Grass Mann Algebra Book

Grassmann AlgebraExploring extended vector algebra with MathematicaIncomplete draft Version 0.50 September 2009

John Browne

Quantica Publishing, Melbourne, Australia

2009 9 3

Grassmann Algebra Book.nb

2

2009 9 3


3

Table of Contents1 Introduction1.1 Background The mathematical representation of physical entities The central concept of the Ausdehnungslehre Comparison with the vector and tensor algebras Algebraicizing the notion of linear dependence Grassmann algebra as a geometric calculus 1.2 The Exterior Product The anti-symmetry of the exterior product Exterior products of vectors in a three-dimensional space Terminology: elements and entities The grade of an element Interchanging the order of the factors in an exterior product A brief summary of the properties of the exterior product 1.3 The Regressive Product The regressive product as a dual product to the exterior product Unions and intersections of spaces A brief summary of the properties of the regressive product The Common Factor Axiom The intersection of two bivectors in a three dimensional space 1.4 Geometric Interpretations Points and vectors Sums and differences of points Determining a mass-centre Lines and planes The intersection of two lines 1.5 The Complement The complement as a correspondence between spaces The Euclidean complement The complement of a complement The Complement Axiom 1.6 The Interior Product The definition of the interior product Inner products and scalar products Sequential interior products Orthogonality Measure and magnitude Calculating interior products Expanding interior products The interior product of a bivector and a vector The cross product 2009 9 3

1.6 The Interior Product The definition of the interior product Inner products and scalar products Sequential interior products Orthogonality Measure and magnitude Calculating interior products Expanding interior products The interior product of a bivector and a vector The cross product 1.7 Exploring Screw Algebra To be completed 1.8 Exploring Mechanics To be completed 1.9 Exploring Grassmann Algebras To be completed 1.10 Exploring the Generalized Product To be completed 1.11 Exploring Hypercomplex Algebras To be completed 1.12 Exploring Clifford Algebras To be completed 1.13 Exploring Grassmann Matrix Algebras To be completed 1.14 The Various Types of Linear Product Introduction Case 1 The product of an element with itself is zero Case 2 The product of an element with itself is non-zero Examples 1.15 Terminology Terminology 1.16 Summary To be completed


4

2 The Exterior Product2.1 Introduction 2.2 The Exterior Product Basic properties of the exterior product Declaring scalar and vector symbols in GrassmannAlgebra Entering exterior products 2.3 Exterior Linear Spaces Composing m-elements Composing elements automatically Spaces and congruence The associativity of the exterior product Transforming exterior products

2009 9 3

2.3 Exterior Linear Spaces Composing m-elements Composing elements automatically Spaces and congruence The associativity of the exterior product Transforming exterior products 2.4 Axioms for Exterior Linear Spaces Summary of axioms Grassmann algebras On the nature of scalar multiplication Factoring scalars Grassmann expressions Calculating the grade of a Grassmann expression 2.5 Bases Bases for exterior linear spaces Declaring a basis in GrassmannAlgebra Composing bases of exterior linear spaces Composing palettes of basis elements Standard ordering Indexing basis elements of exterior linear spaces 2.6 Cobases Definition of a cobasis The cobasis of unity Composing palettes of cobasis elements The cobasis of a cobasis 2.7 Determinants Determinants from exterior products Properties of determinants The Laplace expansion technique Calculating determinants


5

2.8 Cofactors Cofactors from exterior products The Laplace expansion in cofactor form Exploring the calculation of determinants using minors and cofactors Transformations of cobases Exploring transformations of cobases 2.9 Solution of Linear Equations Grassmann's approach to solving linear equations Example solution: 3 equations in 4 unknowns Example solution: 4 equations in 4 unknowns 2.10 Simplicity The concept of simplicity All (n|1)-elements are simple Conditions for simplicity of a 2-element in a 4-space Conditions for simplicity of a 2-element in a 5-space 2.11 Exterior division The definition of an exterior quotient Division by a 1-element Division by a k-element Automating the division process 2009 9 3


6

2.11 Exterior division The definition of an exterior quotient Division by a 1-element Division by a k-element Automating the division process 2.12 Multilinear forms The span of a simple element Composing spans Example: Refactorizations Multilinear forms Defining m:k-forms Composing m:k-forms Expanding and simplifying m:k-forms Developing invariant forms Properties of m:k-forms The complete span of a simple element 2.13 Unions and intersections Union and intersection as a multilinear form Where the intersection is evident Where the intersections is not evident Intersection with a non-simple element Factorizing simple elements 2.14 Summary

3 The Regressive Product3.1 Introduction 3.2 Duality The notion of duality Examples: Obtaining the dual of an axiom Summary: The duality transformation algorithm 3.3 Properties of the Regressive Product Axioms for the regressive product The unit n-element The inverse of an n-element Grassmann's notation for the regressive product 3.4 The Duality Principle The dual of a dual The Grassmann Duality Principle Using the GrassmannAlgebra function Dual 3.5 The Common Factor Axiom Motivation The Common Factor Axiom Extension of the Common Factor Axiom to general elements Special cases of the Common Factor Axiom Dual versions of the Common Factor Axiom Application of the Common Factor Axiom When the common factor is not simple 2009 9 3

3.5 The Common Factor Axiom Motivation The Common Factor Axiom Extension of the Common Factor Axiom to general elements Special cases of the Common Factor Axiom Dual versions of the Common Factor Axiom Application of the Common Factor Axiom When the common factor is not simple 3.6 The Common Factor Theorem Development of the Common Factor Theorem Proof of the Common Factor Theorem The A and B forms of the Common Factor Theorem Example: The decomposition of a 1-element Example: Applying the Common Factor Theorem Automating the application of the Common Factor Theorem 3.7 The Regressive Product of Simple Elements The regressive product of simple elements The regressive product of (n|1)-elements Regressive products leading to scalar results Expressing an element in terms of another basis Exploration: The cobasis form of the Common Factor Axiom Exploration: The regressive product of cobasis elements 3.8 Factorization of Simple Elements Factorization using the regressive product Factorizing elements expressed in terms of basis elements The factorization algorithm Factorization of (n|1)-elements Factorizing simple m-elements Factorizing contingently simple m-elements Determining if an element is simple 3.9 Product Formulas for Regressive Products The Product Formula Deriving Product Formulas Deriving Product Formulas automatically Computing the General Product Formula Comparing the two forms of the Product Formula The invariance of the General Product Formula Alternative forms for the General Product Formula The Decomposition Formula Exploration: Dual forms of the General Product Formulas 3.10 Summary


7

4 Geometric Interpretations4.1 Introduction 4.2 Geometrically Interpreted 1-elements Vectors Points Declaring a basis for a bound vector space Composing vectors and points Example: Calculation of the centre of mass

2009 9 3

4.2 Geometrically Interpreted 1-elements Vectors Points Declaring a basis for a bound vector space Composing vectors and points Example: Calculation of the centre of mass 4.3 Geometrically Interpreted 2-elements Simple geometrically interpreted 2-elements Bivectors Bound vectors Composing bivectors and bound vectors The sum of two parallel bound vectors The sum of two non-parallel bound vectors Sums of bound vectors Example: Reducing a sum of bound vectors 4.4 Geometrically Interpreted m-Elements Types of geometrically interpreted m-elements The m-vector The bound m-vector Bound simple m-vectors expressed by points Bound simple bivectors Composing m-vectors and bound m-vectors 4.5 Geometrically Interpreted Spaces Vector and point spaces Coordinate spaces Geometric dependence Geometric duality 4.6 m-planes m-planes defined by points m-planes defined by m-vectors m-planes as exterior quotients Computing exterior quotients The m-vector of a bound m-vector 4.7 Line Coordinates Lines in a plane Lines in a 3-plane Lines in a 4-plane Lines in an m-plane 4.8 Plane Coordinates Planes in a 3-plane Planes in a 4-plane Planes in an m-plane The coordinates of geometric entities 4.9 Calculation of Intersections The intersection of two lines in a plane The intersection of a line and a plane in a 3-plane The intersection of two planes in a 3-plane Example: The osculating plane to a curve 4.10 Decomposition into Components The shadow Decomposition in a 2-space Decomposition in a 3-space 2009 9 3 Decomposition in a 4-space Decomposition of a point or vector in an n-space


8


9

4.10 Decomposition into Components The shadow Decomposition in a 2-space Decomposition in a 3-space Decomposition in a 4-space Decomposition of a point or vector in an n-space 4.11 Projective Space The intersection of two lines in a plane The line at infinity in a plane Projective 3-space Homogeneous coordinates Duality Desargues' theorem Pappas' theorem Projective n-space 4.12 Regions of Space Regions of space Regions of a plane Regions of a line Planar regions defined by two lines Planar regions defined by three lines Creating a pentagonal region Creating a 5-star region Creating a 5-star pyramid Summary 4.13 Geometric Constructions Geometric expressions Geometric equations for lines and planes The geometric equation of a conic section in the plane The geometric equation as a prescription to construct The algebraic equation of a conic section in the plane An alternative geometric equation of a conic section in the plane Conic sections through five points Dual constructions Constructing conics in space A geometric equation for a cubic in the plane Pascal's Theorem Pascal lines 4.14 Summary

5 The Complement5.1 Introduction 5.2 Axioms for the Complement The grade of a complement The linearity of the complement operation The complement axiom The complement of a complement axiom 2009 9 3 The complement of unity


10

5.2 Axioms for the Complement The grade of a complement The linearity of the complement operation The complement axiom The complement of a complement axiom The complement of unity 5.3 Defining the Complement The complement of an m-element The complement of a basis m-element Defining the complement of a basis 1-element Constraints on the value of ! Choosing the value of ! Defining the complements in matrix form 5.4 The Euclidean Complement Tabulating Euclidean complements of basis elements Formulae for the Euclidean complement of basis elements Products leading to a scalar or n-element 5.5 Complementary Interlude Alternative forms for complements Orthogonality Visualizing the complement axiom The regressive product in terms of complements Glimpses of the inner product 5.6 The Complement of a Complement The complement of a complement axiom The complement of a cobasis element The complement of the complement of a basis 1-element The complement of the complement of a basis m-element The complement of the complement of an m-element Idempotent complements 5.7 Working with Metrics Working with metrics The default metric Declaring a metric Declaring a general metric Calculating induced metrics The metric for a cobasis Creating palettes of induced metrics 5.8 Calculating Complements Entering a complement Creating palettes of complements of basis elements Converting complements of basis elements Simplifying expressions involving complements Converting expressions involving complements to specified forms Converting regressive products of basis elements in a metric space 5.9 Complements in a vector space The Euclidean complement in a vector 2-space The non-Euclidean complement in a vector 2-space 2009 9 3 The Euclidean complement in a vector 3-space The non-Euclidean complement in a vector 3-space


11

5.9 Complements in a vector space The Euclidean complement in a vector 2-space The non-Euclidean complement in a vector 2-space The Euclidean complement in a vector 3-space The non-Euclidean complement in a vector 3-space 5.10 Complements in a bound space Metrics in a bound space The complement of an m-vector Products of vectorial elements in a bound space The complement of an element bound through the origin The complement of the complement of an m-vector Calculating with vector space complements 5.11 Complements of bound elements The Euclidean complement of a point in the plane The Euclidean complement of a point in a point 3-space The complement of a bound element Euclidean complements of bound elements The regressive product of point complements 5.12 Reciprocal Bases Reciprocal bases The complement of a basis element The complement of a cobasis element The complement of a complement of a basis element The exterior product of basis elements The regressive product of basis elements The complement of a simple element is simple 5.13 Summary

6 The Interior Product6.1 Introduction 6.2 Defining the Interior Product Definition of the inner product Definition of the interior product Implications of the regressive product axioms Orthogonality Example: The interior product of a simple bivector and a vector 6.3 Properties of the Interior Product Implications of the Complement Axiom Extended interior products Converting interior products Example: Orthogonalizing a set of 1-elements 6.4 The Interior Common Factor Theorem The Interior Common Factor Formula The Interior Common Factor Theorem Examples of the Interior Common Factor Theorem The computational form of the Interior Common Factor Theorem 2009 9 3


12

6.4 The Interior Common Factor Theorem The Interior Common Factor Formula The Interior Common Factor Theorem Examples of the Interior Common Factor Theorem The computational form of the Interior Common Factor Theorem 6.5 The Inner Product Implications of the Common Factor Axiom The symmetry of the inner product The inner product of complements The inner product of simple elements Calculating inner products Inner products of basis elements 6.6 The Measure of an m-element The definition of measure Unit elements Calculating measures The measure of free elements The measure of bound elements Determining the multivector of a bound multivector 6.7 The Induced Metric Tensor Calculating induced metric tensors Using scalar products to construct induced metric tensors Displaying induced metric tensors as a matrix of matrices 6.8 Product Formulae for Interior Products The basic interior Product Formula Deriving interior Product Formulas Deriving interior Product Formulas automatically Computable forms of interior Product Formulas The invariance of interior Product Formulas An alternative form for the interior Product Formula The interior decomposition formula Interior Product Formulas for 1-elements Interior Product Formulas in terms of double sums 6.9 The Zero Interior Sum Theorem The zero interior sum Composing interior sums The Gram-Schmidt process Proving the Zero Interior Sum Theorem 6.10 The Cross Product Defining a generalized cross product Cross products involving 1-elements Implications of the axioms for the cross product The cross product as a universal product Cross product formulae 6.11 The Triangle Formulae Triangle components The measure of the triangle components Equivalent forms for the triangle components 2009 9 3


13

6.11 The Triangle Formulae Triangle components The measure of the triangle components Equivalent forms for the triangle components 6.12 Angle Defining the angle between elements The angle between a vector and a bivector The angle between two bivectors The volume of a parallelepiped 6.13 Projection To be completed. 6.14 Interior Products of Interpreted Elements To be completed. 6.15 The Closest Approach of Multiplanes To be completed.

7 Exploring Screw Algebra7.1 Introduction 7.2 A Canonical Form for a 2-Entity The canonical form Canonical forms in an n-plane Creating 2-entities 7.3 The Complement of 2-Entity Complements in an n-plane The complement referred to the origin The complement referred to a general point 7.4 The Screw The definition of a screw The unit screw The pitch of a screw The central axis of a screw Orthogonal decomposition of a screw 7.5 The Algebra of Screws To be completed 7.6 Computing with Screws To be completed

2009 9 3


14

8 Exploring Mechanics8.1 Introduction 8.2 Force Representing force Systems of forces Equilibrium Force in a metric 3-plane 8.3 Momentum The velocity of a particle Representing momentum The momentum of a system of particles The momentum of a system of bodies Linear momentum and the mass centre Momentum in a metric 3-plane 8.4 Newton's Law Rate of change of momentum Newton's second law 8.5 The Angular Velocity of a Rigid Body To be completed. 8.6 The Momentum of a Rigid Body To be completed. 8.7 The Velocity of a Rigid Body To be completed. 8.8 The Complementary Velocity of a Rigid Body To be completed. 8.9 The Infinitesimal Displacement of a Rigid Body To be completed. 8.10 Work, Power and Kinetic Energy To be completed.

9 Grassmann Algebra9.1 Introduction 9.1 Grassmann Numbers Creating Grassmann numbers Body and soul Even and odd components The grades of a Grassmann number Working with complex scalars 2009 9 3

9.1 Grassmann Numbers Creating Grassmann numbers Body and soul Even and odd components The grades of a Grassmann number Working with complex scalars 9.3 Operations with Grassmann Numbers The exterior product of Grassmann numbers The regressive product of Grassmann numbers The complement of a Grassmann number The interior product of Grassmann numbers 9.4 Simplifying Grassmann Numbers Elementary simplifying operations Expanding products Factoring scalars Checking for zero terms Reordering factors Simplifying expressions 9.5 Powers of Grassmann Numbers Direct computation of powers Powers of even Grassmann numbers Powers of odd Grassmann numbers Computing positive powers of Grassmann numbers Powers of Grassmann numbers with no body The inverse of a Grassmann number Integer powers of a Grassmann number General powers of a Grassmann number 9.6 Solving Equations Solving for unknown coefficients Solving for an unknown Grassmann number 9.7 Exterior Division Defining exterior quotients Special cases of exterior division The non-uniqueness of exterior division 9.8 Factorization of Grassmann Numbers The non-uniqueness of factorization Example: Factorizing a Grassmann number in 2-space Example: Factorizing a 2-element in 3-space Example: Factorizing a 3-element in 4-space 9.9 Functions of Grassmann Numbers The Taylor series formula The form of a function of a Grassmann number Calculating functions of Grassmann numbers Powers of Grassmann numbers Exponential and logarithmic functions of Grassmann numbers Trigonometric functions of Grassmann numbers Functions of several Grassmann numbers


15

10 Exploring the Generalized Grassmann Product2009 9 3


16

10 Exploring the Generalized Grassmann Product10.1 Introduction 10.2 Geometrically Interpreted 1-elements Definition of the generalized product Case l = 0: Reduction to the exterior product Case 0 < l < Min[m, k]: Reduction to exterior and interior products Case l = Min[m, k]: Reduction to the interior product Case Min[m, k] < l < Max[m, k]: Reduction to zero Case l = Max[m, k]: Reduction to zero Case l > Max[m, k]: Undefined 10.3 The Symmetric Form of the Generalized Product Expansion of the generalized product in terms of both factors The quasi-commutativity of the generalized product Expansion in terms of the other factor 10.4 Calculating with Generalized Products Entering a generalized product Reduction to interior products Reduction to inner products Example: Case Min[m, k] < l < Max[m, k]: Reduction to zero 10.5 The Generalized Product Theorem The A and B forms of a generalized product Example: Verification of the Generalized Product Theorem Verification that the B form may be expanded in terms of either factor 10.6 Products with Common Factors Products with common factors Congruent factors Orthogonal factors Generalized products of basis elements Finding common factors 10.7 The Zero Interior Sum Theorem The Zero Interior Sum theorem Composing a zero interior sum 10.8 The Zero Generalized Sum The zero generalized sum conjecture Generating the zero generalized sum Exploring the conjecture 10.9 Nilpotent Generalized Products Nilpotent products of simple elements Nilpotent products of non-simple elements 10.10 Properties of the Generalized Product Summary of properties

2009 9 3


17

10.10 Properties of the Generalized Product Summary of properties 10.11 The Triple Generalized Sum Conjecture The generalized Grassmann product is not associative The triple generalized sum The triple generalized sum conjecture Exploring the triple generalized sum conjecture An algorithm to test the conjecture 10.12 Exploring Conjectures A conjecture Exploring the conjecture 10.13 The Generalized Product of Intersecting Elements The case l < p The case l p The special case of l = p 10.14 The Generalized Product of Orthogonal Elements The generalized product of totally orthogonal elements The generalized product of partially orthogonal elements 10.15 The Generalized Product of Intersecting Orthogonal Elements The case l < p The case l p 10.16 Generalized Products in Lower Dimensional Spaces Generalized products in 0, 1, and 2-spaces 0-space 1-space 2-space 10.17 Generalized Products in 3-Space To be completed

11 Exploring Hypercomplex Algebra11.1 Introduction 11.2 Some Initial Definitions and Properties The conjugate Distributivity The norm Factorization of scalars Multiplication by scalars The real numbers ! 11.3 The Complex Numbers ! Constraints on the hypercomplex signs Complex numbers as Grassmann numbers under the hypercomplex product 11.4 The Hypercomplex Product in a 2-Space Tabulating products in 2-space The hypercomplex product of two 1-elements The hypercomplex product of a 1-element and a 2-element 2009 9 3 The hypercomplex square of a 2-element The product table in terms of exterior and interior products


18

11.4 The Hypercomplex Product in a 2-Space Tabulating products in 2-space The hypercomplex product of two 1-elements The hypercomplex product of a 1-element and a 2-element The hypercomplex square of a 2-element The product table in terms of exterior and interior products 11.5 The Quaternions " The product table for orthonormal elements Generating the quaternions The norm of a quaternion The Cayley-Dickson algebra 11.6 The Norm of a Grassmann number The norm The norm of a simple m-element The skew-symmetry of products of elements of different grade The norm of a Grassmann number in terms of hypercomplex products The norm of a Grassmann number of simple components The norm of a non-simple element 11.7 Products of two different elements of the same grade The symmetrized sum of two m-elements Symmetrized sums for elements of different grades The body of a symmetrized sum The soul of a symmetrized sum Summary of results of this section 11.8 Octonions To be completed

12 Exploring Clifford Algebra12.1 Introduction 12.2 The Clifford Product Definition of the Clifford product Tabulating Clifford products The grade of a Clifford product Clifford products in terms of generalized products Clifford products in terms of interior products Clifford products in terms of inner products Clifford products in terms of scalar products 12.3 The Reverse of an Exterior Product Defining the reverse Computing the reverse 12.4 Special Cases of Clifford Products The Clifford product with scalars The Clifford product of 1-elements The Clifford product of an m-element and a 1-element The Clifford product of an m-element and a 2-element The Clifford product of two 2-elements 2009 9 3 The Clifford product of two identical elements


19

12.4 Special Cases of Clifford Products The Clifford product with scalars The Clifford product of 1-elements The Clifford product of an m-element and a 1-element The Clifford product of an m-element and a 2-element The Clifford product of two 2-elements The Clifford product of two identical elements 12.5 Alternate Forms for the Clifford Product Alternate expansions of the Clifford product The Clifford product expressed by decomposition of the first factor Alternative expression by decomposition of the first factor The Clifford product expressed by decomposition of the second factor The Clifford product expressed by decomposition of both factors 12.6 Writing Down a General Clifford Product The form of a Clifford product expansion A mnemonic way to write down a general Clifford product 12.7 The Clifford Product of Intersecting Elements General formulae for intersecting elements Special cases of intersecting elements 12.8 The Clifford Product of Orthogonal Elements The Clifford product of totally orthogonal elements The Clifford product of partially orthogonal elements Testing the formulae 12.9 The Clifford Product of Intersecting Orthogonal Elements Orthogonal union Orthogonal intersection 12.10 Summary of Special Cases of Clifford Products Arbitrary elements Arbitrary and orthogonal elements Orthogonal elements Calculating with Clifford products 12.11 Associativity of the Clifford Product Associativity of orthogonal elements A mnemonic formula for products of orthogonal elements Associativity of non-orthogonal elements Testing the general associativity of the Clifford product 12.13 Clifford Algebra Generating Clifford algebras Real algebra Clifford algebras of a 1-space 12.14 Clifford Algebras of a 2-Space The Clifford product table in 2-space Product tables in a 2-space with an orthogonal basis Case 1: 8e1 e1 + 1, e2 e2 + 1< Case 2: 8e1 e1 + 1, e2 e2 - 1< Case 3: 8e1 e1 - 1, e2 e2 - 10 0 0

2.15

2009 9 3


80

10:

The exterior product of elements of odd grade is anti-commutative. a b H- 1L m k b am k k m

2.16

11:

Additive identities act as multiplicative zero elements under the exterior product. :!"#$%$B0, 0 LF, 0 a k k k k m

0>k+ m

2.17

12:

The exterior product is both left and right distributive under addition.

Ka + b O g a g + b gm m k m k m k

2.18

a b + gm k k

ab + agm k m k

13:

Scalar multiplication commutes with the exterior product.

a a b Ka aO b a a bm k m k m k

2.19

A convention for the zerosIt can be seen from the above axioms that each of the linear spaces has its own zero. Apart from grade, these zeros all have the same properties, so that for simplicity in computations we will denote them all by the one symbol 0. However, this also means that we cannot determine the grade of this symbol 0 from the symbol alone. In practice, we overcome this problem by defining the grade of 0 to be the (undefined) symbol 0.

Grassmann algebrasUnder the foregoing axioms it may be directly shown that: 1. L is a field.0

2. The L are linear spaces over L.m 0

3. The direct sum of the L is an algebra. m 2009 9 3


81

1. L is a field.0

2. The L are linear spaces over L.m 0

3. The direct sum of the L is an algebra.m

This algebra is called a Grassmann algebra. Its elements are sums of elements from the L, thusm

allowing closure over both addition and exterior multiplication. For example, we can expand and simplify the product of two elements of the algebra to give another element.%@H1 + 2 x + 3 x y + 4 x y zL H1 + 2 x + 3 x y + 4 x y zLD 1 + 4 x + 6 xy + 8 xyz

A Grassmann algebra is also a linear space of dimension 2n , where n is the dimension of the underlying linear space L. We will sometimes refer to the Grassmann algebra whose underlying1

linear space is of dimension n as the Grassmann algebra of n-space. Grassmann algebras will be discussed further in Chapter 9: Exploring Grassmann Algebra.

On the nature of scalar multiplicationThe anti-commutativity axiomm k k

10 for general elements states that:m

a b H- 1Lm k b a

If one of these factors is a scalar (b = a, say; k = 0), the axiom reduces to:k

aa aam m

Since by axiom 6, each of these terms is an m-element, we may permit the exterior product to subsume the normal field multiplication. Thus, if a is a scalar, a a is equivalent to a a. The latterm m

(conventional) notation will usually be adopted. In the usual definitions of linear spaces no discussion is given to the nature of the product of a scalar and an element of the space. A notation is usually adopted (that is, the omission of the product sign) that leads one to suppose this product to be of the same nature as that between two scalars. From the axioms above it may be seen that both the product of two scalars and the product of a scalar and an element of the linear space may be interpreted as exterior products.

aa aa a am m m

aL0

2.20

2009 9 3


82

Factoring scalarsIn GrassmannAlgebra scalars can be factored out of a product by GrassmannSimplify (alias $) so that they are collected at the beginning. For example:$@5 H2 xL H3 yL Hb zLD 30 b x y z

If any of the factors in the product is scalar, it will also be collected at the beginning. Here a is a scalar since it has been declared as one (by default).$@5 H2 xL H3 aL Hb zLD 30 a b x z GrassmannSimplify works with any Grassmann expression, or lists (to any level) of Grassmann expressions: $BK K 1 + ab ax OF MatrixForm zb 1 - zx

1+ab ax O bz 1 + xz

Grassmann expressionsA Grassmann expression is any well-formed expression recognized by GrassmannAlgebra as a valid element of a Grassmann algebra. To check whether an expression is considered to be a Grassmann expression, you can use the function GrassmannExpressionQ.GrassmannExpressionQ@1 + x yD True GrassmannExpressionQ also works on lists of expressions. Below we determine that whereas the product of a scalar a and a 1-element x is a valid element of the Grassmann algebra, the ordinary multiplication of two 1-elements x y is not. GrassmannExpressionQ@8x y, a x, x yn n n m' n m'

Drop the primes.:#)*+,+B1, 1 LF, a 1 a>n n n m n m

In words, this says that there is a unit n-element which acts as an identity under the regressive product.

The dual of axiom 10Axiom

10 says:

The exterior product of elements of odd grade is anti-commutative.2009 9 3


138

The exterior product of elements of odd grade is anti-commutative.a b H- 1Lm k b am k k m

Replace with , and the grades of elements by their complementary grades. Note that it is only the grades as shown in the underscripts which should be replaced, not those figuring elsewhere in the formula.n-m

a bn-k

H- 1Lm k b an-k

n-m

Replace arbitrary grades m (wherever they occur in the formula) with n|m', k with n|k'. An arbitrary grade is one without a specific value, like 0, 1, or n.a b H- 1LIn-m'M In- k'M b am' k' k' m'

Drop the primes.a b H - 1 L Hn-m L Hn- kL b am k k m

In words this says that the regressive product of elements of odd complementary grade is anticommutative.

Summary: The duality transformation algorithmThe algorithm for the duality transformation may be summarized as follows: 1. Replace with , and the grades of elements and spaces by their complementary grades. 2. Replace arbitrary grades m with n|m', k with n|k'. Drop the primes. An arbitrary grade is one without a specific value, like 0, 1, or n.

3.3 Properties of the Regressive ProductAxioms for the regressive productIn this section we collect the results of applying the duality algorithm above to the exterior product axioms

6 to 12. Axioms 1 to 5 transform unchanged since there are no products involved.

We use the symbol n for the dimension of the underlying linear space wherever we want to ensure the formulas can be understood by GrassmannAlgebra functions (for example, the Dual function to be discussed later). However in general textual comment we will usually retain the simpler symbol n to represent this dimension.

6:

The regressive product of an m-element and a k-element is an (m+k|n)element.

2009 9 3


139

6:

The regressive product of an m-element and a k-element is an (m+k|n)element. :a L, b L> :a b m m k k m k

k+ m -n

L >

3.2

7:

The regressive product is associative.

abm k

g a bgj m k j

3.3

8:

There is a unit n-element which acts as an identity under the regressive product. :!"#$%$B 1 , 1 L F, a 1 a>n n n m n m

3.4

9:

Non-zero n-elements have inverses with respect to the regressive product.

1 1 1 :!"#$%$B , L F, 1 a , &'()**Ba, a ! 0 L F> n n n a a n a

3.5

10:

The regressive product of elements of odd complementary grade is anticommutative. a b H - 1 L Hn-kL Hn- m L b am k k m

3.6

11:

Additive identities act as multiplicative zero elements under the regressive product. :!"#$%$B0, 0 LF, 0 a k k k k m

0 >k+ m -n

3.7

12:

The regressive product is both left and right distributive under addition.

2009 9 3


140

12:

The regressive product is both left and right distributive under addition.

Ka + b O g a g + b gm m k m k m k

3.8

a b + gm k k

ab + agm k m k

13:

Scalar multiplication commutes with the regressive product.

a a b Ka aO b a a bm k m k m k

3.9

The unit n-element The unit n-element is congruent to any basis n-element.The duality algorithm has generated a unit n-element 1 which acts as the multiplicative identity forn

the regressive product, just as the unit scalar 1 (or 1) acts as the multiplicative identity for the0

exterior product (axioms

8 and 8).n 1 n

We have already seen that any basis of L contains only one element. If a basis of L is8e1 ,e2 ,!,en n n n a a n a

Suppose we have an n-element a expressed in terms of some basis. Then, according to 3.10 we cann

express it as a scalar multiple (a, say) of the unit n-element 1.n

We may then write the inverse a-1 of a with respect to the regressive product as the scalar multiplen n

of 1. n 2009 9 3

1 a


142

We may then write the inverse a-1 of a with respect to the regressive product as the scalar multiplen n 1 a

of 1.n

aa1n n

a -1 n

1 a

1n 1 1: a n

We can see this by taking the regressive product of a 1 withn

a a -1 J a 1 N n n n

1 a

1 11 1n n n n

If a is now expressed in terms of a basis we have:n

a b e1 e2 ! en b c 1n n

Hence a-1 can be written as:n

a -1 n

1 1 b c

1n

1

1

b c2n

e1 e2 ! en

Summarizing these results: If a-1 is defined by a a-1 1, thenn n n

aa1n n

a-1 n

1 a

1n

3.14

a b e1 e2 ! enn

a -1 n

1

1

b c2

e1 e2 ! en

3.15

Grassmann's notation for the regressive productIn Grassmann's Ausdehnungslehre of 1862 Section 95 (translated by Lloyd C. Kannenberg) he states:

If q and r are the orders of two magnitudes A and B, and n that of the principal domain, then the order of the product [A B] is first equal to q+r if q+r is smaller than n, and second equal to q+r|n if q+r is greater than or equal to n.Translating this into the terminology used in this book we have:

If q and r are the grades of two elements A and B, and n that of the underlying linear space, then the grade of the product [A B] is first equal to q+r if q+r is smaller than n, and second equal to q+r|n if q+r is greater than or equal to n.

2009 9 3


143

Translating this further into the notation used in this book we have:@ A BD A Bp p q q

p+q 0>m k

The dual theorem states that the regressive product of two elements is zero if the sum of their grades is less than the dimension of the linear space.:a b 0, n - Hk + mL > 0>m k

We can recover the original theorem as the dual of this one.

Example 2This theorem below says that the exterior product of an element with itself is zero if it is of odd grade.:a a 0, m 8OddPositiveIntegersm m

The dual theorem states that the regressive product of an element with itself is zero if its complementary grade is odd.:a a 0, Hn - mL 8OddPositiveIntegersm m

Again, we can recover the original theorem by calculating the dual of this one.

Example 3The following theorem states that the exterior product of unity with itself any number of times remains unity.

2009 9 3


145

11!1a am m

The dual theorem states the corresponding fact for the regressive product of unit n-elements 1 .n

n

1 1 ! 1 a an n m m

Using the GrassmannAlgebra function DualThe algorithm of the Duality Principle has been encapsulated in the function Dual in GrassmannAlgebra. Dual takes an expression or list of expressions which comprise an axiom or theorem and generates the list of dual expressions by transforming them according to the replacement rules of the Duality Principle.

Example 1: The Dual of axiom 10Our first examples are to show how Dual may be used to develop dual axioms. For example, to take the dual of axiom

10 simply enter:k m

DualBa b H- 1Lm k b aFm k

a b H - 1 L H-k+nL H-m +nL b am k k m

Example 2: The Dual of axiom 8Again, to take the Dual of axiom

8 enter:m m

DualB:#)*+,+B1, 1 LF, a 1 a>F0

:!"#$%$B 1 , 1 L F, a 1 a>n n n m n m

Example 3: The Dual of axiom 9To apply Dual to an axiom involving more than one statement, collect the statements in a list. For example, the dual of axiom

9 is obtained as:0 0 0

DualB:#)*+,+Ba-1 , a-1 LF, 1 a a-1 , (-./00Ba, a ! 0 LF>F 1 1 1 :!"#$%$B , L F, 1 a , &'()**Ba, a # 0 L F> n n n a a n a

Note here that we are purposely frustrating Mathematica's inbuilt definition of Exists and ForAll by writing them in script.

2009 9 3


146

Note here that we are purposely frustrating Mathematica's inbuilt definition of Exists and ForAll by writing them in script.

Example 4: The Dual of the Common Factor AxiomOur fourth example is to generate the dual of the Common Factor Axiom. This axiom will be introduced in the next section. Note that the dimension of the space must be entered as n in order for Dual to recognize it as such.CommonFactorAxiom = : a g b g a b g g , m + k + j - n 0>;m j k j m k j j

Dual@CommonFactorAxiomD : a g b g a b g g, - j - k - m + 2 n 0>m j k j m k j j

We can verify that the dual of the dual of the axiom is the axiom itself.CommonFactorAxiom Dual@Dual@CommonFactorAxiomDD True

Example 5: The Dual of a formulaOur fifth example is to obtain the dual of one of the product formulae derived in a later section of this chapter. Again, the dimension of the space must be entered as n.F = a b x Ka x O b + H- 1Lm a b xm k n-1 m n-1 k m k

n-1

;

Dual@FD a b x H - 1 L -m +n a b x + a x bm k m k m k

Again, we can verify that the dual of the dual of the formula is the formula itself.F Dual@Dual@FDD True

Setting up your own input to the Dual function You can use the GrassmannAlgebra function Dual with any expression involving exterior and regressive products of graded, vector or scalar symbols. You can include multiple statements by combining them in a list. The statements should not include any expressions which will evaluate when entered. The dimension of the underlying linear space should be denoted as n. Avoid using scalar or vector symbols as (underscripted) grades if they appear other than as underscripts elsewhere in the expression.

3.5 The Common Factor Axiom2009 9 3


147

3.5 The Common Factor AxiomMotivationAlthough the axiom sets for the exterior and regressive products have been posited, it still remains to propose an axiom explicitly relating the two types of products. The axiom set for the regressive product, which we have created above simply as a dual axiom set to that of the exterior product, asserts that in an n-space the regressive product of an m-element and a k-element is an (m+k-n)-element. Our first criterion for the new axiom then, is that it can be used to compute this new element. Earlier we have also remarked on some enticing correspondences between the dual exterior and regressive products and the dual union and intersection operations of a Boolean algebra. We can already see that the (non-zero) exterior product of simple elements is an element whose space is the 'union' of the spaces of its factors. Our second criterion then, is that the new axiom enable the 'dual' of this property: the (non-zero) regressive product of simple elements is an element whose space is the 'intersection' of the spaces of its factors. This criterion leads us first to return to Chapter 2: Unions and Intersections, where, with an arbitrary multilinear product operation (which we denoted ), we were able to compute the union and intersection of simple elements. In the cases explored the simple elements were of arbitrary grade. We saw there that given simple elements A* C and C B* , we could obtain their union and intersection as (congruent to) A* C B* and C respectively.A B H A * CL HC B* L H A * C B* L C

Since is an arbitrary multilinear product (other than the exterior product), we can satisfy the second criterion by positing the same form for the regressive product.

A B H A * CL HC B* L H A * C B* L C

3.16

What additional requirements on this would one need in order to satisfy the first criterion, that is, that the axiom enables a result which does not involve the regressive product? Axiom gives us the clue.:#)*+,+B 1 , 1 L F, a 1 a>n n n m n m

8 above

If the grade of A* C B* is equal to the dimension of the space n (or in computations, n), then since all n-elements are congruent we can write A* C B* c 1 , where c is some scalar,n

showing that the right hand side is congruent to the common factor (intersection) C.H A * CL HC B* L H A * C B* L C Jc 1 N C c C Cn

2009 9 3


148

An axiom which satisfies both of our original criteria is then8HA* CL HC B* L HA* C B* L C, Grade@A* C B* D nm j j k m j k j

: a g b g a b g g , m + k + j - n 0>m j k j m k j j

As we will see, an axiom of this form works very well. Indeed, it turns out to be one of the fundamental underpinnings of the algebra. We call it the Common Factor Axiom and explore it further below.

Historical NoteThe approach we have adopted in this chapter of treating the common factor relation as an axiom is effectively the same as Grassmann used in his first Ausdehnungslehre (1844) but differs from the approach that he used in his second Ausdehnungslehre (1862). (See Chapter 3, Section 5 in Kannenberg.) In the 1862 version Grassmann proves this relation from another which is (almost) the same as the Complement Axiom that we introduce in Chapter 5: The Complement. Whitehead [1898], and other writers in the Grassmannian tradition follow his 1862 approach. The relation which Grassmann used in the 1862 Ausdehnungslehre is in effect equivalent to assuming the space has a Euclidean metric (his Ergnzung or supplement). However the Common Factor Axiom does not depend on the space having a metric; that is, it is completely independent of any correspondence we set up between L and L . Hence wem n-m

would rather not adopt an approach which introduces an unnecessary constraint, especially since we want to show later that the Ausdehnungslehre is easily extended to more general metrics than the Euclidean.

The Common Factor AxiomWe begin completely afresh in positing the axiom. Let a, b, and g be simple elements with m+k+jm k j

= n, where n is the dimension of the space. Then the Common Factor Axiom states that:

: a g b g a b g g , m + k + j - n 0>m j k j m k j j

3.17

Thus, the regressive product of two elements a g and b g with a common factor g is equal tom j m k k j j j j

the regressive product of the 'union' of the elements a b g with the common factor g (their 'intersection'). If a and b still contain some simple elements in common, then the product a b g is zero, hence,m k m k j

2009 9definition 3 by the above a g b g is also zero. In what follows, we suppose that a b gm j k j m k j

is not zero.


149

If a and b still contain some simple elements in common, then the product a b g is zero, hence,m k m k j

by the definition above a g b g is also zero. In what follows, we suppose that a b gm j k j m k j

is not zero. Since the union a b g is an n-element, we can write it as some scalar factor c, say, of the unit nm k j

element: a b g c 1 . Hence by axiomm k j m j n k j

8 we derive immediately that the regressive productj

of two elements a g and b g with a common factor g is congruent to that factor.

: a g b g g , m + k + j - n 0>m j k j j

3.18

It is easy to see that by using the anti-commutativity axiom 10, that the axiom may be arranged in any of a number of alternative forms, the most useful of which are:

: a g g b a g b g , m + k + j - n 0>m j j k m j k j

3.19

: g a g b g a b g , m + k + j - n 0>j m j k j m k j

3.20

And since for the regressive product, any n-element commutes with any other element, we can always rewrite the right hand side with the common factor first.

: a g g b g a g b , m + k + j - n 0>m j j k j m j k

3.21

Extension of the Common Factor Axiom to general elementsThe axiom has been stated for simple elements. In this section we show that it remains valid for general (possibly non-simple) elements, provided that the common factor remains simple. Consider two simple elements a1 and a2 . Then the Common Factor Axiom can be written for eachm m

as:

2009 9 3


150

a1 g b g a1 b g gm j k j m k j j

a2 g b g a2 b g gm j k j m k j j

Adding these two equations and using the distributivity of and gives:Ka1 + a2 O g b g Ka1 + a2 O b g gm m j k j m m k j j

Extending this process, we see that the formula remains true for arbitrary a and b, providing g ism k j

simple.

: a g b g a b g g g , m + k + j - n 0>m j k j m k j j j

3.22

This is an extended version of the Common Factor Axiom. It states that: the regressive product of two arbitrary elements containing a simple common factor is congruent to that factor. For applications involving computations in a non-metric space, particularly those with a geometric interpretation, we will see that the congruence form is not restrictive. Indeed, it will be quite elucidating. For more general applications in metric spaces we will see that the associated scalar factor is no longer arbitrary but is determined by the metric imposed.

Special cases of the Common Factor AxiomIn this section we list some special cases of the Common Factor Axiom. We assume, without explicitly stating, that the common factor is simple. When the grades do not conform to the requirement shown, the product is zero.

: a g b g 0, m + k + j - n ! 0>m j k j

3.23

If there is no common factor (other than the scalar 1), then the axiom reduces to:

:a b a b 1 L , m + k - n 0>m k m k 0

3.24

The following version yields a sort of associativity for products which are scalar.

2009 9 3


151

: ab g a bgm k j m k j

L , m + k + j - n 0>0

3.25

To prove this, we can use the previous formula to show that each side the equation is equal toa b g 1.m k j

Dual versions of the Common Factor AxiomAs discussed earlier in the chapter, dual versions of formulae can be obtained automatically by using the GrassmannAlgebra function Dual. The dual Common Factor Axiom is:

: a g b g a b g g L, m + k + j - 2 n 0>m j k j m k j j j

3.26

The duals of the three special cases in the previous section are:

: a g b g 0, m + k + j - 2 n ! 0>m j k j

3.27

:a b a b 1 L , m + k - n 0>m k m k n n

3.28

: ab g a bgm k j m k j

L , m + k + j - 2 n 0>n

3.29

Application of the Common Factor AxiomWe now work through an example to illustrate how the Common Factor Axiom might be applied. In most cases however, results will be obtained more effectively using the Common Factor Theorem. This is discussed in the section to follow. Suppose we have two general 2-elements X and Y in a 3-space and we wish to find a formula for their 1-element intersection Z. Because X and Y are in a 3-space, we are assured that they are simple.

2009 9 3


152

X = x1 e1 e2 + x2 e1 e3 + x3 e2 e3 Y = y1 e1 e2 + y2 e1 e3 + y3 e2 e3

We calculate Z as the regressive product of X and Y:Z XY Hx1 e1 e2 + x2 e1 e3 + x3 e2 e3 L Hy1 e1 e2 + y2 e1 e3 + y3 e2 e3 L

Expanding this product, and remembering that the regressive product of identical basis 2-elements is zero, we obtain:Z Hx1 e1 e2 L Hy2 e1 e3 L + Hx1 e1 e2 L Hy3 e2 e3 L + Hx2 e1 e3 L Hy1 e1 e2 L + Hx2 e1 e3 L Hy3 e2 e3 L + Hx3 e2 e3 L Hy1 e1 e2 L + Hx3 e2 e3 L Hy2 e1 e3 L

In a 3 space, regressive products of 2-elements are anti-commutative since H- 1LHn-mL Hn-kL = H- 1LH3-2L H3-2L = - 1. Hence we can collect pairs of terms with the same factors by making the corresponding sign change:Z Hx1 y2 - x2 y1 L He1 e2 L He1 e3 L + Hx1 y3 - x3 y1 L He1 e2 L He2 e3 L + Hx2 y3 - x3 y2 L He1 e3 L He2 e3 L

We can now apply the Common Factor Axiom to each of these regressive products:Z Hx1 y2 - x2 y1 L He1 e2 e3 L e1 + Hx1 y3 - x3 y1 L He1 e2 e3 L e2 + Hx2 y3 - x3 y2 L He1 e2 e3 L e3

Finally, by putting e1 e2 e3 c 1 , we have Z expressed as a 1-element:n

Z = c HHx1 y2 - x2 y1 L e1 + Hx1 y3 - x3 y1 L e2 + Hx2 y3 - x3 y2 L e3 L

Thus, in sum, we have the general congruence relation for the intersection of two 2-elements in a 3space.

Hx1 e1 e2 + x2 e1 e3 + x3 e2 e3 L Hy1 e1 e2 + y2 e1 e3 + y3 e2 e3 L Hx1 y2 - x2 y1 L e1 + Hx1 y3 - x3 y1 L e2 + Hx2 y3 - x3 y2 L e3

3.30

The result on the right hand side almost looks as if it could have been obtained from a crossproduct operation. However the cross-product requires the space to have a metric, while this does not. We will see in Chapter 6 how, once we have introduced a metric, we can transform this into the formula for the cross product. This is an example of how the regressive product can generate results, independent of whether the space has a metric; whereas the usual vector algebra, in using the cross-product, must assume that it does.

2009 9 3


153

Check by calculating the exterior productsWe can check that Z is indeed a common element to X and Y by determining if the exterior product of Z with each of X and Y is zero. We compose the expressions for X, Y, and Z, and then use % (the palette alias of GrassmannExpandAndSimplify) to do the check. GrassmannAlgebra already knows that the special congruence symbol c is scalar, and ComposeBivector will automatically declare the scalar coefficients as ScalarSymbols.!3 ; X = ComposeBivector@xD; Y = ComposeBivector@yD; Z = c HHx1 y2 - x2 y1 L e1 + Hx1 y3 - x3 y1 L e2 + Hx2 y3 - x3 y2 L e3 L; Expand@%@8Z X, Z Y

Application of the formula gives the unsimplified sum.F = SAH#1 @Xf D YL #1 @Xf DE e2 + e3 x2 x1 Hy1 e1 e2 + y2 e1 e3 + y3 e2 e3 L - x1 e1 e3 x3 x1 e3 x3 x1 e3 x2 x1 +

x1 e1 -

Hy1 e1 e2 + y2 e1 e3 + y3 e2 e3 L e2 +

Expansion and simplification givesF = % @FD H- x2 y1 + x1 y2 L e1 e1 e2 e3 + H- x3 y1 + x1 y3 L e2 e1 e2 e3 + H- x3 y2 + x2 y3 L e3 e1 e2 e3

The n-element in this case is e1 e2 e3 . Factoring it out givesF = HH- x2 y1 + x1 y2 L e1 + H- x3 y1 + x1 y3 L e2 + H- x3 y2 + x2 y3 L e3 L e1 e2 e3

Writing this in congruence form gives us the common factor required, and corroborates the result originally obtained by applying the Common Factor Axiom to each term of the full expansion.Z HH- x2 y1 + x1 y2 L e1 + H- x3 y1 + x1 y3 L e2 + H- x3 y2 + x2 y3 L e3 L c He1 H- x2 y1 + x1 y2 L + e2 H- x3 y1 + x1 y3 L + e3 H- x3 y2 + x2 y3 LL

Proof of the Common Factor Theorem2009 9 3


158

Proof of the Common Factor TheoremConsider a regressive product A B where A is given as a simple product of 1-element factors andm k m

m+k = n+j, j > 0. Then A B is either zero, or, by the Common Factor Axiom, has a common factorm k

C. We assume it is not zero. We then express A and B in terms of C as we have done previously.j m k j

A A* Cm m -j j

B C B*k j

k-j

Let the (m-j)-span and (m-j)-cospan of A be written asm

#m-j BAF : A1 , A2 , !, An >m m -j m -j m -j

#m-j BAF :A1 , A2 , !, An >m j j j

then A can be written in any of the formsm

A A1 A1 A2 A2 ! An Anm m -j j m -j j m -j j

Here n is equal to

m j

, or equivalently

m m- j

.

Since the common factor C is a j-element belonging to A it can be expressed as a linearj m

combination of the cospan elements of A.m

C a1 A1 + a2 A2 + + an Anj j j j

The exterior product of the ith (m-j)-span element Ai with C can be expanded to givem -j j

Ai C m -j j a1

Ai a1 A1 + a2 A2 + + an An m -j j j j

Ai A1 m -j j

+ a2

Ai A2 m -j j

+ + ai

Ai Ai m -j j

+ + an

Ai An m -j jm -j

And indeed, because by definition, the exterior product of an (m-j)-span element (say, A2 ) and any of the non-corresponding cospan elements (say, A3 ) is zero, we can write for any ij

Ai C m -j j

ai

Ai Ai m -j j

ai Am

Now return to the Common Factor Axiom in the form

2009 9 3


159

Z AB m k

m -j

A * C C B*j j

k-j

A * C B* C m -j j k -j j

A B* Cm k-j j

Substituting for the common factor in the right hand side givesZ A B* C m k-j j

A B* a1 A1 + a2 A2 + + an Anm k-j j j j

By the distributivity of the exterior and regressive products, and their behaviour with scalars, the scalar factors can be transferred and attached to A.m

Z

Ja1 A N B* A1 + Ja2 A N B* A2 + + Jan A N B* Anm k-j j m k-j j m k-j j

But we have shown earlier thatai A Ai Cm m -j j

Hence substituting givesZ A1 C B* A1 + A2 C B* A2 + + An C B* Anj k-j j m -j j k-j j m -j j k-j j

m -j

Finally, a further substitution of B for C B* gives the final result.k j k-j

Z

A1 B k m -j

A1 + A2 B A2 + + An B An j m -j k j m -j k j n

3.35m -j k j

Ai B Aii=1

The A and B forms of the Common Factor TheoremWe have just proved what we will now call the A form of the Common Factor Theorem (because it works by spanning the first factor in the regressive product). An analogous formula may be obtained mutatis mutandis by factoring B rather than A. Hence B must now be simple. This formulak m k

we will call the B form of the Common Factor Theorem. In the case where both A and B are simple but not of the same grade, the form which decomposes the element of lower grade will generate the least number of terms and be more computationally efficient. Multiple regressive products may be treated by successive applications of the theorem in either of its forms.

2009 9 3


160

The A form of the Common Factor Theoremn

: A B Ai B Ai ,m k i=1 m -j k j

m - j + k - n 0, m n K O> j 3.36

A A1 A1 A2 A2 ! An An ,m m -j j m -j j m -j j

As we have already seen, this is equivalent to the computationally oriented form.n

A B #m-j BAFm k i=1

m PiT

B #m - j B A Fk

m PiT

#m-j BAF : A1 , A2 , !, An >m m -j m -j m -j

#m-j BAF :A1 , A2 , !, An >m j j j

Or, using Mathematica's inbuilt ListabilityA B SBJ#m-j BAF BN #m-j BAFFm k m k m

The B form of the Common Factor Theoremn

: A B A Bi Bi ,m k

i=1

m

m - j + k - n 0, n k > j 3.37

k-j

j

B B1 B1 B2 B2 ! Bn Bn ,k j k-j j k-j j k-j

The computationally oriented form isn

A B A #j BBFm k i=1 m

k PiT

#j BBF

k PiT

#j BBF :B1 , B2 , !, Bn >k j j j

#j BBF : B1 , B2 , !, Bn >k k-j k-j k-j

Or, using Mathematica's inbuilt ListabilityA B SBJA #j BBFN #j BBFFm k m k k

2009 9 3


161

Example: The decomposition of a 1-elementThe special case of the regressive product of a 1-element b with an n-element a enables us ton

decompose b directly in terms of the factors of a. The Common Factor Theorem gives:n n

a b ai b ain i=1 n-1

where:a a 1 a 2 ! a n H - 1 L n-i H a 1 a 2 ! i ! a n L a in

The symbol i means that the ith factor is missing from the product. Substituting in the Common Factor Theorem gives:n

a b H- 1Ln-i HHa1 a2 ! i ! an L bL ain i=1 n

H a 1 a 2 ! a i-1 b a i+1 ! a n L a ii=1

Hence the decomposition formula becomes:

Ha1 a2 ! an L b n

H a 1 a 2 ! a i-1 b a i+1 ! a n L a ii=1

3.38

Writing this out in full shows that we can expand the expression simply by interchanging b successively with each of the factors of a1 a2 !an , and summing the results.

Ha1 a2 ! an L b Hb a2 ! an L a1 + Ha1 b ! an L a2 + ! + Ha1 a2 ! bL anWe can make the result express more explicitly the decomposition of b in terms of the ai by 'dividing through' by a. (The quotient of two n-elements has been defined previously as then

3.39

quotient of their scalar coefficients.)n

bi=1

a 1 a 2 ! a i-1 b a i+1 ! a n a1 a2 ! an

ai

3.40

This expression is, of course, equivalent to that which would have been obtained from Grassmann's approach to solving linear equations (see Chapter 2: The Exterior Product). 2009 9 3


162

This expression is, of course, equivalent to that which would have been obtained from Grassmann's approach to solving linear equations (see Chapter 2: The Exterior Product).

Example: Applying the Common Factor TheoremIn this section we show how the Common Factor Theorem generally leads to a more efficient computation of results than repeated application of the Common Factor Axiom, particularly when done manually, since there are fewer terms in the Common Factor Theorem expansion, and many are evidently zero by inspection. Again, we take the problem of finding the 1-element common to two 2-elements in a 3-space. This time, however, we take a numerical example and suppose that we know the factors of at least one of the 2-elements. Let:x1 3 e1 e2 + 2 e1 e3 + 3 e2 e3 x2 H5 e2 + 7 e3 L e1

We keep x1 fixed initially and apply the Common Factor Theorem to rewrite the regressive product as a sum of the two products, each due to one of the essentially different rearrangements of x2 :x1 x2 Hx1 e1 L H5 e2 + 7 e3 L - Hx1 H5 e2 + 7 e3 LL e1

The next step is to expand out the exterior products with x1 . Often this step can be done by inspection, since all products with a repeated basis element factor will be zero.x1 x2 H3 e2 e3 e1 L H5 e2 + 7 e3 L H10 e1 e3 e2 + 21 e1 e2 e3 L e1

Factorizing out the 3-element e1 e2 e3 then gives:x1 x2 He1 e2 e3 L H- 11 e1 + 15 e2 + 21 e3 L c H- 11 e1 + 15 e2 + 21 e3 L - 11 e1 + 15 e2 + 21 e3

Thus all factors common to both x1 and x2 are congruent to - 11 e1 + 15 e2 + 21 e3 .

Check by calculating the exterior productsWe can check that this result is correct by taking its exterior product with each of x1 and x2 . We should get zero in both cases, indicating that the factor determined is indeed common to both original 2-elements. Here we use GrassmannExpandAndSimplify in its alias form %.!3 ; %@H3 e1 e2 + 2 e1 e3 + 3 e2 e3 L H- 11 e1 + 15 e2 + 21 e3 LD 0 %@H5 e2 + 7 e3 L e1 H- 11 e1 + 15 e2 + 21 e3 LD 0

Automating the application of the Common Factor Theorem2009 9 3


163

Automating the application of the Common Factor TheoremThe Common Factor Theorem can be computed in GrassmannAlgebra by using either the A or B form in its computable form using the appropriate span and cospan. However GrassmannAlgebra also has a more direct function ToCommonFactor which will attempt to handle multiple regressive products in an optimum way.

Using ToCommonFactorToCommonFactor reduces any regressive products in a Grassmann expression to their common factor form.

For example in the case explored above we have:x1 = 3 e1 e2 + 2 e1 e3 + 3 e2 e3 ; x2 = H5 e2 + 7 e3 L e1 ; A; ToCommonFactor@x1 x2 D c H- 11 e1 + 15 e2 + 21 e3 L

The scalar c is called the congruence factor. The congruence factor has already been defined in Section 3.3 above as the connection between the current basis n-element and the unit n-element 1n

(in GrassmannAlgebra input we use n instead of n, in case n is being used elsewhere in your computations).e1 e2 ! en c 1n

Since this connection cannot be defined unless a metric is introduced, the congruence factor remains arbitrary in non-metric spaces. Although such an arbitrariness may appear at first sight to be disadvantageous, it is on the contrary, highly elucidating. In application to the computing of unions and intersections of spaces, perhaps under a geometric interpretation where they represent lines, planes and hyperplanes, the notion of congruence becomes central, since spaces can only be determined up to congruence. In the example above, x1 and x2 were expressed in terms of basis elements. ToCommonFactor can also apply the Common Factor Theorem to more general elements. For example, if x1 and x2 were expressed as symbolic bivectors, the common factor expansion could still be performed.x1 = x y; x2 = u v; ToCommonFactor@x1 x2 D c Hy u v x - x u v yL

The expression uvx represents the coefficient of uvx when uvx is expressed in terms of the current basis n-element (in this case 3-element). As an example, we use the GrassmannAlgebra function ComposeBasisForm to compose an expression for uvx in terms of basis elements, and then expand and simplify the result using the alias % for GrassmannExpandAndSimplify. 2009 9 3


164

As an example, we use the GrassmannAlgebra function ComposeBasisForm to compose an expression for uvx in terms of basis elements, and then expand and simplify the result using the alias % for GrassmannExpandAndSimplify.X = ComposeBasisForm@u v xD He1 u1 + e2 u2 + e3 u3 L He1 v1 + e2 v2 + e3 v3 L He1 x1 + e2 x2 + e3 x3 L % @XD H- u3 v2 x1 + u2 v3 x1 + u3 v1 x2 - u1 v3 x2 - u2 v1 x3 + u1 v2 x3 L e1 e2 e3

Hence in this case uvx is given byu v x H- u3 v2 x1 + u2 v3 x1 + u3 v1 x2 - u1 v3 x2 - u2 v1 x3 + u1 v2 x3 L

Using ToCommonFactorA and ToCommonFactorBIf you wish to explicitly apply the A form of the Common Factor Theorem, you can use ToCommonFactorA. Or if the B form you can use ToCommonFactorB. The results will be equal, but they may look different. We illustrate by taking the example in the section above and expanding it both ways. Again definex1 = x y; x2 = u v; ToCommonFactorA@x1 x2 D c Hy u v x - x u v yL ToCommonFactorB@x1 x2 D c H- v u x y + u v x yL

To see most directly that these are equal, we express the vectors in terms of basis elements.Z = ComposeBasisForm@x1 x2 D He1 x1 + e2 x2 + e3 x3 L He1 y1 + e2 y2 + e3 y3 L He1 u1 + e2 u2 + e3 u3 L He1 v1 + e2 v2 + e3 v3 L

Applying the A form of the Common Factor Theorem givesToCommonFactorA@ZD c He1 Hu3 v1 x2 y1 - u1 v3 x2 y1 - u2 v1 x3 y1 + u1 v2 x3 y1 u3 v1 x1 y2 + u1 v3 x1 y2 + u2 v1 x1 y3 - u1 v2 x1 y3 L + e2 Hu3 v2 x2 y1 - u2 v3 x2 y1 - u3 v2 x1 y2 + u2 v3 x1 y2 u2 v1 x3 y2 + u1 v2 x3 y2 + u2 v1 x2 y3 - u1 v2 x2 y3 L + e3 Hu3 v2 x3 y1 - u2 v3 x3 y1 - u3 v1 x3 y2 + u1 v3 x3 y2 u3 v2 x1 y3 + u2 v3 x1 y3 + u3 v1 x2 y3 - u1 v3 x2 y3 LL

The resulting expression is identical to that calculated by the B form.

2009 9 3


165

ToCommonFactorB@ZD ToCommonFactorA@ZD True

Expressions involving non-decomposable elementsThe GrassmannAlgebra function ToCommonFactor uses the A or B form heuristically. When a common factor can be calculated by using either the A or B form, the A form is used by default. If an expansion is not possible by using the A form, ToCommonFactor tries the B form. For example, consider the regressive product of two symbolic 2-elements in a 3-space. Attempting to compute the common factor fails, since neither factor can be decomposed.A; ToCommonFactorBa bF2 2

ab2 2

However, if either one of the factors is decomposable, then a formula for the common factor can be generated. In the case below, the results are the same; remember that c is an arbitrary scalar factor.:ToCommonFactorBa Hx yLF, ToCommonFactorBHx yL aF>2 2

:c K- y x a + x y aO, c Ky x a - x y aO>2 2 2 2

Example: The regressive product of three 3-elements in a 4-space.ToCommonFactor can find the common factor of any number of elements. As an example, we take the regressive product of three 3-elements in a 4-space. This is a 1-element, since 3 + 3 + 3 - 2 4 # 1. Remember also that a 3-element in a 4-space is necessarily simple (see Chapter 2: The Exterior Product).

To demonstrate this we first declare a 4-space, and then use the GrassmannAlgebra function ComposeBasisForm to compose the required product in basis form.!4 ; X = ComposeBasisFormBa b gF3 3 3

Ha3,1 e1 e2 e3 + a3,2 e1 e2 e4 + a3,3 e1 e3 e4 + a3,4 e2 e3 e4 L Hb3,1 e1 e2 e3 + b3,2 e1 e2 e4 + b3,3 e1 e3 e4 + b3,4 e2 e3 e4 L Hg3,1 e1 e2 e3 + g3,2 e1 e2 e4 + g3,3 e1 e3 e4 + g3,4 e2 e3 e4 L

The common factor is then determined up to congruence by ToCommonFactor.

2009 9 3


166

Xc = ToCommonFactor@XD c2 He1 H-a3,3 b3,2 g3,1 + a3,2 b3,3 g3,1 + a3,3 b3,1 g3,2 - a3,1 b3,3 g3,2 - a3,2 b3,1 g3,3 + a3,1 b3,2 g3,3 L + e2 H-a3,4 b3,2 g3,1 + a3,2 b3,4 g3,1 + a3,4 b3,1 g3,2 a3,1 b3,4 g3,2 - a3,2 b3,1 g3,4 + a3,1 b3,2 g3,4 L + e3 H-a3,4 b3,3 g3,1 + a3,3 b3,4 g3,1 + a3,4 b3,1 g3,3 a3,1 b3,4 g3,3 - a3,3 b3,1 g3,4 + a3,1 b3,3 g3,4 L + e4 H-a3,4 b3,3 g3,2 + a3,3 b3,4 g3,2 + a3,4 b3,2 g3,3 a3,2 b3,4 g3,3 - a3,3 b3,2 g3,4 + a3,2 b3,3 g3,4 LL

Note that the congruence factor c is to the second power. This is because the original expression contained the regressive product operator twice: one c effectively stands in for each basis 4element that is produced during the calculation (3+3+3 = 4+4+1). It is easy to check that this result is indeed a common factor by taking its exterior product with each of the 3-elements. For example:%@Xc Ha3,1 e1 e2 e3 + a3,2 e1 e2 e4 + a3,3 e1 e3 e4 + a3,4 e2 e3 e4 LD Expand 0

3.7 The Regressive Product of Simple ElementsThe regressive product of simple elementsThe regressive product of simple elements is simple. To show this consider the (non-zero) regressive product a b, where a and b are simple, m+k n.m k m k

The 1-element factors of a and b must then have a common subspace of dimension m+k-n = j. Letm k

g be a simple j-element which spans this common subspace. We can then write:j

ab m k

a g m -j j

b g k-j j

a b g g gm -j k-j j j j

The Common Factor Axiom then shows us that since g is simple, then so is the original product ofj

simple elements a b.m k

2009 9 3


167

The regressive product of (n|1)-elementsSince we have shown in Chapter 2: The Exterior Product that all (n|1)-elements are simple, and in the previous section that the regressive product of simple elements is simple, it follows immediately that the regressive product of any number of (n|1)-elements is simple.

Example: The regressive product of two 3-elements in a 4-space is simpleAs an example of the foregoing result we calculate the regressive product of two 3-elements in a 4space. We begin by declaring a 4-space and creating two general 3-elements.!4 ; Z = ComposeBasisFormBx yF3 3

Hx3,1 e1 e2 e3 + x3,2 e1 e2 e4 + x3,3 e1 e3 e4 + x3,4 e2 e3 e4 L Hy3,1 e1 e2 e3 + y3,2 e1 e2 e4 + y3,3 e1 e3 e4 + y3,4 e2 e3 e4 L

The common 2-element is:Zc = ToCommonFactor@ZD c HH- x3,2 y3,1 + x3,1 y3,2 L e1 e2 + H- x3,3 y3,1 + x3,1 y3,3 L e1 e3 + H- x3,3 y3,2 + x3,2 y3,3 L e1 e4 + H- x3,4 y3,1 + x3,1 y3,4 L e2 e3 + H- x3,4 y3,2 + x3,2 y3,4 L e2 e4 + H- x3,4 y3,3 + x3,3 y3,4 L e3 e4 L

We can show that this 2-element is simple by confirming that its exterior product with itself is zero. (This technique was discussed in Chapter 2: The Exterior Product.)%@Zc ZcD Expand 0

Regressive products leading to scalar resultsA formula particularly useful in its interior product form to be derived later in Chapter 6 is obtained by application of the Common Factor Theorem to the productHa1 ! am L b1 ! bmn-1 n-1

where the ai are 1-elements and the bi are (n|1)-elements.n-1

2009 9 3


168

Ha1 ! am L b1 ! bmn-1 n-1

H a 1 ! a m L b j H - 1 L j-1 b 1 ! j ! b mn-1 n-1 n-1

H - 1 L i-1 a i b ji n-1

Ha1 ! i ! am L

H - 1 L j-1 b 1 ! j ! b mn-1 n-1

H - 1 L i+j a i b ji n-1

Ha1 ! i ! am L b1 ! j ! bmn-1 n-1

By repeating this process one obtains finally that:

Ha1 ! a m L b1 ! b mn-1 n-1

DetBai bj Fn-1

3.41

where DetBai bj F is the determinant of the matrix whose elements are the scalars ai bj .n-1 n-1

For the case m = 2 we have:Ha1 a2 L b1 b2n-1 n-1

a1 b1n-1

a2 b2n-1

- a1 b2n-1

a2 b1n-1

This determinant formula is of central importance in the computation of inner products to be discussed in Chapter 6.

Expressing an element in terms of another basisThe Common Factor Theorem may also be used to express an m-element in terms of another basis with basis n-element b by expanding the product b a.n n m

Let the new basis be 8b1 , !, bn < and let b b1 ! bn , then the Common Factor Theoremn

permits us to write:n

b a bi a bin m i=1 n-m m m

where n = K

n O and m

b b1 b1 b2 b2 ! bn bnn n-m m n-m m n-m m

We can visualize how the formula operates by writing a and b as simple products and thenm k

exchanging the bi with the ai in all the essentially different ways possible whilst always retaining the original ordering. To make this more concrete, suppose n is 5 and m is 2: 2009 9 3


169

We can visualize how the formula operates by writing a and b as simple products and thenm k

exchanging the bi with the ai in all the essentially different ways possible whilst always retaining the original ordering. To make this more concrete, suppose n is 5 and m is 2:Hb1 b2 b3 b4 b5 L Ha1 a2 L Ha1 a2 b3 b4 b5 L Hb1 b2 L + Ha1 b2 a2 b4 b5 L Hb1 b3 L + Ha1 b2 b3 a2 b5 L Hb1 b4 L + Ha1 b2 b3 b4 a2 L Hb1 b5 L + Hb1 a1 a2 b4 b5 L Hb2 b3 L + Hb1 a1 b3 a2 b5 L Hb2 b4 L + Hb1 a1 b3 b4 a2 L Hb2 b5 L + Hb1 b2 a1 a2 b5 L Hb3 b4 L + Hb1 b2 a1 b4 a2 L Hb3 b5 L + Hb1 b2 b3 a1 a2 L Hb4 b5 L

Now let the new n-elements on the right hand side be written as scalar factors bi times the new basis n-element b.n

bi a m n-m

bi bn

Substituting givesn n m i=1 n m n n i=1 m

b a bi b bi b bi bi

Since b is an n-element, we can 'divide' through by it to give finallyn

n

a bi bim i=1 m

bi

bi a n- m m bn

3.42

Example: Expressing a 3-element in terms of another basisConsider a 4-space with basis 8e1 ,e2 ,e3 ,e4 < and let a be a 3-element expressed in terms of this3

basis.!4 ; a = 2 e1 e2 e3 - 3 e1 e3 e4 - 2 e2 e3 e4 ;3

Suppose now that we have another basis related to the e basis by:

2009 9 3


170

b1 b2 b3 b4

= 2 e1 + 3 e3 ; = 5 e3 - e4 ; = e1 - e3 ; = e1 + e2 + e3 + e4 ;3

We wish to express a in terms of the new basis. Instead of inverting the transformation to find theei in terms of the bi , substituting for the ei in a, and simplifying, we can almost write the result3

required by inspection.Hb1 b2 b3 b4 L a b1 a Hb2 b3 b4 L - b2 a Hb1 b3 b4 L +3 3 3

b3 a Hb1 b2 b4 L - b4 a Hb1 b2 b3 L3 3

H- 4 e1 e2 e3 e4 L Hb2 b3 b4 L - H2 e1 e2 e3 e4 L Hb1 b3 b4 L + H- 2 e1 e2 e3 e4 L Hb1 b2 b4 L - H- e1 e2 e3 e4 L Hb1 b2 b3 L He1 e2 e3 e4 L HH- 4 b2 b3 b4 L + H- 2 b1 b3 b4 L + H- 2 b1 b2 b4 L + Hb1 b2 b3 LL

But from the transformation we find that% @b1 b2 b3 b4 D - 5 e1 e2 e3 e4

so that the equation becomesHb1 b2 b3 b4 L a 3

Hb1 b2 b3 b4 L 5 HH- 4 b2 b3 b4 L + H- 2 b1 b3 b4 L + H- 2 b1 b2 b4 L + Hb1 b2 b3 LL

1

We can simplify this by 'dividing' through by b1 b2 b3 b4 , to give finallya3

4 5

b2 b3 b4 +

2 5

b1 b3 b4 +

2 5

b1 b2 b4 -

1 5

b1 b2 b3

We can easily check this by expanding and simplifying the right hand side to give both sides in terms of the original e basis.a % B3

4 5

b2 b3 b4 +

2 5

b1 b3 b4 +

2 5

b1 b2 b4 -

1 5

b1 b2 b3 F

True

2009 9 3


171

Expressing a 1-element in terms of another basisFor the special case of a 1-element, the scalar coefficients bi in formula 3.35 can be written more explicitly, and without undue complexity, resulting in a decomposition of a in terms of n independent elements bi equivalent to expressing a in the basis bi .n

ai=1

b1 b2 ! a ! bn b1 b2 ! bi ! bn

bi

3.43

Here the denominators are the same in each term but are expressed this way to show the positioning of the factor a in the numerator. This result has already been introduced in the section above Example: The decomposition of a 1element.

The symmetric expansion of a 1-element in terms of another basisFor the case m = 1, b a may be expanded by the Common Factor Theorem to give:n n

H b 1 b 2 ! b n L a H b 1 b 2 ! b i-1 a b i+1 ! b n L b ii=1

or, in terms of the mnemonic expansion in the previous section:Hb1 b2 b3 ! bn L a Ha b2 b3 ! bn L b1 + Hb1 a b3 ! bn L b2 + Hb1 b2 a ! bn L b3 +! + Hb1 b2 b3 ! aL bn

Putting a equal to b0 , this may be written more symmetrically as:n

H- 1Li Hb0 b1 ! i ! bn L bi 0i=0

3.44

For example, suppose b0 , b1 , b2 , b3 are four dependent 1-elements which span a 3-space, then the formula reduces to the identity:

Hb1 b2 b3 L b0 - Hb0 b2 b3 L b1 + Hb0 b1 b3 L b2 - Hb0 b1 b2 L b3 0

3.45

We can get GrassmannAlgebra to check this formula by composing each of the bi in basis form, and finding the common factor of each term. (To use ComposeBasisForm, we will first have to give the bi new names which do not already involve subscripts.) 2009 9 3


172

We can get GrassmannAlgebra to check this formula by composing each of the bi in basis form, and finding the common factor of each term. (To use ComposeBasisForm, we will first have to give the bi new names which do not already involve subscripts.)A; B = Hb1 b2 b3 L b0 Hb0 b2 b3 L b1 + Hb0 b1 b3 L b2 - Hb0 b1 b2 L b3 ; B = B . 8b0 w, b1 x, b2 y, b3 z< - H w x y zL + w x z y - w y z x + x y z w Bn = ComposeBasisForm@BD - HHe1 w1 + e2 w2 + e3 w3 L He1 x1 + e2 x2 + e3 x3 L He1 y1 + e2 y2 + e3 y3 L He1 z1 + e2 z2 + e3 z3 LL + He1 w1 + e2 w2 + e3 w3 L He1 x1 + e2 x2 + e3 x3 L He1 z1 + e2 z2 + e3 z3 L He1 y1 + e2 y2 + e3 y3 L He1 w1 + e2 w2 + e3 w3 L He1 y1 + e2 y2 + e3 y3 L He1 z1 + e2 z2 + e3 z3 L He1 x1 + e2 x2 + e3 x3 L + He1 x1 + e2 x2 + e3 x3 L He1 y1 + e2 y2 + e3 y3 L He1 z1 + e2 z2 + e3 z3 L He1 w1 + e2 w2 + e3 w3 L ToCommonFactor@Bn D 0 True

Exploration: The cobasis form of the Common Factor AxiomThe Common Factor Axiom has a significantly suggestive form when written in terms of cobasis elements. This form will later help us extend the definition of the interior product to arbitrary elements. We start with three basis elements whose exterior product is equal to the basis n-element. The basis n-element may also be expressed as the cobasis 1 of the unit element 1 of L.0

ei ej es e1 e2 ! en 1 c 1m k p

n

The Common Factor Axiom can be written for these basis elements as:

ei es ej es ei ej es es ,mp

m + k + p n

k

p

m

3.46

k

p

p

From the definition of cobasis elements we have that:ei es H- 1Lm k ejm p k

2009 9 3


173

ej es eik p m

es ei ejp m k

ei ej es 1m k p

Substituting these four elements into the Common Factor Axiom above gives:H- 1Lm k ej ei 1 Hei ej Lk m m k

Or, more symmetrically, by interchanging the first two factors:

ei ej 1 Hei ej L c ei ejm km k m k

3.47

It can be seen that in this form the Common Factor Axiom does not specifically display the common factor, and indeed remains valid for all basis elements, independent of their grades. In sum: Given any two basis elements, the cobasis element of their exterior product is congruent to the regressive product of their cobasis elements.

Exploration: The regressive product of cobasis elementsIn Chapter 5: The Complement, we will have cause to calculate the regressive product of cobasis elements. From the formula derived below we will have an instance of the fact that the regressive product of (n|1)-elements is simple, and we will determine that simple element. First, consider basis elements e1 and e2 of an n-space and their cobasis elements e1 and e2 . The regressive product of e1 and e2 is given by:e1 e2 He2 e3 ! en L H- e1 e3 ! en L

Applying the Common Factor Axiom enables us to write:e1 e2 He1 e2 e3 ! en L H e3 ! en L

We can write this either in the form already derived in the section above:e1 e2 1 H e1 e2 L

or, by writing 1 c 1 as:n

e1 e2 c H e1 e2 L

Taking the regressive product of this equation with e3 gives:

2009 9 3


174

e1 e2 e3 c H e1 e2 L e3 c 1 H e1 e2 e3 L c2 H e1 e2 e3 L

Continuing this process, we arrive finally at the result:

e 1 e 2 ! e m c m -1 H e 1 e 2 ! e m L

3.48

A special case which we will have occasion to use in Chapter 5 is where the result reduces to a 1element.

H - 1 L j-1 e 1 e 2 ! j ! e n c n-2 H - 1 L n-1 e j

3.49

In sum: The regressive product of cobasis elements of basis 1-elements is congruent to the cobasis element of their exterior product. In fact, this formula is just an instance of a more general result which says that: The regressive product of cobasis elements of any grade is congruent to the cobasis element of their exterior product. We will discuss a result very similar to this in more detail after we have defined the complement of an element in Chapter 5.

3.8 Factorization of Simple ElementsFactorization using the regressive productThe Common Factor Axiom asserts that in an n-space, the regressive product of a simple melement with a simple (n|m+1)-element will give either zero, or a 1-element belonging to them both, and hence a factor of the m-element. If m such factors can be obtained which are independent, their product will therefore constitute (apart from a scalar factor easily determined) a factorization of the m-element. Let a be the simple element to be factorized. Choose first an (n|m)-element b whose product withm n- m n-m

a is non-zero. Next, choose a set of m independent 1-elements bj whose products with b are alsom

non-zero. A factorization aj of a is then obtained from:m

a a a1 a2 ! a mm

aj a Kbj b Om n- m

3.50

The scalar factor a may be determined simply by equating any two corresponding terms of the original element and the factorized version. Note that no factorization is unique. Had different bj been chosen, a different factorization would have been obtained. Nevertheless, any one factorization may be obtained from any other by adding multiples of the factors to each factor. 2009 9 3


175

Note that no factorization is unique. Had different bj been chosen, a different factorization would have been obtained. Nevertheless, any one factorization may be obtained from any other by adding multiples of the factors to each factor. If an element is simple, then the exterior product of the element with itself is zero. The converse, however, is not true in general for elements of grade higher than 2, for it only requires the element to have just one simple factor to make the product with itself zero. If the method is applied to the factorization of a non-simple element, the result will still be a simple element. Thus an element may be tested to see if it is simple by applying the method of this section: if the factorization is not equivalent to the original element, the hypothesis of simplicity has been violated.

Example: Factorization of a simple 2-elementSuppose we have a 2-element a which we wish to show is simple and which we wish to factorize.2

a vw + vx + vy + vz + wz + xz + yz2

There are 5 independent 1-elements in the expression for a : v, w, x, y, z, hence we can choose n2

to be 5. Next, we choose b (= b) arbitrarily as xyz, b1 as v, and b2 as w. Our two factors thenn-m 3

become:a1 a b1 b a Hv x y zL2 3 2

a2 a b2 b a H w x y zL2 3 2

In determining a1 the Common Factor Theorem permits us to write for arbitrary 1-elements x and y:Hx y L Hv x y zL Hx v x y zL y - Hy v x y zL x

Applying this to each of the terms of a gives:2

a1 - H w v x y zL v + H w v x y zL z v - z

Similarly for a2 we have the same formula, except that w replaces v.Hx y L H w x y zL Hx w x y zL y - Hy w x y zL x

Again applying this to each of the terms of a gives:2

a2 Hv w x y zL w + Hv w x y zL x + Hv w x y zL y + Hv w x y zL z w + x + y + z

Hence the factorization is congruent to:

2009 9 3


176

a1 a2 Hv - zL H w + x + y + zL

Verification by expansion of this product shows that this is indeed a factorization of the original element.

Factorizing elements expressed in terms of basis elementsWe now take the special case where the element to be factorized is expressed in terms of basis elements. This will enable us to develop formulae from which we can write down the factorization of an element (almost) by inspection. The development is most clearly apparent from a specific example, but one that is general enough to cover the general concept. Consider a 3-element in a 5-space, where we suppose the coefficients to be such as to ensure the simplicity of the element.a a1 e1 e2 e3 + a2 e1 e2 e4 + a3 e1 e2 e53

+ a4 e1 e3 e4 + a5 e1 e3 e5 + a6 e1 e4 e5 + a7 e2 e3 e4 + a8 e2 e3 e5 + a9 e2 e4 e5 + a10 e3 e4 e5

Choose b1 e1 , b2 e2 , b3 e3 , b e4 e5 , then we can write each of the factors bi b as a2 2

cobasis element.b1 b e1 e4 e5 e2 e32

b2 b e2 e4 e5 - e1 e32

b3 b e3 e4 e5 e1 e22

Consider the cobasis element e2 e3 , and a typical term of a e2 e3 , which we write as3

Ha ei ej ek L e2 e3 . The Common Factor Theorem tells us that this product is zero if ei ej ek does not contain e2 e3 . We can thus simplify the product a e2 e3 by dropping3

out the terms of a which do not contain e2 e3 . Thus:3

a e2 e3 Ha1 e1 e2 e3 + a7 e2 e3 e4 + a8 e2 e3 e5 L e2 e33

Furthermore, the Common Factor Theorem applied to a typical term He2 e3 ei L e2 e3 of the expansion in which both e2 e3 and e2 e3 occur yields a 1-element congruent to the remaining basis 1-element ei in the product. This effectively cancels out (up to congruence) the product e2 e3 from the original term.He2 e3 ei L e2 e3 IHe2 e3 L He2 e3 LM ei ei

Thus we can further reduce a e2 e3 to give:3

2009 9 3


177

Thus we can further reduce a e2 e3 to give:3

a1 a e2 e3 a1 e1 + a7 e4 + a8 e53

Similarly we can determine the other factors:a2 a - e1 e3 a1 e2 - a4 e4 - a5 e53

a3 a e1 e2 a1 e3 + a2 e4 + a3 e53

It is clear from inspecting the product of the first terms in each 1-element that the product requires a scalar divisor of a2 1 . The final result is then:a3

1 a1 2

a1 a2 a3

1 a1 2 Ha1 e1 + a7 e4 + a8 e5 L Ha1 e2 - a4 e4 - a5 e5 L Ha1 e3 + a2 e4 + a3 e5 L

Verification and derivation of conditions for simplicityWe verify the factorization by multiplying out the factors and comparing the result with the original expression. When we do this we obtain a result which still requires some conditions to be met: those ensuring the original element is simple. First we declare a 5-dimensional basis and compose a (this automatically declares the coefficients3

ai to be scalar). A; !5 ; ComposeMElement@3, aD a1 e1 e2 e3 + a2 e1 e2 e4 + a3 e1 e2 e5 + a4 e1 e3 e4 + a5 e1 e3 e5 + a6 e1 e4 e5 + a7 e2 e3 e4 + a8 e2 e3 e5 + a9 e2 e4 e5 + a10 e3 e4 e5

To effect the multiplication of the factored form we use GrassmannExpandAndSimplify (in its alias form).

2009 9 3


178

A = % B

1 a1 2

Ha1 e1 + a7 e4 + a8 e5 L Ha1 e2 - a4 e4 - a5 e5 L Ha1 e3 + a2 e4 + a3 e5 LF a1 e1 e2 e3 + a2 e1 e2 e4 + a3 e1 e2 e5 + a4 e1 e3 e4 + a3 a4 a2 a5 a5 e1 e3 e5 + + e1 e4 e5 + a7 e2 e3 e4 + a1 a1 a3 a7 a2 a8 a5 a7 a4 a8 a8 e2 e3 e5 + + e2 e4 e5 + + a1 a1 a1 a1

e3 e4 e5

For this expression to be congruent to the original expression we clearly need to apply the simplicity conditions for a general 3-element in a 5-space. Once this is done we retrieve the original 3-element a with which we began. The simplicity conditions can be expressed by the3

following rules, which we apply to the expression A. We will discuss them in the next section.A . : a3 a4 a1 + + a2 a5 a1 a6 , a5 a7 a1 + a4 a8 a1 a10 >

a3 a7 a1

a2 a8 a1

a9 , -

a

Grass Mann Algebra Book

Documents

Transcript of Grass Mann Algebra Book