An Introduction to Geometric Algebra and Calculus. Alan Bromborsky Army Research Lab (Retired)

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1 An Introduction to Geometric Algebra and Calculus Alan Bromborsky Army Research Lab Retired April 1, 2011

2 2 Introduction Geometric algebra is the Clifford algebra of a finite dimensional vector space over real scalars cast in a form most appropriate for physics and engineering. This was done by David Hestenes Arizona State University in the 1960 s. From this start he developed the geometric calculus whose fundamental theorem includes the generalized Stokes theorem, the residue theorem, and new integral theorems not realized before. Hestenes likes to say he was motivated by the fact that physicists and engineers did not know how to multiply vectors. Researchers at Arizona State and Cambridge have applied these developments to classical mechanics,quantum mechanics, general relativity gauge theory of gravity, projective geometry, conformal geometry, etc.

3 Contents 1 Basic Geometric Algebra Axioms of Geometric Algebra Why Learn This Stuff? Inner,, and outer,, product of two vectors and their basic properties Outer,, product for r Vectors in terms of the geometric product Alternate Definition of Outer,, product for r Vectors Useful Relation s Projection Operator Basis Blades G 3, 0 Geometric Algebra Euclidian Space G 1, 3 Geometric Algebra Spacetime Reflections Rotations Definitions General Rotation Euclidean Case Minkowski Case Expansion of geometric product and generalization of and Duality and the Pseudoscalar Reciprocal Frames Coordinates Linear Transformations Definitions Adjoint Inverse Commutator Product

4 4 CONTENTS 2 Examples of Geometric Algebra Quaternions Spinors Geometric Algebra of the Minkowski Plane Lorentz Transformation Geometric Calculus - The Derivative Definitions Derivatives of Scalar Functions Product Rule Interior and Exterior Derivative Derivative of a Multivector Function Spherical Coordinates Analytic Functions Geometric Calculus - Integration Line Integrals Surface Integrals Directed Integration - n-dimensional Surfaces k-simplex Definition k-chain Definition Algebraic Topology Simplex Notation Fundamental Theorem of Geometric Calculus The Fundamental Theorem At Last! Examples of the Fundamental Theorem Divergence and Green s Theorems Cauchy s Integral Formula In Two Dimensions Complex Plane Green s Functions in N-dimensional Euclidean Spaces Geometric Calculus on Manifolds Definition of a Vector Manifold The Pseudoscalar of the Manifold The Projection Operator The Exclusion Operator The Intrinsic Derivative The Covariant Derivative Coordinates and Derivatives Riemannian Geometry Manifold Mappings

5 CONTENTS The Fundmental Theorem of Geometric Calculus on Manifolds Divergence Theorem on Manifolds Stokes Theorem on Manifolds Differential Forms in Geometric Calculus Inner Products of Subspaces Alternating Forms Dual of a Vector Space Standard Definition of a Manifold Case Study of a Manifold for a Model Universe The Edge of Known Space A BAC-CAB Formulas 105 B Blade Orientation Theorem 107 C References 109

6 6 CONTENTS

7 Chapter 1 Basic Geometric Algebra 1.1 Axioms of Geometric Algebra Let V p, q be a finite dimensional vector space of signature p, q 1 over R. Then a, b, c V there exists a geometric product with the properties - If a 2 0 then a 1 = 1 a 2 a. abc = abc ab + c = ab + ac a + bc = ac + bc aa R 1.2 Why Learn This Stuff? The geometric product of two or more vectors produces something new like the 1 with respect to real numbers or vectors with respect to scalars. It must be studied in terms of its effect on vectors and in terms of its symmetries. It is worth the effort. Anything that makes understanding rotations in a N dimensional space simple is worth the effort! Also, if one proceeds 1 To be completely general we would have to consider V p, q, r where the dimension of the vector space is n = p + q + r and p, q, and r are the number of basis vectors respectively with positive, negative and zero squares. 7

8 8 CHAPTER 1. BASIC GEOMETRIC ALGEBRA on to geometric calculus many diverse areas in mathematics are unified and many areas of physics and engineering are greatly simplified. 1.3 Inner,, and outer,, product of two vectors and their basic properties The inner dot and outer wedge products of two vectors are defined by a b 1 ab + ba a b 1 ab ba ab = a b + a b 1.3 a b = b a 1.4 c = a + b c 2 = a + b 2 c 2 = a 2 + ab + ba + b 2 2a b = c 2 a 2 b 2 a b R 1.5 a b = a b cos θ if a 2, b 2 > Orthogonal vectors are defined by a b = 0. For orthogonal vectors a b = ab. Now compute a b 2 a b 2 = a b b a 1.7 = ab a b ba a b 1.8 = abba a b ab + ba + a b = a 2 b 2 a b = a 2 b 2 1 cos 2 θ 1.11 = a 2 b 2 sin 2 θ 1.12 Thus in a Euclidean space, a 2, b 2 > 0, a b 2 0 and a b is proportional to sin θ. If e and e are any two orthonormal unit vectors in a Euclidean space then e e 2 = 1. Who needs the 1?

9 1.4. OUTER,, PRODUCT FOR R VECTORS IN TERMS OF THE GEOMETRIC PRODUCT9 1.4 Outer,, product for r Vectors in terms of the geometric product Define the outer product of r vectors to be ε i 1...i r 1...r Thus and a 1... a r 1 r! a 1... a j + b j... a r = is the mixed permutation symbol ε i1...ir 1...r a i1... a ir 1.13 i 1,...,i r a 1... a j... a r + a 1... b j... a r 1.14 a 1... a j a j+1... a r = The outer product of r vectors is called a blade of grade r. a 1... a j+1 a j... a r Alternate Definition of Outer,, product for r Vectors Let e 1, e 2,..., e r be an orthogonal basis for the set of linearly independent vectors a 1, a 2,..., a r so that we can write a i = α ij e j 1.16 j Then a 1 a 2... a r = α 1j1 e j1 α 2j2 e j2... α rjr e jr j 1 = j 2 j r j 1,...,j r α 1j1 α 2j2... α rjr e j1 e j2... e jr 1.17 Now define a blade of grade n as the geometric product of n orthogonal vectors. Thus the product e j1 e j2... e jr in equation 1.17 could be a blade of grade r, r 2, r 4, etc. depending upon the number of repeated factors.

10 10 CHAPTER 1. BASIC GEOMETRIC ALGEBRA If there are no repeated factors in the product we have that e j1... e jr = ε j 1...j r 1...r e 1... e r 1.18 Due to the fact that interchanging two adjacent orthogonal vectors in the geometric product will reverse the sign of the product and we can define the outer product of r vectors as a 1... a r = 1...r α 1j1... α rjr e 1... e r 1.19 j 1,...,j r ε j1...jr = det α e 1... e r 1.20 Thus the outer product of r independent vectors is the part of the geometric product of the r vectors that is of grade r. Equation 1.19 is equivalent to equation This can be proved by substituting equation 1.17 into equation 1.13 to get a 1... a r = 1 r! = 1 r! = 1 r! i 1,...,i r i 1,...,i r ε i1...ir 1...r α i1 j 1... α irjr e j1... e jr 1.21 j 1,...,j r ε i1...ir 1...r j 1,...,j r ε j1...jr 1...r j 1,...,j r ε j 1...j r 1...r α i1 j 1... α irj r e 1... e r 1.22 ε j 1...j r 1...r det α e 1... e r 1.23 = det α e 1... e r 1.24 We go from equation 1.22 to equation 1.23 by noting that ε i1...ir 1...r α i1 j 1... α irjr is just det α i 1,...,i r with the columns permuted. Multiplying det α by ε j 1...j r 1...r gives the correct sign for the determinant with the columns permuted. If e 1,..., e n is an orthonormal basis for vector space the unit psuedoscalar is defined as I = e 1... e n 1.25 In equation 1.24 let r = n and the a 1,..., a n be another orthonormal basis for the vector space. Then we may write a 1... a n = det α e 1... e n 1.26 Since both the a s and the e s form orthonormal bases the matrix α is orthogonal and det α = ±1. All psuedoscalars for the vector space are identical to within a scale factor of ±1. 2 Likewise a 1... a n is equal to I times a scale factor. 2 It depends only upon the ordering of the basis vectors.

11 1.6. USEFUL RELATION S Useful Relation s 1. For a set of r orthogonal vectors, e 1,..., e r e 1... e r = e 1... e r For a set of r linearly independent vectors, a 1,..., a r, there exists a set of r orthogonal vectors, e 1,..., e r, such that a 1... a r = e 1... e r 1.28 If the vectors, a 1,..., a r, are not linearly independent then a 1... a r = The product a 1... a r is call a blade of grade r. The dimension of the vector space is the highest grade any blade can have. 1.7 Projection Operator A multivector, the basic element of the geometric algebra, is made of of a sum of scalars, vectors, blades. A multivector is homogeneous pure if all the blades in it are of the same grade. The grade of a scalar is 0 and the grade of a vector is 1. The general multivector A is decomposed with the grade projection operator A r as N is dimension of the vector space: A = As an example consider ab, the product of two vectors. Then We define A A 0 for any multivector A N A r 1.30 r=0 ab = ab 0 + ab Basis Blades The geometric algebra of a vector space, V p, q, is denoted G p, q or G V where p, q is the signature of the vector space first p unit vectors square to +1 and next q unit vectors square to 1, dimension of the space is p + q. Examples are:

12 12 CHAPTER 1. BASIC GEOMETRIC ALGEBRA p q Type of Space 3 0 3D Euclidean 1 3 Relativistic Space Time 4 1 3D Conformal Geometry If the orthonormal basis set of the vector space is e 1,..., e N, the basis of the geometric algebra multivector space is formed from the geometric products since we have chosen an orthonormal basis, ei 2 = ±1 of the basis vectors. For grade r multivectors the basis blades are all the combinations of basis vectors products taken r at a time from the set of N vectors. Thus the number basis blades of rank r are N r, the binomial expansion coefficient and the total dimension of the multivector space is the sum of N r over r which is 2 N G 3, 0 Geometric Algebra Euclidian Space The basis blades for G 3, 0 are: Grade e 1 e 1 e 2 e 1 e 2 e 3 e 2 e 1 e 3 e 3 e 2 e 3 The multiplication table for the G 3, 0 basis blades is 1 e 1 e 2 e 3 e 1 e 2 e 1 e 3 e 2 e 3 e 1 e 2 e e 1 e 2 e 3 e 1 e 2 e 1 e 3 e 2 e 3 e 1 e 2 e 3 e 1 e 1 1 e 1 e 2 e 1 e 3 e 2 e 3 e 1 e 2 e 3 e 2 e 3 e 2 e 2 e 1 e 2 1 e 2 e 3 e 1 e 1 e 2 e 3 e 3 e 1 e 3 e 3 e 3 e 1 e 3 e 2 e 3 1 e 1 e 2 e 3 e 1 e 2 e 1 e 2 e 1 e 2 e 1 e 2 e 2 e 1 e 1 e 2 e 3 1 e 2 e 3 e 1 e 3 e 3 e 1 e 3 e 1 e 3 e 3 e 1 e 2 e 3 e 1 e 2 e 3 1 e 1 e 2 e 2 e 2 e 3 e 2 e 3 e 1 e 2 e 3 e 3 e 2 e 1 e 3 e 1 e 2 1 e 1 e 1 e 2 e 3 e 1 e 2 e 3 e 2 e 3 e 1 e 3 e 1 e 2 e 3 e 2 e 1 1 Note that the squares of all the grade 2 and 3 basis blades are 1. The highest rank basis blade in this case e 1 e 2 e 3 is usually denoted by I and is called the pseudoscalar.

13 1.8. BASIS BLADES G 1, 3 Geometric Algebra Spacetime The multiplication table for the G 1, 3 basis blades is 1 γ 0 γ 1 γ 2 γ 3 γ 0 γ 1 γ 0 γ 2 γ 1 γ γ 0 γ 1 γ 2 γ 3 γ 0 γ 1 γ 0 γ 2 γ 1 γ 2 γ 0 γ 0 1 γ 0 γ 1 γ 0 γ 2 γ 0 γ 3 γ 1 γ 2 γ 0 γ 1 γ 2 γ 1 γ 1 γ 0 γ 1 1 γ 1 γ 2 γ 1 γ 3 γ 0 γ 0 γ 1 γ 2 γ 2 γ 2 γ 2 γ 0 γ 2 γ 1 γ 2 1 γ 2 γ 3 γ 0 γ 1 γ 2 γ 0 γ 1 γ 3 γ 3 γ 0 γ 3 γ 1 γ 3 γ 2 γ 3 1 γ 0 γ 1 γ 3 γ 0 γ 2 γ 3 γ 1 γ 2 γ 3 γ 0 γ 1 γ 0 γ 1 γ 1 γ 0 γ 0 γ 1 γ 2 γ 0 γ 1 γ 3 1 γ 1 γ 2 γ 0 γ 2 γ 0 γ 2 γ 0 γ 2 γ 2 γ 0 γ 1 γ 2 γ 0 γ 0 γ 2 γ 3 γ 1 γ 2 1 γ 0 γ 1 γ 1 γ 2 γ 1 γ 2 γ 0 γ 1 γ 2 γ 2 γ 1 γ 1 γ 2 γ 3 γ 0 γ 2 γ 0 γ 1 1 γ 0 γ 3 γ 0 γ 3 γ 3 γ 0 γ 1 γ 3 γ 0 γ 2 γ 3 γ 0 γ 1 γ 3 γ 2 γ 3 γ 0 γ 1 γ 2 γ 3 γ 1 γ 3 γ 1 γ 3 γ 0 γ 1 γ 3 γ 3 γ 1 γ 2 γ 3 γ 1 γ 0 γ 3 γ 0 γ 1 γ 2 γ 3 γ 2 γ 3 γ 2 γ 3 γ 2 γ 3 γ 0 γ 2 γ 3 γ 1 γ 2 γ 3 γ 3 γ 2 γ 0 γ 1 γ 2 γ 3 γ 0 γ 3 γ 1 γ 3 γ 0 γ 1 γ 2 γ 0 γ 1 γ 2 γ 1 γ 2 γ 0 γ 2 γ 0 γ 1 γ 0 γ 1 γ 2 γ 3 γ 2 γ 1 γ 0 γ 0 γ 1 γ 3 γ 0 γ 1 γ 3 γ 1 γ 3 γ 0 γ 3 γ 0 γ 1 γ 2 γ 3 γ 0 γ 1 γ 3 γ 1 γ 2 γ 3 γ 0 γ 2 γ 3 γ 0 γ 2 γ 3 γ 0 γ 2 γ 3 γ 2 γ 3 γ 0 γ 1 γ 2 γ 3 γ 0 γ 3 γ 0 γ 2 γ 1 γ 2 γ 3 γ 3 γ 0 γ 1 γ 3 γ 1 γ 2 γ 3 γ 1 γ 2 γ 3 γ 0 γ 1 γ 2 γ 3 γ 2 γ 3 γ 1 γ 3 γ 1 γ 2 γ 0 γ 2 γ 3 γ 0 γ 1 γ 3 γ 3 γ 0 γ 1 γ 2 γ 3 γ 0 γ 1 γ 2 γ 3 γ 1 γ 2 γ 3 γ 0 γ 2 γ 3 γ 0 γ 1 γ 3 γ 0 γ 1 γ 2 γ 2 γ 3 γ 1 γ 3 γ 0 γ 3 γ 0 γ 3 γ 1 γ 3 γ 2 γ 3 γ 0 γ 1 γ 2 γ 0 γ 1 γ 3 γ 0 γ 2 γ 3 γ 1 γ 2 γ 3 γ 0 γ 1 γ 2 γ 3 1 γ 0 γ 3 γ 1 γ 3 γ 2 γ 3 γ 0 γ 1 γ 2 γ 0 γ 1 γ 3 γ 0 γ 2 γ 3 γ 1 γ 2 γ 3 γ 0 γ 1 γ 2 γ 3 γ 0 γ 3 γ 0 γ 1 γ 3 γ 0 γ 2 γ 3 γ 1 γ 2 γ 1 γ 3 γ 2 γ 3 γ 0 γ 1 γ 2 γ 3 γ 1 γ 2 γ 3 γ 1 γ 0 γ 1 γ 3 γ 3 γ 1 γ 2 γ 3 γ 0 γ 2 γ 0 γ 3 γ 0 γ 1 γ 2 γ 3 γ 2 γ 3 γ 0 γ 2 γ 3 γ 2 γ 0 γ 2 γ 3 γ 1 γ 2 γ 3 γ 3 γ 0 γ 1 γ 0 γ 1 γ 2 γ 3 γ 0 γ 3 γ 1 γ 3 γ 0 γ 1 γ 3 γ 3 γ 0 γ 1 γ 2 γ 0 γ 1 γ 2 γ 3 γ 0 γ 1 γ 0 γ 2 γ 1 γ 2 γ 0 γ 1 γ 2 γ 0 γ 1 γ 1 γ 3 γ 0 γ 3 γ 0 γ 1 γ 2 γ 3 γ 2 γ 3 γ 1 γ 2 γ 3 γ 0 γ 2 γ 3 γ 2 γ 3 γ 0 γ 2 γ 2 γ 3 γ 0 γ 1 γ 2 γ 3 γ 0 γ 3 γ 1 γ 1 γ 2 γ 3 γ 3 γ 0 γ 1 γ 3 γ 1 γ 3 γ 1 γ 2 γ 0 γ 1 γ 2 γ 3 γ 2 γ 3 γ 1 γ 3 γ 0 γ 0 γ 2 γ 3 γ 0 γ 1 γ 3 γ 3 γ 0 γ 3 γ 0 γ 3 1 γ 0 γ 1 γ 0 γ 2 γ 1 γ 2 γ 3 γ 1 γ 2 γ 0 γ 1 γ 2 γ 1 γ 2 γ 1 γ 3 γ 0 γ 1 1 γ 1 γ 2 γ 0 γ 2 γ 3 γ 0 γ 0 γ 1 γ 2 γ 2 γ 0 γ 2 γ 2 γ 3 γ 0 γ 2 γ 1 γ 2 1 γ 0 γ 1 γ 3 γ 0 γ 1 γ 2 γ 0 γ 1 γ 0 γ 1 γ 0 γ 1 γ 2 γ 1 γ 2 γ 3 γ 0 γ 2 γ 3 γ 0 γ 1 γ 3 1 γ 2 γ 3 γ 1 γ 3 γ 0 γ 3 γ 3 γ 0 γ 1 γ 3 γ 1 γ 0 γ 0 γ 1 γ 2 γ 2 γ 3 1 γ 1 γ 2 γ 0 γ 2 γ 2 γ 0 γ 2 γ 3 γ 2 γ 0 γ 1 γ 2 γ 0 γ 1 γ 3 γ 1 γ 2 1 γ 0 γ 1 γ 1 γ 1 γ 2 γ 3 γ 0 γ 1 γ 2 γ 2 γ 1 γ 0 γ 3 γ 0 γ 2 γ 0 γ 1 1 γ 0 γ 0 γ 1 γ 2 γ 3 γ 1 γ 2 γ 0 γ 2 γ 0 γ 1 γ 3 γ 2 γ 1 γ 0 1

14 14 CHAPTER 1. BASIC GEOMETRIC ALGEBRA 1.9 Reflections We wish to show that a, v V ava V and v is reflected about a if a 2 = 1. Figure 1.1: Reflection of Vector 1. Decompose v = v +v where v is the part of v parallel to a and v is the part perpendicular to a. 2. av = av + av = v a v a since a and v are orthogonal. 3. ava = a 2 v v is a vector since a 2 is a scalar. 4. ava is the reflection of v about the direction of a if a 2 = Thus a 1... a r va r... a 1 V and produces a composition of reflections of v if a 2 1 = = a 2 r = 1.

15 1.10. ROTATIONS Rotations Definitions First define the reverse of a product of vectors. If R = a 1... a s then the reverse is R = a 1... a s = a r... a 1, the order of multiplication is reversed. Then let R = ab so that RR = abba = ab 2 a = a 2 b 2 = R R 1.32 Let RR = 1 and calculate RvR 2, where v is an arbitrary vector. RvR 2 = RvR RvR = Rv 2 R = v 2 RR = v Thus RvR leaves the length of v unchanged. Now we must also prove Rv 1 R Rv 2 R = v 1 v 2. Since Rv 1 R and Rv 2 R are both vectors we can use the definition of the dot product for two vectors Rv 1 R Rv 2 R = 1 Rv1 R Rv 2 R + Rv 2 R Rv 1 R 2 = 1 Rv1 v 2 R + Rv 2 v 1 R 2 = 1 2 R v 1v 2 + v 2 v 1 R = R v 1 v 2 R = v 1 v 2 RR = v 1 v 2 Thus the transformation RvR preserves both length and angle and must be a rotation. The normal designation for R is a rotor. If we have a series of successive rotations R 1, R 2,..., R k to be applied to a vector v then the result of the k rotations will be R k R k 1... R 1 vr 1R 2... R k Since each individual rotation can be written as the geometric product of two vectors, the composition of k rotations can be written as the geometric product of 2k vectors. The multivector that results from the geometric product of r vectors is called a versor of order r. A composition of rotations is always a versor of even order.

16 16 CHAPTER 1. BASIC GEOMETRIC ALGEBRA General Rotation The general rotation can be represented by R = e θ 2 u where u is a unit bivector in the plane of the rotation and θ is the rotation angle in the plane. 3 The two possible non-degenerate cases are u 2 = ±1 { e θ 2 Euclidean plane u u = 2 = 1 : cos θ 2 + u sin θ } 2 Minkowski plane u 2 = 1 : cosh θ 2 + u sinh θ Decompose v = v + v v where v is the projection of v into the plane defined by u. Note that v v is orthogonal to all vectors in the u plane. Now let u = e e where e is parallel to v and of course e is in the plane u and orthogonal to e. v v anticommutes with e and e and v anticommutes with e it is left to the reader to show RR = Euclidean Case For the case of u 2 = 1 Figure 1.2: Rotation of Vector RvR = cos θ + e e sin 2 θ v + v v cos 2 θ + e e sin 2 θ 2 Since v v anticommutes with e and e it commutes with R and RvR = Rv R + v v e A is defined as the Taylor series expansion e A = j=0 A j j! where A is any multivector.

17 1.11. EXPANSION OF GEOMETRIC PRODUCT AND GENERALIZATION OF AND 17 So that we only have to evaluate θ Rv R = cos + e e sin 2 θ v cos 2 θ + e e sin 2 Since v = v e Rv R = v cos θ e + sin θ e and the component of v in the u plane is rotated correctly. θ Minkowski Case For the case of u 2 = 1 there are two possibilities, v 2 > 0 or v2 < 0. In the first case e2 = 1 and e 2 = 1. In the second case e2 = 1 and e2 = 1. Again v v is not affected by the rotation so that we need only evaluate Rv R = cosh θ 2 + e e sinh θ v cosh 2 θ + e e sinh 2 θ 2 Note that in this case v = v 2 and Rv R = { v 2 > 0 : v cosh θ e + sinh θ e v 2 < 0 : v cosh θ e sinh θ e } Expansion of geometric product and generalization of and If A r and B s are respectively grade r and s pure grade multivectors then A r B s = A r B s r s + A r B s r s A r B s minr+s,2n r+s 1.39 A r B s A r B s r s 1.40 A r B s A r B s r+s 1.41 Thus if r + s > N then A r B s = 0, also note that these formulas are the most efficient way of calculating A r B s and A r B s. Using equations 1.28 and 1.39 we can prove that for a vector a and a grade r multivector B r a B r = 1 2 ab r 1 r B r a 1.42

18 18 CHAPTER 1. BASIC GEOMETRIC ALGEBRA a B r = 1 2 ab r + 1 r B r a 1.43 If equations 1.42 and 1.43 are true for a grade r blade they are also true for a grade r multivector superposition of grade r blades. By equation 1.28 let B r = e 1... e r where the e s are orthogonal and expand a r a = a + α j e j 1.44 where a is orthogonal to all the e s. Then 4 ab r = j=1 r 1 j 1 α j e 2 je 1 ĕ j e r + a e 1... e r j=1 = a B r + a B r 1.45 Now calculate r B r a = 1 r j α j e 2 je 1 ĕ j e r 1 r 1 a e 1... e r j=1 r = 1 r 1 1 j 1 α j e 2 je 1 ĕ j e r a e 1... e r j=1 = 1 r 1 a B r a B r 1.46 Adding and subtracting equations 1.45 and 1.46 gives equations 1.42 and Duality and the Pseudoscalar If e 1,..., e n is an orthonormal basis for the vector space the the pseudoscalar I is defined by I = e 1... e n 1.47 Since one can transform one orthonormal basis to another by an orthogonal transformation the I s for all orthonormal bases are equal to within a ±1 scale factor with depends on the ordering of the basis vectors. If A r is a pure r grade multivector A r = A r r then A r I = A r I n r e 1... e j 1 ĕ j e j+1... e r = e 1... e j 1 e j+1... e r

19 1.13. RECIPROCAL FRAMES 19 or A r I is a pure n r grade multivector. Further by the symmetry properties of I we have IA r = 1 n 1r A r I 1.49 I can also be used to exchange the and products as follows using equations 1.42 and 1.43 a A r I = 1 aar I 1 n r A r Ia = 1 aar I 1 n r 1 n 1 A r ai = 1 2 aa r + 1 r A r a I 1.52 = a A r I 1.53 More generally if A r and B s are pure grade multivectors with r+s n we have using equation 1.40 and 1.48 A r B s I = A r B s I r n s 1.54 = A r B s I n r+s 1.55 = A r B s r+s I 1.56 = A r B s I 1.57 Finally we can relate I to I by I = 1 nn 1 2 I Reciprocal Frames Let e 1,..., e n be a set of linearly independent vectors that span the vector space that are not necessarily orthogonal. These vectors define the frame frame vectors are shown in bold face since they are almost always associated with a particular coordinate system with volume element E n e 1... e n 1.59 So that E n I. The reciprocal frame is the set of vectors e 1,..., e n that satisfy the relation The e i are constructed as follows e i e j = δ i j, i, j = 1,..., n 1.60 e j = 1 j 1 e 1 e 2... ĕ j... e n E 1 n 1.61

20 20 CHAPTER 1. BASIC GEOMETRIC ALGEBRA So that the dot product is using equation 1.53 since En 1 I e i e j = 1 j 1 e i e 1 e 2... ĕ j... e n En = 1 j 1 e i e 1 e 2... ĕ j... e n En = 0, i j 1.64 and e 1 e 1 = e 1 e 2... e n En = e 1 e 2... e n En = Coordinates The reciprocal frame can be used to develop a coordinate representation for multivectors in an arbitrary frame e 1,..., e n with reciprocal frame e 1,..., e n. Since both the frame and it s reciprocal span the base vector space we can write any vector a in the vector space as a = a i e i = a i e i 1.68 where if an index such as i is repeated it is assumes that the terms with the repeated index will be summed from 1 to n. Using that e i e j = δ j i we have a i = a e i 1.69 a i = a e i 1.70 In tensor notation a i would be the covariant representation and a i the contravariant representation of the vector a. Now consider the case of grade 2 and grade 3 blades: e i a b = a e i b b e i a e i a e i b b e i a = ab ba = 2a b e i a b c = a e i b c b e i a c + c e i a b e i a e i b c b e i a c + c e i a b = ab c ba c + ca b = 3a b c

21 1.15. LINEAR TRANSFORMATIONS 21 for an r-blade A r we have the proof is left to the reader Since e i e i = n we have e i e i A r = ra r 1.71 e i e i A r = e i e i A r e i A r = n r Ar 1.72 Flipping e i and A r in equations 1.71 and 1.72 and subtracting equation 1.71 from 1.72 gives e i A r e i = 1 r n 2r A r 1.73 In Hestenes and Sobczyk 3.14 it is proved that e k r... e k 1 e j1... e jr = δ j 1 k 1 δ j 2 k 2... δ jr k r 1.74 so that the general multivector A can be expanded in terms of the blades of the frame and reciprocal frame as A = A ij k e i e j e k 1.75 where i<j< <k A ij k = e k e j e i A 1.76 The components A ij k are totally antisymmetric on all indices and are usually referred to as the components of an antisymmetric tensor Linear Transformations Definitions Let f be a linear transformation on a vector space f : V V with f αa + βb = αf a + βf b a, b V and α, β R. Then define the action of f on a blade of the geometric algebra by and the action of f on any two A, B G V by f a 1... a r = f a 1... f a f αa + βb = αf A + βf B 1.78

22 22 CHAPTER 1. BASIC GEOMETRIC ALGEBRA Since any multivector A can be expanded as a sum of blades f A is defined. This has many consequences. Consider the following definition for the determinant of f, det f. f I = det f I 1.79 First show that this definition is equivalent to the standard definition of the determinant again e 1,..., e N is an orthonormal basis for V. Then But so that or f I = f e r = N a rs e s 1.80 s=1 N N a 1s1 e s... a NsN e s s 1 =1 = s N =1 s 1,...,s N a 1s1... a NsN e s1... e sn 1.81 e s1... e sn f I = det f = = ε s 1...s N 1...N e 1... e N 1.82 ε s1...sn 1...N a 1s 1... a NsN I 1.83 s 1,...,s N ε s1...sn 1...N a 1s 1... a NsN 1.84 s 1,...,s N which is the standard definition. Now compute the determinant of the product of the linear transformations f and g or det fg I = fg I = f g I = f det g I = det g f I = det g det f I 1.85 det fg = det f det g 1.86 Do you have any idea of how miserable that is to prove from the standard definition of determinant?

23 1.15. LINEAR TRANSFORMATIONS Adjoint If F is linear transformation and a and b are two arbitrary vectors the adjoint function, F, is defined by a F b = b F a 1.87 From the definition the adjoint is also a linear transformation. For an arbitrary frame e 1,..., e n we have e i F a = a F e i 1.88 So that we can explicitly construct the adjoint as F a = e i e i F a = e i a F e i = e i F e i e j a j 1.89 so that F ij = F e i e j is the matrix representation of F for the e 1,..., e n frame. However F a = e i F e j e i aj 1.90 so that the matrix representation of F is F ij = F e j e i. If the e 1,..., e n are orthonormal then e j = e j for all j and F ij = F ji exactly the same as the adjoint in matrices. Other basic properties of the adjoint are: F a = e i a F e i = e i e i F a = F a 1.91 and F G a = e i e i F G a = e i a F G e i = e i F a G e i = e i e i G F a = G F a 1.92 so that F = F and F G = G F. A symmetric function is one where F = F. As an example consider F F F F = F F = F F 1.93

24 24 CHAPTER 1. BASIC GEOMETRIC ALGEBRA Inverse Another linear algebraic relation in geometric algebra is f 1 A = If I 1 A det f A G V 1.94 where f is the adjoint transformation defined by a f b = b f a a, b V and you have an explicit formula for the inverse of a linear transformation! 1.16 Commutator Product The commutator product of two multivectors A and B is defined as A B 1 AB BA An important theorem for the commutator product is that for a grade 2 multivector, A 2 = A 2, and a grade r multivector B r = B r we have A 2 B r = A 2 B r + A 2 B r + A 2 B r 1.96 From the geometric product grade expansion for multivectors we have A 2 B r = A 2 B r r+2 + A 2 B r r + A 2 B r r Thus we must show that A 2 B r r = A 2 B r 1.98 n Let e 1,..., e n be an orthogonal set for the vector space where B r = e 1... e r and A 2 = α lm e l e m so we can write A 2 B r = Now consider the following three cases n l<m=2 1. l and m > r where e l e m e 1... e r = e 1... e r e l e m 2. l r and m > r where e l e m e 1... e r = e 1... e r e l e m l<m=2 α lm e l e m e 1... e r 1.99

25 1.16. COMMUTATOR PRODUCT l and m r where e l e m e 1... e r = e 1... e r e l e m For case 1 and 3 e l e m commute with B r and the contribution to the commutator product is zero. In case 3 e l e m anticommutes with B r and thus are the only terms that contribute to the commutator. All these terms are of grade r and the theorem is proved. Additionally, the commutator product obeys the Jacobi identity A B C = A B C + B A C This is important for the geometric algebra treatment of Lie groups and algebras.

26 26 CHAPTER 1. BASIC GEOMETRIC ALGEBRA

27 Chapter 2 Examples of Geometric Algebra 2.1 Quaternions Any multivector A G 3, 0 may be written as A = α + a + B + βi 2.1 where α, β R, a V 3, 0, B is a bivector, and I is the unit pseudoscalar. The quaternions are the multivectors of even grades A = α + B 2.2 B can be represented as B = αi + βj + γk 2.3 where i = e 2 e 3, j = e 1 e 3, and k = e 1 e 2, and i 2 = j 2 = k 2 = ijk = The quaternions form a subalgebra of G 3, 0 since the geometric product of any two quaternions is also a quaternion since the geometric product of two even grade multivector components is a even grade multivector. For example the product of two grade 2 multivectors can only consist of grades 0, 2, and 4, but in G 3, 0 we can only have grades 0 and 2 since the highest possible grade is 3. 27

28 28 CHAPTER 2. EXAMPLES OF GEOMETRIC ALGEBRA 2.2 Spinors The general definition of a spinor is a multivector, ψ G p, q, such that ψvψ V p, q v V p, q. Practically speaking a spinor is the composition of a rotation and a dilation stretching or shrinking of a vector. Thus we can write where R is a rotor RR = 1. Letting U = R ψ we must solve ψvψ = ρrvr 2.5 UvU = ρv 2.6 U must generate a pure dilation. The most general form for U based on the fact that the l.h.s of equation 2.6 must be a vector is U = α + βi 2.7 so that UvU = α 2 v + αβ Iv + vi + β 2 IvI = ρv 2.8 Using vi = 1 n 1n 2 2 Iv, vi = 1 n 1 I v, and II = 1 q we get α 2 v + αβ n 1n 2 2 Iv + 1 n+q 1 β 2 v = ρv 2.9 If n 1 n 2 2 is even β = 0 and α 0, otherwise α, β 0. For the odd case ψ = R α + βi 2.10 where ρ = α n+q 1 β 2. In the case of G 1, 3 relativistic space time we have ρ = α 2 +β 2, ρ > Geometric Algebra of the Minkowski Plane Because of Relativity and QM the Geometric Algebra of the Minkowski Plane is very important for physical applications of Geometric Algebra so we will treat it in detail.

29 2.3. GEOMETRIC ALGEBRA OF THE MINKOWSKI PLANE 29 Let the orthonormal basis vectors for the plane be γ 0 and γ 1 where γ0 2 = γ1 2 = 1. 1 geometric product of two vectors a = a 0 γ 0 + a 1 γ 1 and b = b 0 γ 0 + b 1 γ 1 is so that and and Thus since I 2i = 1. Then the ab = a 0 γ 0 + a 1 γ 1 b 0 γ 0 + b 1 γ = a 0 b 0 γ a 1 b 1 γ a 0 b 1 a 1 b 0 γ 0 γ = a 0 b 0 a 1 b 1 + a 0 b 1 a 1 b 0 I 2.13 a b = a 0 b 0 a 1 b a b = a 0 b 1 a 1 b 0 I 2.15 I 2 = γ 0 γ 1 γ 0 γ 1 = γ 2 0γ 2 1 = e αi = = α i I i i=0 i=0 i! α 2i 2i! + α 2i+1 I 2i + 1! i= = cosh α + sinh α I 2.19 In the Minkowski plane all vectors of the form a ± = α γ 0 ± γ 1 are null a 2 ± = 0. One question to answer are there any vectors b ± such that a ± b ± = 0 that are not parallel to a ±. a ± b ± = α b ± 0 b ± 1 = 0 b ± 0 b ± 1 = 0 b ± 0 = ±b ± 1 Thus b ± must be proportional to a ± and the are no vectors in the space that can be constructed that are normal to a ±. Thus there are no non-zero bivectors, a b, such that a b 2 = 0. Conversely, if a b 0 then a b 2 > 0. Finally for the condition that there always exist two orthogonal vectors e 1 and e 2 such that we can state that neither e 1 nor e 2 can be null. 1 I = γ 0 γ 1 a b = e 1 e

30 30 CHAPTER 2. EXAMPLES OF GEOMETRIC ALGEBRA 2.4 Lorentz Transformation We now have all the tools needed to derive the Lorentz transformation with Geometric Algebra. Consider a two dimensional time-like plane with with coordinates t 2 and x 1 and basis vectors γ 0 and γ 1. Then a general space-time vector in the plane is given by x = tγ 0 + x 1 γ 1 = t γ 0 + x 1γ where the basis vectors of the two coordinate systems are related by and R is a Minkowski plane rotor α R = sinh 2 so that and γ µ = Rγ µ R µ = 0, α + cosh γ 1 γ Rγ 0 R = cosh α γ 0 + sinh α γ Rγ 1 R = cosh α γ 1 + sinh α γ Now consider the special case that the primed coordinate system is moving with velocity β in the direction of γ 1 and the two coordinate systems were coincident at time t = 0. Then x 1 = βt and x 1 = 0 so we may write tγ 0 + βtγ 1 = t Rγ 0 R 2.26 Equating components gives Solving for α and t t in equations 2.28 and 2.29 gives 2 We let the speed of light c = 1. t t γ 0 + βγ 1 = cosh α γ 0 + sinh α γ cosh α = t t 2.28 sinh α = t t β 2.29 tanh α = β 2.30

31 2.4. LORENTZ TRANSFORMATION 31 t t = γ = 1 1 β Now consider the general case of x, t and x, t giving tγ 0 + xγ 1 = t Rγ 0 R + x Rγ 1 R 2.32 = t γ γ 0 + βγ 1 + x γ γ 1 + βγ Equating basis vector coefficients recovers the Lorentz transformation t = γ t + βx x = γ x + βt 2.34

32 32 CHAPTER 2. EXAMPLES OF GEOMETRIC ALGEBRA

33 Chapter 3 Geometric Calculus - The Derivative 3.1 Definitions If F a if a multivector valued function of the vector a, and a and b are any vectors in the space then the derivative of F is defined by b F lim ɛ 0 F a + ɛb F a ɛ 3.1 then letting b = e k be the components of a coordinate frame with x = x k e k we are using the summation convention that the same upper and lower indices are summed over 1 to N we have e k F = lim ɛ 0 F x j e j + ɛe k F x j e j ɛ 3.2 Using what we know about coordinates gives or looking at as a symbolic operator we may write F = e j F x j = ej j F 3.3 = e j j 3.4 Due to the properties of coordinate frame expansions F is independent of the choice of the e k frame. If we consider x to be a position vector then F x is in general a multivector field. 33

34 34 CHAPTER 3. GEOMETRIC CALCULUS - THE DERIVATIVE 3.2 Derivatives of Scalar Functions If f x is scalar valued function of the vector x then the derivative is f = e k k f 3.5 which is the standard definition of the gradient of a scalar function remember that in an orthonormal coordinate system e k = e k. Using equation 3.5 we can show the following results for the gradient of some specific scalar functions f = x a, x k, xx f = a, e k, 2x Product Rule Let represent a bilinear product operator such as the geometric, inner, or outer product and note that for the multivector fields F and G we have so that k F G = k F G + F k G 3.7 F G = e k k F G + F k G = e k k F G + e k F k G 3.8 However since the geometric product is not communicative, in general The notation adopted by Hestenes is The convention of the overdot notation is F G F G + F G 3.9 F G = F G + F Ġ 3.10 i. In the absence of brackets, acts on the object to its immediate right ii. When the is followed by brackets, the derivative acts on all the the terms in the brackets. iii. When the acts on a multivector to which it is not adjacent, we use overdots to describe the scope. Note that with the overdot notation the expression Ȧ makes sense!

35 3.4. INTERIOR AND EXTERIOR DERIVATIVE Interior and Exterior Derivative The interior and exterior derivatives of an r-grade multivector field are simply defined as don t forget the summation convention A r A r r 1 = e k k A r 3.11 and Note that A r A r r+1 = e k k A r 3.12 A r = e i i e j j A r = e i e j i j A r = since e i e j = e j e i, but i j A r = j i A r. A r = e i i ej j A r = e i e j i j A r = ±e i e j i j A ri = ±e i e j i j A r I = ± e i e j i j A r I = Where indicates the dual of a multivector, A = AI I is the pseudoscalar and A = ±A I since I 2 = ±1, and we use equation 1.53 to exchange the inner and outer products. Thus for the general multivector field A built from sums of A r s we have A = 0 and A = 0. If φ is a scalar function we also have φ = e i i e j j φ = e i e j i j φ = Another use for the overdot notation would in the case where f x, a is a linear function of its second argument fx, αa + βb = αfx, a + βfx, b and a is a general function of position

36 36 CHAPTER 3. GEOMETRIC CALCULUS - THE DERIVATIVE ax = a i x e i. Now calculate fx, a = e k fx, a = ek xk x f x, a i x e k i 3.16 = e k a i x fx, e x k i 3.17 = e k ai x fx, e i + a i e k k x fx, e i 3.18 k = e k f x, a + a i e k x k x fx, e i 3.19 k Defining f a a i e k x fx, e i = e k k x Then suppressing the explicit x dependence of f we get f a a = f a e k f x k fx, a k a=constant Other basic results examples are x A r = ra r 3.22 x A r = n r A r 3.23 A r ẋ = 1 r n 2r A r 3.24 The basic identities for the case of a scalar field α and multivector field F are αf = α F + α F 3.25 if f 1 and f 2 are vector fields αf = α F + α F 3.26 αf = α F + α F 3.27 f 1 f 2 = f 1 f 2 f 2 f and finally if F r is a grade r multivector field F r I = F r I 3.29 where I is the psuedoscalar for the geometric algebra.

37 3.5. DERIVATIVE OF A MULTIVECTOR FUNCTION Derivative of a Multivector Function For a vector space of dimension N spanned by the vectors u i the coordinates of a vector x are the x i = x u i so that x = x i u i summation convention is from 1 to N. Curvilinear coordinates for that space are generated by a one to one invertible differentiable mapping from x 1,..., x N θ 1,..., θ N where the θ i are called the curvilinear coordinates. If the mapping is given by x θ 1,..., θ N = x i θ 1,..., θ N u i then the basis vectors associated with the transformation are given by e k = x θ = xi k θ u k i 3.30 The one critical relationship that is required to express the geometric derivative in curvilinear coordinated is e k = θk x i ui 3.31 The proof is e j e k = xm θ j = xm θ j = xm θ j θ k x u n m u n 3.32 θ k x n δn m 3.33 θ k x m 3.34 = θk θ j = δk j 3.35 We wish to express the geometric derivative of an R-grade multivector field F R in terms of the curvilinear coordinates. Thus F R = u i F R x i = u i θk FR x i θ k = ek F R θ k 3.36 Note that if we start by defining the e k s the reciprocal frame vectors e k can be calculated geometrically we do not need the inverse partial derivatives. Now define a new blade symbol by e [i1,...,i R ] = e i1... e ir 3.37 and represent an R-grade multivector function F by F = F i 1...i R e [i1,...,i R ] 3.38

38 38 CHAPTER 3. GEOMETRIC CALCULUS - THE DERIVATIVE Then Define F = F i 1...i R θ k e k e [i1,...,i R ] + F i 1...i R e k θ k e [i 1,...,i R ] 3.39 C { } e [i1,...,i R ] e k θ e k [i 1,...,i R ] 3.40 Where C { e [i1,...,i R ]} are the connection multivectors for each base of the geometric algebra and we can write F = F i 1...i R e k e θ k [i1,...,i R ] + F i 1...i R C { } e [i1,...,i R ] 3.41 Note that all the quantities in the equation not dependent upon the F i 1...i R can be directly calculated if the e k θ 1,..., θ N is known so further simplification is not needed. In general the e k s we have defined are not normalized so define e k = ê k = e k e k e 2 k and note that ê 2 k = ±1 depending upon the metric. Note also that since ê j ê k = e j e j ê k = e k e k 3.44 ek e k = δ j e j k e k = δj k 3.45 so that if F R is represented in terms of the normalized basis vectors we have and the geometric derivative is now F R = F i 1...i R R ê [i1,...,i R ] 3.46 F = F i 1...i R θ k ê k e k ê[i 1,...,i R ] + F i 1...i R Ĉ { } ê [i1,...,i R ] 3.47 and Ĉ { } ê k ê [i1,...,i R ] = e k θ k ê[i 1,...,i R ] 3.48

39 3.5. DERIVATIVE OF A MULTIVECTOR FUNCTION Spherical Coordinates For spherical coordinates the coordinate generating function is: x = r sin θ u z + cos θ cos φ u x + sin φ u y 3.49 so that e r = cos θ cos φ u x + sin φ u y + sin θ u z 3.50 e θ = r sin θ cos φ u x + sin φ u y + cos θ u z 3.51 e φ = r cos θ sin φ u x + cos φ u y 3.52 where e r = 1 e θ = r e φ = r cos θ 3.53 and ê r = cos θ cos φ u x + sin φ u y + sin θ u z 3.54 ê θ = sin θ cos φ u x + sin φ u y + cos θ u z 3.55 ê φ = sin φ u x + cos φ u y 3.56 the connection mulitvectors for the normalize basis vectors are Ĉ {ê r } = 2 r 3.57 cos θ Ĉ {ê θ } = r sin θ + 1 r êr ê θ 3.58 Ĉ {ê φ } = 1 cos θ r êr ê φ + ê φ r sin θêθ 3.59 cos θ Ĉ {ê r ê θ } = + 1 r sin θêr r êθ 3.60 Ĉ {ê r ê φ } = 1 r êφ cos θ r sin θêr ê θ ê φ 3.61 Ĉ {ê θ ê φ } = 2 r êr ê θ ê φ 3.62 Ĉ {ê r ê θ ê φ } =

40 40 CHAPTER 3. GEOMETRIC CALCULUS - THE DERIVATIVE For a vector function A using equation 3.41 and that A = A + A 1 A = A θ cos θ + φ A φ + 1 2A r + θ A θ + r A r r sin θ r 3.64 = 1 r 2 r r 2 A r 1 + θ sin θ A θ + φ A φ r sin θ 3.65 A = I A 3.66 θ A φ 1 = + A φ cos θ φ A θ ê r 3.67 r r sin θ φ A r + r sin θ Aφ r ra φ ê θ 3.68 A θ + r + ra θ θa r ê φ 3.69 r A = 1 θ sin θ A φ φ A θ ê r 3.70 r sin θ r sin θ φa r r ra φ ê θ r ra θ θ A r ê φ 3.72 r These are the standard formulas for div and curl in spherical coordinates. 3.6 Analytic Functions Starting with G 2, 0 and orthonormal basis vectors e x and e y so that I = e x e y and I 2 = 1. Then we have r = xe x + ye y 3.73 Map r onto the complex number z via = e x x + e y y 3.74 z = x + Iy = e x r 3.75

41 3.6. ANALYTIC FUNCTIONS 41 Define the multivector field ψ = u + Iv where u and v are scalar fields. Then u ψ = x v v e x + y x + u e y 3.76 y Thus the statement that ψ is an analytic function is equivalent to ψ = This is the fundamental equation that can be generalized to higher dimensions remembering that in general that ψ is a multivector rather than a scalar function! To complete the connection with complex analysis we define z = x Iy z = 1 2 x I y, z = 1 2 x + I y 3.78 so that z z = 1, z z = 0 z z = 0, z 3.79 z = 1 An analytic function is one that depends on z alone so that we can write ψ x + Iy = ψ z and equivalently ψ z z = x + I ψ = 1 y 2 e x ψ = Now it is simple to show why solutions to ψ = 0 can be written as a power series in z. First z = e x r r = e x e x x + e r ye x y = e x e x e x + e y e x e y = e x e x = so that z z 0 k = k e x r z 0 z z 0 k 1 =

42 42 CHAPTER 3. GEOMETRIC CALCULUS - THE DERIVATIVE

43 Chapter 4 Geometric Calculus - Integration 4.1 Line Integrals If F x is a multivector field and x λ is a parametric representation of a vector path curve then the line Integral of F along the path x is defined to be F x dx dλ dλ = F dx lim n n i=1 F i x i 4.1 where x i = x i x i 1, F i = 1 2 F x i 1 + F x i 4.2 if x n = x 1 the path is closed. Since dx is a vector, that is F x dx dλ dx F x, a more general dλ line integral would be F x dx G x dλ = F x dx G x 4.3 dλ The most general form of line integral would be L λ x; x dλ = L dx 4.4 where L a = L a; x = is a multivector-valued linear function of a. The position dependence in L can often be suppressed to streamline the notation. 43

44 44 CHAPTER 4. GEOMETRIC CALCULUS - INTEGRATION 4.2 Surface Integrals The next step is a directed surface integral. Let F x be a multivector field and let a surface be parametrized by two coordinates x x 1, x 2. Then we can define a directed surface measure by A directed surface integral takes the form F dx = dx = x x 1 x x 2 dx1 dx 2 = e 1 e 2 dx 1 dx F e 1 e 2 dx 1 dx In order to construct some of the more important proof it is necessary to express the surface integral as the limit of a sum. This requires the concept of a triangulated surface as shown Each Figure 4.1: Triangulated Surface triangle in the surface is described by a planar simplex as shown The three vertices of the planar simplex are x 0, x 1, and x 2 with the vectors e 1 and e 2 defined by so that the surface measure of the simplex is e 1 = x 1 x 0, e 2 = x 2 x X 1 2 e 1 e 2 = 1 2 x 1 x 2 + x 2 x 0 + x 0 x with this definition of X we have F dx lim n n k=1 F k X k 4.9 where F k is the average of F over the k th simplex.

45 4.3. DIRECTED INTEGRATION - N-DIMENSIONAL SURFACES 45 Figure 4.2: Planar Simplex 4.3 Directed Integration - n-dimensional Surfaces k-simplex Definition In geometry, a simplex or k-simplex is an k-dimensional analogue of a triangle. Specifically, a simplex is the convex hull of a set of k + 1 affinely independent points in some Euclidean space of dimension k or higher. For example, a 0-simplex is a point, a 1-simplex is a line segment, a 2-simplex is a triangle, a 3- simplex is a tetrahedron, and a 4-simplex is a pentachoron in each case including the interior. A regular simplex is a simplex that is also a regular polytope. A regular k-simplex may be constructed from a regular k 1-simplex by connecting a new vertex to all original vertices by the common edge length k-chain Definition Algebraic Topology A finite set of k-simplexes embedded in an open subset of R n is called an affine k-chain. The simplexes in a chain need not be unique, they may occur with multiplicity. Rather than using standard set notation to denote an affine chain, the standard practice to use plus signs to separate each member in the set. If some of the simplexes have the opposite orientation, these are prefixed by a minus sign. If some of the simplexes occur in the set more than once, these are prefixed with an integer count. Thus, an affine chain takes the symbolic form of a sum with integer coefficients.

46 46 CHAPTER 4. GEOMETRIC CALCULUS - INTEGRATION Simplex Notation If x 0, x 1,..., x k is the k-simplex defined by the k + 1 points x 0, x 1,..., x k. This is abbreviated by x k = x 0, x 1,..., x k 4.10 The order of the points is important for a simplex, since it specifies the orientation of the simplex. If any two adjacent points are swapped the simplex orientation changes sign. The boundary operator for the simplex is denoted by and defined by x k k 1 i x 0,..., x i,..., x k k i=0 To see that this make sense consider a triangle x 3 = x 0, x 1, x 2. Then x 3 = x 1, x 2 2 x 0, x x 0, x 1 2 = x 1, x x 2, x x 0, x each 2-simplex in the boundary 2-chain connects head to tail with the same sign. Now consider the boundary of the boundary 2 x 3 = x 1, x x 2, x x 0, x 1 2 = x 1 1 x x 2 1 x x 0 1 x 1 1 = We need to prove is that in general 2 x k = 0. To do this consider the boundary of the i th term on thr r.h.s. of equation 4.11 letting A k 2 ij = x 0,..., x i,..., x j,..., x k k 1. Then x 0,..., x i,..., x k k 1 = i = 0 : 0 < i < k : i = k : j=0 k j=1 i 1 1 j A k 2 ij + 1 j 1 A k 2 ij k j=i+1 k 1 1 j A k 2 ij j=0 1 j 1 A k 2 ij 4.14

47 4.3. DIRECTED INTEGRATION - N-DIMENSIONAL SURFACES 47 The critical point in equation 4.14 is that the exponent of 1 in the second term on the r.h.s. is not j, but j 1. The reason for this is that when x i was removed from the simplex the vertices were not renumbered. We can now express the boundary of the boundary in terms of the following matrix elements B k 2 ij = 1 i+j A k 2 ij as 2 x k = k j=1 1 j 1 A k 2 0j i=1 j=0 k k j=0 k 1 i i 1 j A k 2 ij + = k j=1 k 1 i 1 + i=1 k 1 B k 2 0j + j=0 B k 2 ij j=0 B k 2 kj k 1 k i=1 j=i+1 1 j A k 2 kj k j=i+1 1 j 1 A k 2 ij B k 2 ij = The consider B k 2 ij as a matrix i-row index, j-column index. The matrix is symmetrical and in equation 4.15 you are subtracting all the elements above the main diagonal from the elements below the main diagonal so that 2 x k = 0 and the boundary of a boundary of a simplex is zero. Now add geometry to the simplex by defining the vectors and the directed volume element e i = x i x 0, i = 1,..., k, 4.16 X = 1 k! e 1 e k 4.17 We now wish to prove that x k dx = X 4.18 Any point in the simplex can be written in terms of the coordinates λ i as x = x 0 + k λ i e i 4.19 i=1

48 48 CHAPTER 4. GEOMETRIC CALCULUS - INTEGRATION with restrictions First we show that or k 0 λ i 1 and λ i i=1 dx = x k e 1 e k dλ 1 dλ k = X x k 4.21 dλ 1 dλ k = 1 x k k! 4.22 define Λ j = 1 j i=1 λi Note that Λ 0 = 1. From the restrictions on the λ i s we have x k dλ 1 dλ k = Λ0 0 dλ 1 Λ1 0 Λk 1 dλ 2 dλ k To prove that the r.h.s of equation 4.23 is 1/k! we form the following sequence and use induction to prove that V j is the result of the first j partial Integrations of equation 4.23 V j = 1 j! Λ k j j 4.24 Then V j+1 = = = = Λk j 1 0 Λk j 1 dλ k j V j dλ k j 1 Λk j 1 λ k j j 0 j! 1 [ Λk j 1 λ k j ] j+1 Λ k j 1 j + 1 j! 0 1 j + 1! Λ k j 1 j so that V k = 1/k! and the assertion is proved. Now let there be a multivector field F x that assumes the values F i = F x i at the vertices of the simplex and define the interpolating function f x = F 0 + k λ i F i F i=1

49 4.3. DIRECTED INTEGRATION - N-DIMENSIONAL SURFACES 49 We now wish to show that To prove this we must show that x k f dx = 1 k + 1 k F i X = F X 4.27 i=0 x k λ i dx = 1 k + 1 X, λi 4.28 To do this consider the integral equation 4.25 with V j replaced by λ k j V j Λk j 1 0 dλ k j λ k j V j = = Λk j 1 0 dλ k j 1 j! λk j Λ k j 1 λ k j j 1 j + 2! Λ k j 1 j Note that since the extra λ i factor occurs in exactly one of the subintegrals for each different λ i 1 the final result of the total integral is multiplied by a factor of since the weight of the total k+1 integral is now 1 and the assertion equation 4.28 and hence equation 4.27 is proved. k+1! Now summing over all the simplices making up the directed volume gives volume F dx = lim n The most general statement of equation 4.30 is volume L dx = lim n n i=1 F i X i 4.30 n L i X i 4.31 where L F n ; x is a position dependent linear function of a grade-n multivector F n and L i is the average value of L dx over the vertices of each simplex. i=1 An example of this would be L F n ; x = G x F n H x 4.32 where G x and H x are multivector functions of x.

50 50 CHAPTER 4. GEOMETRIC CALCULUS - INTEGRATION 4.4 Fundamental Theorem of Geometric Calculus Now prove that the directed measure of a simplex boundary is zero x k = x 0,..., x k k = Start with a planar simplex of three points x 0, x 1, x 2 2 = x 1, x 2 1 x 0, x x 0, x so that x 0, x 1, x 2 2 = x 2 x 1 x 2 x 0 + x 1 x 0 = We shall now prove equation 4.33 via induction. First note that i = 0 : x i k 1 = 0 < i k 1 : 1 k 1 x 0 k 2 x k x 1 1 k 1 x i k 2 x k x and so that where x k k 1 = 1 k 1! x 1 x 0 x k 1 x x k = 1 k 1 1 i x i k 1 k 2 x k x 0 + C i=1 C = 1 k 1 x 0 k 2 x k x k x k k if we let δ = x 0 x 1 we can write C = 1 k 1 x 0 k 2 x k x k 1 x 0 k 2 δ + 1 k x k k

51 4.4. FUNDAMENTAL THEOREM OF GEOMETRIC CALCULUS 51 Then Thus However so that and 1 x 0 k 2 δ = k 2! x 2 x 1 x k 1 x 1 δ = k 2! x 2 x 0 + δ x k 1 x 0 + δ δ = k 2! x 2 x 0 x k 1 x 0 δ 4.42 = 1k 2 k 2! δ x 2 x 0 x k 1 x = 1k 1 k 2! x 1 x 0 x 2 x 0 x k 1 x k 1 x 0 k 2 δ = 4.45 = 1k k 1! x 1 x 0 x 2 x 0 x k 1 x = 1 k x k k x k 1 k x k k 1 = 1 k 1 x 0 k 2 δ 4.48 C = 1 k 1 x 0 k 2 x k x k 1 1 = 1 i x i k 1 k 2 x k x 0 i=0 1 = x k 1 k 1 x k x 0 = We have proved equation Note that to reduce equation 4.49 we had to use that for any set of vectors δ, y 1,..., y k we have δ y 1 + δ y k + δ = δ y 1 y k 4.51

52 52 CHAPTER 4. GEOMETRIC CALCULUS - INTEGRATION Think about equation It s easy to prove δ δ = 0! Equation 4.33 is sufficient to prove that the directed integral over the surface of simplex is zero x k ds = k 1 i dx = x k = i=0 The characteristics of a general volume are: x i k 1 1. A general volume is built up from a chain of simplices. 2. Simplices in the chain are defined so that at any common boundary the directed areas of the bounding faces are equal and opposite. 3. Surface integrals over two simplices cancel over their common face. 4. The surface integral over the boundary of the volume can be replaced by the sum of the surface integrals over each simplex in the chain. If the boundary of the volume is closed we have ds = lim n n a=1 ds a = Where ds a is the surface Integral over the a th simplex. Implicit in equation 4.53 is that the surface is orientated, simply connected, and closed. The next lemma to prove that if b is a constant vector on the simplex x k then b x ds = b x k = b X 4.54 x k The starting point of the lemma is equation First define b = k b i e i, 4.55 i=1 where the e i s are the reciprocal frame to e i = x i x 0 so that x x 0 = k λ i e i, 4.56 i=1

53 4.4. FUNDAMENTAL THEOREM OF GEOMETRIC CALCULUS 53 and Substituting into equation 4.28 we get i=1 x k b i = b e i, 4.57 k λ i b i = b x x i=1 k b i λ i dx = b x x 0 dx = 1 k + 1 x k Rearranging terms gives k b x dx = 1 b x i x 0 + k + 1 b x 0 X x k k + 1 i=1 k = b k + 1 x i X i=0 k b e i X 4.59 = b x X 4.60 Now using the definition of a simplex boundary equation 4.11 and equation 4.60 we get b x ds = 1 k x k k i=0 i=1 1 i b x x i + + x k x i k The coefficient multiplying the r.h.s. of equation 4.61 is 1 k and not 1 k + 1 because both x x i + + x k and x i k 1 refer to k 1 simplices the boundary of x k is the sum of all the simplices x i k with proper sign assigned. Now to prove equation 4.54 we need to prove one final purely algebraic lemma k 1 i b x x i + + x k x i k 1 = i=0 Begin with the definition of the l.h.s. of equation k 1! b e 1 e k 4.62 C = k 1 i b x x i + + x k x i k 1 = i=0 k C i 4.63 i=0

54 54 CHAPTER 4. GEOMETRIC CALCULUS - INTEGRATION where C i is defined by { i = 0 : b x1 + + x k x 0 C i = k 1 0 < i k : 1 i b x x i + + x k x i k 1 } 4.64 now define E k = e 1 e k so that using equation 1.61 from the section on reciprocal frames and 1 i 1 e i E k = e 1 ĕ i e k, 0 < i k 4.65 C 0<i k = The main problem is in evaluating C 0 since x 0 k 1 = using e i = x i x 0 reduces equation 4.67 to x 0 k 1 = 1 k 1! b x x i + + x k e i E k k 1! x 2 x 1 x k x k 1! e 2 e 1 e k e but equation 4.68 can be expanded into equation The critical point in doing the expansion is that in generating the sum on the r.h.s. of the first line of equation 4.69 all products containing x 1 of course all terms in the sum contain x 1 exactly once since we are using the outer product are put in normal order by bringing the x 1 factor to the front of the product thus requiring the factor of 1 i in each term in the sum. or e 2 e 1 e k e 1 = e 2 e k = = k 1 i e 1 e 2 ĕ i e k i=2 k 1 i 1 e 1 e 2 ĕ i e k i=1 k e i E k 4.69 i=1 x 0 k 1 = 1 k 1! k e i E k 4.70 i=1

55 4.4. FUNDAMENTAL THEOREM OF GEOMETRIC CALCULUS 55 from equation 4.69 we have C = = = 1 k 1! 1 k 1! 1 k 1! k b x i x 0 e i E k i=1 k b e i e i E k i=1 k b i e i E k i=1 1 = k 1! be k 1 = k 1! b E k + b E k 1 = k 1! b E k 4.71 and equation 4.62 is proved which means substituting equation 4.62 into equation 4.61 proves equation The Fundamental Theorem At Last! The simplicial coordinates, λ i, can be expressed in terms of the position vector, x, and the frame vectors of the simplex, e i = x i x 0. Let the vectors e j be the reciprocal frame to e i e i e j = δ j i. Then λ i = e i x x and let f x be an affine multivector function of x equation 4.26 which interpolates, F, a differentiable multivector function of x on the simplex. Then But f x ds = x k = k F i F 0 e i x x 0 ds x k k F i F 0 e i X 4.73 i=1 i=1 f x λ i = F i F

56 56 CHAPTER 4. GEOMETRIC CALCULUS - INTEGRATION so that the surface integral of equation 4.73 can be rewritten f x ds = x k = k F i F 0 e i X i=1 k i=1 f λ i ei X = f X 4.75 If we now sum equation 4.75 over a chain of simplices realizing that the interpolated function f x takes on the same value over the common boundary of two adjacent simplices, since f x is only defined by the values at the common vertices. In forming a sum over a chain, all of the internal faces cancel and only the surface integral over the boundary remains. Thus f x ds = a f X a 4.76 with the sum running over all simplices in the chain. Taking the limit as more points are added and each simplex is shrunk in size we obtain the first realization of the fundamental theorem of geometric calculus V F ds = V F dx 4.77 We can write dx instead of dx since the vector is totally within the vector space defined by dx so that dx = 0. The method of proof used can also be applied to the form V ds G = A more general statement of the theorem is as follows: V dx Ġ 4.78 Let L A k 1 = L A k 1 ; x be a linear functional of a multivector A k 1 of grade k 1 and a general function of position x which returns a general multivector. The linear interpolation approximation of L over a simplex is defined by: L A L A; x 0 + k λ i L A; x i L A; x i=1 Then the integral over a simplex is Note that since integration is a linear operation, a summation,

57 4.5. EXAMPLES OF THE FUNDAMENTAL THEOREM 57 the integral can be placed inside L A since L A is linear in A k L ds = L ds; x 0 + L λ i ds; x i x k = i=1 k L e i X; x i L e i X; x 0 i=1 = L X k i=1 L λ i ds; x Taking the limit of a sum of simplicies gives L ds = V V L dx Examples of the Fundamental Theorem Divergence and Green s Theorems As a specific example consider L A = JAI 1 where J is a vector, I is the unit pseudoscalar for a n-dimensional vector space, and A is a multivector of grade n 1. Then equation 4.81 gives J dxi 1 = J dx = JdSI V V we have dx = I dx where dx is the scalar measure of the volume. The normal to the surface, n, is defined by n ds = dsi where ds is the scalar valued measure over the surface. With this definition we get J dx = n J ds 4.84 V Now using the form of the Fundamental Theorem of Geometric Calculus in equation 4.78 and let G be the vector J in two-dimensional Euclidean space and noting that since da is a pseudoscalar for 2-D ds is the boundary measure and da is the volume measure it anticommutes with vectors in two dimensions we get - dsj = daj = J da 4.85 A A A V V

58 58 CHAPTER 4. GEOMETRIC CALCULUS - INTEGRATION In 2-D Cartesian coordinates da = Idxdy and ds = ni ds so that dsj = J Idxdy or Letting J = P e x + Qe y and n = n x e x + n y e y we get - so that A A A A A A nij ds = J Idxdy A nji ds = J Idxdy 4.87 A nj ds = J dxdy 4.88 A nj = n J + n x Q n y P e x e y 4.89 Q J = J + x P e x e y 4.90 y n J ds = A n x Q n y P ds e x e y = but dy = n x ds and dx = n y ds so that P dx + Qdy = A which is Green s theorem in the plane. A A A Jdxdy 4.91 Q x P dxdye x e y y 4.92 Q x P dxdy 4.93 y Cauchy s Integral Formula In Two Dimensions Complex Plane Consider a two dimensional euclidean space with vectors r = xe x + ye y. Then the complex number z corresponding to r is z = e x r = x + ye x e y = x + yi 4.94 z = re x = x ye x e y = x yi 4.95 zz = z z = x 2 + y 2 = r

59 4.5. EXAMPLES OF THE FUNDAMENTAL THEOREM 59 Thus even grade multivectors correspond to complex numbers since I 2 = 1 and the reverse of z, z corresponds to the conjugate of z. Even grade multivectors commute with ds, since ds is proportional to I. Let ψ r be an even multivector function of r, then ψds = dsψ = r ψdλ 4.97 λ but the complex number z is given by z = e x r and e x dsψ = ψdz = e x ψds Thus if a function ψ is analytic, ψ = 0 and ψdz = 0. Now note that this will be proved for the N-dimensional case in the next example r a 2 = 2πδ r a 4.99 r a where a = e x a. Now let ψ = r a r a 2 e xf e x r so that e x dsψ = e x = = = dz dz r a ds r a 2 e xf e x r r a r a 2 e xf e x r z a z a z a f z f z dz z a

60 60 CHAPTER 4. GEOMETRIC CALCULUS - INTEGRATION and f z z a dz = e x r a r a 2 e xf e x r ds = e x 2πδ r a e x f e x r + f e x r = 2πIfa + = 2πIfa + = 2πIfa + e x f e x r z a z a 2 I ds 1 e x f e x r z a I ds x + I f z y r a r a 2 e x I ds 1 I ds z a If f z = 0 we have - f z dz = 2πIf a z a which is the Cauchy integral formula. If f z is not zero we can write the more general relation - f f 2πIf a = z a dz 2 1 I ds z z a since and z = 1 2 x + I y z = 1 2 x I y Green s Functions in N-dimensional Euclidean Spaces Let ψ be an even multivector function or let N be even so that ψ commutes with I. The analog of an analytic function in N-dimensions is ψ = 0. The Green s function of the operator is S N = 2π N/2 /ΓN/2 is the hyperarea of the unit radius sphere in N dimensions G x; y = lim ɛ 0 x y S N x y N + ɛ

61 4.5. EXAMPLES OF THE FUNDAMENTAL THEOREM 61 So that To prove equation we need to use lim x G x; y = δ x y ɛ 0 x x y = N and x x y M = M x y x y M 2 So that x G = 1 1x } 1 x { x x y N + ɛ y + x y N + ɛ x y S N = N x y N 1 S 2 + N x y N + ɛ x y N + ɛ = N ɛ S 2 = x G = Ġ x = Ġ x N x y N + ɛ so what must be proved is that N lim ɛ 0 S N First define the volume B τ τ > 0 by ɛ x y N + ɛ 2 = δ x y x B τ x τ and let y = 0 and calculate Note that we use dv since in our notation dv = I dv and the oriented dv is not needed in this proof. Also r = x. N B S N ɛ x N + ɛ 2 dv = 0 Nɛr N 1 r N + ɛ 2 dr ɛ = 0 r N + ɛ 2 d r N [ ] ɛ = r N + ɛ 0 =

62 62 CHAPTER 4. GEOMETRIC CALCULUS - INTEGRATION Thus the first requirement of a delta function is fulfilled. Now let φx be a scalar test function on the N-dimensional space and S a point set in the space and define the functions max φ, S = {max φx x S} and min φ, S = {min φx x S} and calculate the integral N S N B τ ɛ 2 dv = x N + ɛ τ 0 [ = ɛ r N + ɛ 2 d r N ɛ r N + ɛ = 1 ɛ τ N + ɛ ] τ and note that N S N B B τ ɛ 2 dv = x N + ɛ ɛ τ N + ɛ Thus τ > 0 we have N lim ɛ 0 S N B B τ ɛ 2 dv = x N + ɛ and N lim ɛ 0 S N B τ ɛ 2 dv = x N + ɛ Thus min φ, B B τ N ɛ S 2 dv N B B τ x N + ɛ N ɛφx S 2 dv N B B τ x N + ɛ max φ, B B τ N ɛ S 2 dv N B B τ x N + ɛ

63 4.5. EXAMPLES OF THE FUNDAMENTAL THEOREM 63 and min φ, B τ N ɛ S 2 dv N B τ x N + ɛ N ɛφx S 2 dv N B τ x N + ɛ max φ, B τ N ɛ S 2 dv N B τ x N + ɛ Thus and N lim ɛ 0 S N N min φ, B τ lim ɛ 0 S N B B τ B τ ɛφx 2 dv = x N + ɛ ɛφx 2 dv max φ, B τ x N + ɛ Finally lim lim τ 0 ɛ 0 N S N B τ ɛφx 2 dv = φ x N + ɛ and we have proved equation since lim maxφ, B τ minφ, B τ = τ 0 Now in the fundamental theorem of geometric calculus let LA = GAψ so that remember that ψ commutes with dv and x ψ = 0 L x dv = Ġ x dv ψ + G x dv ψ = Ġ x ψ + G x ψ I dv = Ġ x ψi dv = x GψI dv 4.123

64 64 CHAPTER 4. GEOMETRIC CALCULUS - INTEGRATION and V GdSψ = δx y ψx I dv V = ψy I = Iψy or ψy = I 1 S N because ψ is a monogenic function ψ = 0. V x y dsψx N x y

65 Chapter 5 Geometric Calculus on Manifolds 5.1 Definition of a Vector Manifold The definition of a manifold that we will use is - A vector manifold generalized surface is a set of points labeled by vectors lying in a geometric algebra of arbitra dimension and signature. If we consider a path in the surface x λ, the tangent vector is defined by x x λ x λ 0 + ɛ x λ 0 λ = lim 5. ɛ 0 λ0 ɛ and the path length s λ2 λ 1 x x dλ The Pseudoscalar of the Manifold Now introduce a set of paths in the surface all passing through the same point x. These paths define a set of tangent vectors {e 1,..., e n }. We assume these paths have been picked so that the vectors are independent and form a basis for the tangent space at point x. The outer product of 65

66 66 CHAPTER 5. GEOMETRIC CALCULUS ON MANIFOLDS the tangent vectors form the pseudoscalar, I x, for the tangent space Thus I x e 1 e 2 e n e 1 e 2 e n 5.3 I 2 = ±1 5.4 We require that for any point on the manifold the denominator of equation 5.3 is nonzero. We also assume that for all points on the manifold that I x is continuous, differentiable, singled valued, and has the same grade everywhere The Projection Operator Define the projection operator P A operating on any multivector A in the embedding multivector space as P A A I x I 1 x 5.5 We can show that P A extracts those components of A that lie in the geometric algebra defined by I x. Since P A is linear in A if we show that if P A r projects correctly for an r-grade multivector it will do so for a general mulitvector. If n is the dimension of the tangent space and A r is a pure grade multivector we can write equation 5.5 as P A r = A r I x r n I 1 x 5.6 Now consider the blades that make up the components of A r. They will either consist only of blades formed from the tangent vectors e i or they will contain at least one basis vector that is not a tangent vector. In the first case and A r I x r n = A r I x 5.7 P A r = A r 5.8 In the second case there is no component of A r I x of grade r n and P A r = This is easily seen if one constructs at point x an orthogonal basis o i o j = δ ij o 2 i for the tangent space {o 1,..., o n } and an orthogonal basis for the remainder of the embedding space {o n+1,..., o m }. Then any component blade of A r is of the form A i 1,i 2,...,i r r o i1 o i2... o ir 5.10

67 5.1. DEFINITION OF A VECTOR MANIFOLD 67 where i 1 < i 2 < < i r. If i j n 1 j r then A i 1,i 2,...,i r r o i1 o i2... o ir I = A i 1,i 2,...,i r r o i1 o i2... o ir I 5.11 and A i 1,i 2,...,i r r o i1 o i2... o ir I I 1 = A i 1,i 2,...,i r r o i1 o i2... o ir 5.12 and P A i 1,i 2,...,i r r o i1 o i2... o ir = A i 1,i 2,...,i r r o i1 o i2... o ir 5.13 If any i j > m then A i 1,i 2,...,i r r o i1 o i2... o ir I contains no grade r n and P A i 1,i 2,...,i r r o i1 o i2... o ir = The Exclusion Operator For an arbitrary multivector A the exclusion operator P A is defined by P A A P A The Intrinsic Derivative Given a set of tangent vectors {e i } spanning the tangent space the derivative intrinsic to the manifold is defined everywhere by Also note that e i e i = P 5.16 P = 5.17 When we write P or P the or is not differentiating the I x in the P operator anymore than is differentiating dx in the fundamental theorem of Geometric Calculus. We also note that if the vector a is in the tangent space that a = a 5.18 and that a is a scalar operator that gives the directional derivative in the a direction. Also since it is scalar it satisfies Leibniz s rules without using the dot notation remember the convention that if parenthesis are not present the operator precedence is dot product then wedge product then geometric product. a AB = a A B + A a B 5.19

68 68 CHAPTER 5. GEOMETRIC CALCULUS ON MANIFOLDS The Covariant Derivative The operator is entirely within the tangent space and if the general multivector function A x is also entirely within the tangent space, it is still possible even likely that A is not entirely within the tangent space. We need a covariant derivative D that will result in a multivector entirely within the tangent space. This can be done by defining so that a DA x P a A x 5.20 a A = P a A + P a A = a DA + P a A 5.21 Again since a D is a scalar operator we have a D AB = P a AB = a DA B + A a DB 5.22 A component expansion of D is given in the usual way by do not forget the summation convention D = e i e i D 5.23 and and if α x is a scalar function on the manifold then DA r = e i e i DA r = P A r 5.24 D A r DA r r D A r DA r r α x = Dα x 5.27 because in equation 5.27 no basis vectors are differentiated. To relate and D if the function operated on is not a scalar first construct a normalized basis {e i } for the tangent space at point x. Then I = e 1 e 2... e n and I 2 = ± and since a and a D are scalar operators we can move them across the wedge products without any problems n a I I 1 = e 1... a De i + P a e i... e n I = a DI I 1 + i=1 n 1 i 1 P a e i e 1... ĕ i... e n I i=1 = a DI I 1 + P a e i e i 5.31

69 5.1. DEFINITION OF A VECTOR MANIFOLD 69 We go from equation 5.30 to equation 5.31 by using equation 1.61 on page 33 in the section on reciprocal frames. Since a D I is a grade n multivector in the tangent space it must be proportional to I and thus commute with I so that a I I = I a I. Also I 1 = ±I so that we have Thus a DI I 1 = ± a DI I = ± 1 a DI I + I a DI 2 = ± 1 a D I 2 2 = a I = P a e i e i I S a I 5.33 Where S a is the shape tensor associated with the manifold. Since S a is a bivector we can write A B = AB BA /2 a I = I S a 5.34 since S a I = S a I = and by equation 1.96 page 58. Given that a x and b x are vector fields on the manifold both are in the tangent space at point x, form the expressions in equation 5.36 remember that for any three vectors u, v, and w we have u v w = u v w u w v b S a = b e i P a e i = b e i P a e i b P a e i e i = b j e j e i P a e i = b j δjp i a e i = P a b i ė i 5.36 but P a b = P a b i e i = P a ḃi e i + a b i ė i = P a b i ė i 5.37

70 70 CHAPTER 5. GEOMETRIC CALCULUS ON MANIFOLDS and Thus b S a = P a b 5.38 a b = P a b + P a b = a Db + b S a 5.39 and using the fact that the dot product of a vector and bivector are antisymmetric Now consider the expression a D b 1... b r = = a Db = a b + S a b 5.40 r b 1... a b i + S a b i... b r i=1 r b 1... a b i... b r + i=1 r b 1... S a b i... b r i=1 = a b 1... b r + 1 r b 1... S a b i b i S a... b r 2 i=1 = a b 1... b r S a b 1... b r b 1... b r S a r 1 r 1 b 1... S a b i... b r b 1... b i S a... b r + i=2 i=2 1 2 b 1... b r 1 S a b r b 1 S a b 2... b r = a b 1... b r S a b 1... b r b 1... b r S a r 1 r 2 b 1... S a b i... b r b 1... b i S a... b r i=3 = a b 1... b r + S a b 1... b r 5.42 To get from equation 5.41 to equation 5.42 note that in the sums in parenthesis in equation 5.41 the i th term in the first sum cancels the i th + 1 term in the second sum. i=2

71 5.2. COORDINATES AND DERIVATIVES 71 Since any multivector is a linear superposition of terms containing b 1... b r with 1 r n and a scalar we have a DA = a A + S a A 5.43 Where a x and b x are vector fields on the manifold write a b = a P b = a Ṗ b + P a b = a Ṗ b + a Db 5.44 Now substitute equation 5.43 into equation 5.44 to get a Ṗ b = b S a Coordinates and Derivatives In a region of the manifold we introduce local coordinates x i and define the frame vectors as e i = x x i 5.46 From the definition of is follows that e i = x i. The {e i } are referred to as tangent vectors and the reciprocal frame {e i } as the cotangent vector or 1-forms. The covariant derivative along a coordinate vector, e i D, satisfies and defines both D i and S i. The tangent frame vectors satisfy Using the P operator on equation 5.48 gives while using P gives e i DA = D i A = e i A + S e i A i A + S i A 5.47 i e j j e i = i j j i x = D i e j D j e i = e i S j = e j S i 5.50 For arbitrary vectors a and b in the tangent space equation 5.50 becomes a S b = b S a 5.51

72 72 CHAPTER 5. GEOMETRIC CALCULUS ON MANIFOLDS In terms of the coordinate vectors the shape tensor becomes and Then S a = e k P a e k 5.52 S i = e k P e i e k = e k P e k e i 5.53 e i = e k k e i = e k P k e i + P k e i = D e i + S i 5.54 Letting a = a i e i be a constant vector in the tangent space gives the general result Additionally Thus a = D a + S a 5.55 a = P a = Ṗ a + P a = D a + Ṗ a 5.56 Ṗ a = S a 5.57 Note that if a and b are any two vectors in the embedding space then P a b = P a P b and if φ x is a scalar function on the manifold we have but = 0 so φ = P φ = P φ + Ṗ φ = P P φ + Ṗ φ Since S a for any vector a lies outside the manifold we have Letting φ x = x i x, then since x i x is a scalar function so that for a general vector a = a i x e i we have = P φ + Ṗ φ 5.58 φ = S φ 5.59 D Dφ = D Dx i = D e i = D a = D a j e j = e i e j i a j = 1 2 ei e j i a j j a i 5.62 Equation 5.62 is isomorphic to the definition of the exterior derivative of differential geometry.

73 5.3. RIEMANNIAN GEOMETRY Riemannian Geometry We shall now relate the shape tensor to the metric tensor and Christoffel connection. The metric tensor is defined by g ij e i e j 5.63 and the Christoffel connection by so that the components of the covariant derivative are given by Γ i jk D j e k e i 5.64 a Db e i = a j D j b k e k e i = a j j b i + Γ i jkb k 5.65 The Γ i jk can be expressed in terms of the g ij by considering the following relations. First, the Γ i jk are symmetric in the j and k indices. Γ i jk Γ i kj = D j e k D k e j e i = Second, the curl of the basis vectors is given by equation 5.61 By equation 5.66 we can write D e i = D g ij e j = Dg ij e j 5.67 Γ i jk = 1 2 ei D j e k + D k e j = 1 2 ei e j D e k + e k D e j 5.68 Now apply equation A.5 Appendix A and equation 5.67 to each term in equation 5.68 to get e j D e k + e k D e j = e j D e k + e k D e j + e j ė k Ḋ + e k ė j Ḋ = e j D e k + e k D e j + D g jk = e j Dg kl e l + e k Dg jl e j + D g jk 5.69 Now apply equation A.2 Appendix A to e j Dg kl e l and e k Dg jl e j giving in the first case e j Dg kl e l = e j Dg kl e l e j e l Dg kl = e j D g kl e l δ l jdg kl = D j g kl e l Dg kj = j g kl e l g kj 5.70

74 74 CHAPTER 5. GEOMETRIC CALCULUS ON MANIFOLDS so that equation 5.68 becomes Γ i jk = 1 2 ei j g kl e l + k g jl e l g kj = 1 2 ei j g kl e l + k g jl e l e l l g kj = 1 2 gil j g kl + k g jl l g kj 5.71 which is the standard formula for the Γ i jk. Now define the commutator bracket [A, B] of the multivectors A and B by note there is no 1 2 factor [A, B] AB BA 5.72 Now form equation A.2 Appendix A and use the Jacobi identity equation to reduce the double commutator products on the r.h.s. of the equation [D i, D j ] A = i j A + S j A + S i j A + S j A j i A + S i A S j i A + S i A = i S j j S i A + S i S j A 5.73 However see equations 5.33 and 5.34, page 153 so that S i = i I I 1, and i S j j S i = i j I I 1 + j i I I 1 = j I i I I 2 i I j I I 2 = j I I 1 i I I 1 i I I 1 j I I 1 = S j S i S i S j = 2S i S j 5.74 where we have used that I 1 and the partial derivatives of I commute to reduce the second line of equation 5.74 [D i, D j ] A = S i S j A 5.75 The commutator of the covariant derivatives defines the Riemann tensor R a, b P S b S b 5.76 Since R a, b is a bilinear antisymmetric function of a and b we may write R a b = P S b S a 5.77 or R e i e j = P S e j S e i 5.78

75 5.3. RIEMANNIAN GEOMETRY 75 Since both S a and S b are bivectors we can use equation A.10 Appendix A to reduce S b S a S b S a = e k P b e k e l P a e l = e k P a e l P b e k e l e k e l P b e k P a e l + P b e k e l e k P a e l P b e k P a e l e k e l 5.79 In equation 5.79 the first and third terms are zero. The second term is entirely outside the tangent space and the fourth term is entirely inside the tangent space. Also note that since the second term consists of bivectors that are entirely outside the tangent space that term commutes with all multivectors A in the tangent space so that the commutator of the second term with A is zero. Thus the Riemann tensor reduces to or R a b = P b e u P a e v e u e v 5.80 R e i e j = P j e u P i e v e u e v 5.81 To calculate the Riemann tensor in terms of the Christoffel symbols note that [D i, D j ] e k = S j S i e k 5.82 = S j S i e k 5.83 = R e i e j e k 5.84 Equation 5.82 comes from letting A = e k in equation Equation 5.83 comes from the fact that S j S i is a bivector and that the commutator product of a bivector and a vector is the same as the dot product. Finally we have that S j S i e k = P S j S i e k since from equation 5.79 P S j S i = e m e l P e j e m P e i e l so that P S j S i e k = 0. Thus R e i e j e k = [D i, D j ] e k = D i Γ a jk e a Dj Γ a ike a = i Γjk a ea + Γ a jkd i e a j Γ a ik e a Γ a ikd j e a = i Γ a jk + Γ b jkγib a ea j Γ a ik Γ b ikγjb a ea 5.85 so R e i e j e k e l = i Γ a jk + Γ b jk Γ a ib δ l a j Γ a ik + Γ b ikγ a jb δ l a = i Γ l jk + Γ b jkγ l ib j Γ l ik Γ b ikγ l jb 5.86

76 76 CHAPTER 5. GEOMETRIC CALCULUS ON MANIFOLDS Using equation 5.81 we have R e i e j e k e l = P j e u P i e v e u e v e k e l 5.87 Using equation A.9 Appendix A to reduce e u e v e k e l gives e u e v e k e l = g ul δk v g vl δk u 5.88 Substituting equation 5.88 into equation 5.87 gives R e i e j e k e l = P j e k P i e v g vl P j e u P i e k g ul = P j e k P i e l P j e l P i e k 5.89 because P i e v g vl = P g vl i e v = P i g vl e v i g vl e v = P i g vl e v = P i e l 5.90 Finally R l ijk = P j e k P i e l P j e l P i e k = i Γ l jk + Γ b jkγ l ib j Γ l ik Γ b ikγ l jb 5.91 Which is the standard form of the Riemann tensor in terms of the Christoffel symbols. Note that v R ijkl = Rijk g vl = P j e k P i e l P j e l P i e k 5.92 From equation 5.92 and equation 5.48 i e j = j e i we can see that the symmetries of the covariant Riemann tensor are R ijkl = R jikl, R ijkl = R ijlk, and R ijkl = R klij To prove the first Bianchi identity form R e i e j e k and use equation 5.49 D j e k = D k e j to get R e i e j e k = D i D j e k D j D i e k = D i D k e j D j D k e i = [D i, D k ] e j [D j, D k ] e i + D k D i e j D j e i = R e i e k e j R e j e k e i 5.93

77 5.4. MANIFOLD MAPPINGS 77 now defining the function F a, b, c by F a, b, c a R b c + c R a b + b R c a = However F a, b, c is a linear function of a, b, and c. Also F b, a, c = F a, b, c and F a, c, b = F a, b, c so since F is antisymmetric in all arguments we may write F a, b, c = F a b c 5.95 Thus equation 5.94 contains n n n 2 n 1 n 2 3 = scalar coefficients. Since the Riemann 6 tensor is a bivector valued function of a bivector the degrees of freedom of the tensor is no more 2 n n 1 than and equation 5.94 reduces the degrees of freedom by n n 2 3 so that the total degrees of freedom of the Riemann tensor is 2 n n 1 n n = n2 n Manifold Mappings One way of illuminating the connection between geometric calculus on manifolds and the standard presentation of differential geometry is to study the effects of mappings from one manifold to another including mapping of the manifold onto itself. Let f : M M define a mapping from the manifold M to M. For our purposes f is a diffeomorphism. That is f and all of its derivatives are continuous and invertible. Thus we can show that the tangent spaces of M and M are of the same dimension. We shall denote the image of x M as x so that f x = x. Then if x λ defines a curve on M, then f x λ = x λ defines a curve on M. In summary - M: Manifold embedded in vector space V M V s i : Coordinates of manifold M such that x s 1,..., s r M V, dim V > r u i : Basis vectors for embedding space V x = x i u i M x λ: Trajectory in M such that x λ = x s 1 λ,..., s r λ M : Manifold embedded in vector space V u i: Basis vectors for V f: Diffeomorphism from V to V x : Element of M as an image of x M x = f x M V f i : Component of f in V such that f x = f i x u i

78 78 CHAPTER 5. GEOMETRIC CALCULUS ON MANIFOLDS Since dx is in the tangent space of M and remember that the dot product comes before the dλ geometric product dx dλ = d f x λ dλ dx = dλ f x ds i x = dλ s i ek k ds i = dλ e i e k k ds i = dλ δk i k f j x u j f j x u j f j x u j = dsi f j dλ s i u j 5.96 is in the tangent space of M. Thus if a is in the tangent space of M at point x then a = a f x = f a; x = f a 5.97 is in the tangent space of M at point x and the frame basis vectors for M at point x map from one tangent space to the other via e i = e i f x = f e i ; x = f e i 5.98 where in the rhs of equations 5.97 and 5.98 we suppress the position dependence, x, in the linear functional f. Since f x is invertible for all derivatives there is no e i such that f e i = 0 the dimension of the image tangent space is the same as the original. Note that we actually have e i x and e i x. An example is shown in figure 5.1 The cotangent frame, e i, in M is mapped to e i in M via the adjoint of the inverse e i = f 1 e i This is simply proved by noting remember that by definition a f b = b f a e i e j = f e i f 1 e j = e j f 1 f e i = e j e i = δ j i 5.100

79 5.4. MANIFOLD MAPPINGS 79 M : { x = cos s 2 cos s 1 u 1 + sin s 1 u 2 + sin s 2 u 3 : s 2 s 2 max < π } 2 e 2 e 1 e 2 e 1 M f x M : M { f x = cos s 1 u 1 + sin s 1 u 2 + tan s 2 u 3 : s 2 s 2 max < π } 2 e 1 x = f e 1 = sin s 1 u 1 + cos s 1 u 2 e 2 x = f e 2 = u 3 cos 2 s 2 Figure 5.1: Mercator mapping of sphere to cylinder manifold. Top and bottom of sphere excluded.

80 80 CHAPTER 5. GEOMETRIC CALCULUS ON MANIFOLDS The exterior product of two tangent vectors, e i e j, maps to e i e j e i e j = f e i e j since f is linear in tangent space arguments. Likewise if I is the unit pseudoscalar for the tangent space of M at f x I 2 = ±1 and I the unit pseudoscalar for the tangent space of M at x I 2 = ±1 then f I = det f I Since f is invertible the tangent space and its image are isomorphic we can apply the definition of determinant in equation 1.79 to equation For cotangent vectors e i and e j e i e j f 1 e i f 1 e j = f 1 e i e j Since the derivative of a scalar field, φ, is a cotangent vector φ = e i i φ and φ x = φ x we can write = f 1 = f 1 e i i = e i i Since D is also a cotangent vector we also have For the directional derivative of a scalar field or for a vector field a φ = D = f 1 D f a f 1 φ = f 1 f a φ = a φ = a φ a b = a f b The covariant derivative is constructed using the projection operator, P, which contains a contraction with the pseudoscalar I. Thus the covariant derivative depends upon the metric encoded by I x. Consider the following operation where a and b are tangent vectors from now on the parenthesis around the dot operands are implied and using equations 5.44 and 5.45 we get a b b a = a Db b Da a S b + b S a = a Db b Da 5.108

81 5.4. MANIFOLD MAPPINGS 81 since a S b = b S a by equation Now define the Lie derivative of a with respect to b by We will show that First note that a f b b f a = f a b b a + L a b a b b a L a b L a b = f L a b Note that since f a is the differential of f x we have i e j j e i = 0 0 = i j j i f x = i f e j j f e i a ḟ b b ḟ a = i ḟ e j j ḟ e i + f i e j j e i 0 = i e j j e i so that i e j j e i = 0 and i ḟ e j j ḟ e i = 0. Thus a ḟ b b ḟ a = a k e k e i i ḟ b j e j b k e k e j j ḟ a i e i = a i i ḟ b j e j b j j ḟ a i e i and and by equations and we have and the Lie derivative maps simply under f. = a i b j i ḟ e j b j a i j ḟ e i = a i b j i ḟ e j j ḟ e i = a f b b f a = f a b b a Since e k e j = δ k j and e k e i = δ k i we have using equation L a b = f L a b i e k e j j e k e i = 0 e k i e j j e i + e j i e k e i j e k = 0 e j i e k e i j e k = 0 f e j i f 1 e k f e i j f 1 e k =

82 82 CHAPTER 5. GEOMETRIC CALCULUS ON MANIFOLDS Equation inplies that P f 1 f 1 e k = 0 as can be shown by expanding f 1 f 1 e k = f 1 e i i f 1 e k = e i i e k Since equation is a grade two multivector we can expand it in terms of the blades components e l e m as follows from equations 1.75, 1.76, A.9, and the fact that i is a scalar operator that commutes with. f 1 f 1 e k = e i i e k P f 1 f 1 e k = e m e l e i i e k e l e m = e m e l e i i e k e l e m = e m i e k e l e i e m e i e l i e k e l e m = δl i e m i e k δm i e l i e k e l e m = e m l e k e l m e k e l e m But the coefficient of e l e m is the same as equation so that P f 1 f 1 e k = 0. Note that since we expanded f 1 f 1 e k in terms of e l e m cotangent vectors in M it was automatically projected into the tangent space of M at x = f x. Since D = P f 1 we have D e k = D f 1 e k = and using the product rule the exterior derivative of a blade formed from cotangent vectors is just move the term in the product being differentiated to the front using the alternating property of the product D e k 1... e kr = Now expand a grade-r multivector field A x on M in terms of the cotangent vectors Then A x = A i1 i 2...i r x e i 1 e i 2... e ir D A x = DA i1 i 2...i r x e i 1 e i 2... e ir since the exterior derivate of a scalar field is the same as the covariant derivative and that the exterior derivative of a basis blade is zero. Likewise D A x = D A i1 i 2...i r x e i 1 e i 2... e ir 5.123

83 5.5. THE FUNDMENTAL THEOREM OF GEOMETRIC CALCULUS ON MANIFOLDS 83 so that for a general multivector field A x on M we have D A D A = f 1 D A The Fundmental Theorem of Geometric Calculus on Manifolds The fundamental theorem of geometric calculus is simply implemented on manifolds. In figure 5.2 a simplicial decomposition of a surface defined by a close curve on a spherical manifold is shown. The only difference between the derivation of the fundamental theorem of geometical calculus Figure 5.2: Simplical decomposition for surface defined by closed curve on spherical manifold. on a vector space which we have proved and on a manifold is that the pseudo-scalar is not a constant but is now a function of position, I x, on the manifold. If we are not on a manifold the pseudo scalars corresponding to the oriented volume of each simplex are proportional to one another they are equal when normalized so have the same orientation. For a manifold the orientation of I x can change with the position vector x. In the case of figure 5.2 I x is a bi-vector defined by the tangent space tangent plane for each point on the sphere.

84 84 CHAPTER 5. GEOMETRIC CALCULUS ON MANIFOLDS Now consider the directed volume element in the case of figure 5.2 an area element for each simplex in figure 5.2 given by X = 1 k! e 1... e k k = 2 for figure 5.2. As the volume area of X 0, X I x. In equation 4.75 where on the l.h.s. of the equation we are integrating over the boundary of the simplex where f is a linear approximation to an arbitrary multivector function. f x = f X x k consider the operator X where we have left the dot on the to emphasize that it is not differentiating the X. On a given simplex = e i λ i where the λ i s are the simplical coordinates equation 4.19 and the e i s are the reciprocal vectors to the e i s that define the simplex. In the case of a manifold as the volume of X 0 the e i s since the e i s define the same subspace as the e i s pseudoscalar that is proportional to I x so that e i is the projection of the geometric derivative,, from the embedding vector space of λi the manifold to the tangent space of the manifold at point x. Thus in the case of a manifold as the volume of X 0 we have = e i λ. i Note that X = X + X but X = 0 since is within the subspace tangent space defined by X. The fundamental theorem of geometric calculus as applied to a manifold is L ds = L dx = L dx V V One can write dx = I x dx and note that or in equation do not differentiate I x. V Divergence Theorem on Manifolds Let L A = JAI 1 where J is a vector field in the tangent space of the manifold and substitute into equation to get JdSI 1 = J dxi 1 + J dxi V V

85 5.5. THE FUNDMENTAL THEOREM OF GEOMETRIC CALCULUS ON MANIFOLDS 85 Now reduce equation by noting that nds = I ds where n is the outward normal to the surface element ds. We can define an outward normal in n-dimensional manifold since the grade of ds is n 1 and it defines a subspace of the tangent space of dimension n 1 for which a unique normal exists. This gives Also ds = n I ds n2 JdSI 1 = ds = ds n 2 n 2 JnII 1 Jn = J n ds n J dxi 1 = J I dx I 1 = J dx J dxi 1 = J dx = J dx = J dx Finally, since both I 1 and dx are proportional to I J I 1 dx = ± J II dx = ± 1 J 2 II + II dx = ± 1 J I 2 dx 2 = so that the divergence theorem is V n n J ds = J dx V Stokes Theorem on Manifolds Assume that the manifold tangent space dimension is s + 1 and B r is a grade r multivector field. For clarity we denote the grade of volume and surface elements with a subscript on the

86 86 CHAPTER 5. GEOMETRIC CALCULUS ON MANIFOLDS d s so that dx = d s+1 X and ds = d s S. Now let L A = B r A s r so that the application of equation gives B r d s S s r = Ḃr ds+1 X V V s r grade s r {}}{ grade r + 1 {}}{ B r d s S = Ḃ r + Ḃ r }{{} d s+1x V V grade r 1 s r = 1 r B r d s+1 X + Ḃr d V }{{} s+1 X V }{{} lowest grade is s r s r lowest grade is s r + 2 = 1 r B r d s+1 X However V B r d s+1 X = D B r d s+1 X since d s+1 X = I s+1 x d s+1 X and the dot product of any component of B r that is not in the tangent space defined by I s+1 x is zero so that B r d s S = 1 r D B r d s+1 X V The divergence theorem is recovered when r = s 1. This is important for constructing conservation theorems in curved spaces. V s r 5.6 Differential Forms in Geometric Calculus Inner Products of Subspaces Start by considering products of the form where A r and B r are r-grade blades A r B r = a 1... a r b 1... b r Both A r and B r define r-dimensional subspaces of the vector space via the equations x A r = 0 and x B r = 0. We show that if A r and B r do not define the same subspace then A r B r = 0.

87 5.6. DIFFERENTIAL FORMS IN GEOMETRIC CALCULUS 87 Assume that they do not define the same subspaces, but that the intersection of the subspaces has dimension s < r so that one can have an orthogonal basis for A r of {e 1,..., e s, e s+1,..., e r } and an orthogonal basis for B r of { e 1,..., e s, e s+1,..., e r}. Then the geometric product of A r and B r is A rb r = αe r... e s+1 e s... e 1 βe 1... e s e s+1... e r = αβe e 2 s es+1... e 1 e s+1... e r where the quantity in parenthesis is a scalar and the other factors a blade of grade 2 r s. Thus A r B r = A rb r = If A r and B r define the same r-dimensional subspace let {e 1,..., e r } be an orthogonal basis for the subspace and expand A r and B r in terms of the orthogonal basis vectors and reciprocal orthogonal basis vectors respectively - and a i = a i e j e j b i = b i e j e j Let the matrices of the coefficients in equations and be denoted by [a i e j ] and [b i e j ]. Then we can expand A r and B r as and A r = det [ a i e j] e 1... e r B r = det [b i e j ] e 1... e r and since the determinant of a matrix and determinant of the transpose of matrix are equal A r B r = det [ a i e j] det [b i e j ] e r... e 1 e 1... e r But = det [ a i e j] det [b i e j ] = det [ a i e j] det [b i e j ] T = det [ a i e j] det [b j e i ] = det [ a i e k b j e k ] a i b j = a i e k e k b j e l e l = a i e k b j e l δ l k = a i e k b j e k 5.146

88 88 CHAPTER 5. GEOMETRIC CALCULUS ON MANIFOLDS So that A r B r = det [a i b j ] From our derivations we see that if A r is a general r-grade multivector not a blade we can always find a r-grade blade such that A r b 1... b r = a 1... a r b 1... b r Now consider the relation between the basis {a 1,..., a n } and the reciprocal basis {a 1,..., a n } for an n-dimensional vector space where in equation r n and 1 i l, j m n a i1... a ir a j 1... a jr = det [ a il a jm] = det [ δ jm i l ] where i 1 < i 2 < < i r and j 1 < j 2 < < j r. Equation is zero unless i l = j l for all 1 l r so that a i1... a ir a j 1... a jr = δ j 1 i 1 δ j 2 i 2... δ jr i r since if the index ordering condition is satisfied for the i l s and the j m s and there is an index, l, such that i l j l then the two blades do not define the same subspace and the inner product is zero Alternating Forms An alternating r-form is a mapping α r v 1,..., v r : V } {{ V } R r times where v 1,..., v r V and α r is linear in all of its arguments v i and is alternating in that α r v 1,..., v i, v i+1,..., v r = α r v 1,..., v i+1, v i,..., v r 1 i < r In the geometric algebra a simple realization of an alternating r-form is α r v 1,..., v r = A r v 1... v r where A r is a grade-r multivector. Since the grade of A r and the grade of v 1... v r are the same the inner product results in a scalar and also since v 1... v r is a blade it alternates sign upon exchange of adjacent vectors.

89 5.6. DIFFERENTIAL FORMS IN GEOMETRIC CALCULUS 89 The basic operations of the Algebra of Forms are addition/subtraction of two r-forms and the exterior product of a r and s-form. The sum of two r-forms is defined by α r + β r v 1,..., v r A r + B r v1... v r for the exterior product let β s w 1,..., w s = B s w 1... w s so that the exterior product is defined by an r + s form α r v 1,..., v r β s w 1,..., w s A r B s v 1... v r w 1... w s Thus α r β s = 1 rs β s α r The interior product of an r and s-form is defined by r > s an r s form note that we are distinguishing the inner,, and exterior,, products for forms from the geometric algebra products by using boldface symbols A r-form is simple is A r is a blade A r = a 1... a r. β s α r B s A r v s+1... v r Dual of a Vector Space Let V be a n-dimensional vector space. The set of all linear maps f : V R is denoted V and is a vector space called the dual space of V. V is a n-dimensional vector space since for f, g : V R, x V, α R and we define f + g x f x + g x αf x αf x x Then by the linearity of f x and g x, f + g x, αf x, and 0 x are also linear functions of x and V is a vector space. Let e i be a basis for V and define e i V by e i e j = δ j i so that if x = x i e i V then e i x = e i x j e j = x j 5.162

90 90 CHAPTER 5. GEOMETRIC CALCULUS ON MANIFOLDS so that for any f V f x = f x i e i = x i f e i Now assume that so the e i are linearly independent. Now assume f V. Then f x = f x i e i Thus 0 = a i e i e j = a i δ j i = a j = x i f e i = f e i e i x = f e i e i x f = f e i e i and the e i form a basis for V. If the e i form an orthonormal basis for V. Remember that orthonomal implies orthogonal and orthogonal implies that a dot product is defined since the dot product is used to define orthogonal. Then e i x = e i x Standard Definition of a Manifold Let M n be any set 1 that has a covering of subsets, M n = U V... such that 1. For each subset U there is a one to one mapping φ U : U R n where φ U U is an open subset of R n. 2. Each φ U U V is an open subset of R n. 3. The overlap maps f UV = φ V φ 1 U : φ U U V R n This section is based upon sections 1.2 and 2.1 in The Geometry of Physics, An Introduction Second Edition, by T. Frankel

91 5.6. DIFFERENTIAL FORMS IN GEOMETRIC CALCULUS 91 or equivalently the compound maps are differentiable. φ U U V φ 1 U M n φ V R n Take a maximal atlas of coordinate patches {U, φ U, V, φ V,...} and define a topology for M n by defining that a subset W M n is open if for any p W there is a U, φ U such that p U W. If the resulting topology for M n is Hausdorff and has a countable base we say M n is an n- dimensional differentiable manifold look it up since I don t know what it means 2. Figure 5.3 show the relationships between the Manifold and the coordinate patch mappings. Abstract set M n not necessarily embedded in a vector space. M n U U V V φ U φ V R n φ U U V φ V U V φ U U φ V φ 1 U φ V V Figure 5.3: Fundamental Manifold Mappings. 2 G. Simmons,Topology and Modern Analysis, McGraw-Hill,1963

92 92 CHAPTER 5. GEOMETRIC CALCULUS ON MANIFOLDS If f : M n R is a real valued function on the manifold M n it is differentiable if f U = f φ 1 U is differentiable with respect to the coordinates {x 1,..., x n } for any coordinate patch U, φ U. The real scalars {x 1,..., x n } denoted by the tuple x = parenx 1, dots, x n form a coordinate system for φ U U R n. In the future we shall simply say that f is differentiable if f U is differentiable. Likewise we will shall usually omit the process of replacing f by its composition f φ 1 U, thinking of f as directly expressible as a function f x = f x 1,..., x n of any local coordinates. Now let p U V M n and Then x U = x 1 U,..., x n U = φu p and x V = x 1 V,..., x n V = φv p. x U = φ U φ 1 V x V = x U x V and x V = φ V φ 1 U x U = x V x U. Let X U = XU 1,..., Xn U and consider the linear approximation to φ V φ 1 U x U + hx U where h is a small number. Then in the linear approximation so that [ x φ V φ 1 i U x U + hx U = x V x U + h V x j U ] X T U = x V + hx V XV i = xi V x j X j U U [ ] x i where V is the Jacobian matrix of the coordinate transformation from one coordinate patch x j U to another and X T U is the transpose colume vector of the tuple X U row vector. X U is a tangent vector contravariant vector or just a vector if it transforms as shown in equation The basic concept is that in the differential neighborhood of x both φ 1 U x U + hx U and φ 1 V x V + hx V in M n are the same to first linear order. The derivative operator D X f is defined for a real function f : M n R by the limit let f U = f φ 1 U to show that we are using the representation of f attached to the coordinate patch U f U x U + hx U f U x U D XU f lim h 0 h but for any two coordinated patches U and V where φ 1 U x U = φ 1 V D XU f U = D XV f V if equation holds so that we can write x V = p U V we have D X f = D XU f U = D XV f V 5.178

93 5.6. DIFFERENTIAL FORMS IN GEOMETRIC CALCULUS 93 and the value of D X f does not depend upon the coordinate patch representation of the vector X. Thus we have in terms of the components x i U and Xi U D X f = D XU f U = f U x i XU i = f U xu x p i Xi p From now on we will not explicitly denote the coordinate patch U in our expressions, but imply it from context since always will refer to a representation in some coordinate patch U. If x i p M n denote the tangent space of an n-dimensional manifold at point p as M n p. Again let f : M n R, from equation is is easily shown that D αx f = αd X f D X+Y f = D X f + D Y f so that the operators D X form a vector space with a basis of the n scalar operators vector space denoted M n p is the tangent space of M n at point p M n. x i. This A vector field on an open set U M n will be the differentiable assignment of a vector X to each point p U; in terms of local coordinates where the X i x are differentiable functions of x. X p = X i x x i The dual space to the tangent space M n p is the vector space of all linear maps M n p R and is denoted M n p and is called the cotangent space. Let f : M n R. The differential of f at p M n is the element of M n p defined by v p M n p In local coordinates df v i df v v p f x i Now consider the differential of a coordinate function x i, dx i x j = v i p p xi = xi x j = δi j 5.185

94 94 CHAPTER 5. GEOMETRIC CALCULUS ON MANIFOLDS so that dx i v j and the dx i form a basis for M n p. x j = v j dx i = v j δ i x j j = v i

95 Chapter 6 Case Study of a Manifold for a Model Universe We wish to construct a spatially curved closed isotropic Minkowski space with 1, 2, and 3 spatial dimensions. To do this consider a Minkowski space with one time dimension e 2 0 = 1 and 2, 3, or 4 spatial dimensions unit vector squares to 1 as indicated below The vector manifolds will designated by e 2 0 = e 2 1 = e 2 2 = e 2 3 = e 2 4 = Vector Function Spatial Dimensions Coordinates Components X 1 1 τ, ρ e 0, e 1, e 2 X 2 2 τ, ρ, θ e 0, e 1, e 2, e 3 X 3 3 τ, ρ, θ, φ e 0, e 1, e 2, e 3, e 4 The condition that τ does parametrize the time coordinate of the manifold is that e τ e τ = Xi τ Xi τ = since then the time coordinate is given by all spatial coordinates are fixed eτ e τ dτ = τ

96 96 CHAPTER 6. CASE STUDY OF A MANIFOLD FOR A MODEL UNIVERSE Then the manifolds in 1, 2, and 3 spatial dimensions are defined by X 1 = t τ e 0 + r τ X 2 = t τ e 0 + r τ X 3 = t τ e 0 + r τ cos ρ r e 1 + sin ρ ρ cos e 1 + sin e 2 r r ρ cos e 1 + sin r ρ cos θe 2 + sin θe 3 r ρ cos θe 2 + sin θ cos φe 3 + sin φe 4 r Where r τ is the spatial radius of the manifolds. In order for equation 6.2 to be satisfied for all the three manifolds we must have e τ e τ = 2 dt dτ 2 dr = dτ For the specific case of r τ = 1 2 τ 2 the derivatives and t τ are given by tτ rτ Figure 6.1: Expansion Radius of Model Universe, r τ, where τ is the time coordinate.

97 97 dr dτ = τ dt dτ = 1 + τ 2 t τ = 1 τ 1 + τ sinh 1 τ Spatial coordinates for the model universes are given by ρ a spatial radius and the angular coordinates θ and φ if required. A visualization of the 1-D spatial dimensional manifold is shown in figure 6.2. e τ and e ρ are e τ = X1 τ and e ρ = X1 ρ 6.8 and metric tensors for the three cases are g µν i = Xi µ Xi ν 6.9 where µ, ν = {τ, ρ, θ, φ}. If we define h τ, ρ as h τ, ρ = dr ρ dτ r τ 6.10 Then the differential arclength given by the metric tensor, g µν, is ds 2 = g µν dx µ dx ν 6.11 and the metric tensors for the three cases are using sympy to do the algebra 1 h g µν 1 2 h = h 1 g 3 µν = 1 h 2 h 0 g µν 2 = h 1 0 ρ r sin r 1 h 2 h 0 0 h ρ r sin 0 r ρ 2 r sin sin θ r. 6.14

98 98 CHAPTER 6. CASE STUDY OF A MANIFOLD FOR A MODEL UNIVERSE e1 e0 τ e2 ρ eρ Figure 6.2: 1D Universe e2 ρ eτ τ e0 e1