On the Fourier coefficients at cusps of a modular form

Question

Given a holomorphic modular form $f(z)$ of integral weight for $\Gamma_{1}(N)$ with integer Fourier coefficients (at $i\infty$), is it true that the Fourier coefficients at other cusps of $f(z)$ are all algebraic integers? Can anyone help me with this problem? Highly appreciated!

Answer 1

In general no. Take any newform $f$ of weight $k$ on $\Gamma_0(N)$. Then $f$ is an eigenvector of the Atkin-Lehner involution $W_N$, with eigenvalue $\pm 1$. Then by definition of $W_N$, we have \begin{equation*} f |_k \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix} (\tau) = i^{-k} N^{-k/2} (W_N f)\bigl(\frac{\tau}{N}\bigr) = \pm i^k N^{-k/2} f\bigl(\frac{\tau}{N}\bigr), \end{equation*} whose first Fourier coefficient is not an algebraic integer.

More conceptually, the $q$-expansion of modular form is obtained by evaluating algebraic modular forms (in the sense of Katz, see $p$-adic Interpolation of Real Analytic Eisenstein Series) at the Tate curve. The $N$-torsion subgroup of Tate curve is isomorphic to $\mu_N \times (\mathbb{Z}/N\mathbb{Z})$ as a group scheme over $\mathbb{Z}((q^{1/N}))$. To get an action of $\mathrm{GL}_2(\mathbb{Z}/N\mathbb{Z})$, one needs to identify $\mu_N$ with $\mathbb{Z}/N\mathbb{Z}$, and this can only be done over $\mathbb{Z}\bigl[\zeta_N, \frac{1}{N}\bigr]$, because the discrete Fourier transform has denominators.