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Say $f$ is a newform of weight $k$ and level $\Gamma_1(N)$. $f$ is called CM if, for example, there is an imaginary quadratic field $K$ such that for all $p\nmid N$ which are inert in $K$, the $p$th Fourier coefficient $a_p$ of $f$ is 0. (Ribet's article Galois representations attached to eigenforms with Nebentypus is a nice reference for this material). Specific examples include the newforms attached to CM elliptic curves. All examples arise as inductions of algebraic Hecke characters of $K$ of type $(k-1,0)$.

Is there an effective bound in terms of $k$ and $N$ (or other basic invariants of $f$) on how many $a_p$ you have to check to know whether or not $f$ is CM? Or, is there an effective bound on the discriminant of the associated $K$ and the conductor of the associated algebraic Hecke character? What if we assume GRH?

A related question was asked here by Mike Bennett, but no answer has been given.

My motivation is simply to be able to computationally check if a given newform is CM using, say, SAGE. Thanks.

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  • $\begingroup$ Well, the form itself is completely determined by the first $O(kN^3)$ (or something like that) coefficients... $\endgroup$
    – Dror Speiser
    Commented Sep 18, 2011 at 19:08
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    $\begingroup$ Assuming GRH for Rankin-Selberg L-functions allows you to distinguish a modular form from very few terms in its Fourier expansion, on the order of $(\log k q)^2$. This should be in Chapter 5 of Iwaniec-Kowalski but I don't have the book on hand. $\endgroup$
    – Matt Young
    Commented Sep 18, 2011 at 20:12
  • $\begingroup$ @Matt Young: Yup, it's there, in section 5.8 (where your $q$ is my $N$). There's an absolute constant $C$ lying around, but maybe if I traced back everything I could explicitly compute it. Thanks. $\endgroup$
    – Rob Harron
    Commented Sep 18, 2011 at 20:52

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If the form is CM then it will be isomorphic to a quadratic twist of itself. So I think what I'd do with a form which I suspect is or is not CM is to just twist by all the (finitely many ) possible quadratic characters that could be involved and then to check to see if $f$ is the same as its twist, which one can do by proving that the difference is zero using the standard bound (1+degree of $\omega^k$ on the modular curve).

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  • $\begingroup$ Re the discriminant and conductor: if $N$ is the level of $f$ and $f$ is new, then $N$ is the product of $|d|$ ($d$ the discriminant of $K$) and the norm of the conductor of the grossencharacter. This just comes from basic facts about conductors of representations (if I got them right!) $\endgroup$
    – Kevin Buzzard
    Commented Sep 18, 2011 at 20:02
  • $\begingroup$ Ah, yes, for some reason I got myself confused looking at the analytic conductors, but it's clear on the Galois side. Thanks! I'll just add that a proof of the standard bound you mention is available in Murty's paper Congrences between modular forms available at mast.queensu.ca/~murty/congr.dvi . $\endgroup$
    – Rob Harron
    Commented Sep 18, 2011 at 20:42
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Given a modular form with CM type vanishing behavior of its coefficients $a_p$ (as described in Ribet) it is often not difficult to find a Hecke character whose L-series agrees with those of your given form. Sturm's bound in terms of the weight $k$ and the level $N$ then tells you how many terms you have to consider in order to ensure that the two forms are the same. (See Sturm 1987, On the congruence of modular forms.)

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  • $\begingroup$ Sturm's work seems to only check for congruences between modular forms. Am I missing something? $\endgroup$
    – Rob Harron
    Commented Sep 18, 2011 at 20:45
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    $\begingroup$ "Sturm's Bound" seems now to be a general way of saying "the trivial bound coming from algebraic geometry". It's the same in char 0 and char $p$ (with the same proof), and just says "if a section of a degree $d$ line bundle has a zero of order at least $d+1$ somewhere, it's identically zero". $\endgroup$
    – Kevin Buzzard
    Commented Sep 18, 2011 at 20:47
  • $\begingroup$ Ah, that makes sense. Thanks, again! $\endgroup$
    – Rob Harron
    Commented Sep 18, 2011 at 20:54
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What is actually meant by "If the form is CM then it will be isomorphic to a quadratic twist of itself." (K. Buzzard)?

Because if this should really mean (and the rest of the answer sounds like that) that the form is equal to it's twist then I have the impression that this is not correct. If the conductor $M$ of the character $\chi$ divides the level $N$ of $f$, then a coefficient $a_p$ of $f$ with $p \mid M$ might be non-zero according to Ribet's definition. But after the twist this will be $0$.

I think there are also easy counter-examples. For instance take the unique normalized cusp-form for $\Gamma_0(7)$ and character $\left( \frac{-7}{\cdot} \right)$ the Kronecker character. This form is constructed from a hecke character of $Q(\sqrt{-7})$, as far as I can see. And $a_n = 0$ if $\left( \frac{-7}{n} \right) = -1$. But $a_7 \neq 0$.

What am I missing?

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  • $\begingroup$ So, you start with a normalized newform f and then you twist by some $\chi$ and let $f_\chi$ be the associated newform. Then, $f=f_\chi$ for some $\chi$ if $f$ is CM. $\endgroup$
    – Rob Harron
    Commented Dec 7, 2012 at 3:41
  • $\begingroup$ Aha - thanks! So you mean that if $f=\sum_{n=1}^\infty c(n)q^n$, I don't take the form [ g = \sum_{n=1}^\infty \chi(n)c(n)q^n ] but I write $g$ as $g=f_\chi + g'$, where $f_\chi$ is a newform? This would make perfect sense, of course. (But it is a bit confusing that people write in several places things like $f$ has CM if and only if it is equal to a twist of itself.) $\endgroup$
    – Stephan E.
    Commented Dec 7, 2012 at 13:26

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