l 1 mod N. Then Q l contains the Nth roots of unity, hence the Nth cyclotomic field. 14 / 19

61 Reduction mod l Let l be a prime with l 1 mod N. Then Q l contains the Nth roots of unity, hence the Nth cyclotomic field. We can consider ρ as a representation of G N on Q l [A]. 14 / 19

62 Reduction mod l Let l be a prime with l 1 mod N. Then Q l contains the Nth roots of unity, hence the Nth cyclotomic field. We can consider ρ as a representation of G N on Q l [A]. In fact, G acts on Z l [A]. 14 / 19

63 Reduction mod l Let l be a prime with l 1 mod N. Then Q l contains the Nth roots of unity, hence the Nth cyclotomic field. We can consider ρ as a representation of G N on Q l [A]. In fact, G acts on Z l [A]. dim C C[A] G N equals the Z l -rank of Z l [A] G N. 14 / 19

64 Reduction mod l Let l be a prime with l 1 mod N. Then Q l contains the Nth roots of unity, hence the Nth cyclotomic field. We can consider ρ as a representation of G N on Q l [A]. In fact, G acts on Z l [A]. dim C C[A] G N equals the Z l -rank of Z l [A] G N. Reduction mod l: have a short exact sequence of G-modules: 0 lz l [A] Z l [A] r F l [A] 0, where r denotes the reduction map r(f ) : a f (a) + lz l. 14 / 19

65 Reduction mod l Theorem Suppose that (N, l) (2, 3). Then dim Ql Q l [A] G N = dim Fl F l [A] G N. 15 / 19

66 Proof 16 / 19

67 Proof Obtain the long exact sequence in cohomology 0 lz l [A] G N Z l [A] G N r F l [A] G N H 1 (G N, lz l [A]). 16 / 19

68 Proof Obtain the long exact sequence in cohomology 0 lz l [A] G N Z l [A] G N r F l [A] G N H 1 (G N, lz l [A]). If G N is a unit in Z l, then H 1 (G N, lz l [A]) = {0}. 16 / 19

69 Proof Obtain the long exact sequence in cohomology 0 lz l [A] G N Z l [A] G N r F l [A] G N H 1 (G N, lz l [A]). If G N is a unit in Z l, then H 1 (G N, lz l [A]) = {0}. Of course l 1 mod N implies that l > N 16 / 19

70 Proof Obtain the long exact sequence in cohomology 0 lz l [A] G N Z l [A] G N r F l [A] G N H 1 (G N, lz l [A]). If G N is a unit in Z l, then H 1 (G N, lz l [A]) = {0}. Of course l 1 mod N implies that l > N G N = N 3 p N p 2 1 p 2 16 / 19

71 Proof Obtain the long exact sequence in cohomology 0 lz l [A] G N Z l [A] G N r F l [A] G N H 1 (G N, lz l [A]). If G N is a unit in Z l, then H 1 (G N, lz l [A]) = {0}. Of course l 1 mod N implies that l > N p 2 1 p 2 G N = N 3 p N Thus, if l G N, there is a prime p N, such that l p + 1 or l p / 19

72 Proof Obtain the long exact sequence in cohomology 0 lz l [A] G N Z l [A] G N r F l [A] G N H 1 (G N, lz l [A]). If G N is a unit in Z l, then H 1 (G N, lz l [A]) = {0}. Of course l 1 mod N implies that l > N p 2 1 p 2 G N = N 3 p N Thus, if l G N, there is a prime p N, such that l p + 1 or l p 1. However, p 1 < N < l and thus the only possibility is l = p + 1 and N = p. 16 / 19

73 Proof Obtain the long exact sequence in cohomology 0 lz l [A] G N Z l [A] G N r F l [A] G N H 1 (G N, lz l [A]). If G N is a unit in Z l, then H 1 (G N, lz l [A]) = {0}. Of course l 1 mod N implies that l > N p 2 1 p 2 G N = N 3 p N Thus, if l G N, there is a prime p N, such that l p + 1 or l p 1. However, p 1 < N < l and thus the only possibility is l = p + 1 and N = p. This only leaves the case N = 2 and l = 3, which we excluded. 16 / 19

74 Proof Obtain the long exact sequence in cohomology 0 lz l [A] G N Z l [A] G N r F l [A] G N H 1 (G N, lz l [A]). If G N is a unit in Z l, then H 1 (G N, lz l [A]) = {0}. Of course l 1 mod N implies that l > N p 2 1 p 2 G N = N 3 p N Thus, if l G N, there is a prime p N, such that l p + 1 or l p 1. However, p 1 < N < l and thus the only possibility is l = p + 1 and N = p. This only leaves the case N = 2 and l = 3, which we excluded. Exercise: Does the theorem hold for (2, 3)? We didn t find any counterexamples. 16 / 19

75 Some dimensions Table: d = dim C C[A] G for some 2-modules of even signature s A d A d A d A d s s = 0 s = 4 s = / 19

76 Some dimensions Table: Dimension d = dim C C[A] G for some 3-modules of signature s A d A d A d A d s s = 6 s = 2 s = / 19

77 Some dimensions Table: Dimension d = dim C C[A] G for some 3-modules of signature s A d A d A d A d s = 6 s = 2 s = 6 s = / 19

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