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)?

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    $\begingroup$ I think you're missing the trivial zeros in your formulation. $\endgroup$ – Chappers Aug 2 '17 at 21:38
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    $\begingroup$ 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). $\endgroup$ – Chappers Aug 2 '17 at 21:44
  • $\begingroup$ Yes, I should have written "$\Xi(s)=0$". $\endgroup$ – Sylvain Julien Aug 2 '17 at 21:44
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    $\begingroup$ 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 $1-\bar{c}$. $\endgroup$ – Chappers Aug 2 '17 at 22:13
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    $\begingroup$ 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". $\endgroup$ – Sylvain Julien Aug 2 '17 at 23:27

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