7
$\begingroup$

It is known that "Assuming the generalized Riemann hypothesis (GRH) if VP = VNP then PH collapses to second level". Why would one think of a relation between VP,VNP and the Riemann hypothesis. Where does a statement about the zeros of the Riemann zeta function appear in a statement about complexity classes ? I have not read the original result. It would be nice if anyone could provide some intuition.

$\endgroup$
  • $\begingroup$ What rings is this statement about? Anyway, the VP and VNP classes are really more "algebraic geometry classes" than "computational complexity classes", so that's why. $\endgroup$ – Emil Jeřábek supports Monica May 8 '17 at 8:20
  • 1
    $\begingroup$ How many algebraic geometers can write down the proper definition of VNP ;-) $\endgroup$ – Markus Bläser May 8 '17 at 12:31
6
$\begingroup$

Valiant's classes are defined over some field. They can use arbitrary constants from that field. To draw some conclusion about Boolean complexity classes, one needs to replace these arbitrary constants by small discrete constants. Here GRH comes into play, since it ensures the existence of enough primes with certain properties.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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