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Let $ \Delta : =\{z\in\mathbb{C},\Re(z)=1/2\} $ be the critical line that appears in the Riemann Hypothesis, and let $\Gamma_{\Delta}$ the circle on the Riemann sphere $S^{2}$ that transforms into $\Delta$ through stereographic projection.

Can the Riemann sphere be endowed with the structure of a Riemannian manifold whose metric admits $\Gamma_{\Delta}$ as a geodesic?

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Yes. For there's a projective transformation that takes some great circle to this line (you can find it by conjugating the map $z \mapsto z + \frac{1}{2}$ by stereographic projection to get a map from $S^2$ to itself (i.e., "project to $\Bbb C$; add $\frac{1}{2}$; project back to $S^2$"). Call the inverse of this map from $S^2$ to $S^2$ by the name $F$.

Then let $g$ be the usual metric on $S^2$, and define a metric $G$ whose value at a point $P \in S^2$ is computed by $$ G_P(u, v) = g(F_{*}(P)(u), F_{*}(P)(u)) $$ where $u$ and $v$ are tangent vectors to $S^2$ at $P$, and $F_{*}$ is the derivative, so that $F_{*}(P)$ is a map from $T_PS^2 -> T_{F(P)}S^2$.

The function $G$ is a Reimannian metric on $S^2$ with your line as a geodesic.

(By the way, this is just a slight variant of @AlfredYerger's answer).

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  • $\begingroup$ Denoting by P the center of this circle on the Riemann sphere, its seems that the affix of its image through the stereographic projection is the golden ratio. Is it known ? $\endgroup$ – Sylvain Julien Aug 2 '17 at 16:18
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Sure it can. The axes are both great circles on the Riemann sphere, so instead of using the usual stereographic projection, first just translate so that the critical line becomes the vertical line $x = 0$. This shifted plane then can also be stereographically projected to the sphere, and will have the line $x = 1/2$ as a great circle.

More generally, any affine transformation is an automorphism of the plane, so we can arrange that any particular line be a geodesic on the sphere.

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