6 Inversion

Proposition 6.1   When cycle $\pi$ is described in $\mathbb{P}_{\mathbb{C}}\mathbb{C}^{2}$ by $^{\mathbf{t}}\!Z\,.\,\Pi\,.\,\overline{Z}=0$ and anti-homography $\phi\in\mathbb{G}_{circ}^{-}$ is described by $\Xi\doteq\phi\left(Z\right)=\Psi\,.\,\overline{Z}$ then cycle $\phi\left(\pi\right)$ is described by $^{\mathbf{t}}\!\,\Xi\,.\,^{\mathbf{t}}\!\,\Psi^{-1}\,.\,\overline{\Pi}\,.\,\overline{\Psi^{-1}}\,.\,\overline{\Xi}=0$. This can be summarized as :

$$\hat{\phi}\left(\Psi\right)\, =\,^{\mathbf{t}}\!\,\Psi^{-1}\,.\,\overline{\Pi}\,.\,\overline{\Psi^{-1}}\qquad in\;\,\mathbb{P_{C}GL\left(C^{\mathrm{2}}\right)}$$ (6.1)

Proof. Computation is straightforward and it can be checked that $\hat{\phi}\left(\Psi\right)$ inherits the hermitian property from $\Psi$, as required in (1.2). $\qedsymbol$

Proposition 6.2   Let $\boxed{\phi}$ be the matrix describing $\hat{\phi}$ as an action over apexes. Then following properties are verified :

$$\boxed{\phi}=R^{-1}\,.\,\operatorname{tens}\left(^{\mathbf{t}}\!\... ...R}\,\left\vert\det\Psi\right\vert\in\mathbb{P_{R}GL\left(R^{\mathrm{4}}\right)}$$ (6.2)

$$^{\mathbf{t}}\!\,\boxed{\phi}\,.\,Mink\,.\,\boxed{\phi}=Mink\quad... ...(\lambda^{2}-1\right)\left(\lambda-\mu\right)\left(\lambda-\frac{1}{\mu}\right)$$ (6.3)

where $\mu+1/\mu=\left(\left\vert\alpha\right\vert^{2}+\left\vert\delta\right\vert^{2... ...ta\overline{\gamma}+\gamma\overline{\beta}\right)/\left\vert\det\Psi\right\vert$.

Proof. Direct computation from (1.2) and (3.1). Formula (6.2) comes from the very definition of a tensor product apart from $\left\vert\det\Psi\right\vert$. This real factor has been introduced to obtain a better looking result in $\mathbb{R}^{4}$ since only the ratio between eigenvalues has a meaning in $\mathbb{P}_{\mathbb{R}}\mathbb{R}^{4}$. Formula (6.3a) enforces the fact that a point-circle goes to a point-circle. $\qedsymbol$

Proposition 6.3   When $\mu\in\mathbb{R}$, $\mu\neq\pm1$, the corresponding eigenvectors are two orthogonal cycles $\pi_{1},\,\pi_{2}$ and their intersections $\omega_{3},\,\omega_{4}$. When iterating $\phi$, the action on the $isoptic\left(\omega_{3},\,\omega_{4}\right)$ cycles is involutary while the $isotomic\left(\omega_{3},\,\omega_{4}\right)$ cycles are converging towards one of the two points.

Proof. Let $\lambda_{j}$ be the all different eigenvalues and $X_{j}$ their eigenvectors. We have :

$$^{\mathbf{t}}\!X_{j}\,.\,Minc\,.\,X_{k}=^{\mathbf{t}}\!X_{j}\,.\,... ...hi}\,.\,X_{k}=\lambda_{j}\,\lambda_{k}\,^{\mathbf{t}}\!X_{j}\,.\,Minc\,.\,X_{k}$$

Therefore all $mink\left(X_{j},\,X_{j}\right)$ are null except from $mink\left(X_{i},\,X_{i}\right)$ $i=1,2$ and $mink\left(X_{3},\,X_{4}\right)$. From $mink\left(X_{3},\,X_{3}\right)=0$, $\omega_{3}$ is a point. From rank, $\omega_{3}\neq\omega_{4}$ so that each $\pi_{i}$ contains at least two points and are real cycles (Fig. 6(a)). $\qedsymbol$

Fig. 6: Anti-homographic transformations

[ $\mu \approx 2$ ]

[ $\mu \notin \mathbb{R}$ ]

Remark 6.4   When $\mu \notin \mathbb{R}$, we have the same orthogonality relations and eigenvectors $\pi_{1},\,\pi_{2}$ are two circles centered at the same point $\omega_{0}$ (with $r_{2}=i\, r_{1}$), while $\pi_{3},$$\pi_{4}$ are the umbilical lines through $\omega_{0}$ and $\phi^{2}$ is a rotation around $\omega_{0}$ (Fig. 6(b)).

Definition 6.5   An inversion is an involutary element of $\mathbb{G}_{circ}^{-}.$

Theorem 6.6   The fixed cycles of an inversion $\phi$ are a proper cycle $\pi_{0}$ together with all the cycles orthogonal to $\pi_{0}$ -including the $\left\{ \omega\right\}$ where $\omega\in\pi_{0}$. The apex of $\pi_{2}=\phi\left(\pi_{1}\right)$ can be computed as :

$$X_{2}=\left(Id-2\,\boxed{p}\right)\,.\,X_{1}$$ (6.4)

where projector $\boxed{p}$ is defined in Proposition 3.5. This apex can also be constructed by drawing line $\left(X_{0}X_{1}\right)$. When $\pi_{1}$ is a point-circle then $S_{2}=shadow\left(\omega_{2}\right)$ is the second intersection of this line with $\mathbb{S}$. Otherwise, $X_{2}$ is the point of $\left(X_{0}X_{1}\right)$ such that division $\left(X_{0},\, X_{p},\, X_{1},\, X_{2}\right)$ is harmonic where $X_{p}$ = $\left(X_{0}\, X_{1}\right)\cap polar\left(X_{0}\right)$.

Relative to a line $\pi_{0}$, $\phi$ is the ordinary symmetry extended with $\phi\left(\infty\right)=\infty$. Relative to a circle, $\omega_{0}$ and $\infty$ are inverses and, otherwise, $\omega_{2}\doteq\phi\left(\omega_{1}\right)$ is characterized by $\omega_{1},\,\omega_{2}$ aligned with $\omega_{0}$ and $\overline{\omega_{0}\omega_{1}}.\overline{\omega_{0}\omega_{2}}=r_{0}^{2}$.

Proof. By definition, the minimal polynomial of $\phi$ is $\lambda^{2}-1$ so that a basis of eigenvectors can be found in $\mathbb{R}^{4}$. For projective reasons, choice $\mu=1$ can be done in (6.3). Since $X_{0}\in NullSpace\left(\phi+1\right)$, $\pi_{0}$ cannot belong to its polar plane (aka $NullSpace\left(\phi-1\right)$) and $\pi_{0}$ is a proper cycle. Formula (6.4) is $\left(1-p\right)x+p\,\mapsto\left(+1\right)\times\left(1-p\right)x+\left(-1\right)\times p\, x$, implying that $X_{0},\, X,\,\phi\left(X\right)$ are aligned. From this key result, everything follows.

When $\pi_{0}$ is an imaginary circle, we have $\overline{\omega_{0}\omega_{1}}.\overline{\omega_{0}\omega_{2}}=-\left\vert r_{0}\right\vert^{2}$ and there are no more (real) fixed points, while all other results still hold. $\qedsymbol$

Proposition 6.7   When inversion $\psi_{0}$ respective to cycle $\pi_{0}$ permutes the two proper cycles $\pi_{1},\pi_{2}$ then $\pi_{0}$ is one of the two cycles $\pi_{3},\,\pi_{4}$ defined by $X_{3},X_{4}\doteq\pm X_{1}\sqrt{mink\left(X_{2},\,X_{2}\right)}\pm X_{2}\sqrt{mink\left(X_{1},\,X_{1}\right)}$. Conversely, when $\pi_{1},\pi_{2}$ are not tangent to each other, $\pi_{3},\,\pi_{4}$ are two proper cycles that permute $\pi_{1}$ and $\pi_{2}$. When $\pi_{1},\pi_{2}$ are secant, both inversion cycles are real and otherwise there are one of each kind. Moreover, cycles $\pi_{3},\,\pi_{4}$ are ever orthogonal to each other.

Proof. If $\pi_{0}$ is such a cycle, then $X_{2}\simeq X_{1}-2\, X_{0}\,mink\left(X_{1},\,X_{0}\right)/mink\left(X_{0},\,X_{0}\right)$ and $\pi_{0}$ belongs to $P\left(\pi_{1},\,\pi_{2}\right)$ since $mink\left(X_{1},\,X_{0}\right)=0$ would implies $\psi_{0}\left(\pi_{1}\right)=\pi_{1}$. A straightforward computation shows that only the given weights are possible. Finally, the next coming Proposition 3.8 will prove that one of the $\pi_{3},\,\pi_{4}$ degenerates to a point-circle if and only if $\pi_{1},\pi_{2}$ are tangent. $\qedsymbol$

Remark 6.8   When $\pi_{1},\pi_{2}$ are circles, $\omega_{3},\omega_{4}$ are their centers of homothety while $X_{3},\, X_{4}\simeq\pm r_{2}\, X_{1}\pm r_{1}\, X_{2}$. When $\pi_{1},\pi_{2}$ are lines, $\pi_{3},\,\pi_{4}$ are their bisectors.

Construction 6.9   Obtain the inversion cycles $\pi_{3}\left(k\right),\,\pi_{4}\left(k\right)$ of cycles $\pi_{i}\left(k\right)$ centered at $\omega_{i}$ with radius $k\, r_{i}$ where $i=1,2$ and $k>0$ is a parameter.

Let $\Delta$ be the line through $\omega_{1},\omega_{2}$ and $\gamma$ its shadow on $\mathbb{S}$. All the involved cycles belong to bundle $\left(\left\{ \omega_{1}\right\} ,\,\left\{ \omega_{2}\right\} ,\,\left\{ \infty\right\} \right)$, so that everything occurs in the plane $E$ of small circle $\gamma$. Define $\delta=\left\vert\omega_{1}\omega_{2}\right\vert$, $\Lambda=x_{1}y_{2}-x_{2}y_{1}$ and consider matrix $Fig$ :

$$Fig\doteq\left(\begin{array}{cccc} y_{1}-y_{2} & x_{2}-x_{1} & \L... ...{1}\right) & \Lambda\left(x_{1}-x_{2}\right) & \delta^{2} & 0\end{array}\right)$$

Its first line is $^{\mathbf{t}}\!\,apex\left(\Delta\right)\,.\,Mink$, the second is $\overrightarrow{\omega_{2}\,\omega_{1}}$ and the third is their cross-product. Therefore matrix $Fig$ acts over elements of $\mathbb{R}^{3}\times\left\{ 1\right\}$ by sending elements of $E$ onto plane $\left\langle 0,u,v\right\rangle$ : $S_{\infty}$ goes to $\left\langle 0,0,-\delta^{2}\right\rangle$, while the half altitude points of $\gamma$ are sent to $\left\langle 0,\pm\delta^{2}/\sqrt{\delta^{2}+\Lambda^{2}},0\right\rangle$. A further normalization leads to Fig. 7.

Fig. 7: Inversion cycles of concentric circles

By definition, both families $\pi_{1}\left(k\right),$ $\pi_{2}\left(k\right)$ are concentric pencils. This is also the case for families $\pi_{3}\left(k\right),$ $\pi_{4}\left(k\right)$ since $\omega_{3},\,\omega_{4}$ depend only on the ratio of radii of circles $\pi_{1},\,\pi_{2}$. Therefore, the apex of a $\pi_{i}\left(k\right)$ lies on the line from $S=S_{\infty}$ to the corresponding $S_{i}$ ($1,2$ : solid line, $3,4$ : broken line on Fig. 7).

For a given $k$, cycles $\pi_{i}\left(k\right)$ belong to the same pencil and their apexes are on a same line $\Delta_{k}$. Apexes $s_{1},\, s_{2},\, t_{3},\, t_{4}$ are relative to $k\simeq0$ : $\pi_{1},\,\pi_{2}$ are external and $\pi_{3}$ imaginary. Apexes $u_{i}$ are relative to an intermediate value of $k$ : $\Delta_{u}$ is external to $\gamma$ so that $\pi_{1},\,\pi_{2}$ are secant and $\pi_{3},\,\pi_{4}$ are both real. Finally, apexes $v_{i}$ are relative to a greater value of $k$ : one of the $\pi_{1},\,\pi_{2}$ encloses the other and $\pi_{4}$ is imaginary.

Exercise   Identify the circle whose apex is $K$ (where all $\Delta_{k}$ are concurrent).