6
$\begingroup$

It is well known that the Gauss sum of a Dirichlet character modulo $N$

$$G(\chi)=\sum_{a=1}^N\chi(a)e^{2\pi ia/N}$$

Moreover,$$\vert G(\chi)\vert=\sqrt{N}$$

when $\chi$ is primitive.

Question :If $G(\chi)=-\sqrt{N}$, the root number of L-function associated to the character $\chi$ is -1 or -i, which is not likely to occur. How to rule out such possibilities?

Improve this question
$\endgroup$
2
  • 3
    $\begingroup$ The background to your question is incorrect: the functional equation relates $L(s,\chi)$ and $L(1-s,\overline{\chi})$, so you'd get an automatic zero at $s = 1/2$ from the root number being $-1$ only when $\chi = \overline{\chi}$, meaning $\chi$ is quadratic. For quadratic $\chi$ the root number is 1, so you learn nothing about a possible zero at $s=1/2$. And if $\chi$ is not quadratic then no value of the root number could force $L(1/2,\chi)$ to be $0$ since $\overline{\chi}$ is not the same character as $\chi$. $\endgroup$
    – KConrad
    Jun 29, 2013 at 0:18
  • $\begingroup$ @KConrad: Thanks for your correction. $\endgroup$
    – Y. Zhao
    Jun 29, 2013 at 0:40

2 Answers 2

Reset to default
6
$\begingroup$

Here is a remark, if $\chi$ is a quadratic character. Then there is a remarkable result of Gauss, saying that $$ G(\chi, N)=\begin{cases} \sqrt{N}, \;\text{ if } N\equiv 1 (4) \cr 0, \; \text{ if } N\equiv 2 (4) \cr i\sqrt{N} \; \text{ if } N\equiv 3 (4) \cr (1+i)\sqrt{N} \;\text{ if } N\equiv 0 (4) \end{cases} $$ So this rules out that $G(\chi, N)=-\sqrt{N}$. It took Gauss over $4$ years to prove this. A proof is in section $9.10$ of Tom Apostol's book on analytic number theory.

For characters of higher orders $k>4$, I believe it is an open problem, which roots of unity can occur. If $\chi$ is a real, primitive Dirichlet character, then only $1$ and $i$ can occur (i.e., $\sqrt{N}$ or $i\sqrt{N}$).

Improve this answer
$\endgroup$
1
  • $\begingroup$ The modulo p(p is a prime) case seems to have been settled by Chowla and Mordell $\endgroup$
    – Y. Zhao
    Jun 29, 2013 at 1:01
1
$\begingroup$

OK. The corollary 3 of T Funakura combined with the result of Gauss rules out such possibilities.

P.S.The corollary 3 of T.Funakura states that if $G(\chi)/\sqrt{N}$ is a fourth root of unity, then $\chi$ is a real character.

Improve this answer
$\endgroup$
0

Your Answer

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

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