MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

For classical modular forms, the correspondence between the form having CM by an imaginary quadratic field $K$ and it being induced from a Hecke character on $K$ is well-known. (Ribet's paper is a standard reference.)

I am looking for a reference for the analogous result for Hilbert modular forms over a totally real field F. In particular, if the form has CM then it arises from a Hecke character on a quadratic imaginary extension $K$ (over $F$.) I believe, for the converse, Yoshida/Hida is the reference. Thanks

share|cite|improve this question
Just be clear, is your definition of CM that there is some totally imaginary quadratic extension $K$ of $F$ such that for a set of prime ideals of $F$ of density 1 the coefficient $a_\mathfrak{p}$ is zero if and only if $\mathfrak{p}$ is inert in $K/F$? – Rob Harron May 15 '12 at 15:14
@Rob: Yes, this is the definition. – unramified May 15 '12 at 15:20
I don't know a reference for this result. Personally, I'd take the fact that a modular form is induced from a Hecke character of $K$ to be the definition of CM, but that doesn't help answer your question. Have you tried generalizing Ribet's argument? Otherwise, this seems like the type of thing that would show up somewhere in a paper/book of Hida's. – Rob Harron May 15 '12 at 16:12
FWIW non-algebraic Hecke characters can give rise to HMFs with weights which aren't congruent mod 2 and hence don't have associated Galois representations. This presents an obstruction to proving the result using arguments on the Galois side in this generality, which presumably you can try to get around by using some symmetric square argument. – Kevin Buzzard May 15 '12 at 21:14

This result is better understood in terms of automorphic representations. Let $F$ be an algebraic number field, and let $\pi$ be an automorphic representation of $\mathrm{GL}_2\left(\mathbb{A}_F\right)$, where $\mathbb{A}_F$ denotes the ring of adèles of $F$.

Suppose that there exists a nontrivial unitary Hecke character $\omega$ of $F^{\times} \backslash \mathbb{A}_F^{\times}$ such that $\pi \otimes (\omega \circ \det) \cong \pi$. Then $\omega$ must necessarily be quadratic, and the representation $\pi$ is said to be a monomial representation. This is what you call CM, but this labelling only really makes sense when $F = \mathbb{Q}$ and $\pi$ corresponds to a holomorphic modular form, for the reasons outlined in Ribet's paper. When $\pi$ corresponds to a Maaß form, I have seen such form called of CM-type, but this seems a little incongruous.

Let $E$ be the quadratic extension of $F$ associated to $\omega$ via class field theory. Then the following statement is Proposition 6.5 of L-indistinguishability for $\mathrm{SL}(2)$ by Labesse and Langlands:

If $\pi$ is a monomial automorphic representation, then there exists a Hecke character $\chi$ of $E^{\times} \backslash \mathbb{A}_E^{\times}$ such that $\pi \cong \pi(\chi)$.

Here $\pi(\chi)$ denotes the cuspidal automorphic representation of $\mathrm{GL}_1\left(\mathbb{A}_E\right)$ associated to $\chi$.

While I do not believed it is mentioned in this paper, it is worth noting the following:

The monomial automorphic representation $\pi$ is cuspidal if and only if $\chi$ does not factor through the norm map (that is, there does not exist some Hecke character $\widetilde{\chi}$ of $F^{\times} \backslash \mathbb{A}_F^{\times}$ for which $\chi = \widetilde{\chi} \circ N_{E/F}$).

The proof of these results involve automorphic representations instead of Galois representations, and in particular generalise to Maaß forms (when $F = \mathbb{Q}$ and $\pi_{\infty}$ is a principal series representation) and Hilbert modular forms (when $F$ is a totally real field and $\pi_v$ is a discrete series representation for each archimedean place $v$).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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