Basis for modular forms with Nebentypus character

(Sorry for my poor english..)

Let $$N$$ be a positive integer and $$\chi$$ be a Dirichlet character modulo 4N. I already know that the $$\mathbb{C}$$-vector space $$S_{k}(\Gamma_1(N))$$ has a basis $$\{F_1,\cdots F_n\}$$ with the Fourier coefficients of $$F_i$$ (at infinity) are rational.

My questions are..

1. Does the $$\mathbb{C}$$-vector space $$S_{k}(\Gamma_0(N),\chi)$$ also has a basis $$\{G_1,\cdots G_m \}$$ with the Fourier coefficients of $$G_i$$ (at infinity) are rational?

2. Does the $$\mathbb{C}$$-vector space $$S_{k+\frac{1}{2}}(\Gamma_0(4N),\chi)$$ also has a basis $$\{G_1,\cdots G_m \}$$ with the Fourier coefficients of $$G_i$$ (at infinity) are in $$K$$? ($$K$$ is a number field.)

EDIT : How about number field $$K$$ instead of rational number? In other words, does the $$\mathbb{C}$$-vector space $$S_{k}(\Gamma_0(N),\chi)$$ also has a basis $$\{G_1,\cdots G_m \}$$ with the Fourier coefficients of $$G_i$$ (at infinity) are in $$K$$? ($$K$$ is a number field.)

This is not true in general: one can prove that if $$f \in M_k(\Gamma_0(N),\chi)$$ is a non-zero modular form with Nebentypus character $$\chi$$ (without any newform assumption), then the field generated by the Fourier coefficients of $$f$$ contains $$\mathbf{Q}(\chi)$$, see e.g. Corollary 3.6 in this preprint.

On the other hand, if you consider a Galois orbit of Dirichlet characters, and consider the (direct) sum of the spaces of cusp forms corresponding to these characters, then the answer is yes because this space is stable under $$\mathrm{Aut}(\mathbb{C})$$ (acting as usual on the Fourier expansion).

EDIT. In weight 2, the first example appears in level $$N=13$$. The space $$S_2(\Gamma_1(13))$$ is $$2$$-dimensional and decomposes as the direct sum of two lines associated to the Nebentypes $$\chi$$ and $$\overline{\chi}$$ respectively, where $$\chi$$ is the Dirichlet character modulo 13 defined by $$\chi(2)=e^{2\pi i/6}$$. Each of these lines is generated by a newform with coefficients in $$\mathbb{Q}(e^{2\pi i/6})$$, so there is no way to get a modular form with rational coefficients inside these eigenspaces.

EDIT 2. Regarding the question whether $$S_k(\Gamma_0(N),\chi)$$ admits a basis of modular forms with coefficients in some number field $$K$$, an obvious necessary condition is that $$K$$ contains the values of $$\chi$$ (by Corollary 3.6 mentioned above). Conversely, if $$K$$ contains the values of $$\chi$$, then the operator $$\pi_\chi = \sum_{\lambda \in (\mathbb{Z}/N\mathbb{Z})^\times} \bar{\chi}(\lambda) \langle \lambda \rangle$$ commutes with $$\mathrm{Aut}(\mathbb{C}/K)$$ because the diamond operators $$\langle \lambda \rangle$$ are defined over $$\mathbb{Q}$$. So if you take a basis of $$S_k(\Gamma_1(N))$$ consisting of modular forms with rational coefficients and apply $$\pi_\chi$$, you get a generating family of $$S_k(\Gamma_0(N),\chi)$$ with coefficients in $$K$$, as desired.

For half-integral weight modular forms, I have not checked the details but I think these properties still hold, see Theorem 2.6 in this preprint.

EDIT 3. Here is a sketch of an argument for half-integral weight modular forms. I assume that your definition of $$S_{k+1/2}(\Gamma_0(4N),\chi)$$ is the same as in Shimura's Annals paper "On modular forms of half-integral weight", so that the Jacobi theta function $$\theta(q)=\sum_{n \in \mathbb{Z}} q^{n^2}$$ is modular for $$\Gamma_0(4)$$ with trivial character. Let $$f \in S_{k+1/2}(\Gamma_0(4N),\chi)$$ with Fourier coefficients in $$K_f$$. Then $$f \theta \in S_{k+1}(\Gamma_0(4N),\chi)$$, so by the above the field generated by the coefficients of $$f \theta$$ contains $$\mathbb{Q}(\chi)$$. Since $$\theta$$ is invertible in $$\mathbb{Z}[[q]]$$, the same is true for $$f$$. Conversely, if $$K$$ is a number field containing the values of $$\chi$$ then you can proceed as above to show that $$S_{k+1/2}(\Gamma_0(4N),\chi)$$ has a basis of modular forms with coefficients in $$K$$. The key point is that $$S_{k+1/2}(\Gamma_1(4N))$$ has a basis of modular forms with rational coefficients, which you can prove using the map $$f \mapsto f \theta$$.

• How about number field $K$ instead of rational? – ililiil Oct 1 '18 at 11:44
• @ililiil I addressed your new questions in my answer. – François Brunault Oct 1 '18 at 23:59
• @ François Brunault I really appreciate your kind (and nice) answer. I think that it is beautiful method..! However I do not understand half-integral weight case even if I look at the article you sent me. Could you explain some more detailed ideas? – ililiil Oct 2 '18 at 5:59
• @ililiil See my Edit 3 for a sketch of an argument (unfortunately I don't know a reference, so you will have to sketch some details yourself). – François Brunault Oct 2 '18 at 8:11
• @ François Brunault Thank you vey much..!! – ililiil Oct 2 '18 at 9:24