Action of automorphisms on Eisenstein series

Question

Let $ f \in \mathcal{M}_{k}(\Gamma) $ and $ \sigma \in \textit{Aut}(\mathbb{C}) $. Suppose $$ f = \sum_{n=0}^{\infty}a_{n}q^{n} .$$ Then we define the action of $ \textit{Aut}(\mathbb{C}) $ on $ \mathcal{M}_{k}(\Gamma) $ by $$ f^{\sigma} = \sum_{n=0}^{\infty}a_{n}^{\sigma}q^{n} .$$ Does anybody know of a proof that this action preserves the space $ \mathcal{E}_{k}(\Gamma) $, i.e. that the image of any Eisenstein series under the action of $ \textit{Aut}(\mathbb{C}) $ is again an Eisenstein series.

Answer 1

I assume your $\Gamma$ is a congruence subgroup (I don't know if the statement is true otherwise).

If $\Gamma$ is congruence, then it suffices to assume $\Gamma = \Gamma_1(N)$ for some $N$. For such groups one can write down an explicit basis of the space $\mathcal{E}_k(\Gamma)$, indexed by triples $(\chi_1, \chi_2, t)$ where the $\chi_i$ are Dirichlet characters and $t$ is an integer, satisfying various elementary conditions. This is described in any decent modular forms textbook, e.g. Diamond & Shurman or Miyake.

From this description (and the fact that the conjugate of a Dirichlet character under an automorphism is still a Dirichlet character) it follows easily that any automorphism of $\mathbf{C}$ permutes the elements of this basis, so the space is $\mathcal{E}_k(\Gamma)$ is preserved by any such automorphism.