$\R^{n+2}$ equipped with the scalar product $(\ref{sp})$, the volume form $dV$ and the positive half cone $\mathcal{C}_+$. Differentiating the $\R^{2,n}$-valued maps $$\mathcal{B}_j:B \in \mathrm{A}^{\uparrow}_+(2,n)\to B_j\in \R^{2,n},\quad j=0,\dots,n+1,$$ yields $$d\mathcal{B}_j=\sum_{i=0}^{n+1}\mu_j^i \mathcal{B}_i,$$ where $\mu^i_j$ are left-invariant 1-forms. The conditions $\langle B_j,B_i\rangle= \mathtt{m}_{ji}$ imply that $\mu =(\mu^i_j)$ takes values in the Lie algebra $$\mathfrak{a}(2,n)=\{X\in \mathfrak{gl}(n+2,\R) / ^tX\cdot \mathtt m + \mathtt m\cdot X=0\}.$$ of $\mathrm{A}^{\uparrow}_+(2,n)$. Consequently we can write $$\mu = \left(
          \begin{array}{cccc}
            \mu^0_0 & -\mu^1_{n+1} & \mu_{n+1} & 0 \\
            \mu^1_0 & 0 & ^t\mu_1 & \mu^1_{n+1} \\
            \mu_0 & \mu_1 & \widetilde{\mu} & \mu_{n+1} \\
            0 & -\mu^1_0 & ^t\mu_0 & -\mu^0_0 \\
          \end{array}
        \right),$$ where $$\mu_0=^t(\mu^2_0,\dots, \mu^n_0),\quad \mu_1=^t(\mu^2_1,\dots, \mu^n_1),\quad
\mu_{n+1}=^t(\mu^2_{n+1},\dots, \mu^n_{n+1})$$ and $^t\widetilde{\mu}+\widetilde{\mu}=0$. The left-invariant 1-forms $\mu^0_0$, $\mu^j_0$, $\mu^j_{n+1}$, $j=1,\dots, n$ and $\mu^i_j$, $1\le i<j=1,\dots n$ are linearly independent and span the dual of the Lie algebra $\mathfrak{a}(2,n)$. They satisfy the [*Maurer-Cartan equations*]{} $$\label{MC}d\mu^i_j=-\sum_{k=0}^{n+1}\mu^i_k\wedge \mu^k_j,\quad  i,j=0,\dots, n+1.$$

Let $\mathbb{M}^{1,n-1}$ be [*Minkowski n-space*]{}, i.e., the affine space $\R^n$ with the Lorentzian scalar product $$(\mathbf{p},\mathbf{q})=-p_1q_1+p_2q_2+\dots + p_nq_n.$$ The map $$\label{j}
\mathbf{j}(\mathbf{p})= [^t(1,p_1,\dots,p_n,\frac{1}{2}(\mathbf{p},\mathbf{p}))]\in \mathcal{E}^{1,n-1}$$ is a conformal embedding whose image is said the [*Minkowski-chamber*]{} of $\mathcal{E}^{1,n-1}$ Let $\mathrm{P}^{\uparrow}_+(1,n-1)=\mathbb{M}^{1,n-1}\rtimes \mathrm{SO}^{\uparrow}_+(1,n-1)$ be the restricted Poincaré group of $\mathbb{M}^{1,n-1}$. For each $(\mathbf{p},\mathrm{L})\in \mathrm{P}^{\uparrow}_+(1,n-1)$ we put $^{*}\mathbf{p}=(-p_1,p_2,\dots, 