8
$\begingroup$

Let $f$ be a newform of level $\Gamma_1(N)$ and character $\chi$ which is not induced by a character mod $N/p$. I learned from these notes by Ribet and Stein that $|a_p|=p^{(k-1)/2}$ where $k$ is the weight of $f$. So I wonder

1, the proof of this statement,

2, is $a_p$ for $p|N$ in the number field $K_f$ of $f$, i.e the number field generated by all $a_q$ for $(q,N)=1$?

Now I see that 2 is true. Why I thought it's wrong is because $p^{(k-1)/2}$ is of degree 2, which will implies in this case $2|[K_f:\mathbb{Q}]$. Seems strange...

$\endgroup$
2
  • $\begingroup$ Should it be $n\nmid N$ in your last sentence? $\endgroup$
    – Olivier
    Commented Oct 25, 2015 at 8:59
  • $\begingroup$ Sorry for confusion. 1 and 2 are only for characters not induced from a character mod $N/p $ $\endgroup$
    – user42690
    Commented Oct 25, 2015 at 14:43

4 Answers 4

5
$\begingroup$

Here is an answer to both questions.

First question. The quoted result is a special case of Theorem 9.1.10 in the Ribet-Stein notes, which in turn is identical to Theorem 3 in Li: Newforms and functional equations (in fact Ribet-Stein emphasize that they follow Li's treatment closely).

As you can see, Li uses classical Atkin-Lehner theory and also refers to an older work of Ogg's from 1969. I am sure the same result could be deduced by looking at the underlying local newform at the prime $p$ in the Kirillov model, see e.g. Schmidt's excellent notes for that purpose. (Also, I realize that Ben Linowitz answered much the same 2 hours ago.)

Second question. Let $S_k^\text{new}(\Gamma_1(N))$ denote the set of primitive newforms in $S_k(\Gamma_1(N))$, and note that this is a finite set. By Proposition 2.7 in Deligne-Serre: Formes modulaires de poids 1, for any $f\in S_k^\text{new}(\Gamma_1(N))$ and for any field automorphism $\sigma$ of $\mathbb{C}$, there exits an $f^\sigma\in S_k(\Gamma_1(N))$ such that $a_p(f^\sigma)=a_p(f)^\sigma$. It follows that for any $f\in S_k^\text{new}(\Gamma_1(N))$, the $a_p(f)$'s generate a number field $K_f$ (of finite degree) over $\mathbb{Q}$.

Now let $S$ be any subset of primes with Dirichlet density strictly less than $1/8$ (e.g. $S$ is finite), and consider the subfield $K_{f,S}$ of $K_f$ generated by the $a_p(f)$'s with $p\not\in S$. I claim that $K_{f,S}=K_f$. Indeed, if $\sigma$ fixes $K_{f,S}$, then $f^\sigma=f$ by Ramakrishnan's version of strong multiplicity one for $\mathrm{GL}_2$, hence $\sigma$ fixes $K_f$ as a whole, and the claim follows by Galois theory. In particular, $K_f$ is generated by the $a_p(f)$'s with $p\nmid N$, and the remaining $a_p(f)$'s with $p\mid N$ lie in this number field as well.

$\endgroup$
5
  • $\begingroup$ Thanks for providing interesting reference. But I'm still interested in the question whether $a_p$ for $p|N$ is in the number field of $f$! I'm just afraid it's trivial and I can not see it! $\endgroup$
    – user42690
    Commented Oct 26, 2015 at 3:41
  • $\begingroup$ I know all these coefficients generate a number field. Indeed we only need to prove the Hecke algebra is a finite $\mathbb{Z}$-algebra. I make question 2 more clear now :D. $\endgroup$
    – user42690
    Commented Oct 26, 2015 at 4:54
  • 2
    $\begingroup$ @user42690: I have now answered both questions. $\endgroup$
    – GH from MO
    Commented Oct 26, 2015 at 10:56
  • 2
    $\begingroup$ Oh! That's it! It is indeed an easy argument. But because I a priori thought 2 is wrong, so I keep asking! Thanks! $\endgroup$
    – user42690
    Commented Oct 26, 2015 at 12:44
  • 3
    $\begingroup$ @user42690: Glad we clarified this. Interestingly, I come from the analytic side, but I only looked at Li's paper recently after reading about it in Deligne-Serre. Note also that the importance of multiplicity one is emphasized by Deligne-Serre, although the cute stronger version of Ramakrishnan's was not available at the time. This is why I like MO: we can learn from each other by asking and answering questions! $\endgroup$
    – GH from MO
    Commented Oct 26, 2015 at 12:53
5
$\begingroup$

I think you can find the proof you want in:

Perhaps you can find it more thoroughly explained in Shimura's book, or on more concrete notes on Atkin-Lehner.

As for your second question, the first part is way too broad, see for example Ken Ono's "The Web of Modularity: Arithmetic of the Coefficients of Modular Forms and q-series".

Not so sure anymore about the number field generated by those coefficients.

$\endgroup$
1
  • $\begingroup$ I don't think the statement about eigenvalues in your second paragraph is true, it depends on characters. And multiplicity one (the version I know) only says that $a_p$ is determined by $a_q$ with $q|N$. Oh I see, you only mean the trivial character right? Maybe it's trivial, could you give a one line proof? $\endgroup$
    – user42690
    Commented Oct 25, 2015 at 15:06
3
$\begingroup$

A proof that $|a_p|=p^\frac{k-1}{2}$ in the situation you are considering is given in the proof of Theorem 3 of Winnie Li's paper Newforms and Functional Equations, where it is deduced from a result of Ogg. (She also considers the case in which the character of the newform in question is also a character mod $N/p$.)

$\endgroup$
2
$\begingroup$

This is really a comment, but the system won't let me post one. I think there's a confusion and that the absolute value is $p^((k-2)/2)$ where $k$ is the weight. For example there's a level $\Gamma_0 (2)$ newform of weight 8, $f=(\eta(z)\eta(2z))^8$, with $f^3 =\delta(z)\delta(2z)=q^3-24q^4+...$ Then $f=q-8q^2+...$, the $U_2$ eigenvalue is -8, and the exponent is $(8-2)/2$.

EDIT: The confusion was mine--I missed the condition on the character. My example illustrated case (iii) of Li's Theorem 3, while the question related to case (ii). I'd be grateful for an explicit example from the databases of case (ii) to further clear up my confusion.

$\endgroup$
1
  • 2
    $\begingroup$ Here's an example. There's a normalized newform of level 21 and weight 2, with character primitive mod 21, such that $a_2(f) = 0$ and $a_3(f) = (-3-\sqrt{-3})/2$. It's easy to see that the absolute value of $a_3(f)$ is $\sqrt{3}$ as expected. $\endgroup$ Commented Oct 26, 2015 at 16:24

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .