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

Dear friends,

I have some trouble finding a precise definition of what a modular form with complex multiplication. Could anyone provide such a definition and references for the study of CM modular forms and its main properties? I would be grateful

share|cite|improve this question

Let $f$ be a newform of level $N$ and weight $k\geq 2$. We say $f$ has CM by the quadratic field $K$ if there exists a quadratic extension $K/\mathbb Q$ such that if $\eta_{K/\mathbb Q}$ is the quadratic character whose kernel is $G_{K}$ then the automorphic representation $\pi(f)$ of $\operatorname{GL}(2,\mathbb A_{\mathbb Q})$ is isomorphic to $\pi(f)\otimes\eta_{K/\mathbb Q}$. If this is true, then $K$ has to be an imaginary quadratic extension. More generally, if $F$ is a totally real field and $\pi$ is an automorphic representation (EDIT: as wccanard points out, here again the condition that the weight should be greater than $2$ has to be included) of $\operatorname{GL}(2,\mathbb A_{F})$ isomorphic to $\pi\otimes\eta_{K/F}$ for $K/F$ quadratic then $K$ is a CM extension (a totally imaginary quadratic extension of $F$).

As Marc Palm writes, when $f$ has CM by $K$ there exists a character $\chi$ of $\mathbb A_{K}^{\times}/K^{\times}$ such that for all finite place $v$, the $L$-factor $L_{v}(f,s)$ of $f$ is equal to the product $\underset{w|v}{\prod}L_{w}(\chi,s)$ of $L$-factors of $\chi$ over places of $K$ above $v$. A highbrow version of this last statement is that $\pi(f)$ is isomorphic to the automorphic induction of $\chi$ from $K$ to $F$.

share|cite|improve this answer
Thanks, this is a great answer – mod78 Feb 13 '13 at 9:34
Let me add a reference. You can read Motives and automorphic forms: the potentially abelian case, available on L.Fargues webpage. This is a modern exploration of the topic (which contains much much more than the answer to your question). – Olivier Feb 13 '13 at 10:31
Olivier -- your assertion about $K$ being CM is false in the generality that you write it, in the totally real case (at least at the time I am writing this comment); there are for example Hilbert modular forms of parallel weight 1, and also automorphic representations of Artin type attached to non-holomorphic $\pi$s, which are isomorphic to a quadratic twist of themselves where the associated quadratic extension is not $CM$. In fact in the totally real case $K$ may be neither totally real nor totally imaginary. – user30035 Feb 13 '13 at 20:49
Dear wccanard (!), I did write that $k$ should be greater than 2, but now I realize I did not repeat the condition when passing to $F$. Thanks for pointing this out. – Olivier Feb 14 '13 at 8:48
OK I'm happy :-) If $K$ is any quadratic extension of $F$, totally real or totally imaginary or otherwise, and if $\chi$ is a grossencharacter of $K$ then of course you can automorphically induce $\chi$ up to $GL(2)/F$ by standard converse theorems and Hecke/Tate. The point is that if $K$ isn't CM then you have far less choice about what you can do at infinity because of units. Kevin – user30035 Feb 14 '13 at 20:46

There is a more down-to-earth definition. A newform $f=\sum_{n=1}^\infty a_n q^n$ of level $N$ and weight $k$ has CM if there is a quadratic imaginary field $K$ such that $a_p=0$ as soon as $p$ is a prime which is inert in $K$. The field $K$ is then unique (if the weight $k \geq 2$), and one says that $f$ has CM by K.

A quick way to see the uniqueness of $K$, as well as other basic properties, is to consider the $\ell$-adic ($\ell$-an auxiliary prime) Galois representation of dimension 2 attached to $f$ constructed by Deligne. If $\rho: G_{\mathbb Q} \rightarrow GL_2(\bar {\bf Q}_\ell)$ is that representation, one has tr $ \rho (Frob_p) = a_p$ (Eichler-Shimura) for all prime $p$ not dividing $N\ell$ (and $\rho$ is unramified at these primes, I should have said first). So for $p$ inert in $K$, tr $\rho (Frob_p) =0$, hence we deduce by Chebotarev and a little thought that $tr \rho=0$ on the complement on $G_K$ in $G_{\mathbb Q}$, and then by computing the hermitian product of the character of $\rho$ with itself, that the restriction of $G_{\mathbb Q}$ to $G_K$ is reducible, hence that by Frobenius reciprocity that $\rho$ is induced from a character of $G_K$. Again some elementary group theory/representation theory tells you that there is a unique subgroup $G'$ of index $2$ in $G_{\mathbb Q}$ such that $\rho_{|G'}$ is reducible, except when the projective image of $\rho$ in $K_4=(\mathbb Z/2)^2$, which is excluded because for $k \geq 2$ the projective image of $\rho$ is infinite. Hence the uniqueness of $K$, and many information gotten in the way on $\rho$. In weight $k=1$, the theory is roughly the same except from the very special modular forms whose projective image of $\rho$ is $K_4$, which have CM by two quadratic imaginary fields $K$ and $K'$, and also by a third field $K''$ with is quadratic real, in the sense that $\rho_{|G_{K''}}$ is also reducible (but then we say that $f$ has RM by $K''$, not CM).

share|cite|improve this answer

One indirect way to define it would be that it is the Inverse Mellin tranform of the Hasse-Weil L-function of an elliptic curve with complex multiplication.

For more direct things look at the references suggested on pg. 118 and pg.166ff. in Shimura's "Abelian Varities with CM"

share|cite|improve this answer
Is that also true for modular forms of weight $>2$? – mod78 Feb 13 '13 at 9:21
Ah okay, my answer applies only to weight 2 things. I leave it though, since Olivier's perfect anwer addresses mine. – Marc Palm Feb 13 '13 at 9:56

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.