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Technische Universität München Zentrum Mathematik Technische Universität München Zentrum Mathematik Prof. Dr. Dr. Jürgen Richter-Gebert, Bernhard Werner Projective Geometry SS 2018 https://www-m10.ma.tum.de/bin/view/Lehre/SS18/PG1SS18/WebHome Solutions for Worksheet 11 (2018-07-02) — Classwork— Question 1. A concrete Cayley-Klein geometry a) Given the following dual conic B, determine the corresponding primal conic A. 00 0 11 B = @0 0 0A 1 0 1 b) Investigate whether the conic described by the primal-dual pair (A; B) is degenerate or not. If it is degenerate, give coordinates for the components into which it decomposes, both primal and dual. c) Now this conic (A; B) shall be used as the fundamental conic of a Cayley-Klein geometry. Describe the resulting measurements for distances and angles respectively as “hyperbolic”, “parabolic” or “elliptic”. 1 d) Now assume cang = 2 . Characterize under which conditions the angle between two lines will be real. e) The cross ratio which can be constructed most easily is the one for harmonic sets. Use this to construct two lines which enclose a well-defined and real angle. Base your construction on a coordinate system into which the conic (A; B) has been drawn. Also, give an explicit description of the value of the angle you are constructing. Solution: a) The primal conic can be computed as the adjugate matrix of the dual conic: 00 0 01 00 0 01 A = B4 = @0 −1 0A ∼ @0 1 0A 0 0 0 0 0 0 b) The conic is degenerated. The primal conic A has rank 1 and represents the double line l = (0; 1; 0)T , i.e. the x-axis. The dual conic has rank 2 and degenerates into 2 points X = (0; 0; 1)T and Y = (2; 0; 1)T . This can easily be verified by adding a suitable anti-symmetric matrix. 001 001 021 l = @1A X = @0A Y = @0A 0 1 1 c) Distance measuring is Euclidean, i.e. parabolic, because of the double line. Angle measuring is hyperbolic, due to the 2 different points. d) Two lines different from l confine a real angle iff their intersections with l are not separated by X and Y . I.e. when both lie inside of X and Y or outside. In these cases, the cross-ratio is positive and the logarithm is real. 1 e) One can simply construct two points A and B which are in harmonic position together with X and Y . Doing so, X and Y should not form a pair, as otherwise the angle is complex. Connecting these points A and B with an arbitrary additional point P allows us to precisely determine the angle between the connecting lines A _ P and B _ P . Together with the tangents to the conic, i.e. the connections of P with X and Y , they form an harmonic set of lines. As X and Y shall not form a pair, the cross-ratio relevant for the angle computation is not −1, but takes one of 1 the other values possible for harmonic sets. I.e. either 2 or 2 . This reciprocal gets transformed into a change of the sign of the angles—but when we only consider the absolute value of the angle, we can ignore. The absolute angle is ln(2) c ln(2) = ≈ 0;3466: ang 2 In particular, this does not depend on the position of P . So, the blue and red angles shown below are both equal to this value. P P A B X Y Question 2. Hyperbolically equidistant points Let X; Y be the two reference points on a line to make hyperbolic distance measurements. Consider three more points A; B; C on this line. a) Give a conditions on suitable cross-ratios which describes that jA; BjX;Y = jB; CjX;Y and deduce a bracket polynomial equation that describes the same. b) In the picture below, construct a point C fulfilling the above distance relation. c) Construct two more steps with the same size in the same direction and try to re-use as much as possible of your previous auxiliary construction. X A B Y 2 Solution: a) In order to have equal distances, we need equal cross-ratios. I.e. (A; B; X; Y ) = (B; C; X; Y ): Then we make the same computations as for the angles in RP2 seen in the lecture: (A; B; X; Y ) = (B; C; X; Y ) , (A; X; B; Y ) = (B; X; C; Y ) [AB][XY ] [BC][XY ] , = [AY ][XB] [BY ][XC] , [AB][BY ][XC] = [AY ][BC][XB] b) The above means that (AC; BB; XY ) forms a quad set. So, we can make the usual quad set construction. c) We can use the lines through X and Y in the construction again. And also, the “right-hand one” of B and the one through C. Then, we iterate... X A B C Y X A B C Y 3 X A B C Y — Homework— Question 3. Dual partner Given the following matrix 01 1 11 A = @1 1 1A 1 1 1 as well as some other matrices 00 0 11 0 1 0 −11 0 0 −1 1 1 00 0 01 B1 = @0 0 1A B2 = @ 0 0 0 A B3 = @−1 0 1 A B4 = @0 0 0A 1 1 2 −1 0 1 1 1 −2 0 0 0 a) Decide which of these matrices Bi will form a valid primal-dual pair together with A. Justify your answer using a suitable computation. b) Describe these primal and dual conics graphically, both if they form a valid pair and if they do not. c) Describe which Cayley-Klein geometries will result from each of the valid pairs. 4 Solution: a) The matrix has rank 1, so we cannot compute its dual directly. The matrices (A; B) form a primal/dual pair iff A · B = λ · E. And here λ will have to 0, as A has no inverse. 01 1 41 00 0 01 00 0 01 A · B1 = @1 1 4A A · B2 = @0 0 0A A · B3 = @0 0 0A 1 1 4 0 0 0 0 0 0 So, B2 and B3 are dual matrices compatible with A. With B4 we obivously get the zero matrix from A · B, too, but it does not represent a conic. b) A represents—as a primal matrix with rank 1—a degenerate conic consisting of a double line. B3 has rank 2 and therefore represents two distinct points on this line. B1 has rank 1 and gives, hence, a double point on this line. B1 has rank 2 and so it represents two points. But they do not lie on A, as otherwise B1 would be a dual partner to A. 011 001 0−11 0−11 A = @1A · (1; 1; 1) B1 ^= @0A · (1; 1; 1) B2 ^= @ 0 A · (−1; 0; 1) B3 ^= @ 0 A · (0; −1; 1) 1 1 1 1 B1 B1 B2 B3 A A A B3 c) B2 leads—with a double point on a double line—to a Galilean geometry. Measuring both distances and angles are parabolic/Euclidean. B3 leads—with two real points on a double line—to a pseudo-Euclidean geometry. Measuring distances is para- bolic/Euclidean and measuring angles is hyperbolic. Question 4. Very degenerate a) Choose a degenerate conic where both primal and dual matrix have rank 1. State the matrices explicitly, and draw the resulting conic in a coordinate system. b) With respect to this Cayley-Klein geometry, construct a sequence of at least five equidistant points on an arbitrary line. c) Also construct a pencil of at least five equiangular lines through a common point. I.e., two consecutive lines of that sequence should form the same angle with one another. 5 Solution: The picture containing all steps can be found after the full discussion of the single parts. a) As the fundamental object has to be drawn and used for constructions, it should be finite. A possible choice is 01 0 01 00 0 01 A = @0 0 0A B = @0 0 0A 0 0 0 0 0 1 This conic is the y-axis as a double line with the origin as a double point on it. b) The construction of projective scales or projectively equidistant sequences of points can be done as we are used 2 to in RP . The starting points are P0 and P1 and the vanishing point P , in which their connecting line and the reference conic intersect. 1 c) For any given point Q there is exactly one tangent to the fundamental object. Namely, the connection g to the origin. Drawing any line not going through Q, one can use its intersection with the tangent as the point1 at infinity. One can now construct a sequence of equidistant points w.r.t. this point. Consecutive connecting lines gi of these points with Q have the same angle, since, first, the tangent takes on the role of the point at infinity for the angle measurement and, second, the cross-ratio of the lines equals the cross-ratio of the intersection points. 6 A P0 P1 P2 P3 P4 P ∞ B g4 g3 g2 g1 g0 g ∞ Q 7.
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