On harmonic weak Maass forms of half integral weight

We show that certain space of vector valued harmonic weak Maass forms of half integral weight is isomorphic to a space of scalar valued ones whose Fourier coefficients are supported on suitable progressions. This kind of result for holomorphic modular forms was proved by Eichler and Zagier.

Let L be the lattice 2mZ equipped with the quadratic form Q(x) = x 2 /4m. Then its dual L ′ equals Z. Let ρ L denote the Weil representation associated to the discriminant form (L ′ /L, Q), andρ L its dual representation. We denote by M k+ 1 2 ,ρ L the space of C[L ′ /L]valued holomorphic modular forms of weight k + 1 2 and type ρ L . Then Eichler and Zagier (Γ 0 (4m)) ≃ H k+ 1 2 ,ρ L ≃ J cusp k+1,m .
In order to state our main results more precisely, we let k be an integer and fix m = 1 or a prime. We denote by H k+ 1 2 (Γ 0 (4m)) the space of harmonic weak Maass forms of weight k + 1 2 for Γ 0 (4m) (see Section 2.1). Then it is known (see, for instance, [5] Here Γ(a, y) = ∞ y e −t t a−1 dt denotes the incomplete Gamma function. We define a subspace H + Let H k+ 1 2 ,ρ L denote the space of C[L ′ /L]-valued harmonic weak Maass forms of weight k + 1 2 and type ρ L (see Section 2.2). We denote the standard basis elements of the group algebra C[L ′ /L] by e γ for γ ∈ L ′ /L. Suppose that the discriminant form (L ′ /L, Q) is given Then the level of L equals 4m, and b + − b − ≡ 1 mod 8. For example if we take with Q(X) = −m det(X), then (b + , b − ) = (2, 1) and For a given f ∈ H + , and s(γ) = 1 if γ ≡ 0, m mod 2m, and 2 otherwise. Theorem 1. With the notation as above we have the following.
(1) If k is even, then the map f → F defines an isomorphism of H + (1) For a given vector valued modular form F = γ F γ e γ the map F → f will be the inverse isomorphism where f (τ ) := γ F γ (4mτ ).
(3) This kind of result for weakly holomorphic modular forms of integral weight is proved by Bruinier and Bundschuh [4]. Let p be an odd prime. We write M ! k (Γ 0 (p), ( · p )) for the space of weakly holomorphic modular forms of weight k for Γ 0 (p) with Nebentypus ( · p ). For ǫ ∈ {±1} we define the subspace Let L be the lattice so that the discriminant group L ′ /L is isomorphic to Z/pZ. On L ′ /L the quadratic form is equivalent to Q(x) = αx 2 /p for some α ∈ Z/pZ − {0}. Put ǫ = ( α p ). Then Bruinier and Bundschuh [4,Theorem 5] showed that M !ǫ k (Γ 0 (p), ( · p )) is isomorphic to M ! k,ρ L . For integral weight case Bruinier and Bundschuh's argument can be applied to the spaces of harmonic weak Maass forms (see [6,Theorem 1.2]). However, for half integral weight case, we need another argument because Eichler and Zagier's argument depends on the dimension formulas for the spaces of holomorphic modular forms (see the proof of [7, Theorem 5.6]).
It is essential that our proof of Theorem 1 relies on some nontrivial properties of the Weil representation.
Next we show that the spaces in Theorem 1 (2) are isomorphic to the space of harmonic Maass-Jacobi forms recently developed by Bringmann and Richter [2].
Let L be the lattice 2mZ equipped with the positive definite quadratic form Q(x) = x 2 /4m.
Then the space J k,m of Jacobi forms of weight k and index m is isomorphic to the space Hence, we can decompose φ(τ, z) by a linear combination of the theta functions as with any r ∈ Z, r ≡ µ mod 2m, and Using the same argument in [7, Theorem 5.1], the 2m-tuples (h µ ) µ (2m) satisfies the desired transformation formula for vector valued harmonic weak Maass forms. Now the remaining thing is to check ∆ k− 1 2 h µ = 0. By definition, φ(τ, z) vanishes under the action of the Casimir element C k,m (see [2, p. 2305]), and the action of C k,m on functions inĴ cusp k,m agrees with that of . Using the fact that θ m,µ (τ, z) is in the heat kernel, that is, one can conclude by a direct computation that Hence, we get the following theorem: Theorem 2. Let k be even, and m = 1 or a prime. Then .
holomorphic Jacobi forms of weight k and index m. So one can guess that there must be a similar isomorphism as above when k is odd. To do that, we need to introduce a new definition of Maass Jacobi forms which include skew holomorphic Jacobi forms. We hope this can be done by following and modifying the work of Bringmann and Richter [2].
(2) In the argument of the proof, the growth condition doesn't matter. Namely, if we (Γ 0 (4m)) so that they have at most linear exponential growth at for k even and m = 1 or a prime. Here,Ĵ k,m is the corresponding bigger space (see Section 2.3).

Preliminaries
2.1. Scalar valued modular forms. Let τ = x + iy ∈ H, the complex upper half plane, with x, y ∈ R. Let k ∈ 1 2 Z − Z, and m a positive integer. Put ε d := ( −1 d ) 1 2 . Recall that weakly holomorphic modular forms of weight k for Γ 0 (4m) are holomorphic functions f : H → C which satisfy: (ii) f has a Fourier expansion of the form and analogous conditions are required at all cusps.
A smooth function f : H → C is called a harmonic weak Maass form of weight k for Γ 0 (4m) if it satisfies: (ii) ∆ k f = 0, where ∆ k is the weight k hyperbolic Laplace operator defined by We denote the space of these harmonic weak Maass forms by H k (Γ 0 (4m)). This space can be denoted by H + k (Γ 0 (4m)) in the context of [5], which is the inverse image of S 2−k (Γ 0 (4m)) under the certain differential operator ξ k . We have M ! k (Γ 0 (4m)) ⊂ H k (Γ 0 (4m)). The polynomial P f ∈ C[q −1 ] is called the principal part of f at the corresponding cusps. We denote byρ L the dual representation of ρ L .
Let k ∈ 1 2 Z. A holomorphic function f : H → C[L ′ /L] is called a weakly holomorphic modular form of weight k and type ρ L for the group Mp 2 (Z) if it satisfies: (ii) f is meromorphic at the cusp ∞.
Here condition (ii) means that f has a Fourier expansion of the form e γ such that f (τ ) = P f (τ ) + O(e −εy ) as y → ∞ for some ε > 0.
We denote by H k,ρ L the space of these C[L ′ /L]-valued harmonic weak Maass forms. This space is denoted by H + k,L in [5], which is the inverse image of S 2−k,L − under ξ k . We have M ! k,ρ L ⊂ H k,ρ L . Similarly we define the space H k,ρ L . In particular f ∈ H k,ρ L has a unique decomposition

Harmonic
Maass-Jacobi forms. The most part of this session we follow the notation given in [2].

Definition 3. A function φ : H × C → C is a harmonic Maass-Jacobi form of weight k and
index m if φ is real-analytic in τ ∈ H and z ∈ C, and satisfies the following conditions: cτ +d +λ 2 τ +2λz) = φ(τ, z).
Let J k,m be the space of harmonic Maass-Jacobi forms of weight k and index m, which are holomorphic in z. In fact, we are interested in the subspace J cusp k,m consisting of the elements φ ∈ J k,m whose Fourier expansion is of the form The space J cusp k,m is in fact the inverse image of J sk,cusp 3−k,m under the certain differential operator ξ k,m (see [2, Remarks (1) on p. 2307]).

Proof of Theorem 1
We will only give a proof of (1) because exactly the same argument can be applied. We first prove that for a given f ∈ H + We will prove (3.1) for m = 1 or a prime by following his argument. For details we refer to [8, pp. 735-737]. For j prime to 4m put We choose b, d ∈ Z so that jd − 4mb = 1. Then one finds that and M = j b 4m d and J(M, τ ) denotes the automorphy factor for the theta function n∈Z q n 2 , that is, Here ( · · ) is the usual Jacobi symbol. This implies that where j −1 denotes an integer which is the inverse of j in (Z/4mZ) × . On the other hand we have by definition that Replacing τ by τ /4m in (3.2) and (3.3) one has the following identity: Let R be a 2m × 2m matrix defined by In order to show F ∈ H k+ 1 2 ,ρ L we first need to prove (3.1), i.e.     . . .
Since F γ = F −γ the above identity is equivalent to for some matrix B with 2m columns of which the first m + 1 ones are linearly independent.
For instance, in [8] B was chosen as and checked that its rank is m + 1 if m ∈ S − {2}. Here  ℓ is the ℓth largest element in In this paper we take B as CA where . Proof. The βγ-th entry b βγ of the 2m × 2m matrix B is given by From this one can easily infer that B has rank 2ϕ(m), and its first 2ϕ(m) columns are linearly independent. Since ℓ(ϕ(4m)), γ(2m) from the Gauss sum formula (see [8, p. 736]), the identity (3.4) is equivalent to This implies that the identity W 4m for all (j, 4m) = 1.
F ∈ H k+ 1 2 ,ρ L for m = 2. The same procedure as given in [8,Remark 3.2] can be applied to the case when m = 2, so we omit the detailed proof. Now we consider the converse. For a given F = γ(2m) F γ e γ ∈ H k+ 1 2 ,ρ L we define It is straightforward to verify that f satisfies the condition c ± f (n) = 0 unless (−1) k n ≡ mod 4m by inspecting the Fourier expansion of F . Also ∆ k+ 1 2 f = 0 by (3.5). Thus it suffices to show that for all ( a b c d ) ∈ Γ 0 (4m). We may assume that d > 0 by multiplying ( −1 0 0 −1 ) if necessary.
In what follows, our results hold for arbitrary m > 0. In fact one may apply our argument even for somewhat general discriminant forms (L ′ /L, Q).
Lemma 5. For any n ∈ Z one has ρ L 1 0 Proof. Since ( 1 0 n 1 ) = ( 1 0 1 1 ) n it suffices to prove that ρ L 1 0 The last equality is from Milgram's formula (see [1]), that is, First notice that We first consider the case a > 0. The consistency condition implies that a 4mb c/4m d We claim that (3.7) holds true for the case a < 0. If c = 0, then a = d = −1 and thereby it is straightforward to verify (3.7). So we assume that c = 0. If we choose x ∈ Z so that a + xc > 0, then from the elementary identity This completes the proof of Theorem 1.