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Sep 1, 2019 at 2:13 comment added GH from MO @user1728: In this case, one considers an $L$-function $L(s)$ which is holomorphic on $\mathbb{C}$ and real-valued on $\mathbb{R}$. The Riemann Hypothesis for $L(s)$ implies that $L(s)>0$ for $s>1/2$, and hence $L(1/2)\geq 0$. So if we can prove the nonnegativity of $L(1/2)$ unconditionally, then we could confirm an important prediction of GRH. I say "important", because nonnegativity usually has deep applications, while proving it requires a similarly deep understanding of the arithmetic structure behind $L(s)$. For infinitely many $L$-functions, $L(1/2)\geq 0$ is a theorem!
Sep 1, 2019 at 1:44 comment added user1728 Why do you consider nonnegativity of a central $L$-value evidence towards GRH?
Aug 31, 2019 at 8:03 comment added GH from MO @MustafaSaid: I reacted to "are we just relying on numerical evidence". The answer is a clear "no".
Aug 31, 2019 at 8:02 comment added Mustafa Said But where is the heuristic argument? The GUE connection is more along the lines of what I’m looking .
Aug 31, 2019 at 7:58 comment added Sylvain JULIEN And not only are there infinitely many zeta zeros on the critical line, but the proportion thereof has been shown to exceed 41%.
Aug 31, 2019 at 7:58 comment added GH from MO @MustafaSaid: I guess my point is that many powerful consequences of the Riemann hypothesis have been verified independently. Some of these address the distribution of the zeros directly. In my view, these results are much stronger evidence than the numeric verification of the Riemann hypothesis up to a certain height. Add to these that the algebraic geometric analogue of the Riemann hypothesis is a theorem thanks to Weil, Deligne, etc. I am talking about infinitely many $L$-functions in each of the bullet points in my anwer.
Aug 31, 2019 at 7:44 history answered GH from MO CC BY-SA 4.0