$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}{\| #1 \|}$$ $$\newcommand{\inner}{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$

# 12.3: Auxiliary statements

$$\newcommand{\vecs}{\overset { \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}{\| #1 \|}$$ $$\newcommand{\inner}{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}{\| #1 \|}$$ $$\newcommand{\inner}{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$

One may compare the conformal model with a telescope — it makes it possible to see the h-plane from the Euclidean plane. Continuing this analogy further, we may say that the following lemma will be used to aim the telescope at any particular point in the h-plane.

Lemma $$\PageIndex{1}$$

Consider an h-plane with a unit circle as the absolute. Let $$O$$ be the center of the absolute and $$P$$ be another h-point. Suppose that $$P'$$ denotes the inverse of $$P$$ in the absolute.

Then the circle $$\Gamma$$ with the center $$P'$$ and radius $$\dfrac{\sqrt{1-OP^2}}{OP}$$ is perpendicular to the absolute. Moreover, $$O$$ is the inverse of $$P$$ in $$\Gamma$$. Proof

Follows from Exercise 10.5.2.

Assume $$\Gamma$$ is a circline that is perpendicular to the absolute. Consider the inversion $$X \mapsto X'$$ in $$\Gamma$$, or if $$\Gamma$$ is a line, set $$X \mapsto X'$$ to be the reflection across $$\Gamma$$.

The following observation says that the map $$X \mapsto X'$$ respects all the notions introduced in the previous section. Together with the lemma above, it implies that in any problem that is formulated entirely in h-terms we can assume that a given h-point lies in the center of the absolute.

Theorem $$\PageIndex{1}$$

The map $$X \mapsto X'$$ described above is a bijection from the h-plane to itself. Moreover, for any h-points $$P$$, $$Q$$, $$R$$ such that $$P\ne Q$$ and $$Q\ne R$$, the following conditions hold:

1. The h-line $$(PQ)_h$$, h-half-line $$[PQ)_h$$, and h-segment $$[PQ]_h$$ are transformed into $$(P'Q')_h$$, $$[P'Q')_h$$, and $$[P'Q']_h$$ respectively.

2. $$\delta(P',Q')=\delta(P,Q)$$ and $$P'Q'_h=PQ_h$$.

3. $$\measuredangle_h P'Q'R' \equiv -\measuredangle_h PQR$$.

It is instructive to compare this observation with Proposition [prop:reflection].

Proof

According to Theorem 10.5.1, the map sends the absolute to itself. Note that the points on $$\Gamma$$ do not move, it follows that points inside of the absolute remain inside after the mapping. Whence the $$X \mapsto X'$$ is a bijection from the h-plane to itself.

Part (a) follows from Theorem 10.3.1 and Theorem 10.6.1.

Part (b) follows from Theorem 10.2.1.

Part (c) follows from Theorem 10.6.1.

Lemma $$\PageIndex{2}$$

Assume that the absolute is a unit circle centered at $$O$$. Given an h-point $$P$$, set $$x=OP$$ and $$y=OP_h$$. Then

$$y = \ln \dfrac{1 + x}{1 - x}$$                 and                     $$x = \dfrac{e^y - 1}{e^y + 1}.$$

Observe that according to lemma, $$OP_h \to \infty$$ as $$OP \to 1$$. That is if $$P$$ approaches absolute in Euclidean sense, it escapes to infinity in the h-sense.

Proof Note that the h-line $$(OP)_h$$ forms a diameter of the absolute. If $$A$$ and $$B$$ are the ideal points as in the definition of h-distance, then

$$\begin{array} {l} {OA = OB = 1,} \\ {PA = 1 + x,} \\ {PB = 1 - x.} \end{array}$$

In particular,

$$y = \ln \dfrac{AP \cdot BO}{PB \cdot OA} = \ln \dfrac{1 + x}{1 - x}.$$

Taking the exponential function of the left and the right hand side and applying obvious algebra manipulations, we get that

$$x=\dfrac{e^y-1}{e^y+1}$$.

Lemma $$\PageIndex{3}$$

Assume the points $$P$$, $$Q$$, and $$R$$ appear on one h-line in the same order. Then

$$PQ_h + QR_h = PR_h.$$

Proof

Note that

$$PQ_h + QR_h = PR_h$$

is equivalent to

$\delta (P, Q) \cdot \delta (Q,R) = \delta (P, R).$

Let $$A$$ and $$B$$ be the ideal points of $$(PQ)_h$$. Without loss of generality, we can assume that the points $$A$$, $$P$$, $$Q$$, $$R$$, and $$B$$ appear in the same order on the circline containing $$(PQ)_h$$. Then

$$\begin{array} {rcl} {\delta (P, Q) \cdot \delta (Q, R)} & = & {\dfrac{AQ \cdot BP}{QB \cdot PA} \cdot \dfrac{AR \cdot BQ}{RB \cdot QA} =} \\ {} & = & {\dfrac{AR \cdot BP}{RB \cdot PA} =} \\ {} & = & {\delta (P, R).} \end{array}$$

Hence 12.3.1 follows.

Let $$P$$ be an h-point and $$\rho>0$$. The set of all h-points $$Q$$ such that $$PQ_h=\rho$$ is called an h-circle with the center $$P$$ and the h-radius $$\rho$$.

Lemma $$\PageIndex{4}$$

Any h-circle is a Euclidean circle that lies completely in the h-plane.

More precisely for any h-point $$P$$ and $$\rho\ge 0$$ there is a $$\hat\rho\ge 0$$ and a point $$\hat P$$ such that

$$PQ_h= \rho \Leftrightarrow \hat{P}Q= \hat{\rho}$$

for any h-point $$Q$$.

Moreover, if $$O$$ is the center of the absolute, then

1. $$\hat{O}=O$$ for any $$\rho$$ and

2. $$\hat{P} \in (OP)$$ for any $$P\ne O$$.

Proof

According to Lemma $$\PageIndex{2}$$, $$OQ_h\z= \rho$$ if and only if

$$OQ= \hat{\rho}=\dfrac{e^{\rho}-1}{e^{\rho}+1}.$$

Therefore, the locus of h-points $$Q$$ such that $$OQ_h= \rho$$ is a Euclidean circle, denote it by $$\Delta_{\rho}$$. If $$P \ne O$$, then by Lemma $$\PageIndex{1}$$ and the main observation (Theorem $$\PageIndex{1}$$) there is inversion that respects all h-notions and sends $$O \mapsto P$$.

Let $$\Delta_{\rho}'$$ be the inverse of $$\Delta_{\rho}$$. Since the inversion preserves the h-distance, $$PQ_h=\rho$$ if and only if $$Q \in \Delta_{\rho}'$$.

According to Theorem 10.3.1, $$\Delta_\rho'$$ is a Euclidean circle. Let $$\hat P$$ and $$\hat\rho$$ denote the Euclidean center and radius of $$\Delta_\rho'$$.

Finally, note that $$\Delta_\rho'$$ reflects to itself across $$(OP)$$; that is, the center $$\hat P$$ lies on $$(OP)$$.

Exercise $$\PageIndex{1}$$

Assume $$P$$, $$\hat P$$, and $$O$$ are as in the Lemma $$\PageIndex{1}$$ and $$P \ne O$$. Show that $$\hat{P} \in [OP]$$.

Hint

Let $$X$$ and $$Y$$ denote the point of intersections of $$(OP)$$ and $$\Delta_{rho}'$$. Consider an isometry $$(OP) \to \mathbb{R}$$ such that $$O$$ corresponds to 0. Let $$x, y, p$$, and $$\hat{p}$$ denote the real number corresponding to $$X, Y, P$$, and $$\hat{P}$$.

We can assume that $$p > 0$$ and $$x < y$$. Note that $$\hat{p} = \dfrac{x + y}{2}$$ and

$$\dfrac{(1 + x) \cdot (1 - p)}{(1 - x) \cdot (1 + p)} = \dfrac{(1 + p) \dot (1 - y)}{(1 - p) \cdot (1 + y)}$$.

It remains to show that all this implies $$0 < \hat{p} < p$$.