# Can GRH for complex primitive Dirichlet characters fail with a single non-trivial zero off the critical line?

Can GRH for complex primitive Dirichlet character fail with a single non-trivial zero off the critical line?

For real characters this is impossible because the non-trivial zeros are in quadruples.

On the other hand this paper in Russian constructs a function very similar to zeta with exactly one quadruple of zeros off the critical line.

The motivation: GRH implies a certian efficiently computable sum over zeros of $L(s,\chi)$ must be $0$ and experimentally it is (indistinguishable from) $0$ and I wonder if this result is trivial even if GRH is false. IMO the sum should not be zero if the answer to the question is 'yes' or some cancellation doesn't happen.

• At least tuples, because of the symmetries under conjugation, or? – Marc Palm Mar 21 '13 at 10:24
• @Marc From the functional equation if $\rho$ is zero of $L(s,\chi)$ then $1-\rho$ is zero of $L(s,\overline{\chi})$. My tests show both $(\rho,1-\rho)$ need not be zeros of $L(s,\chi)$. lmfdb.org appears to confirm this, check the graph on the critical line: lmfdb.org/L/Character/Dirichlet/5/2 – joro Mar 21 '13 at 10:46
• Btw, I didn't downvote. I was more concerned about the symmetry $L(\overline{s},\chi) = \overline{ L( s, \overline{\chi})}.$ – Marc Palm Mar 21 '13 at 13:27
• @Mark thanks! Tuples means the answer is "no". Please answer so the question is answered. – joro Mar 21 '13 at 16:16
• I suggest to simply change the conjecture to tuples instead. btw. I cant rule out real zeros violating RH. All zeros on the critical line occur in tuples. – Marc Palm Mar 21 '13 at 16:29

If $\rho$ is a zero of $L(s,\chi)$, then $\overline\rho$ is a zero of $L(\overline s,\chi)=\overline{L(s,\overline\chi)}$, hence also of $L(s,\overline\chi)$. Therefore $1-\overline\rho$ is a zero of $L(s,\chi)$. Thus nontrivial zeros of Dirichlet L-functions are symmetric in the critical line. In particular, if GRH fails for a character, it fails for at least two zeros.