Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

If RH is not true, we have that Weil's explicit formula still holds:

$$ \sum_{\gamma} h(\gamma) = h(i/2)+h(-i/2)-2 \sum_{n=1}^{\infty} \frac{ \Lambda(n)}{ \sqrt n}g(logn)+\frac{1}{2\pi} \int_{-\infty}^{\infty}h(r)\Psi(1/4+ir/2)dr.$$

My question is if there are formulae that take into account ONLY the Riemann zeros with real part 1/2 and not the others.. thanks.

share|improve this question
add comment

1 Answer

Absent a proof of RH (or something close to RH), there are going to be some significant obstructions to the existence of such a formula, due to the possibility of two zeroes on the critical line that are very close together. A small perturbation of the zeta function could then move these two zeroes off of the critical line (much as perturbing the polynomial $z^2-\varepsilon^2$ to $z^2+\varepsilon^2$ moves two nearby real zeroes $\pm \varepsilon$ into two complex zeroes $\pm i \varepsilon$). The zeta function and its perturbation would be more or less indistinguishable from each other by the sort of contour integration operations that are traditionally used in Weil-type explicit formulae. As a consequence, such formulae cannot distinguish between two nearby zeroes on the critical line, and two nearby zeroes slightly away from the critical line, which already rules out a lot of options for a formula that only captures the critical line behaviour. Indeed, if one restricts attention to formulae which depend analytically with respect to analytic perturbations of the underlying complex function, I would imagine that there are no interesting formulae which only involve the real zeroes.

A model problem would be to try to locate an analytic formula relating the coefficients of a real polynomial with the purely real zeroes of that polynomial. Even in the quadratic case, it seems clear that no non-trivial formula exists, because one side of the formula would develop some sort of singularity (such as failure of analyticity) as one transitioned from $z^2-\varepsilon^2$ to $z^2+\varepsilon^2$, and the other side of the formula would not.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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