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Date Created: 10/25/15
Chapter 9 Poincar s Disk Model for Hyperbolic Geometry 91 Saccheri s Work Recall that Saccheri introduced a certain family of quadrilaterals Look again at Section 73 to remind yourself of the properties of these quadrilaterals Saccheri studied the three different possibilities for the summit angles of these quadrilaterals Hypothesis of the Acute Angle HAA The summit angles are acute Hypothesis of the Right Angle HRA The summit angles are right angles Hypothesis of the Obtuse Angle HOA The summit angles are obtuse Saccheri intended to show that the rst and last could not happen hence he would have found a proof for Euclid s Fifth Axiom He was able to show that the Hypothesis of the Obtuse Angle led to a contradiction This result is now know as the Saccheri Legendre Theorem Theorem 73 He was unable to arrive at a contradiction when he looked at the Hypothesis of the Acute Angle He gave up rather than accept that there was another geometry available to study It has been said that he wrote that the Hypothesis of the Acute Angle must be false because God wants it that way77 92 The Poincare Disk Model When we adopt the Hyperbolic Axiomthen there are certain rami cations H The sum of the angles in a triangle is less than two right angles to All similar triangles that are congruent ie AAA is a congruence criterion 03 There are no lines everywhere equidistant from one another q If three angles of a quadrilateral are right angles then the fourth angle is less than a right angle 01 If a line intersects one of two parallel lines it may not intersect the other a Lines parallel to the same line need not be parallel to one another 1 Two lines which intersect one another may both be parallel to the same line 99 100 CHAPTER 9 POINCARE S DISK MODEL FOR HYPERBOLIC GEOMETRY How can we visualize this Surely it cannot be by just looking at the Euclidean plane in a slightly different way We need a model with which we could study the hyperbolic plane If it is to be a Euclidean object that we use to study the hyperbolic plane 2 then we must have to make some major changes in our concept of point line andor distance We need a model to see what 9252 looks like We know that it will not be easy but we do not want some extremely dif cult model to construct We will work with a small subset of the plane but give it a different way of measuring distance There are three traditional models for 9 They are known as the Klein model the Poincare Disk model and the Poincare Half Plane model We will start with the Disk model and move to the Half Plane model later There are geometric isomorphisms between these models it is just that some properties are easier to see in one model than the other The two Poincare models tend to give us the opportunity to do computations more easily than the Klein model 7 though the Klein model is somewhat easier to describe In order to give a model for 2 we need to decide on a set of points then determine what lines are and how to measure distance For Poincare s Disk Model we take the set of points that lie inside the unit circle ie the set 952 m2 y2 lt1 Note that points on the circle itself are not in the hyperbolic plane However they do play an important part in determining our model Euclidean points on the circle itself are called ideal points omega points vanishing points or points at in nity Note Poincare himself thought of this set as the set of all complex numbers with length less than 1 9252 z E C lt 1 We will see why this is important when we study the Poincare half plane model A unit circle is any circle in the Euclidean plane is a circle with radius one De nition 91 Given a unit circle P in the Euclidean plane points of the hyperbolic plane are the points in the interior of P Points on this unit circle are called omega 2 points of the hyperbolic plane If we take P to be the unit circle centered at the origin then we would think of the hyperbolic plane as 9252 zy 2 y2 lt 1 and the omega points are the points 9 2 y2 1 The points in the Euclidean 9 plane satisfying z y x2 y2 gt 1 are called vltraideal points Figure 91 Poincare line We now have what our points will be We see that we are going to have to modify our concept of line in order to have the Hyperbolic Axiom to hold De nition 92 Given a unit circle P in the Euclidean plane lines of the hyperbolic plane are arcs of circles drawn orthogonal1 to P and located in the interior of P 1Circles are orthogonal to one another when their radii at the points of intersection are perpendicular MATH 6118 090 Spring 2008 92 THE POINCARE DISK MODEL 101 921 Construction of Lines This sounds nice7 but how do you draw them Start with a circle P centered at O and consider the ray E4 where A lies on the circle7 P H Construct the line perpendicular to 4 at A to Choose a point P on this perpendicular line for the center of the second circle and make PA the radius of a circle centered at P OJ Let B denote the second point of intersection with circle P Then the arc AB repre sents a line in this model F Figure 92 Poincare lines through A Now7 how do you construct these lines in more general circumstances There are three cases we need to consider Case I A7 B E P Case II A E P and B lies inside P Case III A and B both lie insidil Case I Construct rays PA and PB where P is the center of the circle P Construct the lines perpendicular to 17A and 17B at A and B respectively Let Q be the point of intersection of those two lines The circle 9 centered at Q with radius QA intersects P at A and B The line between A and B is the arc of 9 that lies inside P MATH 6118 090 Spring 2008 102 CHAPTER 9 POINCARE S DISK MODEL FOR HYPERBOLIC GEOMETRY Note that this arc is clearly orth gtogonal to P by its construction Case II Construct r P and PB where P is the center of the circle P Construct the line perpendicular to PA at A Draw segment AB and construct its perpendicular bisector Let Q be the point of intersection of this line and the tangent line to P at A The circle 9 centered at Q with radius QA contains A and B The line containing A and B is the arc of 9 that lies inside P This arc7 as constructed is orthogonal to P at A We want to see that it is orthogonal at the other point of intersection with the circle Let that point of intersection be X Then7 X E P means that PA E PX Since X lies on our second circle it follows that QX E QA Since PQ E PQ7 we have that APAQ E APXQ7 which means that APXQ is a right angle7 as we wanted to show A Case III Construct the ray PA and then construct the line perpendicular to PA at A This intersects P in points X and Y Construct the tangents to P at X and at Y These tangent lines intersect at a point C The circle 9 centered at Q is the circle passing through A7 B7 and C The line containing A and B is the arc of 9 that lies inside F From our construction7 we have that APXC N APAX and it follows that lPAHPCl lPXlZ r2 Now7 Q lies on the perpendicular bisectors of AC and AB as Q is the circumcircle for AABC There is a point T on the circle 9 so that the tan gent line to Q at T passes through P Construct the line through P and Q which intersects the circle in two points G1 and G2 so that G1 lies between P and Q Now7 lPTlZ lPQlZ lQTlZ lPQl lQTl lPQl lQTl lPQl 910 Pm 920 lPGlllPGgl which by Theorem 537 lPAHPCl T Figure 93 Poincare line in Case 111 Therefore7 T lies on the circle P and P and Q are orthogonal at that point A similar argument shows that they are orthogonal at the other point of intersection 922 Distance Now7 this area inside the unit circle must represent the in nite hyperbolic plane This means that our standard distance formula will not work We introduce a distance metric by 2dr d p 1 7 72 where p represents the hyperbolic distance and r is the Euclidean distance from the center of the circle Note that dp a 00 as r a 1 This means that lines are going to have in nite extent MATH 6118 090 Spring 2008 92 THE POINCARE DISK MODEL 103 The relationship between the Euclidean distance of a point from the center of the circle and the hyperbolic distance is T Qdu 1r 1 pOmloglt1irgt2tanh r7 or r tanh 8 Now7 for those of you who don t remember ever having seen this function tanhz7 we give a little review The hyperbolic trigonometric functions coshz and sinhm are de ned by m 7 ix sinhz e 2e 1 1 coshz sinhm em 7 e m e2m 7 1 tanhltgt 7 coshz 7 em 1 e m 7 e2m 139 We will study these in more depth later Now7 we can use this to de ne the distance between two points on a Poincare line Given two hyperbolic points A and B7 let the Poincare line intersect the circle in the omega points P and Q De ne AP A AP B ABPQgt A 7Q BPBQ AQ BP to be the cross ratio of A and B with respect to P and Q where AP denotes the the Euclidean arclength De ne the hyperbolic distance from A to B to be olA7 B log lAB7 PQl We will prove the following later Theorem 91 If a point A in the interior of P is located at a Euclidean distance r lt 1 from the center 0 its hyperbolic distance from the center is given by 1 r d A O lo 7 7 g 17 T Lemma 91 The hyperbolic distance from any point in the interior ofP to the circle itself is in nite 923 Parallel Lines It is easy tcgtspe that the Hyperbolic Axiom works in this model Given a line A B Ed a point D AB7 then we can draw at least two lines through D that do not intersect AB Call these two lines through D lines l1 and lg Noticeltnow how two omur previous results do not hold7 as we remarked earlier We have that AB and l1 and AB and lg are parallel7 but l1 and lg are not parallel Note also that lggtersects one of a pair of parallel lines l17 but does not intersect the other parallel line AB MATH 6118 090 Spring 2008 104 CHAPTER 9 POINCARE S DISK MODEL FOR HYPERBOLIC GEOMETRY Figure 94 Multiple parallels through A As we now know the hyperbolic plane has two types of parallel lines The de nition that we will give here will depend explicitly on the model that we have chosen Consider the hyperbgc line AB which intersects the circle 2 in the ideal points A and 9 Take a pointD AB Construct the line through A and D Since this line does not intersect the line AB inside the circle these two hyperbolic lines are parallel However they seem to be approaching one another as we go 77to in nity nce ther e are two 77ends77 of the Poincare line AB there are two of these lines The line AB and DA are horoparallel The de ning property is as follows De nition 93 LetP 6 AB Consider the collection of lines DP as P goes to Q or A The rst line through D in this collection that does not intersect AB in f2 is the horoparallel line to AB in that direction Drop a perpendicular from D to AAB and label this point of intersection M Angles AADM and AQDM are called angles of parallelism Theorem 92 The angles of parallelism associated with a given line and point are congru ent PROOF Assume not ie assume AADM 7 AQDM Then one angle is greater than the other Without loss of generality we may assume that AADM lt AQDM r Then there is a point E in the interng of AQDM such that AADM AEDM The line ED must intersect directicm Let the point of intersec gon be F Choose I 0 AB since DD is the limiting parallel line to AB in that G on AB on the opposite side of DM from F so that J FM GM Then AGMD E AFMD This implies that AGDM AAFDM AADM This means that DQ intersects AB at G This contradicts the condition that DD is limiting parallel to AB Thus the angles of Flgure 95 lemng Parallel parallelism are congruent I Pomcare Llnes Theorem 93 The angles of parallelism associated with a given line and point are acute MATH 6118 090 Spring 2008 92 THE POINCARE DISK MODEL 105 PROOF Assume not7 ie7 assume that AMDQAgt 90 ien there is a point E interior to AMDQ so that AMDE 90 Then7 since DE and AB are perpendicular to the same line7 they are parallel Thus7 DE does not intersect AB which contradicts the condition that DD is the limiting parallel line If the angle of parallelism is 90quot then we can show that we have Euclidean geometry Thus7 in 9252 the angle of parallelism is acute Theorem 94 Lobaclyevskifs Theorem Given a point P at a hyperbolic distance p from a hyperbolic line AB ie olP7 M p the angle of parallelism 0 associated with the line and the point satis es e p tan 2 hmsI and lim 00 paO 2 pace Note then that PROOF The proof of this is interesting in that we play one geometry against the other in order to arrive at our conclusion A Figure 96 Lobachevskii s Theorem Figure 97 After the rst translation We are given a line ABAand a point P not on the line Construct the line through P which is perpendicular to AB Call the point of intersection R as in Figure 96 Then we have that p dPR We can translate P to the center of the unit circle and translate our line to a line so that our line perpendicular to AB is a radius of F as we have done in Figure 97 Construct the radii from P to the ideal points A and B and construct the line s tangent to F at these points These tangent lines intersect at a point Q which lies on PR Now7 since we have moved our problem to the center of the circle7 we can use our previous result to see that if r is the Euclidean distance from P to R then we have 710 1 7 P i g 17 T7 or rewriting this we have 1 7 51 gory1 1 MATH 6118 090 Spring 2008 106 CHAPTER 9 POINCARE S DISK MODEL FOR HYPERBOLIC GEOMETRY Now7 we are talking about Euclidean distances with r and using our Euclidean right triangles with radius 1 we have that 1 7 sin0 rQPiQRQPiQAseczQPAitaanPAsec07tan0 0 cos Now7 algebra leads us to 177 17 7 cos0sin071 cos0sin071cos0sin01 cos07sin01cos0sin01 cos202cos0sin0sin20i1 cos202cos07sin201 7 2sin0cos0 7 sin0 T 2cos202cos0 T1cos0 2singcosg e 924 Hyperbolic Circles Now7 if we have a center of a circle that is not at the center P of the unit circle 2 we know that the hyperbolic distance in one direction looks skewed with respect to the Euclidean distance That would lead us to expect that a circle in this model might take on an elliptic or oval shape We will prove later that this is not the case In fact7 hyperbolic circles embedded in Euclidean space retain their circular appearance 7 their centers are offset Theorem 95 Given a hyperbolic circle with radius R the circumference C of the circle is given by C 27139 sinhR 925 Common Figures in the Disk Model What do some of the common gures7 with which we have become accustomed7 look like in the Poincare Disk Model MATH 6118 090 Spring 2008 92 THE POINCARE DISK MODEL 107 Figure 98 Saccheri quadrilateral in the Poincare Disk 1 Figure 99 Acute Triangle Figure 910 Obtuse Triangle MATH 6118 090 Spring 2008
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