Take the 2-minute tour ×

I understand that the Mellin transform of a modular form is expected to satisfy RH when it is an eigenform of all Hecke operators, in which case it has an Euler product. Now about when the form is not an eigenform: Is it known a case where the zeros are all in the critical strip?

share|improve this question

2 Answers 2

up vote 3 down vote accepted

When it is not a Hecke-Eigen form, the Hecke L-series connected with the modular form does not have an Euler product. However it can still be written as a linear combination of Hecke L-series that have Euler-products. Thus the situation will resemble the case of linear combinations of Dirichlet L-series. In particular we can use joint Voronin universality to obtain $\gg T$ zeroes in any strip $1/2<\sigma_1<\Re(s)<\sigma_2 < 1$, $-T < \Im(s)< T.$ These results follows from the analytic properties of the Rankin-Selberg zeta-function (which gives "independence results" for Hecke eigenvalues of primes attached to different cusp forms). You should be able to find results of this kind in Steuding's SLN "Value Distribution of L-functions".

Also the results of Davenport-Heilbronn for the Hurwitz zeta-function can be proved, i.e. For any $\epsilon>0$ there exists $\gg T$ zeroes with $1<\Re(s)<1+\epsilon$ and $-T < \Im(s) < T $.

Stronger results corresponding to results of Karatsuba, Bombieri, Hejhal and Selberg for Dirichlet L-function that holds close to the critical line likely also holds. I think the russian school (Irina Rezvyakova) has proved results in this direction.

share|improve this answer
Am I right that you only give an argument for RH is not true for that kind of function? This is not the question. –  Marc Palm Jan 14 '13 at 15:09
In the first section I state that. In the second section I mention the analogue of the Davenport-Heilbronn theorem that there are zeroes for Re(s)>1. It should follow in a similar way as the case of the Hurwitz zeta-function for rational parameter. –  Johan Andersson Jan 14 '13 at 15:13
+1, ah okay thanks. –  Marc Palm Jan 14 '13 at 15:14

Perhaps I don't understand the question, but in its current form the answer is no.

A general modular form of fixed weight will be a linear combination of Hecke eigenforms of that weight. The Gamma factors will imply that there are trivial zeros outside the critical stripe.

But I guess you have an L-function with symmetric functional equation in mind? The Mellin transform is a linear operator, so the result is a linear combination of Hecke L functions times a Gamma factors. Assuming that you find some combination non-vanishing out the complement of closure of the critical stripe via some estimates I don't know, you will need a good argument for non-vanishing on the boundary of the critical stripe, where the function is expected to be universal.

What is the motivation for the question?

share|improve this answer
I was thinking if the following may be true: if L is the Mellin transform of a modular form , and L has all its zeros in the critical strip, then the modular form is an eigenform. –  user30637 Jan 14 '13 at 16:51
Okay, that it was Johan Andersson's answer seems to imply. –  Marc Palm Jan 14 '13 at 17:10

Your Answer


