Technische Universität München Zentrum Mathematik

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1 Technische Universität München Zentrum Mathematik Prof. Dr. Dr. Jürgen Richter-Gebert, Bernhard Werner Projective Geometry SS 2016 www-m10.ma.tum.de/projektivegeometriess16 Solutions for Worksheet 6 ( ) Question 1. Bracket polynomials Classwork (For this exercise,) A bracket poynomial is a real polynomial in 3 3-determinants, in which the columns are indeterminate/variable points of RP 2. I.e. expression which are build via addition, multiplication and scalar multiplication from determinants of the form [XY Z] with X, Y, Z P R. The easiest example for them is, of course, [ABC] for arbitrary A, B, C P R. This bracket polynomial is zero if and only if the points A, B, Care collinear. a) Let A through F be six points in general position. The equation [ABC][DEF ] [ABD][CEF ] = 0 is satisfied if and only if the lines (A B), (C D) and (E F ) meet in a common point. Prove this connection by dealing with the special case where at least two of the lines are collinear and the determinants can be interpreted as oriented triangle areas. Argue why this special case does not restrict the generality of the proof. b) Find a bracket polynomial which is zero if and only if the point K lies on the line ( (A B) (C D) ) ( (E F ) (G H) ). You may assume that the points are in sufficiently general position so that this line is uniquely defined. c) In the preceding subtasks, degenerated situations have been explicitely excluded by the assumption of a general position of points. What significance does the polynomial retain if this assumption no longer holds and degenerate situations do occur? a) We can shift the intersection point on the line at infinity with an adequate projective transformation, making the lines in question parallel. This allows us to reformulate the geometric statement: The third line shall be parallel to the given two. In order to be able to talk about oriented triangle areas, we should additionally assume that all points are finite and are represented by vectors with z = 1. Let A B and E F be parallel lines with distance h. We denote the distance of point C to A B by c, the distance of D to this line by d. Then the determinants are 0 = [ABC][DEF ] [ABD][CEF ] = A B c E F (h d) A B d E F (h c) = c(h d) d(h c) = ch cd dh + dc = (c d)h As h is not zero, because of the general position of all points, the equation is true iff c = d. I.e. the points C and D have the same distance to A B and their Join is parallel to that line. The bracket polynomial is multi-homogeneous, i.e. the choice of representatives does not restrict the generality of the setup. Furthermore, shifting one point onto the line at infinity is achievable by a projective transformation, which does not change the validity of the equation. So the proof holds in general. 1

2 b) The expression in part a) can be inferred from the following consideration: The Meet P = (A B) (C D) is exactly the linear combination of A and B that is collinear to C and D. P = λa + µb 0 = [CDP ] = λ[cda] + µ[cdb] λ = [CDB] µ = [CDA] P = [CDB]A [CDA]B We can express the second intersection point Q in the same way and formulate the collinearity as an additional determinant: P = [CDB]A [CDA]B Q = [GHF ]E [GHE]F 0 = [P QK] = [[CDB]A [CDA]B, [GHF ]E [GHE]F, K] = [CDB][GHF ][AEK] [CDB][GHE][AF K] [CDA][GHF ][BEK] + [CDA][GHE][BF K] c) The variant which represents a construction might not be true anymore, as Meets and Joins do not need to be unique. Al that is left is an existential quantification: There are points taking on the roles of P and Q and with which all 5 collinearities in the instruction are fulfilled. But they are note necessarily unique. Question 2. Quadrilateral sets This task deals with so-called quadrilateral sets, often abbreviated as quad sets. The construction to the right will result in six points A through F on a line which satisfy the quad set relation: One starts with 4 points in general position P, Q, R, S, draws all 6 Joins and intersects them with an additional line. The pairs in the quad set are then the intersection points of the last line with the opposite edges of the quadrangle P QRS. The condition for 6 points to form a quad set (A, D; B, E; C, F ) can also be written as an equation using determinants: S R Q P A D F B E C [AE][BF ][CD] = [AF ][BD][CE] a) The von-staudt constructions introduced in the lecture can be seen as a special case of quadrilateral sets. Explicitely name the quad sets in question, and verify the determinant equation by simple computation using standard coordinates. b) Which condition must be satisfied by a bracket polynomial equation in order to make it invariant under both the choice of representatives and projective transformations? c) Dualize the depicted construction to obtain a set of six lines in quad set relation. Note: In the scope of this worksheet, the figure above will be called primal and the one you constructed here will be dual. This choice has no relevance beyond the scope of this worksheet; often the terms are used in exactly the opposite way. d) Intersect the six concurrent lines in the dual construction with another line g in such a way that you obtain six distinct points of intersection. Claim: these six points again form a quadrilateral set. Verify this claim by drawing the primal construction for these points, preferrably in a different color. How many real degrees of freedom do you have in creating this construction? 2

3 It is advisable to use the notation (A, D; B, E; C, F ) for quad sets to express clearly which points form pairs. a) Addition is the quad set (X, Y ; 0, X + Y ;, ). Multiplication is the quad set (X, Y ; 1, X Y ; 0, ). [X, X + Y ][0, ][, Y ] = [X, ][0, Y ][, X + Y ] x x + y y 0 1 = x 1 0 y x + y 0 1 (x x y) ( 1) 1 = ( 1) ( y) 1 y = y [X, X Y ][1, ][0, Y ] = [X, ][1, Y ][0, X Y ] x x y y 1 1 = x 1 1 y x y 1 1 (x xy) ( 1) ( y) = ( 1) (1 y) ( xy) y(x xy) = xy(1 y) Both quad sets do fulfil the given determinant equation. b) For the equation to be independent from the choice of representatives, every point must occur for the same number of times in every summand on every side. Then, the scalar multiple will occur on both sides with the same power and, thus, can be cancelled. To be invariant under projective transformations, every summand has to have the same number of determinants, as every determinant will be multiplied with the determinant of the transformation matrix. When every point occurs for the same number of times on both sides, the number of determinants have be the same, too. So, the invariance under projective transformations will hold automatically. Such a bracket polynomial equation is called multi-homogeneous. c) The primal configuration starts with 4 arbitrary points and their 6 connecting lines, intersecting these with an additional line. Dually, this results in 4 arbitrary lines p, q, r, s (in red), their 6 intersetion points (again, red), connecting them with an additional point G (blue). F g p s E c b C B P Q D a A q A B r G d e f D S R E C F 3

4 d) In the above drawing you already find the intersection line g (magenta) and the corresponding verification construction (green). For this construction we have 3 degrees of freedom e.g. choosing one point arbitrarily and another point on a given line. The rest is predetermined. One particular point to bear in mind is the correct assignment. I.e. points form a pair in the dual construction, when they do not lie on a common configuration line. They determine lines which also form a pair. Finally, their intersections with the additional line form new pairs of point. And these have to be treated as such by the verification construction. Another assignment, in general, does not result in a quad set. The pairs in the dual drawing are (A, D; B, E; C, F ), just as in the primal one. Question 3. Quadrilateral sets (continued) Homework If some subtasks from question 2 were not completed in your tutorial, please consider them homework. e) Algebraically, the quad set relation is an equation of bracket polynomials (in 2 2-determinants). Write the condition for a harmonic set in the form of a bracket polynomial equation as well. f) Describe a harmonic set as a special case of a quadrilateral set. Investigate this equivalence algebraically, and also pay attention to any special cases which arise. g) The quad set relation was given as a equation for points in RP 1. How can this be turned into an equation for collinear points in RP 2? Use an approach analogous to that used for the computation of cross ratios of collinear points in the plane. a) Simply expanding the known equation: (A, B; C, D) = [AC][BD] [AD][BC] = 1 [AC][BD] = [AD][BC] b) The verification construction for an harmonic point set is a special case of the given construction in which to pairs of lines intersect on the number line. Thus, we can rewrite the harmonic point set (A, B; C, D) as the quad set (A, A; B, B; C, D). [AC][BD] [AD][BC] = 1 [AC][BD] = [AD][BC] [AB][BD][CA] = [AD][BA][CB] [AB] [AC][BD] = [AB] [AD][BC] When the determinant [AB] does not vanish, both equations are equivalent. When it does vanish, 2 points forming a pair in the harmonic set collapse in the quad set relation even 4 points collapse. In that case, the equation for harmonic points might not hold anymore. The quad set equation, however, does still hold with this zero determinant. c) Every property that is well-defined as a bracket polynomial in RP 1 has to be multi-homogeneous and is therefore projectively invariant. In the primal construction, a certain expression in 2 2-determinants is true and it can be reformulated with 3 3-determinants by looking from an additional point at the given ones. In the dual construction, the same expression is true for 3 3-determinants when we insert an additional line in each bracket. The reason is, that primal and dual computations are identical. When the concurrent lines run through the origin and we when we look at them from the line at infinity, applying Laplace s formula results in the original 2 2-determinants. given the same situation and considering the intersections with the line at infinity, they fulfil the exact same 2 2-expression on the line at infinity. Because of the projective invariance we can let go of the special roles of origin and line at infinity. We get the result, that it does not matter whether we look at a pencil of lines from an additional line i.e. adding this line into all brackets or intersect the pencil with an additional line and express the determinant in terms of homogeneous coordinates on this line. 4

5 Question 4. CP 2 vs. RP 2 Find an incidence configuration that is realisable in CP 2, but not in RP 2. Use at least one von-staudt construction to do this. Of course, there are arbitrarily many such constructions. The easiest one is the verification that i is the square root of 1 or rather, that 1 is the square of i. Constructing square roots with von-staudt constructions is not possible, as they are not unique. Only given the starting points 0, 1, and i, a possible construction is solely schematically(!) given by: That i 2 is indeed 1, can be shown with an additional von-staudt construction or an harmonic-point-construction (as above). As this is a real representation of a complex construction, on line, sooner or later, has to be bent. That this construction is not realisable over R is due to the following fact: The von-staudt construction guarantees that the point labelled with i is always a square root of 1. No matter over which field one wants to depict this configuration. 5