# Definition of CM modular form

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

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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 $F$.

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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).

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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"

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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