# Is there an explicit expression for the imaginary part of some non-trivial zero of zeta,

Is there an explicit expression for the imaginary part of some non-trivial zero of zeta, in terms of well-known constants, such as say $\gamma$ or $\pi$ say ?

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The imaginary part of the first zero is a well-known constant. –  GH from MO Mar 15 '13 at 11:06
Awfully vexed at the previous comment. –  Captain Darling Mar 17 '13 at 22:40
@Captain: My comment was as honest as it could be. –  GH from MO Mar 19 '13 at 12:22
@Greg: Regarding the student, if the Monotone Sequence Theorem was proved in class, then his/her answer is appropriate. Regarding me, I did not (and still do not) understand the question, and I believe it is not my fault. A question should be clear and well-formulated, especially if it is meant at the research level. I hoped that Captain Darling would make an effort. Finally, the imaginary part of the first zeta-zero is a well-known constant. –  GH from MO Mar 29 '13 at 11:57
I never made any statement about clarity being bad. I merely pointed out that GH from MO's original comment could not possibly be useful to any imaginable version of the question asked, and therefore could only be reasonably construed as insulting. Indeed we have the OP's direct testimony that this is, in fact, how it came across. –  Greg Martin Dec 13 '13 at 0:50

Write $\rho = \frac12 + i \gamma$ for a nontrivial zero of a primitive L-function. ("Primitive" means that it can't be written as the product of other L-functions.)

It is generally believed that:

a) If $\gamma\not=0$ then $\gamma$ is transcendental.

b) If $\gamma\not=0$ then $\gamma$ is algebraically independent of every well-known constant and every other zero of every primitive L-function (except when the L-function has real coefficients, in which case $\frac12 - i \gamma$ is also a zero).

As far as I know, nobody has any clue how to prove these conjectures.

Clarification added later: what definition of L-function are we using?

Greg Martin's comment (below) refers to $L(s+ i y)$ where $L(s)$ is an L-function and $y$ is real. While it is true that for some definitions of "L-function" the set of L-functions is closed under that operation, that is not what I intended.

For the L-functions in my answer above, the Euler product axiom can be written as:

There is a Dirichlet character $\chi$, the "central character" of the L-function, such that $$L(s)= \prod_{p \, {\rm prime}} F_p(p^{-s})^{-1},$$ where $F_p$ is a polynomial of the form $$F_p(z)=1-a_p z + \cdots + (-1)^d\chi(p) z^d .$$

Here $d$ is the degree of the L-function. Note that I have normalized the L-function so that the functional equation relates $s$ to $1-s$.

All known L-functions satisfy that axiom, and this formulation tells you how to select the distinguished member of the family $L(s+i y)$.

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Clarification question: does "primitive" exclude, or does it include, non-abelian Artin L-functions with Artin-Brauer factorizations into ratios of products? Or, e.g., is this contingent on unproven things about their actual holomorphy? –  paul garrett Mar 18 '13 at 18:25
At at a colloquium I once heard the comment, "Can we even prove that all the zeros of every primitive L-function are not all rational multiples of each other?" –  Eric Naslund Mar 18 '13 at 18:33
Indeed, the comment Eric alludes to was made (more or less) by Lior Silberman. One has to be careful, though, since $L(s+ic)$ is also a primitive $L$-function for any real $c$ whenever $L$ is a primitive $L$-function without a pole at $s=1$. So one needs to make more modest formulations that don't fall to this loophole: the set of Dirichlet $L$-functions, for example. –  Greg Martin Mar 18 '13 at 18:54
If an Artin L-function was genuinely a ratio, and not a product, of other L-functions (in other words, if it had infinitely many poles) then it does not deserve to be called an L-function! For this discussion I think it makes sense to assume all reasonable conjectures. –  David Farmer Mar 18 '13 at 19:06
I'm totally happy to assume all reasonable conjectures! :) But what I mean is, whether it is presumed (or known?) that an express for an Artin L-function as ratio of product of abelian L-functions, which has no poles at all (say), has no denominator at all (in a reasonable sense)? That is, that the lack of poles will never be due to cancellation of zeros of numerator and denominator? –  paul garrett Mar 18 '13 at 19:15

There is no explicit value for the imaginary part of the n-th zero. However it satisfies a simple transcendental equation for each n, whose solution is well approximated by the Lambert function. See LeClair and Franca on arXiv, math.NT

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