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Mathematics LibreTexts

17.1: Special bijection on the h-plane

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    23689
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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 \(P \in \Gamma_i\).