{\mathbb{C}}^{*}$. Multiplication from the left acts analogously on $v$. Hence by suitable rotations from both sides we obtain $v = \gamma {\thinspace}(1,i,0)^t$ and $w =
\delta {\thinspace}(1,i,0)^t$ with both $\gamma$ and $\delta$ different from zero since $M \neq 0$. Then projectively we can suppose that $M = \left(
\begin{smallmatrix}
    1 & i & 0 \\
    i & -1 & 0 \\
    0 & 0 & 0 
\end{smallmatrix} 
\right)$.

The action of left multiplication by a translation with vector $s \in {\mathbb{R}}^3$ on the boundary point $\beta$ gives (see Remark \[remark:action\]) $$\big(0:vw^t:(-\langle v,s \rangle + \lambda) {\thinspace}w: \mu {\thinspace}v: r + 2\mu
{\thinspace}\langle v,s\rangle
\big) ,$$ and similarly the action by right multiplication with vector $t \in {\mathbb{R}}^3$ gives $$\big(0:vw^t: \lambda {\thinspace}w: (\langle w,t \rangle + \mu) {\thinspace}v: r - 2\lambda
{\thinspace}\langle w,t \rangle \big) .$$ This shows that we can achieve $\lambda = \mu = 0$ by multiplication by translations from both sides (for example, since we reduced to the situation $v
= (1, i,0)^t$, one can take $s_1 = \mathrm{Re} \lambda$, $s_2 = \mathrm{Im}
\lambda$ and $s_3$ to be arbitrary, where $s = (s_1, s_2, s_3)^{t}$, and similarly for $t$). It also shows that the matrix $N$ is invariant 