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Probably an easy question: let $f$ be an eigenform in $S_k^{\text{new}}(\Gamma_0(N),\chi)$ assumed to be with rational fourier coefficients. Then $\chi$ is necessarily trivial or quadratic. But more precisely, $\chi$ seems to be unique: if $k$ is even, then $\chi$ must be the trivial character, and if $k$ is odd then $N$ must be of the form $N=-Df^2$ with $D$ a negative discriminant, and then $\chi(n)=(D/n)$. Can someone explain why this is true?

More generally, in even weight and nontrivial quadratic character the irreducible Galois orbits all have even dimension (equivalently, the irreducible factors of the characteristic polynomial of Hecke operators all have even degree). Is this true, and why?

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For $f \in \mathcal{S}_k^{\ast}(q,\chi)$, we have that \[\lambda_f(n) \langle f, f\rangle = \langle T_n f, f \rangle = \chi(n) \langle f, T_n f \rangle = \chi(n) \overline{\lambda_f}(n) \langle f, f\rangle\] for $(n,q) = 1$, so that \[\lambda_f(n) = \chi(n) \overline{\lambda_f}(n).\] It follows that for $\lambda_f(n)$ to be real, we must have that either $\chi(n) = 1$ or $\lambda_f(n) = 0$.

  • If $\chi$ is trivial, then $\lambda_f(n)$ is always real. Of course $\chi(-1) = 1$, so that $k$ must be even.
  • If $\chi$ is quadratic and $\lambda_f(n) = 0$ whenever $\chi(n) \neq 1$, then one can show that $f$ is associated to a Hecke Größencharakter $\psi$ coming from some imaginary quadratic field $K = \mathbb{Q}(\sqrt{D})$ with $D$ a negative fundamental discriminant. Of course, $\chi$ is the quadratic character associated to this field, and as $D$ is negative, $\chi(-1) = -1$, so that $k$ must be odd. For more about the conductor, see this answer.
  • On the other hand, if $\lambda_f(n) = 0$ whenever $\chi(n) \neq 1$, then $f \otimes \chi = f$. By a result of Labesse and Langlands (see this answer), any newform satisfying this is the automorphic induction of a Hecke Größencharakter $\psi$ of some field $K$. Since this is automorphic induction from $\mathrm{GL}_1$ to $\mathrm{GL}_2$, $K$ must be quadratic, and of course must be imaginary (for otherwise this would be a Maaß newform).
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I only answer the first part of the question. As it is well-known, $\chi$ and $k$ have the same parity. In particular, if $\chi$ is trivial then $k$ must be even. Conversely, if $\chi$ is non trivial then $f$ has CM by its own Nebentypus $\chi$ because $f$ is assumed to have rational (hence real) coefficients. Therefore $\chi$ is the quadratic character associated with an imaginary quadratic field $K$ of (negative) discriminant $D$ (see Ribet, Galois representations attached to eigenforms with Nebentypus, Modular Functions of One Variable V, LNM Vol. 601). In particular, we have $\chi(-1)=-1$ and $k$ is odd.

Let us now assume that we are in that latter case, i.e. $k$ is odd. By a result of Hecke and Shimura (loc. cit. pp. 34-36), $f$ is the newform attached to a Hecke character $\psi$ of $K$ with conductor, say, $\mathfrak{m}$ and infinite type $k-1$. (That is, if we view $\psi$ as a homomorphism from the group of prime-to-$\mathfrak{m}$ fractional ideals of $K$ into $\mathbf{C}^*$, then $\psi(\alpha \mathcal{O}_K)=\alpha^{k-1}$ for all $\alpha\in K^*$ with $\alpha\equiv 1\pmod{\mathfrak{m}}$; here $\mathcal{O}_K$ denotes the integer ring of $K$.) Moreover, in the notation of the question, we have $N=-\mathcal{N}(\mathfrak{m})D$ where $\mathcal{N}$ denotes the norm map from $K$ to $\mathbf{Q}$.

If $\mathfrak{p}$ is a prime of $\mathcal{O}_K$, denote by $e_{\mathfrak{p}}$ the exponent of the conductor $\mathfrak{m}$ of $\psi$ at $\mathfrak{p}$. By Proposition 6.1 and Table 1 of Schütt's paper CM newforms with rational coefficients (Ramanujan Journal 19 (2009), 187-205), for all prime ideals $\mathfrak{p}$, we have $e_{\mathfrak{p}}=e_{\overline{\mathfrak{p}}}$ and $e_{\mathfrak{p}}$ even if $\mathfrak{p}$ ramifies in $K$. (For a more detailed proof of these results, see chapter II of the author's dissertation available on his homepage.) This in turn implies that $\mathcal{N}(\mathfrak{m})=N/(-D)$ is a square, as desired.

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The Hecke field $K_f$ of a newform $f$ is always stable under complex conjugation: this follows from the identity $\bar{f} = f \otimes \bar{\chi_f}$, where $\chi_f$ is the Nebentypus character of $f$ (one can show further that $K_f$ is either totally real, or a CM field). Now if $f$ has even weight and $\chi_f$ is quadratic, then from Peter Humphries' answer we know that $K_f$ cannot be real, hence the degree of $K_f$ is even.

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