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 $r\leq 2$.

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.

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.

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.


1 Answer 1


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.

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.

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.

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.

  • $\begingroup$ That is the conclusion I reached, yeah, especially after some conversations with people and a couple of failed attempts to prove something. In the application I had in mind (which actually required Artin L-functions is addition to Dirichlet), I ended up just using that it is simple to check in each case. $\endgroup$
    – rlo
    Mar 4, 2012 at 17:00
  • $\begingroup$ The one possible scenario in which one might be able to say something here is if there happens to be a sequence of Siegel zeroes existing in our universe, as this can be used to repel the zeroes of various L-functions from each other (see my previous comment at mathoverflow.net/a/125301/766 ). Of course, we don't believe that Siegel zeroes actually exist, so this implication may end up being vacuous anyway. $\endgroup$
    – Terry Tao
    Jul 3, 2021 at 16:34

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