Let $f=\sum_{n=1}^\infty a_nq^n$ be a newform of level $N$ and weight $k\ge 2$. Suppose that $f$ is a CM modular form in the sense of §3 of Ribet's paper Galois representations attached to eigenforms with nebentypus: i.e. there exists a quadratic character $\varphi$ such that $$a_p = \varphi(p)a_p$$for all primes $p$ in a set of primes of density $1$. Let $K$ be the quadratic field cut out by $\varphi$.

Is there a down-to-earth explanation of why $K$ must be an imaginary quadratic field?

The proof given in theorem 4.5 of the above paper depends quite heavily on the properties of the Serre group $S_{\mathfrak m/K}$.

  • $\begingroup$ Are you asking about form with "inner twists" (i.e. are non CM modular forms but a conjugate is isomorphic to a twist) or about CM forms? Regarding inner twists, the field might be real (see Section 3.8 of math.berkeley.edu/~ribet/Articles/annalen_253.pdf) $\endgroup$
    – A. Pacetti
    Mar 8, 2016 at 22:07
  • $\begingroup$ @A.Pacetti I'm specifically asking about CM forms. $\endgroup$ Mar 8, 2016 at 22:09

2 Answers 2


As Joel says, $a_p=0$ if $p$ is inert in $K$, so this means that the Galois representation attached to $f$ must be induced from a 1-dimensional representation of the absolute Galois group of $K$. If $K$ were real quadratic then the grossencharacter corresponding to this 1-dimensional representation would have to be equal to $|.|^n$ on both the $\mathbf{R}_{>0}$'s at the infinite places. But this means that the associated Galois representation has finite order up to twist (both Hodge-Tate weights are equal) so the weight would have to be 1.


If you ask about CM forms, here is another argument: the Galois representation attached to your form is odd (i.e. in complex conjugation has determinant $-1$ and trace $0$). When you restrict the representation to the CM field, the representation is isomorphic to a sum of two characters, one character and its Galois conjugate (i.e. $\chi(\tau) \oplus \chi(\sigma \tau \sigma^{-1})$ where $\sigma$ is an extension of the Galois group generator of the quadratic extension). In particular both characters behave the same at complex conjugation, so the determinant of the restricted Galois representation is trivial at complex conjugation which implies the quadratic field is imaginary quadratic.

  • $\begingroup$ Your argument seems to work in weight 1 where the result is false (there are 2-dimensional finite image continuous odd irreducible Galois representations induced from a character of a real quadratic field). The problem is that $\chi$ can be different on the two places above infinity so I can't understand this argument even in higher weight. $\endgroup$
    – znt
    Mar 12, 2016 at 21:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.