Conformal geometric algebra

Remarks

From $e_\pm \cdot e_\pm = \pm 1$ one gets $n_0 \cdot n_0 = n_\infty \cdot n_\infty = 0$ and $n_0 \cdot n_\infty = -1$. Hence $X \cdot X = \abs{x}^2 - \abs{x}^2 = 0$, so points become null vectors, and $X \cdot Y = x \cdot y - \tfrac{1}{2}(\abs{x}^2 + \abs{y}^2) = -\tfrac{1}{2} \abs{x - y}^2$. Example: for $c \in \R^n$ and $r > 0$, put $S = C - \tfrac{1}{2} r^2 n_\infty$ with $C$ the image of $c$. Since $X \cdot n_\infty = -1$, one gets $X \cdot S = \tfrac{1}{2}(r^2 - \abs{x - c}^2)$, which vanishes exactly when $x$ lies on the sphere with centre $c$ and radius $r$. Likewise, for a unit vector $m \in \R^n$ and $\delta \in \R$, the vector $m + \delta n_\infty$ satisfies $X \cdot (m + \delta n_\infty) = x \cdot m - \delta$, and it represents the plane $x \cdot m = \delta$. Non-example: the vector $e_- \in \R^{n+1,1}$ is not the image of a point, since $e_- \cdot e_- = -1 \ne 0$. The versors of the algebra, the products of invertible vectors, act on $\R^{n+1,1}$ by conjugation, up to sign. They induce exactly the Möbius transformations of $\R^n \cup \set{\infty}$, the maps generated by translations, rotations, dilations and inversions in spheres.