calculations in hyperbolic geometry

 Here, we look again at equations for hyperbolic lines, and, in particular, to establish a condition for three points to lie on a single hyperbolic line, i.e. to be collinear. We begin by looking at lines on the complex plane. Theorem 1 (1) If α and β are distinct points, then the equation of the line αβ is (zβ*-z*β)+(αz*-α*z)+(βα*-β*α) = 0. (2) Three points α, β, γ are collinear if and only if (γβ*-γ*β)+(αγ*-α*γ)+(βα*-β*α) = 0. proof (1) z lies on αβ if and only if (z-α) and (β-α) are parallel, i.e. (z-α)/(β-α) is real, i.e. (z-α)(β-α) = (z*-α*)/(β*-α*). Cross-multipling and simlifying, we get the stated equaton. (2) If two are equal, then all are collinear, and the stated condition can readily be checked. Otherwise, we may assume that α ≠ β. Then γ lies on αβ if and only if it satisfies the equation in (1). This gives the stated condition. An important mapping We know that a hyperbolic line has an equation of the form azz*-δ*z-δz*+a = 0, with a real and |δ| > |a|. Then w = 2z/(1+zz*) satisfies δ*w+δw*=2a, which is a line. This suggests that we look at the map k(z) = 2z/(1+zz*). Lemma The mapping k (1) maps D injectively to D, (2) fixes points of C, (3) maps hyperbolic lines to segments of lines in D. proof (1) If w = 2z/(1+zz*), then 1-|w| = 1-2|z|/(1+|z|2) = (1-|z|2)/(1+|z|2). Thus, if |z| < 1, |w| < 1. If z,z'εD and k(z)=k(z'), then z(1+z'z'*) = z'(1+zz*), so (z-z') =zz'(z*-z'*). If z ≠ z' then taking moduli, we get 1 = |zz'|. As z,z'εD, |z|, |z'| < 1, so we have a contradiction. Thus, we must have z = z', so k is injective. (2) If |z|=1, then zz*=1, so k(z) = 2z/(1+1) = z. (3) follows from the remarks preceding the lemma, and (1). Theorem 2 (1) If α and β are distinct points of D, the hyperbolic line through α and β has equation (1+zz*)(βα*-β*α) +(1+α*α)(zβ*-z*β) +(1+β*β)(αz*-α*z) = 0. (2) Three points α, β, γ lie on a hyperbolic line if and only if (1+γ*γ)(βα*-β*α) +(1+α*α)(γβ*-γ*β) +(1+β*β)(αγ*-α*γ) = 0 proof These follow quickly from the Lemma and Theorem 1. From the Lemma, points α, β, γ lie on a hyperbolic line if and only if k(α), k(β), k(γ) lie on a euclidean line. The parts of the present result folow from those of Theorem 1 on multiplying through by the product (1+α*α), (1+β*β), (1+γ*γ). Note that the coefficients which occur are of the form uv*-u*v, so are purely imaginary. We can get a real equation/condition by multiplying through by i. alternative description We can also observe that α, β, γ are on a euclidean line if and only if there are real r,s,t, not all zero, with rα+sβ+tγ=0 and r+s+t=0. This is fairly easy to check. Each point γ on αβ has the form γ=kα+(1-k)β. Thus we can take r=k, s=-1, t=1-k. As in Theorem 2, we get the corresponding result for hyperbolic lines, namely α, β, γ lie on a hyperbolic line if and only if rα/(1+α*α)+sβ/(1+β*β)+tγ/(1+γ*γ) = 0, with real r,s,t as above.