Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

A little background: Let $f(z)=\sum_{n=1}^{\infty} a(n) e^{2\pi i nz}$ be a classical holomorphic cuspidal eigenform on $\Gamma_1(N)$, of weight $k \geq 2$ normalized with $a(1)=1$. The Ramanujan conjecture is the assertion that $|a(n)| \leq n^{\frac{k-1}{2}}d(n)$. This is a theorem, due to several people, but the main steps in its proof are the following two:

  1. Show that $f$ can be "realized" in the etale cohomology of a suitable variety. In fact $f$ can be found in $H^{k-1}$ of a $k-2$-fold self-product of the universal elliptic curve over $X_1(N)$. This step is due to Eichler, Igusa, Kuga/Sato, and Deligne (who showed how to desingularize the variety). This reduces the Ramanujan conjecture to the Weil conjectures.

  2. Prove the Weil conjectures. This is of course due originally to Deligne, although if I understand correctly, there are now several other proofs (a p-adic proof by Kedlaya, etc.)

Now, there is another approach to the Ramanujan conjecture, essentially through Langlands functoriality. In particular, if we knew for all $n$ that the L-functions $L(s,\mathrm{sym}^n f)$ were holomorphic and nonvanishing in the halfplane $Re(s)\geq 1$, the Ramanujan conjecture would follow. This observation is, I believe, due to Serre Langlands.

Nowadays these analytic properties are known, by the potential modularity and modularity lifting results of Barnet-Lamb/Clozel/Gee/Geraghty/Harris/Shephard-Barron/Taylor. However, the proofs of potential modularity and modularity lifting seem to utilize the Ramanujan conjecture. Hence my question:

Can the recent proofs of potential modularity for symmetric powers of $GL2$ modular forms be modified so they do not assume the Ramanujan conjecture, hence giving a new proof of the Ramanujan conjecture?

(If I got any of the history or attributions wrong, please correct me!)

share|improve this question
I think that Langlands claims the deduction of Ramanujan from functoriality for symmetric powers to be his own. One could also note that the same argument appears in Deligne's proof of the Riemann hypothesis. –  Emerton May 10 '10 at 15:48
Thanks! Corrected. –  David Hansen May 10 '10 at 16:06
"...another approach to the Ramanujan conjecture, essentially through Langlands functoriality. ... if we knew for all $n$ that the $L$-functions $L(sym^n f)$ were holomorphic and nonvanishing in the halfplane $Re(s)\ge 1$, the Ramanujan conjecture [follows]..." I don't know about functorality - a variant of the weaker claim is in a paper of Ogg springerlink.com/content/k24u6q718u3v770w Ogg, A. P. A remark on the Sato-Tate conjecture. Invent. Math. 9 1969/1970 198--200. Hs shows a holomorphic continuation past 1/2-line (follows from functorality?) implies no zeros on 1-line. –  Junkie May 11 '10 at 2:42
Actually, I am now confused by your normalization of the half-line. Are the coefficients bounded by 1, so that the symmetry line is $Re(s)=1/2$, and the edge is $Re(s)=1$. This is normalization that Ogg uses. –  Junkie May 11 '10 at 2:48
Junkie, yes, that is the normalization I am using. –  David Hansen May 11 '10 at 13:06
show 1 more comment

1 Answer

The claimed analytic properties of $L(s,sym^n f)$ surely imply Sato-Tate, and this is due to Serre, but I don't see how they imply Ramanujan. What Langlands observed is that the automorphy of $L(s,sym^n f)$ on GL(n+1) (assumed for all $n$) implies Ramanujan, but of course analytically this is a much stronger assumption (as far as we know), e.g. all these $L$-functions and most of their twists should be entire with a functional equation (cf. the converse theorems of Cogdell and Piatetski-Shapiro).

share|improve this answer
On the other hand, the following is clear: if in a fixed right half-plane each $L(s,sym^n f)$ is given by an absolutely convergent Euler-product, then Ramanujan holds. This is because the assumption implies that all the Euler factors are holomorphic in a fixed right half-plane which is equivalent to Ramanujan. It is also worth mentioning that an elaboration of this idea (pole of an Euler factor or gamma factor forces a zero of the remaining partial $L$-function), originally due to Serre, lead to nontrivial bounds for cusp forms on GL(n), see the work of Luo-Rudnick-Sarnak. –  GH from MO Jan 2 '11 at 12:55
To deduce Ramanujan: You need to write $L(s,\mathrm{sym}^n(f)\otimes \mathrm{sym}^n(f))$ as a product of $L(s,\mathrm{sym}^j(f))$'s for various j's (use Clebsch-Gordon) and then appeal to the fact that a Dirichlet series with nonnegative coefficients converges absolutely up to its first pole. –  David Hansen Jan 2 '11 at 22:12
That's a very good point, thanks! Does it also work when the nebentypus of $f$ is nontrivial? –  GH from MO Jan 2 '11 at 23:30
Sure, just argue with $L(s,\mathrm{sym}^n(f) \otimes \mathrm{sym}^n(\overline{f}))$ instead, and use the fact that $\overline{f}=f\otimes \overline{\chi}$ where $\chi$ is the nebentypus. –  David Hansen Jan 17 '11 at 22:18
OK, but then we need the relevant properties for $L(s,\text{sym}^n f\otimes\chi^m)$ for all $n$ and many $m$'s, right? I am not familiar with Clebsch-Gordon, and I don't know if the recent work on potential modularity covers these character twists as well. –  GH from MO Jan 17 '11 at 22:33
add comment

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.