5 The south plane

In this section, the horizontal plane trough the south pole will be examined. From Proposition 2.9, this plane is the locus of apexes of two kind of objects : the point-circle $\left\{ \infty\right\}$ and all the straight lines. It will be seen that changing only the meaning of the south pole into "the line at infinity" can shift all the previous results into the polar reciprocity between straight lines drawn in $\mathbb{R}^{2}\times\left\{ 0\right\} \times\left\{ 1\right\}$ and points lying in $\mathbb{R}^{2}\times\left\{ -1\right\} \times\left\{ 1\right\}$.

Fig. 5: Tangential equation of a circle in the south plane

Construction 5.1   Construct the tangential equation of a cycle $\pi_{0}$.

Let $\omega_{1}$ be a point on $\pi_{0}\neq\left\{ \omega_{0}\right\}$. All cycles tangent to $\pi_{0}$ at $\omega_{1}$ belong to pencil $\left(\pi_{0},\,\left\{ \omega_{1}\right\} \right)$ and their apexes are on the projective line $\left(\mathcal{A}_{0}\,\mathcal{S}_{1}\right)$. Finally, the required locus is the cone $\Gamma_{0}$ through $\mathcal{A}_{0}$ tangent to $\mathbb{S}$ along $shadow\left(\pi_{0}\right)$, as shown in Fig. 1(b). For a point-circle $\left\{ \omega_{0}\right\}$, $\Gamma_{0}$ is the projective plane tangent to $\mathbb{S}$ at $\mathcal{S}_{0}$ and for an imaginary cycle, $\Gamma_{0}=\left\{ \pi_{0}\right\}$.

Intersection of $\Gamma_{0}$ with the south plane is the locus of apexes for all the straight lines tangent to $\pi_{0}$. Therefore "tangential" equations are not only abstract objects, but can be seen and touched.

Proposition 5.2   The tangential equation (in the south plane) of a cycle $\pi_{0}$ (living in the equatorial plane) is :

$${\displaystyle \left(U,\, V,\, W\right)\Psi}\left(\negthickspace\... ...a\left(d+c\right) & b\left(d+c\right) & -\left(d+c\right)^{2}\end{array}\right)$$

The case of degeneracy are : (i) point-circle $\left\{ \omega_{0}\right\}$, giving twice the apex line $Ux_{0}+Vy_{0}-W=0$ of pencil "all lines through $\omega_{0}$" and (ii) straight line, giving twice the apex line $U\sin\tau_{0}-V\cos\tau_{0}=0$ of pencil "all lines parallel to $\pi_{0}$". Otherwise, a parametrization of the conic is :

$${\displaystyle \left(\negthickspace\negthickspace\begin{array}{c}... ...ckspace\negthickspace\right)\sqrt{{\it r_{0}}^{2}-{\it\rho_{0}}^{2}}\sin\alpha}$$

where $\alpha$ has to be changed in $i\,\alpha$ when $(0,0)$ and $\omega_{0}$ are separated by circle $\pi_{0}$.

Exercise 5.3   What if we prefer $\mathbb{P}_{\mathbb{R}}\mathbb{R}^{3}$ as model for plane geometry rather $\mathbb{S}=\mathbb{P}_{\mathbb{C}}\mathbb{C}^{2}$ ?

Solution 1   Let $\omega$ a point in $\mathbb{R}^{2}$, distinct from $O$ (cf Fig. 4). Let $A=shadow\left(M\right)$. The meridian through $N,\, A,\, S$ is the shadow of the line $\left(O\,\omega\right)\subset\mathbb{C}$. Its apex is at infinity, namely $u\doteq apex\left(O\,\omega\right)=\left(y,-x,0,0\right)$. The projective line $\left(S\, u\right)$ is the line of apexes of a pencil of straight lines. Containing $\left(O\,\omega\right),$ this pencil is the pencil of all the lines parallel to $\left(O\,\omega\right)$.

Let $v=\left(x,\, y,\,0,\,0\right)$. The projective line $\left(S\, v\right)$ is the conjugate of $\left(S\, u\right)$ relative to $\mathbb{S}$ and therefore $\left(S\, v\right)$ is the line of apexes of the pencil of all parallel straight lines that are orthogonal to $\left(O\,\omega\right)$.

Calling "south plane" the set of all the $\left[a_{i},\, b_{i},\, c_{i},\, d_{i}\right]$ such that $c_{i}+d_{i}=0$, i.e. the regular plane $c=-1$ in $\mathbb{R}^{3}$ together with its projective infinite points $\left(x,\, y,\,0,\,0\right)$ we have a copy of $\mathbb{P}_{\mathbb{R}}\mathbb{R}^{3}$ where each line $\Delta\,:\, a\, X+b\, Y=1$ is represented by $\left(a,\, b\right)$, the inverse of $\left(a/\left(a^{2}+b^{2}\right),\, b/\left(a^{2}+b^{2}\right)\right)$, i.e. by the conjugate of $\Delta$ relative to the unit circle.

When solving a problem in the ordinary plane, it is often useful to introduce points at infinity. When only one infinity point is required (telling us that a cycle is a straight line), stereographic projection onto $\mathbb{S}$ is the method. When an infinity of points at infinity is required (telling us the direction of a straight line), ordinary projection onto the "south plane" gives us the pole-polar duality.

It can be noticed that, in the "south plane", the line $\left(a,b,0,0\right)$ (all of the points at infinity of $\mathbb{P}_{\mathbb{R}}\mathbb{R}^{3}$) is the polar of point $S$ (the infinity point of $\mathbb{P}_{\mathbb{C}}\mathbb{C}^{2}$).