2 Apexes and pencils

Definition 2.1   The apex of a cycle $\pi_{k}$, noted $\mathcal{A}_{k}$ is the projective class in $\mathbb{P}_{\mathbb{R}}\mathbb{R}^{4}$ of the $\left[a_{k},\, b_{k},\, c_{k},\, d_{k}\right]\in\mathbb{R}^{4}$ described by (1.7).

$$\mathcal{A}_{k}\simeq\left[2x_{k},\,2y_{k},\,1-\rho_{k}^{2}+r_{k}^{2},\,1+\rho_{k}^{2}-r_{k}^{2}\right]$$ (2.1)

Notation 2.2   When $d_{k}\neq0$, $A_{k}=X_{k}/d_{k}$ will be the representative of $\mathcal{A}_{k}$ in $\mathbb{R}^{3}\times\left\{ 1\right\}$, while $S_{k}$ remains the point of $\mathbb{S}$ associated by Definition 1.3 to the center $\omega_{k}$ of $\pi_{k}$.

Definition 2.3   When point $\omega_{j}$ is not at infinity, we define $\Omega_{j}$ by $\Omega_{j}\doteq\left[x_{j},y_{j},0,1\right]\in\mathbb{R}^{3}\times\left\{ 1\right\}$. This point represents the apex of circle $X^{2}+Y^{2}-2x_{j}X-2y_{j}Y+1=0$, which is imaginary when $\left\vert z_{j}\right\vert<1$.

Remark 2.4   There are many reasons to "imagine" such circles, **** among them.

Definition 2.5   The stereographic projection is the mapping $S$ defined on $\overline{\mathbb{C}}$ by $\infty\mapsto S_{\infty}\doteq\left[0,0,-1\right]$ and otherwise $\omega_{j}\mapsto S_{j}$ where $S_{j}$ is the representative in $\mathbb{R}^{3}\times\left\{ 1\right\}$ of the apex of point-circle $\left\{ \omega_{j}\right\}$. In other words :

$$S_{j}\in\mathbb{R}^{3}\times\left\{ 1\right\} ,\quad S_{j}\simeq\left[-2x_{j},-2y_{j},\,1-\rho_{j}^{2},\,1+\rho_{j}^{2}\right]$$

Proposition 2.6   Stereographic projection is a one to one correspondence between $\overline{\mathbb{C}}$ and sphere $\mathbb{S}\,:\, x^{2}+y^{2}+z^{2}=1$. When $\omega_{j}\neq\infty$, point $S_{j}$ is the unique intersection of the line through $S_{\infty}$ and $\Omega_{j}$ with $\mathbb{S}\setminus S_{\infty}$.

Proof. $S_{j}\in\mathbb{S}$ comes from $4\left(x^{2}+y^{2}\right)+\left(1-\rho^{2}\right)^{2}=\left(1+\rho^{2}\right)^{2}$ and alignment from :

$$\left[-2x_{j},-2y_{j},\,1-\rho_{j}^{2},\,1+\rho_{j}^{2}\right]=-2... ...x_{j},y_{j},0,1\right]+\left(\rho_{j}^{2}-1\right)\left[0,0,-1,1\right]\qedhere$$ (2.2)

$\qedsymbol$

Proposition 2.7   There are $48$ isometries of $\mathbb{R}^{3}$ that leaves globally invariant the directions of the three axes. Considered as $\mathbb{S}$-actions they can be induced by $\mathbb{P}_{\mathbb{C}}\mathbb{C}^{2}$-actions. The following table gives a set of generators (last three columns), while center of $G_{48}$ contains only $z$ and $-1/\overline{z}$.

$${\displaystyle \begin{array}{ccccccc} \phi & z & -1/\overline{z} ... ... & \left(a,-b,c\right) & \left(-b,-a,c\right) & \left(c,b,-a\right)\end{array}}$$

Proof. $\left\vert G_{48}\right\vert=6\times8$ is obvious. Generators $h_{1},\, h_{2},\, h_{3}$ have been chosen to have the simplest expression as $\mathbb{P}_{\mathbb{C}}\mathbb{C}^{2}$-actions. The best proof of $span\left(h_{i}\right)=G_{48}$ is therefore a direct inspection by computer (e.g., replacing $h_{2}$ by $i\,\overline{z}$ would be wrong). $\qedsymbol$

Proposition 2.8   Stereographic projection is conformal.

Proof. The simplest, the best. Start from $M=\left(\rho\,\cos\tau,\,\rho\,\sin\tau\right)\in\mathbb{R}^{2}$ and obtain $S\in\mathbb{R}^{3}$. Let $\Omega$ be the center of the sphere. By definition, vector $\overrightarrow{u}\doteq\overrightarrow{\Omega S}$ is unitary, radial. Vector $\mathrm{\partial}\overrightarrow{u}/\mathrm{\partial}\rho$ is along the local meridian and $\overrightarrow{v}\doteq\frac{1}{2}\left(1+\rho^{2}\right)\partial\overrightarrow{u}/\partial\rho$ is unitary while $\partial\overrightarrow{u}/\partial\tau$ is along the local parallel directed by $\overrightarrow{w}=\overrightarrow{u}\wedge\overrightarrow{w}$ and $\left(1+\rho^{2}\right)\partial\overrightarrow{u}/\partial\tau=2\rho\overrightarrow{w}$. Since $\Psi\doteq\left(\overrightarrow{u},\,\overrightarrow{v},\,\overrightarrow{w}\right)$ is an orthogonal matrix, we have :

$$\left\Vert \overrightarrow{\mathrm{d}S}\right\Vert =\left\Vert \P... ...t\Vert =\frac{2}{1+\rho^{2}}\left\Vert \overrightarrow{\mathrm{d}M}\right\Vert$$

To complete the proof, it can be noticed that angles at $z=\infty\in\mathbb{P}_{\mathbb{C}}\mathbb{C}^{2}$ are defined as angles at $z=0$ under the action of mapping $z\mapsto-1/\overline{z}$ in while its counterpart in $\mathbb{R}^{3}$, namely $A\mapsto-A$ is clearly conformal. $\qedsymbol$

Proposition 2.9   Given an element $\mathcal{A}$ of $\mathbb{P}_{\mathbb{R}}\mathbb{R}^{4}$ with $\mathcal{A}\simeq\left[a,\, b,\, c,\, d\right]$, then either :

1. $\mathcal{A}\simeq[0,\,0,\,-1,\,1]$. $\mathcal{A}$ is the apex of $\infty\in\mathbb{P}_{\mathbb{C}}\mathbb{C}^{2}$.
2. $d+c=0$. $\mathcal{A}$ is the apex of the ordinary line $a\, X+b\, Y+c=0$.
3. $d+c\neq0$ and $a^{2}+b^{2}+c^{2}-d^{2}\geq0$. $\mathcal{A}$ is the apex of the ordinary circle $x_{0}=a/\left(c+d\right)$, $y_{0}=b/\left(c+d\right)$, $r_{0}=\sqrt{a^{2}+b^{2}+c^{2}-d^{2}}/\left(c+d\right)$.
4. $d+c\neq0$ and $a^{2}+b^{2}+c^{2}-d^{2}<0$. $\mathcal{A}$ is the apex of an imaginary circle.

Proof. Direct inspection. $\qedsymbol$

Definition 2.10   The pencil $P\left(\pi_{1},\pi_{2}\right)$ generated by two different cycles $\pi_{1}$ and $\pi_{2}$ is the family of all the cycles $\pi$ whose equation can be written as :

$$\pi\left(\omega\right)=\lambda_{1}\pi_{1}\left(\omega\right)+\lambda_{2}\pi_{2}\left(\omega\right)$$ (2.3)

Proposition 2.11   All points $\mathcal{A}$ of a given projective line are apexes of cycles (may be imaginary circles) that belong to the same pencil and conversely. This characteristic line will be referred as the apex line of the pencil.

Proof. Obvious from (2.3) and the definition of an apex. $\qedsymbol$

Construction 2.12   Construct the center $\omega_{0}$ of a given circle $\pi_{0}$.

Fig. 1: Circle, shadow, apex

[From circle to shadow]

[From shadow to apex]

Start from circle $\pi_{0}$ and draw each line $\left(S_{0}S_{1}\right)$ where point $\omega_{1}$ belongs to $\pi_{0}$ and obtain a cone (Fig. 1(a)). This cone intersects the sphere $\mathbb{S}$ along a circle $\gamma$ and this circle is the shadow of $\pi_{0}.$ Start now from circle $\gamma$ and, at each point $S_{1}\in\gamma$ , draw the tangent to $\mathbb{S}$ that is orthogonal to $\gamma$ and obtain another cone (Fig. 1(b)). The apex of the cone is precisely $X_{0}=apex\left(\pi_{0}\right)$, eponymous property.

Conclude by drawing projective line $\left(S_{\infty}X_{0}\right)$. This lines intersects the sphere $\mathbb{S}$ at $S_{0}=shadow\left(\omega_{0}\right)$ and the equatorial plane exactly at $\Omega_{0}$ i.e. "quite at $\omega_{0}$".

Remark 2.13   When $\omega_{0}$ is known, construction of the apex can be done using only the meridian plane containing $\omega_{0}$.

Proposition 2.14   When $\omega_{0}$ is an ordinary point, $\left(S_{0}S_{\infty}\right)$ is the apex line of the pencil "all circles centered at $\omega_{0}$", including the point-circle $\left\{ \infty\right\}$.

Proof. Eqn. (2.1) states that $\mathcal{A}_{k}\simeq\left[2x_{0},\,2y_{0},\,1-\rho_{0}^{2},\,1+\rho_{0}^{2}\right]+r_{k}^{2}\,\left[0,0,-1,+1\right]$. $\qedsymbol$

Proposition 2.15   When a pencil contains at least two ordinary non concentric circles $\pi_{1}$ and $\pi_{2}$ then all the circles $\pi$ of the pencil have their centers on the same straight line.

Proof. From Proposition 2.14, all apexes $\mathcal{A}_{0}$ and shadows of centers belong to the well defined projective plane $\Pi\simeq\operatorname{span}\left(X_{1},X_{2},X_{\infty}\right)$. These shadows belongs therefore to the circle $\left(\gamma\right)\doteq\mathbb{S}\cap\Pi$. Containing the south pole, $\gamma$ is the shadow of a straight line. $\qedsymbol$