Distinct simple zeros of Dirichlet L-functions - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T02:01:51Z http://mathoverflow.net/feeds/question/75960 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/75960/distinct-simple-zeros-of-dirichlet-l-functions Distinct simple zeros of Dirichlet L-functions rlo 2011-09-20T15:21:56Z 2012-03-02T17:53:17Z <p>Given a finite set of distinct primitive Dirichlet characters, $\chi_1, \dots, \chi_r$, is it known that the product of the L-functions, $$L(s):=\prod_{i=1}^r L(s,\chi_i),$$ has a simple zero? It's conjectured that all of the zeros of the $L(s,\chi_i)$ are distinct and simple, but I don't know what is known unconditionally except in the case that $r\leq 2$.</p> <p>It's known that each $L(s,\chi)$ has infinitely many simple zeros (in fact, a positive proportion of its zeros are simple and lie on the half-line), which immediately answers the question in the case that $r=1$. If $r=2$, the answer to my question seems to be provided by work of Conrey, Ghosh, and Gonek (Simple zeros of the zeta function of a quadratic number field, I. Invent. Math., MR0860683). They prove that the Dedekind zeta function associated to a quadratic field has $\gg T^{6/11}$ simple zeros with imaginary part up to $T$, all arising from $\zeta(s)$. It appears that their method can be adapted to consider the product of any two Dirichlet L-functions, and this is confirmed by a statement of Bombieri and Perelli (Distinct zeros of L-functions. Acta Arith., MR1611193), who additionally write that $r=2$ is the limit of the Conrey-Ghosh-Gonek method.</p> <p>I have not been able to find any work which applies to my question in the case that $r\geq 3$. The paper of Bombieri and Perelli referenced above discusses counting distinct zeros of more general L-functions, but it is not obvious to me how to isolate the simple zeros in their argument.</p> <p>I also don't know if the fact that I'm only looking for a single simple zero of $L(s)$ saves me anything. That is, I don't know of techniques that detect the existence of such a zero without proving that there are an infinite number. Nevertheless, this could prove to be useful, since it seems entirely possible that showing that $L(s)$ has infinitely many simple zeros when $r\geq 3$ could be quite difficult.</p> http://mathoverflow.net/questions/75960/distinct-simple-zeros-of-dirichlet-l-functions/90061#90061 Answer by David Farmer for Distinct simple zeros of Dirichlet L-functions David Farmer 2012-03-02T17:53:17Z 2012-03-02T17:53:17Z <p>You probably guessed this from the lack of responses, but the answer is that there is no hope of making progress on this question using current methods.</p> <p>One measure of the complexity of an L-function is its degree, where the Riemann zeta function and Dirichlet L-functions have degree 1, the L-function of a holomorphic cusp form has degree 2, the standard L-function of a GL(n) automorphic form has degree n, etc. The precise definition of degree is the number of $\Gamma$-factors in the functional equation, where a $\Gamma(s+A)$ counts as two $\Gamma$-factors.</p> <p>These days we have very few tools for dealing with degree 3 and higher. Your product of Dirichlet L-functions is like one degree $r$ L-function, and so you are stuck once $r$ is bigger than 2. In particular, the fact it is a product doesn't seem to help much.</p> <p>It also doesn't help that you want only one simple zero, unless you have a particular explicit case in mind, in which case you can show it by direct calculation.</p>