AFFINE SPINOR DECOMPOSITION IN THREE-DIMENSIONAL AFFINE GEOMETRY?

Chengran WU()

Academy of Mathematics and Systems Science,Chinese Academy of Sciences,Beijing 100190,China

E-mail:Wuchengran@amss.ac.cn

Hongbo LI()?

Academy of Mathematics and Systems Science,University of Chinese Academy of Sciences,Chinese Academy of Sciences,Beijing 100190,China

E-mail:hli@mmrc.iss.ac.cn

Abstract Spin group and screw algebra,as extensions of quaternions and vector algebra,respectively,have important applications in geometry,physics and engineering.In threedimensional projective geometry,when acting on lines,each projective transformation can be decomposed into at most three harmonic projective reflections with respect to projective lines,or equivalently,each projective spinor can be decomposed into at most three orthogonal Minkowski bispinors,each inducing a harmonic projective line reflection.In this paper,we establish the corresponding result for three-dimensional affine geometry:with each affine transformation is found a minimal decomposition into general affine reflections,where the number of general affine reflections is at most three;equivalently,each affine spinor can be decomposed into at most three affine Minkowski bispinors,each inducing a general affine line reflection.

Key words spin group;spinor decomposition;affine transformation;line geometry;affine line reflection

Dedicated to Professor Banghe LI on the occasion of his 80th birthday

1 Introduction

Spin group and screw algebra,collectively as the Liegroup and Liealgebra for 3-d Euclidean geometry,have important applications in geometry,physics and engineering,in particular in robotics [1],[2],[3],computer vision and computer graphics [4].They are important for difficult problems involving inverse kinematics of serial robots [6],planar structure analysis of parallel robots [7],kinematics of multi-arm robot systems [8],trajectory planning of industrial robots [9],etc.

On the one hand,elements of spin group and screw algebra have compact algebraic forms for efficient algebraic manipulations.On the other hand,these algebraic forms are“closer”to geometry than coordinates and polynomials because of intrinsic invariance.Because of this,geometric transformations in such algebraic forms have important applications in geometric reasoning.In particular,decomposing a complicated geometric transformation into simple synthetic geometric constructions is easier in the language of spin group and screw algebra.

For example,in 3-d Euclidean geometry,a rigid-body motion is the composition of a rotation and a translation;its normal form,according to Chasles’Theorem [10],isa screw motion,which is the composition of a rotation about an axis and a translation along the same axis [11].A rotation can be induced by a quaternion,and a translation can be induced by a dual vector,so a rigid body motion can be induced by the product of the quaternion and the dual vector,which is a dual quaternion.This discovery was made by Clifford [13],[14].Quaternions and dual quaternions are all spinors,or elements of spin groups.

Any nonzero quaternion can be decomposed into the Clifford product of two invertible vectors.In geometrical terms,any rotation about the origin of R3can be decomposed into two reflections with respect to two planes passing through the origin(Figure 1(a)).Similarly,any screw motion can be decomposed into at most two reflections each about a line(Figure 1(b)).A reflection about a line(called a line reflection)is just a rotation of 180?about the line.Algebraically,the spinor inducing the screw motion is decomposed into the Clifford product of at most two orthogonal Minkowski bispinors,each inducing a line reflection.

Figure 1 (a)A 3-d rotation about axis L of angle θ is the composition of two reflections with respect to planes A and B,respectively.(b)A reflection about line L is a 180?rotation about axis L;it changes point p into point p′.

It is natural to consider extending spin group and screw algebra from Euclidean geometry to projective geometry.The first extension to projective geometry was made in [16,17],where,via the exponential map,every special linear group was covered by a spin group [18].In this approach,3-d projective geometry needs to be investigated in an 8-d inner-product space,and computations based on this embedding are relatively complicated [19].Alternatively,in [20],another approach was proposed,where 3-d projective geometry was investigated in the Clifford algebra over the 6-d space R3,3,and where projective lines were used as the basic geometric entities to establish projective geometry.Following the latter approach,[21]investigated some typical projective transformations and their spinor generators.

Recently,[15]proved that any 3-d projective transformation of positive determinant can be decomposed into at most three harmonic projective line reflections(Figure 2).This decomposition providesan explicit synthetic geometric construction of a general projective trans formation via harmonic projective line reflections.In algebraic form,a harmonic projective line reflection is induced by an orthogonal Minkows kibispinor,so any spinor inducing a projective transformation of positive determinant can be decomposed into the Clifford product of at most three orthogonal Minkowski bispinors.For every such a spinor,[15]presented a minimal decomposition in an explicit form.

Figure 2 Harmonic projective reflection with respect to a pair of non-intersecting projective lines L1,L2:for any point w in space,if w/∈L1∪L2,there is a unique projective line L passing through point w and intersecting with both lines L1 and L2;letting u1 and u2,respectively,be the two points of intersection,then the image point w′of w under the projective reflection is the unique point on line L satisfying(w′u1/w′u2)/(w u1/w u2)=?1;namely,points w′,w,u1,u2 are harmonic on the line.

Affine transformation is also an important geometric transformation.Now that any 3-d projective transformation can be decomposed into at most three projective line reflections,a 3-d affine transformation as a projective transformation preserving the plane at infinity can obviously be decomposed into at most three projective line reflections.However,this decomposition is not satisfying.One would expect the three projective line reflections to be affine,but this is not always true.It remains an open problem whether any 3-d affine transformation can be decomposed into at most three affine line reflections,and if so,how such a minimal decomposition can be computed.

This paper solves the above problem.A general affine line reflection is an affine transformation that is at the same time a projective reflection of cross ratioμ0 with respect to an affine line and a line at infinity(Figure 3).We prove that any 3-d affine transformation of positive determinant can be decomposed into at most three general affine line reflections,and for each such a transformation,we provide a minimal decomposition in an explicit form.In algebraic form,every spinor inducing such a transformation is the Clifford product of at most three affine Minkowski bispinors,each inducing a general affine line reflection.

Figure 3 General affine reflection with respect to affine line L,line at infinity n⊥,with simple ratio μ <0:for any affine point p/∈L,its image is p′,which is obtained as follows:draw a line p q that is perpendicular to direction n and intersects with line L,with q as the point of intersection,then p′is the point on line p q such that the signed ratio of directed line segments p′q and p q is μ .

As to 3-d affine transformations of negative determinant,they are the coset not containing the identity of the subgroup of positive-determinant affine transformations.Each such transformation is the composition of a fixed 3-d mirror reflection and a 3-d affine transformation of positive determinant,so the decomposition of such an affine transformation can be induced from that of the positive-determinant one.

The rest of this paper is arranged as follows:in Section 2,some preliminaries regarding the spin group for 3-d projective line geometry are introduced.In Section 3,the Main Theorem of the paper is presented,together with several lemmas.To prove the Main Theorem,in Section 4,a classification of affine Jordan formsis provided.In Section 5,a formal proof of the theorem is given.

2 Preliminaries

The algebraic model of three-dimensional projective line geometry is based on the Grassmann algebra over the inner-product space R3,3,the latter being a 6-dimensional real linear space equipped with an inner product with a matrix form diag(1,1,1,?1,?1,?1).The space R3,3can be constructed from R4as follows:lete0,e1,e2,e3be an orthonormal basis of R4,and let

For any nonzero vectorX∈R3,3,ifX·X=0,>0,<0,thenXis called a null vector,positive vector,or a negative vector,respectively.The positive and negative vectors in R3,3are all the invertible vectors.The inverse of invertible vectorXisX?1:=X/(X·X).

Any vector of R3,3is of the formU=(x1e01+x2e02+x3e03)+(y1e23+y2e31+y3e12).Introduce two vectorsx=(x1,x2,x3),y=(y1,y2,y3)∈R3,and write the components ofUas

represents linexy,whose components are called the Plücker coordinates of the line.It is easy to verify thatL·L=0,soLis a null vector.Conversely,by Plücker’s Theorem,any null vector of R3,3represents a 3-d projective line,with its components providing the Plücker coordinates of the line.

LetU∈R3,3be an invertible vector.ThenUacts on R3,3by the following signed adjoint action:

Here the multiplication among vectors on the right side is the Clifford product.All invertible vectors of R3,3generate a group Pin(3,3)under the Clifford product,called the pin group.All elements of Pin(3,3)that are Clifford products of even number of invertible vectors generate a subgroup Spin(3,3),called the spin group,whose elements are called spinors.Any element of Pin(3,3)acts on R3,3by the signed adjoint action,where the sign is positive if and only if the element is in Spin(3,3).

The signed adjoint action induces an orthogonal transformation in R3,3;in particular,it changes null vectors into null vectors,or in geometrical terms,it changes 3-d lines into 3-d lines.Obviously,both±1∈Spin(3,3)induce the identity transformation of R3,3,which in turn induces the identity transformation in 3-d projective geometry.Let

Then±I3,3∈Spin(3,3),and both of these induce the negation transformation in R3,3,which changes anyX∈R3,3into?X.On the other hand,for any null vectorX∈R3,3,any scalarλ0,λXandXrepresent the same 3-d projective line,so±I3,3again induce the identity transformation in 3-d projective geometry.

Starting from 3-d projective lines as basic geometric entities,projective points can be constructed as the intersection of three incident projective lines,and projective planes can be constructed as the extension of three coplanar projective lines.When viewed in the Grassmann space over R3,3,a projective point correspondsto a null 3-spaceof R3,3,and is represented by a 3-vector of the Grassmann space.Similarly,a projectiveplanealso correspondsto a null 3-space of R3,3,and is represented by a 3-vector of the Grassmann space.The set of null 3-spaces of R3,3,when equipped with the induced topology of the Grassmann space,has two connected components,one being the set of null 3-spaces representing 3-d points,the other being the set of null 3-spaces representing 3-d planes.

As to the orthogonal transformations in R3,3induced by the signed adjoint action of Pin(3,3),whenU∈Spin(3,3),then∈SO(3,3);whenU/∈Spin(3,3),thenchanges all 3-d points into 3-d planes,and changes all 3-d planes into 3-d points,so it is a projective polarity instead of a projective transformation.Hence Spin(3,3)quadruple coversPSL(4),the subgroup of 3-d projective transformations of positive determinant.Because of this,we sometimes call Spin(3,3)the projective spin group,and call its elements projective spinors.In particular,we call elements that are equal to the Clifford product of two invertible vectors bispinors.

A bispinorU=XY∈Spin(3,3),whereX,Y∈R3,3,is said to be Minkowski if 2-spaceX∧Yis Minkowski;it is said to be orthonormal ifX·Y=0.In R3,3,2-spaceX∧Yis Minkowski if and only if

Minkowski bispinors are important in geometric construction,because a Minkowski 2-space has two null 1-subspaces representing a pair of non-intersecting 3-d projective lines,and the projective transformation induced by a Minkowskibispinor is alwaysa projective line reflection,which can be constructed synthetically by the two lines.

For example,for any unit vectorx∈R3,vectorz∈R3,and any nonzero scalarμ,

is an orthogonal Minkowski bispinor.The two vectors are invertible and orthogonal because

and the 2-space they span is Minkowski because there are two null vectorsin the 2-space,one representing the lineLin directionxand passing through pointz,the other representing the line at infinity normal to directionx.

Any point at infinity has homogeneous coordinates of the form(0,y),wherey∈R3is nonzero.A line at infinity is composed of all the points at infinity of an affine plane,and any two parallel planes meet at the same line at infinity.Thus a line at infinity can be represented by a nonzero vector normal to all the parallel planes meeting at the line at infinity.In(2.8),bispinorUinduces the Euclidean line reflection with respect to lineL,namely the rotation of angle 180?about axisL(Figure 1(b)).

In the projective case,for two non-intersecting 3-d projective lines,L1,L2,the bispinor inducing theprojectivereflection of cross-ratioμwith respect to the two linescan beconstructed as follows:letL1,L2be represented by null vectors of R3,3,still denoted by the same symbols.Since the two lines are not coplanar,at most one of these can be at infinity,andL1·L20.There are two cases.

This is obviously Minkowski because of the two null vectors.Since

if eitherx·y=0,which is the case of Euclidean line reflection,orμ=?1,which is the original case of affine line reflection,then the bispinor is orthogonal,other wise it is non-orthogonal.The general affine line reflection has the following 4×4 matrix form:

Geometrically,as shown in Figure 3,for any pointp∈R3,ifp∈L1,its image under the general affine line reflection is stillp.Ifp/∈L1,first draw a plane passing through pointpand a line at infinityL2,and obtain a pointqas the intersection of the plane with lineL1.Then the image point ofpis the pointp′on linep qsatisfyingp′q/p q=μ.

the bispinor is orthogonal if and only ifμ=?1,namely,the projective line reflection is harmonic.

Geometrically,as shown in Figure 2,for any pointw∈R3,ifw∈L1∪L2,its image under the projective line reflection is stillw.Ifw/∈L1∪L2,draw the common perpendicularLof non-coplanar linesL1,L2,and lettingu1,u2be the two feet on the two lines respectively,the image point of pointwis the pointw′on lineLsatisfying that

A spinor in Spin(3,3)inducing a screw motion is called a Euclidean spinor.According to [20],any screw motion is the composition of two Euclidean line reflections.For 3-d projective transformations of positive determinant,according to [15],any 3-d projective transformation of positive determinant is the composition of at most three harmonic projective line reflections.

In fact,Spin group Spin(3,3)has two connected components,Spin0(3,3)and Spin1(3,3),where±1∈Spin0(3,3),and±I3,3∈Spin1(3,3).According to [15],the elements of Spin0(3,3)are equal to the Clifford product of even number of positive vectors and even number of negative vectors,and the elements of Spin1(3,3)are equal to the Clifford product of odd number of positive vectors and the odd number of negative vectors.It is a classical result that Spin0(3,3)is isomorphic toSL(4).To investigate the geometric construction of 3-d projective transformations of positive determinant,it suffices to study the decomposition of the corresponding elements of Spin0(3,3)into the Clifford product of Minkowski bispinors.

A 3-d projective point can be represented by a nonzero vector of R4,and the representation is unique up to scale.Similarly,a 3-d projective plane can be represented by a nonzero vector of(R4)?,the dual of vector space R4;the representation is also unique up to scale.For an orthonormal basis {ei,i=0,1,2,3},the induced orthonormal basis of(R4)?is {i=0,1,2,3},where

Denote byPSP(4)the set of 3-d projective polarities of determinant 1.Any element ofPSP(4)can be represented by a matrix inSL(4)with respect to the basis {ei,i=0,1,2,3}and the dual basis {i=0,1,2,3},and the representation is unique up to sign.This is the 4×4 matrix representation ofPSP(4).Similarly,any element of the groupPSL(4)of 3-d projective transformations of determinant 1 can be represented by a matrix inSL(4),which is unique up to sign.

The signed adjoint action of Pin(3,3)on R3,3can be extended to a linear transformation on the 3-spacesover R3,3.When the 3-spacesaretaken asthe set of 3-d points and planes,then any element of Pin(3,3)induces a linear transformation in the 8-d vector space R4⊕(R4)?.This is the 8×8 matrix representation of Pin(3,3).The 4×4 matrix representation of Pin(3,3)refers to the matrix form of the elements ofPSL(4)∪PSP(4)induced by Pin(3,3).The relation between the two representations is the following:

Given two invertible vectors,U,V∈R3,3,the 8×8 matrix representation of bispinorUVis

From the diagonal elements we get the 4×4 matrix representation of bispinorUVas follows:

In general,forU=U1U2···U2r∈Spin(3,3),where theUi∈R3,3are invertible vectors,the 4×4 matrix representation ofUis

From now on,in a matrix with vacant components,the entries at the components are always assumed to be 0.Consider affine transformations.With respect to an orthonormal basis {ei,i=0..3}of R4,a 3-d affine transformation has a matrix form∈GL(4),where we have the scalarα0,the vectort∈R3,and the matrixB∈GL(3).Such a matrix is called an affine matrix.All affine transformations of determinant±1 form the 3-d affine transformation groupA(3).The affine transformations of determinant 1 form the special affine transformation subgroupSA(3).All spinors of Spin(3,3)inducing 3-d affine transformation form a subgroup Aspin(3,3),whose elements are called affine spinors.In a manner similar to Spin(3,3),group Aspin(3,3)also has two connected components,and the component containing the identity is denoted by Aspin0(3,3).

The following lemma from [15]characterizes affine bispinors:

is affine if and only ifx1×x2=0.

ProofBy direct computation,we get that

The conclusion then follows. □

By the above lemma,the two vector componentsx1andx2of affine bispinor(2.21)are linear dependent and nonzero,so we can rescale them to getx1=x2=x.Sety=y1andz=y2.We introduce the following short-hand notation for affine bispinor(2.21):

The 4×4 matrix representation of the affine line reflection induced by(2.23)is

After rescalingx,y,z,the above matrix can be assumed to be inSL(4).By direct computation,the following lemma from [15]can be easily verifi ed:

Lemma 2.2The characteristic polynomial of matrix

As a corollary,the determinant of the matrix equals(x·z)2(x·y)2=1,andx·zandx·yare two multiplicity-2 eigenvalues of the matrix.

where det(B)=1.We call this subgroup the 3-d unary special affine transformation group,denoted byUSA(3).

From the above observations,we get the conclusion that a general affine matrix cannot be decomposed into the product of affine matrices that are induced by affine orthogonal bispinors.This is a major difference between projective matrices and affine matrices.

3 Main Theorem

The following is the main theorem of this paper:

Theorem 3.1Any 3-d affine transformation of positive determinant is the composition of at most three general affine line reflections.In algebraic form,any element of Aspin0(3,3)can be decomposed into the Clifford product of at most three affine Minkowski bispinors.

In [15],it was proven that any 3-d projective transformation of positive determinant is the composition of at most three projective line reflections of cross ratio?1,namely,harmonic projective reflections.In Euclidean geometry,since any orthogonal reflection with respect to a line is a harmonic projective reflection,the general conclusion on projective transformations when restricted to screw motions is also true.Algebraically,any projective spinor in Spin0(3,3)is the Clifford product of at most three orthogonal Minkowski bispinors.

For 3-d affine transformations,the story is different.While any affine transformation is the composition of at most three general affine line reflections,generally it cannot be decomposed into affine line reflections where the simple ratio equals?1.Algebraically,any spinor in Aspin0(3,3)is the Clifford product of at most three affine Minkowski bispinors,but generally it is not equal to the Clifford product of affine orthogonal Minkowski bispinors.A spinor in Aspin0(3,3)is equal to the Clifford product of affine orthogonal Minkowski bispinors if and only if its 4×4 matrix form is(2.25),namely,the matrix is in the subgroupUSA(3).

The idea for proving the Main Theorem is as follows:

1.For any elementA∈SA(3),change it into affine Jordan form,a normal form that is similar to the usual Jordan form,with the difference that in the similar transformationA=MBM?1,it is required thatM,B∈A(3).The similar matrix induced by affine matrixMis called an affine similar transformation,and affine matricesA,Bare said to be affine similar.

2.Make classification of all possible affine Jordan forms of elements ofSA(3),and for each affine Jordan form,determine whether it is in the parametrized set of matrices that are the product of 1,2,or 3 matrices of the form(2.24).If this is true for all the affine Jordan forms ofSA(3),then the theorem is proven,and as a byproduct,a decomposition into a minimal number of matrices,each representing a general affine line reflection,is obtained.

The fact that the theorem is true for all the affine Jordan forms ofSA(3)is sufficient for it to be true for all elements ofSA(3),and this is guaranteed by the following two lemmas:

Lemma 3.2For anyM∈SA(3),if the affine Jordan form ofMis induced byraffine bispinors,then so isM.

ProofLetNbetheaffine Jordan form ofM,and letA∈A(3)satisfyM=ANA?1.Assume that the rearerbispinorsp1q1,···,prqr∈Spin(3,3)such thatN=mr(p1q1)···mr(prqr),and each mr(piqi)is an affine matrix.Then

The bispinor is Minkowski if and only if the determinant of the inner product matrix is negative.

Lemma 3.2 and Lemma 3.3 indicate that if the affine Jordan form ofM∈SA(3)as an affine transformation is induced byraffine Minkowski bispinors,then so isM.

The rest of this paper is devoted to proving the Main Theorem.There are four key steps.First,obtain the set of all possible affine Jordan forms of elements ofSA(3);denote the set byASF(3).Second,find amongASF(3)all the element srepresenting general affinelinereflections;denote the subset byASF1(3).Third,find amongASF(3)ASF1(3)all the elements that are the product of two matrices,each representing a general affine line reflection;denote the subset byASF2(3).Fourth,for each element ofASF(3)(ASF1(3)∪ASF2(3)),determine whether it can be decomposed into the product of two matrices,one being affine similar to an element ofASF1(3),the other being affine similar to an element ofASF2(3).

4 Affine Jordan Form

angular affine matrix can be found which induces an affine similar transformation that convertst2,t3to zero.We simply assume thatt2=t3=0 inM.

Ift1=0,thenMis already a real Jordan form.Ift1/0,chooseN=diag(1,,1,1),and thenNMN?1is a real Jordan form whose first row is identical to that ofM.

Case 4 Ahas two complex eigenvalues,ae±i θ,and one real eigenvalueλof multiplicity 2.

Subcase 5.1a=0.Ift1=t2=0,thenMis already a real Jordan form.Ift1,t2are not both equal to 0,then,similarly to Case 3,Mcan be converted to a real Jordan form whose

first row is identical to that ofM.For example,if botht1,t2/0,we can choose

Subcase 5.2a=1.Choose

Then,similarly to Case 3,the right side of(4.6)is affine similar to a real Jordan form whose first row is identical to that ofM.

Case 6λis the unique eigenvalue ofA.

Subcase 6.1a=b=0.Ift1=t2=t3=0,thenMis already a real Jordan form,moreover an affine similar transformation can be easily found to changeMinto a real Jordan form,whose first row is identical to that ofM.

Subcase 6.2{a,b}={0,1}.Ifa=0,b=1,an affine matrix can be found that switchesa,bby a similar transformation.Thus we simply assume thata=1 andb=0.Similarly to(4.6),t2can be converted to zero by an affine similar transformation,so we also assume thatt2=0.

Subcase 6.3a=b=1.Similarly to(4.5),t2andt3can both be converted to zero.As fort1,it is either 0 or can be converted to 1,just as in Case 3. □

The following are all possible affine Jordan forms of the elements ofSA(3)?PSL(4):

SinceN∈SA(3),μ1μ2μ3μ4=1 in(4.8)andμ1μ2r2=1 in(4.9).There are,all together,the following possibilities:

Case 1 Nhas a real eigenvalue of multiplicity 4.

ThenNtakes the form(4.8),whereμ1=±1.RescaleNso thatμ1=1.Letkbe the number of nonzero elements in {a,b,c}.Then 0≤k≤3.

?Whenk=0,thenNis the identity matrix,and belongs to typeMA 1.

?Whenk=1,ifa=1,thenNis of typeMB 6,whereμ=1.If eitherc=1 orb=1,then eitherNis of typeMB 4,whereμ=1,or it can be changed into typeMB 4by affine similar transformation.

?Whenk=2,ifa=c=1,thenNis of typeMA 3up to sign.Ifa=b=1,thenNis of typeMB 14.Ifb=c=1,thenNis of typeMB 13.

?Whenk=3,Nis of typeMB 15.

Case 2 Nhas a real eigenvalue of multiplicity 3,and a real eigenvalue of multiplicity 1.

ThenNtakes the form(4.8),where eigenvalueμ1has either multiplicity 1 or multiplicity 3.Letkbe the number of nonzero elements in {a,b,c}.Then 0≤k≤2.

?Whenk=0,Nis of typeMC1,where eitherμ1=μ2=μ3orμ2=μ3=μ4.

?Whenk=1,ifa=1,thenNis of typeMC5.Ifc=1,thenNis of typeMC4.Ifb=1,thenNcan be changed into typeMC 4by an affine similar transformation.

?Whenk=2,then eithera=b=1,orb=c=1.In the first case,Nis of typeMC 7.In the second case,Nis of typeMC6.

Case 3 Nhas two multiplicity-2 eigenvalues.

The two eigenvalues must be real,becauseNis affine.ThusNtakes the form(4.8),where eigenvalueμ1has multiplicity 2,and the other eigenvalue is one ofLetkbe the number of nonzero elements in {a,b,c}.Then 0≤k≤2.

?Whenk=0,Nis of typeMA 1orMA 2.

?Whenk=1,ifa=1,thenNis of typeMB 6orMB 7.Ifc=1,thenNis of typeMB 4orMB 5.Ifb=1,thenNcan be changed into typeMB 4orMB 5by an affine similar transformation.

?Whenk=2,Nis of typeMB 11orMB 12.

Case 4 Nhas one multiplicity-2 eigenvalue and two multiplicity-1 eigenvalues.

The two multiplicity-1 eigenvalues are either both real or both complex,soNtakes either the form(4.8)or the form(4.9).WhenNtakes the form(4.9),it is of one of the following types:MB 3,MB 10(up to sign),MC2orMC3.

WhenNtakes the form(4.8),letkbe the number of nonzero elements in {a,b,c}.Then 0≤k≤1.

?Whenk=0,Nis either of typeMB 1,where one and only one ofμ1, μ2equals±1,or of typeMC1,where none ofμ1, μ2, μ3equals±1.

?Whenk=1,ifa=1,thenNis of typeMB 9(up to sign)orMC5.Ifc=1,thenNis of typeMB 8(up to sign)orMC 4.Ifb=1,thenNcan be changed intoMB 8(up to sign)orMC4by an affine similar transformation.

Case 5 Nhas four multiplicity-1 eigenvalues.

Two of the four eigenvalues may be complex,soNtakes either the form(4.8)or the form(4.9).In the former case,Nis of typeMC1.In the latter case,Nis of typeMB 3orMC2. □

5 Affine Spinor Decomposition

In this section,we prove that fori=1,2,3,any element ofASFi(3)is induced byiaffine bispinors,but cannot be induced by anyj

5.1 ASF1(3)by 1 affine bisp inor but not 1 affine Minkow ski bispinor

ASF1(3)consists of three types:MA 1,MA 2andMA 3.The identity transformationI4is a special case ofMA 1,whereμ=±1.

Theorem 5.1Any element ofASF1(3)?PSL(4)is induced by an affine bispinor.Furthermore,except forI4andMA 3ofASF1(3),any element ofASF1(3)is induced by an affine Minkowski bispinor.

ProofWe investigateMA 1,MA 2andMA 3one by one.Let {e1,e2,e3}be an orthonormal basis of R3.

ForMA 1,if it is induced by some affine bispinorwherex,y,z∈R3,then by Lemma 2.2 and(2.24),we get thatx·z=μandx·y=±μ?1.Choose the positive sign.If(2.24)equalsMA 1,then

Thus bothx,z?yare in the 1-space spanned bye1=(1,0,0)T.

We only need to find a special solution to the above equations in unknown vectorsx,y,z∈R3.Choosex=e1.Then it is easy to get thaty=μ?1e1,z=μe1.Substituting these into(2.24),we get the affine transformation induced bywhich is justMA 1.ThusMA 1is induced by one affine bispinor.

The inner product matrix of affine bispinorwhose determinant is 4?(μ+μ?1)2=?(μ?μ?1)2≤0.Therefore ifμ±1,the bispinor is Minkowski.Ifμ=±1,thenMA 1becomes the identity transformation,which is not an affine line reflection,so it cannot be induced by any affine Minkowski bispinor.On the other hand,the identity transformation is obviously the composition of two identical affine line reflections,so it can be induced by two affine Minkowski bispinors.

ForMA 2,similarly toMA 1,the following equations on unknownsx,y,z∈R3can be set up:

ForMA 3,whose eigenvalues are all 1,just as before,the following equations can be set up:

In the next subsection,we will show thatMA 3can be induced by two affine Minkowski bispinors.

Theorem 5.2IfM∈PSL(4)is induced by an affine bispinor,then eitherMis diagonalizable with two eigenvalues of multiplicity 2,or it has affine Jordan formMA 3.

ProofAs the properties stated in the theorems are invariant under affine similar transformations,Mitself can be assumed to be of an affine Jordan form.Suppose thatMis induced by affinebispinorfor someunknownsx=(x1,x2,x3)T,y=(y1,y2,y3)T,z=(z1,z2,z3)T.ThenMis of the form(2.24).By Lemma 2.2,Mhas two multiplicity-2 eigenvalues.Without loss of generality,we assume that the first two diagonal elements ofMare identical,and thus so are the last two diagonal elements.(2.24)as an affine Jordan form has the following possibilities:

From the second row of the above real Jordan form,we get thataλ=λ,x1(z2?y2)=x1(z3?y3)=0.Thusa=1 andx1=0.In the last two rows ofM,the diagonal entries are equal,and the matrix is lower triangular,so

By the above theorem,no element ofASF2(3)∪ASF3(3)is induced by only one affine bispinor.

5.2 ASF2(3)and MA 3 by 2 affine Minkowski bispinors

ASF2(3)contains 15 types:MB 1toMB 15.

Theorem 5.3Every element ofASF2(3)is induced by two affine Minkowski bispinors.Moreover,I4andMA 3ofASF1(3)are also induced by two affine Minkowski bispinors.

ProofWe first investigate the 15 types ofASF2(3),then investigateMA 3.

Let the remaining degree-2 factor off(λ)beg(λ).By direct computation,based on the matrix product on the right-hand side of(5.14),we get that

Equations(5.14),(5.15)and(5.16)provide a general approach to decomposing an affine matrix into two matrices,one being of typeMA 2,the other being affine similar to a matrix of typeMA 2.

Now thatNis affine similar toMB 3,we can simply choosem1=μ,m2=1,andθ′=θ.Substituting these into(5.15),(5.16),we get three quadratic equations in unknownsu,v,w.There are many solutions to the equations,and we only need to obtain one set of real solutions.The idea of simplifying the problem is to change the three quadratic equations into three linear ones by specifying some coordinates of the 3 vector unknowns.

We choose

wherev1,v3,w1,w2are undetermined coordinates.Substituting these into(5.15)and(5.16),we get

The following is a set of real solutions for the above system:

The corresponding matrixNequals

It is easy to verify by Maple that the above affine matrix is affine similar toMB 3.ThusMB 3can be induced by two affine Minkowski bispinors.

wherex,y,z,u,v,w,α,t,Bare all unknown.

Expanding the matrix product on the right side of(5.21),we get that

Let Π1be the subspace spanned byx,uin R3.By the last equation(5.22),it is easy to verify thatBx,Bu∈Π1.Similarly,letΠ2be the subspace spanned byz?y,w?vin R3.Again by(5.22),it is easy to verify thatBT(z?y),BT(w?v)∈Π2.In other words,Π1and Π2are invariant subspaces ofBandBT,respectively.

Now thatNis affine similar toMB 4,we have thatα=μ,and that the Jordan form ofBis

Chas the following invariant 2-spaces:e2∧e3,ande1∧e3.Similarly,CThas two invariant 2-spaces:e2∧e3,ande1∧e2.

Comparing the invariant subspaces of matrixCand its transpose with those of matrixB,we can make the specification that the two matrices have the same invariant 2-spaces,or,more accurately,that Π1=e1∧e3,Π2=e2∧e3.Then vectorsx,uare ine1∧e3,and vectorsz?y,w?vare ine2∧e3.For simplicity,we simply set thatz,y,w,v∈e2∧e3.

From(x·z)(u·w)=μ,we can make the following simple specifi cation:

Then(5.24)becomes

Substituting these into(5.21),we get that

Direct verifi cation by Maple shows that the above matrix is affine similar toMB 4,soMB 4is induced by two affine Minkowski bispinors.

ForMB 5,whose only difference fromMB 4is that the two identical diagonal entriesμ?1are replaced by?μ?1,the construction of matrixNand its decomposition is similar to that ofMB 4.The result is that by setting

we get an affine matrix

which is affine similar toMB 5.

The vectors in(5.29)satisfy

Whenμ±1,MB 5is induced by two affine Minkowski bispinors.Whenμ=±1,MB 5has an affine Jordan form of typeMB 8(up to sign).The investigation ofMB 8will soon follow.

ForMB 6,which can be obtained from block diagonal matrixMB 4by interchanging the two diagonal blocks,the construction of affine matrixNis similar,and(5.22)still holds.Nowα=μis unchanged,but the Jordan form ofBis diag(μ , μ?1, μ?1).Choose

Suppose thatB=diag(μ , μ?1, μ?1).The matrix has the following invariant 1-spaces:the 1-space spanned bye1,and the 1-space spanned byp2e2+p3e3,for allp2,p3∈R that are not both equal to zero.As the space spanned byx,uis invariant,setx=u.Similarly,set vectorsz?y,w?vto be linearly dependent.It is easy to verify that under the above specifi cations,the following set is a real solution to(5.32)and(5.22):

The corresponding matrix obtained by(5.21)is

Direct verification by Maple shows that the above matrix is affine similar toMB 6,soMB 6is induced by two affine Minkowski bispinors.

ForMB 7,which can be obtained fromMB 6by replacing each diagonal entryμ?1with?μ?1,the construction of matrixNis similar.By setting

from(5.21)we get that

Direct verification by Maple shows that the above matrix is affine similar toMB 7,soMB 7is induced by two affine bispinors.

Choose the same specification(5.17)ofx,y,zas in the caseMB 3.Then(5.15)and(5.16)are updated to the following linear equations:

Direct verification by Maple shows that the above matrix is affine similar toMB 8,soMB 8is induced by two affine Minkowski bispinors.

ForMB 9,which is much the same asMB 6except for the diagonal entries,equations(5.21)and(5.22)are set up just the same.In(5.21),α=1,and the Jordan form of matrixBis diag(1, μ , μ?1).For simplicity,chooseB=diag(1, μ , μ?1).Further more,choose

On the one hand,B?I3=diag(0, μ?1, μ?1?1),so the diagonal matrix has image spacee2∧e3.On the other hand,by(5.41),the last equation of(5.22)becomes

Thus the image space ofB?I3is spanned by vectorsx,u;it is also spanned by vectorsz?y,w?vby symmetry.We simply set all the vectorsx,y,z,u,v,wto be in 2-spacee2∧e3.

By multiplying(5.42)from the right byu,or multiplying(5.42)from the left by(z?y)T,we get the following two equations:

from which we obtain that

Substituting these into(5.42),we get that

Let

Substituting these into matrix equation(5.45),we get only one independent equation:

Thus among the four parameters,α2,α3,β2,β3,three are free.

Choose

Then

To obtainz,y,w,v,from the second equation of(5.22),namely

together with the specific values ofx,z?y,w?v,we get that

Substituting these into(5.21),we get a matrixNthat is exactlyMB 9,soMB 9is induced by two affine Minkowski bispinors.

ForMB 10,which can be obtained fromMB 9by replacing the diagonal block diag(μ , μ?1)with a 2-d rotation matrix of angleθ,the construction of matrixNis similar to that ofMB 9.Choose the inner product list(5.41).Then equations(5.42),(5.44)and(5.45)remain valid.Now thatB?I3=we can still assume the coordinates(5.46)foru,z?y.Substituting these into(5.45),we get only one independent constraint:

After factorization,this becomes

Choose the following vector values:

From(5.44),we get that

Substituting these into(5.50),we get that

From(5.21),we obtain a matrixNthat is exactlyMB 10,soMB 10is induced by two affine Minkowski bispinors.

ForMB 11,which is block diagonal and whose blocks are both of the formwhereλ=μ , μ?1,similarly toMB 3,the following linear equations are set up:

The following is a set of real solutions:

Substituting these into(5.21),we get that

Direct verifi cation by Maple shows that the above matrix is affine similar toMB 11,soMB 11is induced by two affine Minkowski bispinors.

ForMB 12,which can be obtained fromMB 11by replacing eachμ?1in the diagonal with?μ?1,the construction of matrixNis similar.The result is that by setting

From(5.21),we get that

Direct verification by Maple shows that the above matrix is affine similar toMB 12,soMB 12is induced by two affine bispinors.

In(5.60),

Choose the following vector values:

By(5.44),we have that

Substituting these into(5.50),we get that

From(5.21),we get a matrixNthat is exactlyMB 12,soMB 12is induced by two affine Minkowski bispinors.

ForMB 13,which is block diagonal where the first block is composed of a single entry 1,and the second entry is a 3×3 Jordan block with eigenvalue 1,the construction of matrixNis once again similar to that ofMB 9.Choose the inner products(5.41).Then(5.42)to(5.45)are still valid.

Now

the image space ofB?I3ise2∧e3,and the image space ofBT?I3ise1∧e2.By(5.42),x,u∈e2∧e3,andz?y,w?v∈e1∧e2.

Let

Substituting these into(5.45),we get only one independent constraint:

Choose the following vector values:

By(5.44),we have that

Substituting these into(5.50),wheret=(0,0,0)T,we get that

From(5.21),we get a matrixNthat is exactlyMB 13,soMB 13is induced by two affine Minkowski bispinors.

ForMB 14,which can be obtained fromMB 13by interchanging the two diagonal blocks,the construction of matrixNis similar to that ofMB 9.Choose the inner products(5.41).Then(5.42)to(5.45)remain valid.Now thatB?I3=we can still assume the coordinates(5.68).Substituting these into(5.45),we get only one independent constraint:α3β2=0.

Choose the following vector values:

By(5.44),we have that

Substituting these into(5.50),wheret=(1,0,0)T,we get that

From(5.21),we get a matrixNthat is exactlyMB 14,soMB 14is induced by two affine Minkowski bispinors.

ForMB 15,which is a single Jordan block with eigenvalue 1,the construction of matrixNis also similar to that ofMB 13.Equations(5.67)to(5.69)are unchanged.Choose the same vector values as(5.70).Then we get(5.71)again.Substituting these into(5.50),where we now have thatt=(1,0,0)T,and we get that

From(5.21),we get a matrixNthat is exactlyMB 15,soMB 15is induced by two affine Minkowski bispinors.

from(5.21)weget a matrixNthat is exactlyMA 3,soMA 3is induced by two affine Minkowski bispinors. □

Theorem 5.4Any element ofSA(3)that is induced by two affine Minkowski bispinors must have two eigenvalues whose product equals±1.

ProofBy the given hypothesis,M=Letf(λ)be the characteristic polynomial ofM.By direct computation,weget thatλ1=(x·z)(u·w)andλ2=(x·y)(u·v)are two roots off(λ)=0.On the other hand,by Lemma 2.2,(x·z)2(x·y)2=(u·v)2(u·w)2=1.Thus=1,namely,λ1λ2=±1. □

By Theorem 5.4,no element ofASF3(3)can be induced by less than three affine bispinors.

5.3 ASF3(3)by 3 affine Minkow ski bispinors

Theorem 5.5Any element ofASF3(3)is induced by three affine Minkowski bispinors.

ProofForMC1,if the signs ofμ1,μ2andμ3are the same,then set them to be positive.Suppose that there are positive indeterminatesν1,ν2,ν3such thatMC1=N1N2,where

ThenN1is of typeMB 1,N2has an affine Jordan form of typeMA 2,and

Obviously,

area set of real solutions for(5.79),soMC1can be induced by three affine Minkowskibispinors.

where the first matrix factor is of typeMB 3,and the second matrix factor is of typeMA 1.ThusMC2can be induced by three affine Minkowskibispinors whenμ2/±1.Whenμ2=±1,MC2becomesMB 3,which is forbidden by the definition ofASF3(3).

ForMC3,it is easy to verify thatMC3is affine similar to

where the first matrix factor is of typeMB 10,and the second matrix factor is of typeMA 1.ThusMC 3can be induced by three affine Minkowski bispinors whenμ/±1.Whenμ=±1,MC3becomesMB 10,which is forbidden by the definition ofASF3(3).

ForMC4,it is easy to verify thatMC4is affine similar toN1N2,where

SinceN1andN2are of typeMB 8andMA 1,respectively,MC4can be induced by three affine Minkowski bispinors whenμ2±1.Whenμ2=±1,MC 4becomesMB 8,which is forbidden by the definition ofASF3(3).

ForMC5,similarly toMC4,it is easy to verify thatMC4is affine similar toN1N2,where

SinceN1andN2are of typeMB 9andMA 1,respectively,MC5can be induced by three affine Minkowski bispinors whenμ1/±1.Whenμ1=±1,MC5becomesMB 9,which is forbidden by the definition ofASF3(3).

By Theorem 5.1,Theorem 5.3 and Theorem 5.5,for every affine Jordan form ofSA(3),an inducing affine spinor is found that is the Clifford product of at most 3 affine Minkowski bispinors.By Lemma 3.2 and Lemma 3.3,the result can be extended to every element ofSA(3).This completes the proof of the Main Theorem.