of this decomposition is called “the unitary frame”.

3.  Define $f: {\widetilde{\Sigma}}\longrightarrow {\mathbb{H}}^3$ via the Sym-Bobenko formula: $$f(z) = F(z,e^{-q}){F(z,e^{-q})}^*  =: { \mathrm{Sym}}_q F(z)$$ where $F(z,{\lambda}_0):=F(z)({\lambda}_0)$.

Then $f$ is a CMC $H>1$ ($H=\coth q$) conformal immersion from $\widetilde{\Sigma}$ to ${\mathbb{H}}^3$. Its Gauss map (in the direction of the mean curvature vector) is given by $$N(z) = F(z,e^{-q})\sigma_3{F(z,e^{-q})}^* =: {\mathrm{Nor}}_qF(z)$$ where $\sigma_3$ is defined in . The differential of $f$ is given by $$\label{eqdf}
df(z) = 2\sinh(q)b(z)^2 F(z,e^{-q}) \begin{pmatrix}
0 & \beta(z,0)\\
{\overline{\beta(z,0)}} & 0
\end{pmatrix}  F(z,e^{-q})^*$$ where $b(z)>0$ is the upper-left entry of $B(z)\mid_{{\lambda}=0}$. The metric of $f$ is given by $$ds_f(z) = 2\sinh(q)b(z)^2 \left|\beta(z,0)\right|$$ and its Hopf differential reads $$\label{eqHopf}
    -2\beta(z,0)\gamma(z,0)\sinh q ~ dz^2.$$

The results of this paper hold for any $H>1$. We thus fix now $H>1$ and $q={\mathop \mathrm{arcoth}}H$.

#### Rigid motions.

Let $H\in {\Lambda {\mathrm{SL}}(2,{\mathbb{C}})_\rho}$ and define the new holomorphic frame $\widetilde{\Phi} = H\Phi$ with unitary part $\widetilde{F}$ and induced immersion $\widetilde{f} = { \mathrm{Sym}}_q \widetilde{F}$. If $H\in{\Lambda {\mathrm{SU}}(2)_\rho}$, then $\widetilde{F} = HF$ and $\widetilde{\Phi}$ gives rise to the same immersion as $\Phi$ up to an isometry of ${\mathbb{H}}^3$: $$\widetilde{f}(z) = H(e^{-q})\cdot f(z).$$ If $H\notin {\Lambda {\mathrm{SU}}(2)_\rho}$, this transformation is called a “dressing” and may change 