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Hi I am curious about the growth of coefficients of cusp forms. I am aware of Ramanujan-Petersson conjecture/theorem in general terms but I was hoping for a more detailed description of precise statements on the growth of all the cusp form coefficients and whether these were tight. Also, for the Fourier coefficients of Hasse-Weil L-series of elliptic curves (meaning what can we say about these without making use to modularity). Any good references would be very useful also. Thanks!

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You probably mean holomorphic forms, but it's worth commenting that this paper on the arXiv seems to claim to prove the Ramanujan conjecture for Maass forms. – Stopple Oct 28 '10 at 15:20
Does anyone have any ideas about the correctness of this paper? I haven't looked at it but it seems to have been revised every month, which seems strange. – Kimball Oct 28 '10 at 16:15
I myself have some confusion about the correctness... When the first version(s) appeared several months ago, I looked at them in some detail, and was unable to see how the same argument wouldn't apply to situations where the same conclusion is not valid... Perhaps something was implicit that I didn't see. – paul garrett Jun 21 '11 at 17:03

It's worth distinguishing between the prime coefficients $a_p$, and the coefficients $a_n$ for general $n$. Let's look at $a_p$ first.

Firstly: for elliptic curves, it is fairly easy and elementary to prove that $|a_p| < 2\sqrt{p}$ (Hasse's inequality). This should be in any decent textbook e.g. Silverman. And this is the best bound you're going to get, because for any fixed elliptic curve the set $\{ a_p / (2\sqrt{p}) : p\ prime\}$ is dense in $(-1, 1)$, and in fact obeys a very specific distribution. (This distribution depends on whether or not E has complex multiplication. If E does not have CM, it the Sato-Tate distribution, a long-standing conjecture recently settled by Barnet-Lamb, Geraghty, Harris and Taylor. In the CM case you get a different, simpler distribution; thanks KConrad for pointing this out.)

For general modular eigenforms of weight $k$, the "right" bound is $|a_p| \le 2p^{(k-1)/2}$, but this is a very deep theorem (it follows from Deligne's work on the Weil conjectures). There is an easy elementary argument that gives $a_p = O(p^{k/2})$: this is in Miyake's book (corollary 2.1.6 if I recall correctly). By purely analytic methods you can push this a bit further, e.g. Rankin proved $a_p = O(p^{k/2 - 1/5})$ if I remember correctly, but you can't get the "right" bound this way.

For general $n$, the relations giving the $a_n$ in terms of the $a_p$ mean that the Deligne bound $|a_p| \le 2p^{(k-1)/2}$ turns into something like $a_n \le n^{(k-1)/2} d(n)$, where $d(n)$ is the number of divisors of $n$; this is $O(n^{(k-1)/2 + \epsilon})$ for any $\epsilon > 0$ (but it is not $O(n^{(k-1)/2})$).

(EDIT: For Hasse's inequality it's obvious that we have strict inequality $|a_p| < 2\sqrt{p}$ simply because $a_p \in \mathbf{Z}$ and $2\sqrt{p}$ isn't. But for general modular forms one only gets the non-strict inequality $|a_p| \le 2p^{(k-1)/2}$, not a strict inequality as I originally claimed (thanks to François for pointing this out); strict inequality is conjectured to hold if $k > 1$, and Coleman and Edixhoven have shown that this would follow from some standard conjectures on varieties over finite fields. If $k = 1$ then in fact $a_p = 2$ for a positive density set of primes, so equality definitely can occur in this case.)

(EDIT: Just to emphasize, if you have an elliptic curve over $\mathbf{Q}$, the Hasse bound $|a_p| < 2\sqrt{p}$ "follows from" the Deligne bound and the fact that $E$ is modular, but this would be a ridiculously laborious way of proving that: the direct elementary proof of Hasse's inequality is vastly easier than using modularity. In fact (some parts of) Deligne's proof can be interpreted as "trying to adapt Hasse's proof to a general algebraic variety", so the flow of information here is the other way.)

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For general $n$, one has in fact the slightly more precise estimation $|a_n| \leq d(n) n^{(k-1)/2}$ (this follows easily from multiplicativity and from the linear recurrence relation satisfied by the $a_{p^k}$ for $k \geq 0$). – François Brunault Oct 28 '10 at 9:00
By the way, has the strict inequality $|a_p|<2p^{(k-1)/2}$ been proved in general? Because this is equivalent to the semi-simplicity of Frobenius, which is still conjectural, isn't it ? – François Brunault Oct 28 '10 at 9:02
Sorry, you are quite right, the strict upper bound is easy in Hasse's inequality because $a_p \in \mathbf{Z}$ but is not known in general. I will edit my answer to take this into account. – David Loeffler Oct 28 '10 at 10:49
Ok. For the "precise" bound for $a_n$, I have given some details in my answer below. – François Brunault Oct 28 '10 at 11:25
The reference to Sato-Tate doesn't work when E has CM. – KConrad Oct 28 '10 at 15:01

Let me focus on how to deduce a bound for all coefficients $a_n$ assuming the Deligne bound on $a_p$ (this is a standard argument).

Let $f$ be a newform of weight $k$, level $N$ and Nebentypus character $\psi$ modulo $N$. Assume the Deligne bound $|a_p| \leq 2p^{(k-1)/2}$, and let us prove the inequality $|a_n| \leq d(n) n^{(k-1)/2}$ for every $n \geq 1$ (here $d(n)$ is the number of positive divisors of $n$).

Since both sides are multiplicative in $n$, it suffices to consider the case $n=p^m$, where $p$ is prime. Put $u_m=a_{p^m}$. By looking at the Euler factor of $f$ at $p$, we know the following formal identity :

$\sum_{m \geq 0} u_m X^m = \frac{1}{1-a_p X +\psi(p) p^{k-1} X^2}$.

Now there are two main cases :

1) $p$ divides $N$. Then $\psi(p)=0$ and $u_m = a_p^m$ for every $m \geq 0$. It can be shown, using purely analytical arguments, that $|a_p| \leq p^{(k-1)/2}$ (see Theorem 3 in Li's article "Newforms and functional equations", Math. Ann., 1975). Thus we get $|u_m| \leq p^{m(k-1)/2}$ which gives the desired inequality (and in fact, a stronger one).

2) $p$ doesn't divide $N$. Put $1-a_p X +\psi(p) p^{k-1} X^2 = (1-\alpha X)(1-\beta X)$. In the case $k \geq 2$, the Weil conjectures proved by Deligne imply that $|\alpha|=|\beta|=p^{(k-1)/2}$ (this still holds if $k=1$, by a theorem of Deligne and Serre). There are two subcases :

2a) $\alpha \neq \beta$. Solving the linear recurrence relation satisfied by $u_m$ gives

$u_m = \frac{\alpha^{m+1}-\beta^{m+1}}{\alpha-\beta} = \alpha^m+\alpha^{m-1} \beta + \ldots + \beta^m$.

In particular $|u_m| \leq (m+1) p^{m(k-1)/2}$ which is what we want.

2b) $\alpha = \beta = \frac{a_p}{2}$. In this case $u_m = (m+1) \alpha^m$ for every $m \geq 0$, and the inequality also follows.

Remarks : It is conjectured that case 2b never happens if $k \geq 2$ (semi-simplicity of crystalline Frobenius). Note also that in case 2a, we can write $u_m = \beta^m (1+\lambda+ \ldots + \lambda^m)$, where $\lambda := \alpha/\beta$ satisfies $|\lambda|=1$ and $\lambda \neq 1$. Thus in fact we get a bound of the form $|u_m| \leq C \cdot p^{m(k-1)/2}$, where the constant $C$ depends only on the argument of $\alpha/\beta$. Finally, note that we also have the strict bound $|a_p|<2p^{(k-1)/2}$ in case 2a.

EDIT : case 2b can indeed happen in the case $k=1$ (thanks to David for pointing this out).

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Thanks for the sketch, it's very useful. In the second paragraph, you mean to say that $d(n)$ is the number of positive divisors of $n$, right? – aghitza Sep 28 '15 at 11:49
Yes -- corrected, thanks. – François Brunault Sep 29 '15 at 12:10

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