Möbius maps preserve circlines

hbghlyj

complex.pdf page 10

Proposition 2.14. Möbius maps take circlines to circlines. It is enough to check this for translations, dilations and inversions. The first two are easy and left as exercises; it remains to show that inversion preserves circlines.
We consider the case of circles first. Suppose we have a circle $\lvert z-a\rvert=r$. Case 1: $r≠\lvert a\rvert$ and $a≠0$. Then 0 is not on the circle and under inversion it becomes the set of points $\left\{z \in \mathbb{C}:\left\lvert\frac{1}{z}-a\right\rvert=r\right\}$, or equivalently $\left\lvert z-\frac{1}{a}\right\rvert=$ $\frac{r}{\lvert a\rvert}\lvert z\rvert$. By Lemma 1.4 this is also the equation of a circle (note $\frac{r}{\lvert a\rvert}≠1$ ). Case 2: $r=\lvert a\rvert$ and $a≠0$. Then 0 is on the circle and under inversion it becomes the set of points $\left\{z \in \mathbb{C}:\left\lvert z-\frac{1}{a}\right\rvert=\frac{r}{\lvert a\rvert}\lvert z\rvert\right\}$ together with $\infty$. The first set is a line (Lemma 1.4) and so this is a circline. Case 3: $a=0$. Under inversion, the circle $\rvert z\rvert=r$ becomes $\lvert z\rvert=\frac{1}{r}$, still a circle.
Now we look at lines (plus $\infty$). Suppose we have a line $\lvert z-a\rvert=\lvert z-b\rvert$ together with the point $\infty$. (Recall from Lemma 1.3 that this is the general form for a line.) Note that the line is the perpendicular bisector of the segment $\overline{a b}$, so by extending this segment if necessary we can always choose $a, b≠0$. Case 1: $\lvert a\rvert≠\lvert b\rvert$. Then 0 does not lie on the line and under inversion it becomes the set of points $\left\{z \in \mathbb{C} \backslash\{0\}:\lvert a\rvert\left\lvert\frac{1}{a}-z\right\rvert=\lvert b\rvert\left\lvert \frac{1}{b}-z\right\rvert\right\}$. The point $\infty$ maps to 0 . This set of points is a circle, by another application of Lemma 1.4. Case 2: $\lvert a\rvert=\lvert b\rvert$. Then the line passes through 0 . Under inversion it maps to $\left\{z \in \mathbb{C} \backslash\{0\}:\lvert a\rvert\left\lvert \frac{1}{a}-z\right\rvert=\lvert b\rvert\left\lvert \frac{1}{b}-z\right\rvert\right\} \cup\{\infty\}$. The point $\infty$ maps to 0 . By Lemma 1.3 this is a line through 0 together with the point $\infty$, and hence a circline.

Czhang271828

Here are some corollaries you might encounter in the study of Riemann geometry, algebra, modular form, etc.

Corollary 1: Möbius maps preseve the geodesics in upper plane $\mathbb H$ with metric $\mathrm ds^2=\dfrac{\mathrm dx^2+\mathrm dy^2}{y^2}$.
Corollary 2: The holomorphic automorphism group of $\mathbb H$ is isomorphic to $\mathrm{PSL}(2,\mathbb R)$.
Corollary 3: The holomorphic automorphism group of $\mathbb C\cup\{\infty\}$ is isomorphic to the group of Möbius transformations.
Corollary 4: For every transcendental extension $F(u)/F$, the group of automorphism fixing $F$ is the group of Möbius transformations on $F$. For instance, the automorphism group of the field $\mathbb Q[\pi]$ fixing $\mathbb Q$ is $\{\pi\mapsto (a\pi +b)/(c\pi +d)\mid ad-bc\neq 0\}$.

hbghlyj

warwickmaths.com/wp-content/uploads/2020/07/80_-M%C3%B6bius-Transformations.pdf

4.2 Standard Form
For this section, mostly taken from [2, pp. 40], we assume $m \neq\rm id$. Depending on the number of fixed points, we now try to find class representatives for each conjugacy class.

Suppose $m$ has 1 fixed point $z^*$ in $\widehat{\mathbb{C}}$. Let $w$ be any element of $\widehat{\mathbb{C}} \backslash\left\{z^*\right\}$. $z^*$ is the only fixed point, so $\left(z^{*}, w, m(w)\right)$ is a triple of distinct points in $\widehat{\mathbb{C}}$ so there exists⁷ $ p \in \text{Möb}(\widehat{\mathbb{C}})$ that takes this triple to $(\infty, 0,1)$. So $\left(p \circ m \circ p^{-1}\right)(\infty)=p\left(m\left(z^{*}\right)\right)=p\left(z^{*}\right)=\infty$. Hence $\infty$ is a fixed point of this composition.
Therefore we can write $\left(p \circ m \circ p^{-1}\right)(z)=a z+b$ for some $a \neq 0$. (Previously we have shown $m(\infty)=\infty \Leftrightarrow c=0$ ). From the above Lemma, $p \circ m \circ p^{-1}$ also must have one fixed point which we have shown is $\infty$. So there are no solutions in $\mathbb{C}$ to the equation $\left(p \circ m \circ p^{-1}\right)(z)=a z+b=z$ which means that $a=1$. Additionally, $\left(p \circ m \circ p^{-1}\right)(0)=p(m(w))=1$, then $b=1$.

Therefore, Möbius transformations $m$ with exactly one fixed point are conjugate to $n_1: z \mapsto z+1$. We call such transformations parabolic, and $n_1$ its standard form.
________________
⁷ Note that even though a Möbius transformations uniquely exists taking $(z^∗, w, m(w))\mapsto(∞, 0, 1)$, $p$ is not unique.
This arises from the choice we had for $w$.

hbghlyj

Suppose instead that $m$ has 2 fixed points $z_{1}^{*}, z_{2}^{*}$ in $\widehat{\mathbb{C}}$. Take $q \in \text{M}\ddot{\text o}\text{b}(\widehat{\mathbb{C}})$ satisfying $q\left(z_{1}^{*}\right)=0, q\left(z_{2}^{*}\right)=\infty$. Because of the way we chose $q,\left(q \circ m \circ q^{-1}\right)(\infty)=q\left(m\left(z_{2}^{*}\right)\right)=q\left(z_{2}^{*}\right)=\infty$ so $\left(q \circ m \circ q^{-1}\right)(z)=a z+b$ for some $a \neq 0$ as before, and $\left(q \circ m \circ q^{-1}\right)(0)=q\left(m\left(z_{1}^{*}\right)\right)=q\left(z_{1}^{*}\right)=0$, so $b=0$. Also $a \neq 1$ because $\rm id$ is in a conjugacy class of its own.

Therefore, Möbius transformations $m$ with exactly 2 fixed points are conjugate to $n_{2}: z \mapsto a z$ for some $a \in \mathbb{C} \backslash\{0,1\}$. We call $a$ the multiplier of $m$, and $n_2$ its standard form.

Like before (with $p$), this $q$ is not unique. However we see below that this does not really matter, from this lemma from [2, pp. 41].

Lemma 4.8. Suppose $m \in \text{M}\ddot{\text o}\text{b}(\widehat{\mathbb{C}})$ has 2 fixed points $x, y \in \widehat{\mathbb{C}}$, and that $q_{1}, q_{2} \in \text{M}\ddot{\text o}\text{b}(\widehat{\mathbb{C}})$ satisfy $q_{1}(x)=q_{2}(x)=0, q_{1}(y)=q_{2}(y)=\infty$ then $q_{1} \circ m \circ q_{1}^{-1}=q_{2} \circ m \circ q_{2}^{-1}$. (So the multiplier $a$ is the same.) Furthermore, if $q \in \text{M}\ddot{\text o}\text{b}(\widehat{\mathbb{C}})$ satisfies $q(x)=\infty, q(y)=0$ then the multiplier of $q \circ m \circ q^{-1}$ is $\frac{1}{a}$

Corollary 4.9. The multiplier is defined up to its inverse. Furthermore, if $m(z)=a z, T(z)=\frac{1}{z}$, then $T \circ m \circ T^{-1}(z)=\frac{1}{a} z=m^{-1}(z)$

This tells us that the standard form for a non-parabolic transformation (i.e. any transformation with 2 fixed points) is unique up to the multiplicative inverse of the multiplier–and furthermore, that these sit in the same conjugacy class, in particular being conjugate by the inversion map.

We can classify further depending on the multiplier. Recall $a \in \mathbb{C} \backslash\{0,1\}$.
If $|a|=1$, for some $\theta \in(0,2 \pi)$ we can write $a=e^{i \theta}$, so the transformation is a rotation counterclockwise by angle $\theta$. We call $m$ elliptic and say that $q \circ m \circ q^{-1}(z)=e^{i \theta} z$ is its standard form.

The final case is if $|a| \neq 1$. So $a=r e^{i \theta}$ for $r \in \mathbb{R} \backslash\{0,1\}, \theta \in[0,2 \pi)$. $m$ is said to be loxodromic with $q \circ m \circ q^{-1}(z)=r e^{i \theta} z, \theta \neq 0$ is its standard form. In the special case where $\theta=0$ (a pure dilation) then we say $m$ is hyperbolic–having standard form $q \circ m \circ q^{-1}(z)=r z$.

Observe that if $m$ is loxodromic, then it is the composition of dilation by $r$ and a rotation by $\theta$, both about 0, performed in either order; hence its standard form is just the composition of hyperbolic and elliptic transformations.

hbghlyj

Now there is a fast method of determining the standard form of any given Möbius transformation. First we need to introduce the following function

Definition 4.11. For any $m=\frac{a z+b}{c z+d} \in \text{M}\ddot{\text o}\text{b}(\widehat{\mathbb{C}})$ we define $\tau: \text{M}\ddot{\text o}\text{b}(\widehat{\mathbb{C}}) \rightarrow \mathbb{C}$, by $\tau(m)=(a+d)^{2}$. The respective map for $[A] \leftrightarrow m$ from $\operatorname{PSL}(2, \mathbb{C}) \rightarrow \mathbb{C}$ is $\operatorname{tr}^{2}:[A] \mapsto(a+d)^{2}$.

The symbol $\tau$ is chosen for this function as we are motivated by the trace of a matrix. Note that if we had defined instead $\tau(m)=a+d$, then this would be ill defined, due to the dual representation of each map, however $\tau$ is well defined as $((-a)+(-d))^{2}=(a+d)^{2}$.
Using this new function, in [2, pp. 45] you can find the proof for the following:
Proposition 4.12. Let $m \in \text{M}\ddot{\text o}\text{b}(\widehat{\mathbb{C}}) \backslash\{\mathrm{id}\}$. Then:
(a) $m$ is parabolic $\Leftrightarrow \tau(m)=4$
(b) $m$ is elliptic $\Leftrightarrow \tau(m) \in \mathbb{R}$ and $0 \leq \tau(m)<4$
(c) $m$ is hyperbolic $\Leftrightarrow \tau \in \mathbb{R} \backslash(-\infty, 4]$
(d) $m$ is loxodromic $\Leftrightarrow \tau \in \mathbb{C} \backslash[0,4]$
Now we have a very useful tool that enables us to immediately determine the type of a given transformation. We can go further and see that $\tau$ determines the conjugacy class entirely.
Lemma 4.13. If $f, g, p \in \text{M}\ddot{\text o}\text{b}(\hat{\mathbb{C}})$, then $\tau(f \circ g)=\tau(g \circ f)$ and $\tau\left(p \circ f \circ p^{-1}\right)=\tau(f)$.
Proof. (Sketch)
To see $\tau(f \circ g)=\tau(g \circ f)$, use the correspondence with matrices and that $\operatorname{tr}(A B)=\operatorname{tr}(B A)$ for matrices $A, B$. Then, $\tau\left(p \circ f \circ p^{-1}\right)=\tau\left(p^{-1} \circ p \circ f\right)=\tau(f)$.  $\square$

Due to the second property in 4.13, it it enough to just consider the values of $\tau$ on the standard forms.
Now take $f, g \in \text{M}\ddot{\text o}\text{b}(\widehat{\mathbb{C}})$, satisfying $\tau(f)=\tau(g)$. First, if $\tau(f)=4=\tau(g)$, then $f$ and $g$ are parabolic. So they have the same standard form $n_{1}: z \mapsto z+1$. So $f \sim g$.
Now if $\tau(f) \neq 4$, let $\alpha$ be the multiplier of $f$, $\beta$ the multiplier of $g$. So
$$
\tau(f)=\tau(g) \Rightarrow\left(\sqrt{\alpha}+\frac{1}{\sqrt{\alpha}}\right)^{2}=\left(\sqrt{\beta}+\frac{1}{\sqrt{\beta}}\right)^{2} \Rightarrow \alpha+\frac{1}{\alpha}+2=\beta+\frac{1}{\beta}+2 \Rightarrow \alpha+\frac{1}{\alpha}=\beta+\frac{1}{\beta}
$$
which has solutions $\alpha=\beta$ and $\alpha=\frac{1}{\beta}$. Then by corollary 4.9, we can see that the standard forms are either equal (so trivially conjugate) or conjugate by the inversion map, so (as conjugation is an equivalence relation), $f \sim g$.

Coupling this with 4.13, we get that $f \sim g \Leftrightarrow \tau(f)=\tau(g)$. So the multiplier is completely determined by $\tau$ and vice versa. We state this as a proposition:
Proposition 4.14. $f \sim g \Leftrightarrow \tau(f)=\tau(g)$. In terms of matrices, given $A, B$ - matrices corresponding to $f, g$ resp. , then $[A]$ is similar⁸ to $[B] \Leftrightarrow \operatorname{tr}(A)^2=\operatorname{tr}(B)^{2}$
_____________________
⁸ We defined what similarity of elements of $\mathrm{PSL}(2,\mathbb{C})$ means at the start of 4.1 .

hbghlyj

Möbius 变换的共轭分类与上半双曲空间的等距

[Film] Möbius Transformations Revealed
2007 ‧ Short
YouTube 3 Jan 2010
YouTube 9 Jul 2012

multiplier 为 $\alpha$ 的 Möbius 变换对应到$SL(2,\Bbb C)$里的矩阵的特征值是$\sqrt\alpha,\frac1{\sqrt\alpha}$

hbghlyj

We alternatively can think about Möbius transformations as iterative maps in terms of their fixed points, as is done in [6, pp. 169].

Let $m \in \text{M}\ddot{\text o}\text{b}(\hat{\mathbb{C}})$ be a non-identity map have 2 distinct fixed points, $z_{+}, z_{-}$($m$ is any non-parabolic transformation). Now take $q=\frac{z-z_{+}}{z-z_{-}}$ as in section 4.2, a map satisfying $q\left(z_{+}\right)=0, q\left(z_{-}\right)=\infty$. We know
that $q \circ m \circ q^{-1}=n$, where $n: z \mapsto a z$ for some multiplier $a$. Rewriting, as $q \circ m=n \circ q$, and $w=m(z)$ (the image under $m$) we see:
$$
\frac{w-z_{+}}{w-z_{-}}=a \frac{z-z_{+}}{z-z_{-}}
$$
Now call $w^{(k)}=m^{k}(z)$, the image of $z$ under $k$ applications of $m$, and $z_{0}$ the starting point. Then $m=q^{-1} \circ n \circ q$, $m^{k}=\left(q^{-1} \circ n \circ q\right)^{k}=q^{-1} \circ n^{k} \circ q$, so $q \circ m^{k}=n^{k} \circ q$. Thus
$$
\frac{w^{(k)}-z_{+}}{w^{(k)}-z_{-}}=a^{k} \frac{z_{0}-z_{+}}{z_{0}-z_{-}}=a^{k} C
$$
where $C$ is a constant.
We can then solve for $w^{(k)}$ to yield:
$$
w^{(k)}=\frac{z_{+}-a^{k} C z_{-}}{1-a^{k} C}
$$
For $z_{0} \neq z_{+}, z_{-}$, if $|a|>1$, then we see that the $a^{k}$ term dominates, so $w^{(k)} \mapsto z_{-}$, as $k \mapsto \infty$, and if $|a|<1$, then $a^{k} \mapsto 0$ as $k \mapsto \infty$, so $w^{(k)} \mapsto z_{+}$. Note that if $|a|=1$, the limit does not exist. This agrees geometrically, as elliptical maps just rotate points around the fixed points. In the cases where $z_{0}=z_{+}$ or $z_{0}=z_{-}$, then we know these stay put as $k \mapsto \infty$, being fixed points.
So for non-parabolic transformations, points either flow from one fixed point to the other, or oscillate around them. In particular, for hyperbolic transformations, these points move along circles through the 2 fixed points.
If instead $m(z)=\frac{a z+b}{c z+d}$ is parabolic and has one fixed point $z^{*}=\frac{a-d}{2 c} \in \hat{\mathbb{C}}$, then
If $z_{*}$ is finite the normal form is$$\frac{1}{w-z_{*}}=\frac{1}{z-z_{*}} \pm c \quad\text{ so }\quad \frac{1}{w^{(k)}-z_{*}}=\frac{1}{z^{(0)}-z_{*}} \pm k c$$where sign of $c$ is positive if $a+d=2$, negative if $a+d=-2$.
If $z_{*}=\infty$, then we know that $c=0$, and as $a d-b c=1$, then $a=d=\pm 1$ so $m(z)=z \pm b$. Hence the normal form is $^{\color{blue}{12}}$
$$
w=z \pm b \quad \text { so } \quad w^{(k)}=z^{(0)} \pm k b
$$
In both of these cases we can see that $\lim _{k \to \infty} w^{(k)}=z_{*}$.
So for parabolic transformations, all points move towards the fixed point.
___________________
$^{\color{blue}{12}}$ We also know $b\ne0$, as $m\ne\rm id$.

hbghlyj

和不动点法求数列通项有关. 都是分式线性变换的迭代.

hbghlyj

If you can, cast your mind back to when we described the inversion in terms of a rotation of the Riemann sphere. Here we see that there was nothing special about the inversion - we can do it for any Möbius transformation! Additionally, these are the only maps for which this property holds!
This is shown in [3], where the following elegant result is stated and proven.
Theorem 5.1. A complex mapping is a Möbius transformation if and only if it can be obtained by
stereographic projection of the complex plane onto an admissible sphere in $\Bbb R^3$, followed by a rigid motion of the sphere in $\Bbb R^3$ which maps it to another admissible sphere, followed by stereographic projection back to the plane.
An sphere in $\Bbb R^3$ is admissible if its north pole, $N$, lies above the $xy$-plane. For any admissible sphere, a stereographic projection can be defined as before: for each point $P$ on the sphere, draw the line through $N$ and $P$ and map this to the intersection of this line with the $xy$ plane. So again $N$ is identified with $∞$, and the remainder of the sphere with $C$. The proof runs as follows:
Proof. (Sketch) We use the fact (similar to 3.5) that any Möbius transformation $f$ can be built from (in this order), a translation, inversion, dilation, rotation, and a final translation: we can find $\alpha, \beta \in \mathbb{C}, \rho,\theta \in \mathbb{R}$ such that
$$\tag1
f(z)=\frac{\rho r^{i \theta}}{z+\alpha}+\beta
$$
Therefore, we just have to show that for each of these types of map there exists an admissible sphere $S$ and rigid motion $T$, such that $S^{\prime}=T S$ is also admissible and we can write each as $P_{S^{\prime}} \circ T \circ P_{S}$.

For any translation, we can choose $S$ as any admissible sphere, and $T$ to be the same translation extended to $\mathbb{R}^{3}$. $S^{\prime}$ will always be admissible as there is no change in vertical coordinates.

For the other 3 maps, let $S$ be the unit sphere. $T$ for the rotation is also just the rotation extended to $\mathbb{R}^{3}$. The dilation is more interesting, and we take the corresponding motion as a translation vertically by $\rho-1$. To produce the inversion, as we showed earlier we take $T$ to be the rotation around the real axis by an angle of $\pi$.

Then to write (1) in the form $f=P_{S^{\prime}} \circ T \circ P_{S}$, we take $S$ as the unit sphere with centre $-\alpha$, $T$ to be the composition of a translation by $\alpha$, a rotation by $\pi$ about the real axis, a rotation by $\theta$ about the axis orthogonal to the $x y$-plane, a translation upwards by $\rho-1$, and finally a translation by $\beta$.

hbghlyj

The Geometry of Möbius Transformations, John Olsen
mathpages.com/home/kmath464/kmath464.htm
oocities.org/titus_piezas/Cubic.pdf
Möbius Transformations Revealed, Douglas N. Arnold and Jonathan Rogness
www-users.cse.umn.edu/~arnold/moebius/