17.1: Special bijection on the h-plane

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Consider the conformal disc model with the absolute at the unit circle $\Omega$ centered at $O$. Choose a coordinate system $(x,y)$ on the plane with the origin at $O$, so the circle $\Omega$ is described by the equation $x^2+y^2=1$.

$截屏2021-03-01 上午10.40.49.png$
The plane thru $P, O$, and $S$.

Let us think that our plane is the coordinate $xy$-plane in the Euclidean space; denote it by $\Pi$. Let $\Sigma$ be the unit sphere centered at $O$; it is described by the equation

$x^2+y^2+z^2=1.$

Set $S=(0,0,-1)$ and $N=(0,0,1)$; these are the south and north poles of $\Sigma$.

Consider stereographic projection $\Pi\to\Sigma$ from $S$; given point $P\in\Pi$ denote its image in $\Sigma$. Note that the h-plane is mapped to the north hemisphere; that is, to the set of points $(x,y,z)$ in $\Sigma$ described by the inequality $z>0$.

For a point $P'\in \Sigma$ consider its foot point $\hat P$ on $\Pi$; this is the closest point to $P'$.

Note that the composition $P \leftrightarrow P' \leftrightarrow \hat{P}$ of these two maps gives a bijection from the h-plane to itself. Further note that $P=\hat{P}$ if and only if $P\in \Omega$ or $P=O$.

Exercise $\PageIndex{1}$

Suppose that $P\leftrightarrow \hat P$ is the bijection described above. Assume that $P$ is a point of h-plane distinct from the center of absolute and $Q$ is its inverse in the absolute. Show that the midpoint of $[PQ]$ is the inversion of $\hat{P}$ in the absolute.

Hint

$截屏2021-03-01 上午11.23.55.png$

Let $N, O, S, P, P'$ and $\hat{P}$ be as on the diagram above.

Note that $QQ = \dfrac{1}{x}$ and therefore we need to show that $O\hat{P} = 2/(x + \dfrac{1}{x})$. To do this, show and use that $\triangle SOP \sim \triangle SP'N \sim \triangle P'\hat{P}P$ and $2 \cdot SO = NS$.

Lemma $\PageIndex{1}$

Let $(PQ)_h$ be an h-line with the ideal points $A$ and $B$. Then $\hat P,\hat Q\in[AB]$.

Moreover,

\[\dfrac{A \hat{Q} \cdot B \hat{P}}{\hat{QB}\cdot \hat{P}A} = (\dfrac{AQ\cdot BP}{QB\cdot PA})^2. \]

In particular, if $A,P,Q,B$ appear in the same order, then

$PQ_h=\dfrac{1}{2} \cdot \ln \dfrac{A\hat{Q}\cdot B\hat{P}}{\hat{Q}B\cdot \hat{P}A}.$

Proof

Consider the stereographic projection $\Pi \to \Sigma$ from the south pole $S$. Note that it fixes $A$ and $B$; denote by $P'$ and $Q'$ the images of $P$ and $Q$;

According to Theorem 16.3.1c,

\[\dfrac{AQ\cdot BP}{QB\cdot PA}=\dfrac{AQ'\cdot BP'}{Q'B\cdot P'A}.\]

By Theorem Theorem 16.3.1e, each circline in $\Pi$ that is perpendicular to $\Omega$ is mapped to a circle in $\Sigma$ that is still perpendicular to $\Omega$. It follows that the stereographic projection sends $(PQ)_h$ to the intersection of the north hemisphere of $\Sigma$ with a plane perpendicular to $\Pi$.

Suppose that $\Lambda$ denotes the plane; it contains the points $A$, $B$, $P'$, $\hat P$ and the circle $\Gamma=\Sigma\cap\Lambda$. (It also contains $Q'$ and $\hat{Q}$ but we will not use these points for a while.)

$截屏2021-03-01 上午11.16.24.png$
The plane $\Lambda$.

Note that

$A,B,P'\in\Gamma$,
$[AB]$ is a diameter of $\Gamma$,
$(AB)=\Pi\cap\Lambda$,
$\hat{P} \in [AB]$
$(P'\hat{P} \perp (AB).$

Since $[AB]$ is the diameter of $\Gamma$, by Corollary 9.8, the angle $AP'B$ is right. Hence $\triangle A\hat PP' \sim \triangle AP'B \sim \triangle P'\hat PB$. In particular

$\dfrac{AP'}{BP'}=\dfrac{A \hat{P}}{P'\hat{P}}=\dfrac{P'\hat{P}}{B\hat{P}}.$

Therefore

\[\dfrac{A\hat{P}}{B\hat{P}}=(\dfrac{AP'}{BP'})^2.\]

The same way we get that

\[\dfrac{A\hat{Q}}{B\hat{Q}}=(\dfrac{AQ'}{BQ'})^2.\]

Finally, note that 17.1.2+17.1.3+17.1.4 imply 17.1.1

The last statement follows from 17.1.1 and the definition of h-distance. Indeed,

$\begin{aligned} PQ_h&:=\ln\frac{A Q\cdot B P}{QB\cdot PA}= \\ &=\ln\left(\frac{A \hat Q\cdot B \hat P}{\hat QB\cdot \hat PA}\right)^{\frac12}= \\ &=\tfrac12\cdot\ln\frac{A \hat Q\cdot B \hat P}{\hat QB\cdot \hat PA}.\end{aligned}$

Exercise $\PageIndex{2}$

Let $\Gamma_1$, $\Gamma_2$, and $\Gamma_3$ be three circles perpendicular to the circle $\Omega$. Let $[A_1B_1]$, $[A_2B_2]$, and $[A_3B_3]$ denote the common chords of $\Omega$ and $\Gamma_1$, $\Gamma_2$, $\Gamma_3$ respectively. Show that the chords $[A_1B_1]$, $[A_2B_2]$, and $[A_3B_3]$ intersect at one point inside $\Omega$ if and only if $\Gamma_1$, $\Gamma_2$, and $\Gamma_3$ intersect at two points.

$截屏2021-03-01 上午11.21.52.png$

Hint: Consider the bijection $P \leftrightarrow \hat{P}$ of the h-plane with absolute $\Omega$. Note that $\hat{P} \in [A_iB_i]$ if and only if $P \in \Gamma_i$.

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Exercise \(\PageIndex{1}\)

Lemma \(\PageIndex{1}\)

Exercise \(\PageIndex{2}\)