与反演可交换的Möbius变换

hbghlyj

近代欧氏几何学 [美]约翰逊 著

§ 80 定理 设两个已知点 $P, Q$ 关于圆 $c$ 互为反演, $P$, $Q$ 及圆 $c$ 关于另一个圆 $b$ 的反形为 $P^{\prime}, Q^{\prime}$ 及圆 $c^{\prime}$, 则 $P^{\prime}, Q^{\prime}$ 关于圆 $c^{\prime}$ 互为反演.
过 $P, Q$ 任作两个圆 $j, k$, 则它们都与圆 $c$ 正交, 它们关于 $b$ 的反形 $j^{\prime}, k^{\prime}$ 与 $c^{\prime}$ 正交. 因此 $j^{\prime}$ 与 $k^{\prime}$ 的交点 $P^{\prime}, Q^{\prime}$ 关于 $c^{\prime}$ 互为反演.
这个定理可以解释成“反演性经反演后不变”. 即两个互为反形的图形, 连同它们的反演圆, 受到一个反演的作用, 所得的图形仍互为反形.

一般地, Möbius变换保反演性:hyper-transf.pdf

Theorem 17.9 (The Symmetry Principle). If a Möbius transformation $f$ maps circle $c$ to circle $c^{\prime}$, then it maps points symmetric with respect to $c$ to points symmetric with respect to $c^{\prime}$.
Proof: Let $z_1, z_2, z_3$ be on $c$. Since
\begin{aligned}
\left(f\left(z^*\right), f\left(z_1\right), f\left(z_2\right), f\left(z_3\right)\right) &=\left(z^*, z_1, z_2, z_3\right) \\
&=\overline{\left(z, z_1, z_2, z_3\right)} \\
&=\overline{\left(f(z), f\left(z_1\right), f\left(z_2\right), f\left(z_3\right)\right)}
\end{aligned}the result follows.  $_\square$

hbghlyj

两点关于单位圆反演, 经变换$f(z)=β\frac{z-\alpha}{\bar{\alpha} z-1}$后仍然关于单位圆反演.
证明: 变换$T(z)=\frac1{\bar z}$是关于单位圆反演
$$f\left(T(z)\right)=β\frac{T(z)-\alpha}{\bar{\alpha} T(z)-1}=β\frac{1-\alpha\bar z}{\bar{\alpha}-\bar z}$$
所以
$$T\left(f\left(T(z)\right)\right)=\frac1{\barβ\frac{1-\bar\alpha z}{\alpha-z}}=\frac{z-\alpha}{\bar{\alpha} z-1}=f(z)$$
即$f∘T=T∘f$

hbghlyj

hbghlyj 发表于 2022-11-12 02:22
两点关于单位圆反演, 经变换$f(z)=\frac{z-\alpha}{\bar{\alpha} z-1}$后仍然关于单位圆反演.
证明: 变换$T ...

Möbius transformation > Elliptic transforms > circular transform
The simplest possibility of a fractional multiple means $α = π/2$, which is also the unique case of $\displaystyle \operatorname {tr} {\mathfrak {H}}=0$, is also denoted as a circular transform; this corresponds geometrically to rotation by 180° about two fixed points. This class is represented in matrix form as: $\displaystyle {\begin{pmatrix}0&-1\\1&0\end{pmatrix}}.$