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By E. M. Friedlander, M. R. Stein

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More generally if the flow associated with P has the vector direction a ∈ ‫ރ‬, the pairs of horocycles are {z : Im az ¯ = ±s, s = 0}. The prefix “horo” comes form the Greek word for “limit”. Fix a point O ∈ ‫ވ‬3 . Take the hyperbolic sphere σx centered at x ∈ ‫ވ‬3 and passing through O. As x → ζ ∈ ∂ ‫ވ‬3 , the limit of σx is the horosphere at ζ passing through O. We now take up the study of triangles. As already mentioned the area of a triangle is equal to the “angle deficit” π − θi , where the θi are the vertex angles; see Exercise 1-6 for a proof.

Thus z and z¯ are symmetric in ‫ޒ‬, while z and 1/¯z are symmetric in the unit circle centered at the origin. Verify that the formula for symmetric points ζ, ζ ∗ with respect to the circle {|z − a| = R} is ζ∗ −a = R2 . ζ¯ − a¯ This map extends to a reflection about the corresponding hyperbolic plane in ‫ވ‬3 . (i) Prove that a Möbius transformation maps points symmetric in a circle/line to points symmetric in the image circle/line. Hence the extension to ‫ވ‬3 preserves symmetry in planes. ) (ii) If C1 , C2 are disjoint circles and is the region they bound on ‫ޓ‬2 , show how to find a Möbius transformation that sends to an annulus centered at z = 0 with C2 sent to the outer circle of radius 1.

The four hyperbolic planes so obtained pairwise intersect in the six lines. The common exterior of these four planes is a four sided solid called an ideal tetrahedron. It is uniquely determined up to isometry by its four “ideal” vertices z 1 , z 2 , z 3 , z 4 . Now using the upper half-space model, send any one of the vertices to ∞. the three faces meeting at ∞ now become vertical planes. The cross section obtained by intersecting with any sufficiently high horizontal plane {t = N } is a euclidean triangle.

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