1
    \end{pmatrix}\in{\mathrm{SU}}(2).$$ Apply an isometry by setting $$\label{eqftildeSs}
    \widetilde{f}_{\mathbb{S}}(z) := R(q)^{-1}\cdot f_{\mathbb{S}}(z)$$ and compute $$\widetilde{f}_{\mathbb{S}}(z) = \frac{1}{1+|z|^2}\begin{pmatrix}
e^{-q}+ e^{q}|z|^2 & 2z\sinh q \\
2{\overline{z}}\sinh q & e^{q} + e^{-q}|z|^2
\end{pmatrix} = (\cosh q){\mathrm{I}}_2 + \frac{\sinh q}{1+|z|^2} \begin{pmatrix}
|z|^2-1 & 2z \\
2{\overline{z}} & 1-|z|^2
\end{pmatrix}$$ i.e. $$\label{eqftilde_Ss=gammav_Ss}
    \widetilde{f}_{\mathbb{S}}(z)= {\mathrm{geod}}({\mathrm{I}}_2,v_{\mathbb{S}}(z))(q)$$ with ${\mathrm{geod}}$ defined in and where in the basis $(\sigma_1,\sigma_2,\sigma_3)$ of $T_{{\mathrm{I}}_2}{\mathbb{H}}^3$, $$\label{eqVecteurvS}
v_{\mathbb{S}}(z) := \frac{1}{1+|z|^2}\left( 2{\mathop \mathrm{Re}}z, 2 {\mathop \mathrm{Im}}z, |z|^2-1 \right)$$ describes a sphere of radius one in the tangent space of ${\mathbb{H}}^3$ at ${\mathrm{I}}_2$ (it is the inverse stereographic projection from the north pole). Hence, $\widetilde{f}_{\mathbb{S}}(\Sigma)$ is a sphere centered at ${\mathrm{I}}_2$ of hyperbolic radius $q$ and $f_{\mathbb{S}}\left(\Sigma\right)$ is a sphere of same radius centered at ${\mathrm{geod}}({\mathrm{I}}_2, \sigma_3)(q)$.

One can compute the normal map of $f_{\mathbb{S}}$: $$N_{\mathbb{S}}(z) := {\mathrm{Nor}}_q F_{\mathbb{S}}(z) = R(q)\cdot \widetilde{N}_{\mathbb{S}}(z)$$ where $$\begin{aligned}
    \widetilde{N}_{\mathbb{S}}(z) &:= {\mathrm{Nor}}_q\left(\widetilde{F}_{\mathbb{S}}(z)\right) = \frac{1}{1+|z|^2}\begin{pmatrix}
    e^{-q}- e^{q}|z|^2 & -2z\cosh q \\
    -2{\overline{z}}\cosh q & e^{-q}|z|^2 - e^{q}
    \end{pmatrix}\\
    &= -(\sinh q)  {\mathrm{I}}_2 - (\cosh q)  v_{\mathbb{S}}(z) =   -\dot{{\mathrm{geod}}}\left( {\mathrm{I}}_2,v_{\mathbb{S}}(z) \right)(q).\end{aligned}$$ Note that this implies that the normal map ${\mathrm{Nor}}_q$ is oriented by the mean curvature vector.

Delaunay surfaces {#sectionDelaunaySurfaces}
-----------------

#### The data.

Let $\Sigma={\mathbb{C}}^*$, $\xi_{r,s}(z,{\lambda}) = A_{r,s}({\lambda})z^{-1}dz$ where $$\label{eqArs}
    A_{r,s}({\lambda}) := \begin{pmatrix}
    