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Can you give a non-trivial example of an integer weight cusp form which does not lie in the old subspace and it has $a_p=0$ for all primes $p$?

If such a form cannot exist then why?

edited May 21 2010 at 5:56

asked May 21 2010 at 5:50

Idoneal
1,169●6●11

1

Naïve heuristics suggest that no such modular form exists, and surely no such eigenform exists, as you very well know. I thought one could use results about eigenform as an input to prove the general case, but I failed. The (first) difficulty I encountered is that for an arbitrary modular forms, it is not obvious how to relate a_p with the Hecke operator T_p. – Olivier May 21 2010 at 9:27

Can you elaborate on your first comment? What heuristics suggest no such modular form exist? – Idoneal May 21 2010 at 10:59

A common heuristic is to regard the coefficients a_n as random with respect to congruence. So a_p already has a very slight chance of being divisible by p, let alone zero. Like I said, this is very naïve and just a suggestion. Outside of eigenforms, all I seem to be able to do is give rough estimates of the proportion of non-zero coefficients. They wouldn't tell you anything about coefficients at primes. – Olivier May 21 2010 at 11:42

That said, if your question has some definite purpose, perhaps it could help if you explained how such a modular forms would (or would not) help. Presumably, if you have a construction that produces a certain output specifically at primes, this construction tells you something about a_p, and this is exactly what I lack in my (very amateur) understanding of the problem. – Olivier May 21 2010 at 11:46

Idoneal, please add some context to your question. – S. Carnahan♦ May 21 2010 at 12:49

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David Hansen · Accepted Answer · 2010-05-21 15:21:31Z UTC

11

Write $f=\sum c_i f_i$ as a sum over new eigenforms. Your condition is thus equivalent to $\sum c_i \lambda_i(p)=0$ for all $p$. Taking the absolute value squared of this and summing over $p\leq X$ gives

$0=\sum_{i,j}c_i \overline{c_j} \sum_{p\leq X} \lambda_i(p)\overline{\lambda_j(p)}$.

By the pnt for Rankin-Selberg L-functions, the inner sum over primes is $\sim X (\log{X})^{-1}$ if $i=j$, and is $o(X (\log{X})^{-1})$ otherwise. Taking $X$ very large we obtain $0=cX(\log{X})^{-1}+o(X(\log{X})^{-1})$, so contradiction.

answered May 21 2010 at 15:21

David Hansen
8,409●2●18●48

	Very nice David! I wasn't thinking in this direction at all. So in fact you have proved the result for $a_p = 0$ for a set of positive density. – Idoneal May 21 2010 at 15:42
	I'm not sure if that's true - all I've proven is that $\sum_{p \leq X} \|a_p\|^2 \sim cX (\log{X})^{-1}$. It's not obvious to me how to deduce that $a_p$ can't vanish a small but positive percent of the time - all this shows, I think, is that $a_p$ can't vanish a hundred percent of the time! :) – David Hansen May 21 2010 at 15:49
	Yes, you are right. I wasn't careful. In fact for CM forms a_p=0 half the time. By the way, I wonder if there is a way to prove this using linear algebra and the multiplicity one principle as Olivier suggested. – Idoneal May 21 2010 at 16:00

Jonas Kibelbek · Answer 2 · 2010-08-12 01:46:17Z UTC

It is only possible to write f as a sum over Hecke eigenforms, as David does, in a space of congruence modular forms (i.e., forms on a congruence subgroup of SL₂(ℤ) ). On a noncongruence subgroup, the Hecke operators send all genuinely noncongruence forms to 0. (G. Berger, Hecke operators on noncongruence subgroups)

Modular forms with prime Fourier coefficients zero

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