8
$\begingroup$

In page 6, RH Equivalence 5.3. An equivalence of the Riemann Hypothesis says that

$$\sum_{\rho} \frac{1}{|\rho|^2} =\sum_{\rho} \frac{1}{\rho (1{-}\rho)}= 2 + \gamma - \log 4\pi$$ where $\rho$ is over nontrivial zeros of the Riemann zeta function. It's not hard to see that RH implies $$\sum_{\rho} \frac{1}{|\rho|^2}= 2 + \gamma - \log 4\pi.$$ But conversly I don't see how the equality above implies RH and there is no reference in page 6, RH Equivalence 5.3.

$\endgroup$
1
  • $\begingroup$ $$2+\gamma -\log (4 \pi )=-\left(\lim_{s\to 1} \, \left(\frac{\zeta (s) \zeta (s)}{2 \zeta (2 s-1)}+\left(1-\frac{1}{2^{s-1}}\right) \zeta (s)+\frac{\zeta (s)}{2 \zeta (s-1)}+\frac{1}{\zeta (s-1)}\right)\right)$$ $\endgroup$ Jul 21, 2023 at 17:46

1 Answer 1

12
$\begingroup$

Note that if $1/2< \sigma <1, t \in \mathbb R$ one has $\frac{2\sigma-1}{\sigma^2+t^2} < \frac{2\sigma-1}{(1-\sigma)^2+t^2}$.

By a little manipulation, one gets:

$\frac{2\sigma}{\sigma^2+t^2} + \frac{2(1-\sigma)}{(1-\sigma)^2+t^2} < \frac{1}{\sigma^2+t^2} + \frac{1}{(1-\sigma)^2+t^2} $

But if RH is false and there is $\rho=\sigma+it, 1/2<\sigma<1$, the above gives that

$2\Re{\frac{1}{\rho}}+2\Re{\frac{1}{1-\rho}}=2\Re{\frac{1}{\rho (1-\rho)}} < \frac{1}{|\rho|^2}+\frac{1}{|1-\rho|^2}$, so if we group together the four terms $\frac{1}{\rho (1-\rho)}, {\frac{1}{\bar \rho (1-\bar \rho)}}$ corresponding to the four roots $\rho, 1-\rho, \bar \rho, 1-\bar \rho$ we get that their sum is srictly less than the sum of the corresponding reciprocal of the respective four roots square modulus, so RH false implies $\sum_{\rho} \frac{1}{|\rho|^2} >\sum_{\rho} \frac{1}{\rho (1{-}\rho)}$ and the equivalence is established

Edit later - per comments - note that if $\Re \rho =1/2$ then $\bar \rho=1-\rho$ so roots group naturally in pairs only and $\frac{1}{\rho (1-\rho)}=\frac{1}{|\rho|^2}$ so the corresponding (two) terms on both sides of the equality $\sum_{\rho} \frac{1}{|\rho|^2} =\sum_{\rho} \frac{1}{\rho (1{-}\rho)}$ are equal

When there is a root with $\Re \rho_0 >1/2, t>0$ the roots group into four as noted $\rho_0, \bar \rho_0, 1-\rho_0, 1-\bar \rho_0$ and now the corresponding terms in $\sum_{\rho} \frac{1}{|\rho|^2}$ are $2(\frac{1}{|\rho_0|^2}+\frac{1}{|1-\rho_0|^2})$, while the terms in the sum $\sum_{\rho} \frac{1}{\rho (1{-}\rho)}$ are also four and since they are conjugate in pairs add to $2(2\Re{\frac{1}{\rho}}+2\Re{\frac{1}{1-\rho}})=4\Re{\frac{1}{\rho (1-\rho)}}$ and the inequality above applies

$\endgroup$
4
  • 1
    $\begingroup$ It's actually enough to group $\rho$ and $1-\rho$, because $\frac{1}{|\rho|^2}+\frac{1}{|1-\rho|^2}\geq 2\mathrm{Re}\,\frac{1}{\rho(1-\rho)}$ and equality is attained iff $\mathrm{Re}\,\rho=\frac12$ $\endgroup$ Aug 1, 2021 at 15:17
  • $\begingroup$ It seems that the argument is not complete since it may happen that some of $\rho$ have real part 1/2 and some with $\sigma\neq 1/2$. In Conrad's answer it assumes that all $\rho$ have real part $\sigma\neq 1/2$. $\endgroup$
    – Beta
    Aug 1, 2021 at 16:25
  • $\begingroup$ And also I'm a bit confused on how to group together. Could you please show the detailed calculation? Thanks very much! $\endgroup$
    – Beta
    Aug 1, 2021 at 16:27
  • 1
    $\begingroup$ No, it is enough for one root to have real part greater than $1/2$ as for roots with real part $1/2$ trivially $1-\rho=\ bar \rho$ so the corresponding terms which now appear only in pairs are equal; $\endgroup$
    – Conrad
    Aug 1, 2021 at 17:30

Your Answer

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

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