Hyperbolic Geometry|Introduction The basics of the half-plane model of hyperbolic geometry are explored. Now that we know a lot about general geometry, what can we do that is not Euclidean| where we don't need the parallel postulate? In fact, what can we do to show that the parallel postulate is necessary in order to distinguish Euclidean geometry from something else? Imagine living in a 2-dimensional universe that is in the half-plane y > 0, but where distances are warped. The stretching factor is inversely proportional to the distance from the x-axis. So, if it takes one minute to talk from (0; 1) to (1; 1) along the horizontal line segment then it would take only 30 seconds to walk from (0; 2) to (1; 2) while going from (0; 1=3) to (1; 1=3) takes 3 minutes. This may look strange from the outside, but remember that we are inside this crazy plane, so our rulers and our legs shrink and stretch so that these distances look perfectly normal to us. This is called the hyperbolic plane. One of the more interesting things to note is that you really shouldn't travel along horizontal line segments to get from one place to another. This is because if you bow your trip a little north, you may stretch your legs to be long enough to more than make up for the extra (Euclidean) length you are covering. In fact, it turns out the geodesics in this wacky plane are pieces of circles! Specifically, if we are given two points (x1; y1) and (x2; y2): • If x1 = x2 so that the points are separated only vertically, then the geodesic between them is the vertical line segment, and the distance between them is j ln (y2=y1)j. • Otherwise, the geodesic between them is the arc of a circle centered on the x-axis that passes through both points. If this circle is centered at (c; 0) and has (Euclidean) (x1 − c − r)y2 radius r, then the distance between the points is ln . (x2 − c − r)y1 These formulas are not hard to prove with calculus, but not terribly easy either. We have Z b px02 + y02 to compute lengths of curves with the (parametric) arclength formula dt. An a y alternate formula for the distance is as follows. If the points are P and Q, and R and S are the two points where the circle through these points, centered on the x-axis meet the x-axis, PR · QS then the distance between P and Q is ln . This can be extended to vertically QR · PS separated points by taking R as the point at the base of the vertical line and S to be at infinity, and \cancelling" the two infinities. Believe it or not, this geometry satisfies the axioms of neutral geometry! 1. There is a unique geodesic through any two points. 2. Given a geodesic and a point on the geodesic and a distance, it is possible to find a point on the geodeic at the given distance from the original point. 3. Given a point and a segment with that point at one end, it is possible to draw a circle|to find a set of points all at that distance from the given point. 1 4. All right angles are equal to each other. To measure angles, we simply measure the angle between tangent lines. So we are mea- suing Euclidean angles, whcih satisfy the axiom. So axiom 4 is easy. Axiom 1 is easy, also. To explain axiom 2, we let a point slide along a geodesic. As it nears the given point, the distance drops to zero, while as it gets farther away the distance grows without bound. In fact, if we hold P fixed and let Q very, the distance between them grows toward infinity as Q nears the x-axis. Axiom 3 requires some ingenuity. Given point P (x; y) and distance r, draw the Euclidean circle with center (x; y cosh (r)) and radius y sinh (r). I claim that all the points on this circle are equidistance from (x; y). This means that all the theorems we have proved work! Consider something like the SAS theorem. We could draw triangles on very differently sized geodesics, and they are still congruent. For example, the two triangles below are both isosceles right trianlges with legs of length one. 2.