Geometry 1 LECTURE NOTES. Thilo Rörig

Transcription

1 Geometry LECTURE NOTES Thilo Rörig April 0, 204

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3 Contents 0 Introduction 0. Historical Remarks Course Content Projective Geometry 5. Introduction to projective spaces Homogeneous coordinates The theorems of Pappus and Desargues Projective transformations Cross-ratio (Doppelverhältnis) Complete quadrilateral and quadrangle Moebius pairs of tetrahedra The Fundamental Theorem of Real Projective Geometry Duality Conic sections Polarity: Pole - Polar Relationship Quadrics Orthogonal Transformations Hyperbolic Geometry Lorentz Spaces Hyperbolic Spaces Projective model of H n Lines in the hyperbolic plane Distances in the hyperbolic plane Intersections of H 2 with planes Stereographic projection Sphere inversion Poincaré disc/ball model Relation between Kleins and Poincaré model The Poincaré half-space(/-plane) model Hyperbolic triangles Hyperbolic motions in H Klein s Erlangen Program 83 4 Möbius Geometry Two-dimensional Möbius Geometry Relation between projective and Möbius transformations Spherical model of Möbius geometry

4 4.4 Projective model of Möbius geometry Index 87

5 0 Introduction 0. Historical Remarks The word geometry is of Greek origin (γεωμετρία) and means measurement of the earth. But the first accounts were found in Egypt and Mesopotamia (today Iraq) around 2000 BC. It was known how to calculate simple areas and volumes and people of that time already had an approximation of π. But no general statements/theorems and proofs were found. Around 600 BC in Greece, Thales was the first to deduce theorems from a set of axioms. At about 300 BC, Euclid wrote the book Elements in which he states the five axioms of what is today called Euclidean geometry. His fifth axioms is known as the parallel postulate and can be phrased as follows: Given a line and a point not on this line, there is a unique line through the point parallel to the line. In the following centuries people tried to prove this postulate from the other four but were not successful. As we know today, there exist other geometries that violate this postulate. These geometries are called Non-Euclidean Geometries: e.g. spherical geometry, hyperbolic geometry (developed independently by Lobachevsky and Bolyai; 9th century), projective geometry (Poncelet; 9th century). Another important development was the introduction of coordinates in the 7th century by Descartes and Fermat. The geometry starting from axioms is called synthetic geometry and the geometry based of coordinates analytic geometry. 0.2 Course Content In this course we will consider projective, hyperbolic and Moebius geometry. And if time permits we might also have a look at Lie, Laguerre, or Pluecker geometry. The aim is to provide you with lots of geometric images and their analytic counterparts. The continuation of this course is a speciality of the TU Berlin and is called Geometry II: Discrete Differential Geomtry. It is concerned with intelligent discretizations of smooth theories, e.g. from differential geometry and dynamical systems. The recently established SFB/TR Discretization in Geometry and Dynamics will enlarge the opportunities to do further research in this direction. Example 0. (Perspective example). We start with an example painters have to deal with when capturing a scene in a painting, cf. Albrecht Dürer: Mann beim Zeichnen einer Laute, 525. Albrecht Dürer: Mann beim Zeichnen einer Laute, 525

6 0 Introduction We will consider the projection of a spatial scene onto a plane but the central projection of one plane onto another. So consider two planes E, E R 3 that are not parallel and a point P R 3 that does not lie on the planes as shown in the following picture. Fig. : Central projection of the plane E to the plane E through the point P. The projection π : E E does the following: let A be a point on the plane E. Then A = π(a) = line(p, A) E Problem: We would like to have a bijective projection map, but the points on the line l E are not projected to any point on E, since the plane spanned by l E and P is parallel to E. Further the points on l E lack a preimage, since the plane spanned by l E and P is parallel to E. The lines l E resp. l E are the vanishing lines of E resp. E. Idea: Introduce two additional lines l and l to E and E, respectively. The points on l E will be mapped to l and the points on l are mapped to the points on l E. The lines l and l are called the lines at infinity. Done in the right way, this yields a bijective projection π : E l E l. Parallel lines in E are mapped to concurrent lines in E. The intersection point Q of the images of parallel lines lies on the vanishing line l E. The point Q is the image of the intersection point at infinity Q l of the original parallel lines. Example. We are now able to construct a perspective drawing of a chessboard given the image of the first square. The chessboard contains two families of obvious parallel lines, namely the horizontal and vertical directions. But the diagonals of the chessboard are also parallel. Hence we obtain four points on the vanishing line on E, one for each parallel family. These allow us to construct the perspective projection. Fig. 2: Construction of a perspective image of a chessboard. 2

7 0.2 Course Content Question. Can you construct the perspective image of the first square given the position of the chessboard, your position (the eye point), and the position of the plane projected onto? Now let us turn to the analytic treatment of the above example. Let the two planes and the point be the following: x E = x 2 R 3 x 3 = 0, E x = x 2 R 3 x = 0, P = 0. x 3 As the planes are two dimensional subspaces we assign two coordinates to each of the points on either of the planes, i.e. A = ( a ) a 2 and A = ( a ) a. We calculate the coordinates 2 of A from A by considering the line through A and P and its intersection point with E. Thus we need to solve: a a 0 a 2 + λ 0 a 2 = a 0 0 Hence we obtain x 3 a + λ( a ) = 0 λ = a + a. a = a 2 + a ( a 2 ) = a 2 + a + a a 2 = 0 + a ( 0) = a and thus + a + a A = π(a) = + a ( ) a2 a. a 2 (0.) Whenever there is a fractional linear transformation it( is worth ) to introduce homogeneous b coordinates to linearize the equations, i.e. introduce b2 with b 3 0 for ( a 2 ) (unique b 3 ( ) b up to non-zero scaling) with a = b b3 and a 2 = b ( 2 b3. Similarly, b 2 with b b 3 0 for a ) a 2 3 with a = b b and a 3 2 = b 2 b. If we then look back at Equation 0. we obtain the following 3 relation between the homogeneous coordinates: b b 3 b 2 b 3 = a = a 2 + a = b 2/b 3 + b /b 3 = b 2 b + b 3 = a 2 = a + a = b /b 3 + b /b 3 = b b + b 3 b = b 2, b 2 = b, b 3 = b + b 3. and Hence in homogeneous coordinates the projection has become a linear (!) map: b b b 2 b 0 0 b 2 = 0 0 b 2. 0 b 3 b 3 But what happens to the vanishing line l E? The points on the vanishing line have coordinates ( ( ) ) b a 2. Introducing homogeneous coordinates this amounts to b2 with b 3 0 and 3 b 3 b 3

8 0 Introduction b b 3 =. Thus b + b 3 = 0 and in particular b 0. For the image of such a point under the projection these conditions imply: 0 0 b b 2 b b 2 = b = b. 0 b + b 3 0 b 3 The image point does not correspond to a point on the plane, because the last coordinate is zero contradicting the assumption b 3 0 for the homogeneous coordinates of points on E. But since b 0 these coordinates may be interpreted as homogeneous coordinates of a line, namely, the line at infinity l added to the plane E. Question. What are the images of the line l if we assign homogeneous coordinates with b 0 to the points on the line? ( b b20 ) 4

9 Projective Geometry. Introduction to projective spaces Let us start with a general definition for an arbitrary projective space. Yet in this lecture we will almost entirely deal with real projective spaces. Definition... Let V be a vector space over an arbitrary field F. The projective space P (V ) is the set of -dimensional vector subspaces of V. If dim(v ) = n +, then the dimension of the projective space is n. (a) A -dimensional projective space is a projective line. (b) A 2-dimensional projective space is a projective plane. Canonical analytic description: Two vectors v and ṽ span the same -dimensional subspace if there exist λ 0 such that v = λṽ. This yields an equivalence relation on V \ {0}: v ṽ : λ 0 : v = λṽ. The points of the projective space P (V ) are equivalence classes of : P (V ) = (V \ {0})/. We write [v] P (V ) and call v a representative vector of the line {λv λ F }. Let us have a look at different models for real projective space RP 2 = P (R 3 ). Subspace model Consider all -dimensional subspaces l = {λv v R 3, v 0, λ R} of R 3. Fig. 3: Subspace model Sphere model We may normalize non-zero vectors to become unit vectors. Then there are only two vectors representing the same line. If we now introduce an equivalence relation v ṽ : v = ±ṽ for v, ṽ S n we obtain RP n = S n /. 5

10 Projective Geometry Hemisphere model Consider the upper hemisphere H 2 + = Fig. 4: Sphere model {( x ) 2 x 3 } R 3 x =, x 3 0. Then all nonhorizontal lines have a unique representative vector in H 2 + and we only need to identify points on the equator H 2 + {x x 3 = 0}. Examples of lower dimension: Fig. 5: Hemisphere model RP 0 = P (R ) = S 0 / = { } is just a single point. RP = P (R 2 ) = S / = S. This is most obvious using the hemisphere model, since the hemisphere is just a line and the two points on the equator are identified (bottom of the picture below). But we can also twist an S in a way such that opposite points come to lie on top of each other (top of picture below). 6

11 .2 Homogeneous coordinates Fig. 6: Real projective line.2 Homogeneous coordinates Consider a basis {b,..., b n+ } of the vector space V and a vector v V, v 0 with coordinates (x,..., x n+ ), i.e. v = n+ i=0 x ib i. Then (x,..., x n+ ) are homogeneous coordinates of [v] P (V ). These coordinates are not unique. They depend on the choice of basis and are then only unique up to a scalar factor: [v] = [x,..., x n+ ] = [λx,..., λx n+ ], for λ 0. Homogeneous coordinates of RP n. Let [v] RP n with coordinates [v] = [x,..., x n+ ]. Since the homogeneous coordinates are only unique up to a scalar factor we do the following: if x n+ 0 then [v] = [x,..., x n+ ] = [ ] x x n,...,, = [y,..., y n+ ] R n. x n+ x n+ The coordinates (y,..., y n+ ) are affine coordinates for [v] with x n+ 0. if x n+ = 0 then [v] = [x,..., x n, 0] RP n. So we obtain the following decomposition RP n = R n RP n. This can be iterated and RP n may again be decomposed into R n and RP n 2, and so on. The part isomorphic to R n (e.g. x n+ 0) is called the affine part of RP n and the other part isomorphic to RP n (e.g. x n+ = 0) is called the part at infinity. So in case of RP we have a point at infinity, in case of RP 2 there is an entire projective line at infinity. The following picture shows the decomposition of RP 2 into parts R 2 (the red x 3 = plane) and the line at infinity RP corresponding to the lines in the green x 3 = 0 plane. Of course, the x 3 = 0 plane may itself be decomposed into an affine part, the dark blue line x 2 = (and x 3 = 0) and the point at infinity represented by the light blue line x 2 = 0 (and x 3 = 0). 7

12 Projective Geometry Remark.2.. Fig. 7: Affine and infinite part of the real projective plane To introduce an affine coordinate on a projective space we may choose an arbitrary linear form l(x) = n+ i= a ix i on V and decompose the points of the projective space depending on whether l(x) = 0 or l(x) 0. In the decomposition above the linear form is l(x) = x n+. A projective space should be thought of as a homogeneous object (to be made precise). So there are no preferred directions (points of projective space) as the hemisphere model or the introduction of an affine coordinate might suggest. The same definitions are used in the complex case and yield complex projective spaces: CP = C { } = Ĉ is the Riemann sphere or complex projective line. Definition.2.2. A projective subspace of the projective space P (V ) is a projective space P (U) where U is a vector subspace of V. If the dimension of P (U) is k (the corresponding vector subspace has dimension k + ), the we call P (U) a k-plane. A 0-plane is a point, a -plane a line, a 2-plane a plane, and if P (V ) has dimension n a (n )-plane is a hyperplane. Proposition.2.3. Through any two distinct points in a projective space there passes a unique projective line. Proposition.2.4. In a projective plane two distinct projective lines intersect in a unique point. These two propositions can be proved using linear algebra. Further properties of vector subspaces may be transferred to properties of projective subspaces: The intersection of two projective subspaces P (U ) and P (U 2 ) is P (U ) P (U 2 ) = P (U U 2 ). The projective span or join of two projective subspaces P (U ) and P (U 2 ) is P (U + U 2 ). From the dimension formula of linear algebra we obtain a dimension formula for projective subspaces: dim(u + U 2 ) = dim U + dim U 2 dim(u U 2 ), dim P (U + U 2 ) = dim P (U ) + dim P (U 2 ) dim P (U ) P (U 2 ). 8

13 .3 The theorems of Pappus and Desargues.3 The theorems of Pappus and Desargues Definition.3.. Let P (V ) be a projective space of dimension n. Then n + 2 points in P (V ) are said to be in general position if no n + of them are contained in a (n )- dimensional projective subspace. In terms of linear algebra this implies that no n + representative vectors are linearly dependend, i.e. every n + are linearly independent. Example.3.2. If n = we have to consider n + 2 = 3 points on a projective line. These three points are in general position as long as they are disjoint. Fig. 8: Points in general position on a projective line If n = 2 we have to consider n + 2 = 4 points in a projective plane. The four points are in general position if no three are collinear. Fig. 9: Points in general position in a projective plane Question. The above pictures only show an affine part of the respective projective spaces. What happens if some of the points lie at infinity? For points in general position there exists a canonical choice for the representative vectors described in the following lemma. Lemma.3.3. Let A,..., A n+2 be n + 2 points in general position in an n-dimensional projective space P (V ). Then there exist representative vectors v i V with A i = [v i ] for i =,..., n + 2 such that: n+2 i= v i = 0. Moreover, this choice is unique up to a common scalar multiplicative factor, i.e., if ṽ i = µ i v i with µ i 0 for i =,..., n + 2 such that n+2 i= ṽi = 0 then µ = µ 2 =... = µ n+2. Proof. Existence: Let w i with A i = [w i ] be representative vectors of the points A i for i =,..., n + 2. Since dim V = n + the n + 2 vectors are linearly dependent. So there exist λ i for i =,..., n + 2 not all equal to zero such that n+2 i= λ i w i = 0. 9

14 Projective Geometry Since the points A i are in general position all λ i are non-zero. So setting v i = λ i w i we obtain A i = [v i ] with n+2 i= v i = 0. Uniqueness: Let ṽ i = µ i v i with µ i 0 be a different set of representative vectors with n+2 i= ṽ i = 0. From this equation we obtain the following system of n + homogeneous equations in n + 2 variables: n+2 i= µ i v i = 0. Since the points A i are in general position the above system has rank n + and hence a -dimensional set of solutions. But (,..., ) is a solution of the system and hence there exists µ 0 with (µ,..., µ n+2 ) = µ(,..., ). So µ =... = µ n+2 = µ. Theorem.3.4 (Pappus AC). Let A, B, C and A, B, C be two collinear triples of distinct points in the real projective plane. Then the points are collinear. A = (BC ) (B C) B = (AC ) (A C) C = (A B) (AB ) Fig. 0 Proof. Without loss of generality let A, B, B and C be four points in general position. (Question: What happens if the points are not in general position?). Choose representative vectors [a] = A, [b] = B, [b ] = B, and [c ] = C such that a + b + b + c = 0. Since a, b, c are linearly independent we obtain homogeneous coordinates: 0 A = 0, B =, C 0 = 0, B =

15 .3 The theorems of Pappus and Desargues Then C = y with y 0 since C A, and A = with z since A B. 0 z The lines spanned by the six points are the following: AB x = x 2 x 2 = x 3 x 3, A x 0 s B = x 2 s + t = s + t x 3 z 0 sz, AC x = x 2 x 2 = 0 x 3, A x s + t C = x 2 s + t y = s + ty x 3 z 0 sz, BC x = x 2 x = 0, B x s + t C = x 2 s + t y = s + ty 0 s. x 3 For the points of intersection A, B, and C this implies: C = AB A B = z, B = AC A y 0 C = 0, A = BC B C = y. z yz Now we construct a linear dependence to show that the three points are collinear: 0 y y + y z y 0 + ( y) z = z(y ) + ( y)z = 0. yz z z + yz + ( y)z So A, B, and C are collinear. Remark.3.5. The Pappus configuration is very symmetric in the following perspective: It consists of nine lines and nine points and each line contains three points and each point lies on three lines. Hilbert showed in 899 that Pappus Theorem corresponds to the commutativity of multiplication. That is, a synthetic geometry that satisfies Pappus Theorem corresponds to the geometry of a projective space that is the projectivation of a vectorspace over a field. Definition.3.6. Two triangles = (A, B, C ) and 2 = (A 2, B 2, C 2 ) are in perspective w.r.t. a point S if x 3 S = (A A 2 ) (B B 2 ) (C C 2 ) The triangles are in perspective w.r.t. a line l if lie on l. C = (A B ) (A 2 B 2 ) B = (A C ) (A 2 C 2 ) A = (B C ) (B 2 C 2 )

16 Projective Geometry Fig. : Perspective triangles Theorem.3.7 (Desargues). Let = (A, B, C ) and 2 = (A 2, B 2, C 2 ) be two triangles. Then and 2 are in perspective w.r.t. a point if and only if they are in perspective w.r.t. a line. Fig. 2: Desargues configuration Proof. : We may assume that A, A 2, S and B, B 2, S and C, C 2, S are in general position on their respective lines. By a Lemma from Lecture 3 we may choose representative vectors ] ] [ ] A = [a, A 2 = [a 2, S = s such that a + a 2 + s = 0 ] ] [b [b 2 B = C =, B 2 = ] [c 2 [c ], C 2 = such that b + b 2 + s = 0 such that c + c 2 + s = 0 We have: a + a 2 = b + b 2 = c + c 2 = s. The points of intersection of corresponding sides are: ] ] (A C ) (A 2 C 2 ) = [a c = [c 2 a 2 ] ] (A B ) (A 2 B 2 ) = [b a = [a 2 b 2 ] ] (B C ) (B 2 C 2 ) = [c b = [b 2 c 2 These three points lie on one line, because (a c ) + (b a ) + (c b ) = 0 2

17 .3 The theorems of Pappus and Desargues Fig. 3: Desargues configuration : Starting with triangles = (A, B, C ) and 2 = (A 2, B 2, C 2 ) in perspective w.r.t. the line l = (A 3 B 3 ) = (A 3 C 3 ) = (B 3 C 3 ) we observe that the triangles (A, A 2, B 3 ) and (B, B 2, A 3 ) are in perspective w.r.t. C 3. Then the forward direction implies that C = (A B 3 ) (B A 3 ) C 2 = (B 2 A 3 ) (A 2 B 3 ) S = (A A 2 ) (B B 2 ) are collinear. Hence the triangles and 2 are in perspective w.r.t. S. 3d picture: If we consider two triangles in RP 3 in perspective w.r.t. a point P, then each triangle spans a plane and these planes intersect in a line. The intersection points of corresponding sides of the triangles have to lie on the intersection line of the planes. And hence the triangles are in perspective w.r.t. this line. Fig. 4: Desargues configuration in 3D Remark.3.8 (Symmetry of Desargues configuration). Consider 5 points A, A 2,..., A 5 in RP 3 in general position. Further let l ij = (A i A j ) be the ten lines spanned by these points and let E ijk = (A i A j A k ) be the ten planes spanned by these points. Every line is contained in three planes and every plane contains three lines. Now intersect this configuration with a plane E different from E ijk and introduce the following labels: p ij = l ij E g ij = E klm E with {i, j, k, l, m} = {, 2, 3, 4, 5}. 3

18 Projective Geometry Fig. 5: Symmetric labelling of Desargues configuration This labelling shows that there are ( 5 2) pairs of perspective triangles..4 Projective transformations Let V, W be two vectorspaces over the same field and of the same dimension and F : V W a linear isomorphism. In particular ker(f ) = {0}, so F maps -dimensional subspaces to -dimensional subspaces. Hence F induces a map from P (V ) to P (W ). Definition.4.. A projective transformation f from P (V ) to P (W ) is a map definied by a linear isomorphism F : V W such that for all [v] P (V ). f ([v]) = [F (v)] Proposition.4.2. Two linear isomorphisms F, F : V W induce the same projective transformation if and only if F = λ F for some λ 0. Proof. : If F = λ F then for all [v] P (V ). f([v]) := [F (v)] = [λ F (v)] = [ F (v)] =: f([v]) : Let f : P (V ) P (W ) with f ([v]) = [F (v)] = [ F (v) ]. Hence for every [v] P (V ) there exists λ v 0 such that F (v) = λ v F (v). To show: λv = λ for some λ 0 and all [v] P (V ). Let {b, b 2,..., b n+ } be a basis of V. Then there exists λ i 0 with For [ n+ i= F (b i ) = λ i (F )(bi ) i =, 2,..., n +. ] b i there exists λ 0 such that n+ i= λ i F (bi ) = n+ i= F (b i ) = F ( n+ i= b i ) = λ F ( n+ ) n+ b i = i= i= λ F (b i ) 4

19 .4 Projective transformations which is equivalent to n+ i= (λ i λ) F (b i ) = 0. Since F is a linear isomorphism, the set { F (b i ) : i =, 2,..., n + } is a basis of V and in particular linearly independent. It follows λ i = λ for all i =, 2,..., n + and F = λ F. Definition.4.3. The real projective linear group P Gl(n, R) is definied as the quotient of the group of linear isomorphisms (the general linear group) Gl(n, R) = {F : R n R n : F linear isomorphism} = {A: R n R n : n n matrices with det(a) 0}. by the normal subgroup of non-zero multiples of the identity matrix: P Gl(n, R) = Gl(n, R)/ {λ I n : λ 0}. This is the group of projective transformations from RP n to RP n. Example.. In terms of affine coordinates the projective transformations from RP to RP are fractional linear transformations. Fig. 6: Projective transformations of the projective line Central projections are projective transformations. Theorem.4.4. Let A, A 2,..., A n+2 and B, B 2,..., B n+2 be points in general position in the n-dimensional projective spaces P (V ) und P (W ). There exists a unique projective transformation f : P (V ) P (W ) such that f(a i ) = B i. Proof. Existence: Let a i V such that A i = [a i ] for all i =,..., n + and A n+2 = [ n+ a i ]. Analogously there are b i V such that B i = [b i ] for all i =,..., n + and i= [ n+ B n+2 = b i ]. Define a linear isomorphism F : V W using the bases {a i } and {b i } i= by setting F (a i ) = b i for all i =,..., n +. Then ( n+ ) n+ F (a n+2 ) = F a i = i= i= F (a i ) = n+ i= b i = b n+2. Hence f([v]) = [F (v)] is a projective transformation with f(a i ) = B i i =,..., n + 2 Uniqueness:... 5

20 Projective Geometry.5 Cross-ratio (Doppelverhältnis) We are looking for invariants with respect to projective transformations. )] Definition.5.. Let P i = [v i ] = [( xi projective line RP. Then the cross-ratio of these points is y i, i =,..., 4, be four distinct points on a cr(p, P 2, P 3, P 4 ) = det(v v 2 ) det(v 3 v 4 ) det(v 2 v 3 ) det(v 4 v ) = x y 2 x 2 y x 3 y 4 x 4 y 3 x 2 y 3 x 3 y 2 x 4 y x y 4 = y y 2 ( x y x 2 y 2 ) y 3 y 4 ( x 3 x 4 y 3 y 4 ) y 2 y 3 ( x 2 y 2 x 3 y 3 ) y 4 y ( x 4 y 4 x y ). the cross-ratio depends on the choice of representative vector. If y i 0, introduce affine coordinates u i = x i y i. This yields If for example y = 0 (hence u = ), then cr(p, P 2, P 3, P 4 ) = u u 2 u 2 u 3 u 3 u 4 u 4 u. cr(p, P 2, P 3, P 4 ) = u 2 u 2 u 3 u 3 u 4 u 4 = u 3 u 4 u 2 u 3. Cancelling yields the correct result, since cr(p, P 2, P 3, P 4 ) = x y 2 (x 3 y 4 x 4 y 3 ) (x 2 y 3 x 3 y 2 )( x y 4 ) = x y 2 y 3 y 4 (u 3 u 4 ) x y 2 y 3 y 4 (u u 3 ) = (u 3 u 4 ) u 2 u 3. Lemma.5.2. The cross-ratio of four distinct points in RP does not depend on the choice of bases in the vector space R 2 (i.e. it does not depend on the choice of homogeneous coordinates). )] ) [( xi ( xi Proof. Let P i = with = A ỹ i ỹ i ( ) xi y i for some A GL(2, R). Then in our formular for the cross-ratio we can pull out and cancel the determinant of A. Thus the cross-ratio of these points is given by ) ) ( x x det 2 ( x3 x det 4 ỹ ỹ 2 ỹ 3 ỹ 4 cr(p, P 2, P 3, P 4 ) = ) ) ( x2 x det 3 ( x4 x det ỹ 3 ỹ 4 ỹ 4 ỹ ( ( )) ( ( )) x x det A 2 x3 x det A 4 y y 2 y 3 y 4 = ( ( )) ( ( )) x2 x det A 3 x4 x det A y 3 y 4 y 4 y ( ) ( ) x x det 2 x3 x det 4 y y 2 y 3 y 4 = ( ) ( ). x2 x det 3 x4 x det y 3 y 4 y 4 y 6

21 .5 Cross-ratio (Doppelverhältnis) Corollary.5.3. If f : RP RP is a projective transformation, then for pairwise distinct P i RP. cr(p, P 2, P 3, P 4 ) = cr(f(p ), f(p 2 ), f(p 3 ), f(p 4 )) Proof. As above, f([v]) = [Av] with A GL(n, R). Theorem.5.4. Projective transformations map projective lines to projective lines and preserve cross-ratios for arbitrary four distinct points mapped from one projective line to another. Proof. Linear isomorphisms map 2-dim subspaces to 2-dim subspaces. If we restrict the projective transformation to a line we can use the above lemma and corollary. Definition.5.5. Let l, l 2, l 3, l 4 be four concurrent lines (l l 2 l 3 l 4 = P ) in RP 2 that are distinct. Then the cross-ratio of the four lines is cr(l, l 2, l 3, l 4 ) = cr(p, P 2, P 3, P 4 ), where l is a projective line with P / l and P i = l i l. Fig. 7 The cross-ratio does not depend on the line l. Proposition.5.6. Let P, P 2, P 3, P 4 be four distinct points on a projective line and f : RP RP a projective transformation, that maps P 2, P 3, P 4 to points with affine coordinates 0,,, respectively. Then the affine coordinate of f(p ) is cr(p, P 2, P 3, P 4 ). Proof. Let q be the affine coordinate of f(p ). then q = q 0 0 = cr(q, 0,, ) = cr(f(p ), f(p 2 ), f(p 3 ), f(p 4 )) = cr(p, P 2, P 3, P 4 ). q 7

22 Projective Geometry Remark.5.7. The cross-ratio of four distinct points is never 0, or. Proposition.5.8. Let P, P 2, P 3, P 4 and Q, Q 2, Q 3, Q 4 be two sets of distinct points on a projective line (or two projective lines). Then there exists a projective transformation f : RP RP with f(p i ) = Q i if and only if cr(p, P 2, P 3, P 4 ) = cr(q, Q 2, Q 3, Q 4 ). Proof. projective transformations preserve cross-ratios. Let f : RP RP be a projective transformation with f(p 2 ) = Q 2, f(p 3 ) = Q 3 and f(p 4 ) = Q 4. Then f(p ) = Q. Show Q = Q. We have cr(p, P 2, P 3, P 4 ) = cr(f(p ), f(p 2 ), f(p 3 ), f(p 4 )) = cr( Q, Q 2, Q 3, Q 4 ). But that implies cr( Q, Q 2, Q 3, Q 4 ) = cr(q, Q 2, Q 3, Q 4 ). (.) If we introduce coordinates Q 2 = [( )] 0, Q 3 = [( )], Q 4 = [( )] 0 and Q = [( )] q, Q = )] [( q, from (.) we obtain q = q. This implies Q = Q. The cross-ratio depends on the order of the points! cr(p, P 2, P 3, P 4 ) = u u 2 u 2 u 3 u 3 u 4 u 4 u. The value of the cross-ratio does not change, if we swap two distinct pairs of points: cr(p, P 2, P 3, P 4 ) = cr(p 2, P, P 4, P 3 ) = cr(p 3, P 4, P, P 2 ) = cr(p 4, P 3, P 2, P ) From the 24 permutations of the two points we only need to calculate the cross-ratio for permutations of the last three entries: cr(p, P 2, P 3, P 4 ) = cr(q, 0,, ) = q, cr(p, P 2, P 4, P 3 ) = cr(q, 0,, ) = q q, cr(p, P 3, P 2, P 4 ) = cr(q,, 0, ) = q, cr(p, P 3, P 4, P 2 ) = cr(q,,, 0) = q q, cr(p, P 4, P 2, P 3 ) = cr(q,, 0, ) = q and cr(p, P 4, P 3, P 2 ) = cr(q,,, 0) = q. 8

23 .6 Complete quadrilateral and quadrangle.6 Complete quadrilateral and quadrangle Definition.6. (complete quadrilateral). A configuration consisting of four lines in the projective plane no three through one point and the six intersection points, one for each pair of lines, form a complete quadrilateral. Fig. 8: Complete quadrilateral Definition.6.2 (complete quadrangle). Let A, B, C and D be four points in general position in a projective plane. The complete quadrilateral consists of these four points and six lines one for each pair of points. The lines spanned by disjoint sets of points are called opposite. Fig. 9: A complete quadrilateral with opposite lines shown in the same color. Definition.6.3 (quadrangular set). Let l be a line not through any of the points A, B, C and D. We obtain three pairs of intersection points P i, Q i for i =, 2, 3 with the three pairs of opposite lines. The set {P, Q ; P 2, Q 2 ; P 3, Q 3 } is called a quadrangular set. Note the symmetry of the configuration with respect to the exchange of P i and Q i and the permutation of the indices, 2, 3. Fig. 20: Complete quadrangle and quadrangular set 9

24 Projective Geometry Definition.6.4 (multi-ratio). Let P, P 2,..., P 6 be six points on a projective line with representative vectors P i = [v i ] for i =,..., 6. The multi-ratio of these six points is defined by m(p, P 2, P 3, P 4, P 5, P 6 ) = det(v v 2 ) det(v 3 v 4 ) det(v 5 v 6 ) det(v 2 v 3 ) det(v 4 v 5 ) det(v 6 v ). Remark.6.5. For the multi-ratio we have the following properties: The multi-ratio is well-defined and independent of the choice of homogeneous coordinates. The multi-ratio is a projective invariant. For the calculation we may use affine coordinates and if one of the coordinates is infinite, we may cancel them as we did in the calculation of the cross-ratio. The multi-ratio can be calculated via cross-ratios: m(p, P 2, P 3, P 4, P 5, P 6 ) = ( ) cr(p, P 2, P 3, P 4 ) cr(p, P 4, P 5, P 6 ). Theorem.6.6. A quadrangular set of points is characterized by the relation m(p, Q 2, P 3, Q, P 2, Q 3 ) =. Any five points of a quadrangular set determine the sixth point. Proof. We will use the relation between croos-ratio and multi-ratio to prove the relation. So m(p, Q 2, P 3, Q, P 2, Q 3 ) = ( ) cr(p, Q 2, P 3, Q ) cr(p, Q, P 2, Q 3 ). The central projection of the line CD to the line l from A and D yields the following two equations: CD A l: cr(d, C, Q, R) = cr(p, Q 3, P 2, Q ) = CD B l: cr(r, D, C, Q ) = cr(p, Q 2, P 3, Q ). Thus we obtain for the multiratio: cr(p, Q, P 2, Q 3 ) m(p, Q 2, P 3, Q, P 2, Q 3 ) = ( ) cr(p, Q 2, P 3, Q ) cr(p, Q, P 2, Q 3 ) = cr(d, C, Q, R) cr(d, C, Q, R) =. If five points are given we can easily see for example using affine coordinates that the relation m(p, Q 2, P 3, Q, P 2, Q 3 ) = determines the affine coordinate of the sixth point, and hence the point itself. Theorem.6.7 (Pappus Theorem on quadrangular sets). Let {P, Q ; P 2, Q 2 ; P 3, Q 3 } be a quadrangular set of points on a projective line. Let A, B and C be three distinct points such that ABP, ACP 2 and BCP 3 are collinear. Then the lines Q C, Q 2 B and Q 3 A intersect in one point. In other words, we may construct a complete quadrilateral for a quadrangular set. 20

25 .6 Complete quadrilateral and quadrangle Proof. Let D = Q C Q 2 B and Q 3 = AD l. Then both {P, Q ; P 2, Q 2 ; P 3, Q 3 } and {P, Q ; P 2, Q 2 ; P 3, Q 3 } are quadrangular sets. But since five of the points coincide we have Q 3 = Q 3 and hence the points A, D, and Q 3 lie on one line. Fig. 2: Pappus theorem on quadrangular sets Theorem.6.8 (complete quadrilateral). Let l, l 2, l 3 and l 4 be four lines in the projective plane such that no three go through one point. Further let A = l l 2, B = l 2 l 3, C = l 3 l 4, D = l 4 l and P = l l 3 and Q = l 2 l 4 be the six points of the complete quadrilateral. Define l = P Q and X = AC l and Y = BD l. Then cr(p, X, Q, Y ) =. Fig. 22: Complete quadrilateral Remark.6.9. We say that the pair {P, Q} seperates the pair {X, Y } harmonically or Y is the harmonic conjugate of X with respect to {P, Q} if cr(p, X, Q, Y ) =. Computational proof. We can apply a projective transformation of the plane to map the four points A, B, C and D onto the unit square with vertices 0 0 0, 0,,. Then the line l is the line at infinity and we can easily calculate the cross-ratio and see it is equal to. 2

26 Projective Geometry Fig. 23: Normalized complete quadrilateral Proof via multi-ratio. Consider the multi-ratio with P = Q = P, P 3 = Q 3 = Q, Q 2 = X, P 2 = Y, where p is the affine coordinate of P, q the one of Q etc. Then we have = m(p, X, Q, P, Y, Q) = p x q p y q x q p y q p = p x x q q y = cr(p, X, Q, Y ). y p Fig. 24: Complete quadrilateral and quadrangular set To give a third proof, we introduce the notion of projective involutions. Definition.6.0. Consider a projective map f : RP RP. f is called a projective involution if f id but f 2 = id. f can be defined by two pairs of points A, B and C, D such that: f(a) = B, f(b) = A, f(c) = D, f(d) = C. Any projective involution f has either 0 or 2 fixed points. (homework) 22

27 .7 Moebius pairs of tetrahedra If f is a projective involution with two fixed points P, Q, then cr(p, X, Q, f(x)) = cr(p, f(x), Q, X) = Hence cr(p, X, Q, f(x)) = ± for all X P, Q. cr(p, X, Q, f(x)). Since a cross-ratio of four distinct points cannot become zero, we obtain a characteristic equation for a projective involution with two fixed points: cr(p, X, Q, f(x)) = for all X P, Q. Proof of the theorem on complete quadrilateral via projective involutions. Consider a projective transformation f : RP RP with A f B B f A C f D D f C. This map is a projective involution. So f l is also an involution with fixed points P and Q. Further f(x) = Y. This implies cr(p, X, Q, Y ) =..7 Moebius pairs of tetrahedra Definition.7.. A pair of tetrahedra is called a Moebius pair if the vertices of one tetrahedron lie in the planes spanned by the faces of the second and the vertices of the second lie in the face-planes of the first. Theorem.7.2. If four vertices of the first tetrahedron lie in the four face-planes of the second and three of the vertices of the second tetrahedron lie in the three face-planes of the first then the fourth vertex lies in the fourth plane. ( Geometry II, consistency of nets. Bobenko, Suris: Discrete Differential Geometry) Proof. Let us introduce the following labelling motivated by a (combinatorial) cube: Fig. 25: Labelling of a Moebius pair of tetrahedra The vertices of two tetrahedra T and T 2 are the following T : P, P 2, P 3, P 23 T 2 : P, P 2, P 3, P

28 Projective Geometry The face-planes are labelled in the following way: T : Π P, P 2, P 3 T 2 : Π 2 P, P 2, P 23 Π 2 P, P 2, P 23 Π 3 P, P 3, P 23 Π 3 P, P 3, P 23 Π 23 P 2, P 3, P 23 Π 23 P 2, P 23, P 3 Π P, P 2, P 3. Now seven vertices lie on seven planes: P i Π i, P Π, P ij Π ij. To show: P 23 Π 23. The plane Π contains the vertices P, P, P 2 and P 3. The intersections of the planes Π i and Π ij with Π form a complete quadrangle. This yields a quadrangular set on the line Π Π 23. Fig. 26: Quadrangular set Now we have a look at the intersection of the planes Π i and Π ij with Π 23. The lines Π i Π 23 yield a triangle on the vertices P 2, Π 3, and Π 23. Fig. 27 By the Theorem of Pappus on quadrangular sets the lines Π ij Π 23 through the points Π ij intersect in one point on Π 23. This point is P 23 and it lies on Π

29 .8 The Fundamental Theorem of Real Projective Geometry Fig. 28: Moebius pair of tetrahedra Definition.7.3. A Koenigs cube is a combinatorial cube with six flat faces (i.e. each sets P, P i, P j, P ij and P i, P ij, P ik, P 23 lie in a plane) such that the vertices P, P 2, P 3, P 23 and P, P 2, P 3, P 23 each lie in a plane. Theorem.7.4. If the red vertices (P, P 2, P 3, P 23 ) lie in a plane and all the faces are flat, then the blue vertices (P, P 2, P 3, P 23 ) lie in a plane. Proof. Homework..8 The Fundamental Theorem of Real Projective Geometry Theorem.8. (Fundamental Theorem of real projective geometry). Let f : RP n RP n, n 2, be a bijective map that maps lines to lines. Then f is a projective transformation. Remark.8.2. The theorem does not hold for arbitrary fields. For example f : CP n CP n (n 2) with z z f. =. z n+ is a bijective map, mapping complex projective lines to complex projective lines but it is not a projective transformation of CP n. Generaliziation for arbitrary fields: A bijective map f : P (V ) P (V ), where V is a vectorspace over some field that maps lines to lines is induced by an almost linear map F : V V, i.e. i) F (v + w) = F (v) + F (w) for all v, w V, z n+ ii) F (λv) = α(λ)f (v) for all λ F, v V, where α: F F is a field automorphism. The only field automorphism of R is the identity and of C the complex conjugation. 25

30 Projective Geometry Lemma.8.3. Let f : RP 2 RP 2 be a bijective map that maps lines to lines. Then if A, B, C, D l RP 2 with cr(a, B, C, D) = then cr(f(a), f(b), f(c), f(d)) =. Proof. Use theorem on complete quadrilateral. Fig. 29 This implies cr(f(a), f(b), f(c), f(d)) =. Lemma.8.4. Let f : RP RP be a bijective map such that f(0) = 0, f() =, f( ) = and cr(a, B, C, D) = implies cr(f(a), f(b), f(c), f(d)) =. Then f = id. Corollary.8.5. Let f : RP RP bijective, such that cr(a, B, C, D) = implies cr(f(a), f(b), f(c), f(d)) =. Then f is a projective transformation. Proof. Let g : RP RP be a projective transformation with g(f(0)) = 0, g(f()) =, g(f( )) =. Then by Lemma.8.4, g f = id. Thus f = g is a projective transformation. Proof of Lemma.8.4. We look at different points on RP that have a cross-ratio of : i) cr(x, x+y, y, ) = implies cr(f(x), f(x+y ), f(y), ) = and by that we have 2 x+y f(x) f( 2 ) f( x+y 2 =. This gives us f(x+y ) f(y) 2 (f(x) + f(y)). 2 ii) cr(0, x, 2x, ) = implies cr(0, f(x), f(2x), ) =, i.e. therefore f(2x) = 2f(x). iii) (i) and (ii) give us f(x + y) = f(x) + f(y). 2 ) f(x) = f(x+y 2 )+f(y) and thus f(x+y 2 ) = f(x) f(x) f(2x) = and iv) We have 0 = f(0) = f(x x) = f(x+( x)) = f(x)+f( x) and hence f( x) = f(x). v) Now, by (iii) and (iv) we obtain f(kx) = kf(x) for all k Z. vi) (v) implies f(qx) = qf(x) for all q Q. In particular, for x = we have f(q) = q for all q Q. To show f(r) = r for all r R, we will prove the monotonicity of f on R. vii) By cr( x,, x, x 2 ) = we obtain cr(f( x),, f(x), f(x 2 )) =. viii) For f(x 2 ) = f(x) 2 for x > 0. This yields f(x) > 0 using (vii) ix) If y x > 0, then f(y x) > 0 which implies f(y) f(x) > 0 and hence f(y) > f(x). Since f is a monotonically increasing function which is the identity on Q, f is the identity. 26

31 .9 Duality Proof of the fundamental theorem (for n = 2). Let f : RP 2 RP 2 be a bijective map, mapping lines to lines. Let P = [, 0, 0] T, P 2 = [0,, 0] T, P 3 = [0, 0, ] T, P 4 = [,, ] T and g : RP 2 RP 2 be a projective transformation such that g(f(p i )) = P i. Fig. 30 Let X RP 2 not on the line P P 2. Then the lines P i P j are mapped onto themselves since g f fixes the P i and maps lines to lines. In particular, g(f(p 2 P 3 )) = P 2 P 3 and g(f(p P 3 )) = P P 3. Let Q = P P 4 P 2 P 3 and Q 2 = P 2 P 4 P P 3. By Lemma.8.3 g f preserves cross-ratio of and by Corollary.8.5 the map g f restricted to the lines P P 3 and P 2 P 3, resp., is a projective transformation. Since g f has three fixed points on each of the lines P P 3 and P 2 P 3 we see that g f P P 3 and g f P2 P 3 is the identity. Let X = P X P 2 P 3 and X 2 = P 2 X P P 3 then X = P X P 2 X 2 and g(f(x i )) = X i, i =, 2 and X = P X P 2 X 2. So g(f(x)) = X since g(f(p X )) = P X and g(f(p 2 X 2 )) = P 2 X 2. Hence f = g is a projective transformation (if X P P 2 choose different affine coordinates or note that the point P 3 P 4 P P 2 is fixed by g f and g f restricted to the line P P 2 is again the identity). Theorem.8.6. Let U RP n be a subset that contains an open ball B R n RP n (for the decomposition RP n = R n RP n for some affine coordinate) and f : U RP n an injective map mapping lines to lines, that is, if l is a line with l U then there exists a line l such that f(u l) l. Then f is the restriction of a projective transformation..9 Duality Let l be a line in RP 2. Then the line can be described by one homogeneous equation: x x 2 l a x + a 2 x 2 + a 3 x 3 = 0, x 3 where the a i are unique up to a scalar multiple λ 0. We can take in a way that will be explained in detail, the point [a, a 2, a 3 ] T as homogeneous coordinates for the line l. The lines in RP 2 yield another projective plane with homogeneous coordinates [a, a 2, a 3 ] T. This is what we call the dual projective plane (RP 2 ). If we fix one point [x, x 2, x 3 ] T RP 2, the set of lines through this point corresponds to a line in (RP 2 ). 27

32 Projective Geometry Fig. 3: Duality in the real projective plane Note: This is the intuition, let s see the formal definition. Definition.9.. Let V be a vector space of finite dimension over some field F. The dual vector space is defined as V := {ϕ: V F : ϕ is linear}. Note: If dim(v ) =, we have to add continuity as a property for the linear function ϕ. Let us now consider bases. If {b,..., b n } is a basis for V, then {b,..., b n} is a basis for V (this is called dual basis) where b i (b j ) = δ ij. It is easy to check that this is a basis for V. This yields dim(v ) = dim(v ). Remark.9.2. To identify V with (V ) we don t need a basis of V. A vector v V can be interpreted as a linear functional of V as V (V ), v v( ) with v(ϕ) = ϕ(v) for any ϕ V. But this construction cannot be made in infinite dimensional vector spaces and in general V = (V ) does not hold. Definition.9.3. If f : V W is a linear map, then the dual map f is defined as with f (ψ)(v) = ψ(f(v)) for v V. f : W V, ψ f (ψ) Definition.9.4. Let U V a vector subspace, then the annihilator of U is defined as and is a subspace of V. U := {ϕ V : ϕ(u) = 0 for all u U} For the dimensions of U and U we have that dim(u) + dim(u ) = dim(v ). Remark.9.5. The following properties hold: 28

33 .9 Duality (U U 2 ) = U + U 2, (U + U 2 ) = U U 2. In an euclidean vector space (a vector space of finite dimension over R endowed with some inner product) one can identify V and V canonically and up to these identification U = U. Definition.9.6. Let P (V ) a projective space. Then we define the dual projective space as P (V ). Example A point [v] P (V ) corresponds to a hyperplane P (span(v) ) P (V ). 2. Points [ψ] P (span(v) ) correspond to hyperplanes through [v] P (V ). 3. More generally, the dual of a k-plane in P (V ) corresponds to an (n k )-plane if dimp (V ) = n. 4. Points in the dual (n k )-plane correspond to hyperplanes containing the given k-plane. Dictionary for Duality of RP 3 RP 3 point line plane line in a plane intersection of line and plane (RP 3 ) plane line point line through a point plane spanned by a line and a point Remark.9.8. To every incidence configuration we may associate a dual configuration. Example.9.9. Desargues configuration. Cross-ratio of lines through a point (again). Fig. 32: Desargues configuration and its dual 29

34 Projective Geometry Fig. 33: Cross-ratio of four lines through a point Proposition.2. The map which maps a point g p to the intersection point of g l is a projective map. Proof. Let {b, b 2, b 3 } be a basis of R 3, such that P = [b ] and l = P (span{b 2, b 3 }). This yields a dual basis {b, b 2, b 3 } of R3 with b i (b j) = δ ij. The line in (RP 2 ) corresponding to the point P is P = P ({b } ) = P (span{b 2, b 3 }). Let g = [sb 2 + tb 3 ] p be a point on the line P. A point x l can be written as Q = [xb 2 + yb 3 ]. To have Q g, we need 0 = (sb 2 + tb 3)(xb 2 + yb 3 ) = sx + ty. Solving this for x and y yields the following point of intersection Q = g l = [ tb 2 + sb 3 ], so the map P l is given by [( ) ( )] 0 s g l = 0 t for g = [sb 2 + tb 3 ]. Hence, it is a projective transformation and preserves cross-ratio..0 Conic sections Conic sections are solution sets of the following quadratic equations {( ) } u R 2 au 2 + 2buv + cv 2 + du + ev + f = 0. v Theorem.3. Any quadratic equation in Euclidean coordinates can be transformed by an Euclidean motion ( x y ) = A ( u v ) + b, A O(2, R), b R 2 to one of the following forms:. ( x a )2 + ( y b )2 =, ellipse Fig

35 .0 Conic sections 2. ( x a )2 ( y b )2 =, hyperbola Fig y = ax 2, parabola 4. ( x a )2 ( y b )2 = 0, pair of crossing lines Fig x 2 = a 2, two parallel lines Fig x 2 = 0, a double line 7. ( x a )2 + ( y b )2 = 0, a point 8. ( x a )2 + ( y b )2 =, empty set Concept of the proof. Diagonalize the matrix ( a b b c ) describing the quadratic part of the above equation and then calculate translation vector b. 3

36 Projective Geometry These are called conic sections, since they arise as intersections of cone of revolution (or a cylinder) with an affine plane. The cone of revolution is defined by x y R 3 x 2 + y 2 z 2 = 0 z. Fig. 38 Some facts about conic sections:. Ellipse(a > b), principle axes a and b. F,2 = (± a 2 b 2, 0), F,2 are focal points. r + r 2 = constant. 2. Hyperbola F 2 = (± a 2 + b 2, 0), F,2 are focal points. r r 2 = constant. 3. Parabola F = (0, +a ), F is focal point. r r 2 = Proposition.4. A plane cutting through the upper half of the cone of revolution and all its generating lines yields a ellipse. 32

37 .0 Conic sections Proof. Consider two spheres S, S 2 touching the cone and the plane, let p be a point of intersection points of the plane and the cone. We have F P = Y P and F 2 P = Y 2 P, since the tangents here have the equal length, hence F P + F 2 P = Y P + Y 2 P = Y Y 2 = const. Fig. 39 Theorem.5 (Optical properties). Rays emitted from one focal point of the ellipse arrive at the other focal point of the ellipse. Fig

38 Projective Geometry Proof. Consider an ellipse with foci F and F 2, and a point on the ellipse P. Then construct the point F on the line F 2P such that F P = F P. Now let l be the angle bisector of the angle F P F. Fig. 4 We will show that l is the tangent line. Let X l. Then F X = F X. But by the triangle inequality we obtain F 2 X + F X F 2 F = F P + F 2 P with equality if and only if X = P. So l is the tangent and the angles are equal. Rays emitted from one focal point of the hyperbola are reflected such that the reflected lines go through the other focal point. Fig. 42 Rays emitted from the focal point of the parabola become parallel after reflection. 34

39 .0 Conic sections Fig. 43 In the following, we shall consider conic sections from the projective point of view. Definition.0.. Let V be a vector space over R (or C). A map b: V V R is a symmetric bilinear form, if i) b(v, w) = b(w, v) v, w V ii) b(α v + α 2 v 2, w) = α b(v, w) + α 2 b(v 2, w) for all α, α 2 R, v, v 2, w V. b is non-degenerate, if b(v, w) = 0 w V v = 0. The corresponding quadratic form is defined by the q(v) = b(v, v) v V. The symmetric bilinear form is determined by the corresponding quadratic form via the polarisation identity b(v, w) = (q(v + w) q(v) q(w)). 2 If {v,..., v n } is a basis of V, then we can associate a matrix to a bilinear form by B = (b ij ) i,j=,...,n with b ij := b(v i, v j ). This yields n n n n n n b(v, w) = b x i v i, y j v j = b(v i, v j )x i y j = b ij x i y j = v T Bw, i= j= i= j= i= j= where v = (x,..., x n ) T and w = (y,..., y n ) T. For the quadratic form this implies n q(v) = v T Bv = b ij x i x j. Remark.0.2. Quadratic forms correspond to homogeneous polynomial of degree two. i,j= 35

40 Projective Geometry Theorem.0.3 (Sylvester). For a given quadratic form q on a real vector space there is a basis such that p p+q q(v) = x 2 i x 2 i. i= i=p+ If a quadratic form q c is given on a complex vector space, there is a basis such that p q c (v) = zi 2. i= The triple (p, q, n p q) is called the signature of the quadratic form q. The signature is invariant with respect to change of basis. q is non-degenerate, if and only if p + q = n. Definition.0.4. If q is a non-zero quadratic form on R 3. Then is a conic. C = {[x] RP 2 q(x) = 0} This definition does not depend on the choice of representative vectors. Every quadratic form on R 3 corresponds to a symmetric 3 3-matrix. So it is defined by six real values. Every quadratic form corresponds to a point in R 6. As q and λq define the same conic C, a conic corresponds to a point in RP 5. According to the Sylvester s Theorem there exists a basis, such that for some λ i {, 0, }. Non-degenerate conics: q(v) = λ x 2 + λ 2 x λ 3 x 2 3 i) q(v) = x 2 + x2 2 + x2 3, i.e. signature (+++). In this case the conic defined by q is the empty set: {[x] RP 2 q(x) = 0} = {[x] RP 2 x 2 + x 2 2 = } =. ii) q(v) = x 2 + x2 2 x2 3, i.e. signatur hyperbola or a parabola. (++-). The corresponding conic is an ellipse, a Degenerate conics: iii) q(v) = x 2 + x2 2, i.e. signature (++0). The correspondig conic is a single point in RP2. iv) q(v) = x 2 x2 2, i.e. signature (+-0). The corresponding conic consists of two lines in RP 2 (x = x 2, x = x 2 ). v) q(v) = x 2, i.e. signature (+00). Here we have a single line in RP2. Depending on the affine coordinate we obtain different affine images of the conic with signature (++-): 36

41 .0 Conic sections Fig. 44 Theorem.0.5. Let P, P 2, P 3, P 4, P 5 be five points in RP 2, then there exists a conic through P,..., P 5. Moreover: i) If no four points lie on a line, the conic is unique. ii) If no three points lie on a line, the conic is non-degenerate. Lemma.0.6. If three collinear points are on a conic, then the conic contains the whole line spanned by these points. Proof. Let P = [v ], P 2 = [v 2 ], P 3 = [v + v 2 ] be three collinear points and q a quadratic form defining some conic containig these points. Thus 0 = b(v, v ) = q(v ) and 0 = q(v 2 ) = b(v 2, v 2 ) as P and P 2 lie on the conic. Also, 0 = q(v + v 2 ) = b(v + v 2, v + v 2 ) = b(v, v ) + 2b(v, v 2 ) + b(v 2, v 2 ), since P 2 lies on it as well. This yields b(v, v 2 ) = 0. So for Q = [sv + tv 2 ] we obtain and Q lies on the conic. q(sv + tv 2 ) = s 2 b(v, v ) + 2stb(v, v 2 ) + t 2 b(v 2, v 2 ) = 0 Remark.0.7. If a conic in RP 2 contains a line, then it is degenerate. Proof of Theorem.0.5. Existence: Let P i = [v i ] with v i R 3 for all i =,..., 5. Let q : R 3 R 3 be a quadratic form. Then an incidence P i C := {[x] RP 2 q (x) = 0} yields an homogeneous equation for the corresponding matrix entries. q (v i ) = 0 is equivalent to v T i Bv i = 0. The system q (v i ) = 0 for all i =,..., 5 has at least a one dimensional space of solutions. So there exists a quadratic form, such that the conic contains all P i. First statement: If four of the P i lie on a line, then the conic contains a line and is degenerate. In this case there exists a one parameter family of conics containing the P i. If three points P, P 2, P 3 lie on a line l, but P 4, P 5 / l. Then by Lemma.0.6, l C and C is degenerate. So the conic consists of the lines l and P 4 P 5. In particular C is unique. Second statement and second part of (i): Now let no three of the points be colinear and q and q 2 are two quadratic forms with q (v i ) = q 2 (v 2 ) = 0. But then the quadratic form q = q + λq 2 also satisfies q (v i ) = 0 for all i =,..., 5. The determinant det (q + λq 2 ) is a polynomial in λ of degree 3. So it has a zero, i.e. there exists some λ 0 such that 37

42 Projective Geometry det (q + λ 0 q 2 ) = 0 and q + λ 0 q 2 is degenerate. Thus three points must be colinear. Thus q = q 2 is unique and non-degenerate. Fig. 45 Definition.0.8. A pencil of conics is a line in RP 5, which is the space of conics in RP 2. Let [q ] and [q 2 ] be two conics, then the conic [q] is in the pencil, if there exist homogeneous ( ) T, coordinates λ λ 2 such that [q] = [λ q + λ 2 q 2 ]. Let P, P 2, P 3, P 4 RP 2 be 4 points in general position. Then the conics through these four points build a pencil. Fig. 46 The pencil contains three degenerate conics: q = l 2 l 34, q 2 = l 3 l 24, q 3 = l 4 l 23, where l ij : R 3 R is the linear function vanishing on the lines defining P i and P j. Now for an arbitrary fifth point P 5 P i, i =,..., 4, there exists a unique conic through P,..., P 5 by the last theorem. The conics of the pencil are given by homogeneous coordinates (λ λ 2 ) with [q] = [λ q + λ 2 q 2 ]. The fifth point P 5 = [v 5 ] lies on the conic if q (v 5 ) = 0 λ q (v 5 ) + λ 2 q 2 (v 5 ) = 0 λ = q 2 (v 5 ) λ 2 = q (v 5 ). The conic containing P 5 in the pencil is [q] = [q 2 (v 5 ) q q (v 5 ) q 2 ]. Summary: We now have found the following identifications: Quadratic forms symmetric bilinear forms symmetric matrices points inr 6 38

43 . Polarity: Pole - Polar Relationship Conis quadratic forms up to a skalar multiple points in P ( R 6) = RP 5 Pencils of conics lines in RP 5. Theorem.0.9 (Pascal s Theorem). Let A, B, C, D, E and F be six points on a conic. Consider the hexagon defined by these vertices. Then the intersection points of opposite sides are colinear. I.e. there exists a line containing G = AB DE, I = BC EF, H = CD AF. Fig. 47 Proof. Consider the two pencils of the conics through A, B, C, D and A, D, E, F. Then both pencils contain the conic as there holds q = λ l AB l CD + λ 2 l AD l BC = µ l AF l DE + µ 2 l AD l EF which is equivalent to λ l AB l CD µ l AF l DE = l AD (µ 2 l EF λ 2 l BC ). Claim: The line containing the intersection points s given by µ 2 l EF λ 2 l BC.. G = [v G ] is on the line, since l AD (v G ) 0, but l AB (v G ) = 0 and l DE (v G ) = 0. Hence (µ 2 l EF λ 2 l BC ) (v G ) = Analogously, H is on the line. 3. I = [v i ] is on the line, since l EF (v I ) = 0 and l BC (v I ) = 0. Remark.0.0. Pappus theorem corresponds to Pascal s theorem on a degenerated conic.. Polarity: Pole - Polar Relationship Let b: V V F be a symmetric non-degenerate bilinear form. Definition... Let U V be a vector subspace. Then U := {v V b(u, v) = 0, u U}. 39

44 Projective Geometry If U = {u,..., u k } then U := (span U) = {v V b(u i, v) = 0, for all i =,..., k}. We call U the orthogonal complement of U. Remark..2. dim U = dim U 0 = n k, if dim U = k and dim V = n. Example..3. (i) Define b: R 2 R 2 R by ) )) b (( x y, ( x2 y 2 = x x 2 + y y 2. This yields usual orthogonality and scalar product. (ii) Define b: R 2 R 2 R by ) )) b (( x y, ( x2 y 2 = x x 2 y y 2. In this case, orthogonality can be visualized as follows: Fig. 48 We have ( ) {( ) (( ) ( )) } x = R 2 x 0 = b, = x 0 y 0 y ( ) {( ) } x = y x y = 0 x = y ( ) {( ) } x = 2 y x 2y = 0. For example, take x = 2, y =. 40

45 . Polarity: Pole - Polar Relationship If b is indefinite, (i.e. signature contains + s and s), then there exist vectors v with b(v, v) = 0. These are called isotropic vectors. In example 2, if we restrict to the subspace {λ ( ) λ R}, the non-degenerate bilinear form restricted to this subspace is degenerate. Polarity in RP 2. Using the orthogonal complement, we define a map that maps points in RP 2 to a line via [p] [p ] and lines to points, i.e. l [l ]. Definition..4. The line [p ] corresponding to a point [p] is the polar (or polar line) of [p] and [p] is the pole of its polar [p ]. Geometry of pole-polar relationship: Up to projective transformations, there is only one non-degenerate, non-empty conic in RP 2. It corresponds to a non-degenerate symmetric indefinite bilinear form with signature (+ + -), (- - +). Definition..5. A line in RP 2 is a tangent to a conic if it has exactly one point in common with the conic or lies entirely in the conic. Fig. 49 Proposition..6. The polar line of a point [p] on a conic is a tangent at this point. Fig. 50 4

46 Projective Geometry Proof. Let C = {[x] RP 2 b(x, x) = 0}. The polar line to [p] is P ({p} ) = {[x] RP 2 b(p, x) = 0}. p lies on the polar, since b(p, p) = 0 [p] C. now let [q] RP 2 with b(p, q) = 0. The polar line is given by: [λp + q] (except [p]). b(λp + q, λp + q) = λ 2 b(p, p) + 2λb(p, q) + b(q, q) = b(q, q). If b(q, q) = 0, then b(λp + q, λp + q) = 0 for all λ R and thus the polar is contained in C. If b(q, q) 0, then b(λp + q, λp + q) 0 and the polar is tangent at [p]. Example..7. In affine coordinates we get the conic x y b x 2 ), y 2 = x y + x 2 y 2 x 3 y 3. x 3 y 3 C = u = x x 3, u 2 = x 2 x 3 {( u u 2 ) } u 2 + u 2 2 =. A parametrization is given by γ(α) = ( ) u u 2 = ( ) cos α. sin α The tangent at γ(α 0 ) is for setting yields { ( )} {γ(α 0 ) + tγ sin α0 (α 0 )} = γ(α 0 ) + t, cos α 0 cos α p = sin α U x cos α x = y b sin α, y = 0 z z and the condition is equivalent to x cos α + y sin α z = 0. Hence the tangent line to [p] is given by cos α = λ sin α sin α + µ cos α λ, µ R 0. 42

47 . Polarity: Pole - Polar Relationship Fig. 5 Proposition..8. Let P, P 2 RP 2 and l non-degenerate conic, then the polar lines of P and P 2 intersect in the pole of the line P P 2. Proof. Let P( = )[p ], P( 2 = )[p 2 ] and P = [p]. Then we have b (p, p) = 0 and b (p 2, p) = 0, since P P p P p 2. Thus: So the line P P 2 is the polar line of P. b(λp + µp 2, p) = 0, λ, µ R. Construction of the polar line for points outside and inside the conic section. a) If P is outside the conic, then the polar line can be constructed using the two tangents touching the conic as shown in the picture. Fig

48 Projective Geometry b) If P is inside the conic, then we consider two arbitrary lines l and l 2 through P. The polar line l of P is the line through P and P 2 that are the poles of l and l 2. Fig. 53 Theorem..9 (Polar Triangle). Let A, B, C, D be four points in general position on a non-degenerate conic in RP 2. Then the intersection points of pairs of opposite sides of the complete quadrangle formed by A, B, C and D form a polar triangle, i.e. X is the pole of Y Z, Y is the pole of XZ and Z is the pole of XY. Fig. 54 If we choose the basis B = {x, y, z} with X = [x], Y = [y], Z = [z], then the above is equivalent to the matrix of the corresponding bilinear form being diagonal. Additionally, the statement holds for arbitrary non-degenerate conics in the pencil through A, B, C and D. Proof. Normalize: 44

49 . Polarity: Pole - Polar Relationship Fig. 55 Now, Then A = B = C = D =. 0 0 X = 0 Y = 0 Z =. 0 0 If b b 2 b 3 B = b 2 b 22 b 23 b 3 b 23 b 33 then we obtain by evaluating the bilinear form on the points the following equations:. For A on the conic, we get b + b 22 + b 33 2b 2 2b 3 + 2b 23 = For B on the conic, we get b + b 22 + b b 2 + 2b 3 + 2b 23 = For C on the conic, we get b + b 22 + b 33 2b 2 + 2b 3 2b 23 = For D on the conic, we get b + b 22 + b b 2 2b 3 2b 23 = 0. By subtracting the second, third and the fourth equation, resp., from the first, we obtain 4b 2 4b 3 = 0 4b 3 4b 23 = 0 4b 2 4b 23 = 0 Thus, b 2 = b 3 = b 23 = 0. We now have b 0 0 B = 0 b b 33 45

50 Projective Geometry By summing up all four equations, we get b + b 22 + b 33 = 0, so B w.r.t basis {x, y, z} is diagonal. In the above calculation we only used that the points A, B, C and D are on the conic. So X, Y, Z build a polar triangle for an arbitrary non-degenerate conic in the pencil. The pencil can be generated by B = and B 2 = Corollary..0. Consider a non-degenerate conic C, a point P not on C and a line l through P intersection C in two points X and Y. Let Q be the intersection of l with the polar line of P, then CR (P, X, Q, Y ) =. Proof. Use the theorem on the polar triangle and the theorem on the complete quadrilateral to obtain cr(p, X, Q, Y ) =. Fig. 56 Then a central projection with center Z yields the desired result. Dual conics and Brianchon s Theorem Let C be a non-degenerate non-empty conic in RP 2. 46

51 . Polarity: Pole - Polar Relationship Fig. 57 Then the set of tangents to C yields a conic in ( RP 2). (homework) Theorem.. (Brianchon s Theorem). Let A, B, C, D, E, F be a hexagon cirumscribed around a conic (i.e. AB, BC,... are tangents),then the diagonals AD, BE, CF intersect in one point. Proof. Dualize it and then use Pascal s theorem. Fig. 58 Note that the cyclic order of the points on the conic is preserved. In the above picture this is not the case! The order of the points/tangents was intentionally changed to obtain a nice picture for both of the theorems. Use your favourite interactive geometry software to study what happens when you change the order of the points in the hexagons. 47

52 Projective Geometry.2 Quadrics Definition.2.. Let q be a quadratic form on R n or C n. quadric is Q = {[v] RP n : q(v) = 0} A conic is a quadric in RP 2. Then the corresponding How many indefinite non-degenerate quadratic forms exists in R n+ up to change a basis? Classification by the signature: ( ), ( ),..., (+... ), (... ) the assumption of indefiniteness leaves n indefinite quadratic forms ( ),..., (+... ). The number of non-empty non-degenerate quadrics in RP n is n 2 Example.2.2. n = 2: In RP 2 there exist only one indefinite non-degenerate quadric/conic up to projective transformations. It has signature (+ + ) or ( +), resp. n = 3: In RP 3 there exist 3 2 = 2 different quadrics up to projective transformations: (+ + + ), ( +), Q = {[v] RP 3 : v 2 + v2 2 + v2 3 v2 4 = 0}, (+ + ), ( ++), Q 2 = {[v] RP 3 : v 2 + v2 2 v2 3 v2 4 = 0}. Fig. 59. is a unit sphere (or ellipse, paraboloid or 2-sheeted hyperboloid) depending on the choice of affine coordinate. To obtain a sphere as affine image of the quadric we can choose the affine coordinates u = v v 4, u 2 = v 2 v 4, and u 3 = v 3 v is one-sheeted hyperboloid or hyperbolic paraboloid. Degenerate Quadratic forms/quadrics Let b be a degenerate bilinear form. Then the space {u U : b(u, v) = 0 v V } = ker(b) is a subspace of V. Consider a subspace U V such that V = ker(b) U, then b : U defines a non-degenerate quadric Q in P (U ). The quadric Q defined by b is the union of lines through points in the nondegenerate quadric Q defined by b U and points in P (ker(b)) if Q (see Exercise 8.). Example.2.3 (degenerate conic). Consider the following bilinear form in RP 2 : b( x x x 2, x 2 x 3 x 3 ) = x 2 x

53 .3 Orthogonal Transformations Then ker(b) = span{e 3 } and R 3 = ker(b) span{e, e 2 }. Projectively, P (ker(b)) is a point }{{} U and the quadric defined by b U in P (U ) consists of two points. So the degenerate/singular conic definined by b in RP 2 consists of two crossing lines. Fig. 60 Proposition.2.4. A non-degenerate non-empty quadric Q RP n corresponding bilinear form up to a non-zero scalar multiple. determines the Proof. Let b and b be two linear forms defining Q. Consider an orthonormal basis w.r.t. b, i.e. {e,..., e p, f,..., f q }, such that: b(e i, e i ) = for all i =,..., p; b(f j, f j ) = for all j =,..., q; and b(e i, e k ) = b(f j, f l ) = b(e i, f j ) = 0 for all i, k =,..., p with i k and j, l =,..., q with j l. Consider the vectors e i ± f j. Then b(e i ± f j, e i ± f j ) = b(e i, e i ) ± 2b(e i, f j ) + b(f j, f j ) = 0 }{{}}{{}}{{} = =0 0 = b(e i ± f j, e i ± f j ) = b(e i, e i ) ± 2 b(e i, f j ) + b(f j, f j ) ± 2 b(e i, f j ) = b(e i, e i ) b(f j, f j ) b(e i, f j ) = 0 b(e i, e i ) = b(f j, f j ). So setting b(e, e ) = λ 0 implies that b(e i, e i ) = λ for all i =,..., p and b(f j, f j ) = λ for all j =,..., q..3 Orthogonal Transformations Definition.3.. Let b be a non-degenerate symmetric bilinear form on R n+. Then F : R n+ R n+ is orthogonal (wrt. b) if b(f (v), F (w)) = b(v, w) v, w R n+. 49

54 2 Hyperbolic Geometry The group of orthogonal transformations for a bilinear form of signature (p, q) with p + q = n + is denoted by O(p, q). If q = 0 we obtain the "usual" group of orthogonal transformations O(n + ) = O(n +, 0). If Q RP n is a non-degenerate quadric defined by b and F : R n+ R n+ is an orthogonal transformation, then the map f : RP n RP n with [x] f([x]) = [F (x)] maps the quadric onto itself: f(q) = Q. Proposition.3.2. If the signature of a non-degenerate quadric Q is (p, q) with p q and p 0 q, then any projective transformation with f(q) = Q is induced by an orthogonal transformation F w.r.t. b that defines Q. Proof. Let b be the bilinear form defining Q and F the linear transformation defining the projective transformation f. Then b(v, w) := b(f (v), F (w)) is another bilinear form defining Q. According to the previous proposition there exists λ 0 such that b = λ b. If λ > 0 then λ F is an orthogonal transformation, since b( F (v), F (w)) = λ λ λ b(f (v), F (w)) = λ b(v, w) = b(v, w). So we need to show that λ > 0. If {e,..., e n+ } is an orthogonal basis of R n+ w.r.t. b then {F (e ),..., F (e n+ } is also orthogonal with If b(e i, e i ) = then b(e i, e i ) = λ b(e i, e i ) = b(e i, e i ) = λ. But the signature is invariant w.r.t. to change of basis (i.e. w.r.t. F). So since p q we have that the λ is positive. If p = q (neutral signature), for example p = q = 2 for a quadric in RP 3, then there exists a projective transformation preserving the quadric not induced by an orthogonal transformation. Let b(x, x) = x 2 + x 2 2 x 2 3 x 2 4. Then the map: f : RP 3 RP 3, x x 2 x 3 x 4 x 3 x 4 x x 2 yields b(f (v), F (w)) = b(x, x) = x x 4 2 x 2 x 2 2. So f preserves the quadric but it is not induced by an orthogonal transformation F. 2 Hyperbolic Geometry 2. Lorentz Spaces The vector space R p+q with the bilinear form p p+q x, y p,q = x i y i x i y i i= i=p+ 50

55 2.2 Hyperbolic Spaces is the Lorentz space R p,q. For our purposes the space R n, with scalar product (nondegenerate symmetric bilinear form) x, y n, = is the most important case. n x i y i x n+ y n+ i= Reminder The orthogonal transformations are denoted by O(p, q) = {f GL(p + q, R) f(v), f(w) p,q = v, w v, w R p+q }. Example 2... R 2, R 3 with x, y 2, = x y + x 2 y 2 x 3 y 3. Fig. 6 x, x 2, = x 2 + x2 2 x2 3 < 0 (= in picture) time-like vectors x, x 2, = 0 light-like vectors x, x 2, > 0 (= in picture) space-like vectors 2.2 Hyperbolic Spaces Definition The n-dimensional hyperbolic space is n H n := {x R n, x, x n, = x 2 i x 2 n+ =, x n+ > 0}. i= 5