10
$\begingroup$

Let $\chi$ be a real nonprincipal Dirichlet's character modulo $m$.

In my answer to the question on $L(1,\chi)$, I explain a trick for showing that $L(1,\chi)>0$ on the simplest examples of the real characters modulo 3 and 4. The proof goes as follows: one takes $$ f(x)=\sum_{n=1}^\infty\chi(n)x^n=\frac1{1-x^m}\sum_{j=1}^{m-1}\chi(j)x^j $$ and uses Abel's theorem to write $$ L(1,\chi)=\int_0^1f(x)dx; $$ since the corresponding function $f(x)$ is positive on $(0,1)$, the latter integral has to be positive.

Clearly, $1-x^m>0$ on $(0,1)$, so that the required positivity of $f(x)$ reduces to the positivity of the polynomial $$ g_\chi(x)=\sum_{n=1}^{m-1}\chi(n)x^n $$ on $(0,1)$. Trying to verify on how generalizable is this method for $m>3$, I was quite surprised to see that it works perfectly further; for example, $$ g(x)=x(1-x)(1-x^2)>0 \quad\text{if } m=5 $$ or $$ g(x)=x(1-x)(1+x^2+2x^3+3x^4+2x^5+x^6+x^8)>0 \quad\text{if } m=11. $$ Honestly saying, the positivity is not so obvious in many other examples (for example, $m=19$) but nevertheless it is always holds for small values $m\le30$.

Question. Given an integer $m>2$ and a real nonprincipal character $\chi$ modulo $m$, is it true that $g_\chi(x)>0$ for $x\in(0,1)$? If not, are there (in)finitely many $m$ for which the positivity does not take place? Is the above strategy for showing $L(1,\chi)\ne0$ discussed in the literature?

$\endgroup$

2 Answers 2

5
$\begingroup$

These are called Fekete polynomials, and you can find out a great deal about them here. Unfortunately they tend to have lots of real zeros in $(0,1)$ when $m$ is large.

$\endgroup$
8
  • $\begingroup$ Thanks, David! Never heard of them... Feteke or Fekete? And is $L(1,\chi)>0$ for some $m$? $\endgroup$ May 26, 2010 at 1:00
  • $\begingroup$ fekete, sorry! I'm not sure I understand your second question. $\endgroup$ May 26, 2010 at 1:01
  • 1
    $\begingroup$ Oh, no, the inequality $L(1,\chi)>0$ for $\chi$ real goes back to Dirichlet, and is an immediate consequence of his famous class number formula. $\endgroup$ May 26, 2010 at 1:46
  • 1
    $\begingroup$ Wadim and David: there's an easier reason why L(1,chi) can't be negative for real nontrivial chi: the function L(s,chi) for s > 0 is obviously real-valued. As s --> infty it tends to 1, which is positive, and we know L(s,chi) is nonzero for s > 1 by the Euler product. Therefore by continuity L(s,chi) > 0 for s > 1, hence by taking a limit from the right L(1,chi) is definitely not negative. It's not 0 either (harder!), so L(1,chi) > 0. $\endgroup$
    – KConrad
    May 26, 2010 at 2:22
  • 1
    $\begingroup$ It is not the same, but Chowla then Rosser chased around vaguely similar ideas (weighted sums) to show that $L(s,\chi)>0$ for real $s>0$ for various real $\chi$. Chowla: matwbn.icm.edu.pl/ksiazki/aa/aa1/aa119.pdf Rosser: ams.org/journals/bull/1949-55-10/S0002-9904-1949-09306-0/… $\endgroup$
    – Junkie
    May 26, 2010 at 7:17
5
$\begingroup$

As David pointed out, it is known that for large m these polynomials have many zeroes. This is very unfortunate for the following reason: If you assume $g_{\chi}$ is non-negative then it follows by Mellin inversion that the L-function $L(s,\chi)$ can not have a Siegel zero (or, more generally, any zero on the positive real axis).

$\endgroup$
1
  • $\begingroup$ Thanks, Marc! The heuristics is already indicated in David's reference, but only for the Legendre symbol. I acknowledge your contribution by upvoting. $\endgroup$ May 26, 2010 at 1:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.