Poincar´e’s Disk Model for Hyperbolic Geometry

1 Chapter 9 Poincar e s Disk Model forHyperbolic Saccheri s WorkRecall that Saccheri introduced a certain family of quadrilaterals. Look again at Section remind yourself of the properties of these quadrilaterals. Saccheri studied the threedifferent possibilities for the summit angles of these of the Acute Angle(HAA) The summit angles are acuteHypothesis of the Right Angle(HRA) The summit angles are right anglesHypothesis of the Obtuse Angle(HOA) The summit angles are obtuseSaccheri intended to show that the first and last could not happen, hence he would havefound a proof for Euclid s Fifth Axiom. He was able to show that the Hypothesis of theObtuse Angle led to a contradiction. This result is now know as the Saccheri-LegendreTheorem (Theorem ). He was unable to arrive at a contradiction when he looked atthe Hypothesis of the Acute Angle. He gave up rather than accept that there was anothergeometry available to study. It has been said that he wrote that the Hypothesis of theAcute Angle must be false because God wants it that way.

2 The Poincar e Disk ModelWhen we adopt theHyperbolic Axiomthen there are certain ramifications:1. The sum of the angles in a triangle is less than two right All similar triangles that are congruent, is a congruence There are no lines everywhere equidistant from one If three angles of a quadrilateral are right angles, then the fourth angle is less than aright If a line intersects one of two parallel lines, it may not intersect the Lines parallel to the same line need not be parallel to one Two lines which intersect one another may both be parallelto the same 9. POINCAR E S DISK Model FOR Hyperbolic GEOMETRYHow can we visualize this? Surely it cannot be by just lookingat the Euclidean plane ina slightly different way. We need a Model with which we could study the Hyperbolic it is to be a Euclidean object that we use to study the Hyperbolic plane,H2, then wemust have to make some major changes in our concept of point, line, and/or need a Model toseewhatH2looks like.

3 We know that it will not be easy, but wedo not want some extremely difficult Model to construct. We will work with a small subsetof the plane, but give it a different way of measuring are three traditional models forH2. They are known as the Klein Model , thePoincar e Disk Model , and the Poincar e Half-Plane Model . We will start with the Disk modeland move to the Half-Plane Model later. There are geometric isomorphisms between thesemodels, it is just that some properties are easier to see in one Model than the other. Thetwo Poincar e models tend to give us the opportunity to do computations more easily thanthe Klein Model though the Klein Model is somewhat easier order to give a Model forH2, we need to decide on a set of points, then determinewhat lines are and how to measure distance. For Poincar e s Disk Model we take the set ofpoints that lie inside the unit circle, , the setH2={(x, y)|x2+y2<1}.Note that points on the circle itself arenotin the Hyperbolic plane. However they do playan important part in determining our Model .

4 Euclidean points on the circle itself are calledideal points,omega points,vanishing points, orpoints at infinity.[Note: Poincar e himself thought of this set as the set of allcomplex numbers with lengthless than 1H2={z C|kzk<1}.We will see why this is important when we study the Poincar e half plane Model .] PQ Figure : Poincar e lineAunitcircle is any circle in the Euclidean plane is acircle with radius a unit circle in the Euclideanplane, points of the Hyperbolic plane are the points in theinterior of . Points on this unit circle are called omegapoints ( ) of the Hyperbolic we take to be the unit circle centered at the origin,then we would think of the Hyperbolic plane asH2={(x, y)|x2+y2<1}and the omega points are the points ={(x, y)|x2+y2= 1}. The points in the Euclideanplane satisfying{(x, y)|x2+y2>1}are now have what our points will be. We see that weare going to have to modify our concept of line in order tohave the Hyperbolic Axiom to a unit circle in the Euclidean plane, lines of the Hyperbolic planeare arcs of circles drawn orthogonal1to and located in the interior of.

5 1 Circles are orthogonal to one another when their radii at thepoints of intersection are 6118-090 Spring THE POINCAR E DISK Construction of LinesThis sounds nice, but how do you draw them?1. Start with a circle centered atOand consider the ray OA, whereAlies on thecircle, .2. Construct the line perpendicular to Choose a pointPon this perpendicular line for the center of the second circle andmakeP Athe radius of a circle centered LetBdenote the second point of intersection with circle . Then the arcABrepre-sents a line in this : Poincar e lines throughANow, how do you construct these lines in more general circumstances? There are threecases we need to I:A, B Case II:A andBlies inside Case III:AandBboth lie inside .Case I: Construct rays P Aand P BwherePis the center of the circle . Construct the linesperpendicular to P Aand P BatAandBrespectively. LetQbe the point of intersection ofthose two lines. The circle centered atQwith radiusQAintersects atAandB.

6 Theline betweenAandBis the arc of that lies inside .MATH 6118-090 Spring 2008102 CHAPTER 9. POINCAR E S DISK Model FOR Hyperbolic GEOMETRYNote that this arc is clearly orthogonal to by its II: Construct rays P Aand P BwherePis the center of the circle . Construct theline perpendicular to P AatA. Draw segmentABand construct its perpendicular the point of intersection of this line and the tangent lineto atA. The circle centered atQwith radiusQAcontainsAandB. The line containingAandBis the arcof that lies inside .This arc, as constructed is orthogonal to atA. We want to see that it is orthogonal atthe other point of intersection with the circle. Let that point of intersection beX. Then,X means thatP A =P X. SinceXlies on our second circle it follows thatQX = Q =P Q, we have that P AQ = P XQ, which means that P XQis a rightangle, as we wanted to III: Construct the ray P Aand then construct the line perpendicular to P Aat A. Thisintersects in points X and Y.

7 Construct the tangents to atXand atY. These tangentlines intersect at a pointC. The circle centered atQis the circle passing throughA,B,andC. The line containingAandBis the arc of that lies inside .TG2G1 QCXPABF igure : Poincar e line in Case IIIFrom our construction, we have that P XC P AXand it follows that|P A||P C|=|P X|2=r2. Now,Qlies onthe perpendicular bisectors ofACandABas is the circumcircle for ABC. Thereis a pointTon the circle so that the tan-gent line to atTpasses the line throughPandQwhich intersects the circle in two pointsG1andG2so thatG1lies ,|P T|2=|P Q|2 |QT|2= (|P Q| |QT|) (|P Q|+|QT|)= (|P Q| |QG1|) (|P Q|+|QG2|)=|P G1||P G2|which by Theorem ,=|P A||P C|=r2 Therefore,Tlies on the circle and and are orthogonal at that point. A similarargument shows that they are orthogonal at the other point DistanceNow, this area inside the unit circle must represent the infinite Hyperbolic plane. Thismeans that our standard distance formula will not work.

8 We introduce a distance metricbyd =2dr1 r2where represents the Hyperbolic distance andris the Euclidean distance from the centerof the circle. Note thatd asr 1. This means that lines are going to have 6118-090 Spring THE POINCAR E DISK MODEL103 The relationship between the Euclidean distance of a point from the center of the circleand the Hyperbolic distance is: = r02du1 u2= log(1 +r1 r)= 2 tanh 1r,orr= tanh , for those of you who don t remember ever having seen thisfunction tanh(x), wegive a little review. The Hyperbolic trigonometric functions cosh(x) and sinh(x) are definedby:sinh(x) =ex e x2cosh(x) =ex+e x2andtanh(x) =sinh(x)cosh(x)=ex e xex+e x=e2x 1e2x+ will study these in more depth , we can use this to define the distance between two points on a Poincar e line. Giventwo Hyperbolic pointsAandB, let the Poincar e line intersect the circle in the omega pointsPandQ. Define(AB, P Q) =AP/AQBP/BQ=AP BQAQ BP,to be thecross ratioofAandBwith respect toPandQ, whereAPdenotes the theEuclidean arclength.

9 Define the Hyperbolic distance fromAtoBto bed(A, B) = log|AB, P Q|.We will prove the following a pointAin the interior of is located at a Euclidean distancer <1from the centerO, its Hyperbolic distance from the center is given byd(A, O) = log1 +r1 Hyperbolic distance from any point in the interior of to the circle itselfis Parallel LinesIt is easy to see that the Hyperbolic Axiom works in this Model . Given a line ABand apointD / AB, then we can draw at least two lines throughDthat do not intersect these two lines throughDlines 1and 2. Notice now how two of our previousresults do not hold, as we remarked earlier. We have that ABand 1and ABand 2areparallel,but 1and 2are not parallel. Note also that 2intersects one of a pair of parallellines ( 1), but does not intersect the other parallel line ( AB).MATH 6118-090 Spring 2008104 CHAPTER 9. POINCAR E S DISK Model FOR Hyperbolic Geometry D Figure : Multiple parallels throughAAs we now know, the Hyperbolic plane has two types of parallellines.

10 The definitionthat we will give here will depend explicitly on the Model that we have chosen. Considerthe Hyperbolic line ABwhich intersects the circle in the ideal points and . TakeapointD / AB. Construct the line through andD. Since this line does not intersect theline ABinside the circle, these two Hyperbolic lines are , they seem to beapproaching one another as we go to infinity . Since there are two ends of the Poincar eline AB, there are two of these lines. The line ABand D arehoroparallel. The definingproperty is as AB. Consider the collection of lines DPasPgoes to or . Thefirst line throughDin this collection that does not intersect ABinH2is the horoparallelline to ABin that a perpendicular fromDto ABand label this point of intersectionM. Angles DMand DMare calledangles of angles of parallelism associated with a given line and point are congru-ent. D MFigure : Limiting ParallelPoincar e LinesProof: Assume not, , assume DM6= one angle is greater than the other.