multiplication by $\sigma'$ provides $$\sigma' {\thinspace}\sigma \; = \; \big( h: M'M: x: M'y: r \big)$$ while right multiplication by $\sigma'$ provides $$\sigma {\thinspace}\sigma' \; = \; \big( h: M M': M'x: y: r \big)$$

Boundary of $X$. {#compactification:boundary}
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The *boundary* of $X$ is defined as $B := X \setminus {\mathrm{SE}_{3,{\mathbb{C}}}}$. It is the closed subset of $X$ cut out by the linear equation $h = 0$.

For any point $(h: M: x: y: r) \in B$, we have $$M M^t \; = \; M^t M \; = \; 0, \quad \quad M^t y \; = \; M x \; = \; 0,$$ $$\langle x,x \rangle \; = \; \langle y,y \rangle \; = \; 0 .$$ The first equation shows that $\mathrm{rank}(M) \le 1$. Hence $M = v w^t$ for two suitable vectors $v, w \in {\mathbb{C}}^3$. It should be noted that $v$ and $w$ are not unique: one may multiply $v$ by a nonzero complex number and $w$ by its inverse. We have two cases:

-   if $M = 0$, we can take both $v$ and $w$ to be zero;

-   if $M \neq 0$, then again by the same equation, it follows that $\langle v,v \rangle 