# Is there a hidden connection between RH and the golden ratio?

I realized today that, considering the circle $\Gamma_{\Delta}$ on the Riemann sphere whose image through the stereographic projection is the critical line $\Delta$, the affixes of the images of its poles through the stereographic projection are the golden ratio and its conjugate, that are the roots of the equation $z^{2}-Sz+P$ with $S=1$ and $P=-1$. As those poles are antipodal points, the products of the affixes of the images is $-1$. The fact that their sum is $1$ means that one is the image of the other through the symmetry $s\mapsto 1-s$, which appears in the functional equation of $\zeta$.

RH can be reformulated as follows: there is only one circle on the Riemann sphere whose image through stereographic projection contains all the roots of the equation $\zeta(s)=0$. Hence there should be only one pair of antipodal points corresponding to the poles of such a circle.

Does this mean that the meaning of RH is that there's essentially only one possible symmetry (namely $s\mapsto 1-s$) giving rise to a functional equation allowing the analytic continuation of $\zeta$ (and other L-functions as well)?

• I think you're missing the trivial zeros in your formulation. – Chappers Aug 2 '17 at 21:38
• And there are functions with a functional equation of similar form, but for which an equivalent of RH does not hold (search for Davenport–Heilbronn function). – Chappers Aug 2 '17 at 21:44
• Yes, I should have written "$\Xi(s)=0$". – Sylvain Julien Aug 2 '17 at 21:44
• There's also the conjugation map $z \mapsto \bar{z}$, which preserves zeros since $\zeta(\bar{s}) = \overline{\zeta(s)}$ (this is the Schwarz reflection principle, and works for any meromorphic function which is real (or infinite) on the real line, and in particular any meromorphically continuable function with a real Dirichlet series). These tell you that the nontrivial zeros occur either in pairs (if on the critical line) or quadruples: if $c$ is a zero, so is $1-c$, $\bar{c}$ and $1-\bar{c}$. – Chappers Aug 2 '17 at 22:13
• What I have on mind is that the functions that fulfill RH only have $s\mapsto 1-s$ and the complex conjugation as symmetries and that their vanishing forces those two symmetries to coincide, thus leading to $s\mapsto 1-\bar{s}$, hence my phrasing "only one possible symmetry". – Sylvain Julien Aug 2 '17 at 23:27