Geo-Metric-Affine-Projective Computing

Transcription

1 Geo-Metric-Affine-Projective Computing Speaker: Ambjörn Naeve, Ph.D., Senior researcher Affiliation: Centre for user-oriented IT-Design (CID) Dept. of Numerical Analysis and Computing Science Royal Institute of Technology (KTH) Stockholm, Sweden -address: web-site: cid.nada.kth.se/il

2 Projective Drawing Board (PDB) PDB is an interactive program for doing plane projective geometry that will be used to illustrate this lecture. PDB has been developed by Harald Winroth and Ambjörn Naeve as a part of Harald s doctoral thesis work at the Computational Vision and Active Perception (CVAP) laboratory at KTH. PDB is avaliable as freeware under Linux.

3

4

5 Geometric algebra in n-dim Euclidean space Underlying vector space V n with ON-basis e 1,...,e n. Geometric algebra: G= G G( V n ) has 2 n dimensions. A multivector is a sum of k-vectors: A k-blade = blade of grade k : B = b b b n k M n = k = 0 M k A k-vector is a sum of k-blades: M = A + B +K k k k Note: B 0 b, K, b 1 Hence: k k 1 2 K are linearly independent. the grade of a blade is the dimension of the subspace that it spans. k

6 Blades correspond to geometric objects blade of grade equivalence class of directed equal orientation and 1 line segments 2 surface regions 3 3-dim regions length area volume k k-dim regions k-volume

7 Pseudoscalars and duality Def: A n-blade in G n is called a pseudoscalar. A pseudoscalar: P = p p K p 1 2 n A unit pseudoscalar: I = e e K e 1 2 n The bracket of P: [ P]= PI 1 The dual of a multivector x: Dual( x) = xi 1 Notation: Note: Dual( x) If A is a k-blade, then = A * x * is a (n-k)-blade.

8 The subspace of a blade Fact: Fact: To every non-zero m-blade B = b K b 1 m there corresponds a m-dim subspace B n V with If B= Linspan{ b, K, b } 1 m e, e, K, is an ON-basis for B 1 2 e m m and if b = b e = = Linspan{ b V : b B= 0 }. i k 1 ik k then B= ( det b ) e e K e 1 2 ik = ( ) det b ee ik n for i = 1,,m, e K 1 2 m. m

9 Dual subspaces <=> orthogonal complements Fact: If A is a non-zero m-blade A * = A. Proof: We can WLOG choose an ON-basis for V n such that A= λee Ke 1 2 m We then have A * = AI 1 =± λem+ which implies that and I = ee Ke 1 2 n. 1 K e n A * = A.

10 The join and the meet of two blades Def: Given blades A and B, if there exists a blade C such that A= BC = B C we say that A is a dividend of B and B is a divisor of A. Def: The join of blades A and B is a common dividend of lowest grade. Def: The meet of blades A and B is a common divisor of greatest grade. The join and meet provide a representation in geometric algebra of the lattice algebra of subspaces of V n.

11 Join of two blades <=> sum of their subspaces Def: For two blades A and B with A B 0 Join( AB, ) = A B we can define:. In this case:. Example: 0 A B= A+ B and A B=0 ab, V 3 a b a b a b =0 a b = a + b = { λ a+ µ b: λ, µ R}

12 Meet of blades <=> intersection of subspaces Def: If blades AB then:, 0 and A + B =V * * Meet( AB, ) A B= ( A B) I. In this case: A B= A B. n Note: The meet product is related to the outer product by duality: ( A B) I 1 = ( A B ) II 1 = A B * * * * ( A B) = ( A) ( B) Dual Dual Dual

13 Dual outer product Dualisation: G G x a x * = xi * Dual outer product: 1 G G G * * * G G G x y = (( xi ) ( yi )) I * * * x y = ( x y) 1 1

14 Example: * ( ) A = e e I * 1 2 ( ) B = e e I 2 3 A B= ( A B ) I V 3, I = e e e = eee A= e e = ee, B= e e = ee * * = ( ee ) ( eee) = ( ) = ( ee) ( eee) = e ( ) = e e ( eee ) = e eeeee = e 3 A * = e 3 B Hence: A B= A B e 2 e 1 = A A = B * B

15 Projective geometry - historical perspective 1-d subspace parallell to the ground plane line at infinity horizon point at infinity eye parallell lines point ground plane 1-dim subspace through the eye

16 n-dimensional projective space P n P n = P( V n+1) A point p is a 1-dim subspace p = the set of non-zero subspaces of V n+1. (spanned by a 1-blade a). = a = { λa : λ 0} = a = α, α a 0. A line l is a 2-dim subspace (spanned by a 2-blade B 2 ). l = B = { λb : λ 0} 2 2 = B 2. Let B denote the set of non-zero blades of the geometric algebra G(V n+1 ). B B P n. Hence we have the mapping B

17 The projective plane P 2 0 abx,, V 3 Line: a b P 2 { x 0: x a= 0} a b { x 0: x a b= 0}

18 The intersection of two lines in P 2 AB, {non-zero 2-blades in G } 3. P 2 A B B A

19 Collinear points p q p q r The points p, q, r are collinear if and only if p q r =0.

20 Concurrent lines P R P The lines P,, R are concurrent if and only if ( P ) R=0.

21 Desargues configuration a a b b = b b c c q p A= b c A = b c B= c a B = c a p= A A q= B B C = a b r r = C C C = a b P= a a R= c c

22 Desargues configuration (cont.) P= a a = b b R= c c p= A A q = B B r = C C = ( b c) ( b c ) = ( c a) ( c a ) = ( a b) ( a b ) J = a b c =[ abc] I J = a b c = [ abc ] I Leads to: p q r JJ ( P ) R = 0 if and only if = 0 if and only if p,q,r are collinear P,,R are concurrent

23 Desargues theorem B P C A B A C L R p, q, r are collinear iff P,, R are concurrent.

24 Pascal s theorem Let abca,,,, b be five given points in P 2 Consider the second degree polynomial given by px ( ) = (( a b ) ( a b)) (( b x) ( b c)) (( c a ) ( x a)).. It is obvious that pa ( ) = pb ( ) =0 and easy to verify that pc () = pa ( ) = pb ( ) = 0. Hence: px ( )= 0 must be the equation of the conic on the 5 given points.

25 Verifying that p(a ) = 0 pa ( ) = same line = (( a b ) ( a b)) (( b a ) ( b c)) same point (( c a ) ( a a)) = a b = ( a b) a = 0 = a

26 Pascal s theorem (cont.) Hence, a sixth point c lies on this conic if and only if pc ( ) = (( a b ) ( a b)) (( b c ) ( b c)) (( c a ) ( c a)) =0. Geometric formulation: The three points of intersection of opposite sides of a hexagon inscribed in a conic are collinear. This is a property of the hexagrammum mysticum, which Blaise Pascal discovered in 1640, at the age of 16.

27 Pascal s theorem (cont.) a a 4 1 r b t 5 6 b 2 r = ( a b ) ( a b) s= ( b c ) ( b c) t = ( c a ) ( c a) s 3 c c = = = 1 4 r s t =

28 Pascal s theorem (cont.) L P P R R The three points of intersection of opposite sides of a hexagon inscribed in a conic are collinear.

29 Pappus theorem (ca 350 A.D.) R P P R L If the conic degenerates into two straight lines, Pappus theorem emerges as a special case of Pascal s.

30 Outermorphisms Def: A mapping f : G G is called an outermorphism if (i) f is linear. (ii) f()= t t t R (iii) f( x y) = f( x) f( y) xy, G (iv) f k ( G ) G k k 0

31 The induced outermorphism Let Fact: T : V V denote a linear mapping. T induces an outermorphism T : G G given by Ta ( K a) = Ta ( ) K Ta ( ) T( λ) 1 k 1 = λ and linear extension., λ R k Interpretation: T maps the blades of V in accordance with how T maps the vectors of V.

32 Polarization with respect to a quadric in P n Let T n+ 1 n+ 1 : V V denote a symmetric linear map, which means that Tx ( ) y= xty ( ) xy, n+ V 1. The corresponding quadric (hyper)surface in is given by n+ = { x V 1 : x T( x) = 0, x 0 }., P n Def: The polar of the k-blade A with respect to is the (n+1-k)-blade defined by Pol ( A) T( A) * T( A) I 1.

33 Polarization (cont.) Note: T = id In this case = { x V 1 : x x = 0, x 0 }. 1 Pol ( A) AI A and polarization becomes identical to dualization. n+ * Fact: For a blade A we have (i) (ii) Pol ( Pol ( A)) = A If A is tangent to then Pol (A) is tangent to. Especially: If x is a point on, then Pol (x) is the hyperplane which is tangent to at the point x.

34 Pol ( x) x

35 x Pol ( x)

36 Pol ( x) x

37 Polarity with respect to a conic P 2 y x y x Pol ( x) Pol ( y) Pol ( x) Pol ( y) The polar of the join of x and y is the meet of the polars of x and y = = Pol ( x y) Pol ( x) Pol ( y)

38 Pol ( x) Pol ( y) x x y y

39 Polar reciprocity Let xy, n+ V 1 represent two points in P n. Then we have from the symmetry of T : y T( x) * = y ( T( x) I 1 ) = ( ytx ( )) I 1 = ( x T( y)) I 1 = x 1 ( T( y) I ) = x T( y) *. Hence: y Pol ( x) = 0 i.e. the point y lies on the polar of the point x if and only if x lies on the polar of y. x Pol ( y) = 0

40 Brianchon s theorem Pascal line (1640) 3 2 Brianchon point (1810)

41 The dual map Let f : G G and assume that be linear, I 2 0. G * * f G Def: The dual map f : G G is the linear map given by f ( x) = f( xi) I 1 G * * G f f ( x ) = f ~ f( x) = f G G Note: 1 f( x) = f ( xi ) I = = f ( xi 1 I 2 ) I 1 f ( xi 1 ) 2 2 I I I = f ( xi) I 1 ( = f x)

42 Polarizing a quadric with respect to another Let S : G G and T : G G be symmetric outermorphisms, Fact: P= { x G : x S( x) = 0, x 0} = { x G : x T( x) = 0, x 0} be the corresponding two quadrics. Polarizing the multivectors of the quadric P with respect to the quadric gives a quadric R with equation and we have x and let x ( ToS ot( x)) = 0 P Pol ( x) R,.

43 The reciprocal quadric Polarizing the quadric with respect to the quadric generates the quadric P : : x S( x) = 0 x T( x) = 0 R : x ( ToS ot( x)) = 0 R P

44 Cartesian-Affine-Projective relationships V n x V n+1 V n y The affine part of Projective Space: * x * x = x+ e A n n+ 1 y = ( ye ) 1 y e * n+ 1 well defined e n+1 y y P n V n P n V n n+ 1 V n A n * x ( ye ) n+ 1 1 y Affine space V n+1 V n e 1 x = y * * x * e n Cartesian space

45 The intersection of two lines in the plane a,b,c,d p * ( ) *( ) A,B,C,D P = (A B) (C D) A e 3 C P D B Affine plane R 2 a c p d b Cartesian plane

46 The intersection of two lines (cont.) P = ( A B) ( C D) [ ] [ ] = A B C D A B D C [ A B C ]=( A B C) I 1 0 = does not contain e 3 ( ) = ( a+ e ) ( b+ e ) ( c+ e ) eee ( ) = ( a c) ( b c) ( c+ e ) eee ( ) + = ( a c) ( b c) c eee ( ) (( ) ( ) ) = ( a c) ( b c) ee ( a c) ( b c) e eee = a c b c e eee = αd α βc

47 The intersection of two lines (cont.) In the same way we get β = [ A B D ] Hence we can write p as: p = * P = ( αd βc) * = ( A B D) I 1 = ( a d) ( b d) ee 2 1 = (( αd βc) e ) ( αd βc) e = ( α β) ( αd+ αe βc βe ) e 1 = ( α β) ( αd βc) does not contain e 3 does not contain e 3

48 Reflection in a plane-curve mirror s m(s) = = m m 1 t m(s+ds) = m(s). t 1 = t(s+ds) M p uu 1 = -t(p-m)t in-caustic curvature centre of the mirror r out-caustic = - t 1 (p-m 1 )t 1 q = lim ds (m u) (m 1 u 1 )

49 Reflection in a plane-curve mirror (cont.) Making use of the intersection formula deduced earlier and introducing we get t n = t for the unit mirror normal q m= (( p m) t) t ( p m) n) n 1 2 m p ( m) ( p m) n 2 This is an expression of Tschirnhausen s reflection law.

50 Reflection in a plane-curve mirror (cont.) m Tschirnhausens reflection formula p r u v q ± = u m q m v m

51 References: Hestenes, D. & Ziegler, R., Projective Geometry with Clifford Algebra, Acta Applicandae Mathematicae 23, pp , Naeve, A. & Svensson, L., Geo-Metric-Affine-Projective Unification, Sommer (ed.), Geometric Computations with Clifford Algebra, Chapter 5, pp , Springer, Winroth, H., Dynamic Projective Geometry, TRITA-NA-99/01, ISSN , ISRN KTH/NA/R--99/01--SE, Dissertation, The Computational Vision and Active Perception Laboratory, Dept. of Numerical Analysis and Computing Science, KTH, Stockholm, March 1999.