12.3: Auxiliary statements

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

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

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 $0 < \hat{p} < p$.