$r>s$ (spherical family) or $r<s$ (catenoidal family). One has: $$\label{eqH0spherical}
    H_0({\lambda}) = \frac{1}{\sqrt{2}}\begin{pmatrix}
    1 & -{\lambda}^{-1}\\
    {\lambda}& 1
    \end{pmatrix} \qquad \text{in the spherical case},$$ $$\label{eqH0catenoidal}
    H_0({\lambda}) = \frac{1}{\sqrt{2}}\begin{pmatrix}
    1 & -1\\
    1 & 1
    \end{pmatrix} \qquad \text{in the catenoidal case}.$$ In both cases, $\mu_0 = \frac{1}{2}$, and thus, $D_0$ is the same.

#### A basis of $\Lambda{\mathcal{M}}(2,{\mathbb{C}})_{\rho}$.

Let $R$ and $T'$ given by Proposition \[propDiagA\]. Identify $\Lambda{\mathcal{M}}(2,{\mathbb{C}})_{{\rho}}$ with the free $\Lambda{\mathbb{C}}_\rho$-module ${\mathcal{M}}(2,\Lambda{\mathbb{C}}_{{\rho}})$ and define for all $t\in(-T',T')$ the basis $${\mathcal{B}}_t = H_t\left(E_{1},E_{2},E_{3},E_{4}\right)H_t^{-1}=:\left(X_{t,1},X_{t,2},X_{t,3},X_{t,4}\right)$$ where $$E_1 = \begin{pmatrix}
    1&0\\
    0&0
    \end{pmatrix},\quad E_2 = \begin{pmatrix}
    0&1\\
    0&0
    \end{pmatrix}, \quad E_3 = \begin{pmatrix}
    0&0\\
    1&0
    \end{pmatrix}, \quad E_4 = \begin{pmatrix}
    0&0\\
    0&1
    \end{pmatrix}.$$ For all $t\in(-T',T')$, write $$\label{eqCt0}
    C_t(0) = \begin{pmatrix}
    tc_1(t) & {\lambda}^{-1}tc_2(t)\\
    tc_3(t) & -tc1(t)
    \end{pmatrix} = \sum_{j=1}^{4}t\widehat{c}_j(t)X_{t,j}.$$ The functions $c_j, \widehat{c}_j$ are ${\mathcal{C}}^1$ with respect to $t\in(-T',T')$ and take values in $\Lambda{\mathbb{C}}_{{R}}$. Moreover, the functions $c_i(t)$ holomorphically extend to ${\mathbb{D}}_\rho$.

\[propC3\] There exists a continuous function $\widetilde{c}_3: (-T',T')\longrightarrow\Lambda{\mathbb{C}}_{{R}}$ such that for all ${\lambda}\in{\mathbb{S}}^1$ and $t\in(-T',T')$, $$\widehat{c}_3(t) = t({\lambda}-e^q)({\lambda}-e^{-q})\widetilde{c}_3(t).$$

It suffices to show that $\widehat{c}_3(0)=0$ and that the holomorphic extension of $\widehat{c}_3(t)$ satisfies $\widehat{c}_3(t,e^{\pm q}) = 0$ for all $t$.

To show that $\widehat{c}_3(0)=0$, recall that the monodromy problem is solved for 