Download Hyperbolic Geometry (2nd Edition) (Springer Undergraduate by James W. Anderson PDF

By James W. Anderson

The geometry of the hyperbolic airplane has been an energetic and engaging box of mathematical inquiry for many of the earlier centuries. This publication offers a self-contained creation to the topic, appropriate for 3rd or fourth 12 months undergraduates. the fundamental process taken is to outline hyperbolic strains and increase a typical workforce of adjustments retaining hyperbolic strains, after which examine hyperbolic geometry as these amounts invariant less than this crew of transformations.

Topics coated contain the higher half-plane version of the hyperbolic airplane, Möbius ameliorations, the overall Möbius staff, and their subgroups retaining the higher half-plane, hyperbolic arc-length and distance as amounts invariant less than those subgroups, the Poincaré disc version, convex subsets of the hyperbolic aircraft, hyperbolic quarter, the Gauss-Bonnet formulation and its applications.

This up to date moment variation additionally features:
an extended dialogue of planar versions of the hyperbolic airplane bobbing up from advanced analysis;
the hyperboloid version of the hyperbolic plane;
brief dialogue of generalizations to better dimensions;
many new exercises.

The sort and point of the booklet, which assumes few mathematical necessities, make it a great creation to this topic and gives the reader with an organization seize of the recommendations and strategies of this gorgeous a part of the mathematical panorama.

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Additional resources for Hyperbolic Geometry (2nd Edition) (Springer Undergraduate Mathematics Series)

Example text

When we normalize so that the determinant of n (or equivalently of m) is 1, we need to write αz n(z) = −1 , α and so τ (n) = τ (p ◦ m ◦ p−1 ) = τ (m) = (α + α−1 )2 . In the case in which m is elliptic, so that |α| = 1, write α = eiθ for some θ in (0, π). Calculating, we see that τ (m) = (α + α−1 )2 = eiθ + e−iθ 2 = 4 cos2 (θ). In particular, we have that τ (m) is real and lies in the interval [0, 4). In the case in which m is loxodromic, so that |α| ̸= 1, we write α = ρeiθ for some ρ > 0, ρ ̸= 1, and some θ in [0, π).

Substituting this calculation into the equation for A given above yields βz + βz + γ 1 1 = β (w − b) + β (w − b) + γ a a β β β β w+ = w − b − b + γ = 0. a a a a As βa b + βa b = 2 Re βa b is real and as the coefficients of w and w are complex conjugates, this shows that w also satisfies the equation of a Euclidean line. Hence, f takes Euclidean lines in C to Euclidean lines in C. The proof that f takes Euclidean circles to Euclidean circles is similar and is left as an exercise. 2 Show that the homeomorphism f of C defined by setting f (z) = az + b for z ∈ C and f (∞) = ∞, where a, b ∈ C and a ̸= 0, takes Euclidean circles in C to Euclidean circles in C.

As m◦f fixes ∞ and lies in HomeoC (C), we see that m◦f takes Euclidean lines in C to Euclidean lines in C, and it takes Euclidean circles in C to Euclidean circles in C. Also, if X and Y are two Euclidean lines in C that intersect at some point z0 , and if m ◦ f (X) = X and m ◦ f (Y ) = Y , then m ◦ f (z0 ) = z0 and so z0 is contained in this set Z of points fixed by m ◦ f . For each s ∈ R, let V (s) be the vertical line in C through s and let H(s) be the horizontal line in C through is. Let H be any horizontal line in C.

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