DIFFERENTIAL GEOMETRY: A First Course in Curves and Surfaces

§2. An Introduction to Hyperbolic Geometry 87Γ u uv = E v2E = −1 vΓvv u = − G u2E =0Using the formula (∗) for Gaussian curvature on p. 57, we findK = − 1 ( (2 √ Ev) (√EG+ Gu) )√ = − v2EG v EG u 2Γ v uv = G u2G =0Γv vv = G v2G = −1 v .(− 2 v 3 · v2) v = −v2 2 ·2v 2 = −1.Thus, the hyperbolic plane has constant curvature −1. Note that it is a consequence of Corollary1.6 that the area of a geodesic triangle in H is equal to π − (ι 1 + ι 2 + ι 3 ).What are the geodesics in this surface? Using the equations (♣♣) onp.68, we obtain theequationsu ′′ − 2 v u′ v ′ = v ′′ + 1 v (u′2 − v ′2 )=0.Obviously, the vertical rays u = const give us solutions (with v(t) =c 1 e c2t ). Next we seek geodesicswith u ′ ≠0,sowelook for v as a function of u and obtain (with very careful use of the chain rule)d 2 vdu 2 = d ( ) v′du u ′ = u′ v ′′ − u ′′ v ′u ′2 · 1u ′= 1 ) (v ′2 − u ′2) ( )) 2− v ′ v u′ v ′( 1(u ′u ′3 v(= − 1 ( ) ) (v′ 21+v u ′ = − 1 1+vThis means we are left with the differential equation( )v d2 v dv 2du 2 + = d (v dv )= −1,du du duand integrating this twice gives us the solutionsu 2 + v 2 = au + b.( ) )dv2.duThat is, the geodesics in H are the vertical rays and the semicircles centered on the u-axis, aspictured in Figure 2.1. Note that any semicircle centered on the u-axis intersects each vertical linevuFigure 2.1at most one time. It now follows that any two points P, Q ∈ H are joined by a unique geodesic. If P