Outer Pro duct Factorizati on in Cli ord
Algebra
Hongb o Li, Yihong Wu
Institute of Systems Science
Academia Sinica
Beijing 100080, P.R.China
hli, [email protected]
Abstract
Cli ord algebra plays an imp ortant role in mathematics and physics,
and has various applications in geometric reasoning, computer vision
and rob otics. When applying Cli ord algebra to geometric problems,
an imp ortanttechnique is parametric representation of geometric en-
tities, such as planes and spheres in Euclidean and spherical spaces,
which o ccur in the form of homogeneous multivectors. Computing a
parametric representation of a geometric entity is equivalenttofac-
toring a homogeneous multivector into an outer pro duct of vectors.
Such a factorization is called outer pro duct factorization.
When no signature constraint is imp osed on the vectors whose
outer pro duct equals the homogeneous multivector, a classical result
can b e found in the b o ok of Ho dge and Pedo e 1953 , where a su-
cient and necessary condition for the factorability is given and called
quadratic Pluc ker relations p-relations . However, the p-relations are
generally algebraically dep endent and contain redundancy. In this
pap er we construct a Ritt-Wu basis of the p-relations, which serves
asamuch simpli ed criterion on the factorability. When there are
signature constraints on the vectors, we prop ose an algorithm that
can judge whether the constraints are satis able, and if so, pro duce a
required factorization.
1 Intro duction
Cli ord algebra is an imp ortant to ol in mo dern mathematics and physics.
Because of its invariant representation for geometric computation, Cli ord
algebra is gaining wider and wider recognition in elds like geometric reason-
ing, computer vision and rob otics. This short pap er contributes to solving
an often encountered problem in applying Cli ord algebra, the problem of
outer pro duct factorization.
n
Let K b e a eld whose characteristic 6= 2, and let V be a K -vector space of
n n
dimension n. The Grassmann algebra V generated by V is a graded K -
vector space, whose grades range from 0 to n.Themultiplication is denoted
by\^", called the outer pro duct. An element in the Grassmann algebra
is called a multivector, and an r -graded element is called an r -vector. Any
r -vector is called a homogeneous multivector.
n
Let fe ;e ;:::;e g b e a basis of V . It generates a basis fe ^ ^ e j 1
1 2 n i i
r
1
r n
i < ::: < i ng for the space V of r -vectors. Let A be an r -vector,
1 r r
then
X
: 1 e ^ ^ e A = a
i i r i :::i
r r
1 1
1i <:::
r
1
The list a j 1 i < ::: < i n is called the Pluc ker co ordinates of
i :::i 1 r
r
1
A . In this pap er, weallowany order among the suxes by requiring that
r
a be anti-symmetric with resp ect to its suxes.
i :::i
r
1
The outer pro duct of r vectors is called an r -extensor. For an r -extensor
n
A = a ^ ^ a , where the a's are vectors, a vector x 2V is in the
r 1 r
subspace A spanned bythea's if and only if x ^ A = 0. Therefore wecan
r r
use A to represent the space A . Since an extensor represents a subspace,
r r
an imp ortant question is how to judge whether or not a homogeneous multi-
vector is an extensor, and how to factor an extensor. The factorization will
b e unique up to a sp ecial linear transformation in the subspace.
In the b o ok of Ho dge and Pedo e 1953 , there are two theorems that answer
the ab ove question:
Theorem 1.1. An i-vector A is an extensor if and only if its Pluc ker co-
r
ordinates a = a j 1 l ;:::;l n satisfy the following quadratic
l :::l 1 r
r
1
Pluc ker relations p-relations : for any1 i < ::: < i n and any
1 r 1
1 j <::: 1 r +1 r +1 X F a : 1 a a =0: 2 i :::i ;j :::j i :::i j j :::j j :::j 1 r 1 1 r +1 1 r 1 1 r +1 1 +1 =1 Theorem 1.2. Let A be an r -extensor with Pluc ker co ordinates a j r i :::i r 1 T 1 i ;:::;i n , where a 6=0. Let b = b ;:::;b , where 1 r l :::l j l 1 l n r 1 j j b = a : Then l k l :::l kl :::l r j 1 j 1 j +1 1 r A = a b ^ ^ b : 3 r l :::l 1 r r 1 In the b o ok of Iversen 1992 , where the Grassmann algebra is replaced by a nondegenerate Cli ord algebra, Theorem 1.1 is reformulated elegantly as follows, where the dot denotes the inner pro duct in the Cli ord algebra: Theorem 1.3. An r -vector A is an extensor if and only if for any r 1 - r vector X , r 1 A ^ A X =0: 4 r r r 1 The set of p-relations 2 forms a Grobner basis for a convenient monomial order. Furthermore, it has a structure of Ho dge algebra see DeConcini, Eisenbud and Pro cesi, 1982 . However, the p-relations are generally alge- braically dep endent, and contain redundancy when used as a criterion on the factorability of a homogeneous multivector. In this pap er, we construct a Ritt-Wu basis Wu, 1978 , i. e., an irreducible characteristic set, of 2 . The numb er of relations in the basis is much smaller than that in 2 . The basis not only simpli es the criterion on the factora- bility, but also yields an ecientway to reduce a p olynomial of Pluc ker co ordinates by the Pluc ker relations. n n When V is equipp ed with an inner pro duct, the Cli ord algebra C V n n generated by V cf. Crumeyrolle, 1990 is linearly isomorphic to V . When K = R, a nonzero vector x is said to b e p ositive, or negative, or null, if x x>0, or < 0, or = 0, resp ectively. The sign of x x is called the signature of x.LetA = a ^ ^a , where the a's are mutually orthogonal r 1 r vectors and p of which are p ositive, q of which are negative. The triplet p; q ; r p q is called the signature of A . r The problem of outer pro duct factorization in Cli ord algebra is, given an r -extensor, factor it into the outer pro duct of p p ositivevectors, q negative ones and r p q null ones, if the factorization is p ossible. An algorithm is prop osed in this pap er to solve the problem. 2 Factorization in Grassmann Algebra Let A b e a nonzero r -vector with Pluc ker co ordinates a j 1 l ;:::;l r l :::l 1 r r 1 T n . Assume that A = a ^ ^ a ,wherea = a :::a with resp ect to r 1 r i i1 in n the basis fe ;:::;e g of V .Then 1 n a a 1i 1i r 1 . . . . . . a = : 5 . i :::i . . r 1 a a ri ri r 1 Since A is nonzero, at least one of its Pluc ker co ordinates is nonzero. We r assume that a 6= 0, otherwise we simply change the suxes of the basis 1:::r fe ;:::;e g to achieve this. Denote a = a , a = a . First _ 1 n 0 1:::r 1::: i 1 j i+1 :::r j i we derive the classical p-relations. The r -extensor A can b e represented bytherowvectors of the r n matrix r 0 1 0 1 T a a a 11 1n 1 B C B C . . . . B C B C . . . . A = = : 6 . . . . @ A @ A T a a a r 1 rn r Since a 6= 0, the matrix 0 0 1 a a 11 1r B C . . . B C . . . A = 7 . 0 . . @ A a a r 1 rr 1 is invertible. Multiplying A from the left by A , and using 0 1 0 0 1 a _ a 1i i 1 C B B C 1 C B . 1 . B C . . A = C ; 8 B . 0 . @ A A @ a 0 a a _ ri r i we get 0 1 a a _ _ a _ r +1 r +2 1 1 n 1 0 1 1 T B C b B C 1 a a a 0 0 0 B C B C . . . . 1 . . B C B C . . . . . . A A = = ; 9 . . . . . . 0 B C @ A B C a a T _ _ a _ @ A b r r r +2 r +1 r r n 1 a a a 0 0 0 n where the b's are vectors in V . 1 Geometrically, the multiplication of A from the left by A induces an invert- 0 ible linear transformation in the space represented by A .Therowvectors r 1 of matrix A A represents a factorization of A divided by the determinant r 0 1 1 det A =a of the transformation, i. e., 0 0 A = a b ^ ^ b : 10 r 0 1 r This formula provides a factorization of A into the outer pro duct of vectors r represented by its Pluc ker co ordinates. Nowwe consider the constraints that the Pluc ker co ordinates of A satisfy. r Apply 5 to A in its new form 10 , we get r b b 1i 1i r 1 . . . . . . = a a ; 11 . 0 i :::i . . r 1 b b ri ri r 1 T where b = b :::b ,andb = a =a . The equality b ecomes trivial when _ i i1 in ij 0 j i i ;:::;i equals 1;:::;r or di ers from it by one element. The number of 1 r nontrivial relations is r 1 r 1 r r n r 1: 12 = C C 1 C C n n r r n We can further simplify 11 to quadratic one. Expanding the determinant on the right-hand side of 11 with resp ect to the last row, weget b b b b 1i 1i 1i 1i r 1 j 1 j +1 r P . . . . . . j +r ...... = a 1 b a . . 0 ri i :::i . . . . r j 1 j =1 b b b b r 1 i r 1 i r 1 i r 1 i r 1 j 1 j +1 r P 1 = a a a ; _ i :::i ri :::i 0 r 1 j 1 j +1 r i j j =1 where the second step follows from 11 . Writing a as a ,we i :::i ri :::i i r r 1 j 1 j +1 _ j get r X a a a a = : 13 _ i 0 i :::i r r 1 _ r i j j j =1 Conversely, let A be an r -vectors whose Pluc ker co ordinates satisfy 13 . r T =a . The ab ove pro cedure shows that Let b = b :::b , where b = a _ 0 i i1 in ij j i under the assumption a 6= 0, 11 is equivalent to 13 . Therefore 10 holds 0 and A is an extensor. r n Theorem 2.1. let A be an r -vectors in V with Pluc ker co ordinates a j r l :::l r 1 1 l ;:::;l n . If a 6=0, A is an extensor if and only if 13 holds 1 r 1:::r r for any1 i < ::: < i n, where at least two i's are greater than r . 1 r When expanding the determinant on the right-hand side of 11 with resp ect to di erentrows and columns, we get di erent quadratic relations. The set of all these relations is just the p-relations. The number of p-relations is r X 4 r 2 i+1 i 1 r i + C C : 14 C C C n r +2 n n r +1 n r +i n i=3 Nowwe de ne an order among the Pluc ker co ordinates of A . Let a r l :::l r 1 be the numb er of elements in l ;:::;l that are greater than r , then 1 r 1. if a < a , set a a ; i :::i j :::j i :::i j :::j r r r r 1 1 1 1 2. if a = a , but i ;:::i j ;:::;j in lexical order, set i :::i j :::j 1 r 1 r r r 1 1 a a . i :::i j :::j r r 1 1 Theorem 2.2. Under the ab ove order, the set r X a = a fa a > 1; 1 i < ::: < i ng 15 r j a _ i 0 i :::i 1 r i :::i r r 1 1 _ r i j j j =1 is a Ritt-Wu basis of the p-relations. Proof. Any relation in 2 is reduced by 15 to a p olynomial one in the variables fa ;a j 1 i r;r < j ng. By Theorem 2.1, there is no _ 0 j i constraint among these variables. So any relation in 2 is reduced to 0 = 0 by 15 . The characteristic set 15 is linear with resp ect to every of its leading variables and has only one initial a , and so is irreducible. 0 Example 1. When n =5, r = 2, there are 5 Pluc ker relations: F : a a a a + a a =0; 1;234 12 34 13 24 14 23 F : a a a a + a a =0; 1;235 12 35 13 25 15 23 F : a a a a + a a =0; 16 1;245 12 45 14 25 15 24 F : a a a a + a a =0; 1;345 13 45 14 35 15 34 F : a a a a + a a =0: 2;345 23 45 24 35 25 34 When a 6= 0, a Ritt-Wu basis is F ;F ;F .For n>4 and r =2, 12 1;234 1;235 1;245 4 2 2 contains C relations, while the basis contains C relations. n n 2 3 Factorization in Cli ord Algebra 0 0 0 Let K = R, A b e an extensor with signature p ;q ;o . Consider the problem r of factoring A into the outer pro duct of r vectors whicharep p ositive, q r negative, and o = r p q null ones. Wehave the following criterion on the availability of the signature constraints decided by the triplet p; q ; o : 0 0 0 0 0 0 Theorem 3.1. 1 If p = q =0 orp = o =0,orq = o = 0 , then 0 0 0 p = q = 0 or p = o =0,or q = o = 0 . 2 If p = 0 but q 6= 0 or q =0 0 0 but p 6= 0 , then p = 0 or q = 0 , and o varies from 0 to o . 3 In other cases, p; q ; o can takeanyvalues from 0 to r . Proof. 1 is obvious. Let e ;:::;e b e an orthogonal basis of the subspace 1 r 0 0 0 represented by A .For 2 , when p = 0 but q 6= 0, let the rst q vectors r 0 of the basis b e negative. Then e + e is negativeforany q 1 i 0 0 0 3 , when p ;q 6= 0, let the rst p vectors of the basis b e unit p ositiveand 0 the subsequent q vectors b e unit negative. Then e + e is p ositive, null or i j 0 0 0 0 negative for 1 i p , p resp ectively. This and 2 guarantee that p; q ; o can takeanyvalues from 0 to r . Let A b e the space represented by A . We can rst apply 10 to get r r A = b ^^b , where the b's are vectors. The inner pro duct matrix of the r 1 r b's is M = b b : A sp ecial linear transformation T in A changes M i j r r r T to TMT .From linear algebra, we know that there exists a T that changes 0 M to a diagonal matrix whose rst p diagonal elements are p ositiveand 0 the subsequent q diagonal elements are negative. To obtain the required triplet p; q ; o , we only need to nd a sp ecial linear transformation X in A r T that changes TMT to a matrix whichhas p p ositive, q negativeand o zero, T diagonal elements. In matrix form, let B = b ::: b , then a factorization 1 r of A satisfying the signature constraints can b e realized bytherowvectors r of the matrix XT B. The following algorithm realizes the ab ove idea bycho osing X according 0 0 0 to the triplets p; q ; o and p ;q ;o . When o =0,aninputA of rational r co ecients leads to an output whose vectors are of rational comp onents. The algorithm is implemented with Maple 5 Release 3. Input: A = b ^ ^ b , where the b's are vectors in numeric form. r 1 r Step 1. Construct the inner pro duct matrix M = b b . i j r r Step 2. Diagonalize M with the following transformation: T M 7! TMT ; where det T =1: T can b e chosen to b e the comp osition of some elementary row trans- formations that replace a row a of a matrix with a + b, where is a scalar and b is another row of the matrix. T Let D = diag d ;:::;d =TMT . 1 r 0 > 0, Step 3. Rearrange the diagonal elements in D suchthatd ;:::;d 1 p 0 0 0 0 0 < 0, and d ;:::;d = 0. This can b e realized bya d ;:::;d p +q +1 r p +1 p +q 0 linear transformation T of determinant 1, which is the comp osition of some elementary row transformations that swap tworows of a matrix. 0 0 0 Denote o = r p q . 0 0 0 Step 4. The triplet p ;q ;o is the signature of A . Using Theorem 3.1, if r the constraint p; q ; r is not satis able, output this message and exit. Step 5. There are the following p ossibilities: 0 0 Case 1. If p = q = 0, output A = b ^ ^ b and exit. r 1 r 0 0 0 0 0 Case 2. If p 6=0, q =0 or q 6=0, p = 0 , then p p , q = 0 or 0 q q , p = 0 . De ne 0 0 1 1 0 0 I I p q B B C C 0 0 0 0 0 0 X = J I or J I ; @ @ A A p p q q p p p q q q 0 0 I I o o 0 0 0 0 where I represents the k k unit matrix, J or J isa k p p p q q q T matrix whose rst column is 1 ::: 1 and other columns are zeros. 0 0 0 0 0 Case 3. If p ;q 6=0, p p, q q , then o o. De ne 0 1 I p B C 0 0 I J B C p p p p q B C B C I X = ; q B C B C 0 0 I J 0 @ A q q q q p 0 I o where the J 's are matrices whose nonzero elements are the rst columns; 0 the rst column of J is p p q q q q T 0 0 0 0 d =d ; d =d ;:::; d =d ; p+1 p +1 p+2 p +1 p p +1 0 the rst column of J is 0 q q p q q q T 0 0 0 0 =d : d =d ; d =d ;:::; d 1 p +q +1 1 p +q +2 1 p +q 0 0 0 0 0 Case 4. If p ;q 6=0, p p, q q , then o o. De ne 1 0 0 I p C B 0 C I B q C B C B I ; X = o C B C B 0 0 0 J I A @ p p p p p 0 0 J I 0 0 q q q q q T where the J 's are matrices whose rst columns are 1 ::: 1 and other columns are zeros. 0 0 0 0 0 0 0 Case 5. If p ;q 6=0,p p, q q or p p, q q , and o o, de ne 0 1 0 1 0 I I p p B C B C 0 0 0 B C J I B C I q q q q q q B C B C 0 B C B C 0 0 0 0 J I I J X = or ; 0 0 o o p p p p p o o q B C B C B C B C 0 0 I J 0 I 0 0 @ A @ A o o q o o p 0 0 I I o o where the J 's are matrices whose nonzero elements are the rst columns; 0 0 the rst column of J is q q q q q q T 0 0 0 0 d =d ]+ 1; [ d =d ]+1;:::;[ d [ =d ]+1 ; p+1 p +1 p+2 p +1 p+q q p +1 0 0 the rst column of J is p p p q q q T 0 0 d =d ]+ 1; [ d =d ]+ 1;:::;[ d =d ]+ 1 ; [ p +q +1 1 p +q +2 1 p+q 1 0 the rst column of J is 0 0 o o q q q q T 0 0 0 0 0 0 =d ; d =d ; d =d ;:::; d p +1 p+q q +1 p +1 p+q q +2 p +1 p 0 the rst column of J is 0 0 o o p q q q T 0 0 d =d ; d =d ;:::; d =d : p+q +1 1 p+q +2 1 p +q 1 0 0 0 0 0 0 0 Case 6: If p ;q 6=0,p p, q q or p p, q q , and o o, de ne 0 0 1 1 0 I I p p B B C C 0 0 0 B B I J C I C p p q p p q B B C C B B C C 0 0 0 0 I I J X = or ; q q q q q p B B C C B B C C I I @ @ A A o o 0 0 0 0 J I J I 0 0 0 0 o o o o o o q o o p where the J 's are matrices whose nonzero elements are the rst columns; 0 0 the rst column of J is p p q q q q T 0 0 0 0 [ d =d ]+1; [ d =d ]+ 1;:::;[ d =d ]+1 ; p+1 p +1 p+2 p +1 p p +1 0 0 the rst column of J is q q p q q q T 0 0 0 0 [ d =d ]+1; [ d =d ]+1;:::;[ d =d ]+ 1 ; p +q +1 1 p +q +2 1 p +q 1 0 0 T the rst columns of J and J are 1 :::1 . 0 0 0 0 o o q o o p 0 T Output: A = det T a ^ ::: ^ a , where a is the i-th row of the matrix r 1 r i 0 T XT TB,and B = b :::b . 1 r 2;1 Example 2. Let fe ;e ;e g b e an orthogonal basis of R , e e = e e = 1 2 3 1 1 2 2 e e = 1. Let A = e ^ e +3e ^ e + e ^ e . 3 3 2 1 2 1 3 2 3 By the factorization formula 10 , weget A = e e ^ e +3e ; 2 1 3 2 3 when p; q ; o = 3; 0; 0 , using the ab ove algorithm, weget 3 1 1 21 A = e + e + e ^ 3e e + e ; 2 1 2 3 1 2 3 8 8 8 8 when p; q ; o = 0; 3; 0 , we get 23 5 e e ^ e +3e ; A = e 2 3 2 3 2 1 8 8 when p; q ; o = 0; 0; 3 , we get 1 1 8 10 A = e e ^ e +2e + e : 2 1 3 1 2 3 2 2 3 3 In geometric applications, an extensor in a Minkowskii space plays an imp or- tant role. For example, in the hyp erb oloid mo del of hyp erb olic geometry,a Minkowskii r -extensor represents an r 1 -plane; factoring it into an outer pro duct of negativevectors corresp onds to representing the plane with its p oints. In the homogeneous mo del of Euclidean geometry, a Minkowskii r - extensor represents an r 2 -sphere or plane ; factoring it into an outer pro duct of null vectors corresp onds to representing the sphere or plane with its p oints. An r -extensor can also representan r 1 -bunch of spheres and hyp erplanes; factoring it into an outer pro duct of p ositivevectors corresp onds to representing the bunch with its spheres and hyp erplanes. References [1] Crumeyrolle, A. 1990 : Orthogonal and Symplectic Cli ordAlgebras. Kluwer Academic Publishers. [2] DeConcini, C., Eisenbud E. and Pro cesi C. 1982 : Ho dge algebras, Ast erisque 91. [3] Ho dge, W. V. D. and Pedo e, D. 1953 : Methods of Algebraic Aeometry. Cambridge University Press, Cambridge. [4] Iversen, B. 1992 : Hyperbolic Geometry.Cambridge University Press, Cambridge. [5] Wu, W. T. 1978 : On the decision problem and the mechanization of theorem proving in elementary geometry. Scientia Sinica 21.