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In this question, $k \geq 2$ and $N \geq 1$ are integers.

We consider the space $S = \mathcal{S}_k(\Gamma_1(N))^{new}$ of modular forms of weight $k$ for $\Gamma_1(N)$ (the same question can certainly be asked for $k = 1$ but I think that this MO answer should help). It has an action of Hecke operators $T_p$ (for all primes $p$) and diamonds $\langle d\rangle$ (for all integers $d$ coprime to $N$). Let $T_{\mathbb{Q}}$ be the (commutative) $\mathbb{Q}$-algebra generated by these operators (seen as a subalgebra of the ring of endomorphisms of $S$), and $T_{\mathbb{C}}$ be the sub-$\mathbb{C}$-algebra of the ring of endomorphisms of $S$ generated by these operators.

My question is the following: let $f \in S$ be a Hecke normalized newform. Let $\sigma$ be an automorphism of $\mathbb{C}$ (or of $\overline{\mathbb{Q}}$ if we accept that the eigenvalues of Hecke operators are algebraic). It is well-known (it appears in, say, Shimura's 1976 paper The special values of zeta functions associated with cusp forms) that $f^{\sigma}$ (defined as a formal power series by letting $\sigma$ act on all the coefficients of $f$) is another Hecke eigenform. But why is that?

A possible reformulation involves using the well-known fact that such eigenforms are exactly the $\mathbb{C}$-morphisms $T_{\mathbb{C}} \rightarrow \mathbb{C}$. Now, if $f$ corresponds to the morphism $\mu$, $f^{\sigma}$ corresponds to a $\mathbb{C}$-morphism mapping $T_n$ to $\sigma \circ \mu(T_n)$. But this cannot define a $\mathbb{C}$-morphism $T_{\mathbb{C}}$ unless every $\mathbb{C}$-linear relation between the $T_n$ (and the diamonds) is in fact rational. In other words, we need that $T_{\mathbb{Q}} \otimes_{\mathbb{Q}} \mathbb{C} \rightarrow T_{\mathbb{C}}$ be injective. But I'm not sure if this is easier to prove than the original statement.

In weight $2$, Hecke operators can be interpreted as endomorphisms of the Jacobian of $X_1(N)$ and cusp forms can be interpreted as global differentials on said Jacobian. So we could be done if, for a complex abelian variety $J$, $\mathrm{End}(J) \otimes \mathbb{C} \rightarrow \mathcal{L}(H^0(J,\Omega^1))$ is injective. This is, however, false, because of $J=\mathbb{C}/\mathbb{Z}[i]$ (and other instances of complex multiplication).

If, however, we have a full basis of $\mathcal{S}_k(\Gamma_1(N))$ by cusp forms with rational Fourier coefficients, this means that the image of the Fourier expansion map $\mathcal{S}_k(\Gamma_1(N)) \rightarrow \mathbb{C}[[q]]$ is of the form $V \otimes \mathbb{C}$, where $V$ is a subspace of $\mathbb{Q}[[q]]$ and thus that the Galois conjugate of a modular form for $\mathcal{S}_k(\Gamma_1(N))$ is in $\mathcal{S}_k(\Gamma_1(N))$, and then the result follows easily. But this doesn't look easier than the bolded question.

So where does that assertion come from? Why is the Galois conjugate of a newform a newform of same level?

(You may use that Hecke eigenvalues are algebraic integers, but I would appreciate a reference for the weights $k > 2$).

Aphelli
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    The footnote on P.11 of this set of notes reassures that the proof is indeed non-trivial. – Kenny Lau Nov 14 '21 at 09:47
  • @Kenny Lau: thank you for finding this link! The question really bugged me and I kept feeling I ought to know the answer. That link relieves me a little. Note though that Lemma 3.7 is sufficient (and indeed, I’m making this link in the post) to prove the claim. But I don’t know if that’s the way the proof goes. – Aphelli Nov 14 '21 at 12:42
  • You referenced my old MO question about the weight 1 case. If you read that question carefully, you'll see that the question contains a sketch of the proof in weights $\ge 2$; it's formulated as "weight k >= 2 is easy because xxx, so what about weight 1"? (And, indeed, the accepted answer there uses the weight $\ge 2$ case to prove the weight 1 case.) – David Loeffler Nov 15 '21 at 09:04
  • @David Loeffler: I don’t think I’m asking exactly the same question – why is it enough that Hecke eigenvalues are algebraic integers to show that the systems of eigenvalues are stable under Galois? I referenced your question because I thought that the trick in the accepted answer could deduce the case $k=1$ from the case $k=13$ (for instance). (Although, I would be interested if you could clarify what you meant by “Eichler-Shimura machinery” – I thought the result was “only” about the congruence linking Frobenius and Hecke on the special fiber of a Jacobian?) – Aphelli Nov 15 '21 at 10:16
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    There are several distinct results which get given the name "Eichler--Shimura". I meant the isomorphism between $S_k(\Gamma) \oplus \overline{S_k(\Gamma)}$ and $H^1_{\mathrm{par}}(\Gamma, \mathbf{Z}) \otimes \mathbf{C}$. This isomorphism is Hecke equivariant, and $H^1_{\mathrm{par}}(\Gamma, \mathbf{Z})$ is a finitely-generated $\mathbf{Z}$-module. It follows readily from this that $T_{\mathbf{C}} = T_{\mathbf{Z}} \otimes \mathbf{C}$. – David Loeffler Nov 15 '21 at 10:40
  • @David Loeffler: thank you for the explanation, I understand now (if you want to write an answer, I’ll accept it)! Do you know a reference for this Eichler-Shimura isomorphism (ideally in all weights at least $2$)? – Aphelli Nov 15 '21 at 11:30
  • @Mindlack Wiese has some good lecture notes that cover Eichler-Shimura and more. – Mathmo123 Nov 16 '21 at 10:50

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