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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$\begingroup$ Clearly there needs some amount of rigorisation here. There are infinite series with rational values that sum to the zeroes. There are a number of other infinite processes that may be considered fairly well known. What is meant here by explicit expression? Which constants are well known? I remember a competition question that involved line intersections in a complicated curve where I had to ask the professors if we needed to prove the Jordan curve theorem or could assume it. We had not been taught it formally yet. It pays to require clarity. $\endgroup$– ex0du5Commented Dec 12, 2013 at 21:52
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2 Answers
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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 \begin{equation} L(s)= \prod_{p \, {\rm prime}} F_p(p^{-s})^{-1}, \end{equation} where $F_p$ is a polynomial of the form \begin{equation} F_p(z)=1-a_p z + \cdots + (-1)^d\chi(p) z^d . \end{equation}
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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$\begingroup$ 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? $\endgroup$ Commented Mar 18, 2013 at 18:25
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$\begingroup$ 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?" $\endgroup$ Commented Mar 18, 2013 at 18:33
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1$\begingroup$ 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. $\endgroup$ Commented Mar 18, 2013 at 18:54
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$\begingroup$ 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. $\endgroup$ Commented Mar 18, 2013 at 19:06
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$\begingroup$ 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? $\endgroup$ Commented Mar 18, 2013 at 19:15
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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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