Dimension formulas for vector-valued Hilbert modular forms

Transcription

1 Dimension formulas for vector-valued Hilbert modular forms Fredrik Strömberg (j/w N.-P. Skoruppa) March 29, 2013

2 Possible applications Jacobi forms over number fields Same type of correspondence as over Q (between scalar and vector-valued) Liftings between Hilbert modular forms and Jacobi forms (Shimura lift)

3 Preliminary notation (Number fields) K /Q number field of degree n Embeddings: σ i : K R, 1 i n, Trace and norm: If A = Trα = σ i α, Nα = σ i α. ( ) α β M γ δ 2 (K ) we write A σi = ( ) σi (α) σ i (β). σ i (γ) σ i (δ)

4 More preliminaries There are two important lattices related to K : O K the ring of integers with integral basis 1 = α 1,α 2,...α n O K α 1 Z α n Z, O K the unit group with generators ±1,ε 1,...,ε n 1 O K ±1 ε 1 ε n 1 Λ the logarithmic unit lattice: v i = (ln σ 1 ε i,...,ln σ n 1 ε i ) Λ = v 1 Z v n 1 Z. The volume of Λ is called the regulator Reg(K ). The volume of O K is d K 1 2, d K is the discriminant of K.

5 More preliminary notation Define the ring C K := C Q K Multiplication: (z a,w b) (zw ab) Algebra structure over C and K by identifications K = 1 Q K and C = C Q 1 Also R K := R Q K as a subring of C K. Imaginary part (similarly for real part): I(z a) = I(z) a, Extend embeddings: σ(z a) = zσ(a) For x R we say that x a is totally positive, x a 0 if σ i (x a) > 0, i = 1,2

6 Example Q ( 5 ) In Q( 5 ) we have the fundamental unit ε and its conjugate ε : ε 0 = 1 2 ( 1 + ) 5, ε = ε 1 0 = 1 ( 1 ) 5. 2 And O K Z + ε 0 Z, Λ Z ln with the volume given by ( 1 ( ) ( ) O K = det Λ = ln 1 ( 1 + 5) ) = 5

7 The generalized upper half-plane For r R K and z C K we define z r C K by σ(z r ) = exp(iσ(r)argσ(z) + σ(r)log σ(z) ), σ Subgroups SL 2 (K ) SL(2, R K ) SL(2, C K ) Generalized upper half-plane H K = {z C K : I(z) 0}. Action by SL(2, R K ) on H K : ( a b c d ) z = (az + b)(cz + d) 1.

8 The Hilbert modular group The Hilbert modular group: Γ K = SL 2 (O K ) = {( ) a b c d, a,b,c,d OK, ad bc = } 1 If A = ( a b c d ) ΓK and τ H K then Aτ := (aτ + b)(cτ + d) 1 H K.

9 Cusps of SL 2 (O K ) Cusp: λ = (ρ : σ) P 1 (K ) Fractional ideal a λ = (ρ,σ) Known: λ µ (mod SL 2 (O K )) a λ = (α)a µ The number of cusp classes equals the class number of K. Cusp-normalizing map: ξ,η a 1 λ s.t. ( ρ ξ A λ = σ η A 1 λ SL ( ) 2(O K )A λ = SL 2 a 2 O K ) SL 2 (K ),

10 Vector-valued Hilbert modular forms Let V be a complex SL 2 (O K )-module of rank d < s.t. the kernel of V is a finite index normal subgroup Γ. α Z(SL 2 (O K )) acts with multiplication by 1 k α. Denote the action by (γ,v) γ.v For f O (H K,V) and A SL 2 (O K ) we define (A.f)(z) = A.(f (z))

11 Vector-valued Hilbert modular forms Define M k (V) = {f O (H K,V), A.f = f k A, A SL 2 (O K )} If f M k (V) and f = f i v i then f i M k (Γ) (scalar-valued) S k (V) = { f = f i v i M k (V), : f i S k (Γ) }

12 Main theorem If k Z n with k 2 then: dims k (V) = 1 2 dimv ζ K ( 1) N(k 1) +"elliptic order terms" +"parabolic terms Identity (main) term: ζ K ( 1) (a rational number) Example: ζ Q( 5) = 1 30, ζ Q( 193) ( 1) = , ζ Q( 1009) ( 1) = 211. Finite order ( elliptic ) terms Parabolic ( cuspidal ) term

13 The elliptic terms "elliptic terms" = U 1 U χ V (A) E (A) ±1 A U here U runs through elliptic conjugacy classes and χ V (A) = Tr(A,V), ρ(a σ ) E (A) = 1 k σ σ ρ(a σ ) ρ(a σ ), 1 ρ(a) = 1 ( t + sgn(c) ) t 2 1, t = TrA 2

14 Cuspidal term The cuspidal contribution is the value at s = 1 of the twisted Shimizu L-series dk (( L(s;O K,V) = χ 1 )) a sgn(n(a)) ( 2πi) 2 V 0 1 N(a) s. 0 a O K /U 2 The untwisted L-series (V = 1) is known to have analytic cont. and functional equation ( ) s + n ( ) 1 vol(ok ) s Λ(s) = Γ L(s;O K,1) = Λ(1 s) 2 π n+1 It is easy to see that the L-function for V 1 also has AC. FE is more complicated (cf. Hurwitz-Lerch). If K has a unit of norm 1 then L(s;O K,1) = 0 (conditions on V in general)

15 Notes on the L-series Note that L(s;O K,1) is proportional to L(s,χ) = 0 a O K χ(a) N(a) s where the sum is over all integral ideals of O K and χ(a) = sgn(n(a)). Studied by Hecke, Siegel, Meyer, Hirzebruch and others. Can be expressed in terms of Dedekind sums (Siegel) Proof uses Kronecker s limit formula.

16 Main idea of proof The proof goes in essentially the same way as the usual Eichler-Selberg trace formula.

17 Conjugacy classes Scalar if A = ±1 Elliptic: A has finite order. Parabolic: If A is not scalar but TrA = ±2. Mixed (these do not contribute to the dimension formula).

18 How to find elliptic conjugacy classes? Let A SL 2 (K )\{±1} have trace t. Then TFAE A is of finite order m σ(a) is elliptic in SL 2 (R) for every embedding σ. t = z + z 1 for an m-th root of unity z In this case Q(t) is the totally real subfield of Q(z) and 2[Q(t) : Q] = ϕ(m) where [Q(t) : Q] divides the degree of K since t K.

19 Which orders can appear? If K = Q( D ) then the possible orders are: 3,4,6 (solutions of ϕ(l) = 2), and 5,8,10,12 (solutions of ϕ(l) = 4)

20 Elliptic elements of trace t Lemma ( t2 ) Let a be a fractional ideal and t K be such that K 4 is a cyclotomic field. Then A = ( ) a d + t a b 2 4 c d λ(a) = 2c defines a bijection between the set of elements of SL 2 (a O K ) with trace t and { z = x + t 2 4 2y H K : x O K, y a, x 2 t O K }.

21 Key: Can compute set of representatives for elliptic fixed points Explicit bound on the x,y which can appear.

22 Distance to a cusp Distance to infinity (z, ) = N(y) 1 2 Distance to other cusps λ is a closest cusp to z if (z,λ) = ( A 1 λ z, ). (z,λ) (z,µ), µ P 1 (K ).

23 Reduction algorithm for z H K Find closest cusp λ and set z = x + iy = A 1 λ z. z is SL 2 (O K )-reduced if it is Γ -reduced, where {( ε µ Γ = ), ε O 0 ε 1 K,µ O } K. Local coordinate (wrt. lattices Λ and O K ): ΛY = ỹ B OK X = x where ỹ i = ln y i n Ny.

24 Reduction algorithm Then z is SL 2 (O K )-reduced iff Y i 1 2, 1 i n 1, X i 1 2, 1 i n. If z not reduced we can reduce: Y i by acting with ε = ε k i O K : U (ε) = A 1 ( ε 0 ) λ Aλ : z ε 2k 0 ε 1 i z, Y i Y i + k. X by acting with ζ = a i α i O K : ( ) T (ζ) = A 1 1 ζ λ A 0 1 λ : z z + ζ, X i X i + a i.

25 Remarks Once in a cuspidal neighbourhood reduce in constant time. The hard part is to find the closest cusp. Elliptic points are on the boundary, i.e. can have more than one closest cusp.

26 Finding the closest cusp Let z H K and λ = a c P1 (K ). Then ) (z,λ) 2 = N(y) 1 N (( cx + a) 2 + c 2 y 2. For each r > 0 there is only a finite (explicit!) number of pairs (a,c ) O 2 K /O K s.t. ( z,λ ) r. In fact, for i = 1,...,n we have bounds on each embedding: ) σ i (c) c K r 1 2 σi (y 1 2, Here c K is an explicit constant. σ i (a cx) 2 σ i ( rc 2 K y c 2 y 2)

27 Key Lemma Lemma If K /Q is a number field andα K with Nα = 1 then there exists ε O K such that σ i (αε) r n 1 2 K where r K = max k { } max( σ1 (ε k ),..., σ n (ε k ),1). min( σ 1 (ε k ),..., σ n (ε k ),1) Remark r K 1 always. If K = Q( D ) has a f.u. ε0 with σ 1 (ε 0 ) > 1 > σ 2 (ε 0 ) then r K = σ 1 (ε 0 ) 2.

28 Example Q ( 5 ) The orders which can appear are: 3, 4, 5, 6, 8, 10, 12 The possible traces are: m t ( ) ( ) 10 ε 0 = ( ) ε 0 = 1 2 ( )

29 Example (contd.) A set of reduced fixed points is: order trace fixed pt ell. matrix 4 0 i S = ( ) 0 1 1( iε 0 SE (ε 0 ε ) ) = 0 ε ρ TS = ( ) ( ) 6 1 ρε 0 SE (ε 0 )T ε3 = 0 ε 0 ε ε - 1ε i 2 3 ε0 ST ε 0 = ( ) ε 0 10 ε 1ε ( ) i 2 ε 0 3 ε 0 T ε 0 S = ε Here ρ 3 = 1 and we always choose correct Galois conjugates to get points in H n.

30 Example Q ( 3 ) 1 t z t Y X 1 X 2 Ny a 0 i i a b 0 ε 0 i c 0 i ) (1 2 i a a 1 6b 1 12a i 3 ( 3 2 i ) i

31 Example Q ( 10 ) order 4 We have two cusp classes: c 0 = = [1 : 0] and c 1 = [ 3 : ] Orders: 4 (trace 0) and 6 (trace 1). order label fixed pt close to ( 4 4a 1 ) ± b 2 4 = i ( 4 4c 1 ) ± d e c f ( ) ± c 1 Here 4 ± = ±2i with sign choosen depending on the embedding of 10.

32 Example Q ( 10 ) order 4 label x N (x) y N (y) 4a b c d e f

33 Example Q ( 10 ) Note that if A is the cuspnormalizing map of c 1 then label A 1 z x y ( 4e ) ( 4f 1 ) ±

34 Factoring matrices Given elliptic element A: Find fixed point z Set z 0 = z + ε s.t. z 0 F Γ (well into the interior). w 0 = Az 0 Find pullback of w 0 in to F Γ (make sure w 0 = z 0). Keep track of matrices used in pullback.

35 Example ( ) K = Q 3, z = 1+ ( 3 i 1+ ) 3 A = w 0 = Az 0 (close to 0) w 1 = Sw 0 (close to a 1) w 2 = ST 1 a w 1 w 3 = T 1+a w 2 reduced A = T 1+a ST a 1 S (as a map) A = S 2 T 1+a ST a 1 S (in SL 2 (O K ))

36 Section of a fundamental domain x = (0.00,0.00)

37 Section of a fundamental domain x = (0.05,0.05)

38 Section of a fundamental domain x = (0.10,0.10)

39 Section of a fundamental domain x = (0.15,0.15)

40 Section of a fundamental domain x = (0.20,0.20)

41 Section of a fundamental domain x = (0.25,0.25)

42 Section of a fundamental domain x = (0.30,0.30)

43 Section of a fundamental domain x = (0.35,0.35)

44 Section of a fundamental domain x = (0.40,0.40)

45 Section of a fundamental domain x = (0.45,0.45)

46 Section of a fundamental domain x = (0.50,0.50)

47 Elliptic points of order 4 and 10 x = ( , ) red = order 10 green = order 4