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The Tensor Product and the Wedge Product Phil Lucht Rimrock Digital Technology, Salt Lake City, Utah 84103 last update: April 13, 2016 Maple code is available upon request. Comments and errata are welcome. The material in this document is copyrighted by the author. The graphics look ratty in Windows Adobe PDF viewers when not scaled up, but look just fine in this excellent freeware viewer: http://www.tracker-software.com/pdf-xchange-products-comparison-chart . The table of contents has live links. Most PDF viewers provide these links as bookmarks on the left. Notation 5 1. The Tensor Product 8 1.1 The Tensor Product as a Quotient Space 8 1.2 The Tensor Product in Category Theory 13 2. A Review of Tensors in Covariant Notation 16 2.1 R, S and how tensors transform : Picture A 16 2.2 The metric tensors g and g' and the dot product 21 2.3 The basis vectors en and en 23 2.4 The basis vectors un and un 24 2.5 The basis vectors e'n and u'n and a summary 25 2.6 How to compute a viable x' = F(x) from a set of constant basis vectors en 28 2.7 Expansions of vectors onto basis vectors 30 2.8 The Outer Product of Tensors and Use of 33 2.9 The Inner Product (Contraction) of Tensors 36 Dot products in spaces VV, VW, VVV and VWX 38 2.10 Tensor Expansions 39 (a) Rank-2 Tensor Expansion and Projection 39 (b) Rank-k Tensor Expansions and Projections 40 2.11 Dual Spaces and Tensor Functions 42 (a) The Dual Space V* in Matrix and Dirac Notation 42 (b) Functional notation 43 (c) Basis vectors for the dual space V* 44 (d) Rank-2 functionals and tensor functions 47 (e) Rank-k functionals and tensor functions 50 (f) The Covariant Transpose 53 (g) Linear Dirac Space Operators 54 (h) Completeness 60 3. Outer Products and Kronecker Products 62 3.1 Outer Products Reviewed: Compatibility of Chapter 1 and Chapter 2 62 3.2 Kronecker Products 64 4. The Wedge Product of 2 vectors built on the Tensor Product 72 4.1 The tensor product of 2 vectors in V2 72 4.2 The tensor product of 2 dual vectors in V*2 75 4.3 The wedge product of 2 vectors in L2 78 4.4 The wedge product of 2 dual vectors in Λ2 86 5. The Tensor Product of k vectors : the vector spaces Vk and T(V) 92 5.1 Pure elements, basis elements, and dimension of Vk 92 5.2 Tensor Expansion for a tensor in Vk ; the ordinary multiindex 93 5.3 Rules for product of k vectors 94 5.4 The Tensor Algebra T(V) 95 5.5 Comments about tensors 97 5.6 The Tensor Product of two or more tensors in T(V) 97 6. The Tensor Product of k dual vectors : the vector spaces V*k and T(V*) 102 6.1 Pure elements, basis elements, and dimension of V*k 102 6.2 Tensor Expansion for a tensor in V*k ; the ordinary multiindex 103 6.3 Rules for product of k vectors 103 6.4 The Tensor Algebra T(V*) 104 6.5 Comments about Tensor Functions 105 6.6 The Tensor Product of two or more tensors in T(V*) 105 7. The Wedge Product of k vectors : the vector spaces Lk and L(V) 109 7.1 Definition of the wedge product of k vectors 109 7.2 Properties of the wedge product of k vectors 111 7.3 The vector space Lk and its basis 113 7.4 Tensor Expansions for a tensor in Lk 116 7.5 Expansions for the wedge product of k vectors 119 7.6 Number of elements in Lk compared with Vk. 121 7.7 Multiindex notation 121 7.8 The Exterior Algebra L(V) 123 Associativity of the Wedge Product 123 7.9 The Wedge Product of two or more tensors in L(V) 126 (a) Wedge Product of two tensors T^ and S^ 126 (b) Special cases of the wedge product T^^ S^ 129 (c) Commutivity Rule for the Wedge Product of two tensors T^ and S^ 129 (d) Wedge Product of three or more tensors 131 (e) Commutativity Rule for product of N tensors 134 (f) Theorems from Appendix C : pre-antisymmetrization makes no difference 135 (g) Spivak Normalization 137 8. The Wedge Product of k dual vectors : the vector spaces Λk and Λ(V) 141 8.1 Definition of the wedge product of k dual vectors 141 8.2 Properties of the wedge product of k dual vectors 142 8.3 The vector space Λk and its basis 143 8.4 Tensor Expansions for a dual tensor in Λk 145 8.5 Expansions for the wedge product of k dual vectors 147 8.6 Number of elements in Λk compared with V*k. 149 8.7 Multiindex notation 150 8.8 The Exterior Algebra Λ(V) 151 Associativity of the Wedge Product 151 8.9 The Wedge Product of two or more dual tensors in Λ(V) 154 (a) Wedge Product of two dual tensors T^ and S^ 154 (b) Special cases of the wedge product T^^ S^ 155 (c) Commutivity Rule for the Wedge Product of two dual tensors T^ and S^ 155 (d) Wedge Product of three or more dual tensors 156 (e) Commutativity Rule for product of N dual tensors 159 (f) Theorems from Appendix C : pre-antisymmetrization makes no difference 159 (g) Spivak Normalization 161 9. The Wedge Product as a Quotient Space 163 9.1. Development of Lk as Vk/S. 163 9.2. Development of L as T/I. 165 10. Differential Forms 168 10.1. Differential Forms Defined 168 10.2. Differential Forms on Manifolds 170 10.3. The exterior derivative of a differential form 172 10.4. Commutation properties of differential forms 179 10.5. Closed and Exact, Poincaré and the Angle Form 179 10.6 Transformation Kinematics 182 (a) Axis-Aligned Vectors and Tangent Base Vectors : The Kinematics Package 182 (b) What happens for a non-square tall R matrix? 184 (c) Some Linear Algebra for non-square matrices 187 (d) Implications for the Kinematics Package 189 (e) Basis vectors for the Tangent Space at point x' on M 190 10.7 The Pullback Operator R and properties of the Pullback Function F* 192 10.8 Alternate ways to write the pullback of a k-form 199 10.9 A Change of Notation and Comparison with Sjamaar and Spivak 203 Appendix A: Permutation Support 211 A.1 Rearrangement Theorems and a Determinant Theorem 211 A.2 The Alt Operator 215 A.3 The Sym Operator 219 A.4 More on Alt and Sym and the decomposition of functions 221 A.5 Application to Tensors 223 (a) Alt Equations (translated from Section A.2 above) 224 (b) Sym Equations (translated from Section A.3 above) 225 (c) Alt/Sym and Other Equations (translated from Section A.4 above) 226 A.6 The permutation tensor ε 227 A.7 The wedge-product-of-vectors Alt equation 229 A.8 Application to Tensor Functions 230 (a) Alt Equations (translated from Section A.2 above) 231 (b) Sym Equations (translated from Section A.3 above) 232 (c) Alt/Sym and Other Equations (translated from Section A.4,A.6 and A.7 above) 233 (d) Alt/Sym when there are two sets of indices 234 A.9 The Ordered Sum Theorem 238 A.10 The Alt Operator in Dirac bra-ket notation 239 Appendix B: Direct Sum of Vector Spaces 242 Appendix C: Theorems on Pre-Symmetrization 250 C.1 Theorem One 250 C.2 Theorem Two 253 C.3 Theorem Three 256 C.4 Summary and Generalization 257 Appendix D: A Unified View of Tensors and Tensor Functions 261 D.1 Tensor functions in Dirac notation 261 D.2 Basis change matrix 262 D.3 Transformations of tensors and tensor functions 264 D.4 Tensor Functions and Quantum Mechanics 266 Appendix E: Chapter 10 with x' = F(x) changed to x = φ(t) 268 Overview This document is meant as a sort of User Guide for both tensor products and wedge products. These objects are often glossed over in literature that makes heavy use of them, the assumption being that everything is obvious and not worth describing too much. Our monograph is very focused on these two kinds of products and has little to say about applications such as quantum mechanics, differential forms, the generalized Stokes's Theorem or supersymmetry. We attempt to include both the mathematical view and the engineering/physics view of things, but the emphasis is on the practical. The notation is entirely covariant, which one can say includes the practical for Cartesian spaces where indices can be all down. The study of wedge products is known as the exterior algebra and is credited to Grassmann. Summary Chapter 1 surveys the mathematician's description of the tensor product as a quotient space, and then places the tensor product in the framework of category theory. This approach is resumed much later in Chapter 9 for the wedge product, after the reader is more familiar with that object. Chapter 2 reviews tensor analysis and then introduces a meaning for the tensor product symbol in terms of outer products of tensors. After a quick review of tensor expansions and projections, the last section introduces the notion of a dual space and includes the use of the Dirac bra-ket notation. The notion of a tensor function is introduced. Chapter 3 discusses the theory of Chapter 1 versus the practicality of Chapter 2 in terms of outer products. It then derives the Kronecker product of two matrices in covariant notation. This topic is somewhat tangential to the main development, but is included since it is often not explained very well in the literature. Maple is used to compute a few such Kronecker products. Chapter 4 has four parts involving products of two vectors and their vector spaces: tensor product, dual tensor product, wedge product, and then dual wedge product. Chapters 5,6,7,8 continue this order of presentation but for products of k vectors and then of general tensors. The order is : tensor product (Ch 5), dual tensor product (Ch 6), wedge product (Ch 7) and then dual wedge product (Ch 8). The chapters intentionally have a high degree of parallelism, though details are omitted from the later chapters to reduce repetition. It is the dual tensor chapters which involve tensor functions as the closure of tensor functionals onto a general set of vectors. The tensor-product tensor functions are multilinear, whereas the wedge-product ones are multilinear and totally antisymmetric. Alternate wedge product normalizations are discussed. Chapter 9 returns to the mathematician's world giving a description of the wedge product in terms of quotient spaces. Appendix A explains our permutation notation and the powerful rearrangement theorem used in various proofs throughout the document. The Alt and Sym operator properties are presented in a generic permutation space, and then those generic results are applied to tensors and tensor functions. The permutation tensor ε is given honorable mention, and a few obscure theorems are proved. Appendix B discusses the direct sums of vectors, vector spaces and operators in those spaces. Appendix C shows that when one antisymmetrizes a product of tensors, pre-antisymmetrizing one or more of those tensors makes no difference in the result. Appendix D shows how tensors and tensor functions are the same objects expressed in different bases. Notation This list gives most of the symbols used in the document and should give the reader an idea of the general flavor of the presentation. Vectors are sometimes bolded, sometimes not. Vector functionals and tensors of rank 2 or greater are never bolded. In general, dual objects are given Greek or script symbols. MT transpose of a matrix M det(a,b,c...) determinant of an nxn matrix whose n columns are vectors a,b,c... tensor product of spaces or objects in those spaces direct sum of spaces or objects in those spaces (App B) x Cartesian product, as in VxW with element (v,w) K a real field (such as the reals, or such as binary {0,1} ) ≡ is defined as F(x) a general transformation x' = F(x) Rij the differential of transformation F Alt total antisymmetrization operator Sym total symmetrization operator V real vector space of dimension n a,b vectors in V |a> a Dirac ket a b scalar product of two vectors <a|b> = <b|a> V,v vectors in V |v> vi vector with label i in V |vi> vi component i of the vector v <ei|v> ei basis vector in V |ei> ei dual basis vector in V |ei> ui axis-aligned basis vector in V |ui> W real vector space of dimension n' e'i basis vector in W |e'i> w vector in W |w> a b tensor product of two vectors, a pure element of the vector space V2 = VV eiej tensor product of two basis vectors, basis vector of V2 = VV a ^ b wedge product of two vectors, a pure element of the vector space L2 ei^ ej wedge product of two basis vectors, basis element of L2 = V ^ V T tensor of rank k |T> T = Σii....i Tii....i ei ei ..... ei S tensor of rank k' |S> R tensor of rank k" |R> T^ general element of Lk T^ = Σii....i Tii....i (ei^ ei^ .....^ ei) S^ general element of Lk' R^ general element of Lk" V* dual space to V α,β vector functionals in V* α = <α|, a Dirac bra αi vector functional in V* with label i αi = <αi| λi basis vector in V* λi = <ei | = (ei)T α β tensor product of two dual vector, a pure element of the vector space V*2 = V*V* λiλj basis vector of V*2 = V*V* α ^ β wedge product of two dual vectors, a pure element of the vector space Λ2 λi ^ λj wedge product of dual basis vectors, basis element of Λ2 = V ^ V T tensor functional of rank k <T| T = Σii....i Tii....i λi λi ..... λi S tensor functional of rank k' <S| R tensor functional of rank k" <R| T^ general element of Λk T^ = Σii....i Tii....i (λi^ λi .....^ λi) S^ general element of Λk' R^ general element of Λk" λi(v) dual basis vector tensor function, rank-1 <ei |v> α(v) general rank-1 tensor function <α |v> T(v1,v2) rank-2 tensor function <T | v1, v2> I, J multiindices ΣI Σii...i symmetric sum Σ'I Σi<i<....<i ordered sum Vk vector space of rank-k tensors |T> V*k vector space of dual rank-k tensors = rank-k tensor functionals <T| V*k vector space of rank-k tensor functions T(vI) = <T| vI> (k-multilinear) Lk vector space of totally antisymmetric rank-k tensors Λk vector space of totally antisymmetric dual rank-k tensors Λk vector space of totally antisymmetric rank-k tensor functions (k-multilinear) T(V) V0 V V2 V3 ...... the tensor algebra T(V*) V*0 V* V*2 V*3 ....... dual tensor algebra L(V) L0 L1 L2 L3 + .... exterior tensor algebra Λ(V) Λ0 Λ1 Λ2 Λ3 + .... dual exterior tensor algebra εii...i rank-k permutation tensor P permutation S(P) swaps in a permutation (-1)S(P) swap parity of a permutation 1. The Tensor Product There are two theoretical paths leading to the tensor product. These are briefly summarized in a non-rigorous manner in Sections 1.1 and 1.2 below, after a comment on terminology. Nomenclature: Tensor Product vs Direct Product. The tensor product described below sometimes goes by other names. In quantum mechanics, a system of two particles might be in a quantum state |ψ1> |ψ2> which is an element of a tensor product space V1V2 (as we shall describe below). Some quantum authors refer to this tensor product as a direct product (e.g. Shankar) while others call it a tensor product (e.g. Messiah). It happens that in quantum theory states like |ψ1> reside in a vector space which is also a Hilbert space. Similarly, when a quantum system has a symmetry, such as rotational invariance (e.g., an isolated atom), the quantum states can be classified into certain vector spaces associated with the matrix representations of the symmetry group, and the tensor products of these spaces are usually called direct products. For example, the rotation group has matrix representations called "(n/2)" for any integer n (matrices are n+1 x n+1), and one writes for example 1 (1/2) to indicate the "direct product" of these two spaces. It happens that the matrix for 1 (1/2) can be brought into block diagonal form by a similarity transformation where the diagonal blocks are (1/2) and (3/2), and so one writes 1 (1/2) = (1/2) (3/2). Sometimes the tensor product is called a tensor direct product, which phrase seems associated with the outer product componentization of the tensor product noted in Chapter 3. Sometimes the raw Cartesian product (see below) is called a tensor product, but usually there is some additional structure involved. Generally, the term direct product seems most suitable for the direct product of groups, rings, modules and related objects, whereas in the current document we are discussing the tensor product of vector spaces and of the tensors contained within those spaces. Category Theory mentioned below attempts to put all these products into a uniform framework. 1.1 The Tensor Product as a Quotient Space It does seem odd that one might think of a product VW in terms of a quotient. We shall outline how this path goes in a series of steps. The key results are stated in Steps 7 and 9. 1. Cartesian Product. Start with the inert Cartesian product set VxW with elements (v,w), where in our application the sets V and W are vector spaces. This set VxW is "inert" in the sense that one has no instructions for what can be done with its elements. 2. Space F(VxW). We now endow VxW with an addition operator + and a scalar multiplication operator (indicated by juxtaposition) allowing us to form linear combinations of elements of VxW with scalar coefficients. Let's define F(VxW) to be a space which contains all such linear combinations. A typical element of this F(VxW) space might be 3(v1,w3) - 2.1(v2,w5). Of course (v1,w3) also lies in F(VxW) and one might call this a pure element, whereas 3(v1,w3) - 2.1(v2,w5) is a mixed element. Because the sum of two linear combinations is again a linear combination of the same form, the space F(VxW) is closed under addition. 3. Field K. We usually assume (as above) that the scalars are in the field R of real numbers, but to be more general one can assume the scalars are elements of some arbitrary field traditionally called K (though sometimes F or F). In addition to the reals R, there are various fields having an infinite number of elements (like rational or complex numbers), and there are various fields of having a finite number of elements (the Galois Fields). Footnote: Sometimes the space F(VxW) is described as a "free vector space" which is a set of functions f such that f: VxW → K. Usually such spaces are defined over a discrete set S, and it is not clear how this works when the set S is continuous, this being the case for S = VxW. Moreover, the functions f mapping to K cannot be identified with our linear combinations since for example (v2,v5) is not an element of K. We therefore refrain from giving F(VxW) this moniker and the reader should regard F(VxW) only as we have defined it above. 4. Equivalence Relations and Classes. Now define the following set of "equivalence relations" (v1+v2, w) ~ (v1,w) + (v2,w) for all v1,v2 ϵ V and all w ϵ W (v, w1+w2) ~ (v,w1) + (v,w2) for all v ϵ V and all w1, w2 ϵ W s(v,w) ~ (sv,w) for all v ϵ V and all w ϵ W and all s ϵ K s(v,w) ~ (v,sw) for all v ϵ V and all w ϵ W and all s ϵ K (1.1.1) where ~ means "is equivalent to" and s is a scalar in K. Rewrite these relations as (v1+v2, w) – (v1,w) – (v2,w) ~ 0 (v, w1+w2) – (v,w1) – (v,w2) ~ 0 s(v,w) – (sv,w) ~ 0 s(v,w) – (v,sw) ~ 0 . (1.1.2) We are declaring here that lots of linear combinations in F(VxW) are equivalent to 0. The reason we do this is to make our tensor product space (to be defined below) have "nice properties" (i.e., it is then a vector space). These linear combinations taken together define an "equivalence class" which is equivalent to 0. Call this class N (for null). 5. The Quotient F(VxW)/N. There then exists a space which we shall call F(VxW)/N, or F(VxW) "mod" N. This is a standard structure in equivalence class theory where one takes the quotient of one space S divided by another space of equivalent items in space S, often written S/~. The upshot is that the elements of the new quotient space F(VxW)/N consist of all linear combinations of F(VxW) except that any linear combination which has one of the four forms shown above is filtered out ("modded out") by setting it equal to 0. Example: 3(v3,w4) + (v2, w1+w2) – (v2,w1) – (v2,w2) = an element of F(VxW) 3(v3,w4) = the corresponding element of F(VxW)/N. (1.1.3) 6. The Space VW. We now give this space F(VxW)/N a new name: F(VxW)/N = VW = the tensor product space of V and W . (1.1.4) The elements of VW are linear combinations of elements called vw instead of (v,w) as a reminder that the equivalence class N must be respected. Whereas the comma in (v,w) was a mere separation operator, the in vw is regarded as a new "tensor product multiplication operator" with the properties listed below which, in effect, implement the equivalence relations stated above. 7. Practical Summary. The end result of all this song and dance is the following: The tensor product space VW is the set of all linear combinations of elements (v,w) of the Cartesian product set VxW, written as vw, where the following rules are declared by fiat: (v1+v2) w = (v1w) + (v2w) for all v1,v2 ϵ V and all w ϵ W v (w1+w2) = (vw1) + (vw2) for all v ϵ V and all w1, w2 ϵ W s(vw) = (sv)w for all v ϵ V and all w ϵ W and all s ϵ K s(vw) = v(sw) . for all v ϵ V and all w ϵ W and all s ϵ K (1.1.5) The first two rules state that multiplication is distributive over addition (from right and left), while the last two rules state the scalars work in the expected manner. If these rules were declared for a function f(v,w), they would appear as f(v1+v2,w) = f(v1,w) + f(v2,w) f(v,w1+w2) = f(v,w1) + f(v,w1) s f(v,w) = f(sv,w) s f(v,w) = f(v,sw) (1.1.6) Such a function would then be described as being bilinear because it is linear separately in each argument with the other argument held fixed. One can then regard the rules shown above for as expressing bilinearity for the tensor product space VW. Usually the above scalar and distributive rules are combined into the slightly more compact form, (s1v1+s2v2) w = s1(v1w) + s2(v2w) v (s1w1+s2w2) = s1(vw1) + s2(vw2) (1.1.7) and similarly for a bilinear function, f(s1v1+s2v2,w) = s1f(v1,w) + s2f(v2,w) f(v, s1w1+s2w2) = s1f(v,w1) + s2f(v,w2) . (1.1.8) Normally one denotes scalars by α and β rather than si, but α and β will have other meanings below. But below I immediately revert to α and β, so something needs to change here. 8. vw does not commute. Whereas the + operation within VW is commutative, it should be clear that the operation is not commutative. If v ϵ V and w ϵ W, then vw ϵ VW whereas wv is an element of a completely different space which is WV. Even if W = V, one has vv' ≠ v'v if v≠v'. The fact goes back to the original Cartesian product set VxV where one has (v,v') ≠ (v',v) if v≠v' because (v,v') is an ordered tuplet, not a set {v,v'}. If V = W = R, one would not identify the point (x,y) with the point (y,x) in RxR = R2 if x ≠ y. Another word for commutative is abelian. 9. VW is a vector space. The space VW is a vector space whose vectors are linear combinations of vw. We shall now verify this to be the case. We already know VW is closed under addition since F(VxW) has this property. The + inverse of vw is (-1)(vw). Addition is commutative and associative. Any element of the form 0w or v0 can be taken as the identity for addition (the "zero") since, for example, using the first rule of (1.1.5), 0w = (v - v) w = (vw) + ((-v)w) = (vw) - (vw) = 0 (0 in the space VW) . (1.1.9) There is a scalar multiplicative identity since all fields K have an identity "1": 1(vw) = (vw). "Vector multiplication" is distributive over scalar addition (here the "vector" is vw), (α + β)(vw) = [(α+β)v]w = [αv+βv]w = (αv)w + (βv)w = α(vw) + β(vw). (1.1.10) Multiplication by a scalar is distributive over "vector addition" : α (v1w2 + v3w4) = α (v1w2) + α (v3w4) . (1.1.11) This property we more or less add by fiat to the earlier properties. It is the only reasonable way to do things since elements of VW are linear combinations of pure elements of the form vw. 10. Basis of VW and general elements of VW. In the above verification that VW is a vector space, we used only pure vectors of VW, but general vectors of VW are linear combinations of the pure vectors so we really should rehash the above for general vectors. To do this, we first note that, since V and W are vector spaces, each has a basis, and we call these bases {ei} for V and {e'i} for W. It is not hard to show that the set of elements of the form eje'j forms a basis for VW, so a general vector in VW can be expressed as u = Σij uij (eie'j) . // coefficients uij ϵ field K (1.1.12) The inverse element -u is pretty obvious. Addition u + u' is commutative and u + u' + u" is associative. The zero element is the same. Vector multiplication is still distributive over scalar addition, (α + β)u = (α + β)[ Σij uij (eie'j)] = Σij uij [ (α + β) (eie'j)] = Σij uij [ α (eie'j) + β(eie'j)] = α [ Σij uij (eie'j)] + β [ Σij uij (eie'j)] = α u + β u . (1.1.13) In this manner, all the required properties of a vector space can be verified for general elements of VW. 11. Vector vs Tensor. Since VW is a vector space, it is proper to refer to its elements vw (or linear combinations of same) as "vectors". On the other hand, we shall refer to vw as a "tensor" ( a cross tensor) in the tensor product space VW. In particular, it is a "rank-2 tensor" composed from v and w which are vectors in their respective vector spaces V and W. The word vector must be evaluated in its context. The notion of tensors is developed more in Section 2.4 below. 12. Dimension of VW. As noted above, the basis of the vector space VW consists of elements of the form eie'j . If the dimensions of V and W are n and n', then i takes n values, j takes n' values, and the dimension of the vector space VW is n*n' ( = nn'), the product of the separate vector space dimensions: dim(VW) = n*n' where n = dim(V) and n' = dim(W) (1.1.14) 13. Generalization. The above development is easily generalized to the tensor product of any finite number of vector spaces. One first defines F(V,W,....Z) as linear combinations of elements of the Cartesian product space VxWx..xZ , which elements have the form (v,w,...z). One then defines a large set of equivalence relations analogous to those described above. One ends up with a large set of linear combinations which are all equivalent to 0, and this defines the equivalence class N. One then creates F(V,W,....Z)/N as the space of linear combinations where any pieces which are equivalent to 0 are filtered out. One then defines VW...Z ≡ F(VxWx...xZ)/N = the tensor product of spaces V and W and... and Z. (1.1.15) The tensor product space VW...Z is the set of all linear combinations of elements (v,w,...z) of the Cartesian product space VxWx...xZ, written as vw...z, where the following rules are declared by fiat: (v1+v2)w .... z = v1w .... z + v2w .... z v(w1+w2) .... z = vw1 .... z + vw2 .... z , etc. and s(vw...z) = (sv)w...z = v(sw)...z , etc. s ϵ K (1.1.16) When these rules are written for a function f(v,w,....z) one has, f(v1+v2,w,...z) = f(v1,w,...z) + f(v2,w,...z) f(v,w1+w2,...z) = f(v,w1,...z) + f(v,w2,...z), etc s f(v,w,...z) = f(sv,w...z) = f(v,sw,...), etc. s ϵ K (1.1.17) If there are k factors in the tensor product VW...Z, then the function f has k arguments, and a function obeying all of the above rules is said to be k-multilinear. For k = 2 we have bilinear, for k = 3 we have trilinear, and so on. One can mix in the scalar rule by saying for example f( s1v1+s2v2, w, ...z) = s1f(v1,w,...z) + s2f(v2,w,...z) f(v, s1w1+s2w2, ...z) = s1f(v,w1,...z) + s2f(v,w1,...z), etc. (1.1.18) We can then regard the set of rules shown above as describing k-multilinearity for the tensor product space VW...Z. Written in the second form, (s1v1+s2v2)w ..... z = s1 (v1w .... z) + s2 (v2w .... z) v (s1w1+s2w2) ...z = s1 (vw1 .... z) + s2 (vw2 .... z). etc. (1.1.19) An alternate approach to developing the tensor product of three or more vector spaces is to inductively build up by grouping things. For example VWX = (VW) X = the tensor product of two vector spaces, one of which is VW VWXY = (VWX)Y = the tensor product of two vector spaces, one of which is VWX The results are the same with either approach. 1.2 The Tensor Product in Category Theory Category theory is an attempt to abstract the essence of algebraic structures which apply generally to objects like vector spaces, sets, rings, groups, modules and so on. One encounters certain category diagrams which must allow for flow through the diagram in all possible ways (the diagram must "commute"). A diagram consists of certain objects which are connected by arrows known as morphisms. For our application, these arrows are function mappings between spaces, and two sequential arrows in a path represent function composition in the sense f o g. At a higher level, if the objects in the diagram are themselves categories, the morphism arrows are called functors. For example, for the category C of "all vector spaces over a field K" where the diagram arrows are linear maps, one can regard the equation V2 = VV as lying in the map CxC → C, and this map is then a functor, and the mapping is said to be functorial. Category theory is a relatively recent addition to the house of many mansions. With precursor work done by Emily Noether (whose work shows up in a lot of places), category theory was developed in the early 1940's by Saunders Mac Lane (and others) who then summarized the theory in a text Algebra (1967) with coauthor Garrett Birkhoff. These same authors wrote the classic textbook A Survey of Modern Algebra (1941/1997) which is known to many students as "Birkhoff and MacLane". We give here just an outline of this rather slippery tensor product development. It seems more of a fitting of our conclusions of Section 1.1 into category theory. The reader interested in more detail can look at Chapter 14 "Tensor Products" of Roman's text Advanced Linear Algebra (2007). We start with the following triangle diagram (an example of a category diagram), (1.2.1) In this diagram VxW is the Cartesian product of two vector spaces V and W, exactly as in Section 1.1 above. Elements of VxW are (v,w). There are two mappings f: VxW → X and g: VxW → Y where X and Y are for the moment just spaces. They in turn are linked by a mapping traditionally called τ, so τ : X → Y. One says that "g can be factored through f". The functions f and g are declared bilinear from the get-go. This is analogous to our declared equivalence relations in the approach of Section 1.1. The set of all bilinear mappings f: VxW→X is called homK(V,W; X) where K is the field of scalars. The letters hom stand for homomorphism ("same shape") which is a structure-preserving map. Linear maps (like τ discussed below) preserve vector space structure. The triangle diagram must commute, so we must have g = τ o f (function composition). The space X is our candidate space for the tensor product VW space. One needs to construct the function τ. To do so, use the fact that the diagram commutes to evaluate τ at the pure point vw, τ(vw) = g(v,w). (1.2.2) Now "extend" τ so it applies to linear combinations of vw elements by declaring that, for si ϵ K, τ (s1 vw + s2 v'w' ) = s1 τ(vw) + s1 τ(v'w') = s1 g(v,w) + s2g(v',w') (1.2.3) so τ is now a linear function τ: X→Y. It maps every element of X into an element of Y, and it is unique by its construction. Once we have τ being a unique linear mapping, the "pair" (X, f:VxW→X) becomes a "universal pair". The idea here is that any alternate "pair" like (Y, g:VxW→Y) is equivalent to (X, f:VxW→X) up to the isomorphism implied by τ. In this sense, then, the mapping f:VxW → VW is essentially unique -- it is "universal for bilinearity" -- so the tensor product mapping is well-defined. Function τ is called a mediating morphism, f is called the tensor map, and the elements of VW are tensors. From the top of the triangle one has vw = f(v,w) (1.2.4) since f :VxW→X = VW. Our "rules" of Section 1.1 for operator now derive from the fact that f is a bilinear function: (v1+v2) w = f(v1+v2,w) = f(v1,w) + f(v2,w) = (v1w) + (v2w) v (w1+w2) = f(v,w1+w2) = f(v,w1) + f(v,w2) = (vw1) + (vw2) s(vw) = s f(v,w) = f(sv,w) = (sv)w s ϵ K s(vw) = s f(v,w) = f(v,sw) = v(sw) s ϵ K (1.2.5) We then end up with the same space VW and rules as in the previous quotient development, and we have extra assurance that VW is a unique and well-defined object (it is universal). The above scenario directly generalizes to the tensor product of k vector spaces with the following corresponding category diagram, (1.2.6) Lang (Algebra) for example shows the equivalent of this diagram on page 602 of his Chapter 16 (The Tensor Product). See also Roman's Chapter 14 on the Tensor Product, p 383. 2. A Review of Tensors in Covariant Notation In Chapter 1 we generally avoided mentioning components of vectors and tensor products. But in many ways, "components" is what tensors are all about. Anyone who wants to use tensor analysis to actually do something practical is going to use tensor components. The whole notion of what it means to be a tensor of some rank requires components and component indices. Later when we deal with tensor functions, the components will in morph into the vector arguments of multilinear functions. Tensor analysis provides some very heavy-duty machinery to handle manipulations of tensors and tensor components. A key idea is that a true tensor is something that transforms in a certain manner relative to some defined underlying transformation which below is called x' = F(x). In the following notes, we review this machinery, The review is based on our document Tensor which follows the unusual path of developing tensor analysis in a "developmental notation" where all indices are down and covariant objects have overbars, then later this notation is converted to "standard notation" with the usual up and down indices. It is a very large and complicated world, and below we report out only those facts which are useful for our efforts here. Equation numbers referring to Tensor are followed by a prime ' . Bolding Vectors. For the time being we shall display all vectors in bold font because we feel it helps the reader when dealing with covariant dot products and is compatible with Tensor. However, vector components are not bolded. Thus vector V will have components Va and Va. The exception is when vectors have extra labels, such as for the basis vectors en. It's components are written (en)a and (en)a. Eventually in Section 3.1 where we finally tie back to Sections 1.1 and 1.2 we shall quietly stop bolding vectors and will then be compatible with those earlier sections. Higher rank tensors are never bolded. 2.1 R, S and how tensors transform : Picture A Tensor is in large part based on the following "picture", (1.11)' (2.1.1) Below we shall be thinking of x-space as a vector space V having a set of basis vectors {ei} or {ui}. Then x'-space is a vector space V'. Below we shall use V as a prototype vector in space V, so V ϵ V . The two vector spaces V and V' have the same dimension N. In what follows, repeated indices are implicitly summed (Einstein convention) so for example RabVb means Σb=1N RabVb. Hanging indices like a in RabVb can take any value in the range a = 1,2...N. The implied summation convention reduces symbol clutter especially when there are many summed indices in an equation. Sometimes however we will display sums for emphasis. Figure (2.1.1) summarizes a generally non-linear transformation x' = F(x) between two spaces called x-space on the right (metric tensor g) and x'-space on the left (metric tensor g'). The coordinates of x-space are called x, and those of x'-space are called x'. Quantities in x'-space always have a prime, while those in x-space have no prime. A vector V in x-space has contravariant components Va and covariant components Va. The corresponding components V'a and V'a of V' in x'-space are these, V'a = RabVb Rab ≡ (∂x'a/∂xb) = ∂bx'a (7.5.3)' (7.5.2)' V'a = SbaVb Sba ≡ (∂xb/∂x'a) = ∂'axb (7.5.4)' (2.1.2) As noted, primed equation numbers refer to Tensor. The matrices R and S in (2.1.2) and Fig (2.1.1) are linearizations of the generally non-linear transformation x' = F(x) and x = F-1(x') in the close neighborhood of a selected point x in x-space and x' = F(x) in x'-space. Therefore, R and S are in general functions of x, though we suppress this dependence. R is sometimes call the differential of the transformation x' = F(x). We call it the R matrix. Many texts don't make up a symbol like R for the differential, and so tensor equations such as (2.1.8) below are strewn with partial derivatives of the form Rab = . This is useful in doing chain rules, but otherwise obscures how the indices work. We settled on symbols R and S after rejecting various reasonable alternatives. One should understand that Rab is in general not a simple rotation matrix despite the letter R. From the chain rule, one can see that the matrices R and S are inverses of each other, Sab Rbc = δac . // SR = 1 (7.6.1)' (2.1.3) If desired, the matrix S can be eliminated from the discussion by the fact that (reflect indices in a vertical line between the indices) Sab = Rba Sab = Rba . (7.5.13)' (2.1.4) Then (2.1.2) can be then be written with only R's , V'a = RabVb Rab ≡ (∂x'a/∂xb) = ∂bx'a // V' = RV V'a = RabVb Rab ≡ (∂xb/∂x'a) = ∂'axb . (2.1.5) Here is a table summarizing different forms of the differentials R and S : Sab = Rba = (∂xa/∂x'b) = (∂x'b/∂xa) = ∂'bxa = ∂ax'b Sab = Rba = (∂xa/∂x'b) = (∂x'b/∂xa) = ∂'bxa = ∂ax'b Sab = Rba = (∂xa/∂x'b) = (∂x'b/∂xa) = ∂'bxa = ∂ax'b Sab = Rba = (∂xa/∂x'b) = (∂x'b/∂xa) = ∂'bxa = ∂ax'b . (7.5.16)' (2.1.6) Notice how an upper index in a derivative denominator acts as a lower index and vice versa. The fact that each item can be represented by two partial derivatives follows from (2.1.4), (2.1.2) and the raising and lowering operations described in Section 2.2 below. Vectors which transform under (according to) transformation F as in (2.1.5) are called rank-1 tensors. Here is how the four forms of a rank-2 tensor M transform under F, M'ab = Raa' Rbb' Ma'b' // pure contravariant M'ab = Raa' Rbb' Ma'b' // mixed M'ab = Raa' Rbb' Ma'b' // mixed M'ab = Raa' Rbb' Ma'b' . // pure covariant (7.5.8)' (2.1.7) We sometimes refer to Raa' as the "down-tilt" R matrix, and Raa' as the "up-tilt" R matrix. One sees that a down-tilt R transforms each contravariant (up) index, while an up-tilt R transforms each covariant (down) index. From the above, one can intuit the way an arbitrary rank-n tensor transforms under F. For example T 'abcde = Raa' Rbb' Rcc' Rdd' Ree' Ta'b'c'd'e' . (7.10.1)' (2.1.8) The various forms of the R matrix have these four orthogonality rules, 1: Rba Rbc = δac 2: Rba Rbc = δac // sum is on 1st index 3: Rab Rcb = δac 4: Rab Rcb = δac . // sum is on 2nd index (7.6.4)' (2.1.9) These are just renditions of (2.1.3) with (2.1.4) ( that is to say, SR = RS = 1 ). Using these rules, one can show (proof below) that the inverses of the vector transforms shown above in (2.1.5) are these Va = RbaV'b (7.6.7)' Va = RbaV'b (7.6.8)' (2.1.10) where the summation index b on R is not abutted against the following vector. The inversion of any tensor equation can be obtained instantly using the following simple rule: Inversion Rule: For each R, reflect the indices in a vertical line between the indices. (2.1.11) Examples: V'n = Rnm Vm , inversion is: Vn = Rmn V'm Rn| m → Rmn V'n = Rnm Vm , inversion is: Vn = Rmn V'm Rn| m → Rmn Proof : (2.1.8)#1 V'n = Rnm Vm RniV'n = RniRnmVm = δimVm = Vi Vi = RniV'n Vn = RmnV'm . The proof of the 2nd example follows from application of the tilt reversal rule (2.9.1) to the first example General Proof: Recall from (2.1.4) that Sab = Rba. The vertical line reflection Ra| b → Rba = Sab just changes R into S, and S is the inverse of R as in (2.1.3). So V' = RV gives V = SV' and similarly for higher tensor cases. Exercise: Invert equation (2.1.7) which says M'ab = Raa' Rbb' Ma'b' : Result: Mab = Ra'a Rb'b M'a'b' . The canonical vectors are the differential distances dxi in x-space and dx'i in x'-space (near some point x and corresponding x'). From (2.1.5) we then have, dx'a = Rabdxb // dx' = Rdx dx'a = Rabdxb . (2.1.12) From rule (2.1.11) the inverses of (2.1.12) are, dxa = Rbadx'b dxa = Rbadx'b . (2.1.13) The derivative operator ∂a ≡ ∂/∂xa transforms like any other covariant vector. It is in fact the canonical covariant vector for the transformation F. Thus, from (2.1.5), ∂'a = Rab∂b Rab ≡ (∂x'a/∂xb) = ∂bx'a ∂a = ∂/∂xa ∂'a = ∂/∂x'a ∂'a = Rab∂b Rab ≡ (∂xb/∂x'a) = ∂'axb ∂a = ∂/∂xa ∂'a = ∂/∂x'a (2.1.14) For example, if φ(x) is a scalar field (rank-0 tensor), ∂aφ(x) transforms as a covariant vector under F, and ∂aφ(x) transforms as a contravariant vector. Derivatives of tensors above rank-0 are more complicated, see Comment 1 below. The R matrix is really four matrices, and we have seen two of its forms above. The R object is not a tensor because, as Rab ≡ (∂x'a/∂xb) suggests, R has one foot in x'-space and one foot in x-space. In fact, the first index of R is raised or lowered by g', while the second index is raised or lowered by g: Rab = (∂x'a/∂xb) Rab = Rab' gb'b = (∂x'a/∂xb) // g pulls up the second index of Rab Rab = g'aa'Ra'b = (∂x'a/∂xb) // g' pulls down the first index of Rab Rab = g'aa'Ra'b' gb'b = (∂x'a/∂xb) . // both actions at once (7.5.9)' (2.1.15) Tensor fields We have suppressed the fact that in general everything above is a function of x (or equivalently x'). For example, when we compute Rab ≡ (∂x'a/∂xb) we generally obtain Rab(x). The transformation of a vector from x-space to x'-space was given in (2.1.5) as V'a = RabVb. For general x' = F(x) this really a statement about the transformation of vector fields: V'a(x') = Rab(x)Vb(x). The rank-2 tensor transformation in (2.1.6) really says M'ab(x') = Raa'(x) Rbb'(x) Ma'b'(x) and we are transforming a rank-2 tensor field. In special relativity it happens that x' = F(x) is linear so x'a = Fabxb (usually written with non-bold 4-vectors and Greek indices like x'μ = Λμνxν). In this situation Rab does not depend on x, and one can then have vectors which are not fields like p'μ = Λμνpν (momentum of a point particle) and vectors that are fields like A'μ(x') = ΛμνAν(x) (electromagnetic vector potential). Notice on the x'-space side of the equation that the vector field A'μ is a function of the x'-space coordinate x', while on the x-space side Aμ has argument x. In continuum mechanics and general relativity, Rab is a function of x so everything is a field. In the following examples, matrix R is the linearization of x' = F(x). In the neighborhood of the point x one has dx' = R(x)dx as a "linear fit" to the generally non-linear x' = F(x). If F(x) is a linear transformation, then x' = Fx so dx' = Fdx and then R = F = independent of x. (a) T' a'b'c'd' = Ra'aRb'bRc'cRd'd Tabcd tensor (rank 4, mixed) (b) T' a'b'c'd'(x') = Ra'a(x)Rb'b(x)Rc'c(x)Rd'd(x)Tabcd tensor which is not a field in x-space (c) T' a'b'c'd'(x') = Ra'aRb'bRc'cRd'd Tabcd(x) tensor field, linear F(x) (d) T' a'b'c'd'(x') = Ra'a(x)Rb'b(x)Rc'c(x)Rd'd(x) Tabcd(x) general tensor field (2.1.16) Item (a) is the generic form we use for a transformation of a rank-4 tensor. If F(x) is linear (as in special relativity), then Rab does not depend on position, and it is possible for T'abcd and Tabcd not to be fields. Item (b) is for a non-linear x' = F(x) where Tabcd is not a field. Obviously T'a'b'c'd' must depend therefore on x, and hence x', and so it is a field. In this unusual situation, the entire dependence of T' on x' is induced by the non-linearity of the transformation. Item (c) is more standard, where the transformation is linear and a tensor field is being transformed. This is the case in special relativity. Item (d) is the same thing, but the transformation is non-linear so one has Rij(x). Comments: 1. For situations where Rab is a function of x, it is easy to see why there is trouble with derivatives. One need only consider : V'a(x') = Rab(x)Vb(x) and ∂'b = Rbc(x)∂c so V'a,b(x') ≡ ∂'bV'a(x') // Va,b is just a new notation for the derivative = (Rbc(x)∂c)[Rad(x)Vd(x)] = Rbc(x) Rad(x) (∂cVd(x)) + Rbc(x)(∂cRad(x)) Vd(x) . (2.1.17) It is this second term that causes ∂cVd not to transform as a rank-2 tensor. It only transforms as a tensor if it happens that x' = F(x) is linear so Rab is constant (as in special relativity). This problem is remedied by introducing the covariant derivative Vd;c as discussed in Tensor Appendix F, see for example (F.9.5). This new object then properly transforms as a rank-2 tensor, V'b;a(x') = Rbc(x) Rad(x)Vd;c(x) . (2.1.18) 2. We have chosen to write (∂x'a/∂xb) as Rab as a space-saving notation. This object is often called "the differential" of the transformation x' = F(x) at point x. Tensor deals only with transformations where x and x' have the same number of components N, but the idea (∂x'a/∂xb) as Rab generalizes beyond this restriction. Of course then the matrix Rab is no longer square and the associated linear algebra is more complicated. This situation arises in Chapter 10 below. 3 Although we have used the letter R in Rab, one should not think that R is a rotation. It could be a rotation, but in general it is more complicated, involving both rotation and stretching. It could be a rotation which, although being a rotation, is a different rotation at every point in space, like Ry(θ(x)). 2.2 The metric tensors g and g' and the dot product Within each space (x-space and x'-space in the (2.1.1) Picture A), the metric tensor lowers or raises vector indices, Va = gabVb V'a = g'abV'b Va = gab Vb V'a = g'ab V'b (7.4.4)' (2.2.1) In the same way, the metric tensor lowers or raises any index on any tensor. The contravariant and covariant metric tensors are inverses of each other, gabgbc = gac = δac = δa,c // note that δij = δij = δi,j . (2.2.2) Here the gab lowers the first index on gbc to make gac which is δac = δa,c so gdngup = 1. The objects gab, gab, gba and gab are true tensor objects whereas δac and δa,c are not. It just happens that the value of gac is δac. In writing covariant equations, one should replace δac by gac before attempting to raise index a or lower index c. The metric tensor is a rank-2 tensor like any other rank-2 tensor, and so, looking at the first and last lines of (2.1.6), g'ab = Raa' Rbb' ga'b' g'ab = Raa' Rbb' ga'b' . (7.5.7)' (2.2.3) Any metric tensor is symmetric, gab = gba g'ab = g'ba gab = gba g'ab = g'ba . (5.4.3)' (2.2.4) The metric tensor defines a dot product in each space a b = gijaibj = gijaibj = aibi = aibi x-space a ' b' = g'ija'ib'j = g'ija'ib'j = a'ib'i = a'ib'i x'-space . (2.2.5) The dot product is a scalar (rank-0 tensor) so it must be the same in either space a ' b' = a b . (2.2.6) An exception to this rule is noted for fluid flow, see Tensor Section 5.2. When applied to the canonical differential vector dxi we find dx dx = gijdxidxj = || dx ||2 ≡ (ds)2 x-space dx' dx' = g'ijdx'idx'j = || dx' ||2 ≡ (ds')2 x'-space (2.2.7) Thus from (2.2.6) ds = ds' (invariant distance). In special relativity, ds is called the proper time dτ. The metric tensor gets its name from these last equations. The distance between two points x and y in a metric space is determined by a function called "the metric" d(x,y). A commonly used metric is d(x,y) = ||x-y|| where the metric is defined by the norm. The distance between two close points x and x+dx is then given by d(x,x+dx) = ||x+dx-x|| = || dx || = ds. Squaring, d2(x,x+dx) = || dx ||2 = gijxixj which shows how the metric tensor gij describes the squared metric d(x,x+dx) in the metric space of interest. One might recall that in a raw vector space, there is no distance concept d(v1,v2). A vector space with a metric and an inner product, such as that shown above as is then a Hilbert Space. It happens that the same metric tensor gij has two roles to play: it determines differential distance, and it lowers an index. See Chapter 5 of Tensor for more details. As shown in (2.1.16) (d), for a general transformation everything (including g) is a function of x. For example, Va(x) = gab(x)Vb(x) V'a(x') = g'ab(x')V'b(x') (2.2.1) g'ab(x') = Raa'(x) Rbb'(x) ga'b'(x) . (2.2.3) (2.2.8) 2.3 The basis vectors en and en There are two sets of basis vectors called en and en which exist in x-space (vector space V). The integer n is a label, not a component index. These basis vectors are defined as en = ∂x/∂x'n = ∂'nx tangent base vectors en = ∂x/∂x'n = ∂'nx . reciprocal base vectors (7.13.5)' (2.3.1) where x = F-1(x'). The tangent base vectors en are tangent to the "coordinate lines" in x-space. If in x'-space a particular x'n is allowed to vary while all other x'i are held fixed, the locus in x'-space is a line parallel to the n axis, while the mapping of that line in x-space is the (often curved) coordinate line associated with x'n. Then en = ∂x/∂x'n evaluated at a point x on that coordinate line is a vector in x-space tangent to that coordinate line. For coordinate line examples, see e.g. Tensor (3.2.8)', (3.4.3)' or (3.4.7)' . Meanwhile, the reciprocal vectors en are "dual to" the en in that en em = δnm . In fact we have, en em = g'nm g'in en = ei en em = δnm = g'nm en em = g'nm g'in en = ei (7.18.1)' (2.3.2) The equations on the left imply those on the right which show how to raise and lower basis vector labels. Although the en and en are vectors in x-space, it is the metric tensor of x'-space which raises and lowers. For "duality" see text above Tensor (6.2.8)'. It is shown there that a unique dual basis bn always exists for any given basis bn. For a general transformation F, the tangent base vectors are functions of location and should be written en(x), and of course the same is true for en(x). Looking at (2.3.2), we see that en(x) em(x) = δnm manages to be valid at every point in x-space. On the other hand, g'nm(x) = en(x) em(x) shows that the metric tensor is also a function of x. Example: In polar coordinates (r,θ) one has er = and eθ = r , both of which obviously depend on location in space. The metric tensor is gab = and also depends on spatial location through r. The coordinate lines for θ are circles whose tangents are eθ , while those for r are rays whose tangents are er. These basis vectors are like any other vectors in x-space, and so they have contravariant and covariant components such as (en)i and (en)i . Going back to our definition (2.3.1) we see that en ≡ ∂x/∂x'n (en)i = ∂xi/∂x'n = Rni . // from (2.1.5) Looking at (en)i = Rni we make these observations: (1) the n label of (en)i goes up and down with g' as shown in (2.3.2). (2) the n index of Rni also goes up and down with g' as shown in (2.1.13). (3) the i index of (en)i goes up and down with g as shown in (2.2.1). (4) the i index of Rni also goes up and down with g as shown in (2.1.13). Therefore the equation (en)i = Rni is "covariant", even though it is not a tensor equation (since R is not a tensor), so we can raise and lower indices at will on both sides. Therefore (en)i = Rni (en)i = Rni (en)i = Rni (en)i = Rni . (2.3.4) The en and en also satisfy a "completeness relation", (en)a(en)b = δab (2.3.5) where the implied sum is over the label n. From (2.3.4) this says Rna Rnb = δab which in fact is just orthogonality rule #2 in (2.1.8). This completeness relation is different from the "orthogonality relation" en em = δnm of (2.3.2) which, with x-space components, is written (en)i (em)i = δnm . Here the implied sum is over the component index i. Recall from above that en em = g'nm . (2.3.2) There are two cases that are often of interest: en em = fn δn,m the {en} form an orthogonal basis for V en em = δn,m the {en} form an orthonormal basis for V (2.3.6) Remember that the vectors en exist in x-space, despite the fact that the metric tensor g'nm is for x'-space in our transformation Picture A. 2.4 The basis vectors un and un One can also define a set of "axis-aligned" basis vectors in x-space as follows (un)i = δni = gni (un)i = δni = gni (7.18.3)' (2.4.1) which can be compared with (2.3.4) for en and en. Relations involving the u basis vectors are un um = gnm gin un = ui un um = δnm = gnm un um = gnm gin un = ui . (7.18.3)' (2.4.2) Notice the similarity to the relations for the en and en. The un are dual to the un. Whereas the en and en involve the x'-space metric tensor g', the un and un involve the x-space metric tensor g. We can easily calculate from (2.3.4) and (2.4.1) that en um = (en)i(um)i = Rni δmi = Rnm . (2.4.3) According to (2.3.2), g' raises and lowers the label n on en. According to (2.1.13), g' raises and lowers the first index n on Rnm. Similarly, the label on um and the second index of Rnm are raised and lowered by g. Thus our equation (2.4.3) is "covariant" (even though it is not a true tensor equation), so we can at once write out all four forms of the dot products between the e and u basis vectors on the left below, en um = Rnm en = Rnm um = Rnm um en um = Rnm en = Rnm um = Rnm um en um = Rnm un = Rmn em = Rmn em en um = Rnm un = Rmn em = Rmn em (2.4.4) The left column implies the right column (implied sum on m). For example, for the third line un = Σm Rmn em ek un = Σm Rmn ek em = Σm Rmn δkm = Rkn . (2.4.5) The equations on the right of (2.4.4) show that the R matrix is the "basis change matrix" relating the two different basis of x-space. The basis ui has a special place among possible bases for x-space. Above we call it an "axis-aligned" basis since (un)i = δni so for example in R2 we would have u1* = (1,0) and u2* = (0,1). Here the little asterisk is a notation to show we are talking about contravariant components. To say that (un)i = δni does not say that x-space is Cartesian, since x-space could have any metric tensor gij. What is really being said by the statement (un)i = δni is that the components of vectors (and higher tensors) are being defined in a specific way. If v is a vector, then vi has the following meaning: vi = ui v. Another way to say this is that the vi are the components of v when v is expanded in the u basis: v = Σi viui. In particular, since un is itself a vector, we have un = Σi (un)iui = Σi δniui = un. So the components of vectors are always defined relative to the u basis when u is selected as the "axis aligned basis" with (un)i = δni. The component indices on all tensors of Chapter 2 are referred to the u basis, for example the i and j of the metric tensor gij and of any tensor Mij or Mijk. We shall see below that one could expand vector v on the en basis, but then one gets v = Σi v'iei where v'i = ei v . The coefficient v'i here is of course different from vi since the bases are different. In fact v'i = Rijvj. 2.5 The basis vectors e'n and u'n and a summary Basis vectors e'n and u'n are just mappings of en and un from x-space to x'-space, u'n = R un un = S u'n u'n = R un un = S u'n e'n = R en en = S e'n e'n = R en . en = S e'n (2.5.1) Each of these mappings is like any vector mapping V' = RV. The primed basis vectors exist in x'-space, whereas the unprimed ones exist in x-space. The components of these vectors are easily computed, (u'n)i = Rij (un)j = Rij δnj = Rin // (2.4.1) (e'n)i = Rij (en)j = Rij Rnj = δin = g'in // (2.3.4) and (2.1.9) #3 . (2.5.2) Whereas the un were the axis-aligned basis vectors in x-space, we see that the e'n are the axis-aligned basis vectors in x'-space. Since (u'n)i = Rin from (2.5.2), and since Rin = (∂x'i/∂xn) = ∂nx'i from (2.1.6), we can write u'n = ∂nx' which is similar to the corresponding en = ∂'nx of (2.3.1) and shows that the u'n really are tangent base vectors for the inverse transformation. We summarize : (un)i = δni axis-aligned basis vectors in x-space en = ∂'nx tangent base vectors in x-space (e'n)i = δni axis-aligned basis vectors in x'-space u'n = ∂nx' inverse tangent base vectors in x'-space (2.5.3) The situation is depicted in this drawing, (2.5.4) The figure shows just one base vector of each type. Here red e1 is the tangent base vector for coordinate x'1, whereas blue u'1 is the tangent base vector of the inverse transformation x = F-1(x') for coordinate x1. In each case a light black curve represents a piece of a coordinate line (curve) whose tangent is the tangent base vector. The associated dot products are obtained from (2.4.2) and (2.3.2), u'n u'm = gnm // = un um gin u'n = u'i u'n u'm = gnm u'n u'm = gnm gin u'n = u'i (2.5.5) e'n e'm = g'nm // = en em g'in e'n = e'i e'n e'm = g'nm e'n e'm = g'nm g'in u'n = u'i (2.5.6) and the basis vectors can therefore be raised and lowered as shown on the right by an appropriate metric tensor. So far we have computed these basis vector components, (un)i = gni = δni (2.4.1) (en)i = Rni (2.3.4) (u'n)i = Rin (2.5.2) (e'n)i = g'in = δin (2.5.2) (2.5.7) By raising n, lowering i, or doing both, one arrives at 12 more equations to get this full set of 16, 1 (un)i = gni = δni (en)i = Rni 2 (u'n)i = Rin (e'n)i = g'in = δin 3 (un)i = gni (en)i = Rni 4 (u'n)i = Rin (e'n)i = g'in 5 (un)i = gni (en)i = Rni 6 (u'n)i = Rin (e'n)i = g'in 7 (un)i = gni (en)i = Rni 8 (u'n)i = Rin (e'n)i = g'in (2.5.8) The reason one is allowed to do this follows from the label raising and lowering relations shown on the right side of (2.3.2), (2.4.2),(2.5.5), (2.5.6), and finally from (2.1.15) concerning indices on Rij . Since (u'n)i = Rin , the contravariant u'n vectors are the columns of R**. Since (en)i = Rni , the covariant en vectors are the rows of R** : R** = [u'1*,u'2*, ...u'N* ] = (2.5.9) Fact: One can treat the en as an arbitrary set of basis vectors (2.5.10) Above we started off by assuming some arbitrary transformation x' = F(x) and then the en(x) are the tangent base vectors for this transformation F. From a different viewpoint, one can assume some arbitrary expressions for the en(x) and try to find a corresponding x' = F(x) for which those en(x) are the tangent base vectors. Given the functions en(x), one would know the matrix of functions Rni(x) from (2.5.8) item 1. One could then attempt to integrate (2.1.10) which says dx'a = Rab(x)dxb to find x' = F(x). Let's assume this is all doable so x' = F(x) can always be found. From this point of view, one can regard the above equations concerning the en(x) to apply to an arbitrary set of basis functions en(x). Of course they have to be linearly independent at each value of x. The next section provides a very simple example. 2.6 How to compute a viable x' = F(x) from a set of constant basis vectors en As a first step, select g = 1 so that x-space is the usual Cartesian space and un = un are the usual orthonormal axis-aligned unit vectors of Cartesian space. Suppose we are handed a set of constant-in-space basis vectors en specified by their components relative to the x-space axes. From (2.5.8) item 1, (em)n = Rmn (2.6.1) so we know the matrix Rab (the em are the rows of the R matrix). What is the simplest way to fit this scenario into the tensor environment of Picture A in (2.1.1)? We try a linear transformation of the form x' = F(x) = Fx where F is a constant matrix (independent of x). Since x'a = Fabxb we find Rab ≡ (∂x'a/∂xb) = Fab, and then of course Rab = Fab. So we have found a linear transformation F that works: Fab = Rab. Then the en are the tangent base vectors for this transformation F, and en are the reciprocal base vectors (the dual vectors of en). The metric tensor g'nm can be computed from the dot product (2.3.2), g'nm = en em = (en)i (em)i = (en)i (em)i = RniRmi . (2.6.2) Since g = 1, the second index on R is allowed to move up and down "for free". This g'nm can then be inverted to determine g'nm. The reciprocal base vectors are then given by (2.3.2), en = g'nm em (2.6.3) with components (en)i = g'nm (em)i = g'nm Rmi . (2.6.4) We then rewrite the above equations as (em)n = Rmn // the em are the rows of matrix R, (2.6.1) lower n g'nm = RniRmi = RniRTim = (RRT)nm g' = RRT // (2.6.2) lower i en = g'nm em = hnm em // where we define hnm ≡ gnm // (2.3.2) (en)i = hnmRmi = (hR)ni. // the en are the rows of matrix (hR), (2.6.4) (2.6.5) Exercise: You are handed these three constant vectors en in a 3-dimensional Cartesian x-space, e1 = (2,-1,3) // for example, (e1)2 =( e1)2 = - 1 e2 = (-1,2,4) e3 = (1,3,2) // e3 = 1 u1 + 3 u2 + 2 u3 so R** = // = F; "the em are the rows of matrix R" from (2.6.5) (2.6.6) Use Maple to compute g, h and then (hR). Here we are just implementing the equations in (2.6.5). First enter the three en vectors and construct matrix R**, FIX CODE!!!! ****** Then compute the covariant metric tensor g' = RRT , From this compute the contravariant metric tensor h = g'-1 and then matrix (hR) The rows of (hR) are the vectors en (called En in the code), As a check, we verify that en em = δnm : 2.7 Expansions of vectors onto basis vectors Given vector V in x-space and the corresponding vector V' in x'-space, one may write the following expansions ( Tensor (7.13.10,11) ), 1 V = Σn Vn un where Vn = un V axis-aligned basis 2 V = Σn Vn un where Vn = un V axis-aligned basis 3 V = Σn V'n en where V'n = en V tangent base vector basis 4 V = Σn V'n en where V'n = en V tangent base vector basis 5 V' = Σn V'n e'n where V'n = e'n V' axis-aligned basis 6 V' = Σn V'n e'n where V'n = e'n V' axis-aligned basis 7 V' = Σn Vn u'n where Vn = u'n V' tangent base vector basis 8 V' = Σn Vn u'n where Vn = u'n V' tangent base vector basis (2.7.1) Notice that expansions 7,8 are obtained from 1,2 by applying the R matrix, since V' = RV and u'n = Run. Similarly, expansions 5,6 are obtained from 3,4 Any expansion can be directly verified by dotting the left column into an appropriate basis vector. For example, for expansion 6, using e'm e'n = δmn , V' = Σn V'n e'n e'm V' = e'm (Σn V'n e'n ) = Σn V'n e'm e'n = Σn V'n δmn = V'm . (2.7.2) Notice that each set of coefficients appears twice on the right in (2.7.1), once for V and once for V'. This duplication arise because a b = a' b' for any pair of vectors in either space. For example, Vn = un V = u'n V' appears in lines 1 and 7 (2.7.3) Looking a bit ahead, we shall be extending the notion of a vector expansion to that of tensors of any rank, and the meaning of component indices is still governed by Fact (2.7.5). First, for expansions on the axis-aligned un we write, V = Σn Vn un rank-1 tensor, (2.7.1) line 1 M = Σnm Mnm unum rank-2 tensor T = Σii....i Tii....i (ui ui ..... ui) . rank-k tensor (2.7.8) As discussed in (2.5.10) we may regard the basis vectors en as being an arbitrary basis. The primed tensor components are then the components of the x'-space version of the tensor under the transformation x' = F(x) generated by those arbitrary en. The corresponding expansions are then, V = Σn V'n en rank-1 tensor, (2.7.1) line 3 M = Σnm M'nm enem rank-2 tensor T = Σii....i T' ii....i (ei ei ..... ei) . rank-k tensor (2.7.9) In the above, any pair of tilted matching indices can be tilted the other way. The meaning of the symbol is discussed below. For an expansion on a mixed basis like enum , we will use the following notation M = Σnm [M(e,u)]nm enum . (2.7.10) Using this same notation we could write, M = Σnm M'nm enem = Σnm [M(e,e)]nm enem ≡ Σnm [M(e)]nm enem . (2.7.11) Exercise: Consider these two expansions shown above of the rank-2 tensor M, M = Σab Mab uaub // (2.7.8) (2.7.12) M = Σab M'ab eaeb // (2.7.9) (2.7.13) Verify that the coefficients M'ab and Mab are related as expected. Use the result in (2.4.4) line 2 that em = Rmiui to get for (2.7.12), M = Mab uaub // all implied sums = Mab (Ria ei) (Rjb ej) = (RiaRjbMab) ei ej so comparing to (2.7.13) one concludes that M'ab = RiaRjbMab which is the correct statement that M transforms as a rank-2 tensor as shown in (2.1.7). Confusion about vectors and scalars The following issue is a subtle one that is worth nailing down early on because it can lead to confusions and seeming paradoxes. Consider this fact, un V = u'n V' (2.7.14) Being the dot product of two vectors, this object transforms as a scalar under x' = F(x) as shown in (2.2.6). We might express this fact by writing s(n) = un V s'(n) = u'n V' s(n) = s'(n) n = 1,2....N . (2.7.15) What we have here is a set of N scalars, s(n) for n = 1,2..N, where n is just a label. One would never claim that these scalars s(n) transform as a vector, which would require that s'(n) = Rnm s(m). We don't have such a relation; what we have is s'(n) = s(n). Now it happens that s(n) = s'(n) = Vn where Vn is the component of a vector. Notice that : V'n = RnmVm true : the Vn transform as a vector V'n = Vn false s(n) = s'(n) true : each s(n) = un V transforms as a scalar s'(n) = Rnms(m) false (2.7.16) We can summarize this discussion as follows: Fact: Just because the scalars s(n)= un V take the values Vn does not mean that the scalars s(n) transform as vectors, nor does it mean that the vector components Vn transform as a scalar. (2.7.17) 2.8 The Outer Product of Tensors and Use of Consider two vectors which transform in the usual rank-1 tensor manner relative to some underlying transformation x' = F(x) (for which R is the linearization), a'i = Σj Rij aj b'k = Σm Rkm bm from (2.1.5) (2.8.1) where we temporarily show the summation symbols. Multiplying these equations together gives (a'i)(b'k) = (ΣjRijaj)(ΣmRkmbm) = Σjm RijRkm(ajbm) and then hiding the sums again, (a'ib'k) = RijRkm(ajbm) . (2.8.2) Looking at the first line of (2.1.7), we see that this object is transforming as a rank-2 tensor, therefore it is a rank-2 tensor, and we can write it as M'ik = RijRkmMjm where Mij ≡ aibj . (2.8.3) The rank-2 tensor Mij = aibj is said to be the outer product of two rank-1 tensors (vectors). This idea can be generalized ad infinitum. For example, if K is a rank-2 tensor and v is a vector, then Mijk = Kijvk (2.8.4) is a rank-3 tensor because it transforms as one, using the same argument shown above. Next, consider, Mabcde = KabKcd ve . (2.8.5) If K is a rank-2 tensor and v is a rank-1 tensor, then M is a rank-5 tensor. Of course since this is a "true tensor equation" (a covariant one), indices may be shuffled any way one wants, such as Mabcde = KabKcd ve . (2.8.6) Just imagine applying g** several times to both sides of (2.8.5) to get (2.8.6). There are so many possibilities for creating outer product tensors that one sometimes forgets that not all tensors can be "factored" into products of lower rank tensors. The Symbol Appears subtitle: "the rabbit goes into the hat" In Sections 1.1 and 1.2 we had vw being an element of a "tensor product space" VW and we described two approaches to the development of the meaning of the symbol : quotient space and category theory. Here we provide a third approach to the meaning of which is equivalent to that of the first two approaches. This third approach is geared to dealing with tensor components so there are lots of indices floating around, whereas in Sections 1.1 and 1.2 components were not even mentioned. Recall our previous two equations Mabcde = KabKcd ve . (2.8.5) Mabcde = KabKcd ve . (2.8.6) In order to display the outer product as a unified entity, we had to make up a new symbol M to represent the outer product tensor. We can avoid having to do this by writing M = KKv, so that the symbol in our "third approach" is just a way to name an outer product tensor. The above equations are then (KKv)abcde = KabKcd ve . (2.8.7) (KKv)abcde = KabKcd ve . (2.8.8) Here K and v are tensors, they are not spaces, so this is more like vw than VW. In fact, as a special case we can use this idea to name the outer product of two vectors to be rank-2 tensor ab, (ab)ij = aibj a,b ϵ V . (2.8.9) Notice that is a non-commuting operator: ab ≠ ba . In our "third approach" the symbol exists only within the context of VV, since vectors a and b both belong to the x-space of Picture A (2.1.1) which we identify with vector space V. However, one can extend this meaning of to apply more generally as the outer product of vectors in different vector spaces, (vw)ij = viwj v ϵ V w ϵ W . (2.8.10) If we try to fit this into our notion of tensor transformations, we would need two copies of Picture A, one for U→V and the other for X→W with vector transformations v(V)i = R(V)ijv(U)j R(V)ij = linearization of some transformation x' = F(V)(x) w(W)i = R(W)ijw(X)j . R(W)ij = linearization of some transformation y' = F(W)(y) (2.8.11) Then the transformation of the outer product "tensor" would be written as, (v(V)iw(W)i) = R(V)iaR(W)jb (v(U)aw(X)b) or [(vw)(V,W)] ij = R(V)iaR(W)jb [(vw)(U,X)]ab . UX → VW (2.8.12) One might refer to (vw)(V,W) as a "cross-space rank-2 tensor" (cross tensor). Normally the word "tensor" is used when W = V. Then the above reads, [(vw)(V,V)] ij = R(V)iaR(V)jb [(vw)(U,U)]ab UU → VV or (vw)' ij = RiaRjb (vw)ab . // Picture A (2.1.1) (2.8.13) The outer product thus has the same form in x'-space and in x-space, being a rank-2 tensor, (ab)ij = aibj a,b ϵ V . (ab)'ij = a'ib'j a',b' ϵ V' . (2.8.14) As a final outer product example, consider the outer product of three vectors, Mijk = aibjck a,b,c ϵ V , (2.8.15) Using our naming method for outer products, this becomes (abc)ijk = aibjck (2.8.16) with this obvious extension to the outer product of any number of vectors (abc....)ijk.... = aibjck.... (2.8.17) Associativity of The outer product operator as defined here is an associative operator, because multiplication of real numbers is associative. Consider for example, (ABv)abcde = AabBcdve (multiplication of reals is associative) [(AB)v]abcde = [(AB)abcd] ve = [ AabBcd] ve = AabBcdve . (2.8.19) Adding the parentheses on the second line in (AB)v does not alter the value of the components. This is true for the tensor product of any number of tensors, (T1T2T3....TN)III...I = T1IT2IT3I .... TNI (2.8.20) where each Ii represents a set of indices to go with Ti. For example, (T1(T2T3)....TN)III...I = T1I(T2T3)II ... TNI = T1I[ T2IT3I] ... TNI = T1IT2IT3I .... TNI (2.8.21) Therefore we have, Fact: The operator is associative for any tensor product, so parentheses can be added anywhere in a tensor product. (2.8.22) The associativity of the product of a set of real numbers along with the outer product definition of is what causes the operator to be associative. With the abstract definitions of Chapter 1, associativity of is added by fiat as an axiom. 2.9 The Inner Product (Contraction) of Tensors When any tensor structure contains a pair of implicitly summed indices which are "tilted", one says that those indices are contracted. It is easy to show that, due to the orthogonality rules (2.1.9), such internal index contractions behave as a scalar, which is to say, behave as if they weren't there at all with respect to a transformation. A proof appears in Tensor (7.12.2). Such contractions in a tensor structure reduce the rank of the tensor by two, resulting in an inner product. The contracting sum must occur only on a "tilted pair" of indices. Tilt Reversal Rule: Any such tilted index pair can have its tilt reversed "for free". (2.9.1) Proof: Using (2.2.1) and (2.2.2), [-----a---------a----] = gab gac [-----b---------c----] = gba gac [-----b---------c----] = δbc [-----b---------c----] = [-----b---------b----] = [-----a---------a----] where dashes indicate up or down tensor indices we don't care about. This "tilt reversal rule" applies to any contracted index within a tensor expression. It applies as well in other cases where g raises and lowers things so the above proof still works. The classic example involves expansions of the form (2.5.1) V = V'n en = V'n en . (2.9.2) The tilt can be reversed even though the n on en is a label and not a tensor index. The reason is that en = g'ni ei en = g'ni ei (2.3.2) V'n = g'nbV'b V'n = g'nb V'b . (2.2.1) The standard first example of an inner product is the inner product of two vectors. Consider, Mij = aibj = a rank-2 tensor, which we now contract to form, s = Mii = aibi = aibi = a rank-0 tensor (a scalar) (2.9.3) Using our notation (2.2.5) this is written s = a b // <a | b> in Dirac notation (2.9.4) which is an "inner product" of two vectors. This is of course the inner product / scalar product / dot product which makes our vector space V be a Hilbert space. In this example, creating an "inner product" of the two vectors ai and bj which has rank-0 goes in the opposite direction of the "outer product" that creates Mij = Mij = aibj of rank-2. The term "contraction" is more often applied to reducing the rank of tensors than is "inner product", and perhaps it is best to reserve the term "inner product" for the above dot product of two vectors. Here are other examples of rank reduction by contraction. Define Mabcd ≡ KabQcd = rank-4 tensor (2.9.5) Tac ≡ Mabcb = KabQcb = rank-2 tensor . (2.9.6) In this last example, contraction on the b index happens to occur between the two rank-2 tensors from which M was constructed as an outer product. One more step, S ≡ Taa = KabQab = rank-0 tensor (scalar) . (2.9.7) Using the notation introduced in the previous section, we can write the inner product s = a b as a contraction of the outer product ab s = (ab)ii = (ab)ii and ||a||2 ≡ aiai = (aa)ii . (2.9.8) Similarly (2.9.5,6,7) can be written (KQ)abcd = KabQcd = rank-4 tensor (2.9.9) Tac = (KQ)abcb = rank-2 tensor (2.9.10) S = Taa = (KQ)abab = rank-0 tensor (scalar) . (2.9.11) Dot products in spaces VV, VW, VVV and VWX Recall that (2.2.5) defines the (covariant) dot product of two vectors in V a b = gijaibj = gijaibj = aibi = aibi . x-space = V (2.2.5) It is possible to define an inner product operator for use between two elements of VV : (ab) (cd) ≡ Σij(ab)ij(cd)ij (ab), (cd) ϵ VV (2.9.12) = Σijaibjcidj . With this definition, we have a tiny theorem: Theorem: (ab) (cd) = (ac)(bd) a,b,c,d ϵ V (2.9.13) Proof: (ac)(bd) = (Σi aici)( Σj bjdj) = Σij aicibjdj = Σij aibj cidj = Σij(ab)ij(cd)ij = (ab) (cd) . Suppose dim(V) = n and dim(W) = n'. Then we can extend the above theorem to VW in this way. First define the dot product as, (ab') (cd') ≡ Σi=1nΣj=1n'(ab')ij(cd')ij (vw),(v'w') ϵ VW (2.9.14) = Σijaib'jcid'j . The corresponding Theorem is then Theorem: (ab') (cd') = (a c)(b' d') a,c ϵ V b',d' ϵ W (2.9.15) Proof: (ac)(b'd') = (Σi=1n aici)( Σj=1n b'jd'j) = Σijaicib'jd'j = Σijaib'jcid'j = Σij (ab')ij (cd')ij = (ab') (cd') . In a similar fashion one can show using (2.8.17) that with the following definition, (abc) (def) ≡ Σijk (abc)ijk (def)ijk VVV (2.9.16) one obtains Theorem: (abc) (def) = (a d)(b e)(c f) all vectors ϵ V (2.9.17) with a similar extension to VWX, Theorem: (ab'c") (de'f") = (a d)(b' e')(c" f") a,d ϵ V; b',e' ϵ W; c",f"' ϵ X (2.9.18) 2.10 Tensor Expansions Having a name for the outer product of two vectors allows us to write expansions of tensors of rank greater than 1 in a compact notation. The template is the vector expansion from (2.7.1) line 3, V = Σa V'a ea . (2.10.1) (a) Rank-2 Tensor Expansion and Projection As shown in (2.7.9) and (2,7,13), one can expand a rank-2 tensor on the tangent base vectors as follows, M = Σab M'ab eaeb . (2.10.2) To verify that this is the correct expansion, we take the components of both sides, [M]'ij = [ Σab M'ab eaeb]'ij = Σab M'ab (eaeb)'ij = Σab M'ab (e'a)i (e'b)j // (2.8.14), outer product in x'-space = Σab M'ab δaiδbj // (2.5.8) line 2 used twice = M'ij (2.10.3a) so the expansion is correct. Similarly, M = Σab Mab uaub [M]ij = [ Σab Mab uaub]ij = Σab Mab (uaub)ij = Σab Mab (ua)i (ub)j // (2.8.14), outer product in x-space = Σab M'ab δaiδbj // (2.5.8) line 1 used twice = Mij . (2.10.3b) A convenient notational method for projecting out the coefficients of any tensor expansion is the use of tensor-product-space dot products defined in Section 2.9. To demonstrate, we use a tensor expansion in VW where the basis vectors are en and e'n for V and W, using notation of (2.7.9), M = Σab [M(e,e')]ab eae'b . M ϵ VW . (2.10.4) The appropriate projector is (eie'j), which is just the expansion's basis eae'b with up/down toggled on the indices, and dummy labels like i,j selected. Using this projector one finds, (eie'j) M = Σab [M(e,e')]ab (eie'j) (eae'b) = Σab [M(e,e')]ab (ei ea)(e'j e'b) // theorem (2.9.15) = Σab [M(e,e')]ab δia δjb // dual pairs as in (2.3.2) = [M(e,e')]ij (2.10.5) and indeed, the coefficient is duly projected out of M. Here is more complicated example where M is now a rank-3 tensor, and where we use a perverse mixed basis, M = Σabc [M(e,u',e")]abc ea u'b e"c M ϵ VWX . (2.10.6) The projector is (ei u'j e"k) and we use it to project out the coefficient in (2.10.6) : (ei u'j e"k) M = (ei u'j e"k) Σab [M(e,u',e")]abc ea u'b e"c = Σab [M(e,u',e")]abc (ei u'j e"k) (ea u'b e"c) = Σab [M(e,u',e")]abc (ei ea)(u'j u'b)(e"k e"c) // theorem (2.9.18) = Σab [M(e,u',e")]abc δia δjb δkc // each pair is dual as in (2.3.2) = [M(e,u',e")]ijk . (2.10.7) (b) Rank-k Tensor Expansions and Projections A rank-k tensor T in Vk has this expansion on the er basis, T = Σii....i T'ii....i (ei ei ..... ei) . (2.10.8) To verify, we take components of both sides, [T]jj...j = Σii....i T'ii....i (ei ei ..... ei)jj...j = Σii....i T'ii....i (ei)j (ei)j .....(ei)j // (2.8.17) = Σii....i T'ii....i Rij Rij .....Rij // (2.5.8) line 1 = Σii....i [ Rij Rij .....Rij T'ii....i ] = Tjj...j . (2.10.9) To get the last step, we use the inversion rule (2.1.11) applied to the known tensor transformation T'jj...j = Rji Rji .....Rji Tii....i . (2.10.10) The coefficients T'ii....i can be projected out from T as in (2.10.5), (eiei... ei) T = T'ii...i (2.10.11) with an appropriate generalization of the dot product to the space Vk = VV...V , (v1v2...vk) (w1w2...wk) ≡ Σii....i (v1v2...vk)ii....i (w1w2...wk)ii....i = Σii....i (v1)i(v2)i... (vk)i (w1)i(w2)i... (wk)i // outer products = (v1 w1) (v2 w2) .... (vk wk) . (2.10.12) Using the notion of a multiindex I (an ordinary multiindex), I ≡ i1, i2, .....ik // each ir ranges 1,2....n n = dim(V) (2.10.13) and a shorthand notation for the basis vectors eI ≡ ei ei ..... ei eI ≡ ei ei ..... ei (2.10.14) the expansion (2.10.8) can be stated in the following compact form, T = ΣI T'I eI (2.10.8) (2.10.15) and the coefficients T'I can be projected out according to (2.10.11), eI T = T'I . (2.10.11) (2.10.16) With no comments, we now repeat the above set of steps for the expansion of T on the ur basis: T = Σii....i Tii....i (ui ui ..... ui) (2.10.17) [T]jj...j = Σii....i Tii....i (ui ui ..... ui)jj...j = Σii....i Tii....i (ui)j (ui)j .....(ui)j // (2.8.17) = Σii....i Tii....i δij δij .....δij // (2.5.8) line 1 = Tjj...j (2.10.18) (uiui... ui) T = Tii...i (2.10.19) uI ≡ ui ui ..... ui uI ≡ ui ui ..... ui (2.10.20) T = ΣI TI uI (2.10.21) uI T = TI . (2.10.22) 2.11 Dual Spaces and Tensor Functions We denote dual-space vectors and tensors by Greek or script font letters. The dual space V* is by definition the space of linear functionals over V. If α ϵ V*, we can then write α : V → K α(v) = k ϵ K (2.11.1) where K is any field (but we always use the reals). Since α is a linear functional, α(v) is a linear function. In normal calculus, if f: V → R, one refers to f as a function, and f(v) as that function evaluated at some point in V, though loosely speaking f(v) is also called a function. To emphasize the distinction, we shall refer to f as a "functional" and f(v) as a "function". Comments: Much of the rest of this section will be repeated in later Chapters. We have found that the notations involved can be a major stumbling block, and feel it is important to exercise the notation in many ways to make the reader (and author) feel comfortable with it. As with most endeavors, it is a matter of practice. We also try to explain why certain notations are used. (a) The Dual Space V* in Matrix and Dirac Notation For every column vector v in V, there exists a row vector vT such that (vT)i = vi. For example, for N=2, v = = |v> vT = (a, b) = <v| (2.11.a.1) Here we have snuck in the Dirac bra-ket notation where the ket |v> is a column vector and the bra <v| is the corresponding row vector. The notation vT means that the row vector is the Transpose of the column vector. We now have multiple ways to write the dot (inner, scalar) products of Section 2.9 : v v' = vTv' = (a, b) = aa'+bb' = <v | v' > . (2.11.a.2) Because our vectors have real components, the above can also be written v' v = v'Tv = (a', b') = aa'+bb' = <v' | v > . (2.11.a.3) We mention real components only because in the Dirac notation one has <a|b> = <b|a>* where * means complex conjugation, so if this scalar product is real, then <a|b> = <b|a> . We regard v or |v> as being a vector in the vector space V, while vT or <v| (the row vector) is a vector in the dual space V*. This is really a simple concept. Sometimes the dual-space vector vT = <v| is referred to as the covector of v = | v > . Suppose αT = <α | is a vector in the dual space V* . One can regard this dual-space vector <α | as being a functional which acts on vectors in the space V. Then, α = <α | = functional α(v) = <α | v> = α v = function = a scalar number (2.11.a.4) α : V → K . // K = any real field, such as the real numbers Just as the space of column vectors V is a linear space (a vector space), so also the dual space of row vectors V* is a linear space, so we know that the functional <α | is a linear functional. That in turn implies that the function α(v) is a linear function, which we now show directly: α(sv) = α (sv) = s ( α v ) = s α(v) α(v+v') = α (v+v') = α v + α v' = α(v) + α(v') . (2.11.a.5) (b) Functional notation We have now a slight notational conundrum. We like to write a scalar-valued function F(v) in non-bold font, whereas a vector-valued function would be F(v). Thus we have written α(v) above with a non-bold α, since α(v) is a scalar-valued function. On the other hand, α(v) is really a function of the vector α, so it seems misleading to refer to it as α(v), and we ought to call it α(v) so then α(v) = <α | v> has everything bolded on both sides. But then the functional would have to be called α = <α | . But this contradicts our notation earlier that α is a vector, αT is the transpose, and we should write αT = <α |. If we use α = <α | then we avoid that contradiction. This is what authors end up doing, writing a functional as a scalar entity which for us means an unbolded entity. A possible solution would be to say, fα = <α | = functional fα(v) = <α | v> = α v = function (2.11.b.1) where f is non-bold, and the subscript label α is bold, but then we have introduced a new symbol f which seems superfluous. So the conclusion is this: α = <α| = functional, α(v) = <α | v> = function, and one must understand that α(v) is a function of the vector quantity α. Obviously there is a unique functional α(v) for each vector α in V (and thus for each αT in V*). The spaces V and V* have the same dimension n and are isomorphic to each other in the sense just noted. (c) Basis vectors for the dual space V* Now recall that our axis-aligned x-space basis vectors ui have dual basis vectors ui where ui uj = δij which is the idea of orthogonality in the covariant world (which might be non-Cartesian). In our notations above, ui uj = (ui)T uj = <ui | uj> = δij = ui uj = (ui)T uj = <ui | uj> = δij = δi,j . (2.11.c.1) Since ui and ui are in general different column vectors in V, (ui)T and (ui)T are different row vectors in the dual space V*. Just as the column vectors |ui> and |ui> form two distinct bases for V, the row vectors <ui | and <ui | form two distinct bases for V*. Certainly dim(V) = dim(V*). Definition of λi Above we discussed α = <α| as a vector functional, and α(v) = <α|v> as the corresponding scalar function. Whereas <α| is some general vector in V*, we now consider in its place a basis vector <ui| in V*. With what notation shall we represent this functional? In analogy with α and α(v) we could use ui and ui(v) where the ui is unbolded to indicate a scalar function. Or we could use fu = <ui| and fu(v) = <ui| v> . The first notation is not uncommon (see wiki dual space where ui = ei), while the latter notation is unpleasant. Other common notations are v*i(v) or e*i(v) which for us would be u*i(v). We shall use the following notation, λi ≡ <ui| basis functional in V* // λi = (ui)T so λi(v) = <ui|v> basis function in V*f // λi(v) = (ui)Tv . (2.11.c.2) The λ is unbolded, consistent with α(v). λ is a Greek letter consistent with our plan to use Greek or script letters for dual space objects. The index on λi is up, matching the index on ui in <ui|. Notice that λi(uj) = <ui|uj> = δij . (2.11.c.3) We think of λi ≡ <ui| as being in the dual space V* while λi(v) = <ui|v> lies in a directly corresponding space of functions which we call V*f. Comment: For the dual space basis functionals Sjamaar used symbol λi in his 2006 notes (p 84), but changed to βi in his 2015 update (p 91). Spivak uses φi (p 76). Wiki (dual basis) uses basis vectors vi instead of ei so their λi is called vi. Wiki (dual space) uses ei while Lang Algebra uses fi (p 143). There seems to be no standard notation as in physics where F = ma is universally recognized as Newton's Second Law which would be hard to identify if written G = nb. Probably ui or ui (unbolded) is the most logical choice if the V basis vectors are ui, but it is so easy to confuse functional ui with the vector ui (especially when we drop our bolding of vectors starting in Chapter 3) that we shall stick with λi. Eq. (2.7.1) line 3 gives the expansion of a vector v onto the ui v = Σi vi ui where vi = ui v or |v> = Σi vi |ui> where vi = <ui| v > = ui v . (2.11.c.4) Notice therefore that λi(v) = <ui|v> = ui v = vi . (2.11.c.5) The function λi(v) is sometimes called "the ith coordinate function" since it projects out the ith component the vector v. As summarized in (2.7.17), since each dot product ui v is a scalar, the functions λi(v) i = 1..N transform as scalars despite the fact that the values of these scalars are the components of the vector vi. Transposing (2.11.c.4) produces a vector functional expansion in V*, vT = Σi vi (ui)T or <v| = Σi vi <ui| = Σi vi <ui| . (2.11.c.6) Using Greek letters for dual space objects we write this as α = <α| = Σn αi <ui| = Σn αiλi . (2.11.c.7) Then, α = Σiαiλi functional (2.11.c.8) α(v) = Σiαiλi(v) = Σiαivi = α v function (2.11.c.9) or in bra-ket notation, <α| = Σiαi<ui| functional α(v) = Σiαi<ui|v> =Σiαivi α v = <α |v> function (2.11.c.10) and we replicate the result (2.11.b.1). Definition of λ'i We have defined λi ≡ <ui| as a notation for a certain basis functional in dual x-space. We would like to somehow define an object λ'i which is a basis functional in dual x'-space. How should this be done? One might intuitively feel that one should set λ'i ≡ <u'i| . Or one might think that once λi is defined as above, then the meaning of λ'i is forced upon us by some equation like λ'i = Rijλj . Both these notions are not what we want to do. We are not forced to say λ'i ≡ <u'i| just because λi ≡ <ui| since we are making two separate definitions. And λ'i = Rijλj is complete nonsense for the following reason. The N functionals λi for i = 1..N are each vectors in V*, so {λi} is a set of vectors, not a set of numbers, whereas when one tries to write λ'i = Rijλj one is implying that λj is a set of numbers which form a vector. Recall that the ui are the "axis-aligned" basis vectors in x-space since (ui)j = (ui)j = δi,j. Recall that the e'i are the "axis-aligned" basis vectors in x'-space since (e'i)j = (e'i)j = δi,j. This suggests that the proper definition of λ'i is the following: λ'i ≡ <e'i| (2.11.c.11) One then finds that, for v' a vector in x'-space, λ'i(v') ≡ <e'i| v'> = v'i (2.11.c.12) which is then analogous to λi(v) ≡ <ui| v'> = vi (2.11.c.5) In both cases then λi and λ'i are the "ith coordinate functions", projecting out the ith coordinate from a vector. Since v'i = Rijvj we can certainly write λ'i(v') = Rij λj(v) (2.11.c.13) as a statement relating two vectors of scalars. Notice this does not say λ'i = Rijλj which we already noted above does not even make sense. If we display the fact that Rij in general is Rij(x) then λ'i(v') = Rij(x) λj(v) (2.11.c.14) Since this does not fit into any of the molds shown in (2.1.16), one cannot quite claim that λj(v) transforms as a vector field, but the transformation is similar. We can study (2.11.c.13) in Dirac notation as follows (see below for Dirac notation details). λ'i(v') = <e'i|v'> = <ei|v> = <ei| 1 | v> = <ei|uj >< uj| v> = ei uj λj(v) = Rijλj(v) (2.11.c.15) where the last step comes from (2.4.3). Once we have a functional λ'i, we can define a general rank-1 functional α' in dual x'-space as follows: α' = Σiα'iλ'i functional in V'* (2.11.c.16) α'(v') = Σiα'iλ'i(v') = Σiα'iv'i = α' v' function in V'*f (2.11.c.17) It then follows that α'(v') = α' v' = α v = α(v) (2.11.c.18) and in some sense one could say that α(v) transforms as a scalar field, where v plays the role normally occupied by the position vector x. On the other hand, the vector |v> and the dual vector (functional) α = <α| transform as vectors and so α is a vector functional. We now define α(v) to be a "rank-1 tensor function". Spivak would call it a "1-tensor". We have this seeming contradiction that α(v) is a rank-1 tensor function, yet that function transforms as a rank-0 scalar. The rank-1 description really applies to the functional α = <α| which is in fact a vector and transforms as a vector. When this is closed with the ket |v> one obtains the scalar object α(v) = <α | v>. (2.11.c.19) Vector space names: V, V*, V*f and V', V'*, V'*f (2.11.c.20) Here we have associated vector space names V and V* with x-space in Picture A (2.1.1), while V' and V'* are associated with x'-space. All these spaces have the same dimension N and all are isomorphic. There is a 1-to-1 relationship between V and dual space V* as noted above, and there is a 1-to-1 relationship between V and V' since for every vector v in x-space there is a unique corresponding vector v' = Rv in x'-space. We refer to V* as dual x-space and V'* as dual x'-space. Associated with the dual space V* of functionals is the space V*f of corresponding functions, and similarly for V'* and V'*f. (d) Rank-2 functionals and tensor functions A rank-2 tensor may be represented as T = Σab Tab ua ub (2.11.d.1) V = Σa Va ua where on the second line for comparison we show a general rank-1 tensor (vector). In Dirac notation, we write | ua, ub> ≡ |ua> |ub> ↔ ua ub (2.11.d.2) which represents any of the n2 basis vectors of the tensor product space V2 = VV. We could write this as | ua, ub>2 ≡ |ua>1 |ub>1 to distinguish the fact that some kets are in V1 and others in V2, but the contents of the ket usually make it obvious to which vector space a ket belongs. In Dirac notation, the tensor T is written |T> = ΣabTab |ua> |ub> = ΣabTab | ua, ub> (2.11.d.3) and this is a general element of the space V2. The corresponding rank-2 linear functional in the dual space V*2 is given by <T| = ΣabTab <ua | <ub| = ΣabTab < ua, ub| . (2.11.d.4) This is done in analogy with the vector case |V> = Σa Va |ua> <V| = Σa Va<ua | (2.11.d.5) where we are careful to have the index "tilt" have the form of a contraction, even though we are not really contracting indices on a tensor. The rank-2 functional <T| is linear in both V* spaces of V*V*, so it is called a bilinear functional. If we let (subscript 1 and 2 are labels of two vectors, not components of v ) | v1, v2> ≡ |v1> |v2> (2.11.d.6) represent an arbitrary (but pure) element of V2 = VV, then we may construct T = <T| rank-2 tensor functional T(v1,v2) = <T | v1, v2> rank-2 tensor function (a Spivak "2-tensor") . (2.11.d.7) It follows that T(v1,v2) = <T | v1, v2> = ΣabTab < ua, ub| v1, v2> = ΣabTab <ua| v1> <ub| v2> = ΣabTab (v1)a (v2)b (2.11.d.8) where we have used the fact that the scalar product for elements of V*2 with elements of V2 is the product of two V*-with-V scalar products, as seen for example in (2.9.13). In the last line above we see that the tensor function T(v1,v2) is the contraction of a rank-2 tensor with two rank-1 tensors, and so is a scalar. Thus, T'(v'1,v'2) = T(v1,v2) (2.11.d.9) and a rank-2 tensor function transforms as a "scalar field of two arguments". The "rank-2" description applies to the tensor functional <T| , and when this is closed with an element of V2 the result is a scalar. Note from (2.11.d.8) and (2.4.1) that (ui)a = δia , T(ui,uj) = ΣabTab (ui)a (uj)b = ΣabTab δia δjb = Tij (2.11.d.10) so the tensor function evaluated at the basis vectors gives a corresponding element of the tensor. Using λi = < ui| as defined above in (2.11.c.2), we can rewrite (2.11.d.4) <T| = ΣabTab <ua | <ub| as rank-2 tensor functional T = ΣabTab λa λb (2.11.d.11) which is in analogy to <α | = Σa αa <ua| rank-1 tensor functional α = Σa αa λa . (2.11.d.12) We continue to use script or Greek fonts to represent functionals, such as α and T. Taking the special case of a rank-2 functional which is just λa λb we construct the following rank-2 tensor function, (λa λb)(v1,v2) = < ua, ub| v1, v2> = <ua| v1> <ub| v2> = (v1)a (v2)b = λa(v1) λb(v2) . (2.11.d.13) This function is manifestly linear in both arguments, since λa(v1) is linear, so it is a bilinear function. For example, (λa λb)(v1+v1',v2) = (v1+v1')a (v2)b = (v1)a (v2)b + (v'1)a (v2)b = (λa λb)(v1,v2) + (λa λb)(v'1,v2) . (2.11.d.14) Whereas <T| shown above is a general rank-2 tensor functional, we can consider the special case of a pure rank-2 functional formed from two vector functionals α = <α| and β = <β| . In that case one finds, (α β) = <α| <β| = <α, β | rank-2 functional (α β)(v1,v2) = <α, β | v1, v2> = <α | v1> <β | v2> = α(v1)β(v2) rank-2 tensor function (α β)(ui,uj) = α(ui)β(uj) = αiβj = (α β)ij rank-2 tensor (2.11.d.15) where the very last item is αiβj expressed in the tensor product notation of (2.8.9). Once again, evaluation of a tensor function at two basis vectors creates an element of the tensor. Comment on vertical bars in the Dirac Notation Let |a> be a vector in V, and <b| a vector in the dual space V*. Notice that <b| |a> = <b||a> = <b|a> . (2.11.d.16) The official notation for the scalar product is <b | a> not <b || a> so one replaces the || with | . The same replacement is made for example doing a scalar product between elements of V*2 and V2 <a| <b| |c> |d> = <a,b||c,d> = <a,b | c,d> or <a| <b| |c> |d> = ( <a| |c> ) ( <b| |d> ) = <a|c><b|d> . (2.11.d.17) (e) Rank-k functionals and tensor functions It is a simple matter to generalize from k = 2 to k = k, so the vector space is Vk and the dual space is V*k, Vk ≡ VxVx....xV k factors // Cartesian product of k spaces Vk ≡ VV....V k factors // tensor product of k vector spaces V*k ≡ V*V*....V* k factors // tensor product of k dual spaces . (2.11.e.1) We then have as a most general element of Vk (a rank-k tensor), T = Σii....i Tii....i (ui ui ..... ui ) . T = ΣITI uI (2.11.e.2) with (ui ui ..... ui) = |ui> |ui >..... |ui> = | ui, ui ....., ui >k (2.11.e.3) = | ui, ui ....., ui > . uI ≡ ui ui ..... ui On the right in red we show our equations expressed in the multi-index notation introduced in (2.10.17-22). The letter Z which appears below is used to represent the set of integers 1,2...k. Then the rank-k tensor T in Vk is represented in Dirac notation as |T> = Σii....i Tii....i | ui, ui ....., ui > . |T> = ΣITI |uI> (2.11.e.4) The rank-k tensor functional <T| of V*k is then <T | = Σii....i Tii....i < ui, ui ....., ui | <T| = ΣITI <uI| or (2.11.e.5) T = Σii....i Tii....i λi λi ..... λi . T = ΣITI λI A general pure element of Vk is specified by |v1, v2, ...vk> = |v1> |v2> .... |vk> . |vZ> = |v1> |v2> .... |vk> (2.11.e.6) The corresponding rank-k tensor function is given by T(v1, v2, ...vk) = <T | v1, v2, ...vk> = Σii....i Tii....i < ui, ui, ..., ui | v1, v2, ...vk> = Σii....i Tii....i < ui|v1>< ui|v2> .... < ui|vk> (2.11.e.7) = Σii....i Tii....i (v1)i (v2)i....(vk)i T(vZ) = ΣITI (vZ)I This shows that the rank-k tensor function is a linear combination of the products of the argument components weighted by the components of the corresponding rank-k tensor. Since this is the contraction of a rank-k tensor with k rank-1 tensors, the result transforms as a scalar, so then T'(v'1, v'2, ...v'k) = T(v1, v2, ...vk) . T'(v'Z) = T(vZ) (2.11.e.8) That is to say, the rank-k tensor function transforms as a scalar field, where the term "rank-k" is associated with the functional T = <T| which is an element of the dual space V*k . Finally we see that T(uj,uj, .... uj) = <T | uj,uj, .... uj > = Σii....i Tii....i (uj)i (uj)i....(uj)i = Tjj....j . T(uJ) = TJ (2.11.e.9) From (2.11.e.7) one sees that the tensor function T(v1, v2, ...vk) is manifestly k-multilinear, which is the generalization of linear for k = 1 and bilinear for k = 2. Once can construct a rank-k tensor functional purely from the dual basis vectors, (λiλi ... λi) = <ui| <ui| ... <ui| rank-k tensor functional λI = <uI| (2.11.e.10) (λiλi ... λi)(v1,v2....vk) = λi(v1)λi(v2)....λi(vk) = (v1)i(v2)i ... (vk)i rank-k tensor function λI(vZ) = (vZ)I (2.11.e.11) (λiλi ... λi)(uj,uj, .... uj) = λi(uj)λi(uj)....λi(uj) = (uj)i(uj)i ... (uj)i = δjiδji ... δji evaluated at basis vectors . λI(uJ) = δIJ (2.11.e.12) As an alternative to the most general rank-k tensor functional T and the all-basis-vector rank-k tensor functional (λiλi ... λi), one can consider a "pure" rank-k tensor functional constructed from k dual vectors which we shall call <αi| . In this case we find, <α1, α2....αk| = <α1| <α2| .... <αk| pure rank-k tensor functional = α1 α2... αk = (α1α2...αk) (2.11.e.13) (α1α2...αk)(v1, v2, ...vk) = α1(v1)α2(v2) ....αk(vk) = (α1v1)(α2v2) ....(αkvk) pure rank-k tensor function (2.11.e.14) (α1α2...αk)(uj,uj, .... uj) = α1(uj)α2(uj) ....αk(uj) = (α1 uj) (α2 uj) ... (αk uj) evaluated at ur = (α1)j(α2)j...(αk)j = (α1α2...αk)jj... j outer product notation (2.11.e.15) Hopefully after this long slog, the following paragraph makes some sense to the reader: A rank-k tensor function is the bra-ket closure (inner product) of a rank-k dual tensor functional <T| of V*k with a pure rank-k non-dual tensor |v1,v2...vk> of Vk such that T(v1,v2,...vk) = <T|v1,v2...vk>. The tensor function is k-multilinear in its arguments, and transforms as a scalar field with k vector arguments. When the rank-k tensor function is evaluated at the basis vectors ur, it replicates the non-dual rank-k tensor with which is it associated, which is to say, T(uj,uj, .... uj) = Tjj....j . Spivak on page 75 refers to a rank-k tensor function as a "k-tensor". (2.11.e.16) As we shall see later, the motivation for using tensor functions is their crashingly simple description of the tensor product of an arbitrary rank-k tensor with an arbitrary rank-k' tensor to produce a rank-(k+k') tensor : k<T | v1,v2...vk>k k'< S| vk+1,vk+2...vk+k'>k' = [ k<T k'<S| ] [|v1,v2...vk>k |vk+1,vk+2...vk+k'>k'] = k+k'<TS | v1,v2...vk+k'>k+k' (2.11.e.17) or T(v1,v2,...vk) S(vk+1,vk+2,...vk+k') = (TS)(v1,v2 .... vk+k') . (2.11.e.18) This equation appears below as (6.6.13) and also appears in Spivak page 75. As noted by Benn and Tucker page 2, the relationship between the vector space Vk and the dual vector space V*k is a reciprocal one. One could, as they say, perversely regard V*k as the starting vector space and then Vk would be the dual space of V*k. This amounts to swapping bra ↔ ket in the Dirac notation outlined above. Instead of having a functional α(v) = <α|v>, one would have a functional v(α) = <v|α>. We find that things are hard enough to understand without doing this "perverse" swapping of things right off the bat as they do. They refer to a rank-k tensor as a tensor of degree k, while other authors refer to rank as the order of a tensor. We us the term rank and promise not to confuse it with the different notion of the rank of a matrix which is the number of linearly independent rows or columns, or with various other meanings of the word "rank" in mathematics. (f) The Covariant Transpose Whereas the matrix transpose of a matrix Mab would be (MT)ab = Mba (swap the rows and columns), it is the covariant transpose (MT)ab = Mba that is significant in covariant notation. We quote from Tensor where M is a general rank-2 tensor while R and S are the "differentials" of (2.1.2), (MT)ab = Mba (RT)ab = Rba (ST)ab = Sba (MT)ab = Mba (RT)ab = Rba (ST)ab = Sba (MT)ab = Mba (RT)ab = Rba (ST)ab = Sba (MT)ab = Mba (RT)ab = Rba (ST)ab = Sba . (7.9.3)' (2.11.f.1) Equations in any column can be obtained by lowering one or both indices in the top equation, so that the covariant transform MT is a rank-2 tensor if M is a rank-2 tensor. For all-up or all-down indices, the two kinds of transposes are the same: (MT)ab = (MT)ab = Mba. The transpose always has the indices reflected in a vertical line between the indices. This subject is discussed in Tensor Section 7.9 where all claims are proved. We quote some of the conclusions: det(M) = det(MT) = det(MT) (7.9.7)' (2.11.f.2) RRT = RTR = 1 SST = STS = 1 RS = SR = 1 RT = R-1 = S ST = S-1 = R . (7.9.8)' (2.11.f.3) (g) Linear Dirac Space Operators Consider these three ways of writing the same real number, where M is a matrix sandwiched between vector b on the right and transpose vector a on the left, aT (Mb) M acts to the right ( * * *) [ ] (aTM)b M acts to the left, and note that (aTM) = (MTa)T [ (* * *) ] aTM b can think of M acting either to the right or to the left. (2.11.g.1) In writing these equations, one normally thinks of M as being a matrix Mij = (M[u])ij or M = M[u] . By default, the matrix elements are taken in the axis-aligned ui basis on both left and right (and this applies to all indices as discussed at the end of Section 2.4) so that (ui)TM (uj) = Σa,b (ui)a Mab (uj)b = Σa,b δia Mab δjb = Mij // = (M[u])ij. (2.11.g.2) One could, however, do this in some other basis, for example using the tangent base vectors ei, (ei) TM (ej) = Σa,b (ei)a Mab (ej)b = Σa,b Ria Mab Rjb = (RMRT)ij // = (M[e])ij (2.11.g.3) and the result is a completely different matrix. In this case the matrices are related by a covariant similarity transformation by R M[e] = R M[u]RT . (2.11.g.4) It is useful to think of the object M as being a basis-independent abstract linear operator which, when sandwiched between certain basis vectors, has certain matrix elements. Different types of basis vectors yield different matrices. One could also have mixed basis elements, (ui) TM (ej) = Σa,b (ui)aMab(ej)b = Σa,b δiaMabRjb = (MRT)ij // = (M[u,e])ij (2.11.g.5) so in this case we get M[u,e] = MRT . (2.11.g.6) The abstract operator M only becomes a matrix when it is properly sandwiched between basis vectors. This notion of thinking of the object M as a basis-independent linear operator becomes more pronounced in the Dirac notation. We restate the above equations as follows, all of which evaluate to the same real number, <a| ( M |b>) M acts to the right = <a | Mb > (<a|M ) |b> M acts to the left, and note that <a|M = <MTa | = <MTa | b > <a| M |b> can think of M acting either to the right or to the left. (2.11.g.7) The space between the vertical bars is inhabited by abstract linear operators like M. The matrix elements shown above are then <ui | M | uj> = (M[u])ij = Mij <ei | M | ej> = (M[e])ij = (RMRT)ij <ui | M | ej> = (M[u,e])ij = (MRT)ij (2.11.g.8) To emphasize this notion of abstract operator, we shall write the operator in a different font, so M is a matrix and M is a Dirac-space operator, and then <a| M |b> = <a| ( M |b>) = <a |M b> = a scalar product of two vectors <a| M |b> = (<a|M ) |b> = <MTa | b > = a scalar product of two vectors <ui | M | uj> = (M[u])ij = Mij etc . (2.11.g.9) Here then is a review of the matrix and Dirac notations, a' = (Ma) = (M)a (b')T = (Mb)T = bT MT matrix notation |a'> = |Ma> = M|a> <b'| = <Mb| = <b|MT . Dirac notation (2.11.g.10) Then consider the following claim Fact: <a | M | b> = <b| MT | a> (2.11.g.11) where both M and MT are the names of abstract linear operators. Proof: <a | M | b> ≡ <a | M b> = ai(Mb)i = ai[ Mijbj] = ai Mij bj = bj Mij ai = (bj)T (MT)ji ai = (bj)T [MTa]j = <b| MTa> = <b| MT |a> . Operator M is defined by its action on an arbitrary ket vector M | b> = | M b> Operator MT is defined by its action on an arbitrary ket vector MT | b> = | MT b> Notice in the proof that the covariant transpose MT is the correct transpose to use since Mij = (MT)ji. Exercise: Show that wv is a scalar under any transformation x' = F(x) : w'v' = <w' | v'> = <Rw|Rv> = <w|RTR|v> = <w| 1 |v> = <w|v> = wv . (2.11.g.12) In this example R is a matrix, whereas R is the corresponding Dirac space operator. The statement RTR = 1 (2.11.g.13) is the operator version of our (2.11.f.3) matrix statement RTR = 1 (2.11.g.14) which we verify as follows, (RTR)ac = (RT)abRbc = Rba Rbc = δac // (2.1.9) #1 (2.11.g.15) and which is valid for any transformation differential matrix Rij. Ways of representing M One may represent a Dirac operator M in various ways, for example, M = Σij | ui> Mij <uj| = Σij | ei> [M[e]] ij <ej| = Σij | ui> [M[u,e]] ij <ej| (2.11.g.16) as can be verified by closing with the appropriate basis vectors. For example, for the last line above, <ua | M | eb> = <ua | { Σij | ui> [M[u,e]] ij <ej|} | eb> = Σij <ua | ui> [M[u,e]] ij <ej| eb> = Σij δai [M[u,e]] ij δjb = [M[u,e]] ab . (2.11.g.17) When M = 1 we find 1 = Σij | ui> δij <uj| = Σi | ui><ui| (2.11.g.18) which is just a statement that the | ui> basis is complete, as discussed more below in section (h). We can compare the abstract Dirac operator M with the abstract rank-2 "vector" M, M = Σij | ui> Mij <uj| // Dirac operator M = Σij Mij ui uj // (2.8.10), "vector" in vector space V2 or |M> = Σij Mij | ui> | uj> . (2.11.g.19) The first object M is an operator in the Dirac Hilbert Space V. The second object M or |M> is a vector in the tensor product space V V. M and M are completely different objects, though they both involve the same matrix elements Mij. In each case, we can project out those matrix elements in an appropriate fashion: <ua | M | ub> = <ua | { Σij | ui> Mij <uj|} | ub> = Mab [< ua| < ub| ] | M > = [< ua| < ub| ] Σij Mij | ui> | uj> = Mab . (2.11.g.20) Non-square matrices The above discussion is presented implicitly for a square matrix M, but only small adjustments are needed for it to apply to a non-square matrix. In this case, in aTM b one thinks of vectors a and b as having different dimensions. Perhaps b lies in x-space which is Rn while a lies in x'-space which is Rm with m > n, and then Mij is an m x n matrix. The x'-space V' has n basis vectors |u'i> while the x-space V has m basis vectors |ui>. Then one would have, for example, <ui | M | u'j> = Mij M = Σi=1m Σj=1n | ui> Mij <u'j| |M> = Σi=1m Σj=1n Mij | ui> | u'j> 1' = Σi | u'i><u'i| completeness in V' 1 = Σi | ui><ui| completeness in V (2.11.g.21) This is exactly the situation we shall encounter in Chapter 9 where the matrix R is an m x n matrix. Linearity of M We emphasize that any Dirac operator like M is a linear operator. This is so because the action of M is defined in terms of the matrix M which is of course a linear operator. Specifically, M | s1a + s2b> = | M(s1a + s2b)> // definition of M = | s1Ma + s2 Mb > // matrix algebra = | s1Ma> + | s2 Mb > // the ket vector space V is a linear space = s1| Ma> + s2| Mb > // the ket vector space V is a linear space = s1M| a > + s2M| b > (2.11.g.22) Following the same steps, one finds that M is also linear when it acts to the left on vectors in the dual space V*, < s1a + s2b | M = < MT(s1a + s2b) | = s1 <MTa | + s2 <MTb | = s1 <a |M + s2 <b |M (2.11.g.23) M acting on tensor product spaces Tensor product spaces and wedge product spaces (regular and dual) are described in later Chapters of this document, so our presentation is a little out of order here. We just want to have all material for Dirac operators collected in one place. It is possible to extend the definition of M to describe its action on a tensor product space. Suppose |T> and |S> are elements of V2 = VV (to be discussed in Section 4.1). Calling this extended operator M(2), we first define it to be a linear operator, M(2) [ s1 |T> + s2 |S> ] ≡ s1M(2) |T> + s2M(2) |S> . // definition Then we state instructions for how M(2) acts on a vector in V2, using a general expansion for |T>, M(2) |T> = M(2) [ ΣijTij |ui> |uj> ] = ΣijTij M(2)[ |ui> |uj>] ≡ ΣijTij M|ui> M|uj> // definition = ΣijTij |Mui> |Muj> . In general, if | a> and | b> are vectors in V1, then M(2) [ | a> | b> ] ≡ M| a> M| b> = | Ma> | Mb> . Normally we write M(2) just as M so the nature of M is implied by the space on which it acts. Then M [ s1 |T> + s2 |S> ] ≡ s1M |T> + s1M |S> M[ | a> | b> ] ≡ M| a> M| b> = | Ma> | Mb> . // M acting on V2 (2.11.g.24) In generalizing the above equations to the tensor product space Vn = VV...V (Chapter 5), the first equation above stays the same, where then M on the left side means M(n), while the second equation changes, so M [ s1 |T> + s2 |S> ] ≡ s1M |T> + s1M |S> M[ | v1> | v2> .... |vn> ] // M acting on Vn [ see (5.6.17) ] ≡ M| v1> M| v2> .... M|vn> = | Mv1> | Mv2> .... |Mvn> . (2.11.g.25) Parallel statements apply for the dual space V*n (Chapter 6) [ s1 <T| + s2 <S| ]M ≡ s1 <T|M + s1 <S|M [ <v1| < v2| .... <vn| ] M // M acting on V*n [ see (6.6.18) ] ≡ < v1|M < v2|M .... <vn|M = < MTv1| < MTv2| .... <MTvn| . (2.11.g.26) M acting on wedge product spaces As will be shown in Chapter 7, the last two equation sets above have the same form for action on wedge product spaces, but is replaced by ^, so M [ s1 |T> + s2 |S> ] ≡ s1M |T> + s1M |S> M[ | v1> ^ | v2> ^ .... ^ |vn> ] // M acting on Ln [ see (7.9.d.15) ] ≡ M| v1> ^ M| v2> ^ .... ^ M|vn> = | Mv1> ^ | Mv2> ^ .... ^ |Mvn> . (2.11.g.27) For the dual space Λn, [ s1 <T| + s2 <S| ]M ≡ s1 <T|M + s1 <S|M [ <v1| ^ < v2| ^ ....^ <vn| ] M // M acting on Λn [ see (8.9.d.15) ] ≡ < v1|M ^ < v2|M ^ .... ^ <vn|M = < MTv1| ^ < MTv2| ^ .... ^ <MTvn| . (2.11.g.28) We give the M definitions above for tensor products and wedge products of vectors, but the equation numbers "[ see (...) ]" show the results generalized further to the products of an arbitrary set of tensors. Special Case R As a special case of the general matrix M and its Dirac linear operator M, we can consider the transformation differential matrix R and its corresponding Dirac operator R, where then R|a> = |Ra> . The matrix R can be non-square as was noted above for M, and this situation will arise in Chapter 10. The main point is this: Fact: The operator R is a linear operator with respect to any of the Dirac spaces it acts upon. (2.11.g.29) These spaces could be Vn, V*n, Ln, Λn or any tensor/wedge products of these spaces such as Λn ^ Λm. For activities in x'-space, the vector space names are V'n, V'*n, L'n, Λ'n . (h) Completeness Let bi be a set of basis vectors for vector space V. Then bi is the dual basis and we have δij = bi bj = (bi)T bj = ( * * ) (2.11.h.1) where we show the dot product and vector forrns of the scalar product. By the definition of a basis, any set of basis vectors for vector space V is "complete", which means that those vectors are sufficient to expand any vector v in V, v = Σi=1n vibi where vi = bi v = (bi)Tv . (2.11.h.2) One can then write the above equation as v = Σi=1n bi [ vi ] = Σi=1n [ vi ] = Σi=1n bi [(bi)Tv] = Σi=1n [ ( * * ) ] = Σi=1n [bi(bi)T] v = Σi=1n [ ( * * ) ] = { Σi=1n [bi(bi)T]} v = { Σi=1n [ ] } = M v = . (2.11.h.3) On the right we show the vector/matrix structure of each expression in the simple case of R2. We end up then with v = M v. Since this must be true for any v in V, it must be that M = 1, the identity matrix, so Σi=1n bi(bi)T = 1 (2.11.h.4) which is the official statement that the basis bi is "complete". To make this statement in terms of the components of the basis vectors, we can apply apply (uj)T on the left and (uk) on the right to get (uj)T [ Σi=1n bi(bi)T ] uk = (uj)T 1 uk = (uj)Tuk = uj uk = δjk or Σi=1n (uj)T [bi(bi)T]uk = δjk or Σi=1n [(uj)T bi] [(bi)T]uk] = δjk or Σi=1n [uj bi] [bi uk] = δjk or Σi=1n (bi)j(bi)k = δjk (2.11.h.5) We now repeat the above development in the Dirac notation shown on the right, |v> = Σi=1n vi |bi> where vi = <bi| v> (2.11.h.6) v = Σi=1n bi [ vi ] |v> = Σi=1n |bi> [ vi ] = Σi=1n bi [(bi)Tv] = Σi=1n |bi> [ <bi | v> ] = Σi=1n [bi(bi)T] v = Σi=1n [ |bi> <bi| ] | v> = { Σi=1n [bi(bi)T]} v = { Σi=1n |bi> <bi| } | v> . (2.11.h.7) Completness expressed in Dirac notation is then Σi=1n |bi> <bi| = 1 . (2.11.h.8) where 1 is the Dirac operator form of the matrix identity matrix 1. Applying <uj| on the left and |uk> on the right this becomes Σi=1n <uj| bi> <bi|uk> = <uj|uk> = δik or Σi=1n (bi)j(bi)k = δjk (2.11.h.9) which replicates (2.11.h,5) 3. Outer Products and Kronecker Products 3.1 Outer Products Reviewed: Compatibility of Chapter 1 and Chapter 2 Vectors were unbolded in Chapter 1, but were bolded for clarity in Chapter 2. Here we express all vectors in unbolded notation. Also, we quietly switch from contravariant (upper) to covariant (lower) tensor indices. Chapter 1 developed the idea of the tensor product space VW with elements vw which satisfy a set of bilinear rules (1.1.5), (v1+v2) w = (v1w) + (v2w) for all v1,v2 ϵ V and all w ϵ W (1.1.5) v (w1+w2) = (vw1) + (vw2) for all v ϵ V and all w1, w2 ϵ W s(vw) = (sv)w = v(sw) for all v ϵ V and all w ϵ W and all s ϵ K (3.1.1) The essence of the tensor product is this bilinearity, and there is no requirement to describe the objects vw in more detail. In the formal sense we are done and fini. However, for "engineering purposes", it is useful to add more structure to the tensor product by defining "tensor components", and that was the subject of Chapter 2. We conjured up a way to add components to the theory by defining tensor product components in terms of the outer product of two vectors, (vw)ij ≡ viwj v ϵ V w ϵ W . (2.8.10) (3.1.2) The components vi and wj can be elements of any field K and the juxtaposition of viwj implies multiplication in that field (we have in mind that K = R, the real numbers). The key point: because the function viwj is manifestly bilinear, this extra specification does not conflict with any of the earlier tensor product "rules". For example we can evaluate, [(v1+v2) w]ij = (v1+v2)iwj (v1w)ij + (v2w)ij = (v1)i wj + (v2)i wj . (3.1.3) The first rule of (3.1.1) says the left sides of these two equations must be equal, but we can see that the right sides are also equal, so our "tensor componentization" does not conflict with the no-components theory of Chapter 1. Thus it is that we simply glom this component structure onto the tensor product concepts of Chapter 1. We extended the tensor product idea to include the tensor product of k spaces VW...Z with this associated set of k-multilinear rules, (s1v1+s2v2)w ..... z = s1 (v1w .... z) + s2 (v2w .... z) v (s1w1+s2w2) ...z = s1 (vw1 .... z) + s2 (vw2 .... z). etc. (1.1.16) (3.1.4) Onto this skeleton we hang a component structure again using the outer product of vectors, (vwz....)ijk.... = viwjzk.... (2.8.18) (3.1.5) The function viwjzk.... is manifestly k-multilinear, so this structural enhancement is compatible with the general theory of Chapter 1. We have tried to keep things general up this point by using VW...Z where all the vector spaces can be different, but now we assume they are all the same, Vk ≡ VV....V // k copies, fancy notation Πi=1k V (2.11.24) (3.1.6) and this is our main interest, since the elements are then true "tensors" in the sense of Chapter 2. There is then only one set of basis functions {ei} to worry about, the basis for V. In this case, if a and b transform as vectors, then ab transforms as a rank-2 tensor and thus provides a name for the outer product tensor whose components are aibj . It has already been shown in Section 2.8 (in the components world) how the product can combine tensors into tensors of higher rank using the outer product idea. We had for example for the combination of a vector with two rank-2 tensors, the following rank-5 tensor (KKv)abcde = KabKcd ve . (2.8.7) (3.1.7) Another example would be this, A = ab Aij = (ab)ij = aibj B = cd Bij = (cd)ij = cidj (3.1.8) One can then define the tensor product of A and B in a fairly obvious manner, AB ≡ (ab)(cd) = abcd ϵ V4 . // associative (2.8.22) (3.1.9) This equation has no indices and so is acceptable in the component-free world of Chapter 1. The components are then taken in the following obvious manner, [AB]ijkl = [ abcd]ijkl = aibjckdl = AijBkl . (3.1.10) The above lines shows that AB is in fact a rank-4 tensor constructed by taking the outer product of two rank-2 tensors (or the outer product of four rank-1 tensors). In Section 3.2 we shall have use for an object defined in this strange manner [AB]ik,jl ≡ [AB]ijkl = AijBkl (3.1.11) and we just mention it here in passing. Note that the indices are shuffled relative to the LHS of (3.1.10). Using the same method as above, one can construct a rank-6 tensor from the tensor product of three rank-2 tensors, [ABC]abcdef = AabBcdCef (3.1.12) or from the tensor product of two rank-3 tensors [AB]abcdef = AabcBdef . (3.1.13) In general, one can take the tensor product of any set of tensors to create a new tensor whose rank is the sum of the ranks of the tensors that were combined by the symbol. If A,B,C... are arbitrary tensors, having multiindices I,J,K (for example I = {i1,i2,i3} if A is rank-3), one could write a general formula for the components of the tensor product of any number of pure tensor objects in this manner, ( A B C ....)IJK... = AI BJ CK...... // outer product (3.1.14) The tensor here is A B C ...., and it is the tensor product and the outer product of the individual tensors A,B,C.... The equation specifies its components. 3.2 Kronecker Products The subject here is the tensor product of two linear operators and is included here because it seems therefore to fit into the topic of "tensor products". This is a stand-alone section and nothing in it is referenced in later sections of our document. For that reason, a reader uninterested in Kronecker Products would do well to skip this section and continue into Chapter 4 on the wedge product development. The energetic reader can regard this section as an exercise in using the covariant tensor product machinery of Chapter 2. Let V and X be vector spaces of dimension n and m. Basis(V) = ui Basis(X) = ui Let W and Y be vector spaces of dimension n' and m' Basis(W) = u'i Basis(Y) = u'i . (3.2.1) We imagine that vector spaces V,X,W,Y have metric tensors g, g, g', g' which can be used to raise and lower subscripts in the standard manner shown in (2.2.1). Often one assumes that all these spaces have a Cartesian metric tensor, so up and down indices are the same, but we shall carry out the development below in full covariant notation as part of our "exercise". Rather than use Einstein implied sums, we shall display all sums explicitly in this section. Consider linear operators S and T such that, x = Sv = a vector in X S: V→X xi = Σa=1n Siava i = 1,2..m y = Tw = a vector in Y T:W→Y yj = Σb=1n'Tjbwb j = 1,2..m' . (3.2.2) Notice that on Sia the first index is an X-space index which can be raised and lowered by metric tensor g, whereas the second index on Sia is a V-space index which can be raised and lowered by g. So we can regard Sia as the components of a "cross tensor" involving the spaces X and V. In any equation below, we are free to change the "tilt" of any contracted index pair in the manner of (2.9.1) because such tilted index pairs will always be associated with the same metric tensor. Similar comments apply to Tjb. The linear operator S is represented by matrix Sia which has m rows and n columns (m x n). The linear operator T is represented by matrix Tjb which has m' rows and n' columns (m' x n'). We want to create a meaning for ST which is the tensor product of these two operators S and T. A candidate definition for this meaning is the following, (ST)(vw) = (Sv)(Tw) . // = (xy) ST : VW → XY . (3.2.3) (ST)| vw> = (ST) |v> w> = S|v> T|w> = |x> |y> = |xy> // Dirac notation Consider the following processing steps, (ST)([αv1 + βv2]w) = (S[αv1 + βv2])(Tw) // (3.2.3) = (α Sv1+ βSv2) (Tw) // S:V→X is linear = α (Sv1)(Tw) + β(Sv2)(Tw) // using the first rule in (3.1.1) = α (ST)(v1w) + β (ST)(v2w) . // (3.2.3) used twice (3.2.4) This shows that (ST)(vw) is linear in v. A similar argument shows it is also linear in w. Thus, the operator (ST) as defined above is a bilinear operator on VW, and we confirm the essential characteristic of the tensor product, which is its bilinearity. We accept the candidate definition (3.2.3). ____________________________________________________________________ Exercise: Compute the action of (ST) on a general element of VW . Apply (ST) to a general element of VW using tensor expansion (2.10.4) and then (3.2.3), (ST)[ ΣijFij uiu'j] = ΣijFij (ST)(uiu'j) = ΣijFij (Sui)(Tu'j) . (3.2.5) The action of S on a vector v (and T on w) can be written x = (Sv) = Σa[Sv]aua = Σa(ΣbSabvb)ua = Σab(Sabvb)ua y = (Tw) = Σc[Tw]cu'c = Σc(ΣdTcdwd)u'c = Σcd(Tcdwd)u'c . (3.2.6) Select v = ui and w = u'j in these last two equations to get, (Sui) = ΣabSab(ui)b ua (Tu'j) = ΣcdTcd(u'j)d u'c . (3.2.7) Then the tensor product appearing in (3.2.5) can be written (Sui)(Tu'j) = [ ΣabSab(ui)b ua] [ ΣcdTcd(u'j)d u'c] = Σabcd Sab(ui)bTcd(u'j)d (uau'c) (3.2.8) and so the action of the tensor product operator (ST) is given by. (ST)[ ΣijFij uiu'j] = ΣijFij (Sui)(Tu'j) // (3.2.3) = Σijabcd FijSab(ui)bTcd(u'j)d (uau'c) // (3.2.8) = Σac { Σijbd FijSab(ui)bTcd(u'j)d } (uau'c) // regroup = Σac Gac (uau'c) where Gac = Σijbd FijSab(ui)bTcd(u'j)d . (3.2.9) We have then shown the action of operator ST on a general element of VW : (ST) { ΣijFij uiu'j } = Σac Gac (uau'c) (ST) : VW → XY where Gac = Σijbd Fij Sab (ui)b Tcd (u'j)d . (3.2.10) _______________________________________________________________ It is useful now to consider the component analysis of the action of ST on a pure element of VW in the sense of outer products. Then (xy) = (ST)(vw) = (Sv)(Tw) (3.2.3) so (xy)ii' = [(ST)(vw)]ii' = [(Sv)(Tw)]ii' . (3.2.11) The right side of this last equation can be expanded using (3.1.2) and (3.2.2) to get [(Sv)(Tw)]ii' = (Sv)i(Tw)i' = (Σj Sijvj)(Σj'Ti'j'wj') = Σjj' SijTi'j' vjwj' = Σjj' SijTi'j' (vw)jj' . // = (xy)ii' (3.2.12) so then (3.2.11) may be written [(ST)(vw)]ii' = Σjj' [ SijTi'j'] (vw)jj' . // = (xy)ii' (3.2.13) We now define (ST)ii',jj' ≡ SijTi'j' (3.2.14) The comma is used to distinguish the left side from the rank-4 tensor (ST)ii'jj' = Sii'Tjj' which is a different animal. Since S and T are (cross) tensors, we can raise and lower indices on the right side of (3.2.14) using the appropriate metric tensors as discussed below (3.2.1), and then the left side indices follow since this is a definition. For example. (ST)ii',jj' ≡ SijTi'j' . (3.2.15) This definition was mentioned in (3.1.11) where it was compared to the usual notation used for a rank-4 outer product tensor (ST)iji'j' = SijTi'j'. In (3.2.15) the two first indices of S and T are listed before the comma while the two second indices appear after the comma. Installing (3.2.14) into (3.2.13), one gets [(ST)(vw)]ii' = Σjj'(ST)ii',jj' (vw)jj' . // = (xy)ii' (3.2.16) The structure of this equation suggests that we are multiplying a vector (vw) by a matrix (ST), but the usual summation index is replaced by two summation indices j and j'. In a multiindex notation one might write the above as xI = [(ST)(vw)]I = ΣJ (ST)IJ (vw)J . I = {i,i'} J = {j,j'} (3.2.17) Is there some way to write ST as a standard matrix with two indices instead of four? Start with (3.2.16) written as (xy)ii' = Σjj' (ST)ii',jj' (vw)jj' or (xiyi') = Σjj' (ST)ii',jj' (vjwj'). (ST)ii',jj' = (SijTi'j') . (3.2.18) We want to write this somehow in a form q'r = Σs Mrs qs . (3.2.19) For illustration purposes, assume n = 2 and n' = 3. Then write the components (vjwj') as a single column vector in this obvious manner, where the w component index moves fastest, = = q with components qs where s = 1,2....n*n' . (3.2.20) If vjwj' → qs, one can compute s from j,j' as follows: ( here 3 = n' = dim(W) for this special case ) s = (j-1)3 + j' (s-1) = (j-1)3 + (j'-1) = (j-1) + int() = j-1 and rem () = j'-1 . (3.2.21) Thus for general n' we can compute j and j' from s in this way (integer part and remainder) j = 1+int( ) j' = 1+rem( ) s = 1,2....n*n' . (3.2.22) One can similarly consider xiyi'→ q'r where the column vector q' has m*m' components. The rules here are analogous to those above, i = 1+int( ) i' = 1+rem( ) r = 1,2...m*m' . (3.2.23) Therefore, comparing (3.2.19) and (3.2.18), the desired Mrs is given by Mrs = (ST)ii',jj' = SijTi'j' where i = 1+int( ) j = 1+int( ) s = 1,2....n*n' i' = 1+rem( ) j' = 1+rem( ) r = 1,2...m*m' . (3.2.24) Thus we have reconfigured our multi-index equation xI = ΣJ (ST)I,J (vw)J into an ordinary matrix equation q'r = Σs Mrs qs where Mrs is given as stated above. This matrix Mrs = (ST)ii',jj' = SijTi'j' is known as the Kronecker product of the matrices S and T. The subscripts i,i',j'j' are all functions of r and s as shown in (3.2.24). Symbolically we write this Kronecker product as M = ST. Normally in writing M = ST one would imply Mabcd = SabTcd which is unrelated to the Kronecker product. It is a bit tedious to compute and display one of these M matrices by hand, so we let Maple do it for us. For this example we use the following dimensions m, n, m', n' for the spaces X, V, Y, W : S = m x n = 2 x 3 rows = m*m' = 6 T = m' x n' = 3 x 4 cols = n*n' = 12 (3.2.25) The code simply does what (3.2.24) says to do: (3.2.26) (3.2.27) One should interpret each matrix element of the form SabTcd as SabTcd -- we don't know how to make Maple display things this way. If all metric tensors are Cartesian, then (3.2.27) is correct as is. Staring at the above matrix, one can see that the T submatrix is repeated six times, and one can write this matrix in a shorthand notation as M = where T = . (3.2.28) This provides an easy way to manually construct such matrices. This construction can be understood if we look back at the M matrix definition, Mrs = (ST)ii',jj' = SijTi'j' where i = 1+int( ) j = 1+int( ) s = 1,2....n*n' i' = 1+rem( ) j' = 1+rem( ) r = 1,2...m*m' . (3.2.24) The indices i,j on S select a rectangular subregion of the M matrix due to their integer part definitions. Then within each subregion the i'j' indices run through their full ranges so a copy of matrix T appears in that subregion, multiplied by the Sij for that subregion. One is commonly interested in the case where S: V→V S = n x n matrix T: W→W T = n' x n' matrix (3.2.29) With n = m = 2 and n' = m' = 2 the above code generates this matrix M, (3.2.30) which can be compared with a result quoted on the (current) wiki tensor product page. Some other properties Suppose S1 and S2 are two matrices of the same dimension as S, and T1 and T2 are two matrices of the same dimension as T. Recall from above, (ST)ii',jj' ≡ SijTi'j' . (3.2.14) It then follows that (S1 + S2) T = S1T + S2T S (T1 + T2) = ST1 + ST2 (S1 + S2) (T1 + T2) = S1T1 + S2T1 + S2T1 + S2T2 (3.2.31) which is just the statement that S T is a bilinear operator. To prove the first line using (3.2.14), [(S1 + S2) T]ii',jj' = (S1 + S2)ijTi'j' = (S1)ijTi'j' + (S2)ijTi'j' = (S1T)ii',jj' + (S2T)ii',jj' (3.2.32) Suppose S1 and S2 are both n x n and T1 and T2 are both n' x n' . Then one can write, (S1S2)(T1T2) = (S1T1)(S2T2) (3.2.33) Proof: Again using (3.2.14), [(S1S2)(T1T2)]ii',jj' = (S1S2)ij(T1T2)i'j' = (S1)ik (S2)kj (T1)i'k' (T2)k'j' = (S1)ik (T1)i'k' (S2)kj(T2)k'j' = (S1T1)ii'kk' (S2T2)kk'jj' (3.2.34) so that in multiindex notation, [(S1S2)(T1T2)]IJ = (S1T1)IK(S2T2)KJ . (3.2.35) 4. The Wedge Product of 2 vectors built on the Tensor Product We now back up and reconsider the space VW and its elements vw. The goal of the next two sections is to establish the parallelism between the vector space VW and the "dual" vector space V*W*. Some repetition is used to review and reinforce earlier stated facts. Then Sections 4.3 and 4.4 introduce the wedge product developed in a similar parallel fashion. At the end of each of the four sections below a selection of equations is re-expressed in Dirac notation. 4.1 The tensor product of 2 vectors in V2 Note: the u'i used below are unrelated to the u'i of Chapter 2. Basics. Consider two vector spaces V and W (defined over field K) of dimension n and n'. Let {ui} = basis of V dim(V) = n v = Σi=1n vi ui = general vector in V vi ϵ K {u'i} = basis of W dim(W) = n' w = Σj=1n'wj u'j = general vector in W wj ϵ K ui = |ui > u'i = |u'i > // Dirac notation {uiu'j} = basis for the tensor product space VW dim(VW) = n*n' vw = a pure "vector" in the tensor product space VW vw ≠ wv if v ≠ w : VxW → VW : (v,w) ↦ vw (4.1.1) The last line shows as a mapping → between two sets, while ↦ shows how set elements map. Note that vw ≠ wv. For V≠W, wv does not even make sense since that requires w ϵ V and v ϵ W. For V = W the objects vw and wv are still different unless v = w. Outer Product Revisited. The notion of an outer product was discussed in Sections 2.8 and 3.1. We had for example (where ai and bj are the covariant components of vectors a and b both ϵ V), (a b)ij = aibj // outer product of two vectors (3.1.8) (A B)abcd = AabBcd . // outer product of two rank-2 tensors (3.1.10) The "outer product" of two vectors a and b may be written in vector/matrix notation as follows, (a b)** = abT = ( b1. b2....bn) = (a b)ij = (abT)ij (4.1.2) The same vector/matrix notation used above can also be used to express the "inner product" (dot product) appearing in (2.2.5), with the caveat noted below, a b = aTb = ( a1. a2....an) = Σk=1n akbk = <a | b> (4.1.3) If the a components are contravariant, the b components must be covariant, and vice versa. Chapter 1 Tensor Product Revisited. By convention one represents an element of a tensor product space using the symbol. It is a certain kind of "product" between a vector in one vector space and a vector in another vector space. On can treat as an operator : VxW → (VW) in the sense that (v,w) = (v) (w) = (vw) = element of tensor product space (VW). Certain rules were declared in (1.1.5) which make the tensor product space be a vector space, and which in an intuitive sense just seem "reasonable", (sv) w = v (sw) = s (vw) // s = scalar (ϵ K) v (w1+ w2) = vw1 + vw2 // left distributive property (v1 + v2) w = v1w + v2w . // right distributive property (1.1.5) (4.1.4) In the last two equations, the + on the left represents addition in either W or V, whereas the + on the right side represents addition in VW. These lines say that multiplication "distributes" over addition +. The scalar rule can be combined with the distributive rules to obtain this equivalent rules restatement: v (s1w1+ s2w2) = s1(vw1) + s2(vw2) // s1,s2 = scalar (ϵ K) (s1v1 + s2v2) w = s1(v1w)+ s2(v2w) .// s1,s2 = scalar (ϵ K) (1.1.7) (4.1.5) The above rules in effect say that defines a "bilinear" operation -- it is linear separately in each of its operands. Notice that the following two rules are incorrect: v w = w v // wrong! (unless V = W and v = w) (sv) (sw) = (vw) // wrong! (unless s = 1) As noted in Appendix B the second rule applies to a direct sum . Using the correct "rules" above, one may write v w = ( Σiviui)( Σjwjuj') = Σijviwj (uiu'j) (4.1.6) showing how this pure tensor product vector can be expressed in terms of the basis functions. In Dirac notation |v> |w> = Σijviwj |ui> |uj> . (4.1.6a) General tensors in VW and V2. A general "vector" (rank-2 cross tensor) in WV can be written as a linear combination of the basis vectors, as was shown in (2.10.4), T ≡ Σij Tij uiu'j T ϵ VW . Σij ≡ Σi=1nΣj=1n' (4.1.7) If W = V, we refer to the space VW = VV as V2, and then T ≡ Σij Tij uiuj T ϵ VV = V2 . Σij ≡ Σi=1nΣj=1n (4.1.8) Although we have said T is a "vector" in the abstract sense that a vector space (even a tensor product vector space) has "vectors" as elements, the usual terminology is to say that T is a "rank-2 tensor" in the space V2. Meanings of tensor. The word "tensor" has a weak and a strong meaning. In the weak meaning, a rank-2 tensor is something that has components with two indices like Tij. In the strong meaning, a rank-2 tensor is a set of components Tij which transform in a certain manner with respect to some underlying transformation, T'ab = Raa' Rbb' Ta'b' Picture A T ϵ VV (2.1.6) Covariant expansion forms. The rank-2 tensor T can be expanded in many various as shown in Section 2.10, and each such expansion has its own characteristic coefficients. Here are all four versions of (4.1.6) obtained using the tilt reversal rule (2.9.1) : T ≡ Σij Tij uiu'j T ϵ VW T ≡ Σij Tij uiu'j T ≡ Σij Tij uiu'j T ≡ Σij Tij uiu'j . (4.1.9) Notation Comments: In (4.1.9) one could replace Tij by [T(u,u')]ij to be more precise about the meaning of the coefficients, namely, that they are those which arise when one expands on the basis uiu'j. Then in (4.1.8) one could write Tij as [T(u,u)]ij or [T(u)]ij, but in this case Tij is the "default" notation for expanding on the ui basis vectors as shown back in ****. The basis vectors ui here are those of Chapter 2, but the basis vectors u'i here are not those of Chapter 2 and are here used just to provide a simple notation for the basis of W. Dot Products. One can define a covariant dot product between two elements of V2 in this manner A B ≡ ΣijAijBij = ΣijAijBij = ΣijBijAij = B A . (4.1.11) If A or B is a pure rank-2 tensor, one can write as well (ab)B = ΣijaibjBij A(cd) = ΣijAijcidj (ab)(cd) = Σijaibjcidj = (ac)(bd) . (4.1.12) The last line appears as (2.9.13). Dirac Notation for Section 4.1 . It seemed best not to clutter that above text with these alternate forms. The Dirac version of an equation gets a D subscript on its equation number. ui = |ui> u'i = |u'i> bases for V and W uiu'j = |ui> |u' > basis for the tensor product space VW vw = |v> |w> a pure "vector" in the tensor product space VW (4.1.1)D T = |T> = Σij Tij |ui> |u'j> rank-2 tensor in VW (a b)ij = <ui| <uj| |a> |b> = <ui|a> <uj|b> = aibj outer product a b = <a | b> dot product |v> |w> = Σijviwj |ui> |u'j> expansion of pure vector on basis vectors (4.1.6)D T ≡ |T> = Σij Tij |ui> |u'j> expansion of a rank-2 tensor (4.1.7)D <ua| <u'b| |T> = <ua| <u'b| Σij Tij |ui> |u'j> = Σij Tij <ua|ui><u'b|u'j> = Σij Tij δaiδbj = Tab (2.10.5)D <a| <b| |c> |d> = <a|c><b|d> = Σiaici Σjbjdj = Σijaibjcidj dot product in V2 (4.1.12)D 4.2 The tensor product of 2 dual vectors in V*2 The dual space V* of V was discussed in Section 2.11. Space W* is dual to W. We continue our convention of using Greek or script letters for dual space objects. The current section is basically a generalization of Section 2.11 to the case where V and W are different vector spaces. We show corresponding Section 2.11 equations in italics. Note: the λ'i used below are unrelated to the λ'i of Chapter 2. Basics. Consider the two dual vector spaces V* and W* (defined over field K) of dimension n and n'. Let {λi} = basis of V* dim(V*) = n α = Σi=1n αi λi = general linear functional in V* {λ'i} = basis of W* dim(W*) = n' β = Σj=1n'βj λ'j = general linear functional in W* λi = (ui)T = <ui| λ'i = (u'i)T = <u'i| // matrix and Dirac notation {λiλ'j} = basis for the tensor product space V*W* dim(V*W*) = n*n' αβ = a pure "vector" in the tensor product space V*W* αβ ≠ βα if α ≠ β : V*xW* → V*W* : (α,β) ↦ αβ (4.2.1) Note that αβ ≠ βα. For V*≠W*, βα does not even make sense since that requires β ϵ V* and α ϵ W*. For V* = W* the objects αβ and βα are still different unless α = β. Vector expansions in V* and W*. Linear functionals in V* and W* can be written as linear combinations of the basis functionals, α = Σiαiλi α(v) = Σiαiλi(v) α: V → K (2.11.8) β = Σjβjλ'j β(v) = Σjβjλj'(v) β: W → K . (4.2.2) The middle column shows the corresponding functions α(v) and β(v), and we now have α(ei) = αi (2.11.9) β(e'j) = βj . (4.2.3) Basis tensors in V*W*. The basis functionals for V*W* are the λiλ'j where, (λiλ'j)(v,w) = λi(v)λ'j(w) = scalar * scalar = scalar ϵ K (2.11.15) (4.2.4) where (recall these are called the "ith coordinate functions") λi(v) = vi λ'i(w) = wi (2.11.7) (4.2.5) so that (λiλ'j)(v,w) = viwj . (2.11.16) (4.2.6) This function is manifestly bilinear in its two vector arguments. The Rules for V*W* . The "rules" (1.1.5) for the operator in the space V*W* are the same as those for in the space VW, since V*W* is, after all, a tensor product of two spaces, (sα) β = α (sβ) = s (α β) s ϵ K, α ϵ V* β ϵ W* α (β1+ β2) = α β1 + α β2 // distributive property (α1 + α2) β = α1β + α2β . // same idea as above (4.1.4) (4.2.7) Rank-2 cross-tensor expansion in V*W*. A general functional of the dual tensor product space V*W* can be written T ≡ Σij Tij λiλ'j T ϵ V*W* Σij ≡ Σi=1nΣj=1n' (2.11.17) (4.2.8) The Tij here are exactly the same Tij which appear in the VW expansion (4.1.7), T = ΣijTij uiu'j . Evaluating at a point (v,w) in VxW one gets, T(v,w) = Σij Tij (λiλ'j)(v,w) = Σij Tij λi(v) λj'(w) = Σij Tij viwj (2.11.18) (4.2.9) so one may regard T : VxW → K, and T(v,w) as manifestly bilinear in its arguments. Setting v = ui and w = u'j , one finds that T(ui,u'j) = Tij ϵ K (2.11.19) (4.2.10) This may be compared with α(ui) = αi in (4.2.3). An arbitrary rank-2 tensor functional T can be represented either by its expansion T = ΣijTijλiλ'j or by the corresponding tensor function T(v,w). As a special case, consider α ϵ V*and β ϵ W* as shown above. Then, α β = (Σaαaλa)(Σbβbλ'b) = Σab αaβb λaλ'b ϵ V*W* (2.11.21) (4.2.11) (α β)(v,w) = Σabαaβb(λaλ'b)(v,w) = Σabαaβbλa(v)λb'(w) = Σabαaβbvawb (2.11.22) (4.2.12) which then is just a particular example of (4.2.9). Continuing the above, (α β)(v,w) = Σabαaβbvawb = [Σaαava][Σbαbwb] = α(v) β(w) . (2.11.22) (4.2.13) In particular, (α β)(ui,u'j) = α(ui) β(u'j) = αiβj = (α β)ij . (2.11.23) (4.2.14) If it happens that W = V, then W* = V* and we write V*W* = V*V* = V*2. The equations above then revert to those given in Section 2.11 (referenced in italics above). Dirac Notation for Section 4.2 λi = (ui)T = <ui| λ'i = (u'i)T = <u'i| bases for V* and W* λiλ'j = <ui| <u'j| basis for the dual tensor product space V*W* αβ = <α| <β| pure element of V* W* (4.2.1)D α = <α| = Σiαi<ui| α(v) = <α|v> = Σiαi<ui|v> vector functional expansions β = <β| = Σiβi<u'i| β(v) = <β|v> = Σiβi<u'i|v> (4.2.2)D α(ui) = <α|ui> = αi vector function α(v) evaluated at v = ui β(ui) = <α|ui> = βi vector function β(v) evaluated at v = ui (4.2.3)D (λiλ'j)(v,w) = <ui| <u'j| |v> |w> = <ui| v><u'j|w> = viwj (4.2.4)D (4.2.5)D T = <T| = ΣijTij<ui | <u'j| = Σij Tij λi λ'i rank-2 tensor functional in V*W* (4.2.8)D T(v,w) = <T | |v> |w> = <T |v,w> rank-2 tensor function for V*W* = ΣijTij<ui | <u'j| |v> |w> = ΣijTij<ui |v> <u'j|w> = ΣijTijviwj (4.2.9)D (α β)(v,w) = <α | <β| |v> |w> = <α | v> <β | w> (4.2.13)D The vector spaces V2* and V2*f We can regard both T = <T| and T(v,w) = <T|v,w> as representations of the same rank-2 tensor functional <T| in V*W*. The object T is a bilinear rank-2 tensor functional, whereas T(v,w) is a bilinear rank-2 tensor function (a Spivak 2-tensor). There is a 1-to-1 correspondence between T and T(v,w) . We shall say T ϵ V*W* while T(v,w) ϵ (V*W*)f (f = function), and the two spaces are isomorphic. If W = V, then T ϵ V2* and T(v,w) ϵ V2*f and V2* and V2*f are isomorphic. Fact: The vector space V*2 is equivalent to the vector space V*2f of bilinear functions on V2. (4.2.15) 4.3 The wedge product of 2 vectors in L2 (a) Definition of the wedge product of 2 vectors and the space L2 Momentarily jumping ahead, consider this equation, v ^ w = (vw - wv)/2 . v ϵ V and w ϵ W If V and W are different vector spaces, this makes no sense since the second term wv implies that w lies in the left space V and v lies in the right space W. So in our discussion of wedge products, we require that W = V. This being the case, instead of using letters v and w as representative vectors, we shall use a and b. Then ui are the basis vectors for both component spaces in the tensor product space VV. So, we start off by defining the following "wedge product" ("exterior product") of two vectors a,b ϵ V, a ^ b ≡ (ab - ba)/2 . dim(V) = n (4.3.1) Notice therefore that a ^ b is an element of VV = V2, since it is a linear combination of elements of VV. It is "antisymmetrized" under a ↔ b. Since not all elements of VV can be written this way, the set of elements a ^ b exist in a subset of VV which we shall call L2, so L2 V2. Some authors write L2 as V^V. The above definition trivially implies that a ^ b = - b ^ a a, b ϵ V (4.3.2) and a ^ a = 0 a ϵ V . (4.3.3) In (1.1.5) we stated certain scalar and distributive properties of the operator. These properties are passed through to the wedge ^ operator by the above definition. For example, (sa) ^ b = [ (sa)b - b(sa)]/2 = s [ ab - ba ]/2 = s (a ^ b) s = scalar (a+c) ^ b = [(a+c)b - b(a+c)]/2 = [ab + cb - ba - bc]/2 = [ ab - ba ]/2 + [ cb - bc ]/2 = (a ^ b) + (c ^ b) distributive and similarly for a ^ (sb) and a ^ (b + c). To summarize, we have a set of rules as follows: (sa) ^ b = s (a ^ b) (a+c) ^ b = (a ^ b) + (c ^ b) s ϵ K a ^ (sb) = s (a ^ b) a ^ (b + c) = (a ^ b) + (a ^ c) a,b,c ϵ V (4.3.4) The operator ^ is then seen to be "bilinear" over elements of V: it is separately linear in each operand. To more precisely define the space L2, we claim that the most general element T^ of the space L2 can be written this way, T^ = Σij Tij ui ^ uj . Σij ≡ Σi=1n Σj=1n (4.3.5) where Tij are the expansion coefficients. For example, if Tij = aibj this would be, T^ = Σij aibj ui ^ uj = (Σiaiui) ^ ( Σjbjuj) = a ^ b (4.3.6) and then a ^ b is included in L2 for any vectors a and b in V. We can take the ab component of (4.3.5) as follows T^ab = Σij Tij (ui ^ uj)ab = Σij Tij(uiuj - ujui)ab / 2 = Σij Tij[ (uiuj)ab - (ujui)ab] / 2 = Σij Tij [ uiaujb - ujauib] / 2 = Σij Tij [ δia δjb - δja δib] / 2 = (1/2)[ Tab -Tba] T^ab = - T^ba (4.3.7) This shows that the expansion (4.3.5) can only represent an antisymmetric rank-2 tensor T^. One could rearrange the n2 basis vectors of VV into these two groups, (ui^ uj) = [uiuj - ujui]/2 n(n-1)/2 independent elements in this set (4.3.8) (ui * uj) ≡ [uiuj+ ujui]/2 n(n)/2 independent elements in this set for a total of n(n-1)/2+ n(n)/2 = n2 basis vectors. One would say then that L2 is spanned by just the first set of basis vectors. It was noted above that L2 is a subset of V2. A stronger statement is that L2 is a subspace of V2. First of all, L2 is obviously closed under addition of vectors since Σij Tij ui ^ uj + Σij T'ij ui ^ uj = Σij (Tij+T'ij) ui ^ uj . (4.3.9) And if (a ^ b) is an element of L2 then so is s(a ^ b) = (sα) ^ b ϵ L2 . Finally, since a ^ a = 0, L2 includes the 0 element. So L2 then is a vector space which is a subspace of V2. (b) How big is the space L2 compared to the space V2? Consider this most general element of L2: T^ = Σij Tij (ui ^ uj) = Σi≠ j Tij (ui ^ uj) // (ui ^ ui) = 0 = Σi<j Tij (ui ^ uj) + Σi>j Tij (ui ^ uj) = Σi<j Tij (ui ^ uj) + Σj>i Tji (uj ^ ui) // i↔j in second sum = Σi<j Tij (ui ^ uj) - Σi<j Tji (ui ^ uj) // (uj ^ ui) = - (ui ^ uj) = Σi<j (Tij - Tji) (ui ^ uj) = Σi<j Aij (ui ^ uj) Aij ≡ (Tij - Tji) Aij = - Aji . (4.3.10) Thus, the number of elements in L2 is equal to the number of antisymmetric n x n matrices A one can construct which contain elements of field K. An n x n antisymmetric matrix has only n(n-1)/2 places to insert independent values since the diagonal is all zeros and one triangle is the negative of the other. If the scalar space K contains N elements ( N = ∞ for the reals), one could then construct exactly Nn(n-1)/2 antisymmetric matrices A. Meanwhile, the most general element of V2 can be written T = Σij Tij (ui uj) . (4.1.9) Now each matrix Tij defines an element of V2. Using the same counting method as above, the total number of elements of V2 is Nn2. We conclude that = = (1/2) = (1/2) (1 - ) . (4.3.11) The conclusion is that L2 contains less than half the number of elements in V2. This ratio is of course the same as the (4.3.8) count ratio of L2 basis vectors to V2 basis vectors: [n(n-1)/2] / [n2] = (n-1)/2n. Below we use this terminology, T^ = Σij Tij (ui ^ uj) = the "symmetric expansion" of T^ T^ = Σi<j Aij (ui ^ uj) = the "ordered expansion" of T^ . (c) Wedge products and determinants: the geometry connection From (4.3.6) and (4.3.10) with Tij = aibj we get, a ^ b = Σij aibj (ui ^ uj) = Σi<j (aibj- ajbi) (ui ^ uj) = Σi<j det (ui^uj) Aij = (aibj- ajbi) = det . (4.3.12) The determinants which appear here are 2x2 minors of a matrix having n rows and 2 columns. The two columns are the vectors a and b, each of which has n components. Below that matrix is shown on the left, and some of the 2x2 minors (row i < row j) are shown in gray on the right: (4.3.13) If V = R2 (so n=2) there is only one term in the sum (4.3.12), the one with i=1 and j=2, so a ^ b = det u1^u2 = det(a,b) u1^u2 = [ a1b2 - a2b1] u1^u2 . (4.3.14) If one draws a parallelogram (2-piped) in the x-y plane with edges a and b, one knows that the area of that 2-piped is |a x b| which is then |a1b2 - a2b1| = |det(a,b)|. There is then some connection between the wedge product of two vectors in R2 and the geometry of R2. Later in (7.5.6) we will show that for V = R3 the triple wedge product of three vectors is given by, a ^ b ^ c = det(a,b,c) (u1^ u2^ u3) (4.3.15) and here det(a,b,c) is the volume of the 3-piped spanned by the vectors a,b,c, so again there is a geometry connection. However, for R3 the wedge product of two vectors is more complicated. Using the above expression, we find a ^ b = det u1^u2 + u1^u3 + u2^u3 = [a1b2 - a2b1] u1^u2 + [a3b1 - a1b3] u3^u1 + [a2b3 - a3b2] u2^u3 . (4.3.16) The coefficients are those which appear in the normal "cross product" of two contravariant vectors, a x b = [a1b2 - a2b1] u3 + [a3b1 - a1b3] u2 + [a2b3 - a3b2] u1 . We do not wish, however, to identify for example u1^u2 with u3. After all, u3 is a basis vector in V, whereas u1^u2 is a vector in the tensor product space VV. One can, on the other hand, define a correspondence of sorts where one says (each line in cyclic order, and ↔ means "corresponds to") u1^u2 ↔ u3 u2^u3 ↔ u1 u3^u1 ↔ u2 // = - u1^u3 (4.3.17) in which case one can say a ^ b = [a1b2 - a2b1] u1^u2 + [a3b1 - a1b3] u3^u1 + [a2b3 - a3b2] u2^u3 ↔ a x b = [a1b2 - a2b1] u3 + [a3b1 - a1b3] u2 + [a2b3 - a3b2] u1 (4.3.18) so there is then a correspondence between the wedge product and the cross product in R3. This correspondence was described by Scottish mathematician William Hodge (1903-1975) around 1941 and the relationship ↔ is formalized by the Hodge dual operator * where for example *(u1^u2) = u3 in R3. For Rn with n > 3 there is no cross product of two vectors, but there is a wedge product. With V = R4 for example, using the result (4.3.12) stated above, a ^ b = det u1^u2 + detu1^u3 + detu1^u4 + detu2^u3 +detu2^u4 + detu3^u4 . (4.3.19) There are enthusiastic workers (e.g. Denker) who recommend deep-sixing the cross product altogether and replacing it with the wedge product for the study of topics like angular momentum (Ref...). (d) Components For the tensor product of two basis vectors we have, (uiu'j)rs = (ui)r(u'j)s = δirδjs // VW (uiuj)rs = (ui)r(uj)s = δirδjs // VV = V2 (4.3.20a) For the wedge product (ui^ uj) we have instead, (ui^ uj)rs = (1/2)[uiuj - ujui]rs = (1/2)[(uiuj)rs - (ujui)rs] = (1/2)[(ui)r(uj)s - (uj)r(ui)s] = (1/2) (δir δjs - δis δjr) (ui^ uj)rs = - (ui^ uj)sr = - (uj^ ui)rs . // two forms of antisymmetry (4.3.20b) We now examine the pure wedge product a ^ b using both the symmetric expansion (4.3.5) and the ordered expansion (4.3.10). Using the symmetric double sum expansion form (4.3.5) with Tij = aibj one has from (4.3.6) and (4.3.20), (a ^ b)rs = Σij aibj (ui ^ uj)rs = Σij aibj [δirδjs - δjrδis]/2 = (arbs - asbr)/2 . (4.3.24) Using the ordered double sum expansion (4.3.10) with Tij = aibj, we find instead (a ^ b)rs = Σi<j (aibj- ajbi) (ui ^ uj)rs = Σi<j det (ui ^ uj)rs = (1/2) Σ1≤i<j≤n det [δirδjs - δjrδis] . (4.3.25) For r = s, one clearly has (a ^ b)rs = 0. If r < s, then only the δirδjs term can contribute to the ordered sum, since this will make i < j , otherwise only the second term contributes. Then using θ(Boolean) = 1 if true else 0, we can evaluate as follows, 2(a ^ b)rs = det θ(r<s) - det θ(s<r) = det θ(r<s) + det θ(s<r) // swap rows 2nd term = det [ θ(r<s) + θ(s<r)] = det = arbs - asbr . (4.3.26) Combining the results for r=s and r≠s we get (a ^ b)rs = (arbs - asbr)/2 // T^rs = (Trs- Tsr)/2 = Ars / 2 (4.3.27) in agreement with (4.3.24) which used the symmetric sum. We now repeat this comparison for general elements of L2. Using the symmetric double sum (4.3.5), T^rs = Σij Tij (ui ^ uj)rs = Σij Tij (δir δjs - δis δjr)/2 = (Trs- Tsr)/2 = Ars/2 . (4.3.28) Using the ordered double sum (4.3.10), T^rs = Σi<j Aij (ui ^ uj)rs = Σi<j Aij [δirδjs - δjrδis]/2 . (4.3.29) For r = s one has [..] = 0 so T^rs = 0. Otherwise, 2 T^rs = Ars θ(r<s) - Asrθ(s<r) = Ars θ(r<s) + Arsθ(s<r) = Ars[ θ(r<s) + θ(s<r)] = Ars (4.3.30) with the conclusion that T^rs = Ars/2 for all r,s ϵ (1,n) (4.3.31) which agrees with (4.3.28) using the symmetric expansion. (e) Dot Products Since a^b is an element of V2 as well as of L2, we may use the V2 dot product to write (a^b) (cd) = {(ab-ba)/2} (cd) = (1/2) [ (ab)(cd) - (ba)(cd)] = [(ac)(bd) - (bc)(ad)]/2 // (2.9.13) (4.3.32) with this special case (ui^uj) (cd) = [(uic)(ujd) - (ujc)(uid)]/2 = [cidj - cjdi]/2 . (4.3.33) The dot product of two-vector wedge products is the same as (4.3.32), (a^b) (c^d) = {(ab-ba)/2} {(cd-dc)/2} = = (1/4) [ (ab)(cd) - (ba)(cd) - (ab)(dc) + (ba)(dc) ] = (1/4) [ (ac)(bd) - (bc)(ad) - (ad)(bc) + (bd)(ac) ] = [ (ac)(bd) - (bc)(ad) ]/2 = (a^b) (cd) = (ab) (c^d) (4.3.34) so then the special case is the same as (4.3.33), (ui^uj) (c^d) = [cidj - cjdi]/2 . (4.3.35) Dirac Notation for Section 4.3 (a selection) Section 4.3 (a) |a> ^ |b> = ( |a> |b> - |b> |a> )/2 (4.3.1)D |a> ^ |b> = - |b> ^ |a> (4.3.2)D and |a> ^ |a> = 0 (4.3.3)D T^ = |T^> = Σij Tij |ui> ^ |uj> (4.3.5)D T^ab = < ua| <ub | |T^> = < ua| <ub | Σij Tij |ui> ^ |uj> = Σij Tij <ua| <ub | [ |ui> |uj> - |uj> |ui> ]/2 = Σij Tij [ <ua |ui> <ub |uj> - <ua |uj> <ub |ui> ]/2 = Σij Tij [ δaiδbj- δajδbi]/2 = (Tab - Tba)/2 = - T^ba (4.3.7)D Section 4.3 (b) T^ = |T^> = Σi<j Aij |ui> ^ |uj> (4.3.10)D Section 4.3 (c) |a> ^ |b> = Σij aibj |ui> ^ |uj> = Σi<j det |ui> ^ |uj> (4.3.12)D |a> ^ |b> ^ |c> = det(a,b,c) |u1> ^ |u2> ^ |u3> n = 3 (4.3.15)D Section 4.3 (d) (a ^ b)rs = <ur| <us | |a> ^ |b> = <ur| <us | ( |a> |b> - |b> |a> )/2 = [ <ur| <us | |a> |b> - <ur| <us | |b> |a> ]/2 = [ <ur |a><us |b> - <ur |b><us |a> ]/2 = (arbs - asbr)/2 (4.3.27)D Section 4.3 (e) (ui^uj) (cd) = [(uic)(ujd) - (ujc)(uid)]/2 = [cidj - cjdi]/2 . <ui| ^ <uj| |c> |d> = (1/2)[<ui| <uj| - <uj| <ui| ] |c> |d> = (1/2)[ <ui| c><uj|d> - <uj| c><ui|d> ] = (1/2) [ cidj - cjdi] (4.3.33)D 4.4 The wedge product of 2 dual vectors in Λ2 Section 4.3 considered the wedge product of two vectors in V2. Here we consider the wedge product of two vectors in the dual space V*2. We mimic the approach of Section 4.3, omitting some details, and we match equation numbers even though this leaves some "holes" in the sequence. (a) Definition of the wedge product and the space Λ2 We start off by defining the following wedge product of two vectors (linear functionals) α and β of V*, α ^ β ≡ (α β - β α)/2 . (4.4.1) Notice therefore that α ^ β is an element of V*V* = V*2, since it is a linear combination of elements of V*V*. It is "antisymmetrized" under α ↔ β. Since not all elements of V*V* = V*2 can be written this way, the set of elements α ^ β exist in a subspace of V*2 which we shall call Λ2, so Λ2 V*2. Some authors write Λ2 as V*^V*. The proof that Λ2 is a subspace and not just a subset of V*2 is the same as in the Section 4.3 (a). The above definition trivially implies that α ^ β = - β ^ α α, β ϵ V* (4.4.2) and α ^ α = 0 α ϵ V* . (4.4.3) The "rules" for the ^ operator in Λ2 V*2 are found just as they were for L2 V2, namely : (sα) ^ β = s (α ^ β) (α+γ) ^ β = (α ^ β) + (γ ^ β) α ^ (sβ) = s (α ^ β) α ^ (β + γ) = (α ^ β) + (α ^ γ) (4.4.4) where α,β,γ are vectors in V* and s is a scalar in K. To more precisely define the space Λ2, we claim that the most general element of the space Λ2 can be written this way (that is, Λ2 is the space spanned by the λi ^ λj basis vectors) T^ = Σij Tij λi ^ λj . // λi ^ λj = <ui| ^ <uj| in Dirac notation (4.4.5) For example, if Tij = αiβj this would be T^ = Σij αiβj (λi ^ λj) = (Σiαiλi) ^ ( Σjβjλj) = α ^ β (4.4.6) and then α ^ β is included in Λ2 for any vectors α and β in V*. (b) How big is the space Λ2 compared to the space V*2? Just as in Section 4.3 (b), we can show that T^ = Σij Tij (λi ^ λj) = Σi<j Aij (λi ^ λj) Aij ≡ (Tij - Tji) Aij = - Aji (4.4.10) where Aij is an antisymmetric n x n matrix. Using the same argument presented there, we find = = (1/2) = (1/2) (1 - ) . (4.4.11) (c) Wedge products and determinants From (4.4.6) and (4.4.10) with Tij = αiβj we get, α ^ β = Σij αiβj (λi ^ λj) = Σi<j (αiβj- αjβi) (λi ^ λj) = Σi<j det (λi ^ λj) Aij = (αiβj- αjβi) = det . (4.4.12) If V* = R2 (so n=2) there is only one term in the sum (4.4.12), the one with i=1 and j=2, so α ^ β = det λ1 ^ λ2 = det(α,β) λ1 ^ λ2 = [ α1β2 - α2β1] λ1 ^ λ2 . (4.4.14) Later we will show that for V = R3 the triple wedge product of three vectors is given by, α ^ β ^ γ = det(α,β,γ) (λ1 ^ λ2 ^ λ3) (4.4.15) It does not seem useful to discuss "geometry" in the space of functionals. (d) Tensor Functions In Section 4.3(d) we discussed components (uiu'j)rs and (ui^uj)rs. In the dual world, the corresponding objects are the tensor functions (λiλ'j)(vr,ws) and (λi^ λj)(vr,ws) . For the tensor product of two dual basis vectors we have these tensor functions, (λiλ'j)(vr,ws) = λi(vr)λ'j(ws) = (vr)i(ws)j // V*W*; r and s are vector labels (λiλj)(vr,vs) = λi(vr)λj(vs) = (vr)i(vs)j // V*V* and (λiλ'j)(ur,u's) = (ur)i(u's)j = δriδsj // V*W* (λiλj)(ur,us) = (ur)i(us)j = δriδsj . // V*V* (4.4.20a) For the wedge product (λi^ λj) we have instead, (λi^ λj)(vr,vs) = [(λiλj)(vr,vs) - (λjλi)(vr,vs)]/2 = [λi(vr)λj(vs) - λj(vr)λi(vs)]/2 = (1/2) [ (vr)i(vs)j - (vr)j(vs)i] // Λ2 = V* ^ V* (λi^ λj)(ur,us) = (1/2) [ δriδsj - δrjδsi] (λi^ λj)(vr,vs) = - (λi^ λj)(vs,vr) = - (λj^ λi)(vr,vs) . // two forms of antisymmetry (4.4.20b) We now examine the pure wedge product α ^ β using both the symmetric expansion (4.4.5) and the ordered expansion (4.4.10). Using the symmetric double sum expansion form (4.4.5) with Tij = αiβj one has, (α ^ β)(vr,vs) = Σij αiβj (λi ^ λj)(vr,vs) = Σij αiβj [ λi(vr)λj(vs) - λj(vr)λi(vs)]/2 = Σij αiβj[(vr)i(vs)j - (vr)j(vs)i]/2 = [ α(vr)β(vs) - α(vs)β(vr)]/2 . Evaluated at vr = ur and vs = us this becomes, using (4.2.6), (α ^ β)(ur,us) = Σij αiβj [δirδjs - δjrδis]/2 = (αrβs - αsβr)/2 . (4.4.24) Using the ordered double sum expansion (4.4.10) with Tij = αiβj, we find instead (α ^ β)(vr,vs) = Σi<j (αiβj- αjβi) (λi ^ λj)(vr,vs) = Σi<j det (λi ^ λj)(vr,vs) = Σi<j det [ λi(vr)λj(vs) - λj(vr)λi(vs)]/2 = Σi<j det (1/2) det . (4.4.25) Evaluation at vr = ur and vs = us gives (α ^ β)(ur,us) = Σi<j det (1/2) [δirδjs - δjrδis] . (4.4.26) Repeating the argument (4.3.27) this becomes (α ^ β)(ur,us) = (αrβs - αsβr)/2 // later this will be called [Alt(αβ)]rs (4.4.28) in agreement with (4.4.24) which used the symmetric sum. We now repeat this comparison for general elements of Λ2. Using the symmetric double sum (4.4.5), T^(vr,vs) = Σij Tij (λi ^ λj)(vr,vs) = Σij Tij [ λi(vr)λj(vs) - λj(vr)λi(vs)]/2 (4.4.29) one finds that T^(ur,us) = Σij Tij [δirδjs - δjrδis] / 2 = (Trs - Tsr)/2 = Ars/2 . // see (4.4.10) (4.4.30) Using the ordered double sum (4.4.10), T^(ur,us) = Σi<j Aij (λi ^ λj)(ur,us) = Σi<j Aij [δirδjs - δjrδis]/2 . (4.4.31) For r = s one has [..] = 0 so T^(ur,us) = 0. Otherwise, 2T^(ur,us) = Ars θ(r<s) - Asrθ(s<r) = Ars θ(r<s) + Arsθ(s<r) = Ars[ θ(r<s) + θ(s<r)] = Ars (4.4.32) with the conclusion that T^(ur,us) = Ars/2 for all r,s ϵ (1,n) (4.4.33) in agreement with (4.4.30). Dirac Notation for Section 4.4 (a selection) Section 4.4 (a) <α| ^ <β| = (<α| <β| - <α| <β| )/2 (4.4.1)D <α| ^ <β| = - <β| ^ <α| (4.4.2)D <α| ^ <α| = 0 (4.4.3)D T^ = <T^| = Σij Tij <ui| ^ <uj| // λi = <ui| (4.4.5)D T^ = <T^| = Σijαiβj <ui| ^ <uj| = ( Σiαi<ui| )( Σjβi<uj| ) = <α| ^ <β| (4.4.6)D Section 4.4 (b) T^ = <T^| = Σi<j Aij <ui| ^ <uj| (4.4.10)D Section 4.4 (c) α ^ β = <α| ^ <β| = Σi<j det <ui| ^ <uj| (4.4.12)D Section 4.4 (d) (λi^ λj)(vr,vs) = <ui| ^ <uj| | vr> | vs> = (1/2) (<ui| <uj| - <uj| <ui| ) | vr> | vs> = (1/2) ( <ui|vr><uj| vs> - <ui|vs><uj| vr>) = (1/2) [ (vr)i(vs)j - (vr)j(vs)i] (4.4.20b)D The vector spaces Λ2 and Λ2f Looking at (4.4.20), (4.4.23) and (4.4.29), one sees that (λi^ λj)(vr,vs), (α ^ β)(vr,vs) and T^(vr,vs) are all antisymmetric bilinear functions of the two arguments vr, vs ϵ V . In Section 4.3 we declare that the rank-2 tensor T^ = |T^> is antisymmetric (alternating) if T^ab = - T^ba . In similar fashion, we declare that the rank-2 tensor functional T^ = <T^| is antisymmetric (alternating) if T^(v,v') = - T^(v',v). That is to say, saying that the functional is alternating means that the corresponding tensor function is alternating. Section A.10 writes <T^| = <Alt T | in an attempt to declare the asymmetry of <T^| without reference to the corresponding tensor function. We can regard both T^ = <T^| and T^(v,v') = <T^|v,v'> as representations of the same antisymmetric rank-2 tensor functional <T^| in V* ^ V* = Λ2. The object T^ is an asymmetric bilinear rank-2 tensor functional, whereas T^(v,v') is an asymmetric bilinear rank-2 tensor function (a Spivak 2-tensor). There is a 1-to-1 correspondence between T^ and T^(v,v') . We shall say T^ ϵ Λ2 while T^(v,v') ϵ Λ2f (f = function), and the two spaces are isomorphic. Therefore, Fact: The vector space Λ2 is equivalent to the vector space Λ2f of asymmetric bilinear functions on V2. (4.4.34) This may be compared to our earlier statement for the larger space V*2 = V*V* , Fact: The vector space V*2 is equivalent to the vector space V*2f of bilinear functions on V2. (4.2.15) 5. The Tensor Product of k vectors : the vector spaces Vk and T(V) Our task is now to generalize the tensor product from V2 to Vk, where Vk ≡ VV .... V . // tensor product of k vector spaces, each one is V (5.1) We are setting up for a parallel treatment in Chapter 6 where becomes ^, so certain rather obvious statements will be made here to allow for comparison later with the wedge product. 5.1 Pure elements, basis elements, and dimension of Vk A generic pure ("decomposable") element of Vk is this tensor product of k vectors, v1 v2 ..... vk all vi ϵ V (5.1.1) = |v1> |v2> ... |vk> = |v1,v2, ... vk> . // Dirac notation Since is associative by (2.8.22), one can install parentheses anywhere in (5.1.1) without altering the meaning of the object, for example, v1 (v2 v3) .... vk = v1 v2 v3 .... vk . The basis elements of Vk are ui ui ..... ui = |ui> |ui> ... |ui> = | ui, ui ...ui> . (5.1.2) In (5.1.1) and (5.1.2) the subscripts are labels, not components. The components of these two tensor objects are given by the (2.8.18) outer product form, (v1 v2 ..... vk)jj...j = (v1)j (v2)j .... (vk)j (5.1.3) (ui ui ..... ui)jj...j = (ui)j (ui)j .... (ui)j = δij δij .... δij . (5.1.4) If n = dim(V), the total number of such basis elements is nk, so dim(Vk) = nk. (5.1.5) In the full set of tensor-product basis elements shown in (5.1.2), two or more of the ui might be the same. This will always be the case if k > n where n ≡ dim(V). For example, for k = 3 and n = 2 one such element would be u1 u1 u2 ≠ 0. In Dirac notation, we can write (5.1.3) and (5.1.4) as < uj, uj ...,uj | v1,v2, ...vk> = < uj|v1>< uj|v2>...< uj|vk> = (v1)j (v2)j .... (vk)j (5.1.3a) < uj, uj .....,uj | ui,ui, ... ui> = δij δij .... δij . // <uJ|uI> = δJI (5.1.4a) 5.2 Tensor Expansion for a tensor in Vk ; the ordinary multiindex Note: This section is subset of Section 2.10 (b) with new equation numbers and with Dirac notation equations added at the end. A rank-k tensor T in Vk has this general expansion on the ur basis, T = Σii....i Tii....i (ui ui ..... ui) . (5.2.1) As expected, [T]jj...j = Σii....i Tii....i (ui ui ..... ui)jj...j = Σii....i Tii....i (δij δij .... δij) = Tjj...j . (5.2.2) The coefficients Tii....i can be projected out from T as in (2.10.19), (uiui... ui) T = Tii...i (5.2.3) with an appropriate generalization of the dot product to the space Vk = VV...V, (v1v2...vk) (u1u2...uk) ≡ Σii....i (v1v2...vk)ii....i (u1u2...uk)ii....i = Σii....i (v1)i(v2)i... (vk)i (u1)i(u2)i... (uk)i // outer products = (v1 u1) (v2 u2) .... (vk uk) . (5.2.4) Using the notion of a multiindex I (an ordinary multiindex), I ≡ i1, i2, .....ik // each is ranges 1,2....n n = dim(V) (5.2.5) and a shorthand notation for the basis vectors uI ≡ ui ui ..... ui uI ≡ ui ui ..... ui (5.2.6) the expansion (5.2.1) can be stated in the following compact form, T = ΣI TI uI . (5.2.1) (5.2.7) and the coefficients TI can be projected out according to (5.2.3), uI T = TI . (5.2.3) (5.2.8) The Dirac notation restatements of selected equations above are , |T> = Σii....i Tii....i | ui, ui .....,ui> . (5.2.1a) [T]jj...j = < uj, uj .....,uj | T> = Σii....i Tii....i < uj, uj .....,uj | ui, ui .....,ui> = Σii....i Tii....i (δij δij .... δij) = Tjj...j (5.2.2a) or [T]J = <uJ|T> = ΣITI <uJ|uI> = ΣITI δJI = TJ . (uiui... ui) T = <ui, ui, ...ui | T> = Tii...i // = T(ui, ui, ...ui) (5.2.3a) <v1,v2...vk| u1,u2...uk> = <v1|u1><v1|u1>...<vk|uk> (5.2.4a) |T> = ΣI TI |uI> (5.2.7a) <uI| T> = TI . (5.2.8a) 5.3 Rules for product of k vectors The tensor product of k vectors is "k-multilinear" meaning it is linear in each of its k factors. This was discussed in (1.1.16) and later in (3.1.4). For example, v1(v2 + v'2)v3.....vk = v1v2v3 .....vk + v1v'2v3 .....vk v1(sv2)v3 ..... vk = s(v1v2v3 .....vk) s = scalar (5.3.1) Here we show linearity in the 2nd factor. All the other factors have similar equations. We impose this k-multilinearity by fiat with the result that: Fact: The space Vk is a vector space. (5.3.2) The proof of this fact follows that of the text near (1.1.9). For example, the "0" in Vk is represented by (5.1.1) with one or more vectors being 0, since for example, v10 .....vk = v1(v2 - v2) .....vk = v1v2 .....vk - v1v2 .....vk = 0 . (5.3.3) "Vector multiplication" is distributive over scalar addition (here the "vector" is v1v2 .....vk), as one finds applying the rules (5.3.1), (s1 + s2)(v1v2 .....vk) = [(s1+s2)v1]v2 .....vk = [s1v1+s2v1]v2 .....vk (5.3.4) = s1(v1v2 .....vk)+ s2(v1v2 .....vk) s1,s2 ϵ K and multiplication by a scalar is distributive over "vector addition", s[(v1v2 .....vk) + (v'1v'2 .....v'k)] = s (v1v2 .....vk) + s (v'1v'2 .....v'k) . (5.3.5) All the above equations are meaningful for any positive integer k, regardless of the value n = dim(V). 5.4 The Tensor Algebra T(V) Direct Sums A direct sum of two vector spaces Z = VW is a new vector space and has elements vw. Similarly, a direct sum of three vector spaces Z = VWX is a new vector space with elements vwx. The idea can be applied to any number of vector spaces. Below we use Z = V0V1V2 .... The reader unfamiliar with direct sums will find a detailed description in Appendix B including a simple "tall vector" method of visualizing such spaces. The Tensor Algebra Normally one does not add apples and oranges, so one does not add items of the form ab ϵ V2 to those of the form abc ϵ V3. However, as one writer notes, fruit salad is great, and so we could define a very large vector space of the form T(V) ≡ V0 V V2 V3 ....... = Σk=0∞ Vk . (5.4.1) Here V0 = the space of scalars, V1 = V the space of vectors, V2 = VV = the space of rank-2 tensors, and so on. The most general element t of the space T(V) has the form t = s ΣiTi ui Σij Tij uiuj Σijk Tijk uiujuk + ...... s ϵ K (5.4.2) with all coefficients in a field K. Fact: This large space T(V) is in fact itself a vector space. (5.4.3) We know this is true since T(V) = Σk=0∞ Vk and we showed in (5.3.2) that each Vk is a vector space. For example, the "0" element in T(V) is the direct sum of the "0" elements of all the Vk. See Appendix B for more detail. To show that T(V) is an algebra, we must show that it is closed under both addition and multiplication. It should be clear to the reader that T(V) is closed under addition and has the right scalar rule. For example, if k1 and s are scalars, k1 a bc fgh = sum of 4 elements of T(V) = an element of T(V) s(k1 a bc fgh) = (sk1) (sa) (sb)c f(sg)h = element of T(V) . (5.4.4) This additive closure is of course necessary for T(V) be a vector space. The space is also closed under the multiplication operation . For example (bc)(fgh) = bcfgh = ϵ V5 = ϵ T(V) . // (bc) ϵ V2 , (fgh) ϵ V3 (5.4.5) Here we have used the associative property (2.8.22) applied to vectors. This closure claim is stated more generally below (5.6.7). For later comparison with the corresponding wedge picture, here we have: Object lin comb is Rank(grade) Space s scalar ϵ K 0 V0 a vector 1 V1 ab rank-2 tensor 2 V2 abc rank-3 tensor 3 V3 abcd rank-4 tensor 4 V4 ..... abcd.... rank-k tensor k Vk ..... arbitrary element of T(V) multivector mixed T(V) (5.4.6) Since T(V) is closed under the operations and , it is "an algebra" (the space Vk alone is not an algebra because it is not closed under ). The T(V) algebra is different from that of the reals due to its definition as a direct sum of vector spaces. The elements of T(V) have different "grades" as shown in the right column above, and T(V) is known therefore as a "graded algebra". The grade here is just the tensor rank. Sometimes T(V) is called "the tensor algebra" over V, see for example Benn and Tucker page 3. Any linear combination of a set of tensor products of k-vectors is a rank-k tensor. More generally, a rank-k tensor has the form shown in (5.2.1). A multivector is any linear combination of rank-k tensors for any mixed values of k The dimensionality of the space T(V) is as follows, where n = dim(V), dim[T(V)] = 1 + n + n2 + n3 + ... = ∞ (5.4.7) Here are a few Dirac notation restatements of equations above k1 |a> |b,c> |f,g,h> = sum of 4 elements of T(V) = an element of T(V) (5.4.4a) s(k1 |a> |b,c> |f,g,h>) = (sk1) |sa> |sb,c> |f,sg,h> = element of T(V) . |b,c> |f,g,h> = |b,c,f,g,h> = ϵ V5 = ϵ T(V) . // |b,c> ϵ V2 , |f,g,h> ϵ V3 (5.4.5a) 5.5 Comments about tensors The following fact is doubtless obvious to the reader, but we feel it is worth stating explicitly. First, suppose Tij are the components of a rank-2 tensor. Define Qij ≡ Tji. Then Q is also a rank-2 tensor (although one different from T if T is not symmetric). Here is a formal proof of this claim: transformation (2.1.6) i↔j and a↔b reorder (T = rank-2 tensor) T'ij = RiaRjbTab T'ji = RjbRiaTba = Ria RjbTba Q'ij = T'ji = Ria RjbQab (Q = rank-2 tensor) (5.5.1) In similar fashion the reader can verify the following : Fact: If Tii....i are the components of a rank-k tensor, then Tjj....j are the also components of a rank-k tensor, where the {jn} are any permutation of the {in}. The permuted tensor is in general a different rank-k tensor from the unpermuted one. (5.5.2) Corollary: Any linear combination of permutations of Tii....i is a rank-k tensor. (5.5.3) Example: If Tij is a rank-2 tensor, then so is Aij = (Tij - Tji). Thus, the Aij shown in (4.3.10) is a rank-2 tensor given that Tij is a rank-2 tensor. 5.6 The Tensor Product of two or more tensors in T(V) The tensor algebra T(V) shown in (5.4.1) is closed under both and . It seems evident how one would add two tensors of T(V) of the form (5.4.2), but how would one multiply two tensors? During this set of steps, we try to gracefully transition into multiindex notation by doing a "real-time translation" for each line. Consider two tensors of rank k and k' expanded as in (5.2.1), T = Σii....i Tii....i (ui ui ..... ui) . rank k, T ϵ Vk (5.6.1) ΣITIuI S = Σjj....j Sjj....j (uj uj ..... uj) rank k', S ϵ Vk' . (5.6.2) ΣJSJuJ Multiplying these together with one gets, using the rules (5.3.1), TS = [Σii....iTii....i(ui ui ..... ui)][Σjj....jSjj....j(uj uj ....uj)] [ ΣITIuI] [ΣJSJuJ] (a) = Σii....i Σjj....jTii....i Sjj....j(ui ui ..... ui) (uj uj ..... uj) ΣI,JTISJ(uI) (uJ) (b) = Σii....ijj....jTii....i Sjj....j(ui ui ..... ui uj uj ..... uj) ΣI,JTISJ(uI uJ) (c) = Σii....iii....i Tii....i Sii....i (ui ui ...... ui) . ΣI,I'TISI'(uI uI') (d) = Σii....iii....i [TS]ii....i ii....i (ui ui ...... ui) . ΣI,I'[TS]I,I'(uI uI') (u) = Σii...i [TS]ii...i (ui ui ...... ui) . (5.6.3) ΣI (TS)I uI . // italic I's Comparing lines one sees that I ≡ i1, i2...ik I' ≡ ik+1, ik+2, ....ik+k' I ≡ I, I' = i1,i2...ik+k' uI ≡ (ui ui .... ui) uI' ≡ (ui...ui) uI ≡ (ui ui .... ui) . (5.6.4) Notice that the (2.8.22) associativity of is used going from (a) to (b). In step (c) we renamed the dummy jr summation indices so that j1 = ik+1 , j2 = ik+2 and so on. Step (d) uses the outer product form (3.1.14) to replace TI SI' = (TS)I,I' = (TS)I . The conclusion is that TS = ΣI (TS)I uI I ≡ I, I' = i1,i2...ik+k', uI ≡ (ui ui .... ui) . (5.6.5) Since the uI are basis vectors in Vk+k', we have shown that: T ϵ Vk and S ϵ Vk' TS ϵ Vk+k' T(V) . (5.6.6) Thus we have strengthened the claim made in (5.4.5) that T(V) is closed under the operation . We shall now undertake the tensor product of three tensors T,S,R of ranks k,k',k" by mimicking the above set of steps, but leaning more heavily now on multiindex notation (no training wheels here), TSR = [ΣITIuI][ΣJSJuJ][ΣKSKuK] (a) = ΣI,J,K TISJRK (uI) (uJ) (uK) (b) = ΣI,J,K TISJRK (uI uJ uK ) // associative of used here (d) = ΣI,I',I" TISI'RI" (uI uI' uI") // rename multiindices J→I',K→I" I ≡ i1, i2...ik I' ≡ ik+1, ik+2, ....ik+k' I" ≡ ik+k'+1, ik+k'+2, ....ik+k'+k" uI ≡ (ui.... ui) uI' ≡ (ui... ui) uI" ≡ (ui... ui) (e) = ΣI (TSR)I uI uI ≡ (ui.... ui) I ≡ I, I',I" = i1,i2...ik+k'+k" (5.6.7) Now the outer product form is TISI'RI" = (TSR)I,I',I" = (TSR)I . The conclusion is this: TSR = ΣI (TSR)IuI I ≡ I, I',I" = i1,i2...ik+k'+k" uI ≡ (ui.... ui) (5.6.8) Since the uI are basis vectors in Vk+k'+k", we have shown that: T ϵ Vk and S ϵ Vk' and R ϵ Vk" TSR ϵ Vk+k'+k" T(V) . (5.6.9) To develop a more systematic approach, consider the first three tensors in a product of tensors, T1 = tensor of rank k1 I1 = {i1, i2.....ik} T2 = tensor of rank k2 I2 = {ik+1, ik+2.....ik+k} T3 = tensor of rank k3 I3 = {ik+k+1, ik+k+2.....ik+k+k} . (5.6.10) Define the following "cumulative ranks", κ1 = k1 κ2 = k1+ k2 κ3 = k1+ k2 + k3 ... κN = k1 + k2 + ... + kN = Σi=1N ki . (5.6.11) Then rewrite and extend (5.6.10), T1 = tensor of rank k1 I1 = {i1, i2.....iκ} T2 = tensor of rank k2 I2 = {iκ+1, iκ+2.....iκ} T3 = tensor of rank k3 I3 = {iκ+1, iκ+2.....iκ} ... Ts = tensor of rank ks Is = {iκ+1, iκ+2.....iκ} ... TN = tensor of rank kN IN = {iκ+1,iκ+2.....iκ} . (5.6.12) In this notation, and generalizing the above development for the tensor product of three tensors, we find the following expansion for the tensor product of N tensors of T(V), T1T2...TN = ΣI (T1T2....TN)I uI (5.6.13) where uI = ui ui ..... ui = ui ui ..... uκ and (T1T2....TN)I = T1IT2I .... TNI . The rank of this product tensor is then κ = Σi=1N ki and the tensor is an element of Vκ T(V). In Dirac notation, one rewrites (5.6.13) as | T1,T2....TN> = | T1T2....TN> = |T1>|T2>...|TN> = ΣI (T1T2....TN)I |uI> (5.6.13a) Example 1: The tensor product of two rank-1 tensors. TS = Σii[Ti Si] (ui ui) = Σij TiSj (uiuj) = Σij (TS)ij (uiuj) (TS)ab = Σij TiSj (uiuj)ab = Σij TiSj δiaδjb = TaSb (5.6.14) Example 2: The tensor product of two rank-2 tensors. TS = Σiiii Tii Sii (ui ui ui ui ) (TS )abcd = TabScd (5.6.15) In both examples the evaluation of components produces the expected outer product forms. Special cases of the tensor product TS. Assume T and S have rank k and k'. If S = κ' ϵ K = a scalar, then rank(S) = k' = 0 and (5.6.4) reads, TS = Σii....ijj....jTii....i Sjj....j(ui ui ..... ui uj uj ..... uj) = Σii....iTii....i (κ') (ui ui ..... ui) = κ'T and ST = Σjj....jii....i Sjj....jTii....i(uj uj ..... uj ui ui ..... ui) = Σii....i (κ') Tii....i ( ui ui ..... ui) = κ'T so we find that TS = ST = κ'T . If T = κ and S = κ', the result above would be TS = κκ' and ST = κ'κ and so TS = ST = κκ'. Thus, TS = κS = ST = Sκ = κS if T = κ ϵ V0 TS = Tκ' = ST = κ'T = κ'T if S = κ' ϵ V0 TS = κκ' = ST = κ'κ = κκ' if T,S = κ,κ' ϵ V0 (5.6.16) All equations above can be written in Dirac notation, for example, |T>|S> = [ΣITI|uI>] [ΣI'SI'|uI'>] = ΣI,I'TISI'|uI>|uI'> = ΣI (TS)I |uI> (5.6.5)D |T> ϵ Vk and |S> ϵ Vk' |TS> = |T>|S> = ϵ Vk+k' T(V) . (5.6.6)D Operators on the tensor product space Recall from above the following tensor product space vector, | T1,T2....TN> = | T1T2....TN> = |T1> |T2>... |TN> (5.6.13a) which is an element of the tensor product space Vk Vk ... Vk. The action of a linear operator P on such a tensor product vector is defined in terms of its action in the spaces from which the tensor product is composed, P [ |T1> |T2>... |TN>] = P |T1> P |T2>... P |TN> (5.6.17) 6. The Tensor Product of k dual vectors : the vector spaces V*k and T(V*) Every equation from Chapter 5 can be converted to an appropriate equation of Chapter 6 using this simple set of translation rules: 1. |X> → <X| and <Y| → |Y>. That is, reverse all Dirac bras and kets. 2. Swap lower and upper indices, indices. eg. ui → ui, Tij → Tij (really: reverse all tilts). 3. |vi> → <αi| // use Greek/script names for functionals; vi → αi 4. Vk → V*k // space goes to dual space 5. < T | v1,v2.....vk > = T(v1,v2.....vk) = a tensor function (a new item) (6.1) In general, translation of a Chapter 5 equation to Chapter 6 is most easily done if the Chapter 5 equation is first stated in Dirac notation. We could end Chapter 6 right here, allowing the reader to apply the above rules, but that seems unsportsmanlike, so we proceed with a partial mimicry of Chapter 5. 6.1 Pure elements, basis elements, and dimension of V*k A generic pure ("decomposable") element of V*k is this tensor product of k functionals, α1 α2 ..... αk . all αi ϵ V* (6.1.1) = <α1| <α2| ... <αk| = <α1,α2, ... αk| . // Dirac notation Since is associative by (2.8.22), one can install parentheses anywhere in (6.1.1) without altering the meaning of the object, for example, α1 (α2 α3) .... αk = α1 α2 a3 .... αk . The basis elements of V*k are λi λi ..... λi = <ui| <ui| ... <ui| = < ui, ui ...ui| . (6.1.2) The subscripts in (6.1.1) and the superscripts in (6.1.2) are labels, not components. Equations corresponding to (5.1.3) and (5.1.4) are these (λj λj ..... λj)(v1,v2.....vk) = λj(v1) λj(v2) .....λj(vk) = (v1)j (v2)j .... (vk)j (6.1.3) (λj λj ..... λj)(ui,ui...ui) = λj(ui) λj(ui) .....λj(ui) = δji δji .... δji . (6.1.4) If n = dim(V), the total number of such basis elements is nk, so dim(V*k) = nk. (6.1.5) In the full set of dual tensor-product basis elements shown in (6.1.2), two or more of the λi might be the same. This will always be the case if k > n where n ≡ dim(V*). For example, for k = 3 and n = 2 one such element would be λ1 λ1 λ2 ≠ 0. In Dirac notation, we can write (6.1.3) and (6.1.4) as <λj, λj...λj | v1, v2 ...,vk > = <λj|v1> <λj|v1>... <λj|v1> = (v1)j (v2)j...(vk)j (6.1.3a) <λj, λj...λj | ui, ui ...,ui > = <λj|ui> <λj|ui>...<λj|ui> = δji δji...δji (6.1.4a) 6.2 Tensor Expansion for a tensor in V*k ; the ordinary multiindex We apply our translation rules to (5.2.1a) through (5.2.8a) to get this dense translation of Section 5.2 <T| = Σii....i Tii....i < ui, ui .....,ui| . tensor functional (6.2.1) <T| = ΣI TI <uI| <uI| = < ui, ui .....,ui | (6.2.2) T = Σii....i Tii....i (λi λi ..... λi) . (6.2.3) T = ΣI TI λI λI ≡ λi λi ..... λi (6.2.4) T (uiui... ui) = <T| ui, ui .....,ui> = Tii...i = T(ui, ui .....,ui) (6.2.5) T uI = <T|uI> = TI = T(uI) (6.2.6) T (v1v2... vk) = <T| v, v .....,v> = T(v, v .....,v) tensor function (6.2.7) T vZ = <T|vZ> = T(vz) (6.2.8) 6.3 Rules for product of k vectors The tensor product of k vectors is "k-multilinear" meaning it is linear in each of its k factors. This was discussed in (1.1.16) and later in (3.1.4). For example, α1(α2 + α'2)α3.....αk = α1α2α3 .....αk + α1α'2α3 .....αk α1(sα2)α3 ..... αk = s(α1α2α3 .....αk) s = scalar (6.3.1) Here we show linearity in the 2nd factor. All the other factors have similar equations. We impose this k-multilinearity by fiat with the result that: Fact: The space V*k is a vector space. (6.3.2) Proof: Repeat the discussion of Section 5.3 with all vi→ αi, meaning |vi> → <αi| . 6.4 The Tensor Algebra T(V*) T(V*) ≡ V*0 V* V*2 V*3 ....... = Σk=1∞ V*k . (6.4.1) Here V*0 = the space of scalars, V*1 = V the space of dual vectors, V*2 = V**V = the space of rank-2 dual tensors, and so on (tensor = functional). The most general element t of the space T(V*) has the form τ = s ΣiTi λi Σij Tij λiλj Σijk Tijk λiλjλk + ...... s ϵ K (6.4.2) with all coefficients in a field K. Fact: This large space T(V*) is in fact itself a vector space. (6.4.3) The proof is the same as that shown in Section 5.4 with a,b,c,d,e,f replaced by Greek letters. For example k1 α βκ ρση = sum of 4 elements of T(V*) = an element of T(V*) k2 <α| <β,κ| <ρ,σ,η| (6.4.4) For later comparison with the corresponding dual wedge picture, here we have: Object lin comb is Rank(grade) Space s scalar ϵ K 0 V*0 α dual vector 1 V*1 αβ dual rank-2 tensor 2 V*2 αβγ dual rank-3 tensor 3 V*3 αβγδ dual rank-4 tensor 4 V*4 ..... αβγδ.... dual rank-k tensor k V*k ..... arbitrary element of T(V*) dual multivector mixed T(V*) (6.4.6) All objects listed in the left column are tensor functionals, but we just call them tensors above and below. Any linear combination of a set of tensor products of dual k-vectors is a dual rank-k tensor. More generally, a dual rank-k tensor has the form shown in (6.2.3). A dual multivector is any linear combination of dual rank-k tensors for any mixed values of k . The dimensionality of the space T(V*) is as follows, where n = dim(V*), dim[T(V*)] = 1 + n + n2 + n3 + ... = ∞ (6.4.7) 6.5 Comments about Tensor Functions For every rank-k tensor functional <T| = T in V*k there exists a corresponding tensor function: T(v, v .....,v) = <T| v1,v2...vk> // T(vZ) = <T|vZ> T(ui, ui .....,ui) = <T| ui, ui .....,ui> = Tii...i // (6.2.5) (6.5.1) There is a simple one-to-one relationship between the rank-k tensors |T> of Vk and the rank-k tensor functionals <T| of V*k and the rank-k tensor functions T(vZ) of V*k. These functions are manifestly k-multilinear since | v1,v2...vk> = | v1>| v2>...| vk> is k-multilinear. That is to say, each V space in the tensor product Vk = VV ...V is a linear (vector) space. Fact: The vector space V*k is equivalent to the vector space of k-multilinear functions on Vk. (6.5.2) This is the generalization of Fact (4.2.15) from k = 2 to k = k. The point made in Section 5.5 about tensors remaining tensors if their indices are shuffled around is reflected in the space of tensor functions: if T(v,v .....,v) is a rank-k tensor function, then so is the function T(vi, vi .....,vi) where the arguments are any permutation of v,v .....,v . 6.6 The Tensor Product of two or more tensors in T(V*) Were we to write out the full detailed development of Section 5.6, it would begin as follows : Consider two tensor functionals of rank k and k' expanded as in (6.2.3), T = Σii....i Tii....i λi λi ..... λi rank k, T ϵ V*k S = Σjj....j Sjj....j λj λj ..... λj rank k', S ϵ V*k' . (6.6.1) In multiindex and then Dirac notation these equations say T = ΣI TI λI or <T| = ΣI TI <uI | S = ΣI SI λI or <S| = ΣJ SJ <uJ | (6.6.2) and the tensor product of interest is TS = <T| <S| . (6.6.3) The entire development proceeds as shown in Section 5.6 but with the translation rules outlined at the start of Chapter 6, in particular, that all bra-kets are reversed. One then finds for the tensor product of a rank-k tensor functional with a rank-k' one, TS = ΣI,J TISJ λIλJ = ΣI (TS)I λI (TS)I = TISI' (6.6.4) I = {i1, i2, .. ik+k'} λI ≡ λi λi ..... λi I = {i1, i2, .. ik} I' = {ik+1, ik+2, .. ik+k'} . which compare to the non-dual (5.6.5) ( recall that uI = <uI| and λI = <uI| ) TS = ΣI (TS)I uI . (5.6.5) A triple tensor product is then TSR = ΣI,J,K TISJRK λIλJλK = ΣI (TSR)I λI (TSR)I = TISI'RI" (6.6.5) I = {i1, i2, .. ik+k'+k"} λI ≡ λi λi ..... λi I = {i1, i2, .. ik} I' = {ik+1, ik+2, .. ik+k'} I" = {ik+k'+1, ik+k'+2, .. ik+k'+k"} . For an arbitrary set of tensor functionals Ti of rank ki the tensor product is T1T2...TN = ΣI [(T1)I(T2)I .... (TN)I] λI = ΣI (T1T2....TN)I λI (6.6.6) Ti = tensor functional of rank ki , Ii = multiindex range of ir values for tensor Ti I1 = {i1, i2.....ik}, I2 = {ik+1, ik+2.....ik+k}, etc. I = I1 I2 .... IN = {i1, i2......ik+k+...k} λI ≡ λi λi ..... λi (6.6.7) where the resulting tensor has a rank equal to the sum of the ranks of the combined tensors. In Dirac notation, equation (6.6.6) is written, < T1,T2....TN| = < T1T2....TN | = <T1| <T2| ...<TN| = ΣI (T1T2....TN)I <uI| (6.6.8) which is just the Dirac transpose (plus tilt reversal) of the equation (5.6.13a) of Chapter 5. It is understood here that each bra space fits the rank of its tensor, and one could write <Ti| = k<Ti| to make this fact more explicit. Then (6.6.8) would read k<T1| k<T2 | ... k<TN| = ΣI (T1T2....TN)I κ<uI| κN = k1+k2+...+kN . (6.6.9) Consider now the following generic tensor product ket, |v1,v2...vk>k |vk+1,vk+2...vk+k>k .... = |VI>k |VI>k .... |VI>k = |VI> κ = | v1,v2,v3,v4 ................vκ> . (6.6.10) If we close the bra (6.6.9) with this ket, we obtain the simple rule for the tensor product of the corresponding tensor functions, < T1T2....TN | v1,v2 ...vκ> = [ k<T1| k<T2 | ... k<TN| ] [ |VI>k |VI>k .... |VI>k ] = k<T1|VI>k * k<T2|VI>k .... * k<TN|VI>k = T1(vI) T2(vI) ..... TN(vI) (6.6.11) or (T1T2....TN)(v1,v2 ..............vκ) = (T1T2....TN)(vI,vI ...vI) = T1(vI) T2(vI) ..... TN(vI) . (6.6.12) Example: For N = 2 and k1= k and k2 = k' : (TS)(v1,v2....vk, vk+1....vk+k') = T(v1,v2....vk) S(vk+1,vk+2....vk+k') . (6.6.13) Example: For N = 3 and k1= k and k2 = k' and k3 = k" : (TSR)(v1, v2, ... vk+k'+k") T(v1,v2....vk) S(vk+1,vk+2....vk+k')R(vk+k'+1,vk+k'+2....vk+k'+k") . (6.6.14) Here is a direct proof of (6.6.12), independent of Chapter 5, where we make use of the dense multiindex notation. Let, Ti = tensor of rank ki Ii = multiindex range of ir values for tensor Ti I1 = {i1, i2.....ik}, I2 = {ik+1, ik+2.....ik+k}, etc. (6.6.15) Then we have T1T2...TN = ΣII...I (T1)I(T2)I ...(TN)I λIλI...λI (6.6.16) (T1T2...TN)(vI, vI ... vI) = ΣII...I (T1)I(T2)I ...(TN)I (λIλI...λI)(vI, vI ... vI) = ΣII...I (T1)I(T2)I ...(TN)I (vI)I(vI)I ... (vI)I = [ΣI(T1)I(vI)I] [ΣI(T2)I(vI)I] ... [ΣI(T2)I(vI)I] = T1(vI) T2(vI) .... TN(vI) . (6.6.17) Operators on the tensor product space Recall from (6.6.8) the following tensor product space vector, < T1,T2....TN| = < T1T2....TN | = <T1| <T2| ...<TN| (6.6.8) which is an element of the tensor product space V*k V*k ... V*k. The action of a linear operator Q on such a tensor product vector is defined in terms of its action in the spaces from which the tensor product is composed, [ <T1| <T2| ...<TN| ] Q = <T1| Q <T2| Q ... <TN| Q (6.6.18) This equation is the transpose of (5.6.17) if we set Q = PT . 7. The Wedge Product of k vectors : the vector spaces Lk and L(V) Wedge products and the spaces Lk and L(V) to be defined below were developed by Hermann Grassmann (1809-1877) in the 1840's. The algebra of these spaces is now called the exterior algebra and the wedge products are alternately called exterior products. Grassmann more or less invented the notions of linear algebra and vector spaces -- the so-called "modern algebra" did not exist. Other people were involved, but he was a very major pioneer. His work, naturally, was unappreciated at that time. 7.1 Definition of the wedge product of k vectors We wish to define the wedge product of k vectors vi ϵ V, v1^ v2^ .....^ vk . // |v1> ^ |v2> ^ .....^ |vk> Wedge products of this form (and their linear combinations) inhabit a vector space we call Lk. We now impose the requirement that this wedge product must change sign when any two vectors are swapped. This property is injected into the wedge product theory, it does not fall out from it. One motivation for the requirement relates to geometry. We showed in (4.3.14) that a ^ b = det(a,b) u1^u2 where det(a,b) is the signed area of the 2-piped (parallogram) spanned by a and b. Then b ^ a = [ -det(a,b)] u1^u2 has the same area but of opposite sign. One associates this sign with the "orientation" of the area in exactly the same sense that a x b and b x a represent areas of opposite sign. So b ^ a = - a ^ b reflects the change in orientation, as suggested by these drawings from Suter, a b b ^ a (7.1.1) For R3, as shown in (4.3.15), one associates a^b^c with a 3-piped whose "orientation" is determined by the sign of the volume det(a,b,c), which one can associate with the "handedness" of the 3-piped. For a k-piped it is hard to imagine "handedness", but it is easy to talk about orientation as the sign of det(a,b,c....) where swapping any two vectors changes the sign of the "volume". This sign-change requirement leads to the following candidate definition for the wedge product of k vectors in V (the jr are vector labels), vj^ vj^ .....^ vj = (1/k!) ΣP (-1)S(P) (vj vj ..... vj) = (1/k!) [ (vj vj ..... vj) + all signed permutations ] = Alt(vj vj ..... vj) . (7.1.2) An explanation of the ΣP (-1)S(P) notation is presented in Section A.1: the sum is over all permutations P of [1,2..k], S(P) is the number of index swaps required to get from [1,2..k] to P[1,2..k], and (-1)S(P) is the parity of permutation P. The important Alt operator is described generically in Section A.2 and is then applied to tensors in Section A.5. The definition of the Alt operator on the last line in (7.1.2) is the expression on the right side of the first line. The (1/k!) is present in both our normalization and the Spivak normalization. For the purposes of this section, we simplify things by taking jr → r so (7.1.2) becomes, v1^ v2^ .....^ vk = (1/k!) ΣP (-1)S(P) ( vP(1) vP(2) ..... vP(k)) = (1/k!) [ (v1 v2 ..... vk) + all signed permutations ] . = Alt(v1 v2 ..... vk) = (1/k!) Σii...i εii...i (vi vi ..... vi) ir = 1 to k . (7.1.3) We have added a fourth form for the sum using the permutation tensor εii...i . This tensor is described in Section A.6 and the equivalence of the first and fourth forms above is shown in (A.6.8). From our viewpoint, the (1/k!) normalization factor appearing in (7.1.2) and (7.1.3) is just a convention that many authors use. However, Benn & Tucker (p 11 bottom and p 5 footnote) and Conrad (p 13 top) argue that the (1/k!) is in fact the "correct" normalization to be consistent with more elegant methods of defining the wedge product, as briefly reviewed in our Chapter 9. For other authors like Spivak, the (1/k!) is replaced by 1. The implications of this Spivak normalization are described in Section 7.9(g) below. Notice that when the (1/k!) is present, (7.1.3) gives v1^ v2 = (1/2)( v1v2 - v2v1) which is the form already assumed in (4.3.1) and (4.4.1). Almost everything one does with the wedge product is unaffected by the normalization choice. Our approach here is that v1^ v2^ .....^ vk is defined in terms of v1v2 .....vk . In Section 9.1 it is shown how v1^ v2^ .....^ vk can be defined perhaps more elegantly in the language of modern algebra. Examples v1^ v2 = (1/2!) Σa,b =12 εab va vb // 2! = 2 terms = (v1 v2 - v2 v1)/2 // agrees with (4.3.1) (7.1.4) v1 ^ v2 ^ v3 = (1/3!) Σa,b,c =13 εabc va vb vc // 3! = 6 terms = (v1v2v3 - v1v3v2 + v3v1v2 - v3v2v1 + v2v3v1 - v2v1v3)/6 . (7.1.5) 7.2 Properties of the wedge product of k vectors 1. The sums in (7.1.2) and (7.1.3) have k! terms. (7.2.1) Since there are k! permutations P of {1,2..k} (including the identity permutation) there are k! terms in the ΣP sums in (7.1.2) and (7.1.3). Because εii...i vanishes whenever two or more indices are the same, the ε tensor has k! non-zero components (k for the first index, (k-1) for the second index, and so on). Thus, the second sum in (7.1.3) has k! terms (not kk), just like the first sum. 2. The wedge product is k-multilinear. (7.2.2) It is by-fiat axiom that the wedge product of k vectors is k-multilinear and therefore satisfies these rules, v1^(v2 + v'2)^v3^.....^vk = v1^v2^v3^ .....^vk + v1^v'2^v3^ .....^vk v1^(sv2)^v3^ .....^ vk = s(v1^v2^v3^ .....^vk) s,r = scalar ϵ K or v1^(rv2 + sv'2)^v3^.....^vk = r(v1^v2^v3^ .....^vk) + s(v1^v'2^v3^ .....^vk) . (7.2.3) Here we show the rules just for the 2 position, but k-multilinear means these rules must apply to all the vector positions. These rules cannot be derived from the similar tensor product rules (5.3.1) except in the case k = 2 as was shown in (4.3.4). Our candidate expansions (7.1.2) and (7.1.3) satisfy (7.2.3) because they are k-multilinear: v1^ (rv2 + sv'2) ^ .....^ vk = (1/k!) ΣP (-1)S(P) ( vP(1) [rvP(2)+ sv'P(2)] ..... vP(k)) = (1/k!) ΣP (-1)S(P) { r( vP(1) vP(2) ..... vP(k)) + s( vP(1) v'P(2) ..... vP(k)) } // (5.3.1) = r(1/k!) ΣP (-1)S(P)(vP(1) vP(2) ..... vP(k)) + s(1/k!) ΣP (-1)S(P)(vP(1) v'P(2) ..... vP(k)) = r(v1^v2^v3^ .....^vk) + s(v1^v'2^v3^ .....^vk) . or equivalently Alt(v1(rv2 + sv'2)v3^ .....vk) = r Alt(v1v2v3^ .....vk)+ sAlt(v1v'2v3^ .....vk) . Above we have used the fact that the product is k-multilinear (also by fiat) as declared in (5.3.1). 3. The wedge product changes sign if any vector pair is swapped. (7.2.4) Consider the last line of (7.1.2) which says, Tjj...j = [Alt(F)]jj...j or T = Alt(F) where Tjj...j ≡ vj ^ vj ^ .....^ vj Fjj...j ≡ vj vj ..... vj . But we know from (A.5.9) that Tjj...j ≡ [Alt(F)]jj...j is totally antisymmetric in the indices. Therefore vj^ vj^ .....^ vj is totally antisymmetric in the labels, and so changes sign if any pair of labels is swapped. Comment: From (7.2.2) the pure vector wedge product v1^ v2^ .....^ vk is k-multilinear in the vi, and so is the underlying tensor product v1v2 .....vk. 4. Wedge product of vectors vanishes if any two vectors are the same. Given a sign change (7.2.4) for any pair swap of vectors in the wedge product, we know that v1^ v2^ .....^ vk = 0 if any two (or more) vectors are the same. (7.2.5) Proof: For example, a ≡ v2^ v1^ .....^ vk = - v1^ v2^ .....^ vk = -a; if 1 = 2 then a = -a so a = 0 . 5. Wedge product vanishes if vectors are linearly dependent. (7.2.6) It was just shown that the wedge product vanishes if any two vectors are the same. It is also true that the wedge product v1^ v2^ .....^ vk vanishes if the vectors vi are linearly dependent. Linear dependence means one can write at least one vector in the set as a linear combination of the others, so perhaps v2 = ( Σi≠2 aivi). Then v1^ v2^ .....^ vk = v1^ ( Σi≠2 aivi) ^ .....^ vk = Σi≠2 ai (v1^ vi ^ .....^ vk) . // since ^ is k-multilinear, see (7.2.3) The sum Σi≠2 requires that index i be some other index appearing in (v1^ vi ^ .....^ vk), but then one has two indices the same and by (7.2.5) it follows that (v1^ vi ^ .....^ vk) = 0 for each term in the sum. QED [ Grassmann also invented the notion of linear independence. ] 6. Wedge product vanishes if k > n . (7.2.7) If dim(V) = n, there can be at most n linearly independent vectors in V. If k > n, any set of k vectors vi must be linearly dependent. Thus, by (7.2.6) the wedge product of any set of k vectors must vanish if k > n. Therefore for a given vector space V of dimension n, the only wedge products of interest are those for k = 1,2,3....n. For example, for n = 2 and k = 3 one has e1 ^ e1 ^ e2 = 0. 7. Components. From (7.1.2) we find (vj^ vj^ .....^ vj)ii...i = (1/k!) ΣP (-1)S(P) (vj vj ..... vj)ii...i = (1/k!) ΣP (-1)S(P) (vj)i(vj)i ... (vj)i // outer product form = (1/k!) ΣP (-1)S(P) (vj)i(vj)i ... (vj)i // (A.1.19) with Mab = (vj)i = (1/k!) det[ (vj)i] . // (A.1.19) (7.2.8) As noted earlier, in the Spivak normalization the factor (1/k!) is replaced by 1. Fact: (vj^vj^...^vj)ii...i is totally antisymmetric in both the labels jr and the indices ir. (7.2.9) Proof: We already know from (7.2.4) that (vj^ vj^ .....^ vj)ii...i is totally antisymmetric in the labels jr. For antisymmetry on the ir, we give two arguments. We can take components of (7.1.2) to get (vj^ vj^ .....^ vj)ii...i = [ Alt(vj vj ..... vj)]ii...i which we think of as saying Tii...i ≡ [Alt(F)]ii...i. According to (A.5.9) Tii...i is totally antisymmetric in the ir indices. Alternatively, (vj^ vj^ .....^ vj)ii...i = (1/k!) det[ (vj)i] is totally antisymmetric on the ir because the det changes sign when any two rows or columns are swapped. 8. Associative Property of the wedge product. This topic is addressed below in (7.9.2) where the need first arises. The conclusion there is that the wedge product is fully associative. For example, (v1^ v2)^ v3 = v1^ (v2^ v3) = v1^ v2^ v3 . 7.3 The vector space Lk and its basis Lk is the space whose elements are all linear combinations of wedge products of k vectors of V. (7.3.1) A more precise name for this space is Lk(V) but we call it Lk. Lk is a vector space (7.3.2) Fact (5.3.2) showed that Vk is a vector space where the 0 element could be any Vk element such as v10 .....vk. Lk is a vector space by a similar argument. It is closed under addition, scalars work correctly according to the rules (7.2.3), and the 0 element can be any element such as v1^0^ .....^vk as the reader can verify looking for example at (7.1.5). A key point is that it is the imposition of the k-multilinear wedge product rules (7.2.3) that makes Lk be a vector space. We had a similar situation in Chapter 5 where the imposition of the k-multilinear tensor product rules (5.3.1) made Vk be a vector space. Basis elements for Lk Consider the following objects in Lk obtained by wedging together k basis elements of V, where each ei is selected from the set of n available for V (which has dimension n), (uj ^ uj ^ .... ^ uj) . (7.3.3) Of these putative nk objects, only n*(n-1)*...*(n-k+1) = n!/(n-k)! are non-zero by (7.2.5) because all the others have at least two vectors the same. Thus we can assume that all the labels jr are different. Now there exists a unique permutation P of the all-different labels jr such that [ j1, j2....jk] = P[ i1, i2....ik] where i1 < i2 < ..... < ik . (7.3.4) If this permutation involves S(P) pairwise swaps of indices, then (uj ^ uj ^ .... ^ uj) = (-1)S(P) (ui ^ ui ^ .... ^ ui) where i1 < i2 < ..... < ik (7.3.5) because from (7.2.4) each pairwise swap of vectors in a wedge product creates a minus sign. Since there are k! possible permutations P, there are k! equations like (7.3.5) which relate different objects to the same object (ui ^ ui ^ .... ^ ui) which has i1 < i2 < ..... < ik .Thus, if we want to count the number of independent basis elements of Lk, we have to divide our earlier count of n!/(n-k)! non-vanishing objects by k!. The conclusion is that there are independent basis elements for Lk and they all have this form (ui ^ ui ^ .... ^ ui) where i1 < i2 < ..... < ik basis elements (7.3.6) Examples: (7.3.7) For k = 3 and n ≥ 5, the following k! = 3! basis elements involving u1, u3 and u5 are all equal to the one ordered element u1^u3^u5 with a + or - sign : u1^u3^u5 = (-1)0 u1^u3^u5 = +u1^u3^u5 135 u1^u5^u3 = (-1)1 u1^u3^u5 = - u1^u3^u5 153→135 u3^u1^u5 = (-1)1 u1^u3^u5 = - u1^u3^u5 315 →135 u3^u5^u1 = (-1)2 u1^u3^u5 = +u1^u3^u5 351→315→135 u5^u1^u3 = (-1)2 u1^u3^u5 = +u1^u3^u5 513→153→135 u5^u3^u1 = (-1)3 u1^u3^u5 = - u1^u3^u5 531→513→153→135 For k = 2 and n = 3, the 3 basis elements are u1^u2, u1^u3, u2^u3 and = 3. Fact: (uj ^ uj ^ .... ^ uj) = Alt(uj uj .... uj) u^J = Alt(uJ) (7.3.8) This is a special case of (7.1.2) with v→u. The right shows equivalent multiindex notation. Components of the basis elements for Lk Now reconsider the basis vectors of the vector space Lk , (uj ^ uj ^ .... ^ uj) . (7.3.3) The components are given from (7.2.8) with v = u as [ recall (ui)j = δij ] , (uj^ uj^ .....^ uj)ii...i = (1/k!) ΣP (-1)S(P) δjiδji .... δji = (1/k!) ΣP (-1)S(P) δjiδji ... δji // (A.1.19) = (1/k!) det[ δji] . (e^J)I = (1/k!) det(δJI) (7.3.9) Once again, in Spivak normalization (1/k!) → 1. Then (7.2.9) applied to v = u shows that, Fact: (uj^uj^...^uj)ii...i is totally antisymmetric in both the labels jr and the indices ir. (7.3.10) We saw an example of both antisymmetries for k = 2 back in equation (4.3.20), (ui^ uj)rs = - (ui^ uj)sr = - (uj^ ui)rs . // two forms of antisymmetry (4.3.20) Either form in (7.3.9) can be expressed in our usual informal notation, (uj ^ uj ^ .... ^ uj)ii...i = (1/k!) [ δji δji...δji + signed permutations] . (7.3.11) Example: 3! (uj ^ uj ^ uj)iii = det = det ≡ det ( δjjjiii ) = det(δJI) . // in multiindex notation (7.3.12) 7.4 Tensor Expansions for a tensor in Lk Recall now the tensor expansion for a most-general tensor T in Vk, T = Σii....i Tii....i (ui ui ..... ui) . Tϵ Vk (5.2.1) (7.4.1) It has been shown in (7.3.10) that (ui^ ui^ .....^ ui)jj...j is totally symmetric in both the ir and jr indices. This object then meets the conditions of Fact (A.8.26) which states that AltI = AltJ when applied to such an object. Consider then this most-general object in Lk which has a similar look to (7.4.1) and for which we explicitly display the tensor components, Σii....i Tii....i (ui^ ui^ .....^ ui)jj...j = Σii....i Tii....i AltI[(ui ui ..... ui)jj...j] // (7.3.8) = Σii....i Tii....i AltJ[(ui ui ..... ui)jj...j] // (A.8.26) = AltJ[ Σii....i Tii....i (ui ui ..... ui)jj...j ] // (A.5.10), Alt is linear = AltJ(Tjj...j) // (7.4.1) = Alt(Tjj...j) // no ambiguity = (1/k!) ΣP (-1)S(P) Tjj...j // def of Alt (A.5.3) = [Alt(T)]jj...j // (A.5.3) ≡ [T^]jj...j (7.4.2) where we define this notation, T^ ≡ Alt(T) . (7.4.3) From (7.4.2) we then have the following fully general element of Lk, T^ = Σii....i Tii....i (ui^ ui^ .....^ ui) . T^ ϵ Lk (7.4.4) We refer to this type of expansion as a symmetric expansion, and we know it is redundant since the symmetric sum includes each true basis vector k! times. According to (A.5.9), we know from (7.4.3) that Fact: T^ii....i is a totally antisymmetric tensor. (7.4.5) Therefore, Fact: The space Lk is the space of all totally antisymmetric rank-k tensors T^. To say that T^ is totally antisymmetric means that T^ii....i is totally antisymmetric. (7.4.6) In contrast, the space Vk is the space of all rank-k tensors T, so Lk Vk. See Section 7.7 below. Since the set (ui ^ ui ^ .... ^ ui) with 1 ≤ i1 < i2 < ..... < ik ≤ n forms a complete basis for Lk, as discussed below (7.3.3), it must be possible to express T^ in the following manner T^ = Σ1≤i<i<....<i≤n Aii...i (ui ^ ui ^ .... ^ ui) . (7.4.7) Example: If n = 3 and k = 2, then T^ = Σ1≤i<i≤3 Aii (ui^ ui) = A12 (u1^u2) + A13 (u1^u3) + A23 (u2^u3) . (7.4.8) What then is the connection between the Aii...i of (7.4.7) and the Tii...i of (7.4.4)? Start with the symmetric form (7.4.4), T^ = Σii...i Tii...i (ui ^ ui ^ .... ^ ui) ir = 1,2..n = Σi≠i≠...≠i Tii...i (ui ^ ui ^ .... ^ ui) . // (7.2.5) (7.4.9) Partition the summation space as follows (1 ≤ ir ≤ n), Σi≠i≠...≠i = [ Σi<i<...<i + Σi<i<...<i + many similar reorderings ] . (7.4.10) The total sum can be written in this manner, using the permutation sum notation, Σi≠i≠...≠i = ΣP Σi<i<...<i (7.4.11) where P are the k! permutations of the k integers [1,2,...k]. Using the form (7.4.11), the sum (7.4.9) may be rewritten as, T^ = ΣP Σi<i<...<i Tii...i (ui ^ ui ^ .... ^ ui) . (7.4.12) In (A.9.1) it is shown that, ΣP [Σi<i<...<i] fii...i = Σi<i<...<i [ΣP fi)i...i] . (A.9.1) Within the ΣP permutation sum, the permutation operators P can be moved from the summation index subscripts to the summand index subscripts. One then has from (7.4.12), T^ = Σi<i<...<i ΣP [ Tii...i (ui ^ ui ^ .... ^ui) ] . (7.4.13) But we know from (7.3.5) that (ui ^ ui ^ .... ^ui) = (-1)S(P) (ui ^ ui ^ .... ^ ui) (7.4.14) where S(P) is the number of swaps associated with permutation P. Thus, T^ = Σi<i<...<i [ΣP (-1)S(P) Tii...i] (ui ^ ui ^ .... ^ ui) (7.4.15) which we can compare with the ordered sum (7.4.7), T^ = Σi<i<....<i Aii...i (ui ^ ui ^ .... ^ ui) . (7.4.5) Thus, since the basis is complete, the relation between the A and T coefficients is given by Aii...i = ΣP (-1)S(P) Tii...i i1 < i2 < ..... < ik = [ Tii...i + all signed permutations ] // k! terms = k! [Alt(T)]ii...i . // (A.5.3) def of Alt or A = k!Alt(T) = k! T^. // (7.4.3) (7.4.16) The Aii...i appear in the expansion (7.4.7) only for index values 1 ≤ i1 < i2 < ..... < ik ≤ n, but we can interpret (7.4.16) as defining Aii...i for all index values. Since A = k! T^ , (7.4.5) shows that Fact: Aii...i and T^ii...i are both totally antisymmetric tensors. (7.4.17) Comment: Tii...i and Aii...i are both rank-k tensors, see (5.5.3). Examples: (relating the A and T coefficients) Aab = Tab - Tba // as in (4.3.10) k = 2 Aabc = Tabc - Tacb + Tcab - Tcba + Tbca - Tbac k = 3 (7.4.18) Vector Case. For k = 1, we find that T = ΣiTi ui // (5.2.1) T^ = ΣiTi ui // (7.4.4) T^ = T (7.4.19) so for a vector there is no distinction between T^ and T (and in fact V1 = L1). 7.5 Expansions for the wedge product of k vectors The symmetric expansion for the wedge product of k vectors is very straightforward. Let Tii...i = (v1)i (v2)i ... (vk)i = (v1 v2 ..... vk)ii...i or (7.5.1) T = (v1 v2 ..... vk) . Then the symmetric expansion (7.4.4) gives, T^ = Σii...i Tii...i (ui ^ ui ^ .... ^ ui) (7.4.4) = Σii...i (v1)i (v2)i ... (vk)i (ui ^ ui ^ .... ^ ui) // (7.5.1) (7.5.2) = [Σi(v1)iui] ^ [Σi(v2)i ui] ^ .... ^ [Σi (vk)i ui] = v1 ^ v2 ^ ... ^ vk . (7.5.3) This pure tensor T^ = v1 ^ v2 ^ ... ^ vk is an element of Lk . Expressing v1 ^ v2 ^ ... ^ vk in terms of the ordered expansion is more complicated. One must first compute the tensor A as in (7.4.16), (1/k!) A = Alt(T) = Alt(v1v2.....vk) = T^ = (v1 ^ v2 ^ ... ^ vk) . (7.5.4) Then the ordered expansion (7.4.7) can be written in a battery of ways, v1 ^ v2 ^ ... ^ vk = (a) = Σi<i<....<i Aii...i (ui ^ ui ^ .... ^ ui) // (7.4.7) (b) = Σi<i<....<i k! [Alt(v1v2...vk)]ii...i (ui ^ ui ^ .... ^ ui) . // (7.5.4) (c) = Σi<i<....<i k! [v1 ^ v2 ^ ... ^ vk]ii...i (ui ^ ui ^ .... ^ ui) . // (7.1.3) (d) = Σi<i<....<i det[ (v*)i] (ui ^ ui ^ .... ^ ui) . // (7.2.8) with jr → r (e) = Σi<i<....<i det (ui ^ ui ^ .... ^ ui) . (f) = Σi<i<....<i det (ui ^ ui ^ .... ^ ui) . (g) = Σii...i (v1)i (v2)i ... (vk)i (ui ^ ui ^ .... ^ ui) // (7.5.2) (7.5.5) where we throw in the symmetric sum at the end. Remember that, since generally dim(V) = n > k, the determinant in (f) is a full-width minor of matrix M = [v1, v2.....vk]. If k = n, the minor is the full matrix. Example : Suppose k = n = 3. Then the following sum (form (f)) has only one term in which i1= 1, i2= 2 and i3= 3, v1 ^ v2 ^ v3 = Σ1≤i<i<i<3 det (ui ^ ui ^ ui) = det[v1, v2, v3] (u1 ^ u2 ^ u3) (7.5.6) as quoted in (4.3.15). Example : Here are the above expressions for k = 2 and general n ≥ k : (a) v1 ^ v2 = Σi<i Aii (ui ^ ui) (b) = Σi<i 2! [Alt(v1v2)]ii (ui ^ ui) (c) = Σi<i 2! (v1 ^ v2)ii (ui ^ ui) (d) = Σi<i det[ (v*)i] (ui ^ ui) (e) = Σi<i det (ui ^ ui) (f) = Σi<i det (ui ^ ui) = Σi<i[(v1)i(v2)i - (v2)i(v1)i](ui ^ ui) (g) = Σii (v1)i (v2)i (ui ^ ui) = [Σi(v1)iui] ^ [Σi(v2)i ui] = v1 ^ v2 (7.5.7) Result (f) matches that shown in (4.3.12), a ^ b = Σij aibj (ui ^ uj) = Σi<j (aibj- ajbi) (ui ^ uj) = Σi<j Aij (ui ^ uj) = Σi<j det (ui^uj) Aij = (aibj- ajbi) = det . (4.3.12) 7.6 Number of elements in Lk compared with Vk. We know from (5.1.5) and (7.3.6) that, dim(Vk) = nk // number of basis elements of Vk (5.1.5) dim(Lk) = // number of basis elements of Lk (7.3.6) If the number of elements of field K is N ( N → ∞ for K= reals), then ratio = = = = / nk . (7.6.1) For a given n, this is a strongly decreasing function of k. For example, for n = 10 we can plot the log of the ratio for k = 0 to 10, (7.6.2) 7.7 Multiindex notation In this section, multiindex versions of equations are shown in red. Multiindexing is done in two different ways. First, for the symmetric expansion (7.4.4) : T^ = Σii...i Tii...i (ui ^ ui ^ .... ^ ui) (7.4.4) T^ = ΣI TI u^I where u^I ≡ ui ^ ui ^ .... ^ ui TI ≡ Tii...i and I ≡ {i1, i2,.... ik} with 1 ≤ ir ≤ n = ordinary multiindex, n = dim(V) . (7.7.1) The more significant notation involves the ordered expansion (7.4.7) which has only one term for each linearly independent basis element. Note our use of Σ'I (prime) to indicate an ordered multiindex summation : T^ = Σi<i<....<i Aii...i (ui ^ ui ^ .... ^ ui) (7.4.7) T^ = Σ'I AI u^I where u^I ≡ ui ^ ui ^ .... ^ ui AI ≡ Aii...i and I ≡ {i1, i2,.... ik} with 1 ≤ i1< i2<....< ik ≤ n = ordered multiindex, n = dim(V) . (7.7.2) Here are some unofficial multiindex notations for other equations developed above: T^ = v1 ^ v2 ^ ... ^ vk T^ = (^vZ) (7.5.3) Tii...i = (v1)i (v2)i ... (vk)i ≡ (vZ)I TI = (vZ)I (7.5.1) with the idea that Z = 1,2...k . Continuing on, T^ = Σii...i (v1)i (v2)i ... (vk)i (ui ^ ui ^ .... ^ ui) T^ = ΣI (vZ)I u^I (7.5.2) A = k! Alt(v1v2...vk) A = k! Alt(vZ) (7.5.4) v1 ^ v2 ^ ... ^ vk = Σi<i<....<i k![Alt(v1v2...vk)]ii...i (ui ^ ui ^ .... ^ ui) . (7.5.5b) (^vZ) = Σ'I k! Alt(vZ)I u^I Aii...i = det AI = det(vZI) (7.5.5a+f) v1^ v2^ .....^ vk = Σi<i<....<i det (ui ^ ui ^ .... ^ ui). (7.5.5f) (^vZ) = Σ'I det(vZI) u^I 7.8 The Exterior Algebra L(V) We now construct the graded algebra L(V) in analogy with that of T(V) in (5.4.1). Define a large vector space of the form ( this is "the exterior algebra on V" ) L(V) ≡ L0 L1 L2 L3 + .... // L(V) = Σk=0∞ Lk (7.8.1) Here L0 = the space of scalars, L1 = V the space of vectors, L2 = V ^ V V2 the space of antisymmetric rank-2 tensors (7.4.6), and so on. The most general element of the space L(V) would have the form X = s ΣiTi ui Σij Tij ui^uj Σijk Tijk ui^uj^uk + ..... // symmetric or X = s ΣiTi ui Σi<j Aij ui^uj Σi<j<k Aijk ui^uj^uk + ..... // ordered (7.8.2) The direct sum is described in Appendix B. Associativity of the Wedge Product We have carefully managed to avoid this topic in all that has transpired above. Nothing so far has been assumed concerning associativity of the ^ operator. In (2.8.22) it was stated that the operator is associative, and this was "proved" in our outer product approach to , but for the formal approaches of Chapter 1 it is an axiom that is associative. Once we define the space L(V) above, we must face the issue of wedge products of the form (ui^uj)^uk and more generally (v1^v2)^v3. These products arise when we multiply an element of L2 by an element of L1. Notice that our grandiose expansion (7.1.3) says nothing about (v1^v2)^v3. All it says is this: v1^ v2^ v3 = (v1v2v3 - v1v3v2 + v3v1v2 - v3v2v1 + v2v3v1 - v2v1v3)/6 v1^ v2 = (v1v2 - v2v1)/2 . (7.8.3) The product (v1^v2)^v3 = (1/2)(v1v2 - v2v1) ^ v3 is the wedge product of an antisymmetric rank-2 tensor and a vector and up to this point we have no idea how to evaluate such an creature. Now is the time, then, to add a new axiom to the wedge product theory. We declare that, Fact: The wedge product of k vectors v1^ v2^ .....^ vk can be "associated" in any manner without altering the meaning of the product. By this we mean that parentheses can be added in any manner without altering the object. (7.8.4) What this in effect does is define an array of new objects to be the same as v1^ v2^ .....^ v6 . For example, (v1^ v2)^ v3^ v4^ v5^ v6 ≡ v1^ v2^ v3^ v4^ v5^ v6 v1^ (v2^ v3) ^ v4^ v5^ v6 ≡ v1^ v2^ v3^ v4^ v5^ v6 v1^ (v2^ v3 ^ v4) ^ v5^ v6 ≡ v1^ v2^ v3^ v4^ v5^ v6 v1^ (v2^ v3 ^ v4) ^ (v5^ v6) ≡ v1^ v2^ v3^ v4^ v5^ v6 // multiple (v1^ v2^ v3) ^ (v4^ v5^ v6) ≡ v1^ v2^ v3^ v4^ v5^ v6 (v1^ v2) ^ (v3^ v4) ^ (v5^ v6) ≡ v1^ v2^ v3^ v4^ v5^ v6 . (7.8.5) Given these definitions, it follows that nested parenthesis are also allowed. For example, v1^ (v2^ (v3^ v4)^ v5)^ v6 = v1^ (v2^ v3^ v4^ v5)^ v6 = v1^ v2^ v3^ v4^ v5^ v6 . (7.8.6) Since tensors like T^ can be expanded on (ej ^ ej ^ .... ^ ej), and since one may associate this wedge product arbitrarily as claimed in (7.8.4), one easily shows that : Fact: The wedge product of N general tensors A^^B^^C^.... can be "associated" in any manner without altering the meaning of the product. By this we mean that parentheses can be added in any manner without altering the object. (7.8.7) This fact then extends the claim (7.8.4) made for N vectors, and is exactly analogous to the similar axiomatic statement for associativity made in (2.8.22). Example: In (7.7.1) multiindex notation, consider three L(V) tensors T^, S^, R^ of rank k,k',k" : (T^ ^ S^) ^ R^ = ( (ΣITIu^I) ^ (ΣJSJu^J) ) ^ (ΣKRKu^K) = ΣITIΣJSJ { ( u^I ^ u^J) ^ (ΣKRKu^K) } // rules (7.2.3) = ΣITIΣJSJΣKRK (u^I ^ u^J) ^ (u^K) // rules (7.2.3) again = ΣIJKTISJRK (u^I ^ u^J ^ u^K) // detail shown below = (ΣITIu^I) ^ (ΣJSJu^J) ^ (ΣKRKu^K) // rules (7.2.3) again = T^ ^ S^ ^ R^ . Our example shows that for arbitrary L(V) tensors, (T^ ^ S^) ^ R^ = T^ ^ S^ ^ R^. Here we illuminate the key detail above: (uI ^ uJ) ^ uK = ( (ui^ ui^...^ ui) ^ (uj^ uj^...^ uj' ) ) ^ (uk^ uk^...^ uk" ) = ( ui^ ui^...^ ui ^ uj^ uj^...^ uj' ) ^ (uk^ uk^...^ uk" ) = ui^ ui^...^ ui ^ uj^ uj^...^ uj' ^ uk^ uk^...^ uk" = (ui^ ui^...^ ui) ^ (uj^ uj^...^ uj') ^ (uk^ uk^...^ uk") = (uI ^ uJ ^ uK) . In each step above the rule (7.8.4) for vectors (applied to basis vectors) is used. Having faced up to the issue of associativity, we now resume the discussion of L(V). Fact: This large space L(V) is in fact itself a vector space. (7.8.8) We know this is true since L(V) = Σk=0∞ Lk and we showed in (7.3.2) that each Lk is a vector space. For example, the "0" element in L(V) is the direct sum of the "0" elements of all the Lk. See Appendix B for more detail. To show that L(V) is an algebra, we must show that it is closed under both addition and multiplication. It should be clear to the reader that L(V) is closed under addition and has the right scalar rule. For example, if k1 and s are scalars, k1 a b^c f^g^h = sum of 4 elements of L(V) = an element of L(V) s(k1 a b^c f^g^h) = (sk1) (sb) ^c f^(sg)^h = element of L(V) (7.8.9) This additive closure is of course necessary for L(V) be a vector space. The space is also closed under the multiplication operation ^. For example (b^c)^(f^g^h) = b^c^f^g^h = ϵ L5 = ϵ L(V) . // (b^c) ϵ L2 (f^g^h) ϵ L3 (7.8.10) Here we have used the associative property (7.8.4). This closure claim is stated more generally below (7.9.a.6). One then makes the following definitions with regard to the space L(V), where n = dim(V) : Object Name any blade lincomb: Grade(rank): Space s 0-blade scalar ϵ K 0 L0 a 1-blade vector 1 L1 a^b 2-blade bivector 2 L2 a^b^c 3-blade trivector 3 L3 a^b^c^d 4-blade quadvector 4 L4 ..... a^b^c^d^.... k-blade k-vector k Lk .... a^b^c^d^.... n-blade n-vector n Ln arbitrary element of L(V) multivector mixed L(V) (7.8.11) Since L(V) is closed under the operations and ^, it is "an algebra" (the space Lk alone is not an algebra because it is not closed under ^). The L(V) algebra is different from that of the reals due to its definition as a sum of vector spaces. The elements of L(V) have different "grades" as shown above, so L(V) is a "graded algebra". Sometimes L(V) is called "the exterior tensor algebra" over V. A k-blade is a pure wedge product of k vectors, whereas a k-vector is any linear combination of k-blades. A multivector is any linear combination of k-vectors for any mixed values of k. Note that s1(a^b) s2(c^d) = (s1a)^b (s2c)^d = (a'^b) (c'^d) // 2-blades s1(a^b) s2(c^d^e) = (s1a)^b (s2c)^d^e = (a'^b) (c'^d^e) // multivector so it is also correct to say that a k-vector is any sum of k-blades, and a multivector is any sum of k-vectors. That is, any linear combination can be written as a sum as shown in the above examples. Unlike in Tensor World, in Wedge World the above list (7.8.11) is finite for a given n = dim(V). For k = n there is exactly one linearly independent basis vector which is the ordered wedge product of all the basis vectors of V. For k > n, all wedge products vanish since the vectors in the wedge product are linearly dependent, see (7.2.6). The dimensionality of the space L(V) is as follows, based on (7.8.1) and (B.10)', dim[L(V)] = dim[L0 L1 L2 L3 + ....] = dim(L0) + dim(L1) + dim(L2) + dim(L3) + ... but for dim(V) = n this series truncates with Ln and we find from (7.3.6), dim[L(V)] = 1 + n + + + ... + = Σk=0n = 2n = a finite number (7.8.12) 7.9 The Wedge Product of two or more tensors in L(V) (a) Wedge Product of two tensors T^ and S^ Here we shall mimic the developmental approach used in Section 5.6 for the tensor product. As before, we quietly "break in" the multiindex notation. The symmetric expansions (7.4.4) of T^ and S^ are given by, T^ = Σii....i Tii....i (ui^ ui .....^ ui) rank k, T^ ϵ Lk (7.9.a.1) ΣITIu^I S^ = Σjj....j Sjj....j (uj^ uj .....^ uj) rank k', S^ ϵ Lk' . (7.9.a.2) ΣJSJu^J We form the wedge product of these two tensors in a manner similar to (5.6.3) : T^^S^= [Σii....iTii....i (ui^ ui .....^ ui)]^[ Σjj....j Sjj....j(uj^ uj .....^ uj)] [ ΣITIu^I] ^ [ΣJTJu^J] (a) = Σii....i Σjj....jTii....i Sjj....j(ui^ ui .....^ ui) ^ (uj^ uj .....^ uj) ΣI,JTISJ(u^I) ^ (u^J) (b) = Σii....ijj....jTii....i Sjj....j(ui^ ui .....^ ui^ uj^ uj .....^ uj) ΣI,JTISJ(u^I ^ u^J) (c) = Σii....iii....i[Tii....i Sii....i] (ui^ ui ......^ ui) ΣI,I'TISI'(u^I ^ u^I') (d) = Σii....iii....i[TS]ii...i ii....i (ui^ ui ......^ ui) ΣI,I'[TS]I,I'(u^I ^ u^I') (e) = Σii....i[TS]ii...i(ui^ ui ......^ ui) . (7.9.a.3) ΣI (TS)I u^I Comparing lines one sees that I ≡ i1, i2...ik I' ≡ ik+1, ik+2, ....ik+k' I ≡ I, I' = i1,i2...ik+k' u^I ≡ (ui^ ui ....^ ui) u^I' ≡ (ui^...^ui) u^I ≡ (ui^ ui ....^ ui) (7.9.a.4) Notice that the (7.8.4) vector associativity of ^ is used going from (a) to (b). The conclusion is that T^^ S^ = ΣI (TS)I u^I I ≡ I, I' = i1,i2...ik+k', u^I ≡ (ui^ ui ....^ ui) . (7.9.a.5) Since the u^I are basis vectors in Lk+k', we have shown that: T^ ϵ Lk and S^ ϵ Lk' T^^ S^ ϵ Lk+k' L(V) . (7.9.a.6) Thus we have strengthened the claim made in (7.8.10) that L(V) is closed under the operation ^. Recall now from (7.3.8) the relationship between u^I and uI, (ui ^ ui ^ .... ^ ui) = Alt(ui ui .... ui) u^I = Alt(uI) (7.3.8) and (5.6.5) for the expansion of the tensor product TS, TS = ΣI (TS)I uI I ≡ I, I' = i1,i2...ik+k', uI ≡ (ui ui .... ui) . (5.6.5) Applying Alt to this last equation (with component indices J ), [Alt(TS)]J = AltJ [(TS)J] // (A.5.3b) = AltJ [ ΣI (TS)I(uI)J ] // component J of (5.6.5) TS = ΣI (TS)I eI = ΣI (TS)I AltJ [(uI)J] // (A.5.10) that Alt is linear = ΣI (TS)I AltI [(uI)J] // (A.8.30) since (eI)J has product form = ΣI (TS)I (u^I)J // (7.3.8) = [ΣI (TS)I (u^I)] J = (T^^ S^)J // (7.9.a.5) so we end up with the following elegant and compact way to write the wedge product of two tensors, T^^ S^ = Alt(TS) . // see Sec (g) below for this result in Spivak normalization (7.9.a.7) Concealing the AltI and AltJ details one can get the correct result with this sequence, Alt(TS) = Alt( ΣI (TS)I uI ) = ΣI (TS)IAlt(uI) = ΣI (TS)I(u^I) = T^^ S^ . The components of (7.9.a.7) are, [T^^ S^]J = [Alt(TS)]J = ΣP(-1)S(P) (TS)P(J) // (A.5.3) = ΣP(-1)S(P) TP(J)SP(J') . // see e.g. (5.6.15) (7.9.a.8) This last line is an explicit instruction for computing the components of the tensor T^^ S^ . We have added this new notation, TP(I) ≡ Tii...i for I = i1, i2...ik (7.9.a.9) Example: Let S and T both be rank-2 tensors so k = k' = 2 . Then [T^^ S^]I = [T^^ S^]iiii = (1/4!) ΣP(-1)S(P)TiiSii = (1/24) [ TiiSii - TiiSii + TiiSii - TiiSii + 20 more terms ] (7.9.a.10) Here as elsewhere we show in red the indices to be swapped to make the next term. From (7.9.c.6) below, T^^ S^ = (-1)2*2 S^^ T^ = S^^ T^. (7.9.a.11) (b) Special cases of the wedge product T^^ S^ Assume T^ and S^ have rank k and k'. If S = κ' ϵ K = a scalar, then rank(S) = k' = 0 and (7.9.a.3) (b) reads, T^^ S^ = Σii....ijj....jTii....i Sjj....j (ui^ ui .....^ ui^ uj^ uj .....^ uj) = Σii....iTii....i (κ') (ui^ ui .....^ ui) = κ'T (7.9.b.1) and S^^ T^ = Σjj....jii....i Sjj....j Tii....i( uj^ uj .....^ uj^ ui^ ui .....^ ui) = Σii....i (κ') Tii....i ( ui^ ui .....^ ui) = κ'T (7.9.b.2) so we find that T^S = S^T = κ'T . If T = κ and S = κ', the result above would be T^^S^ = κκ' and S^^T^ = κ'κ and so T^^S^ = S^^T^ = κκ'. Thus, T^^S^ = κ^S^ = S^^T^ = S^^κ = κS^ if T^ = κ ϵ V0 T^^S^ = T^^κ' = S^^T^ = κ'^T^ = κ'T^ if S^ = κ' ϵ V0 T^^S^ = κ^κ' = S^^T^ = κ'^κ = κκ' if T^,S^ = κ,κ' ϵ V0 (7.9.b.3) These special case results are seen to be the same as those for TS shown in (5.6.15). When T is rank-0 or rank-1 we can write T^ = T according to (7.4.17), but we continue to use T^ (c) Commutivity Rule for the Wedge Product of two tensors T^ and S^ Recall the expansion of T^^ S^ from (7.9.a.3) item (b), T^^ S^ = Σii....ijj....jTii....i Sjj....j(ui^ ui .....^ ui^uj^ uj .....^ uj) ΣI,J TI SJ ( u^I ^ u^J) (7.9.c.1) Swapping T↔S, k↔k' and i ↔ j gives the following form for the wedge product S^^T^ , S^^T^ = Σjj....jii....iSjj....j Tii....i (uj^ uj .....^ uj^ui^ ui .....^ ui) ΣJ,I SJTI (u^J ^ u^I) = Σii....ijj....j Tii....i Sjj....j(uj^ uj .....^ uj^ui^ ui .....^ ui) ΣI,J TISJ (u^J ^ u^I) . (7.9.c.2) Equations (7.9.c.1) and (7.9.c.2) are identical except for the last factor involving the basis vectors. Consider the basis vector factor appearing in (7.9.c.2), (u^J ^ u^I) = (uj^ uj .....^ uj^ ui^ ui .....^ ui) . (7.9.c.3) To make this match the basis factor in (7.9.c.1), we have to slide all the red basis vectors to the left through all the black basis vectors. Each time a red passes through a black, we pick up a minus sign due to the rule (7.2.4). Thus, (uj^ uj .....^ uj^ ui^ ui .....^ ui) = (-1)k' ui ^ (uj^ uj .....^ uj^ ui .....^ ui) = (-1)k' (-1)k' ui ^ ui ^ (uj^ uj .....^ uj .....^ ui) = etc. = = [(-1)k']k ( ui^ ui .....^ ui ^ uj^ uj .....^ uj) . (7.9.c.4) Therefore, (u^J ^ u^I) = (-1)kk' (u^I ^ u^J) . (7.9.c.5) Inserting this result into (7.9.c.2) gives S^^ T^ = (-1)kk'T^^ S^ ranks of the two tensors are k and k' . (7.9.c.6) Since the commutivity sign is a function of the ranks (grades) of the tensors, this statement is sometimes referred to as "graded commutivity". The wedge product of two tensors commutes if kk' is even, and anticommutes if kk' is odd. Using (7.9.a.7) the above becomes. Alt(ST) = (-1)kk'Alt(TS) . (7.9.c.7) Example: If k = k' = 1, (-1)kk' = -1 and we recover the simple rule for vectors, S^^T^ = - T^^S^ // S and T are rank-1 tensors (vectors) (7.9.c.8) as first stated in (4.3.2). One must keep in mind that the result S^^T^ = - T^^S^ is not valid for arbitrary tensors S^ and T^. Examples: If k = 0 so T = κ, rule (7.9.c.6) says S^^T^ = T^^S^, consistent with (7.9.b.3) line 1. If k=k'=0 so T = κ and S = κ', rule (7.9.c.6) again says S^^T^ = T^^S^, consistent with (7.9.b.3) line 3. (7.9.c.9) (d) Wedge Product of three or more tensors Mimicking (5.6.7) we write T^^S^^R^ = [ΣITIu^I]^[ΣJ SJu^J]^[ΣK RKu^K] (a) = ΣI,J,K TISJRK (u^I) ^ (u^J) ^ (u^K) (b) = ΣI,J,K TISJRK (u^I ^ u^J ^ u^K ) // associative of ^ used here (d) = ΣI,I',I" TISI'RI" (u^I ^ u^I' ^ u^I") // rename multiindices J→I',K→I" I ≡ i1, i2...ik I' ≡ ik+1, ik+2, ....ik+k' I" ≡ ik+k'+1, ik+k'+2, ....ik+k'+k" u^I ≡ (ui^....^ ui) u^I' ≡ (ui^...^ui) u^I" ≡ (ui^...^ui) (e) = ΣI (TSR)I u^I u^I ≡ (ui^....^ ui) I ≡ I, I',I" = i1,i2...ik+k'+k" (7.9.d.1) The outer product form is TISI'RI" = (TSR)I,I',I" = (TSR)I . The conclusion is then, T^^S^^R^ = ΣI (TSR)Ie^I I ≡ I, I',I" = i1,i2...ik+k'+k" , u^I ≡ (ui^....^ ui) (7.9.d.2) Since the u^I are basis vectors in Lk+k'+k", we have shown that: T^ ϵ Lk and S^ ϵ Lk' and R^ ϵ Lk" T^^S^^R^ ϵ Lk+k'+k" L(V) . (7.9.d.3) We now mimick the sequence of steps above (7.9.a.7) : [Alt(TSR)]J = AltJ [(TSR)J] // (A.5.3b) = AltJ [ ΣI (TSR)I(uI)J ] // component J of (5.6.8) TSR = ΣI (TSR)I eI = ΣI (TSR)I AltJ [(uI)J] // (A.5.10) that Alt is linear = ΣI (TSR)I AltI [(uI)J] // (A.8.30) since (eI)J has product form = ΣI (TSR)I (u^I)J // (7.3.8) = [ΣI (TSR)I (u^I)] J = (T^^ S^^ R^)J // (7.9.d.2) so T^^ S^^ R^ = Alt(TSR) (7.9.d.4) and then [T^^ S^^ R^]I = [Alt(TSR)]I = ΣP (-1)S(P) (TSR)P(I) // (A.5.3) = ΣP (-1)S(P) TP(I)SP(I') RP(I") (7.9.d.5) which gives instructions for how to compute the components of T^^S^^R^ . Using the systematic notation outlined in (5.6.11) and (5.6.12), and generalizing the above development for the wedge product of three tensors, we find the following expansion for the wedge product of N tensors of L(V), (T1)^^(T2)^^...^(TN)^ = ΣI (T1IT2I .... TNI) e^I = ΣI (T1T2....TN)I e^I (7.9.d.6) where u^I = ui^ ui .....^ ui = ui^ ui .....^ ui and (T1T2....TN)I = T1IT2I .... TNI . The rank of this product tensor is then κ = Σi=1N ki and the tensor is an element of Lκ L(V). Notice that if κ > n, the tensor product (7.9.d.6) vanishes since there are then > n factors in e^I so one or more are then duplicated, (T1)^^(T2)^^...^(TN)^ = 0 if κ = Σi=1N ki ≥ n+1 (7.9.d.7) For example, if all the tensors are the same tensor T^ of rank k, then T^N ≡ T^^T^^...^T^ = 0 if Nk ≥ n+1 or N ≥ (n+1)/k (7.9.d.8) If N ≥ (n+1), then N ≥ (n+1)/k for any k ≥ 1. Thus T^N = 0 for any N ≥ n+1 assuming k ≠ 0. (7.9.d.9) Recall (5.6.13), T1T2...TN = ΣI (T1IT2I .... TNI) uI = ΣI (T1T2....TN)I uI . (5.6.13) Repeating the sequence above (7.9.d.4) for a longer product, we find that (T1)^^(T2)^^...^(TN)^ = Alt(T1T2...TN) . (7.9.d.10) Components of this tensor are computed as follows: [(T1)^^(T2)^^...^(TN)^]I = [Alt(T1T2...TN)]I = ΣP(-1)S(P) (T1T2...TN)P(I) // (A.5.3) = ΣP(-1)S(P) T1P(I)T2P(I) ....TNP(I) (7.9.d.11) where T1P(I) ≡ T1ii...i for I1 = i1, i2...iκ T2P(I) ≡ T2ii...i for I2 ={iκ+1, iκ+2.....iκ} etc. // see (5.6.10 thru 12) for details In the Dirac notation of Section 2.11 one can write (7.9.d.10) as | (T1)^> ^ | (T2)^> ^ ... ^ | (TN)^> = Alt ( | T1> | T2> ... | TN> ) (7.9.d.12) It is shown in (C.4.17) that "pre-antisymmetrization makes no difference", so the above may also be written | (T1)^> ^ | (T2)^> ^ ... ^ | (TN)^> = Alt ( | (T1)^> | (T2)^> ... | (TN)^> ) (7.9.d.13) Both sides of this equation are elements of the wedge product space Lk+k+..+k , but they are also both elements of the larger tensor product space Vk Vk ... Vk . The action of linear operator P on a tensor product space vector is defined in the obvious manner, as in (5.6.17), P [ | (T1)^> | (T2)^> ... | (T2)^> ] = P | (T1)^> P | (T2)^> ... P | (T2)^> . (7.9.d.14) In other words, the action of P on the larger space is defined in terms of its action on the spaces which make up the tensor product. This result holds as well for the wedge product of N tensors, P [ | (T1)^> ^ | (T2)^> ^ ... ^ | (T2)^> ] = P | (T1)^> ^ P | (T2)^> ^ ... ^ P | (T2)^> (7.9.d.15) Proof: P [ | (T1)^> ^ | (T2)^> ^ ... ^ | (T2)^> ] = P [Alt ( | (T1)^> | (T2)^> ... | (T2)^> ) ] = Alt [P ( | (T1)^> | (T2)^> ... | (T2)^> ) ] = Alt [ ( P | (T1)^> P | (T2)^> ... P | (T2)^> ) ] = P | (T1)^> ^ P | (T2)^> ^ ... ^ P | (T2)^> . (e) Commutativity Rule for product of N tensors To reduce clutter, in this section we abbreviate e^I by eI and (Ti)^ by Ti. Consider an example where we have a wedge product of 9 tensors. The eI basis function groups are uI ^ uI ^ uI ^ uI ^ uI ^ uI ^ uI ^ uI^ uI (7.9.e.1) which goes with T1 ^ T2 ^ T3 ^ T4 ^ T5 ^ T6 ^ T7 ^ T8 ^ T9 . (7.9.e.2) The sign caused by swapping T3 ↔ T7 will be the same as the sign swapping eI ↔eI in the basis function. We do it one step at a time, first sliding the group eI to the left using (7.9.c.5), uI ^ uI ^ uI ^ uI ^ uI ^ uI ^ uI ^ uI^ uI = uI ^ uI ^ uI ^ uI ^ uI ^ uI ^ uI ^ uI^ uI (-1)kk = uI ^ uI ^ uI ^ uI ^ uI ^ uI ^ uI ^ uI^ uI (-1)(k+k)k = uI ^ uI ^ uI ^ uI ^ uI ^ uI ^ uI ^ uI^ uI (-1)(k+k+k)k = uI ^ uI ^ uI ^ uI ^ uI ^ uI ^ uI ^ uI^ uI (-1)(k+k+k+k)k . (7.9.e.3) Now with this as a starting point, we slide eI to the right, one group at a time, uI ^ uI ^ uI ^ uI ^ uI ^ uI ^ uI ^ uI^ uI = uI ^ uI ^ uI ^ uI ^ uI ^ uI ^ uI ^ uI^ uI (-1)kk = uI ^ uI ^ uI ^ uI ^ uI ^ uI ^ uI ^ uI^ uI (-1)k(k+k) = uI ^ uI ^ uI ^ uI ^ uI ^ uI ^ uI ^ uI^ uI (-1)k(k+k+k) (7.9.e.4) and now we have successfully swapped uI ↔ uI so alsoT3 ↔ T7. The total sign is sign = (-1)m where m = (k6+ k5+ k4+ k3)k7 + (k4+k5+k6)k3 = (k4+k5+k6)(k3+k7)+ k3k7 . (7.9.e.5) Based on this result, we claim that : Fact: In a product of tensors T1^T2^T3.... of rank k1, k2, k3 ... , if two tensors are swapped Tr ↔ Ts (with r < s), the resulting tensor incurs the following sign relative to the starting tensor, sign = (-1)m where m = (kr+1+kr+2 ...+ks-1)(kr+ks) + krks . (7.9.e.6) Corollary: If the sum of the ranks of the two swapped tensor is even, in effect m = krks . (7.9.e.7) Example 1: T1 ^ T2 ^ T3 = (-1)m T2 ^ T1 ^ T3 r = 1 s = 2 m = (0)(k1+k2) + k1k2 = k1k2 (-1)m = (-1)kk (7.9.e.8) which is consistent with (7.9.c.6) saying T1 ^ T2 = (-1)kk T2 ^ T1 . Example 2: T1 ^ T2 ^ T3 = (-1)m T3 ^ T2 ^ T1 r = 1 s = 3 m = (k2)(k1+k3) + k1k3 = k1k2 + k1k3 + k2k3 (-1)m = (-1)kk+kk+kk (7.9.e.9) This result can be obtained as well by direct pairwise swapping using (7.9.c.6) and the associativity of ^, T1 ^ T2 ^ T3 = (-1)kk T2 ^ T1 ^ T3 = (-1)kk (-1)kk3 T2 ^ T3 ^ T1 = (-1)kk (-1)kk (-1)kk T3 ^ T2 ^ T1 (7.9.e.10) Example 3: Suppose all the tensors are vectors with rank = 1. Then the sum of the ranks of any two tensors is 2, which is even, so the Corollary above says m = krks = 1*1 = 1, so swapping any two of these tensors produces a minus sign, phase = (-1)m = - 1 where m ≈ krks = 1*1 = 1 in agreement with the basic vector swap rule (7.2.4). (7.9.e.11) (f) Theorems from Appendix C : pre-antisymmetrization makes no difference We showed above that one can form wedge products of elements of L(V) in this manner (in Spivak normalization, the right sides of these equations incur factorials as shown in Section (g) below), T^^ S^ = Alt(TS) . (7.9.a.7) T^^S^^R^ = Alt(TSR) (7.9.d.4) (T1)^^(T2)^^...^(TN)^ = Alt(T1T2...TN) (7.9.d.7) where the operator Alt acts on the tensor indices which are not displayed in the above compact notation. For example T^^ S^ = Alt(TS) means, in multiindex notation, (T^^ S^)I = AltI [(TS)I] = AltI [ TISI'] = ΣP(-1)P TP(I)SP(I') . A very simple case is the following (recall for vectors that a = a^ ) (a ^ b)ii = Alt [(ab)ii] = Alt [aibi] = ΣP(-1)P ai bi = (1/2) [ aibi - aibi] = (1/2) [(ab)ii - (ba)ii ] = { (1/2) [(ab) - (ba) ]}ii which replicates our Chapter 4 statement that a ^ b = [ab- ba]/2 . (4.3.1) Appendix C uses the rearrangement theorem in three separate Theorems to show that T^^ S^ = Alt(TS) = Alt(T^S) = Alt(TS^) = Alt(T^S^) . (7.9.f.1) Theorem One Theorem Two Theorem Three Recall that T^ ≡ Alt(T) (7.4.3) so that T^ is a totally antisymmetric tensor. What (7.9.f.1) says is that Alt(TS) provides total antisymmetrization on all tensor indices, so pre-antisymmetrizing either or both tensors makes no difference. A similar statement applies to working with totally symmetric tensors. So we have, Alt[TS] = Alt[T^S] = Alt[TS^] = Alt[T^S^] where T^ = Alt(T) S^ = Alt(S) (C.4.1) Sym[TS] = Sym[TsS] = Sym[TSs] = Sym[TsSs] where Ts = Sym(T) Ss = Sym(S) . (C.4.2) These can of course be rewritten as Alt[TS] = Alt[Alt(T)S] = Alt[TAlt(S)] = Alt[Alt(T)Alt(S)] (C.4.3) Sym[TS] = Sym[Sym(T)S] = Sym[TSym(S)] = Sym[Sym(T)Sym(S)] . (C.4.4) Similarly Appendix C shows that T^^S^^R^ = Alt(TSR) = Alt(T^SR) = Alt(TS^R) = Alt(TSR^) = Alt(T^S^R)= Alt(T^SR^)= Alt(TS^R^) = Alt(T^S^R^) . (7.9.f.2) Adding ^ subscripts inside an Alt expression changes nothing. Here is another example: T^^S^^R^ = Alt(TSR) = Alt((TS)R) = Alt((TS)^R) = Alt(Alt(TS)R) . (7.9.f.3) (g) Spivak Normalization Spivak's definition of the Alt operator is the same as ours and the same as Benn & Tucker's, but the latter authors write the Alt operator in an elaborate script font as . Our wedge product normalization is the same as Benn & Tucker's but differs from that of Spivak, a difference which we now explore. Suppose we were to omit the (1/k!) normalization factor in the definition of the wedge product of k vectors, so that (7.1.2) would become vj^ vj^ .....^ vj = 1 ΣP (-1)S(P) (vj vj ..... vj) = 1 [ (vj vj ..... vj) + all signed permutations ] = k! Alt(vj vj ..... vj) . (7.1.2)S In particular, a ^ b = 1[ ab - ba] . // no factor of 1/2 (4.3.1)S We show all factors that are different from our normalization in red. Earlier equations converted to Spivak normalization are shown below with a subscript S added to the earlier equation number. For example, equation (7.2.8) becomes (vj^ vj^ .....^ vj)ii...i = 1 det[ (vj)i] (7.2.8)S and correspondingly (uj^ uj^ .....^ uj)ii...i = 1 det[ δji] = 1 det(δJI) . (7.3.9)S Our Lk basis vectors of (7.3.8) become (ui^ ui^ .....^ ui) = k! Alt(ui ui ..... ui) or u^I = k!Alt(uI) . (7.3.8)S where recall that the Alt operator (A.5.3) always contains an internal factor (1/k!) which is required so AltT = T if the tensor T is already totally antisymmetric. The tensor expansion for T^ ϵ Lk is still given by (7.4.4), T^ = Σii....i Tii....i (ui^ ui^ .....^ ui) . (7.4.4)S and of course the corresponding tensor expansion of T ϵ Vk is also unaltered, T = Σii....i Tii....i (ui ui ..... ui) . (7.4.1) The reader is thus reminded of the difference between tensors T^ and T in our notation. Then the new (7.4.2) is, Σii....i Tii....i (ui^ ui^ .....^ ui)jj...j = Σii....i Tii....i k! AltI[(ui ui ..... ui)jj...j] // (7.3.8)S = Σii....i Tii....i k! AltJ[(ui ui ..... ui)jj...j] // (A.8.26) = k! AltJ[ Σii....i Tii....i (ui ui ..... ui)jj...j ] // (A.5.10), Alt is linear = k! AltJ(Tjj...j) // (7.4.1) = k! Alt(Tjj...j) // no ambiguity = 1 ΣP (-1)S(P) Tjj...j // def of Alt (A.5.3) = k! [Alt(T)]jj...j // (A.5.3) ≡ [T^]jj...j (7.4.2)S with the result (rank T = k, rank S = k') T^ ≡ k!Alt(T) and S^ ≡ k'!Alt(S) . (7.4.3)S The wedge product development of Section 7.9 (a) goes through with no change to give the result T^^ S^ = ΣI (TS)I u^I I ≡ I, I' = i1,i2...ik+k', u^I ≡ (ui^ ui ....^ ui) . (7.9.a.5)S But then we find Alt(TS)J = AltJ(TS)J = ΣI (TS)IAltJ(uI)J // (5.6.5) and (A.5.10) that Alt is linear = ΣI (TS)IAltI(uI)J // use (A.8.26) since (uI)J is totally antisymmetric in I and J = ΣI (TS)I (u^I)J // (7.3.8)S above with k → k+k' = (T^^ S^)J // (7.9.a.5)S above so T^^ S^ = (k+k')! Alt(TS) . (7.9.a.7)S The fact that "pre-antisymmetrizing makes no difference" is unaltered, so we still have Alt(TS) = Alt(Alt(T)Alt(S)) . (C.4.3) Then using T^ ≡ k!Alt(T) and S^ ≡ k'!Alt(T) we end up with T^^ S^ = (k+k')! Alt(TS) = (k+k')! Alt(Alt(T)Alt(S)) = Alt(T^S^) T^ ϵ Lk and S^ ϵ Lk' . (7.9.g.1) By the same analysis, we would find for a triple product in the Spivak normalization, T^^ S^^ R^ = (k+k'+k")! Alt(TSR) = Alt(T^S^R^) T^ ϵ Lk , S^ ϵ Lk', R^ ϵ Lk" . (7.9.g.2) Spivak uses lower-case Greek letters for elements of Lk, so the above two equations appear as Spivak page 79 Spivak page 80 Actually, Spivak never talks about rank-k tensors and Lk, only rank-k tensor functions which he calls "k-tensors" and which we will associate with the dual space Λk in Chapter 8, and that is what the Greek objects are in the above. But if he did talk about rank-k tensors and Lk, the above in red would be his normalization. One advantage of the Spivak normalization is that vector wedge products don't have the annoying 1/k! so that, for example, there is no overall 1/3! in the following, v1 ^ v2 ^ v3 = v1v2v3 - v1v3v2 + v3v1v2 - v3v2v1 + v2v3v1 - v2v1v3 (7.1.5)S The disadvantage, which seems a large one to us, is all the extra factorials in the wedge products of multiple tensors, and the fact that T^ = k!Alt(T) instead of the simpler T^ = Alt(T). 8. The Wedge Product of k dual vectors : the vector spaces Λk and Λ(V) Comment: This Chapter 8 is a partial copy, paste and edit version of Chapter 7 -- a translation from non-dual to dual. See our similar comment at the start of Chapter 6. Since Chapter 7 is so long, here in Chapter 8 we shall delete all material that is basically unchanged from the non-dual Chapter 7. We also delete most "comments" and examples. Since the dual world involves functions as well as functionals, a new Section 8.12 is tacked onto the end, similar to Section 6.6 for the dual tensor product. The equation numbers for Chapter 8 match those of Chapter 7, and deletions thus cause "holes" in the sequence for Chapter 8. [ rewrite this paragraph! ] 8.1 Definition of the wedge product of k dual vectors We wish to define the wedge product of k dual vectors αi ϵ V*, α1^ α2^ .....^ αk . // <α1| ^ <α2| ^ .... ^ <αk| Wedge products of this form (and their linear combinations) inhabit a vector space we call Λk(V) or Λk. We now impose the requirement that this wedge product must change sign when any two vectors are swapped. This property is injected into the wedge product theory, it does not fall out from it. This sign-change requirement leads to the following candidate definition for the wedge product of k vectors in V (the jr are vector labels), αj^ αj^ .....^ αj = (1/k!) ΣP (-1)S(P) ( αP(j) αP(j) ..... αP(j)) = (1/k!) [ (αj αj ..... αj) + all signed permutations ] = Alt(αj αj ..... αj) . (8.1.2) For the purposes of this section, we simplify things by taking jr → r so (7.1.2) becomes, α1^ α2^ .....^ αk = (1/k!) ΣP (-1)S(P) ( αP(1) αP(2) ..... αP(k)) = (1/k!) [ (α1 α2 ..... αk) + all signed permutations ] . = Alt(α1 α2 ..... αk) = (1/k!) Σii...i εii...i (αi αi ..... αi) ir = 1 to k (8.1.3) where (1/k!) is a normalization factor. In Spivak normalization, the (1/k!) factors above are all replaced by 1, see discussion in Section 8.9 (g) below. 8.2 Properties of the wedge product of k dual vectors Where there is no comment on an item, see the the parallel item in Chapter 7. 1. The sums in (8.1.2) and (8.1.3) have k! terms. (8.2.1) 2. The wedge product is k-multilinear. (8.2.2) It is by-fiat axiom that the wedge product of k vectors is k-multilinear and therefore satisfies these rules, α1^(α2 + α'2)^α3^.....^αk = α1^α2^α3^ .....^αk + α1^α'2^α3^ .....^αk α1^(sα2)^α3^ .....^ αk = s(α1^α2^α3^ .....^αk) s,r= scalar ϵ K or α1^(rα2 + sα'2)^α3^.....^αk = r(α1^α2^α3^ .....^αk) + s(α1^α'2^α3^ .....^αk) . (8.2.3) Here we show the rules just for the 2 position, but k-multilinear means these rules must apply to all the vector positions. These rules cannot be derived from the similar tensor product rules (6.3.1). 3. The wedge product changes sign if any vector pair is swapped. (8.2.4) 4. Wedge product of vectors vanishes if any two vectors are the same. Given a sign change (7.2.4) for any pair swap of vectors in the wedge product, we know that α1^ α2^ .....^ αk = 0 if any two (or more) vectors are the same. (8.2.5) 5. Wedge product vanishes if vectors are linearly dependent. (8.2.6) 6. Wedge product vanishes if k > n . (8.2.7) 7. Components. For the dual space, we consider tensor functions in place of tensor components, so (αj^ αj^ .....^ αj)(vi,vi.....vi) α^J (vI) = (1/k!) ΣP (-1)S(P) (αj αj ..... αj)(vi,vi.....vi) // (8.1.2) = (1/k!) ΣP (-1)S(P) (αj)(vi) (αj)(vi) ... (αj)(vi) // (6.6.17) for vectors = (1/k!) det [ αj(vi) ] . // (A.1.17) (1/k!) det [ αJ(vI) ] **(8.2.8a) Evaluating at the basis vectors then gives, (αj^ αj^ .....^ αj)(ui,ui.....ui) = (1/k!) det [ αj*(ui*) ] α^J (uI) = (1/k!) det [ (αj*)i*] // see (2.11.c.9) (1/k!) det [ (αJ)I ] **(8.2.8b) In the last equation, the αr are rank-1 functionals. We know that each such functional is associated with a unique vector αr in V which appears in (2.11.a.4), αr(v) = <αr | v> = αr v. Thus, we can form a tensor in Vk called (αj^ αj^ .....^ αj) . For this rank-k tensor we have (αj^ αj^ .....^ αj)ii..i = (αj^ αj^ .....^ αj)(ui,ui.....ui) // (6.5.1) = (1/k!) det [ (αj*)i*] // (8.2.8b) **(8.2.8c) In the Spivak normalization the factor (1/k!) in equations (8.2.8) is replaced by 1. Fact: ( αj^αj^...^αj)(vi,vi....vi) is totally antisymmetric in both the labels jr and the labels ir. **(8.2.9) Proof: Antisymmetry on the jr follows from (8.2.4), while antisymmetry on ir follows from (8.2.8a) (determinant changes sign if any two rows or columns are swapped). 8. Associative Property of the wedge product. For example, (α1^ α2)^ α3 = α1^ (α2^ α3) = α1^ α2^ a3 . 8.3 The vector space Λk and its basis Λk is the space whose elements are all linear combinations of wedge products of k vectors of V*. (8.3.1) A more precise name for this space is Λk(V) but we just call it Λk. It seems useful at this point to compare our vector space names with those of Spivak: Names of spaces. [ TA = totally antisymmetric = alternating ] us Spivak tensor product spaces Vk -- space of rank-k tensors, T = |T>, Tii....i V*k -- dual space of k-multilinear tensor functionals on V, T = <T| V*kf Tk(V) space of k-multilinear tensor functions on V, T(v) = <T|v> wedge product spaces Lk -- space of TA rank-k tensors, T^, T^ii....i Λk -- dual space of TA k-multilinear tensor functionals on V, T^ Λkf Λk(V) space of TA k-multilinear tensor functions on V, T^(v) (8.3.1a) Sjamaar refers to the last space as AkV (2006) and Ak(V) in his 2015 update. We wanted to end up with the name Λk for the last space to agree with Spivak, Benn & Tucker, Conrad and others, and this led to the non-Greek Lk for the corresponding non-dual wedge space. We now go down the list of items in Section 7.3, adapting them as needed. Again, where there is no comment on an item below, please see the the parallel item in Chapter 7. Multiindex versions of some equations appear on the right below in red. Λk is a vector space (8.3.2) Basis elements for Λk Consider the following objects in Λk obtained by wedging together k basis elements of V*, where each λi is selected from the set of n available for V* (which has dimension n), (λj ^ λj ^ .... ^ λj) . (8.3.3) There are independent basis elements for Λk and they all have this form (λi ^ λi ^ .... ^ λi) where i1 < i2 < ..... < ik basis elements (8.3.6) Fact: (λi ^ λi ^ .... ^ λi) = Alt(λi λi .... λi) λ^I = Alt(λI) (8.3.8) This is just a special case of (8.1.2). Components of the basis elements for Λk . Now reconsider the basis vectors of the vector space Λk , (λj ^ λj ^ .... ^ λj) . (8.3.3) For this dual space Λk, we consider tensor functions in place of tensor components, so we then have these special cases of (8.2.8a) and (8.2.8b), (λj ^ λj ^ .... ^ λj)(vi,vi.....vi) λ^J (vI) = (1/k!) ΣP (-1)S(P) (λj λj ..... λj)(vi,vi.....vi) = (1/k!) ΣP (-1)S(P) (λj)(vi) (λj)(vi) ... (λj)(vi) // (6.6.17) for vectors = (1/k!) ΣP (-1)S(P)(vi)j(vi)j ... (vi)j // (2.11.c.5) = (1/k!) det [ (vi)j ] . // (A.1.19) (1/k!) det [ (vI)J ] **(8.3.9a) Evaluation at the basis vectors then gives, (λj ^ λj ^ .... ^ λj)(ui,ui.....ui) = (1/k!) det [ (ui)j ] λ^J (vI) = (1/k!) det [ δij ] // see (2.6.8) (um)n = δmn (1/k!) det [ δIJ ] **(8.3.9b) Once again, in Spivak normalization (1/k!) → 1 for equations (8.3.9). Looking at det [ (vi)j ] above we immediately conclude that, Fact: (λj ^ λj ^ .... ^ λj)(vi,vi....vi) is totally antisymmetric in both the labels jr and the labels ir. **(8.3.10) We saw an example of both antisymmetries for k = 2 back in equation (4.4.20), (λi^ λj)(vr,vs) = - (λi^ λj)(vs,vr) = - (λj^ λi)(vr,vs) // two forms of antisymmetry **(4.4.20) Equation (8.3.9b) can be expressed in our usual informal notation, (λj ^ λj ^ .... ^ λj)(ui,ui....ui) = (1/k!) [ δji δji...δji + signed permutations] **(8.3.11) Example: [ see (7.3.12) ] 3!(λj ^ λj^ λj)(ui,ui,ui) = 3! λ^J(uI) = det(δJI) = det ( δjjjiii ) **(8.3.12) ok to here 8.4 Tensor Expansions for a dual tensor in Λk Recall now the tensor expansion for a most-general tensor T in V*k , T = Σii....i Tii....i (λi λi ..... λi) T ϵ V*k (6.2.1) (8.4.1) where Tii....i are some general coefficients Consider then the similar-looking most-general object in Λk, evaluated at (vj,vj....vj), Σii....i Tii....i (λi ^ λi .....^ λi)(vj,vj....vj) = Σii....i Tii....i AltI(λi λi ..... λi)(vj,vj....vj) // (7.3.8) = Σii....i Tii....i AltJ(λi λi ..... λi)(vj,vj....vj) // (8.2.9) and (A.8.26) = AltJ[ Σii....i Tii....i (λi λi ..... λi)(vj,vj....vj) ] // (A.5.10), Alt is linear = AltJT(vj,vj....vj) = [AltJ(T)](vj,vj....vj) = [Alt(T)](vj,vj....vj) // (8.4.1) ≡ T^(vj,vj....vj) . (8.4.2) Here we define this functional (dual-space) notation, T^ ≡ Alt(T) (8.4.3) which is really this statement about tensor functions, T^(vj,vj....vj) ≡ [Alt(T)] (vj,vj....vj) . (8.4.3a) From (8.4.2) we then have the following fully general element of Λk, T^ = Σii....i Tii....i (λi ^ λi .....^ λi) . (8.4.4) We refer to this type of expansion as a symmetric expansion, and we know it is redundant since the symmetric sum includes each true basis vector k! times. According to (A.8.9), we know from (8.4.3a) and (8.2.2) that Fact: T^(vi,vi....vi) is a totally antisymmetric k-multilinear tensor function. (8.4.5) Therefore, Fact: The space Λk is the space of all totally antisymmetric k-multilinear rank-k tensors T^. To say that T^ is totally antisymmetric means that T^(vi,vi....vi) is a totally antisymmetric tensor function. (8.4.6) In contrast, the space V*k is the space of all k-multilinear rank-k tensors T, so Λk V*k. See Section 7.7 below. Since the set (λi^ λi .....^ λi) with 1 ≤ i1 < i2 < ..... < ik ≤ n forms a complete basis for Λk, as discussed above in (8.3.6), it must be possible to express T^ in the following manner T^ = Σ1≤i<i<....<i≤n Aii...i (λi ^ λi .....^ λi) . (8.4.7) What then is the connection between the Aii...i of (8.4.7) and the Tii...i of (8.4.4)? The discussion goes exactly as in Chapter 7 and the result is, Aii...i = ΣP (-1)S(P) Tii...i i1 < i2 < ..... < ik = [ Tii...i + all signed permutations ] // k! terms = k! [Alt(T)]ii...i . // (A.5.3) def of Alt or A = k!Alt(T) = k! T^. // (7.4.3) (8.4.16) where T^ = Alt(T) is shown in (7.4.4), not to be confused with tensor functional T^ ≡ Alt(T) in (8.4.3). Since A = k! T^ , (7.4.5) shows that Fact: Aii...i and T^ii...i are both totally antisymmetric tensors. (8.4.17) Vector Case. For k = 1, we find that T = ΣiTi λi // (6.2.1) T^ = ΣiTi λi // (8.4.4) T^ = T (8.8.19) so for a vector there is no distinction between T^ and T (and in fact V*1 = Λ1). 8.5 Expansions for the wedge product of k dual vectors We have generally stopped bolding vectors in V, but in this section we bold the vectors αr to distinguish them from the corresponding functionals αr, where recall (2.11.a.4) that αr(v) = αr v . The symmetric expansion is very straightforward. First, consider this rank-k tensor in Vk , Tii...i = (α1)i (α2)i ... (αk)i = (α1 α2 ..... αk)ii...i // αr ϵ V or (8.5.1a) T = (α1 α2 ..... αk) or |T> = |α1,α2,... αk> . // T ϵ Vk The corresponding rank-k tensor functional is, T = (α1 α2 ..... αk) // αr ϵ V*, T ϵ V*k or (8.5.1b) <T| = <α1,α2,... αk| . Then the symmetric expansion (8.4.4) gives, T^ = Σii...i Tii...i (λi ^ λi .....^ λi) (8.4.4) = Σii...i (α1)i (α2)i ... (αk)i (λi ^ λi .....^ λi) // (8.5.1a) (8.5.2) = [Σi (α1)iλi] ^ [Σi (α2)i λi] ^ .... ^ [Σi (αk)i λi] = α1 ^ α2 ^ ... ^ αk . (8.5.3) This pure tensor functional T^ = (α1 ^ α2 ^ ... ^ αk) is an element of Λk . The corresponding element of Lk is T^ = (α1 ^ α2 ^ ... ^ αk) = Alt(α1 α2 ... αk) as in (7.4.3). Expressing α1 ^ α2 ^ ... ^ αk in terms of the ordered expansion is more complicated. One must first compute the tensor A as in (8.4.16) or (7.4.3), (1/k!) A = Alt(T) = Alt(α1α2.....αk) = T^ = (α1 ^ α2 ^ ... ^ αk) ϵ Λk (8.5.4) Then the ordered expansion (8.4.7) can be written in a battery of ways, α1 ^ α2 ^ ... ^ αk = // α1 ^ α2 ^ ... ^ αk ϵ Λk (a) = Σ1≤i<i<....<i≤n Aii...i (λi ^ λi .....^ λi) // (8.4.7) (b) = Σi<i<....<i k! [Alt(α1α2...αk)]ii...i (λi ^ λi .....^ λi) . // (8.5.4) (c) = Σi<i<....<i k! [α1 ^ α2 ^ ... ^ αk]ii...i (λi ^ λi .....^ λi) . // (8.1.3) (d) = Σi<i<....<i det[ (α*)i] (λi ^ λi .....^ λi) . // (8.2.8c) with jr → r (e) = Σi<i<....<i det (λi ^ λi .....^ λi) . (f) = Σi<i<....<i det (λi ^ λi .....^ λi) . (g) = Σii...i (α1)i (α2)i ... (αk)i (λi ^ λi .....^ λi) // (8.5.2) (8.5.5) where we throw in the symmetric sum at the end. Remember that, since generally dim(V) = n > k, the determinant in (f) is a full-width minor of matrix M = [α1, α2.....αk]. If k = n, the minor is the full matrix. Example : Suppose k = n = 3. Then the following sum (form (f)) has only one term, α1 ^ α2 ^ α3 = Σ1≤i<i<i<3 det[α1, α2, α3] (λi ^ λi ^ λi) = det[α1, α2, α3] (λi ^ λi ^ λi) (8.5.6) as quoted in (4.4.15). Here are the above expressions for k = 2 and general n ≥ k : (a) α1 ^ α2 = Σi<i Aii (λi ^ λi) (b) = Σi<i 2! [Alt(α1α2)]ii (λi ^ λi) (c) = Σi<i 2! (α1 ^ α2)ii (λi ^ λi) (d) = Σi<i det[ (α*)i] (λi ^ λi) (e) = Σi<i det (λi ^ λi) (f) = Σi<i det (λi ^ λi) = Σi<i[(α1)i(α2)i- (α2)i(α1)i](λi ^ λi) (g) = Σii (α1)i (α2)i (λi ^ λi) = [Σi(α1)iλi] ^ [Σi(α2)i λi] = α1 ^ α2 (8.5.7) Result (f) matches that shown in (4.4.12), α ^ β = Σij αiβj (λi ^ λj) = Σi<j (αiβj- αjβi) (λi ^ λj) = Σi<j Aij (λi ^ λj) = Σi<j det (λi ^ λj) Aij = (αiβj- αjβi) = det . (4.4.12) 8.6 Number of elements in Λk compared with V*k. We know from (5.1.5) and (8.3.6) that, dim(V*k) = nk // number of basis elements of V*k (6.1.5) dim(Λk) = // number of basis elements of Λk (8.3.6) If the number of elements of field K is N ( N → ∞ for K= reals), then ratio = = = = / nk . (8.6.1) For a given n, this is a strongly decreasing function of k, see graph in (7.6.2). ok to here 8.7 Multiindex notation In this section, multiindex versions of equations are shown in red. Multiindexing is done in two different ways. First, for the symmetric expansion (8.4.4) : T^ = Σii....i Tii....i (λi^ λi .....^ λi) (7.4.4) T^ = ΣI TI λ^I where λ^I ≡ λi ^ λi ^ .... ^ λi TI ≡ Tii...i and I ≡ {i1, i2,.... ik} with 1 ≤ ir ≤ n = ordinary multiindex, n = dim(V*) . (8.7.1) The more significant notation involves the ordered expansion (8.4.7) which has only one term for each linearly independent basis element. Note our use of Σ'I (prime) to indicate an ordered multiindex summation : T^ = Σ1≤i<i<....<i≤n Aii...i (λi ^ λi .....^ λi) . (8.4.7) T^ = Σ'I AI λ^I where λ^I ≡ λi ^ λi ^ .... ^ λi AI ≡ Aii...i and I ≡ {i1, i2,.... ik} with 1 ≤ i1< i2<....< ik ≤ n = ordered multiindex, n = dim(V*) . (8.7.2) Here are some unofficial multiindex notations for other equations developed above: T^ = α1 ^ α2 ^ ... ^ αk T^ = (^αZ) (8.5.3) Tii...i = (α1)i (α2)i ... (αk)i ≡ (αZ)I TI = (αZ)I (8.5.1a) with the idea that Z = 1,2...k . Continuing on, T^ = Σii...i (α1)i (α2)i ... (αk)i (λi ^ λi .....^ λi) T^ = ΣI (αZ)I λ^I (8.5.2) A = k! Alt(α1α2.....αk) A = k! Alt(αZ) (8.5.4) α1 ^ α2 ^ ... ^ αk = Σi<i<....<i k! [Alt(α1α2...αk)]ii...i (λi ^ λi .....^ λi) . (8.5.5b) (^αZ) = Σ'I k! Alt(αZ)I λ^I Aii...i = det AI = det[(αZ)I] (8.5.5a+f) α1 ^ α2 ^ ... ^ αk = Σi<i<....<i det (λi ^ λi .....^ λi). (8.5.5f) (^αZ) = Σ'I det[(αZ)I]λ^I 8.8 The Exterior Algebra Λ(V) We now construct the graded algebra Λ(V) in analogy with that of T(V) in (5.4.1). Define a large vector space of the form ( this is "the dual exterior algebra on V" ) Λ(V) ≡ Λ0 Λ1 Λ2 Λ3 + .... // Λ(V) = Σk=0∞ Λk(V) (8.8.1) Here Λ0 = the space of scalars, Λ1 the space of dual vectors, Λ2 = Λ ^ Λ V*2 the space of antisymmetric dual rank-2 tensors (8.4.6), and so on. The most general element of the space Λ(V) would have the form X = s ΣiTi λi Σij Tij λi^λj Σijk Tijk λi^λj^λk + ..... or X = s ΣiTi λi Σi<j Aij λi^λj Σi<j<k Aijk λi^λj^λk + ..... (8.8.2) The direct sum is described in Appendix B. Associativity of the Wedge Product See discussion near (7.8.3) and replace ei→ λi and v→α. One then concludes that, Fact: The wedge product of k vectors α1^ α2^ .....^ αk can be "associated" in any manner without altering the meaning of the product. By this we mean that parentheses can be added in any manner without altering the object. (8.8.4) What this in effect does is define an array of new objects to be the same as α1^ α2^ .....^ α6. For example, (α1^ (α2^ α3 ^ α4) ^ (α5^ α6) ≡ α1^ α2^ α3^ α4^ α5^ α6 α1^ (α2^ (α3^ α4)^ α5)^ α6 ≡ α1^ α2^ α3^ α4^ α5^ α6 (8.8.5) Since tensors like T^ can be expanded on (λi ^ λi ^ .... ^ λi), and since one may associate this wedge product arbitrarily as claimed in (8.8.4), one easily shows that : Fact: The wedge product of N general dual tensors A^^B^^C^.... can be "associated" in any manner without altering the meaning of the product. By this we mean that parentheses can be added in any manner without altering the object. (8.8.7) This fact then extends the claim (8.8.4) made for N vectors, and is exactly analogous to the similar axiomatic statement for associativity made in (2.8.22). Example: In (8.7.1) multiindex notation, consider three Λ(V) tensors T^,S^,R^ of rank k,k',k" : (T^ ^ S^) ^ R^ = ( (ΣITIλ^I) ^ (ΣJSJλ^J) ) ^ (ΣKRKλ^K) = ΣITIΣJSJ { ( λ^I ^ λ^J) ^ (ΣKRKλ^K) } // rules (8.2.3) = ΣITIΣJSJΣKRK (λ^I ^ λ^J) ^ (λ^K) // rules (8.2.3) again = ΣIJKTISJRK (λ^I ^ λ^J ^ λ^K) // detail shown above (7.8.8) with ei→λi = (ΣITIλ^I) ^ (ΣJSJλ^J) ^ (ΣKRKλ^K) // rules (8.2.3) again = T^ ^ S^ ^ R^ . Our example shows that for arbitrary Λ(V) tensors, (T^ ^ S^) ^ R^ = T^ ^ S^ ^ R^. Fact: The large space Λ(V) is in fact itself a vector space. (8.8.8) We know this is true since Λ(V) = Σk=0∞ Λk and we showed in (8.3.2) that each Λk is a vector space. For example, the "0" element in Λ(V) is the direct sum of the "0" elements of all the Λk. See Appendix B for more detail. To show that Λ(V) is an algebra, we must show that it is closed under both addition and multiplication. It should be clear to the reader that Λ(V) is closed under addition and has the right scalar rule. For example, if k1 and s are scalars, k1 α β^κ ρ^σ^η = sum of 4 elements of Λ(V) = an element of Λ(V) s(k1 α β^κ ρ^σ^η) = (sk1) (sα) (sβ)^κ ρ^(sσ)^η = element of Λ(V) (8.8.9) This additive closure is of course necessary for Λ(V) be a vector space. The space is also closed under the multiplication operation ^. For example (β^κ)^(ρ^σ^η) = β^κ^ρ^σ^η = ϵ Λ5 = ϵ Λ(V) . // (β^κ) ϵ Λ2 (ρ^σ^η) ϵ Λ3 (8.8.10) Here we have used the associative property (8.8.4). This closure claim is stated more generally below (8.10.6). One then makes the following definitions with regard to the space Λ(V), where n = dim(V*) = dim(V): (the objects in this table are pure multilinear functionals) Object Name any blade lincomb: Grade(rank): Space s dual 0-blade scalar ϵ K 0 Λ0 α dual 1-blade dual vector 1 Λ1 α^β dual 2-blade dual bivector 2 Λ2 α^β^γ dual 3-blade dual trivector 3 Λ3 α^β^γ^δ dual 4-blade dual quadvector 4 Λ4 ..... α^β^γ^δ^.... dual k-blade dual k-vector k Λk .... α^β^γ^δ^.... dual n-blade dual n-vector n Λn arbitrary element of Λ(V) dual multivector mixed Λ(V) (8.8.11) Since Λ(V) is closed under the operations and ^, it is "an algebra" (the space Λk alone is not an algebra because it is not closed under ^). The Λ(V) algebra is different from that of the reals due to its definition as a sum of vector spaces. The elements of Λ(V) have different "grades" as shown above, so Λ(V) is a "graded algebra". Sometimes Λ(V) is called "the dual exterior tensor algebra" over V. A k-blade is a pure wedge product of k vectors, whereas a k-vector is any linear combination of k-blades. A multivector is any linear combination of k-vectors for any mixed values of k. Note that s1( α^β) s2(γ^δ) = (s1α)^β (s2γ)^δ = (α'^β) (γ'^δ) // 2-blades s1( α^β) s2(γ^δ^ε) = (s1α)^β (s2γ)^δ^ε = (α'^β) (γ'^δ^ε) // multivector so it is also correct to say that a k-vector is any sum of k-blades, and a multivector is any sum of k-vectors. That is, any linear combination can be written as a sum as shown in the above examples. Unlike in Tensor World, in Wedge World the above list (8.8.11) is finite for a given n = dim(V). For k = n there is exactly one linearly independent basis vector which is the ordered wedge product of all the basis vectors of V*. For k > n, all wedge products vanish since the vectors in the wedge product are linearly dependent, see (7.2.6). The dimensionality of the space Λ(V) is as follows, based on (8.8.1) and (B.10)', dim[Λ(V)] = dim[Λ0 Λ1 Λ2 Λ3 + ....] = dim(Λ0) + dim(Λ1) + dim(Λ2) + dim(Λ3) + ... but for dim(V*) = n this series truncates with Λn and we find from (7.3.6), dim[Λ(V)] = 1 + n + + + ... + = Σk=0n = 2n = a finite number (8.8.12) Recall from the discussion above (4.4.34) that the space Λ2 of rank-2 tensor functionals is isomorphic to the space Λ2f of rank-2 tensor functions, where we added a subscript f to distinguish these two vector spaces. We apply this similar notation to the full space Λ(V) to obtain this tensor function version of (8.8.1), Λf(V) ≡ Λ0f Λ1f Λ2f Λ3f + .... // Λf(V) = Σk=0∞ Λkf(V) (8.8.13) where now Λf(V) is the space of all multilinear totally antisymmetric (alternating) functions of any number of vector arguments. 8.9 The Wedge Product of two or more dual tensors in Λ(V) (a) Wedge Product of two dual tensors T^ and S^ Rather than translate the many details of this section from Chapter 7, we will skip these details and state the conclusions. The details may be obtained from Section 7.9 by making these simple replacements: ui → λi uI → λI u^I → λ^I Tii....i → Tii....i , TI → TI , T^ → T^ Sii....i → Sii....i , SI → SI , S^ → S^ In subsection (d) below on the product of three tensors, more details are provided. Here then are selected results: Tensor product of two tensors: T^^ S^ = ΣI (TS)I λ^I I ≡ I, I' = i1,i2...ik+k', λ^I ≡ (λi^ λi .....^ λi) . (8.9.a.5) Closure: T^ ϵ Λk and S^ ϵ Λk' T^^ S^ ϵ Λk+k' Λ(V) . (8.9.a.6) Basis relation: (λi^ λi .....^ λi) = Alt(λi λi .... λi) λ^I = Alt(λI) (7.3.8) TS == ΣI (TS)I λI I ≡ I, I' = i1,i2...ik+k', uI ≡ (ui ui .... ui) . (5.6.5) Alt(TS)(vJ) = AltJ(TS)(vJ) = ΣI (TS)I AltJ (λI(vJ)) // (5.6.5) and (A.5.10) that Alt is linear = ΣI (TS)I AltI (λI(vJ)) // use (A.8.26) since λI(vJ) is totally antisymmetric in I and J = ΣI (TS)I λ^I(vJ) // (8.3.8) = (T^^ S^)(vJ) // (8.9.a.5) so T^^ S^ = Alt(TS) . (8.9.a.7) The "components" (tensor functions) are (T^^ S^)(vJ) = Alt(TS)(vJ) = ΣP(-1)S(P) (TS)(vP(J)) = ΣP(-1)S(P) T(vP(J)) S(vP(J')) (8.9.a.8) where J ≡ j1, j2...jk J' ≡ jk+1, jk+2, ....jk+k' J ≡ J, J' = j1,j2...jk+k' (7.9.a.4) This last line is an explicit instruction for computing the "components" of the tensor T^^ S^ . We have added this new notation, T(vP(J)) ≡ T(vj, vj... vj) for J ≡ j1, j2...jk . (8.9.a.9) Example: Let S and T both be rank-2 dual tensors so k = k' = 2 . Then (T^^ S^)(vI) = (T^^ S^)(v1,v2,v3,v4) = (1/4!) ΣP(-1)S(P)T(vi, vi)S(vi, vi) = (1/24) [ T(vi,vi)S(vi,vi) - T(vi,vi)S(vi,vi) + T(vi,vi)S(vi,vi) + 21 more terms] (8.9.a.10) Here as elsewhere we show in red the indices to be swapped to make the next term. From (8.9.c.6) below, T^^ S^ = (-1)2*2 S^^ T^ = S^^ T^. (8.9.a.11) (b) Special cases of the wedge product T^^ S^ Same as Section 7.9 (b) with T^→ T^ and S^→ S^ . Here are the conclusions : T^^S^ = κ^S^ = S^^T^ = S^^κ = κS^ if T^ = κ ϵ V*0 [ V*0 = V0 ] T^^S^ = T^^κ' = S^^T^ = κ'^T^ = κ'T^ if S^ = κ' ϵ V*0 T^^S^ = κ^κ' = S^^T^ = κ'^κ = κκ' if T^,S^ = κ,κ' ϵ V*0 (8.9.b.3) (c) Commutivity Rule for the Wedge Product of two dual tensors T^ and S^ Same as Section 7.9 (c) with T^→ T^ and S^→ S^ and e→λ. Here are some of the translated conclusions: (λ^J ^ λ^I) = (-1)kk' (λ^I ^ λ^J) . dual basis vectors (8.9.c.5) S^^ T^ = (-1)kk'T^^ S^ ranks of the two dual tensors are k and k' . (8.9.c.6) (d) Wedge Product of three or more dual tensors For this section we do a full translation of Section 7.9 (d) : T^^S^^R^ = [ΣITIλ^I]^[ΣJ SJλ^J]^[ΣK RKλ^K] (a) = ΣI,J,K TISJRK (λ^I) ^ (λ^J) ^ (λ^K) (b) = ΣI,J,K TISJRK (λ^I ^ λ^J ^ λ^K) // associative of ^ used here (d) = ΣI,I',I" TISI'RI" (λ^I ^ λ^I' ^ λ^I") // rename multiindices J→I',K→I" I ≡ i1, i2...ik I' ≡ ik+1, ik+2, ....ik+k' I" ≡ ik+k'+1, ik+k'+2, ....ik+k'+k" λ^I ≡ (λi^....^ λi) λ^I' ≡ (λi^...^λi) λ^I" ≡ (λi^...^λi) (e) = ΣI (TSR)I λ^I λ^I ≡ (λi^....^ λi) I ≡ I, I',I" = i1,i2...ik+k'+k" (8.9.d.1) The outer product form is TISI'RI" = (TSR)I,I',I" = (TSR)I . The conclusion is this: T^^S^^R^ = ΣI (TSR)Iλ^I I ≡ I, I',I" = i1,i2...ik+k'+k" , λ^I ≡ (λi^....^ λi) (8.9.d.2) Since the λ^I are basis vectors in Λk+k'+k", we have shown that: T^ ϵ Λk and S^ ϵ Λk' and R^ ϵ Λk" T^^S^^R^ ϵ Λk+k'+k" Λ(V) . (8.9.d.3) Recalling the Chapter 6 result, TSR = ΣI (TSR)I λI I ≡ I, I',I" = i1,i2...ik+k'+k" λI ≡ (λi.... λi) (6.6.5) and (8.3.8) that λ^I = Alt(λI) we find, Alt(TSR) = ΣI (TSR)I Alt(λI) // Alt is linear, see (7.9.d.4) = ΣI (TSR)I λ^I // (8.3.8) = T^^S^^R^ // (8.9.d.2) so T^^S^^R^ = Alt(TSR) (8.9.d.4) and then [T^^S^^R^](vI) = [Alt(TSR)](vI) = ΣP(-1)S(P) (TSR)(vP(I)) // (A.5.3) = ΣP(-1)S(P) T(vP(I))S(vP(I'))R(vP(I")) (8.9.d.5) which gives instructions for how to compute the "components" of T^^S^^R^ . Using the systematic notation outlined in (5.6.10) through (5.6.12), and generalizing the above development for the wedge product of three tensors, we find the following expansion for the wedge product of N tensors of Λ(V), (T1)^^(T2)^^...^(TN)^ = ΣI (T1)I(T2)I .... (TN)I λ^I = ΣI (T1T2....TN)I λ^I (8.9.d.6) where λ^I = λi^ λi .....^ λi = λi^ λi .....^ λi and (T1T2....TN)I = (T1)I(T2)I .... (TN)I . The rank of this product tensor is then κ = Σi=1N ki and the tensor is an element of Λκ Λ(V). Notice that if κ > n, the tensor product (8.9.d.6) vanishes since there are then > n factors in λ^I so one or more are then duplicated, (T1)^^(T2)^^...^(TN)^ = 0 if κ = Σi=1N ki ≥ n+1 (8.9.d.7) For example, if all the tensors are the same tensor T^ of rank k, then T^N ≡ T^^T^^...^T^ = 0 if Nk ≥ n+1 or N ≥ (n+1)/k (8.9.d.8) If N ≥ (n+1), then N ≥ (n+1)/k for any k ≥ 1. Thus T^N = 0 for any N ≥ n+1 assuming k ≠ 0. (8.9.d.9) Recall (6.6.16), T1T2...TN = ΣI (T1IT2I .... TNI) λI = ΣI (T1T2....TN)I λI . (6.6.16) Applying Alt to both sides again with λ^I = Alt(λI) shows that, as in (8.9.d.4), (T1)^^(T2)^^...^(TN)^ = Alt(T1T2...TN) . (8.9.d.10) "Components" (the tensor function) of this tensor are computed as follows: [(T1)^^(T2)^^...^(TN)^](vI) = [Alt(T1T2...TN)](vI) = ΣP(-1)S(P) (T1T2...TN)(vP(I)) // (A.5.3) = ΣP(-1)S(P) T1(vP(I))T2(vP(I)) ...TN(vP(I)) (8.9.d.11) where T1(vP(I)) ≡ T1(vi,vi...vi) for I1 = i1, i2...iκ T2(vP(I)) ≡ T2(vi, vi...vi) for I2 ={iκ+1, iκ+2.....iκ} etc. // see (5.6.10 thru 12) for details In the Dirac notation of Section 2.11 one can write (8.9.d.10) as <(T1)^| ^ < (T2)^| ^...^ < (TN)^| = Alt( <T1| <T2| ... <TN| ) . (8.9.d.12) It is shown in (C.4.17) that "pre-antisymmetrization makes no difference", so the above may also be written <(T1)^| ^ < (T2)^| ^...^ < (TN)^| = Alt( <(T1)^| <(T2)^| ... <(TN)^| ) . (8.9.d.13) Both sides of this equation are elements of the dual wedge product space Λk+k+..+k , but they are also both elements of the larger dual tensor product space V*k V*k ... V*k . The action of linear operator Q on a dual tensor product space vector is defined in the obvious manner, as in (6.6.18), [ <(T1)^| <(T2)^| ... <(TN)^| ] Q = <(T1)^|Q <(T2)^|Q ... <(TN)^|Q . (8.9.d.14) In other words, the action of Q on the larger space is defined in terms of its action on the spaces which make up the tensor product. This result holds as well for the wedge product of N dual tensors, [<(T1)^| ^ < (T2)^| ^...^ < (TN)^|] Q = <(T1)^|Q ^ < (T2)^|Q ^...^ < (TN)^|Q (8.9.d.15) Proof: [<(T1)^| ^ < (T2)^| ^...^ < (TN)^|] Q = [ Alt( <(T1)^| <(T2)^| ... <(TN)^| ) ] Q = Alt [ ( <(T1)^| <(T2)^| ... <(TN)^| ) Q ] = Alt [ ( <(T1)^|Q <(T2)^|Q ... <(TN)^|Q ) ] = <(T1)^|Q ^ <(T2)^|Q ^ ...^ <(TN)^|Q . Equations (8.9.d.14) and (8.9.d.15) are the transposes of (7.9.d.15) and (8.9.d.15) if we set Q = PT . (e) Commutativity Rule for product of N dual tensors The argument of Section 7.9 (e) can be repeated with e→λ. Here we just quote the conclusion: Fact: In a product of tensors T1^T2^T3.... of rank k1, k2, k3 ... , if two tensors are swapped Tr ↔ Ts (with r < s), the resulting tensor incurs the following sign relative to the starting tensor, sign = (-1)m where m = (kr+1+kr+2 ...+ks-1)(kr+ks) + krks . (8.9.e.6) Corollary: If the sum of the ranks of the two swapped tensor is even, in effect m = krks . (8.9.e.7) Example: T1 ^ T2 ^ T3 = (-1)m T3 ^ T2 ^ T1 r = 1 s = 3 m = (k2)(k1+k3) + k1k3 = k1k2 + k1k3 + k2k3 (-1)m = (-1)kk+kk+kk (8.9.e.8) (f) Theorems from Appendix C : pre-antisymmetrization makes no difference We showed above that one can form wedge products of elements of Λ(V) in this manner, T^^ S^ = Alt(TS) . (8.9.a.7) T^^S^^R^ = Alt(TSR) (8.9.d.4) (T1)^^(T2)^^...^(TN)^ = Alt(T1T2...TN) (8.9.d.7) where the operator Alt acts on the vector arguments which are not displayed in the above compact functional notation. For example T^^ S^ = Alt(TS) means, in multiindex notation, (T^^ S^)(vI) = AltI [(TS)(vI)] = AltI [ T(vI)S(vI')] = ΣP(-1)P T(vP(I))S(vP(I')) . A very simple case is the following (recall for vectors that α = α^ ) (α ^ β)(vi,vi) = Alt (αβ)(vi,vi) = Alt α(vi)β(vi) = ΣP(-1)P α(vi) β(vi) = (1/2) [ α(vi)β(vi) - α(vi)β(vi)] = (1/2)[ (αβ)(vi,vi) - (βα)(vi,vi) ] = { (1/2) [(αβ) - (βα) ]}(vi,vi) which replicates our Chapter 4 statement that α ^ β = [αβ- βα]/2 . (4.4.1) The objects here are functionals in Λ(V) which, when closed with a vector set, become tensor functions in Λf(V). For example, as was shown in (2.11.f.7), T^ = <T| ϵ Λk // functional in Λk T^(v1, v2....vk) = <T| v1, v2....vk> // the corresponding tensor function of Λkf . Appendix C uses the rearrangement theorem in three separate Theorems to show that T^^ S^ = Alt(TS) = Alt(T^S) = Alt(TS^) = Alt(T^S^) . (8.9.f.1) Theorem One Theorem Two Theorem Three These three theorems are derived in a generic space with vectors |1,2...k> and so apply to both tensors and tensor functionals, TI and T(vI). Recall that T^ ≡ Alt(T) (8.4.3) so that T^ is a totally antisymmetric tensor functional. What (8.9.f.1) says is that Alt(TS) provides total antisymmetrization on all the (undisplayed) vector argument indices, so pre-antisymmetrizing either or both tensors makes no difference. A similar statement applies to working with totally symmetric tensors. So we have, Alt[TS] = Alt[T^S] = Alt[TS^] = Alt[T^S^] where T^ = Alt(T) S^ = Alt(S) (C.4.1) Sym[TS] = Sym[TsS] = Sym[TSs] = Sym[TsSs] where Ts = Sym(T) Ss = Sym(S) . (C.4.2) These can of course be rewritten as Alt[TS] = Alt[Alt(T)S] = Alt[TAlt(S)] = Alt[Alt(T)Alt(S)] (C.4.3) Sym[TS] = Sym[Sym(T)S] = Sym[TSym(S)] = Sym[Sym(T)Sym(S)] . (C.4.4) Similarly Appendix C shows that T^^S^^R^ = Alt(TSR) = Alt(T^SR) = Alt(TS^R) = Alt(TSR^) = Alt(T^S^R)= Alt(T^SR^)= Alt(TS^R^) = Alt(T^S^R^) . (8.9.f.2) Adding ^ subscripts inside an Alt expression changes nothing. Here is another example: T^^S^^R^ = Alt(TSR) = Alt((TS)R) = Alt((TS)^R) = Alt(Alt(TS)R) . (8.9.f.3) which appears in Spivak p 80 as (C.4.8) (g) Spivak Normalization We won't repeat the discussion of Section 7.9 (g), but the reader can do the translation with the usual rules v → α ei → λi TI → TI etc. Here are the results, where factors shown in red show changes caused by the Spivak notation in which the normalization factor in (8.1.2) is changed from (1/k!) to 1, αj^ αj^ .....^ αj = 1 ΣP (-1)S(P) ( αP(j) αP(j) ..... αP(j)) = 1 [ (αj αj ..... αj) + all signed permutations ] = k! Alt(αj αj ..... αj) . (8.1.2)S In particular, α ^ β = 1[ αβ - βα] . // no factor of 1/2 (4.4.1)S The affected equations are these: (λi^ λi^ .....^ λi) = k! Alt(λi λi ..... λi) or λ^I = k!Alt(λI) (8.3.8)S T^ ≡ k!Alt(T) and S^ ≡ k'!Alt(S) . (8.4.3)S T^^ S^ = (k+k')! Alt(TS) . (8.9.a.7)S T^^ S^ = Alt(T^S^) T^ ϵ Λk and S^ ϵ Λk' . (8.9.g.1) T^^ S^^ R^ = Alt(T^S^R^) T^ ϵ Λk , S^ ϵ Λk', R^ ϵ Λk" . (8.9.g.2) These now correspond exactly with Spivak's wedge product definition for tensor functions, page 79 page 80 In (8.3.1a) we showed a table comparing our notation to that of Spivak. Here are a few more items: us Spivak λi φi dual space basis vectors Λk -- space of k-multilinear alternating tensor functionals Λkf Λk(V) space of k-multilinear alternating tensor functions T^,R^,S^ ω,η,θ typical elements of Λk ( and Λkf) k,k',k" k,l,m ranks (degrees) of the above typical elements σ P permutation operator sgn σ (-1)S(P) permutation parity, S(P) = swap count Sk G set (group) of all permutations of [1,2...k], App. A. (7.9.g.3) Comments: 1. Spivak refers to a totally antisymmetric tensor function as an alternating function which is the traditional terminology in this realm, hence the operator name Alt. 2. Spivak uses all lower indices, whereas we have used covariant notation. 3. The Spivak normalization is compatible with the traditional definition of a "pullback" as described below in Chapter 10. 9. The Wedge Product as a Quotient Space We present here a wedge product "theory section" which really should be part of Chapter 1, but we wanted to have the reader first immersed in the nuts and bolts approach to the wedge product presented in Chapters 4, 7 and 8. As is the case for Chapter 1, this chapter makes no mention of the components of vectors or tensors. 9.1. Development of Lk as Vk/S. The presentation below is based on the paragraph titled Definition 3.1 on page 5 of Conrad. Consider the vector space V defined over some field K (the scalars, normally reals). If V used coefficients in a ring R instead of a field K, V would be called an R-module. Since any field K is also a ring, we can regard our usual V as an R-module (any vector space is also an R-module). Statements about R-modules are more general that statements about vector spaces, so for that reason one sees the R-module moniker in discussions of our current topic. We shall use the bare term module. Thus, the vector space Vk = VV...V can be regarded as a module since its vectors are defined over the field K which is also a ring. The pure elements of Vk have the form v1v2v3.... vk (k factors). Consider the subset S of Vk whose elements have a repeat of one of the vectors. That is, suppose we have vi = vj for some i ≠ j in v1v2v3.... vk. There could be other vectors which are also equal to vi, so at least two vectors are the same. For example, if k = 4 one would say axbx and xxbx were in the subset S. Adding elements of this subset produces another element of the subset, so this subset is itself a module. Thus we are talking about elements of a submodule S of the module Vk. Notice that 0 is an element of S, which can be represented by any element of Vk having one or more vectors being the 0 vector of V, as in (1.1.9). For k = 4, consider this element of Vk, A' = 3 abcd + 5abca - 2 aacd + 3 aaca . (9.1.1) If we were to throw out elements of the set S, we would get A = 3 abcd . (9.1.2) The set of elements of Vk that is generated by adding all elements of set S to A is called the coset of A, usually written [A]. Thus, the coset of A is A + s where s ϵ S. The elements of Vk can be partitioned into an array in this manner, where each row (coset) involves all the si ϵ S : row name coset → [0] 0 0 + s1 0 + s2 ..... [A] A A + s1 A + s2 .... [B] B B + s1 B + s2 .... (9.1.3) ... For example our Vk element A' lies somewhere in the row of this chart labeled on the left by [A]. It turns out that the rows themselves (the cosets) form a module called Vk/S . The elements of this module can be regarded as being those in the first column of the cosets. So A is an element of Vk/S , but A' is not. Strictly speaking, there is an isomorphism between A and [A], but we ignore such details. Fact: To enumerate the elements of the module Vk/S we write down all the elements of Vk and just set to 0 all terms in which a vector is repeated, such as the last three terms of A' above. We thus filter out such terms, they are "modded out", which is why Vk/S is sometimes called Vk mod S. (9.1.4) Define Lk to be Lk ≡ Vk/S . (9.1.5) The fact that Vk elements lying in S (those that have repeated vectors) are "thrown out" (modded out, set equal to 0) is reminiscent of the construction (1.1.4) that F(VxW)/N = VW and certain elements of the full set F(VxW) were similarly modded out (set to 0, such as (v2, w1+w2) – (v2,w1) – (v2,w2)). Elements of Vk are written v1v2v3.... vk. This product is "associative" in that parentheses can be placed any way one wants, such as v1(v2v3).... vk, with no change in value. (9.1.6) Elements of Lk ≡ Vk/S are written v1^v2^v3.... ^vk . This product is declared to be "associative" in that parentheses can be placed any way one wants, such as v1^(v2^v3).... ^vk, with no change in value. (9.1.7) Using this definition of the wedge product of k vectors, we can derive some of its properties. Fact 1: v1^v2^v3.... ^vk = 0 if two (or more) vectors are the same. (9.1.8) Proof: This follows from the definition of Lk ≡ Vk/S and the Fact (9.1.4) stated above. Fact 2: v1^v2 = - v2^v1 (9.1.9) Proof: We know that (v1+v2) ^ (v1+v2) = 0 since this has the form v3 ^ v3 which is 0 by Fact 1. Expanding, 0 = (v1+v2) ^ (v1+v2) = v1^v1 + v1^v2 + v2^v1 + v2^v2 = v1^v2 + v2^v1 so 0 = v1^v2 + v2^v1 and v1^v2 = - v2^v1 QED Fact 3: Swapping any pair of vectors in v1^v2^v3.... ^vk creates a minus sign. (9.1.10) Proof by example: (swap v1 and v3 by making use of Fact 2 three times) : v3^v2^v1.... ^vk = + v3^(v2^v1).... ^vk = - v3^(v1^v2).... ^vk = - (v3^v1)^v2.... ^vk = + (v1^v3)^v2.... ^vk = + v1^(v3^v2).... ^vk = - v1^(v2^v3).... ^vk = - v1^v2^v3.... ^vk QED Fact 4: vj ^ vj ^ .... ^ vj = εjj....j ( v1^ v2^ .....^ vk ) (9.1.11) Proof: Fact 4 is the combination of Fact 3 and Fact 1. Fact 4 appears as (4.6.12). Fact 5: vj ^ vj ^ .... ^ vj = 0 if the vectors are linearly dependent. (9.1.12) Proof: See (4.6.14). In this manner, we can derive all the properties of the wedge product stated in Section 4.6 without having to lean on the construction of the wedge product as a linear combination of tensor products. However, we know that the elements of Lk are linear combinations of the elements of Vk. We have written in (4.6.2) that v1^ v2^ .....^ vk = (1/k!) Σii....i εii....i (vi vi ..... vi) = (1/k!) ΣP (-1)S(P) ( vP(1) vP(2) ..... vP(k)) . (4.6.2) Since in Section 4.6 this linear combination generates all the Facts listed above, and does not contradict any of them, we conclude that this must be the linear combination of Vk elements that equals v1^ v2^ .....^ vk (apart from a possible normalization factor). Alternate Language. Looking at A and A' above, we could say that A and A' are in the same equivalence class so that A ~ A'. Two elements of Vk are in the same equivalence class if they differ by an element of S, so we have A' - A = s ϵ S. The elements of the equivalence class of A are then just the coset [A]. The submodule of Vk called Vk/S is called a quotient module. Using category diagrams, one can consolidate this notion with that of quotient rings and quotient groups. 9.2. Development of L as T/I. Start with the tensor algebra (vector space) shown in (5.4.1), T(V) = V0 V V2 V3 ....... (9.2.1) The elements of the vector space T(V) form a ring with operations and . It is easy to show that T(V) is closed under addition and multiplication and has the other required ring properties. Ideal Example 1: Consider the set S of elements of T(V) which are linear combinations of elements of the form ABC (with coefficients in field K) where B is some fixed element of T(V) and A,C ϵ T(V) are allowed to vary. This set is closed under addition. For example, ABC + A'BC' ϵ S. Since coefficients in K can be absorbed into A, one could just say that the elements of S are sums of elements of the form ABC. One could take any 0BC as the "0" element, and -ABC is the additive inverse. The set S is commutative and associative under addition. Therefore S forms an additive subgroup of the ring T(V). Moreover if we left or right multiply (using ) any element of this set by any element of T(V), the result clearly lies in T(V). Q(ABC) ϵ T(V) (ABC)Q ϵ T(V) . (9.2.2) Therefore this set S is a two-sided ideal of the ring T(V). Ideal Example 2: S = sums of elements of the form ABCDE where elements B and D are fixed and A,C are E varied, all letters being ϵ T(V) . Ideal Example 3: S = sums of elements of the form AxCxE where vector x is fixed and A,C,E ϵ T(V) are varied. This set is the set of all sums of elements of T(V) in which the vector x appears at least twice. Let's call this particular ideal by the name S = I, because this is our ideal of interest. Now suppose we declare the following equivalence relation AxCxE ~ 0 x,A,C,E ϵ T(V) . (9.2.3) Sums of such elements form the ideal I discussed above, and we are in effect setting all elements of this ideal equal to 0. There then exists a subset of T(V) which we shall call T(V)/I, or T(V) "mod" I. This is a standard algebraic structure where one takes the quotient of a ring R divided by a two-sided ideal I of that ring. The upshot is that the elements of the new quotient set T(V)/I consist of all sums of T(V) elements except that any term which matches the form (9.2.3) is filtered out ("modded out") by setting it equal to 0. Example: t' = k1 abcd + k2 ab + k3 bcc + k4 abca = element of T(V) t = k1 abcd + k2 ab = element of T(V)/I (9.2.4) In algebra terminology, adding all elements of the form AxCxE to t generates a coset associated with t called [t], and T(V)/I is in effect the set of all such cosets. Element t' is one element of the t coset. The elements of T(V)/I themselves form a new ring called the quotient ring or factor ring. The ring/ideal situation is quite similar to that discussed above for the module/submodule situation Vk/S. Recall from (7.8.1) that the full wedge (exterior) tensor algebra is given by the direct sum space L(V) = L0 L1 L2 L3 + .... . (9.2.5) The claim then is that L(V) = T(V)/I where I = the ideal of Example 3 above. (9.2.6) This is then the space of all T(V) elements where all terms in which a vector is repeated are set to 0 and thus are not part of L(V). Notice that the quotient of Section 9.1 has a finer granularity. It deals with individual Lk Vk spaces, whereas Section 5.2 deals with the entire L(V) T(V). Many texts refer to Lk as Λk(V) and L(V) as Λ(V). We have reserved the Λ names for the dual spaces. The category theory approaches to Lk and L(V) are similar to the discussion of Section 1.2 with the main point being that Lk and L(V) are "universal" and therefore uniquely defined up to isomorphism. The role played by k-multilinear functions is played by antisymmetric k-multilinear functions. 10. Differential Forms In this chapter we consider aspects of the topic of differential forms from the viewpoint of Chapter 2 on the tensor algebra of transformations, and Chapter 8 on the exterior algebra of wedge products. 10.1. Differential Forms Defined A differential form is in fact just an element of the wedge space Λk(V) described in Chapter 8. Recall that our most general element of Λk(V) was written in symmetric sum notation as (sums run 1 to n = dimV) , T^ = Σii....i Tii....i (λi ^ λi .....^ λi) ir = 1 to n n = dim(V) (8.4.4) T^ = ΣITIλ^I . (10.1.1) This sum is redundant since each basis vector appears k! times. In an ordered sum form, each basis vector of Λk appears only once, T^ = Σ1≤i<i<....<i≤n Aii...i (λi ^ λi .....^ λi) (8.4.7) T^ = Σ'IAIλ^I (10.1.2) where Σ'I indicates the ordered summation. If one is given Tii....i in the first expansion, a viable expression for Aii...i which makes the second expansion valid is this A = k!Alt(T) (8.4.16) (10.1.3) which is a shorthand notation for AI = k! Alt(TI) (10.1.4) which in turn means Aii...i = k! Alt(Tii...i) = Tii....i - Tii....i + all other signed permutations . (10.1.5) Whereas T is an arbitrary rank-k tensor, A obtained from T in (10.1.4) is a totally antisymmetric rank-k tensor. On the other hand, if one is given Aii....i in (10.1.2), it is likely that Aii....i is not a totally antisymmetric tensor. One might have, for example, Aii....i = ∂iBi....i. The sum in (10.1.2) only "senses" the values of Aii....i = AI for values of I which are ordered, and AI for non-ordered I play no role. Given some Aii....i in (10.1.2) a viable expression for Tii....i in (10.1.1) is this, Tii...i = (10.1.6) TI = AI θ(I=ordered) // θ(bool) = 1 if bool true else 0 since then ΣITIλ^I = ΣI[ AI θ(I=ordered) ] λ^I = Σ'I AIλ^I . (10.1.7) We shall take the vector space V in Λk(V) to be V = Rn. Below we shall treat the objects TI and AI as rank-k tensor fields with an argument in Rn, so we will then have for example, Aii...i(x) x ϵ Rn ( "x-space") . (10.1.8) In the usual presentation of the theory of differential forms, the dual-space basis vector λi is given the purely cosmetic name dxi dxi ≡ λi = <ui| = (ui)T , ui = axis-aligned basis vectors of Rn (10.1.9) where λi was a notation introduced in (2.11.c.2). This object dxi is very different from the normal calculus differential dxi, and for that reason we write dxi in a red italic font. For example, one can then write, dxi(v) = λi(v) = <ui| v> = vi . (10.1.10) In contrast, there is no calculus differential object called dxi(v). The differential forms (elements of Λk) shown above in (10.1.1) and (10.1.2) are now written in cosmetic notation as T^ = Σii....i Tii....i ( dxi ^ dxi .....^ dxi) T^ = ΣITI dx^I (10.1.11) T^ = Σ1≤i<i<....<i≤n Aii...i ( dxi ^ dxi .....^ dxi) T^ = Σ'IAI dx^I . (10.1.12) We have used the hat subscript notation to distinguish dual tensors in V*k from those in Λk(V) , λI = λi λi ..... λi // basis vector in dual space V*k λ^I = λi ^ λi .....^ λi // basis vector in dual space Λk(V) . (10.1.13) The traditional names for differential forms are α, β and so on, so we take T^ → α and write our arbitrary differential form (10.1.12) now as α = Σ'I fI(x) λ^I = Σ'I fI(x) dx^I α ϵ Λk(V) V = x-space = Rn (10.1.14) where fI is the more traditional name for AI. Once again, V = Rn, Euclidean space, where the basis vectors ui = |ui> are independent of x, and so the λi = <ui| are also independent of x. Comment: Since our monograph deals with both tensor products and wedge products, we feel it is useful to maintain the distinction between λ^I and λI, or between dx^I and dxI . Most discussions of differential forms involve only wedge products and the corresponding wedge spaces, so they write dx^I as dxI . And of course they don't use our red italic notation, so the final result is just dxI. Furthermore, many presentations don't show the wedge product ^ symbols, so one sees dxI = dx1dx2 which we would write as dx^I = dx1 ^ dx2 . Our use of italics is only to maintain the form/calculus distinction for black and white printed copies of this document. 10.2. Differential Forms on Manifolds Chapter 2 was concerned with the general transformation x' = F(x) where x-space and x'-space both had the same dimension N. Here we shall be considering x-space = Rn and x'-space = Rm with n ≤ m. If we allow x to exhaust some dimension-n region U in x-space, the image x' = F(x) will exhaust some region V in x'-space of dimension n ≤ m. If Fi and its derivatives are "smooth", the region V in x'-space will lie on a "manifold" M which is embedded in x'-space. Here is a crude graphical representation, (10.2.1) Here we have in effect reflected Picture A (2.1.1) left to right to get (10.2.2) One can define a differential form αx' at a point x' ϵ M in this way, αx' = Σ'I fI(x') λ'^I = Σ'I fI(x') dx'^I = Σ'I fI(x') <e'^I| ϵ Λ'k ≡ Λk(Rm) (10.2.3) where λ'^I = <e'^I| is based on (2.11.c.11). Recall that the e'i are axis-aligned basis vectors in x'-space. We think of this differential form αx' as "being in dual x'-space" to which we give the name Λ'k . The manifold M is a "surface" of dimension n within Rm with n ≤ m. The manifold M could be some full chunk of Rm (or all of Rm), in which case it has dimension n = m. If the manifold is a "hypersurface" in Rm it then has dimension n = m-1. In general M is some n-dimensional "surface" embedded within Rm where 1 ≤ n ≤ m. Recall from (2.5.1) and Fig (2.5.4) that the x'-space basis vectors u'i = R ui are the tangent base vectors for the inverse transformation x = F-1(x'). For m > n this inverse transformation only exists for points x' on manifold M, and, for u'i with i = 1 to n, the u'i continue to be tangent base vectors. The remaining u'i for i = n+1 to m can be defined "as needed" to provide a full basis in x'-space Rm . By the definition of M as the mapping image, we know that the first n u'i are "tangent to" the surface M, meaning that tiny arrows ε u'i(x') for ε << 1 lie on/in M at point x'. Since the full basis is by definition complete (elements are linearly independent), the remaining u'i(x') for i = n+1 to m are all "normal to" the surface M. This is all specific to some point x' on M. For example, for a manifold that is a smooth non-self-intersecting 3D curve embedded in R3, one would have u'1 being tangent to the curve at x', and then u'2 and u'3 are both normal to the curve at x'. On the other hand, if M is a 2D surface in R3, u'1 and u'2 will be tangent to the surface M and u'3 will be normal to that surface, all at point x' on M. The set of n linearly independent basis vectors {u'1...u'n} which are tangent to M at x' are first thought of as having their tails right at the point x' on M. When these vectors are translated so their tails are all at the origin, the {u'1...u'n} then span an n-dimensional vector space. This vector space is usually written Tx'M and is called the tangent space to M at point x' on M, dimension n. As with any vector space, there is a corresponding dual space. The dual space to the tangent space is called the cotangent space and it is the set of all rank-n linear functionals of vectors in Tx'M. The name cotangent is like the name covector mentioned below (2.11.a.3) and has nothing to do with the cotangent of any angle. The conglomeration of all the tangent spaces Tx'M on M has the structure of a fiber bundle and is often called the tangent bundle. There is a corresponding dual cotangent bundle. See Spivak [1999] Chapter 3 or wiki on tangent bundles. As one moves from x' to a nearby point x' + dx' on M, the basis vectors in general will move slightly (M is "smooth"). The dual basis vectors u'i of course also move to maintain u'i u'j = δij. Thus we have λ'i = <u'i| also depending on x'. We don't want to write this λ'i as λ'(x') because then we have to write <u'i|v> = (λ'i(x'))(v) which is rather messy (although Spivak uses this kind of notation with x' = p in various places). We hesitate to write the left side αx' in (10.2.3) as α(x') because this makes α look like a function, but it is in fact a differential form. Notice one benefit of the cosmetic notation λ'^I = dx'^I. The dependence on x' can be regarded as being implied by writing dx' instead of say dy. It is customary to abbreviate αx' as just α with the understanding that it is at some point x' on M. In proofs below we sometimes call it α' since it is a differential form in dual x'-space. As already noted, a simple example of a manifold is a non-self-intersecting and "smooth" finite piece of 3D curve hanging in R3 which is defined by some function x' = F(x) where x is a scalar parameter which marks points on the curve. In this case αx' is a differential 1-form defined at every point x' along that curve, and the tangent space at any point x' as noted is one dimensional and contains the tangent vector to the curve at that point on the curve. Our second example of a manifold is a non-self-intersecting and "smooth" finite piece of 2D surface hanging in R3 which is defined by some function x' = F(x) with x = (x1,x2) where every point on the surface is marked by a unique value of x. Perhaps this surface is a piece of a toroidal surface or sphere. In this case αx' is a differential 2-form defined at every point x' on that surface. The tangent space at any point x' on M is 2 dimensional. See Sjamaar Chapter 6 or elsewhere for a formal definition of a manifold and smoothness. A manifold is roughly a smooth "surface" which can be cobbled together from a set of smooth mappings x' = Fi(x) which are said to cover the manifold, the way an atlas of flat maps can cover the entire globe of the Earth. A manifold is a "surface" which is locally smooth in the region of any point x' on the manifold. Since x' = F(x) must be 1-to-1 between the parameter x-space and x'-space, the manifold cannot be self-intersecting. That is to say, such a point of intersection in x'-space must back-map into at least two different points in x-space. Each mapping has some open domain Ui in Rn and one writes Fi: Ui → M and Fi must be 1-to-1 as noted. But (∂Fi/∂tj) : Ui→M must also be 1-to-1 to provide clean differentiability at all points on M and in all directions from any such point. This is often stated as (DFi) must be 1-to-1. We discuss manifolds a bit more in Section 10.10. 10.3. The exterior derivative of a differential form Motivation The exterior derivative dα plays a key role in the theory of differential forms, as does the notion of the boundary ∂M of a manifold M. Although we shall not derive it, Stokes' Theorem for differential forms says ∫M dα = ∫∂M α . Here α is a k-form, and dα is the exterior derivative of α which we shall see below is a (k+1)-form. The main work involved in proving this theorem involves not so much an understanding of dα as it does dealing with an explicit definition of the boundary ∂M in an arbitrary number of dimensions including issues of orientation. The single statement above encompasses a large set of theorems from analysis only one of which bears the specific Stokes' Theorem moniker: ∫M gdx = g(c(1))-g(c(0)) = g(b) - g(a) "Line integral of a Gradient" theorem // α = 0-form, dα = 1-form, M = curve, ∂M = 2 curve endpoints) ∫M (∂xg - ∂yf) dxdy =∫∂M (fdx + gdy) Green's Theorem in the plane // α = 1-form, dα = 2-form, M = planar area, ∂M = its bounding curve ∫M [curl (F n)] dA = ∫∂M Fdx Stokes's theorem of analysis // α = 1-form, dα = 2-form, M = non-planar area, ∂M = curve bounding that area ∫M [div F] dx1dx2..dxn = ∫∂M (F n) dA(n-1) divergence theorem (Gauss's theorem for n=3) // α = (n-1)-form, dα = n-form, M = "volume", ∂M = "surface" bounding that volume The sudden appearance of familiar objects like the grad, curl and divergence is part of the Hodge * dual operator "correspondence" we mentioned below (4.3.18). In that correspondence one has α = f 0-form in Rn α ↔ f dα = df 1-form in Rn dα ↔ f α = f 0-form in Rn α ↔ f *d[*(dα)] = ∂i2f = 2f 0-form on Rn *d[*(dα)] ↔ 2f α = F dx 1-form on Rn α ↔ F *[d(*α)] = (div F) 0-form on Rn *[d(*α)] = div F α = F dx 1-form in R3 α ↔ F *(dα) = [curl F] dx 1-form in R3 *(dα) ↔ curl F where one sees various appearances of the exterior differential operator d on the left side. The action of the Hodge * is this: dx^I = some ordered multi-index wedge product of k dxi in Rn (a basis vector k-form) (*dx^I) = (sign) dx^Ic = ordered wedge product of the missing dxi within Rn (c = complement) Example: dxI = dx2 ^ dx4 in R6 *dxI = (sign) dx1 ^ dx3 ^ dx5 ^ dx6 *(*dx^I) = (-1)kn+k dx^I (sign) = (-1)a+b+..+q (-1)k(k+1)/2 where dxI = dxa ^ dxb ^ ....^ dxq a < b < ... < q Again, we do not derive these facts, but refer the reader to Sjamaar, Buck, Spivak or other sources. Our intention is to provide the reader with some motivation for slogging through the rest of this section on "d". Definition of the Exterior Derivative In Section 10.1 we noted that TI(x) = Tii....i(x) and AI(x) = Aii...i(x) were rank-k tensor fields with respect to some unspecified Chapter 2 transformation x' = F(x) and dx' = Rdx. We now regard these objects as being just scalar-valued functions which happen to have label I. We refer to either of these functions for the moment as f(x). Such a function by itself is a 0-form because it has no λi factors. That is, the object f , f = f ϵ Λ0, (10.3.1) is a differential 0-form (abbreviated 0-form). The exterior derivative of such a 0-form is written df and is defined as df ≡ Σj=1n [∂f(x)/∂xj] λj = Σj=1n [∂jf(x)] λj . (10.3.2) Here we put df in red italic so it won't be confused with a calculus differential df of a function f(x). We could have written the 0-form f as f , but since then f = f there is no reason to do so. The first thing we notice is that, since f is a 0-form, df is a 1-form because the sum is a linear combination of single λj dual basis vectors Using the cosmetic notation defined above, we then write (10.3.2) as, df = Σj=1n [∂jf(x)] dxj . (10.3.3) Now we begin to see the motivation for the cosmetic notation dxj. The above equation looks just like the corresponding calculus equation df = Σj=1n [∂jf(x)] dxj . (10.3.4) In this last equation df(v) would make no sense, but in (10.3.3) one can write df (v) = Σj=1n [∂jf(x)] λj(v) = Σj=1n [∂jf(x)] vj . // (2.11.c.5) (10.3.5) The exterior derivative of a general differential form α has an extremely simple definition. Renaming AI in (10.1.2) to be the more traditional fI, we write α = Σ'I fI(x) λ^I general k-form α ϵ Λk dα ≡ Σ'I (dfI(x)) ^ λ^I = Σ'I ( Σj=1n [∂jfI(x)] λj) ^ λ^I // from (10.3.2) = Σ1≤i<i<...<i≤n Σj=1n [∂jfii...i(x)] λj ^ λi ^ λi ...^ λi . (10.3.6) Since there are now k+1 wedged dual basis vectors λr, this dα must be a (k+1)-form. So, Fact: If α is a k-form, then dα is a (k+1)-form. (10.3.7) We pause to take note of a fact that perhaps seems obvious: Fact: One can compute dα in the same manner for the ordered or the symmetric sum form of α. α = Σ'I AI(x) λ^I dα = Σ'I ( Σj=1n [∂jAI(x)] λj) ^ λ^I // ordered sum (10.3.8) α = ΣI TI(x) λ^I dα = ΣI ( Σj=1n [∂jTI(x)] λj) ^ λ^I // symmetric sum where we assume that the λ^I are constants in x. Proof: The only question here is whether the dα computed on the second line above is the same as the dα computed on the first line. Assume they are different and call the second line dα". Reorder to get, dα = Σj=1n λj ^ ( Σ'I [∂jAI(x)] λ^I) dα" = Σj=1n λj ^ ( ΣI [∂jTI(x)] λ^I) (10.3.9) But write (10.1.1) = (10.1.2) and then apply ∂j to both sides, ΣI TI(x) λ^I = Σ'I AIλ^I ΣI ∂jTI(x) λ^I = Σ'I ∂jAI(x) λ^I and thus the two right-side expressions in (10.3.9) are the same and so α" = α. QED So far we have shown that if α is a k-form, then dα is a (k+1)-form. What can be said about d2α ≡ d(dα) ? One might reasonably think this would be a (k+2)-form, but that is not correct. In fact: Fact: d2α = 0 for any k-form α (differential forms have zero "curvature") . (10.3.10) Since a differential form involves linear functionals, the above Fact seems intuitively reasonable. Proof: The proof is quite simple if we use the redundant symmetric sum (10.1.1) to express α. Then α = ΣITI(x) λ^I dα = ΣI(dTI(x)) ^ λ^I = ΣI ( Σr=1n [∂rTI(x)] λr) ^ λ^I = ΣIΣr=1n [∂rTI(x)] (λr ^ λ^I) d(dα) = ΣI Σr=1n d[∂rTI(x)] (λr ^ λ^I) = ΣI Σr=1n (Σs=1n∂s[∂rTI(x)] λs ) ^ (λr ^ λ^I) = ΣI Σr=1n Σs=1n[ ∂s∂rTI(x)] (λs ^ λr ^ λ^I) = 0 . QED The result is 0 because in the symmetric sum Σrs the object ∂s∂rFI(x) is symmetric under r↔ s while the object (λs ^ λr ^ λ^I) is antisymmetric under r↔s. That is to say, if S is symmetric and A antisymmetric, swap names r↔s use symmetries sum = Σrs SrsArs = Σsr SsrAsr = Σrs (+Srs)(-Ars) = - Σrs SrsArs = - sum = 0 (10.3.11) Expressing dα in standard form Recall from above that α = Σ'I fI(x) λ^I k-form dα = Σ'I ( Σj=1n [∂jfI(x)] λj) ^ λ^I . (k+1)-form (10.3.6) We wish to rewrite dα in a more standardized form. To this end, starting with the k-multiindex I we create a (k+1)-multiindex J as follows I = i1,i2.....ik J = j1,j2.....jk,jk+1 ≡ i1,i2.....ik, j // j = jk+1 J' = j1,j2.....jk = I = i1,i2.....ik (10.3.12) Then one can rewrite the summation appearing in dα above as Σ'I Σj = Σ'J' Σj = Σj<j<...<j Σj . (10.3.13) Then dα = Σj<j<...<j Σj [∂jfjj...j(x)] λj ^ λj ^ λj .... ^ λj . (10.3.14) Because each swap of vectors in a wedge product of same creates a minus sign, λj ^ λj ^ λj .... ^ λj = (-1)k λj ^ λj .... ^ λj ^ λj = (-1)k λ^J (10.3.15) and then dα = (-1)k Σj<j<...<j Σj [∂jfjj...j(x)] λ^J . (10.3.16) The summations appearing above can be written as Σj<j<...<j Σj = Σj<j...<j<j + Σj<j...<j<j + ... + Σj<j<j...<j . (10.3.17) Here we are just exhausting all possible locations that jk+1 can take relative to the other indices. We don't have to worry about cases like jk+1 = j2 because in that case λ^J = 0 and there is no contribution to the sum (10.3.16). One can then write, (-1)kdα = Σj<j...<j<j<j [∂jfjj...j] λj ^ λj .... ^ λj + Σj<j...<j<j<j [∂jfjj...j] λj ^ λj .... ^ λj + Σj<j...<j<j<j [∂jfjj...j] λj ^ λj .... ^ λj ..... + Σjj<j...<j<j [∂jfjj...j] λj ^ λj .... ^ λj (10.3.18) Next, define the following index subscript swap operator S(r,s), S(r,s) Fjj. j...j...j = Fjj. j...j...j . (10.3.19) In each term in (10.3.18) all the summation indices are of course dummy indices and their names can be swapped around at will. Notice that : S(k,k+1)[Σj<j...<j<j<j] = Σj<j...<j<j<j S(k,k+1)S(k-1,k+1)[Σj<j...<j<j<j ] = S(k,k+1)[Σj<j...<j<j<j] = Σj<j...<j<j<j S(k,k+1)S(k-1,k+1)S(k-2,k+1) [ Σj<j...<j<j<j<j] = S(k,k+1)S(k-1,k+1) [ Σj<j...<j<j<j<j] = S(k,k+1) [ Σj<j...<j<j<j<j] = Σj<j...<j<j<j ..... S(k,k+1)S(k-1,k+1)....S(1,k+1)[Σjj<j...<j<j] = Σj<j...<j<j<j (10.3.20) Thus these swap combinations convert each summation to the standard form shown in the first line of (10.3.18). So the next step is to apply the swap combinations not just to the summations, but to the entire lines shown in (10.3.18), since one is allowed to do this without changing each line's value since these are just dummy index swaps. The first effect of doing this is that all the summations forms become that shown on the first line which is just Σ'J. The second effect is that the λ wedge products can be restored to their first-line ordering by adding a minus sign for each swap. For example, S(k,k+1)S(k-1,k+1) λj ^ λj .... ^ λj = (-1)2 λj ^ λj .... ^ λj or S(k,k+1)S(k-1,k+1) λ^J = (-1)2 λ^J . (10.3.21) Since on each line going down the number of swaps increases by 1, we pick up alternating signs. Doing this, one can rewrite (10.3.18) as (-1)kdα = Σ'J [∂jfjj...j] λ^J - Σ'J [S(k,k+1){∂jfjj...j}] λ^J + Σ'J [S(k,k+1)S(k-1,k+1){∂jfjj...j}] λ^J ..... + (-1)k Σ'J [S(k,k+1)S(k-1,k+1)....S(1,k+1){∂jfjj...j}] λ^J . (10.3.22) Here then is the way to write the derivative of a k-form in standard form: α = Σ'I fI(x) λ^I k-form dα = (-1)k Σ'J QJ(x) λ^J (k+1)-form (10.3.23) QJ = [ 1 - S(k,k+1) + S(k,k+1)S(k-1,k+1) - ... +(-1)kS(k,k+1)S(k-1,k+1)...S(1,k+1)] ∂jfjj...j . The general result is admittedly unwieldy and perhaps has some more pleasant form, but we take it as is and consider some simple examples. Example 1: Exterior derivative of a 1-form: α = Σ'I fI(x) λ^I = Σi fi(x) λi QJ = [ 1 - S(1,2) ] ∂jfj = (∂jfj- ∂jfj) dα = (-1)k Σ'J QJλ^J = - Σj<j (∂jfj- ∂jfj) λj ^ λj = Σj<j (∂jfj - ∂jfj) λj ^ λj . (10.3.24a) In cosmetic notation, α = Σi fi(x) dxi dα = Σj<j (∂jfj - ∂jfj) dxj ^ dxj (10.3.24b) dα = Σi<j (∂ifj - ∂jfi) dxi ^ dxj which agrees with Sjamaar p 21 (2-2). Example 2: Exterior derivative of a 2-form: α = Σ'I fI(x) λ^I = Σi<i fii(x) λi ^ λi QJ = [ 1 - S(2,3) + S(2,3)S(1,3) ] ∂jfjj S(2,3)∂jfjj = ∂jfjj S(2,3)S(1,3) ∂jfjj = S(2,3)∂jfjj = ∂jfjj QJ = ∂jfjj - ∂jfjj+ ∂jfjj dα = (-1)k Σ'J QJλ^J = + Σj<j<j (∂jfjj - ∂jfjj+ ∂jfjj ) λi ^ λi ^ λi . (10.3.25a) In cosmetic notation, α = Σi<i fii(x) dxi ^ dxi dα = Σj<j<j (∂jfjj - ∂jfjj+ ∂jfjj ) dxj ^ dxj ^ dxj (10.3.25b) dα = Σi<j<k (∂ifjk - ∂jfik + ∂kfij ) dxi ^ dxj ^ dxk which agrees with Sjamaar p 21 (2-4). Example 3: (last one!) Exterior derivative of a 3-form: α = Σ'I fI(x) λ^I = Σi<i<i fiii(x) λi ^ λi^ λi QJ = [ 1 - S(3,4) + S(3,4)S(2,4) - S(3,4)S(2,4)S(1,4) ] ∂jfjjj S(3,4)∂jfjjj = ∂jfjjj S(3,4)S(2,4)∂jfjjj = S(3,4)∂jfjjj = ∂jfjjj S(3,4)S(2,4)S(1,4)∂jfjjj = S(3,4)S(2,4)∂jfjjj = S(3,4)∂jfjjj = ∂jfjjj QJ = ∂jfjjj - ∂jfjjj + ∂jfjjj - ∂jfjjj dα = (-1)k Σ'J QJλ^J (10.3.26a) = Σj<j<j<j (∂jfjjj - ∂jfjjj+ ∂jfjjj - ∂jfjjj) λi ^ λi ^ λi ^ λi . In cosmetic notation, α = Σi<i<i fiii(x) dxi ^ dxi ^ dxi (10.3.26b) dα = Σj<j<j<j (∂jfjjj - ∂jfjjj+ ∂jfjjj - ∂jfjjj) dxj ^ dxj ^ dxj ^ dxj dα = Σi<j<k<l (∂ifjkl - ∂jfikl + ∂kfijl - ∂lfijk) dxi ^ dxj ^ dxk ^ dxl . We leave it to the reader to deduce a "general rule by inspection" for the series of terms for any k. This might involve rotations of certain subsets of the subscripts. 10.4. Commutation properties of differential forms Recall these three results from Chapter 8 concerning elements of Λ(V), S^^ T^ = (-1)kk'T^^ S^ ranks of the two dual tensors are k and k' . (8.9.c.6) In a product of tensors T1^T2^T3.... of rank k1, k2, k3 ... , if two tensors are swapped Tr ↔ Ts (with r < s), the resulting tensor incurs the following sign relative to the starting tensor, sign = (-1)m where m = (kr+1+kr+2 ...+ks-1)(kr+ks) + krks (8.9.e.6) T^N = 0 for any N ≥ n+1 assuming k ≠ 0 . (8.9.d.9) In the language of differential forms these three results become α ^ β = (-1)kk'β ^ α α = k-form, β = k'-form (10.4.1) α1 ^ α2 ^ ... αr ... αs ... ^ αk = (-1)m α1 ^ α2 ^ ... αs ... αr ... ^ αk where m = (kr+1+kr+2 ...+ks-1)(kr+ks) + krks (10.4.2) αN = 0 for N ≥ n+1 dim(V) = n α = any k-form with k ≥ 1 where αN ≡ α ^ α ... ^ α . (10.4.3) Equations (10.4.1) and (10.4.3) appear in Sjamaar as "2.1 Proposition" on page 19 and the preceding equation. In Sjamaar, Buck and many other source all ^ symbols are suppressed so (10.4.1) is written αβ = (-1)kk'βα and one must understand that these are wedge products in Λ(V). 10.5. Closed and Exact, Poincaré and the Angle Form Closed: If dα = 0 for a k-form α, α is said to be closed. The analogous fact for a function f(x) with df = 0 would be that f(x) = constant. (10.5.1) Exact: Sometimes one finds that a form α can be written α = dβ where β is some other form. If α is a k-form, we know from (10.3.6) that β must be a (k-1)-form. When α = dβ for some form β, α is said to be exact. We showed in (10.3.12) that d2β = 0 for any form β, so it follows that if α = dβ, then dα = 0 and α is closed. Thus we have shown that : (10.5.2) Fact: If α is exact, then α is closed. (10.5.3) In 1D calculus if f = dh/dx one says that dh = f dx is an "exact (perfect) differential" and one then writes !Syntax Error, If(x) dx = !Syntax Error, I() dx = !Syntax Error, Idh = h(a) - h(b) dh = () dx . (10.5.4) In nD calculus if f = h one says that dh = h dx is an exact (perfect) differential. The above integral then becomes a line integral over a smooth curve c having endpoints a and b, !Syntax Error, If(x) dx =!Syntax Error, Ih dx = ∫c dh = h(a) - h(b) where dh = h dx = Σi=1n (∂ih(x))dxi = Σi=1n fi(x) dxi = f(x) dx . (10.5.5) The line integral depends only on the line endpoints a and b, and not on the particular shape of the curve c joining a and b. For a closed curve a = b and we find ∫C dh = h(a) - h(b) dh = h(a) - h(a) = 0 . (10.5.6) In physics if f(x) is a "conservative force field" (like gravity) then h(a) - h(a) = 0 is the work done in moving a particle that senses the field (has mass) around a closed path. We shall see below a similar theorem for α = dβ where β is a 0-form (a function) and α is 1-form: ∫c α = ∫c dβ = g(F(a)) - g(F(b)) if α = dβ (α is exact) (10.5.7) where g is a certain function related to α, and where x' = F(x) describes a curve c in x'-space for scalar x in [a,b]. Thus for a closed curve we get α = dβ = g(F(a)) - g(F(a)) = 0 if α = dβ (α is exact) (10.5.8) In some sense, a 1-form α being exact is like dh being an exact differential. Fact (10.5.3) above says α exact α closed. Is it possibly also true that α closed α exact and so then the two descriptions are one in the same? The answer is "not quite" as expressed in this claim: Poincaré Lemma: If any differential form α on Rn is closed for x in some open star-shaped domain in Rn which includes the origin, then α is exact. (10.5.9) This Lemma appears on p 38 of Spivak from which we quote, and Spivak proceeds to give a detailed proof. In topological language, the star-shaped domain is any domain that is "contractible to a point". Certainly the Lemma is valid for a domain which is an open "cube" or "sphere" (n dimensions) about the origin. The domain need not be convex (as the star shows). The classic example of this theorem involves the so-called angle form defined on R2 with coordinates (x1,x2), α = Σi=12 fi(x)λi where f1(x) = - (x2/r2) r2 = x12 + x22 f2(x) = (x1/r2) (10.5.10) Then dα = Σi dfi(x)λi = Σij (∂jfi) λj ^ λi . Notice that, using the fact that ∂ir = xi/r, (∂1f2) = ∂1(x1/r2) = [ r2 * 1 - x1 (∂1r2)] / r4 = [ r2 - x12r (∂1r)] / r4 = - [r2 - x12r(x1/r)] / r4 = [r2 - 2x12] / r4 = [x12 + x22 - 2x12] / r4 = (x22 - x12) / r4 and (∂2f1) = - ∂2 (x2/r2) = - [ r2 * 1 - x2 (∂2r2)] / r4 = - [ r2 - x22r (∂2r)] / r4 = - [r2 - x22r(x2/r)] / r4 = - [r2 - 2x22] / r4 = - [x12 + x22 - 2x22] / r4 = (x22 - x12) / r4 = (∂1f2) . Thus it turns out that the quantity (∂jfi) is symmetric under i ↔ j. Then by the argument (10.3.14) we get dα = Σij (∂jfi) λj ^ λi = Σij (Sij)(Aji) = 0 α = closed so α is a closed 2-form. As we shall show below, the line integral of α around a circle centered at the origin gives α = 2π. Thus the angle form is not exact because if it were one would have α = 0 as in (10.5.8). So here is a form α which is closed, but which is not exact. The condition of the Poincaré Lemma must therefore be violated, and that is indeed the case since the form α is undefined for r = 0 where f1 and f2 blow up, so α is then defined on R2 punctured at the origin, sometimes written R2/ {0} or R2 - {0}. Thus we can't have any open star-shaped set including the origin for α, so Poincaré's Lemma does not apply. Our plan now is first to define the "pullback" of a differential form, and then in later sections to use the pullback to define the meaning of integration of a differential form over a manifold. But we wish to show how the notion of a pullback fits into the general transformation scenario of Chapter 2, and this requires several digressions before we get to the pullback discussion in Section 10.7 and 10.9. 10.6 Transformation Kinematics Much mathematical hardware accompanies a mapping. In mechanics, the selection of an appropriate set of coordinates and corresponding basis vectors is sometimes referred to as stating the kinematics of a problem (as opposed to the dynamics which involves equations of motion). Here we apply this term loosely to the cloud of equations associated with a mapping. Not all these equations will be used in our analysis, but we like being able to see them all in one place just in case something is needed. In the following Sections we shall move in and out of the Dirac notation of Section 2.11 in a somewhat repetitive fashion intended to make the reader more comfortable with that notation. We feel that the Dirac notation is the safest notation in terms of avoiding wrong interpretations of rank-1 and rank-2 tensor indices. The notion of a pullback is often presented as "something new", but the main point of the following sections is to show that the pullback operator is just the R/R matrix/operator of the underlying transformation. In Chapter 2 we discussed the transformation x' = F(x) from x-space to x'-space using Picture A (2.1.1). The vector transformation and "the differential" (the R-matrix) of the transformation were given by V'a = RabVb Rab ≡ (∂x'a/∂xb) = ∂bx'a ≡ (F)ab ≡ (DF)ab ≡ (DF)ab (2.1.2) dx'a = Rabdxb dx' = Rdx . (2.1.12) (10.6.2) Here V'a = RabVb shows the transformation of a contravariant vector under x' = F(x). In matrix notation one would write V' = RV. Repeated indices are always summed unless otherwise stated. Above we have defined F and DF as alternate names for matrix R because many authors (like Spivak) use this notation. In Tensor (E.4.4) we show that this is in fact a "reverse dyadic notation". Often (DF)ab is written unbolded (DF)ab so then R = (DF) with the idea that a matrix like R is normally not bolded. (a) Axis-Aligned Vectors and Tangent Base Vectors : The Kinematics Package We gather here various facts derived in Chapter 2 which comprise our "kinematics package" for the transformation x' = F(x) . We cosmetically flip Picture A of (2.1.1) left to right. Rn Rm m ≥ n (a) x' = F(x) transformation Rij ≡ (∂x'i/∂xj) = ∂jx'i R = (DF) V' = R V vector Sij ≡ (∂xi/∂x'j) = ∂'jxi (2.1.2) (b) e'i with (e'i)j = δij axis-aligned basis vectors in x'-space (i = 1..m) (2.5.2) ei ei= Se'i tangent base vectors in x-space (i = 1..n) (2.5.1) (c) ui with (ui)j = δij axis-aligned basis vectors in x-space (i = 1..n) (2.4.1) u'i u'i= Rui tangent base vectors in x'-space (i = 1..n) (2.5.1) (u'i)j = Rjk (ui)k (2.11.g8) (d) 1' = | e'i> <e'i| = | e'i> <e'i| = | u'i> <u'i| = | u'i> <u'i| completeness in x'-space 1 = | ei> <ei| = | ei> <ei| = | ui> <ui| = | ui> <ui| completeness in x-space (e) (uj)i = ui uj = <ui | uj > = gij = u'i u'j = <u'i | u'j > (ej)i = ui ej = <ui | ej > = Sij = Rji (e'j)i = e'i e'j = <e'i | e'j > = g'ij = ei ej = <ei | ej > (u'j)i = e'i u'j = <e'i | u'j > = Rij = Sji (2.5.8) (f) ei = g'ij ej e'i = g'ij e'j ui = gij uj u'i = gij u'j (2.3.2),(2.4.2),(2.5.6) ei = g'ij ej e'i = g'ij e'j ui = gij uj u'i = gij u'j (g) <ej | S | e'i> = <e'i | R | ej> = g'ij // <e'i | Rej> = <e'i | e'j> = g'ij from (e) <ej | S | u'i> = <u'i | R | ej> = Sij = Rji // <u'i | Rei> = <u'i | e'j> = Rji from (e) <uj | S | e'i> = <e'i | R | uj> = Rij = Sji // <e'i | Rui> = <e'i | u'j> = Rij from (e) <uj | S | u'i> = <u'i | R | uj> = gij . // <u'i | Rui> = <u'i | u'j> = gij from (e) (h) S = RT Sij = (RT)ij = Rji (2.11.f3), (2.11.f1) R = ST Rij = (ST)ij = Sji (i) S = R-1 R = S-1 RS = SR = 1 (2.11.f3) RRT = RTR = SST = STS = 1. (10.6.a.1) In any equation, any index or label can be raised or lowered on both sides. The object gij is the tensor-correct form of gij = δij = δi,j , allowing for indices to be raised and lowered, see (2.2.2). Here is a sample Dirac notation manipulation using the above information (implied sum in completeness), |ei> = [1] |ei> = | uj> <uj|ei> = | uj>Rij = Rij |uj> or ei = Rijuj . (10.6.a.2) The result ei = Rijuj appears in (2.4.4) showing that Rij is the basis change matrix between these two sets of basis vectors. Notice that an equation like ei = Σj=1n Rijuj is a "vector sum equation" since it has a sum of vectors on the right side. No component indices appear on the vectors in this equation (i and j are labels). As discussed in Section 2.11 (g), abstract operators in the Dirac space will be written in script font. The operator R for example is completely determined by all its matrix elements <e'i| R | uj> = Rij. The identity operator in a Dirac space we then write as 1 for x-space and 1' for x'-space, as appear in the completeness statements of (10.6.a.1) item (d). (b) What happens for a non-square tall R matrix? In Chapter 2 and in Tensor it was assumed that x' = F(x) was an invertible mapping F: RN→RN . Now however we wish to consider the non-invertible mapping x' = F(x) where F: Rn → Rm m > n F: x-space → x'-space x ϵ Rn, x' ϵ Rm F(x) = x' . (10.6.b.1) In Rab = (∂x'a/∂xb) the row index a ranges 1 to m, while column index b ranges 1 to n. Thus the down-tilt R matrix is a "tall" non-square matrix having m rows and n columns with m > n. As outlined in Section 10.2. if we let the variable x exhaust some domain U within x-space, the mapping x' = F(x) generates a "surface" embedded within x'-space = Rm which has dimension n. We assume that the mapping F has sufficient smoothness properties so that this surface can be called a Manifold M. Thus, the mapping x' = F(x) is defined in effect for all x in Rn (or perhaps for a region U in Rn as in Fig (10.2.1), and produces (as its image) the manifold M within Rm . The inverse mapping x = F-1(x') is then only defined for points x' on the manifold M. For such points, the mapping and its inverse are assumed one-to-one. This inverse mapping is a set of n equations which one can presumably write down. The equations represent x = F-1(x') only when x' lies on M. For other values of x', the set of equations still exists but no longer represents the inverse function x = F-1(x'). This point is hopefully clarified by some Examples. Example 1: Let U be a square in R2 x-space with corners (-1,-1) to (1,1). We map this square into R3 using the following map x' = F(x): x'1 = x1 x'2 = x2 x'3 = x' = F(x) (10.6.b.2) The image in R3 x'-space is a partial upper hemispherical surface of radius 2 (see below). What is the inverse mapping x = F-1(x') ? One can take it to be the first two lines above, x1 = x'1 x2 = x'2 x = F-1(x') (10.6.b.3) but the inverse mapping only applies to points x' on the hemisphere. The above two equations of course exist for points x' not on the hemisphere, but they only act as the inverse mapping for points on the hemisphere. Here is Maple code for Example 1. The transformation is first entered and plotted, xp = x' : Maple then computes the "tall" R matrix, Rij ≡ (∂x'i/∂xj), . The S matrix Sij= (∂xi/∂x'j) is computed by hand from (10.6.b.3) and is then entered into Maple. Maple then computes the matrix products RS and SR, Notice that RS ≠ 1 while SR = 1. Example 2: Let U be the same square as in Example 1, but the new mapping is this x'1 = x1 + 2x2 1 x'2 = 2x1 + x2 2 x'3 = x1 + 3x2 3 x' = F(x) (10.6.b.4) The image in R3 x'-space is a tilted plane passing through the origin. We reuse the above Maple code for this example, but don't display the Maple output. What is the inverse mapping x = F-1(x') ? If one solves the first two equations for x1 and x2 the result is x1 = -1/3 x'1 + 2/3 x'2 x2 = 2/3 x'1 - 1/3 x'2 x = F-1(x') (10.6.b.5) and this then can be taken to be the inverse mapping x = F-1(x'). Inserting these expressions into the third equation gives 5/3 x'1 - 1/3 x'2 - x'3 = 0 (10.6.b.6) which is the equation of the tilted image plane passing through the origin whose normal is (5/3,-1/3,-1). On the other hand, if one instead solves the second two equations in (10.6.b.4) one finds x1 = 3/5 x'2 - 1/5 x'3 x2 = - 1/5 x'2 +2/5 x'3 x = F-1(x') . (10.6.b.7) Notice that this inverse mapping is different from (10.6.b.5). When these two expressions are inserted into the first equation of (10.6.b.4), one gets x'1 - 1/5 x'2 - 3/5 x'3 = 0 (10.6.b.8) Multiplication by 5/3 gives (10.6.b.6) so this is, of course, the equation for the same tilted plane. In this Example we find that the inverse equation set x = F-1(x') is not unique. If we work with the first and third equations in (10.6.b.4) we get a third set of inverse equations which we leave to the reader. By visual inspection, the R matrix computed from x' = F(x) (10.6.b.4) is this: R = Rab = (∂x'a/∂xb) = (10.6.b.9) and is the "tall" R matrix for this example. For the two inverse transformations stated in (10.6.b.5) and (10.6.b.7) we compute an S matrix, again by inspection (Maple did the products on the right) S = Sab = (∂xa/∂x'b) = SR = = (10.6.b.10) S = Sab = (∂xa/∂x'b) = SR = = Thus we have found two different "left inverses" S of the tall matrix R. If we try out these S matrices on the right of R, we find RS = = ≠ RS = = ≠ (10.6.b.11) Example 2 serves then to illustrate that a tall R matrix might have multiple left inverses, but those left inverses are not also right inverses. It turns out that there are in fact no right inverses for a tall R, as shown in section (c) below. Before leaving this example, we comment on the "coordinate lines" in x-space using our first inverse solution (10.6.b.5). x1 = -1/3 x'1 + 2/3 x'2 x2 = 2/3 x'1 - 1/3 x'2 x = F-1(x') (10.6.b.5) If we vary only x'1 (keeping the other two coordinates in x'-space fixed) both x1 and x2 vary, and not surprisingly they define a certain line in x-space, and this is the coordinate line in x-space for x'1 . If we instead vary only x'2, again both x1 and x2 vary and they define some other line in x-space, the x'2 coordinate line. If we vary only x'3 , then x1 and x2 do not vary and this coordinate line is just a point! Recall that the tangent base vectors en are tangent to the coordinate lines in x-space. As shown in (10.6.a.1) (c) one has (ej)i = Sij so the tangent base vectors are the columns of S, S = [e1, e2, e3]. Looking at S = for our first inverse solution, we see that the first two tangent base vectors are indeed reasonable tangents to coordinate lines in x-space. Since the third coordinate line is just a point, it can have no tangent base vector, and in fact e3 = (0,0) which "resolves" this problem. (c) Some Linear Algebra for non-square matrices The linear algebra for non-square matrices is a topic often omitted in linear algebra presentations. Here we consider only the special case of two matrices where each has the shape of the transpose of the other, and we cherry-pick a few relevant theorems. As shown below, non-square matrices never have two-sided inverses, so one talks only about the possibility of such a matrix having a "right inverse" or a "left inverse". Consider then the following matrix products where we assume m > n : (10.6.c.1) A nameless matrix rank theorem states the following : Fact: rank(AB) ≤ min{rank(A), rank(B) } . (10.6.c.2) Consider first the upper part of Fig (10.6.c.1). Both S and R each have some rank ≤ n, since this is the smaller matrix dimension. The Fact then says rank(SR) ≤ n. Since SR is an n x n matrix, it could therefore have full rank n, and then it is possible that one could have SR = 1. This says that it is possible for R to have a left inverse S, and for S to have a right inverse R. Another nameless theorem states that if R has full rank n then in fact it has at least one left inverse S, and if S is full rank it has at least one right inverse R. The theorem does not say how to compute these inverses, nor does it suggest how many inverses there might be (a non-trivial problem). So, Fact: tall R has full rank R has at least one left inverse S wide S has full rank S has at least one right inverse R (10.6.c.3) In our Example 2 above, matrix R in (10.6.b.9) has full rank 2, so we know it has at least one left inverse S. We explicitly found two such left inverses S as shown in (10.6.b.10). Since each of these left inverses has R as a right inverse, we know (and confirm) that each S must have full rank 2. Thus, we know (and confirm) that two of the tangent base vectors en are linearly independent (these being columns of S). Now consider the lower part of Fig. (10.6.c.1). Fact (10.6.c.2) says rank(RS) ≤ n, but the matrix RS is m x m. Thus it cannot possibly have full rank m, so it can never be the m x m identity matrix. We may then conclude that R has no right inverses and S has no left inverses: Fact: tall R has no right inverses wide S has no left inverses (10.6.c.4) Corollary: A non-square matrix cannot have a two-sided inverse. (10.6.c.5) If we take S = RT, then the two matrices on the right in the drawing are RTR and RRT. Yet another matrix rank theorem says, Fact: rank(RRT) = rank(RTR) = rank(R). (10.6.c.6) If R has full rank n, then the small matrix RTR has rank n and so is full rank, det(RTR) ≠ 0, and RTR is invertible. But the m x m larger matrix RRT having rank n must have det(RRT) = 0 and is not invertible. Fact: If tall R has full rank n, then (RTR)-1 exists. For any tall R, (RRT)-1 does not exist. (10.6.c.7) With this in mind, another theorem says that if tall R is full rank, then we know one of its left inverses: Fact: If tall R has full rank n, then one left inverse is given by S = (RTR)-1RT . (10.6.c.8) Proof: By the previous fact we know (RTR)-1 exists, so SR = [(RTR)-1RT]R = (RTR)-1 (RTR) = 1 . Fact: If wide S has full rank n, then one right inverse is given by R = ST (SST)-1. . (10.6.c.8) Proof: Reader exercise. We mention in passing two other matrix theorems for arbitrary conforming matrices A,B,C: Fact: (Sylvester's Inequality) rank(A) + rank(B) ≤ rank(AB) + n where n is the conforming dimension (10.6.c.9) Fact: (Frobenius Inequality) rank(AB) + rank(BC) ≤ rank(ABC) + rank(B) (10.6.c.10) (d) Implications for the Kinematics Package The set of relations shown in (10.6.a.1) still stands for F: Rn→ Rm with its tall R matrix, with the exception of the last item (i), (i) S = R-1 R = S-1 RS = SR = 1 RRT = RTR = SST = STS = 1 . (10.6.a.1) This must be replaced by (i) SR = 1 SST = RTR = 1 (10.6.d.1) since RS ≠ 1 and two-sided inverses R-1 and S-1 do not exist for F: Rn→ Rm with m>n. A second implication is that certain items in the kinematics package are no longer unique. We have already seen that Sij is not unique, so anything depending on Sij is also not unique. Here is a list showing which objects are unique, and which are not: Metric tensors gij, gij unique g'ij unique, since g'ij = RiaRjbgab g'ij not unique, since g'ij = RiaRjbgab = SaiSbjgab and Sij not unique Transformation matrices Rij = Sji unique (tall R matrix from x' = F(x)) Rij = Sji unique since Rij = gjaRia and both gja and Ria are unique Rji = Sij not unique Rij = Sji not unique, since Rij = g'ia Raj and g'ia not unique Axis-aligned basis vectors (uj)i unique since (uj)i = gji (e'j)i unique since (e'j)i = g'ij (= δij) (uj)i unique since (uj)i = gji (e'j)i not unique since (e'j)i = g'ij (uj)i unique since (uj)i = gji (e'j)i unique since (e'j)i = g'ij (uj)i unique since (uj)i = gji (e'j)i unique since (e'j)i = g'ij (= δij) Tangent base vectors (ej)i not unique since (ej)i = Rji (u'j)i unique since (u'j)i = Rij (ej)i not unique since (ej)i = Rji (u'j)i not unique since (u'j)i = Rij (ej)i unique since (ej)i = Rji (u'j)i unique since (u'j)i = Rij (ej)i unique since (ej)i = Rji (u'j)i not unique since (u'j)i = Rij (10.6.d.2) (e) Basis vectors for the Tangent Space at point x' on M From (10.6.a.1) we select as a basis for x-space the set of n axis-aligned basis vectors ui, {ui} i = 1,2...n basis for x-space (ui)j = δij components of these basis vectors in x-space . (10.6.e.1) These map into a set of n tangent base vectors u'i in x'-space, u'i = R ui |u'i> = R |ui> (2.5.1) or (u'i)j = Rja (ui)a = Rja δia = Rji i = 1,2...n j = 1,2..m . (10.6.e.2) We know that u'i = R ui because this is the way any vector transforms: v' = R v. Since there are m basis vectors in x'-space, we define the rest of the u'i arbitrarily such that the m basis vectors {u'i} in Rm are linearly independent, so u'i = as needed i = n+1, n+2 .....m . (10.6.e.3) Note in (10.6.e.2) that (u'i)j = Rja(ui)a = Σa=1n Rja(ui)a is a "component sum equation", in contrast with the "vector sum equation" ei = Σj=1nRijuj appearing in (10.6.a.2). To summarize for u'i : u'i = . (10.6.e.4) We show just below that the first n u'i span the tangent space Tx'M. Since the remaining u'i must be selected so that the full set of m u'i is a basis for x'-space, we know that the higher m-n u'i must span the perp space (Tx'M) of the tangent space, and this space is said to have codimension m-n within Rm. Based on (10.6.e.2) that Rji = (u'i)j, one concludes that the columns of R** are the contravariant basis vectors u'i which span Tx'M. Each of these u'i has m components and R** has m rows. R** = [u'1, u'2 ....u'n] . (2.5.9) (10.6.e.5) As long as R** has full rank n, the columns are linearly independent so the u'i form a (complete) basis. We now show that the first n tangent base vectors u'i do in fact span the tangent space Tx'M. Assume that, as x ranges over some portion of x-space, the mapping x' = F(x) describes a "smooth surface" M embedded in x'-space, hopefully a manifold or a piece thereof. If we start at some x and move to x + dx in x-space, we move from some point x' on M to some nearby point x' + dx' on M. By the definition of M, this dx' lies on the surface M and so is tangent to the surface M at x' and thus lies in the tangent space Tx'M of M at point x'. Applying R to each of the n axis-aligned differentials dxi = dxi(ui) in x-space (no i sum), we thereby generate a set of n differential vectors dx'i = Rdxi in x'-space which are in effect a set of short basis vectors which span the tangent space Tx'M. Since dx'i = dx'i (u'i), we may take the basis vectors {u'i, i=1,2..n} as spanning Tx'M. The upper u'i are orthogonal to M and span the perp space (Tx'M) as noted. We know from the fact u'i u'j = δij that the up-label vectors {u'i, i=1,2..n} also form a basis for the tangent space Tx'M. This conclusion can be reached as well by raising all i indices in the previous paragraph. In this case, the set {u'i, i=n+1,n+2..m} are then all orthogonal to the "surface" M. These last paragraphs and (10.6.e.5) have shown that: Fact: The first n x'-space tangent base vectors u'i, which are the columns of full-rank R** , span the tangent space Tx'M at point x' on M, and this is true as well for the u'i . (10.6.e.6) 10.7 The Pullback Operator R and properties of the Pullback Function F* From (10.6.e.2), or just from the fact that vectors transform as v' = Rv, we know that u'i= Rui |u'i> = R |ui> i = 1,2..n . (2.5.1) (10.7.1) One can say that the n axis-aligned basis vectors ui in x-space are "pushed forward" by R to become the tangent-space-spanning vectors u'i in x'-space. Applying S to both sides and using (10.6.d.1) that SR = 1, one finds that ui= S u'i |ui> = S |u'i> i = 1,2..n . (10.7.2) From the package (10.6.a.1) item (h) we know that S = RT and S = RT for the corresponding Dirac operators, so the above may be written, ui= RT u'i |ui> = RT |u'i> i = 1,2..n . (10.7.3) Thus, while operator R "pushes forward" the |ui> to the |u'i>, the operator RT "pulls back" the |u'i> from x'-space into the |ui> in x-space, just reversing the first process. For the label-up u and e basis vectors one then has, u'i = Rui |u'i> = R |ui> i = 1,2..n push forward ui = RT u'i |ui> = RT |u'i> i = 1,2..n pull back e'i = Rei |e'i> = R |ei> i = 1,2..n push forward ei = RT e'i |ei> = RT |e'i> i = 1,2..n pull back . (10.7.4) Here is a picture, reminiscent of Fig (2.5.4) (but reversed left to right), showing the above activity just for the u1 and u'1 basis vectors, (10.7.5) In the dual space of bras (linear functionals) (10.7.4) becomes, according to (2.11.g.10), (u'i)T= (ui)T RT <u'i| = <ui|RT i = 1,2..n push forward (ui)T = (u'i)TR <ui| = <u'i|R i = 1,2..n pull back (e'i)T= (ei)T RT <e'i| = <ei|RT i = 1,2..n push forward (ei)T = (e'i)TR <ei| = <e'i|R i = 1,2..n pull back . (10.7.6) We refer to the R operator acting to the left as the pullback operator. A picture similar to (10.7.5), which has dual x-space (Rn)* on the left and dual x'-space (Rm)'* on the right, would show the push forward <u'1| = <u1|RT in red and the pullback <u1| = <u'1|R in blue. Below we shall have hybrid pictures showing the non-dual spaces but also showing the mapping of linear functionals between the dual-spaces. Recall now the notations used in (8.7.1) for basis vectors in the dual wedge product spaces Λ'k(Rm) and Λk(Rn) , λ'^I ≡ λ'i ^ λ'i ^ .... ^ λ'i = <e'^I| ≡ <e'i| ^ <e'i| ^ ..... ^ <e'i| λ^I ≡ λi ^ λi ^ .... ^ λi = <u^I| ≡ <ui| ^ <ui| ^ ..... ^ <ui| (10.7.7) where the e'i (ui) are axis-aligned basis vectors in x'-space (x-space). Recall also that, <ei| = <e'i|R (10.7.6) ei = Rij uj or |ei> = Rij |uj> <ei| = Rij<uj| = <e'i|R . (2.4.4) Then, <e'^I|R = <e'i|R ^ <e'i|R ^ ..... <e'i| R // (8.9.d.15) = ( Rij<uj| ) ^ ( Rij<uj| ) ^ ... ^ ( Rij<uj| ) // (2.4.4) above = RijRij ...Rij ( <uj| ^ <uj| ^ ... ^ <uj| ) // reorder = ΣJ RIJ <u^J| // multiindex or [λ'^I R] = <e'^I|R = ΣJ RIJ <u^J| = ΣJ RIJ λ^J . (10.7.8) On the last line we write [λ'^I R] where R acts to the left on λ'^I as a reminder of what is happening in the Dirac notation. For k=1 one would write <e'i|R = [λ'iR] . Eq. (10.7.8) shows that the pullback of a k-form basis vector λ'^I = <e'^I| from dual x'-space to dual x-space is a linear combination of k-form basis vectors λ^J = <u^J| in dual x-space which is then some k-form in dual x-space. The above equations are meaningful for k ≥ 1. For k=0, a 0-form in x'-space is just a scalar function f(x'). Since there are no basis vectors involved, there is no shuffling with RIJ and the pullback of the scalar f(x') is just itself. That is to say, there is no distinction between the spaces Λ'0(Rm) = V0 = K and Λ0(Rn) = V'0 = K where K is the field of scalars . However, in x-space we want any object to be expressed in terms of x-space variables, so we write f(x') as f(F(x)) since x' = F(x). Therefore, Fact: The pullback of a 0-form may be written as [f(x') R] = f(x') = f(F(x)) (10.7.9) so R is really the unity operator in this Λ0 = V0 = K (scalars) space. Note that [f(x')g(x') R] = f(x')g(x') = [f(x') R] [f(x') R] . (10.7.10) The pullback of a 0-form (a scalar function) times a k-form basis vector is then, [f(x') λ'^I ] R = < f(x') e'^I | R = ( f(x') < e'^I | ) R // the space Λ'k(Rm) is linear since it is a vector space = f(x') ( < e'^I | R ) // R is a linear operator as in (2.11.g.29) = [f(x') R] [λ'^I R ] // using (10.7.9) and (10.7.8) . (10.7.11) The pullback of an arbitrary k-form is then given by, αx' = Σ'I fI(x')λ'^I ϵ Λ'k // k-form as in (10.1.10) but now for x'-space (10.7.12) [αx'R] = < Σ'I fI(x') e'^I | R = [ Σ'I fI(x') < e'^I | ] R // the space Λk(Rm) is linear since it is a vector space = Σ'I fI(x') [< e'^I | R ] // R is a linear operator as in (2.11.g.29) = Σ'I fI(x') [ λ'^I R ] // λ'^I = < e'^I | = Σ'I fI(F(x)) ΣJ RIJ λ^J // (10.7.8) = ΣJ [ Σ'I fI(F(x)) RIJ ] λ^J // reorder = ΣJ GJ(x)λ^J ϵ Λk where GJ(x) ≡ Σ'I fI(F(x)) RIJ . (10.7.13) Note that αx' is a k-form in dual x'-space, while [αx'R] is a linear combination of the λ^J and therefore is a k-form in dual x-space. This will be rewritten with the ordered sum Σ'J in Section 10.8 below. So, Fact: The pullback of a k-form in Λ'k(Rm) is a k-form in Λk(Rn) . (10.7.14) Finally, for a general k-form scaled by a function g(x'), using the same steps as above, (g(x')αx')R = < g(x') αx' | R = g(x') <αx' | R = [g(x')R ] [αx'R] . (10.7.15) The pullback of a rank-k tensor function is obtained by closing [αx'R] with a vector in the space Vk , [αx'R](v1,v2....vk) = < αx'|R| v1,v2....vk> = < αx'|R [ | v1> | v2> ..... | vk> ] // definition of |v1,v2....vk> = < αx'|[ | Rv1> | Rv2> ..... | Rvk> ] // (5.6.17) = < αx'| Rv1,Rv2....Rvk> = αx'(Rv1,Rv2....Rvk) . (10.7.16) The object αx'(Rv1,Rv2....Rvk) is a rank-k tensor function in Λ'kf(Rm) : the functional αx' lies in Λ'k(Rm) while the k vector arguments v'i = Rvi all lie in Rm . In contrast, the object [αx'R] (v1,v2....vk) is a rank-k tensor function in Λkf(Rn): the functional [αx'R] lies in Λk(Rn) while the k vector arguments vi all lie in Rn. The functional [αx'R] is the pullback of the functional αx'. Equation (10.7.16) says that the pulled-back tensor function [αx'R] in Λkf when evaluated at arguments (v1,v2....vk) is equal in value to the un-pulled-back tensor function αx' in Λ'kf evaluated at arguments (Rv1,Rv2....Rvk). These tensor functions are the objects that Spivak [1965] uses and he refers to them as k-tensors. Presentations which use only tensor functions regard (10.7.16) as the definition of a pullback [αx'R] of a differential k-form αx'. The notation used above with the Dirac operator R acting to the left on a dual space vector is a bit clumsy, so one defines the following pullback function where <α'| is any k-form in Λ'k(Rm), F*(α') ≡ <α' | R // α' ≡ αx' (10.7.17) F* : Λ'k(Rm) → Λk(Rn) . (10.7.18) Recall that the differential matrix R = (DF) and its associated Dirac operator R are specific to the underlying general transformation x' = F(x), so to be more precise we could have written RF and RF. The letter F in the function F* makes this connection explicit. Various equations above can now be recast in terms of the pullback function F* : Some Properties of the F* pullback function (10.7.19) 0 F*(α') ≡ <α' | R = <α' | RF // definition of F*, (10.7.17) 1 F*(f(x')) = f(x') = f(F(x)) // F* on a 0-form, (10.7.9) for k = 0 2 F*(f(x') g(x') ) = F*(f(x')) F*(g(x')) // F* on a product of two 0-forms, (10.7.10) 3 F*(f(x') λ'^I) = F*(f(x')) F*( λ'^I) // F* on 0-form and basis-vector k-form, (10.7.11) 4 F*(λ'^I) = ΣJ RIJ λ^J // F* on a basis-vector k-form, (10.7.8) for k ≥ 1 5 F*(λ'i) = Σj Rij λj // F* on a basis-vector 1-form, k=1 of the above 6 αx' = Σ'I fI(x')λ'^I // general k-form in Λ'k(Rm), (10.7.12) 7 F*(αx') = Σ'I fI(F(x)) ΣJ RIJ λ^J // F* pulling back a general k-form from Λ'k, (10.7.13) 8 F*(g(x') αx') = F*(g(x')) F*(αx') // F* on a 0-form times a general k-form, (10.7.15) 9 [F*(αx')](v1,v2...vk) = αx'(Rv1,Rv2...Rvk) // F* pulling back a rank-k tensor function, (10.7.16) Note that ΣJ in items 4 and 7 is the redundant symmetric sum. In (10.8.2) below we restate items 4 and 7 using the ordered sum Σ'J, and then we restate everything again using cosmetic notation. Other Properties of the F* pullback function Fact: F* is linear, so F*(s1α' + s2β') = s1F*(α') + s2F*(β') where α' and β' are k-forms. (10.7.20) Proof for k>0 : F*( s1α' + s2β') = <s1α' + s2β'| R // (10.7.17) definition of F* = s1(<α' |R) + s2(<β' |R) // R is a linear operator, see (2.11.g.29) = s1F*(α') + s2F*(β') // (10.7.17) definition of F* twice Proof for k=0 : F*( s1f(x') + s2g(x') ) = s1f(x') + s2g(x') // (10.7.16) definition of F* = F*(s1f(x')) + F*(s2g(x')) // (10.7.17) definition of F* twice = s1F*(f(x')) + s2F*(g(x')) // (10.7.19) items 2 and 1 ___________________________________________________________________________________ Fact: F*(α'1 ^ α'2 ^...^ α'N) = F*(α'1) ^ F*(α'2) ^...^ F*(α'N) where α'i is an arbitrary ki-form. (10.7.21) Proof: F*(α'1 ^ α'2 ^....^ α'N) = [ <α'1| ^ <α'2| ^ ... ^ <α'N| ] | R // Dirac notation = [ <α'1| R ^ <α'2| R ^ ... ^ <α'N| R ] // (8.9.d.15) = F*(α'1) ^ F*(α'2) ^...^ F*(α'N) . // (10.7.17) QED The result is valid if one or more of the forms are 0-forms. In this case, the two ^ operators surrounding a 0-form can be replaced by one ^. For example, <α'1| ^ f(x) ^ <α'3| = f(x) <α'1| ^ <α'3| . In vector space notation, one has Λn ^ Λ0 ^ Λm = Λn ^ Λm where Λ0 is the space of scalars. ___________________________________________________________________________________ Fact: F*(dα') = d(F*(α')) where α' ϵ Λ'k is a k-form in x'-space (10.7.22) This Fact says that the pullback function F* commutes with the external derivative operator d. Proof: Show that Left Hand Side = Right Hand Side: LHS: α' = Σ'IfI(x') λ'^I ϵ Λ'k (k-form in x'-space) // (10.7.12) dα' = Σ'I dfI(x') λ'^I = Σ'I Σj=1m [∂'jfI(x')] λ'j ^ λ'^I ϵ Λ'k+1 // (10.3.6) F*(dα') = Σ'I Σj=1m [∂'jfI(x')] F*(λ'j ^ λ'^I) ϵ Λk+1 // (10.7.20) F* linear = Σ'I Σj=1m [∂'jfI(x')] F*(λ'j) ^ F*(λ'^I) . // (10.7.21) product RHS: α' = Σ'IfI(x') λ'^I ϵ Λ'k (k-form in x'-space) // (10.7.12) F*(α') = Σ'I F*(fI(x')) F*(λ'^I) ϵ Λk // (10.7.19) item 3 = Σ'I fI(F(x)) F*(λ'^I) ϵ Λk // (10.7.19) item 1 d(F*(α')) = Σ'I dfI(F(x))) F*(λ'^I) ϵ Λk+1 // (10.3.6) = Σ'I Σj=1m [∂'jfI(x') Σr=1n (∂x'j/∂xr)] λr ^ F*(λ'^I) // (10.3.6) = Σ'I Σj=1m [∂'jfI(x') Σr=1n Rjr] λr ^ F*(λ'^I) // (2.1.2) = Σ'I Σj=1m [∂'jfI(x')] (Σr=1n Rjr λr) ^ F*(λ^I) // regroup = Σ'I Σj=1m [∂'jfI(x')] F*(λ'j) ^ F*(λ'^I) . // (10.7.19) item 5 The LHS and RHS results are the same, so F*(dα') = d(F*(α')) . QED ___________________________________________________________________________________ Corollary: d(F*(dα')) = 0 . (10.7.23) Proof: d(F*(dα')) = d(dF*(α')) = d2 [F*(α')] = 0 // (10.7.22) then (10.3.12) ___________________________________________________________________________________ Fact: F*(G*α) = (G o F)* α where α is a k-form (10.7.24) Proof: This theorem involves two mappings F and G which are composed to form a third H : x" = G(x') x' = F(x) x → x' → x" x → x" F G H x" = G(F(x)) ≡ [G o F ](x) = H(x) H* = (G o F)* dx" = RG dx' dx' = RF dx dx" = RGRFdx x" = H(x) dx" = RHdx so: RH = RGRF and RH = RGRF Using the definition (10.7.17) that F*(β) ≡ <β|RF and G*(α) ≡ <α|RG we find, F*(G*α) = F*(<α|RG ) = (<α|RG ) RF = <α| RGRF = <α|RH = H*(α) = (G o F)* (α) QED If α is a 0-form (a function) α = f(x"), then by (10.7.19) item 1, G*(f(x")) = f(G(x')) so F*(G*α) = F*(G*f(x")) = F*(f(G(x'))) = f(G(F(x))) = f((G o F)(x)) = f(H(x)) = H*(f(x")) = (G o F)*(f(x")) = (G o F)*α QED A Chapter 1 style category diagram for this scenario would be (10.7.25) ___________________________________________________________________________________ The following hybrid drawing shows the forward mapping x' = F(x) between the non-dual spaces, and at the same time the pullback βx = F*(αx') from the dual space Λ'k on the right to dual space Λk on the left, (10.7.26) Here βx is just a made-up name for the pulled back k-form αx'. Recall that x' on αx' means that the k-form αx' = Σ'I fI(x') λ'^I is specific to the point x' on manifold M, while the x on βx means that the k-form βx is specific to the point x in x-space. 10.8 Alternate ways to write the pullback of a k-form The ordered sum form of a k-form pullback Certain expressions above contain the sum ΣJ RIJ λ^J . As shown in Appendix A, because the object RIJ has a "factored form" , this sum can be written as on ordered sum Σ'J as follows ΣJ RIJ λ^J = Σ' J det(RIJ) λ^J // (A.8.36) (10.8.1) where the determinant magically appears. We can then rewrite two items from (10.7.19) : 4 F*(λ'^I) = Σ' J det(RIJ) λ^J // F* on a basis-vector k-form 7 F*(αx') = Σ'I fI(F(x)) Σ' J det(RIJ) λ^J // F* pulling back a general k-form from Λ'k = Σ' J gJ(x)λ^J where gJ(x) ≡ Σ'I fI(F(x)) det(RIJ) . (10.8.2) It is useful to write out (10.8.2) in full detail. using the cosmetic notation λi = dxi , F*(αx') = Σ1≤j<j<....<j≤n Σ1≤i<i<....<i≤m fii...i(F(x)) * det [RIJ] * ( dxi ^ dxi .....^ dxi ) (10.8.3) where RIJ is this kxk matrix, Rij Rij ... Rij RIJ = Rij Rij ... Rij ..... Rij Rij ... Rij (10.8.4) with Rij = (DF)ij = (∂Fi/∂xj) = ∂jFi(x) . // (10.6.2) The object det(RIJ) is a k x k minor of the full "tall" m x n matrix R, so k ≤ n ≤ m in our application. Remember that, due to the ordered sums, all the ir are different, and all the jr are different, so no row or column appears twice in RIJ. The dFi form of a k-form pullback It is customary to define the object shown above in (10.8.1) as a certain k-form, dF^I ≡ ΣJ RIJ λ^J = Σ' J det(RIJ) λ^J . (10.8.5) The motivation for doing this arises from the k = 1 case where the above becomes dFi ≡ Rij λj = Rij dxj (10.8.6) where we replace λj by its cosmetic notation dxj. The above equation then "looks just like" the normal calculus differential one obtains from transformation x' = F(x) so x'i = Fi(x), dFi = (∂Fi/∂xj) dxj = (∂x'i/∂xj) dxj = Rijdxj . (10.8.7) That is to say, dFi = Rijdxj (10.8.7) dFi = Rij dxj (10.8.6) (10.8.8) and in this same cosmetic notation we can rewrite (10.8.5) as dF^I = ΣJ RIJ dx^J = Σ' J det(RIJ) dx^J . (10.8.9) Using this new object, we write (10.8.2) as 4 F*(λ'^I) = dF^I // F* on a basis-vector k-form 7 F*(αx') = Σ'I fI(F(x)) dF^I // F* pulling back a general k-form from Λ'k (10.8.10) This shows that dF^I is just the pullback of a basis vector in Λ'k . In fact, recalling (10.7.6), <ei| = <e'i|R = [λ'iR] = F*(λ'i) = F*(<e'i|) , (10.7.6) and using theorem (10.7.21) we can identify dF^I as an old friend, the wedge product of the dual tangent base vectors en in dual x-space, dF^I = <e^I| ϵ Λk . // recall that λ^I = <u^I| (10.8.11) As an example of (10.8.9) we write for k = 2, dFi ^ dFi = Σ1≤j<j≤n det dxj ^ dxj . (10.8.12) Note that both sides of the above equation are 2-forms in Λ2, whereas dx'i ^ dx'i is a 2-form in Λ'2, so we cannot identify dFi ^ dFi with dx'i ^ dx'i even though x'i = Fi(x). But from (10.8.8), dFi ^ dFi = F*(λ'i^ λ'i) = F*( dx'i ^ dx'i ) (10.8.13) so in fact dFi ^ dFi is just the pullback of dx'i ^ dx'i . This is just restating the basic fact of (10.7.6) that <ei| = <e'i|R so <ei| is the pullback of <e'i| . The reader will hopefully appreciate our use of red italic font to distinguish differential form objects from calculus objects of the same name. Otherwise things can be very confusing, especially in presentations where all ^ symbols are suppressed and where symmetric and ordered sums are both written as ΣI. Summary all in cosmetic notation dx'^I ≡ dx'i ^ dx'i ^ .... ^ dx'i ϵ Λ'k(Rm) dx^I ≡ dxi ^ dxi ^ .... ^ dxi ϵ Λk(Rn) (10.7.7) (10.8.14) Some Properties of the F* pullback function (10.7.19) (10.8.15) 0 F*(α') ≡ <α' | R = <α' | RF // definition of F*, (10.7.17), α' ≡ 1 F*(f(x')) = f(x') = f(F(x)) // F* on a 0-form, (10.7.9) for k = 0 2 F*(f(x') g(x') ) = F*(f(x')) F*(g(x')) // F* on a product of two 0-forms, (10.7.10) 3 F*(f(x') dx'^I) = F*(f(x')) F*(dx'^I) // F* on 0-form and basis-vector k-form, (10.7.11) 4 F*(dx'^I) = ΣJ RIJ dx^J // F* on a basis-vector k-form, (10.7.9) for k ≥ 1 5 F*(dx'i) = Σj Rij dxj // F* on a basis-vector 1-form, k=1 of the above 6 αx' = Σ'I fI(x') dx'^I // general k-form in Λ'k, (10.7.12) 7 F*(αx') = Σ'I fI(F(x)) ΣJ RIJ dx^J // F* pulling back a general k-form from Λ'k, (10.7.13) 8 F*(g(x') αx') = F*(g(x')) F*(αx') // F* on a 0-form and general k-form, (10.7.15) 9 [F*(αx')](v1,v2...vk) = αx'(Rv1,Rv2...Rvk) // F* pulling back a rank-k tensor function, (10.7.16) Other Properties of the F* pullback function (10.8.16) Fact: F* is linear, so F*(s1α' + s2β') = s1F*(α') + s2F*(β') where α' and β' are k-forms. (10.7.20) Fact: F*(α'1 ^ α'2 ^...^ α'N) = F*(α'1) ^ F*(α'2) ^...^ F*(α'N) where α'i is an arbitrary ki-form. (10.7.21) Fact: F*(dα') = d(F*(α')) where α' ϵ Λ'k is a k-form in x'-space (10.7.22) Corollary: d(F*(dα')) = 0 . (10.7.23) Fact: F*(G*α) = (G o F)* α where α is a k-form . (10.7.24) The ordered sum form of a k-form pullback ΣJ RIJ dx^J = Σ' J det(RIJ) dx^J // (A.8.36) (10.8.1) (10.8.17) 4 F*(dx'^I) = Σ' J det(RIJ) dx^J // F* on a basis-vector k-form (10.8.18) 7 F*(αx') = Σ'I fI(F(x)) Σ'J det(RIJ) dx^J // F* pulling back a general k-form from Λ'k = Σ'J gJ(x) dx^J where gJ(x) = Σ'I fI(F(x)) det(RIJ) (10.8.2) (10.8.19) The dFi form of a k-form pullback dF^I ≡ ΣJ RIJ dx^J = Σ' J det(RIJ) dx^J = F*(dx'^I ) = <e^I| (10.8.9, 10,11) (10.8.20) F*( dx'i ^ dx'i ) = dFi ^ dFi = Σ1≤j<j≤n det dxj ^ dxj . ≡ Σ1≤j<j≤n dxj ^ dxj (10.8.12) (10.8.21) dFi = Rij dxj (10.8.8) (10.8.22) F*(dx'^I) = F*(dx'i ^ dx'i ...^ dx'i) = dFi ^ dFi ...^ dFi = dF^I (10.8.10) (10.8.23) F*(αx') = Σ'I fI(F(x)) dF^I (10.8.10) (10.8.24) 10.9 A Change of Notation and Comparison with Sjamaar and Spivak To this point we have maintained the notation of Chapter 2 (and Tensor) for transformations x' = F(x). To compare our results with other sources, we shall now make the following change of notation : → x-space → t-space x'-space → x-space F → φ general transformation name x' = F(x) → x = φ(t) general transformation equation R = (DF) → R = (Dφ) differential matrix F* → φ* pullback function (10.9.1) Confusingly, in Picture A' the left-side space has the name x-space, while in Picture F' this happens to be the name of the right-side space. This is just a coincidental new definition of x-space. It is like changing the names of variables x→t and x'→x so that under this change f(x,x') → f(t,x). This notation is convenient for presenting results, but it is somewhat clumsy for developing those results as we have done above. Having primes and no primes is very efficient compared to the other changes one must make to develop in this new notation. For example one must write λ'i → xλi and λi → tλi. Similarly, one has u'i → xui for the tangent base vectors which span TxM and ui → tui are the axis-aligned basis vectors in t-space ("parameter space"). Just for the record, these changes are all shown in Appendix E. Drawings for the new spaces In terms of the new t-space and x-space, this drawing (a translation of Fig (10.7.5) above) shows the push forward and pull back of the first basis vectors in Rn and Rm, (10.9.2) The next drawing (translation of Fig (10.7.26) above) shows the pullback of a differential k-form αx from xΛk(Rm) to k-form βt in tΛk(Rn), (10.9.3) Here βt is just a made-up name for the pulled back k-form αx. Recall that x on αx means that the k-form αx = Σ'I fI(x) xλ^I is specific to the point x on manifold M, while the t on βt means that the k-form βt is specific to the point t in t-space. Here is a more practical picture for the special case n = 2 and k = 2: (10.9.19) Here the open region U is a unit square [0,1]2 which maps into a patch on a torus. That is, if m = 3 the object on the right is a torus in R3, but we can imagine it to be a torus embedded in Rm for any m ≥ 3. The space of vectors defined on U R2 is a 2-dimensional dual space (R*2)(U). On this space we can define either 1-forms or 2-forms. The above picture suggests a 2-form since the region U is an area, and since we will later associate dt1 ^ dt2 with the calculus differential dt1dt2 which represents an area (we are not there yet). The picture shows the "forward map" x = φ(t), suggesting that forward means left to right in the picture. Then αx is "pulled back" right to left from dual x-space to dual t-space where it becomes βt. One could imagine a set of 16x6 = 96 mappings like the one shown above which would "cover the torus", using one little patch for each mapping (with some small overlap between patches). One would then have an atlas of 96 square maps like that on the left which would serve to cover the surface of Planet Toroid. This is the basic idea of a manifold. In the torus example, one could do the job with only 2 maps. Doing it with a single map does not fly since then some seam curve on the torus would map back to two boundaries of the square and the mapping is then not one-to-one and smooth. Manifold mappings have to be continuous in both mapping directions at every point, and a seam is a place without continuity. The aspect ratio of the 2-cube on the left is not significant. One could change it to be an arbitrary rectangle in t-space and select a φ to make it map to the same small image patch in x-space. Or one could construct a mapping φ which maps the unit 2-cube [0,1]2 to the entire left half of the torus. See Sjamaar. The black arrows on the left are the t-space basis vectors tui (only tu2 is labeled). As shown in (10.8.6), these map according to xui = R tui into basis vectors which are tangent to M, and these vectors then span the tangent space TxM at point x on M. It is clear that the two xui vary as the point x on M is varied. As another example consider this situation with n = 1 and k = 1, (10.9.20) Now the domain in t-space is U = 1-cube [0,1] which maps to a (generally non-planar) red curve which is embedded in Rm . Here αx and βt are 1-forms. The red curve segment V lies on the manifold curve M as shown, just as the patch of the previous example lay on the torus. There is only one basis vector tu in t-space (not shown) and it maps to the unlabeled black arrow on the right which is xu and is of course tangent to the curve at x. We now reproduce the "Summary in all cosmetic notation" given above at the end of Section 10.8 but in terms of this new notation: Summary all in cosmetic notation dx^I ≡ dxi ^ dxi ^ .... ^ dxi basis vector ϵ xΛk(Rm) dt^I ≡ dti ^ dti ^ .... ^ dti basis vector ϵ tΛk(Rn) (10.8.14) (10.9.4) Some Properties of the φ* pullback function R = (Dφ) (10.8.15) (10.9.5) 0 φ*(αx) ≡ <αx | R = <αx | RF // definition of φ*, (10.7.17) 1 φ*(f(x)) = f(x) = f(φ(t)) // φ* on a 0-form, (10.7.9) for k = 0 2 φ*(f(x) g(x) ) = φ*(f(x)) φ*(g(x)) // φ* on a product of two 0-forms, (10.7.10) 3 φ*(f(x) dx^I) = φ*(f(x)) φ*(dx^I) // φ* on 0-form and basis-vector k-form, (10.7.11) 4 φ*(dx^I) = ΣJ RIJ dt^J // φ* on a basis-vector k-form, (10.7.9) for k ≥ 1 5 φ*(dxi) = Σj Rij dtj // φ* on a basis-vector 1-form, k=1 of item 4 6 αx = Σ'I fI(x) dx^I // general k-form in xΛk, (10.7.12) 7 φ*(αx) = Σ'I fI(φ(t)) ΣJ RIJ dt^J // φ* pulling back a general k-form from xΛk, (10.7.13) 8 φ*(g(x) αx) = φ*(g(x)) φ*(αx) // φ* on a 0-form and general k-form, (10.7.15) 9 [φ*(αx)](v1,v2...vk) = αx(Rv1,Rv2...Rvk) // φ* pulling back a rank-k tensor function, (10.7.16) Other Properties of the φ* pullback function (10.8.16) (10.9.6) These five items are translations of (10.7.20) through (10.7.24) : Fact: φ* is linear, so φ*(s1α + s2β) = s1φ*(α) + s2φ*(β) where α and β are k-forms. (10.9.7) Fact: φ*(α1 ^ α2 ^...^ αN) = φ*(α1) ^ φ*(α2) ^...^ φ*(αN) where αi is an arbitrary ki-form. (10.9.8) Fact: φ*(dα) = d(φ*(α)) where α ϵ xΛk is a k-form in x-space (10.9.9) Corollary: d(φ*(dα)) = 0 . (10.9.10) Fact: φ*(ψ*α) = (ψ o φ)* α where α is a k-form . (10.9.11) The ordered sum form of a k-form pullback R = (Dφ) ΣJ RIJ dx^J = Σ' J det(RIJ) dt^J // (A.8.36) (10.8.17) (10.9.12) 4 φ*(dx^I) = Σ' J det(RIJ) dt^J // φ* on a basis-vector k-form (10.8.18) (10.9.13) 7 φ*(αx) = Σ'I fI(φ(t)) Σ' J det(RIJ) dt^J // φ* pulling back a general k-form from xΛk = Σ' J gJ(t) dt^J where gJ(t) = Σ'I fI(φ(t)) det(RIJ) (10.8.19) (10.9.14) The dφi form of a k-form pullback dφ^I ≡ ΣJ RIJ dt^J = Σ'J det(RIJ) dt^J = φ*(dx^I ) = <te^I| (10.8.20) (10.9.15) φ*( dxi ^ dxi ) = dφi ^ dφi = Σ1≤j<j≤n det dtj ^ dtj . ≡ Σ1≤j<j≤n dtj ^ dtj (10.8.21) (10.9.16) dφi = Rij dtj (10.8.22) (10.9.17) φ*(dx^I) = φ*(dxi ^ dxi ...^ dxi) = dφi ^ dφi ...^ dφi = dφ^I (10.8.10) (10.9.18) φ*(αx) = Σ'I fI(φ(t)) dφ^I (10.8.24) (10.9.19) Comparison with Sjamaar Our document was strongly motivated by Sjamaar's excellent notes, so it seems useful to make some connection to those notes. In most of our document we used the transformation x' = F(x) but in Section 10.9 we changed this to be x = φ(t) to bring things closer to Sjamaar and other authors. Sjamaar uses y = φ(x) in his Ch 3 on pullbacks, x = c(t) in Ch 4 on 1-forms, and x = ψ(t) in Ch 5 on integration and Ch 6 on manifolds. He does not stress the notion of an underlying transformation as we have done because he has many more important details to attend to, but he does show y = φ(x) in his figure on page 39. All wedge product symbols ^ are suppressed with the idea that almost all products are wedge products, so one sees equations like dx1dx2 = - dx2dx1. His sum ΣI is almost always an ordered sum which we write as Σ'I. Here then is a sampling of our equations above and how they appear in Sjamaar's 2015 notes : α ^ β = (-1)kk'β ^ α α = k-form, β = k'-form (10.4.1) Sja p 19, "graded commutivity" Fact: d2α = 0 for any k-form α (differential forms have zero "curvature") . (10.3.10) Sja p 22 α = Σ'I fI(x) xλ^I = Σ'I fI(x) dx^I // a k-form (10.1.14) Sja p 19,39 φ*(f(x) dx^I) = φ*(f(x)) φ*(dx^I) (10.9.5) 2 Sja p 39, related to the above φ*(dx^I) = φ*(dxi ^ dxi ...^ dxi) = dφi ^ dφi ...^ dφi = dφ^I (10.8.23) Sja p 39 Fact: φ* is linear, so φ*(s1α + s2β) = s1φ*(α) + s2φ*(β) where α and β are k-forms. (10.9.7) Fact: φ*(α1 ^ α2 ^...^ αN) = φ*(α1) ^ φ*(α2) ^...^ φ*(αN) where αi is an arbitrary ki-form. (10.9.8) Fact: φ*(ψ*α) = (ψ o φ)* α where α is a k-form (10.9.11) Sja p 40 Fact: φ*(dα) = d(φ*(α)) where α ϵ xΛk is a k-form in x-space (10.9.9) Sja p 41 φ*(dx^I ) = dφ^I = Σ'J det((Dφ)IJ) dtJ for x = φ(t) (10.9.15) Sja p 44 for y = φ(x) φ*(αx) = Σ' J gJ(t) dt^J where gJ(t) = Σ'I fI(φ(t)) det(RIJ) for x = φ(t) (10.9.14) Sja p 44 for y = φ(x) dxi ≡ λi = <ui| = (ui)T ui = axis-aligned basis vectors of Rn (10.1.1) Sja p 92 The tensor function pullback For the pullback of a tensor function we have stated [φ*(αx)](v1,v2...vk) = αx(Rv1,Rv2...Rvk) for x = φ(t) (10.9.5) item 9 (10.9.20) where recall [φ*(αx)](v1,v2...vk) = <αx | R | v1,v2...vk>. If R acts to the left, one gets the left side of (10.9.20), while if R acts to the right one gets the right side. Here the function φ*(αx) is the pullback of the function αx and the pulled-back function is associated with t-space, so we wish to write the right side expression entirely in t-space variables. To this end we replace αx by αφ(t) on the right of (10.9.20). The tensor functional [φ*(αx)] is in dual t-space, so we can write it as [φ*(αx)]t similar to the βt appearing in Fig (10.9.3) above. The vectors vi are in t-space Rn. We write R = (Dφ) = (D(t)φ) to show that the derivatives are with respect to t. Then in more detail we can write the above tensor function pullback equation as [φ*(αx)]t(v1,v2...vk) = αφ(t)( [D(t)φ(t)]v1, [D(t)φ(t)]v2... [D(t)φ(t)]vk) // x = φ(t) (10.9.21) where the expression on the right contains only t variables (no x variables), as appropriate for expressing the t-space tensor function [φ*(αx)]t(v1,v2...vk). Translating (10.9.21) according to x = φ(t) → y = φ(x) gives [φ*(αy)]x(v1,v2...vk) = αφ(x)( [D(x)φ(x)]v1, [D(x)φ(x)]v2... [D(x)φ(x)]vk) // y = φ(x) . (10.9.22) It is this equation we then compare to Sjamaar's page 96 equation, (10.9.23) He writes [D(x)φ(x)] as Dφ(x) and [φ*(αy)]x as φ*(α)x . The tensor function pullback equation also appears in Spivak but not quite as we have written it. Spivak says on the top of page 90 and the bottom of page 89, which we interpret to mean [f*(ω)](p)(v1,v2...vk) = ω(f(p)) ( (D(p)f)v1, (D(p)f)v2, ... (D(p)f)vk ) . Replacing ω→α, f→φ and p → x gives [φ*(α)](x)(v1,v2...vk) = α(φ(x))( (D(x)φ)v1, (D(x)φ)v2, ... (D(x)φ)vk ) . We then interpret (x) on the left and (φ(x)) on the right as spatial locations in the respective non-dual spaces, so the above becomes [φ*(α)]x(v1,v2...vk) = αφ(x)( (D(x)φ)v1, (D(x)φ)v2, ... (D(x)φ)vk ) (10.9.24) in agreement with our (10.9.22) and with Sjamaar's form (10.9.23). Sjamaar 2015 refers to φ*α as the pullback of α, but Spivak writing in 1965 does not use the term pullback in his book. Having a name for something is always helpful. Spivak's entire presentation is in terms of tensor functions, there are no functionals per se. He uses our tensor function pullback equation as the definition of a pullback (not calling it by that name). We have tried to define the pullback more generally in terms of the general transformation x = φ(t) so that it has a meaning for vectors, dual vectors (functionals), and tensor functions. Appendix A: Permutation Support Summary Section A.1 describes our permutation notation, presents two rearrangement theorems for the permutation group, and then states the familiar determinant theorem det(M) = det(MT) in permutation notation. Section A.2 describes the action of a permutation operator on a generic function f(1,2...k), and states several theorems concerning multiple permutation operators. At the same time, the Alt operator is defined and various facts are proven concerning this operator. The notion of a totally antisymmetric generic function is directly related to the Alt operator. Section A.3 mimics Section A.2 for the Sym operator in place of the Alt operator. The notion of a totally symmetric generic function is directly related to the Sym operator. Section A.4 states some facts which concern both the Alt and the Sym operators together. Up to this point, the various facts and theorems have taken place in a "generic permutation space" which consists of functions of k arguments which are a permutation of 1,2...k, such as f(2,1,3...k). Section A.5 applies all the previous facts and theorems to the permutation space whose elements are the component indices of a rank-k tensor, so f(1,2,3...k) = Tii...i. The results of Sections A.2, A.3 and A.4 are adapted to the tensor world in subsections (a),(b) and (c). Section A.6 deals with the permutation tensor εii...i and shows how it can provide an alternative to the permutation notation in some situations associated with the Alt operator. Section A.7 adapts the above generic results to the case f(1,2,....k) = (vj vj ..... vj) which is the tensor product of k vectors. Then the wedge product of k vectors is defined in terms of this application of the Alt operator, so that (vj ^ vj ^ .....^ vj) ≡ Alt(vj vj ..... vj). Section A.8 is similar to Section A.5, but the facts and theorems are applied not to tensors, but to "tensor functions", so here f(1,2...k) = T(vi,vi....vi). The permutation space is now the set of subscript labels on the k vector arguments of a tensor function. The results of Sections A.2, A.3 and A.4 are adapted to the tensor function world in subsections (a),(b) and (c). Section A.9 proves an obscure ordered permutation sum theorem that is used in (7.4.12). Section A.10 shows how to represent the Alt or Sym operators in Dirac bra-ket notation. A.1 Rearrangement Theorems and a Determinant Theorem Definition: A permutation P (of order k) reorders the list of integers [1,2,3...k] in some manner to give [i1,i2,i3...ik] . (A.1.1) Including the initial ordering [1,2,3...k], there are k! possible permutations. Fact: ΣP(1) = k! // there are k! equal terms in this sum (A.1.2) The permutation group rearrangement theorem states the following: ΣP f(QP) = ΣP f(PQ) = ΣP f(P) . (A.1.3) Here ΣP is a sum over all k! permutations of [1,2...k], and Q is any one of these permutations. The first two sums are just reorderings or rearrangements of the third sum and so equal the third sum. Proof: This theorem is true because the permutations P of [1,2...k] form a group G : P1P2 = P3 ϵ G // closure (P1P2)P3 = P1(P2P3) // associative P = I // identity exists, permutation that does nothing to [1,2...k] P-1 exists for any P // just the inverse permutation. (A.1.4) It is a fact that for any group G with k elements gi, ga [g1, g2, ....gk] = [ gag1, gag2, ....gagk] = [g'1, g'2, ....g'k] = reordering of [g1, g2, ....gk] [g1, g2, ....gk]ga = [ g1ga, g2ga, ....gkga] = [g"1, g"2, ....g"k] = reordering of [g1, g2, ....gk] . (A.1.5) To show that [g'1, g'2, ....g'k] is a reordering of [g1, g2, ....gk], we have to show that no two elements of [g'1, g'2, ....g'k] are the same. Suppose for example g'1 = g'2 . That would imply gag1 = gag2. Since ga-1 exists in a group for any ga, apply ga-1 to both sides to get ga-1gag1 = ga-1gag2 or g1 = g2. But that contradicts the basic starting point that [g1, g2, ....gk] enumerates the distinct group elements. Therefore Σi f(gagi) = Σif(giga) = Σif(gi) . (A.1.6) This is valid only if the sum is over all elements of the group, which in the rearrangement theorem (A.1.3) means the sum ΣP must be over all permutations P. In any group, if g exists, so does g-1, and it is just some element of the group. For the permutation group P-1 exists and is in fact the permutation which reverses the permutation of P : P[1,2...k] = [i1,i2...ik] [1,2...k] = P-1[i1,i2...ik] PP-1 = P-1P = 1 . (A.1.7) In the above, since [i1,i2...ik] is a permutation of [1,2...k], one can get from [1,2...k] to [i1,i2...ik] by making some number of swaps of the integers in [1,2...k]. Comment: We are following a Maple convention that [a,b,c...] is a "list" where order is significant, whereas {a,b,c...} is a "set" where order is not significant. The swap count S(P) Any two permutations of [1,2,...k] can be linked by a number of pairwise swaps of the integers. For example, if we have P[1,2,...k] = [i1,i2,...ik], one can get from the first integer sequence to the second by doing some number S(P) of pairwise swaps. The integer S(P) is not unique, but whether it is an even or an odd integer is unique, so the factor (-1)S(P) is unique to a particular P (we leave it to the reader to prove this fact) . Sometimes (-1)S(P) is called the parity of permutation P. Example: [1,2,3] → [2,1,3] S(P) = 1 (-1)S(P) = -1 [1,2,3] → [1,3,2] → [2,3,1] → [2,1,3] S(P) = 3 (-1)S(P) = -1 (A.1.8) It seems clear that the number of position swaps to get from [1,2...k] to [i1,i2...ik] is the same as it is going the other direction, so S(P-1) = S(P) . (A.1.9) Finally, consider P1P2[1,2...k] = P[1,2...k] = [i1,i2...ik] . P = P1P2 If P2 causes S2 position swaps and then P1 causes S1 more, then P does S2+S1 total swaps. Thus S(P) = S(P1P2) = S(P1) + S(P2) so (-1)S(PP) = (-1)S(P) (-1)S(P) = (-1)S(PP) . (A.1.10) From the above these trivial corollaries follow : (-1)S(PP) = 1, (-1)S(PQ) = (-1)S(QP). (-1)S(P) (-1)S(P) = 1 (-1)S(P) = (-1)S(Q) (-1)S(PQ) (A.1.11) Fact: ΣP(-1)S(P) = 0 (A.1.12) Proof : By the rearrangement theorem (A.1.3) and then (A.1.11) we know that, for any permutation Q, ΣP(-1)S(P) = ΣP(-1)S(QP) = (-1)S(Q)ΣP(-1)S(P) . Select a Q which has (-1)S(Q) = -1. Then ΣP(-1)S(P) = - ΣP(-1)S(P) ΣP(-1)S(P) = 0. QED Another Rearrangement Theorem Another version of the rearrangement theorem is the following, ΣQ f(Q) = ΣQ f(Q-1) . (A.1.13) Again, this is just a reordering of the sum. Consider, {g1-1, g2-1, ....gk-1} = {g1', g2', ....gk'} = reordering of {g1, g2, ....gk} To show that {g1', g2', ....gk'} is a reordering of {g1, g2, ....gk} we have to show that no two elements are the same. Suppose for example that g1' = g2' . That would say g1-1 = g2-1 which in turn says g1 = g2, but that contradicts the basic starting point that {g1, g2, ....gk} enumerates the distinct group elements. Therefore, Σi f(gi) = Σif(gi-1) (A.1.14) Comment: For continuous groups (like the rotation group SO(3)) , the rearrangement theorems become ∫dg f(gag) = ∫dg f(gga) = ∫dg f(g) ∫dg f(g) = ∫dg f(g-1) (A.1.15) where dg is called the invariant Haar measure. For SO(3) it is dg = dφd(cosθ)dψ (Euler angles). A Determinant Theorem It is well known that the determinant of a kxk matrix M can be written two equivalent ways in which the rows and columns are swapped (for Mab, this is the statement that det(M) = det(MT) ), det(Mab) = Σii...i εii...iMi1Mi2 ...Mik = Σii...i εii...iM1iM2i ...Mki ≡ det(M**) . (A.1.16) In permutation notation the above equations are written, det(Mab) = ΣP (-1)S(P) MP(1)1MP(2)2 ...MP(k)k = ΣP (-1)S(P) M1P(1)M2P(2) ...MkP(k) ≡ det(M**) . (A.1.17) If we start over with the "up-tilt" matrix Mab (mixed rank-2 tensor) then (A.1.16) becomes. det(Mab) = Σii...i εii...iMi1Mi2 ...Mik = Σii...i εii...iM1iM2i ...Mki ≡ det(M**) (A.1.18) which in permutation notation becomes det(Mab) = ΣP (-1)S(P) MP(1)1MP(2)2 ...MP(k)k = ΣP (-1)S(P) M1P(1)M2P(2) ...MkP(k) ≡ det(M**) . (A.1.19) Statements for det(M**) and det(M**) are similar. A Sum Theorem For any particular permutation P of the indices on ii...i , one has Σii...i fii...i = Σii...i fii...i or ΣI fI = ΣI fP(I) // multiindex notation (A.1.20) Proof: Since ΣI is a symmetric sum, one is free to shuffle the dummy summation index names at will, and this shuffle is indicated by permutation P. For example ΣI fI = Σii...i fii...i = Σii...i fii...i = Σii...ifii...i = Σii...i fii...i = ΣI fP(I) where P[1,2...k] = [2,1..k] A.2 The Alt Operator The generic Alt operator acts on a function f of the integers [1,2....k] to create a new function, g = Alt(f), as follows: g(1,2...k) = [Alt(f)](1,2...k) ≡ (1/k!) ΣP (-1)S(P)f( P(1),P(2)...P(k) ) (A.2.1) where f is any function and P are the k! permutations of the set of integers [1,2....k]. Informally we write the above as g(1,2...k) = (1/k!) [ f(1,2...k) - f(2,1...k) + other signed permutations ] Examples: g(1,2) = [Alt(f)](1,2) = (1/2) [ f(1,2) - f(2,1) ] (A.2.2) g(1,2,3) = [Alt(f)](1,2,3) = (1/6) [ f(1,2,3) - f(1,3,2) + f(3,1,2) - f(3,2,1) + f(2,3,1) - f(2,1,3) ] . Now, let R be some permutation of [1,2....k]. We write R[1,2....k] = [R(1),R(2).....R(k)] (A.2.3) where for example R(1) gives the integer into which 1 is converted by the permutation R. We can apply the operator R to a function of 1,2...k in this manner, R f(1,2,...k) = f( R(1),R(2).....R(k) ) . (A.2.4) Fact: Any permutation R is a linear operator, so R( Σiaifi) = Σiai(Rfi) (A.2.5) Proof: Let h(1,2,...k) ≡ Σiaifi(1,2,...k). Then R h(1,2,...k) = h( R(1),R(2).....R(k) ) // (A.2.4) applied to h = Σiaifi( R(1),R(2).....R(k) ) // definition of h = Σiai R fi(1,2,...k) . // (A.2.4) applied to fi QED Now suppose R = QP, the product of two permutations Q and P. Then starting with (A.2.3), (QP) f(1,2,...k) = f( (QP)(1),(QP)(2).....(QP)(k) ) = f( Q(P(1), Q(P(2)).... Q(P(k) ) ≡ f( QP(1), QP(2)... .QP(k) ) . (A.2.6) But from (A.2.4), (QP) f(1,2,...k) = Q { P f(1,2,...k)} = Q f( P(1),P(2).....P(k) ) (A.2.7) Therefore we have shown that Q f( P(1),P(2).....P(k) ) = f( QP(1), QP(2)... .QP(k) ) . (A.2.8) Definition: A function f(1,2..k) is totally antisymmetric if it changes sign when any two arguments are swapped. (A.2.9) Examples: f(1,2) = - f(2,1) f is totally antisymmetric f(1,2,3) = -f(2,1,3) f(1,2,3) = -f(3,2,1) f is totally antisymmetric f(1,2,3) = -f(1,3,1) Fact: f(1,2..k) totally antisymmetric P f(1,2,3..k) = (-1)S(P) f(1,2,3..k) (A.2.10) Proof: [] S(P) is the number of pairwise swaps going from [1,2,3...k] to P[1,2,3...k] = [i1,i2,i3...in]. If f is totally antisymmetric by the definition above, each such swap causes a minus sign, and the product of these minus signs is then (-1)S(P). [] If P = any pairwise swap, (-1)S(P) = -1, so f(1,2..k) is then totally antisymmetric. Fact: The function g(1,2..k) ≡ [Alt(f)](1,2...k) is totally antisymmetric in its arguments. (A.2.11) Proof: Let Q be some permutation of [1,2....k]. Then apply Q to the function g(1,2..k), Q g(1,2..k) = Q { (1/k!) ΣP (-1)S(P)f(P(1),P(2)...P(k)) } // definition of g = (1/k!) ΣP (-1)S(P) Q f(P(1),P(2)...P(k)) // (A.2.5), Q is linear = (1/k!) ΣP (-1)S(P) f(QP(1),QP(2)...QP(k)) // (A.2.8) = (-1)S(Q) (1/k!) ΣP (-1)S(QP) f(QP(1),QP(2)...QP(k)) // (A.1.11) = (-1)S(Q) (1/k!) ΣP (-1)S(P) f(P(1),P(2)...P(k)) // (A.1.3), rearrang. thm. = (-1)S(Q) g(1,2..k) // definition of g By (A.2.10) it follows that g(1,2,..k) is totally antisymmetric. QED Fact: Alt is a linear operator, so Alt(Σiaifi) = Σiai Alt(fi) . (A.2.12) Proof: Let h(1,2,...k) ≡ Σiaifi(1,2,...k). Then [Alt(h)](1,2,...k) = (1/k!) ΣP (-1)S(P)h( P(1),P(2)...P(k) ) // (A.2.1) def of Alt(h) = (1/k!) ΣP (-1)S(P){ Σiaifi(P(1),P(2)...P(k)) } // definition of h = Σiai [ (1/k!) ΣP (-1)S(P) fi(P(1),P(2)...P(k)) ] // reorder sums = Σiai Alt(fi) // (A.2.1) def of Alt(fi) Fact: Alt is a projection operator, so Alt(Alt(f)) = Alt(f) . (A.2.13) Comment: This is why (1/k!) is included in the definition of Alt. Proof: By (A.2.11) we know that Alt is a totally antisymmetric function, and therefore from (A.2.10), P [Alt(f)](1,2...k)] = (-1)S(P)[Alt(f)](1,2...k)] . (A.2.14) Next, consider that [Alt(f)](P(1),P(2)...P(k) ) = P [Alt(f)](1,2...k) // (A.2.4) applied with f→ Alt(f), R→P = (-1)S(P)[Alt(f)](1,2...k) . // (A.2.14) (A.2.15) Now examine Alt(Alt(f)) : [Alt(Alt(f))](1,2...k) = (1/k!) ΣP (-1)S(P)[Alt(f)]( P(1),P(2)...P(k) ) = (1/k!) ΣP (-1)S(P){(-1)S(P)[Alt(f)](1,2...k)]} // (A.2.15) = (1/k!) ΣP[Alt(f)](1,2...k)]} // (-1)S(P)(-1)S(P) = 1 = [Alt(f)](1,2...k)] {(1/k!)ΣP(1)} // reorder factors = [Alt(f)](1,2...k)] {1} . // (A.1.2) QED Fact: If f is a totally antisymmetric function, then Alt(f) = f . (A.2.16) Proof: Alt(f)(1,2...k) = (1/k!) ΣP (-1)S(P)f( P(1),P(2)...P(k) ) // definition of Alt(f) (A.2.1) = (1/k!) ΣP (-1)S(P) P f(1,2,...k) // (A.2.4) with R→P = (1/k!) ΣP (-1)S(P) (-1)S(P) f(1,2,...k) // (A.2.10) = (1/k!) ΣP f(1,2,...k) // (A.1.11) = (1/k!) f(1,2,...k) { ΣP (1) } // reorder = f(1,2,...k) // (A.1.2) Fact: If f is totally antisymmetric, then f(1,2,3...k) =( -1)k-1 f(2, 3,... k-1, k, 1) forward cyclic (A.2.17) f(1,2,3...k) = (-1)k-1 f(k, 1, 2, 3,... k-1) backward cyclic Proof: Let B be the particular permutation which does this: B[1,2,3..k-1,k] = B[2,3,...k-1,k,1] (Backward cyclic). One then has S(B) = k-1 because it takes k-1 swaps to move the 1 from one end to the other. If f is totally antisymmetric, then according to (A.2.10) one has B f(1,2,3..k) = (-1)S(B) f(1,2,3..k) so then f(2,3,... k,1) = B f(1,2,3..k) = (-1)S(B) f(1,2,3..k) = (-1)k-1 f(1,2,3...k) . On the other hand, if F[1,2,3..k-1,k] = [k, 1, 2, 3,... k-1] (Forward cyclic), S(F) = k-1 for the same reason, and then f(k, 1, 2, 3,... k-1) = F f(1,2,3..k) = (-1)S(F) f(1,2,3..k) = (-1)k-1 f(1,2,3...k) . Example: A very commonly used fact is that, for k = 3, (-1)k-1 = (-1)2 = 1 and so f(1,2,3) = f(2,3,1) = f(3,1,2) f totally antisymmetric (A.2.18) Suppose now that function f has k arguments and function g has k' arguments. We then define a meaning for the tensor product in our generic function space as follows Definition: Tensor product: (fg)(1,2,....k+k') ≡ f(1,2...k) g(k+1,k+2....k+k') (A.2.19) A.3 The Sym Operator This section is an obvious copy, paste and edit job on the previous section. We omit what would be (A.3.3) through (A.3.8) since they would be the same as (A.2.3) through (A.2.8). The changes are mainly these: antisymmetric → symmetric (-1)P → 1 Alt → Sym . The Sym operator acts on a function of the integers [1,2....k] to create a new function, g = Sym(f), as follows: g(1,2...k) = [Sym(f)](1,2...k) ≡ (1/k!) ΣPf( P(1),P(2)...P(k) ) (A.3.1) where f is any function and P are the k! permutations of the set of integers [1,2....k]. Informally we write the above as g(1,2...k) = (1/k!) [ g(1,2...k) + g(2,1...k) + other permutations ] Examples: g(1,2) = [Sym(f)](1,2) = (1/2) [ f(1,2) + f(2,1) ] (A.3.2) g(1,2,3) = [Sym(f)](1,2,3) = (1/6) [ f(1,2,3) + f(1,3,2) + f(3,1,2) + f(3,2,1) + f(2,3,1) + f(2,1,3) ] Definition: A function f(1,2..k) is totally symmetric if it is unchanged when any two arguments are swapped. (A.3.9) Examples: f(1,2) = f(2,1) f is totally symmetric f(1,2,3) = f(2,1,3) f(1,2,3) = f(3,2,1) f is totally symmetric f(1,2,3) = f(1,3,1) Fact: f(1,2..k) totally symmetric P f(1,2,3..k) = f(1,2,3..k) (A.3.10) Proof: [] S(P) is the number of pairwise swaps going from [1,2,3...k] to P[1,2,3...k] = [i1,i2,i3...in]. If f is totally symmetric by the definition above, each such swap causes a plus sign, and the product of these plus signs is then 1. [] If P = any pairwise swap, (-1)S(P) = 1, so f(1,2..k) is then totally symmetric. Fact: The function g(1,2..k) ≡ [Sym(f)](1,2...k) is totally symmetric in its arguments. (A.3.11) Proof: Let Q be some permutation of [1,2....k]. Then apply Q to the function g(1,2..k), Q g(1,2..k) = Q { (1/k!) ΣP f(P(1),P(2)...P(k)) } // definition of g = (1/k!) ΣP Q f(P(1),P(2)...P(k)) // (A.2.5), Q is linear = (1/k!) ΣP f(QP(1),QP(2)...QP(k)) // (A.2.8) = (1/k!) ΣP f(P(1),P(2)...P(k)) // (A.1.3), rearrang. thm. = g(1,2..k) // definition of g By (A.3.10) it follows that g(1,2,..k) is totally symmetric. QED Fact: Sym is a linear operator, so Sym(Σiaifi) = Σiai Sym(fi) (A.3.12) Proof: Let h(1,2,...k) ≡ Σiaifi(1,2,...k). Then [Sym(h)](1,2,...k) = (1/k!) ΣP h( P(1),P(2)...P(k) ) // (A.3.1) def of Sym(h) = (1/k!) { Σiaifi(P(1),P(2)...P(k)) } // definition of h = Σiai [ (1/k!) fi(P(1),P(2)...P(k)) ] // reorder sums = Σiai Sym(fi) // (A.3.1) def of Sym(fi) Fact: Sym is a projection operator, so Sym(Sym(f)) = Sym(f) . (A.3.13) Comment: This is why (1/k!) is included in the definition of Sym. Proof: By (A.3.11) we know that Sym is a totally symmetric function, and therefore from (A.3.10), P [Sym(f)](1,2...k)] = [Sym(f)](1,2...k)] . (A.3.14) Next, consider that [Sym(f)](P(1),P(2)...P(k) ) = P [Sym(f)](1,2...k)] // (A.2.4) applied with f→ Sym(f), R→P = [Sym(f)](1,2...k)] . // (A.3.14) (A.3.15) Now examine Sym(Sym(f)) : [Sym(Sym(f))](1,2...k) = (1/k!) ΣP [Sym(f)]( P(1),P(2)...P(k) ) = (1/k!) ΣP {[Sym(f)](1,2...k)]} // (A.3.15) = [Sym(f)](1,2...k)] {(1/k!)ΣP(1)} // reorder = [Sym(f)](1,2...k)] {1} // (A.1.2) QED Fact: If f is a totally symmetric function, then Sym(f) = f . (A.3.16) Proof: Sym(f)(1,2...k) = (1/k!) ΣP f( P(1),P(2)...P(k) ) // definition of Sym(f) (A.3.1) = (1/k!) ΣP P f(1,2,...k) // (A.2.4) with R→P = (1/k!) ΣP f(1,2,...k) // (A.3.10) = (1/k!) f(1,2,...k) { ΣP (1) } // reorder = f(1,2,...k) // (A.1.2) Fact: If f is totally symmetric , then f(1,2,3...k) = f(2, 3,... k-1, k, 1) forward cyclic (A.3.17) f(1,2,3...k) = f(k, 1, 2, 3,... k-1) backward cyclic This is just a special case of (A.3.10) which says Q f(1,2,3..k) = f(1,2,3..k) for any Q, so it certainly true for F = forward cyclic or B = backward cyclic permutations. Example: f(1,2,3) = f(2,3,1) = f(3,1,2) f totally symmetric (A.3.18) For comparison, recall (A.2.18) which said f(1,2,3) = f(2,3,1) = f(3,1,2) f totally antisymmetric (A.2.18) A.4 More on Alt and Sym and the decomposition of functions Fact: The projection operators Alt and Sym are orthogonal, so Alt(Sym(f)) = Sym(Alt(f)) = 0 . (A.4.1) Proof left: Alt(Sym(f)) = Alt( {(1/k!) ΣPf( P(1),P(2)...P(k) ) } = (1/k!) ΣP [Alt(f)]( P(1),P(2)...P(k) ) ] // (A.2.12), Alt is linear = (1/k!) ΣP (-1)S(P)[Alt(f)](1,2...k) // (A.2.15) = {(1/k!) [Alt(f)](1,2...k)} {ΣP (-1)S(P)} // reorder = {(1/k!) [Alt(f)](1,2...k)} {0} // (A.1.12) = 0 Proof right: Sym(Alt(f)) = Sym( {(1/k!) ΣP (-1)S(P)f( P(1),P(2)...P(k) ) } = (1/k!) ΣP (-1)S(P)[Sym(f)]( P(1),P(2)...P(k) ) ] // (A.3.12), Sym is linear = (1/k!) ΣP (-1)S(P)[Sym(f)](1,2...k) // (A.3.15) = {(1/k!) [Sym(f)](1,2...k)} {ΣP (-1)S(P)} // reorder = {(1/k!) [Sym(f)](1,2...k)} {0} // (A.1.12) = 0 We can define a third projection operator this way, Else() ≡ 1 - Alt() - Sym() // projection operator Else(f) = f - Alt(f) - Sym(f) . // applied to f(1,2,3...k) (A.4.2) One can then decompose an arbitrary function f into three pieces, f = Alt(f) + Sym(f) + Else(f) = a + s + e // a = a(1,2....k) etc (A.4.3) where the "else" piece is whatever is left over, which is to say, e ≡ f - a - s. Then consider, Alt(f) = Alt(a + s + e) = Alt(a) + Alt(s) + Alt(e) // (A.2.12), Alt is linear = Alt(Alt(f)) + Alt(Sym(f)) + Alt(Else(f)) // (7.5.14) = Alt(f) + 0 + Alt(Else(f)) // (A.2.13) and (A.3.1) Alt(Else(f)) = 0 (A.4.4) Sym(f) = Sym(a + s + e) = Sym(a) + Sym(s) + Sym(e) // (A.2.12), Alt is linear = Sym(Alt(f)) + Sym(Sym(f)) + Sym(Else(f)) // (7.5.14) = 0 + Sym(f) + Sym(Else(f)) // (A.3.13) and (A.3.1) Sym(Else(f)) = 0 (A.4.5) This verifies that the "else" piece e of a function has neither a totally antisymmetric nor a totally symmetric component. Example 1: For a function f(1,2) one has from (A.2.2) and (A.3.2), a(1,2) = (1/2) [ f(1,2) - f(2,1)] s(1,2) = (1/2) [ f(1,2) + f(2,1) ] e(1,2) = f(1,2) - a(1,2) - s(1,2) = 0 (A.4.6) so the leftover else piece e(1,2) is null. Example 2: On the other hand, for a function f(1,2,3) one has from (A.2.2) and (A.3.2) a(1,2,3) =(1/6) [ f(1,2,3) - f(1,3,2) + f(3,1,2) - f(3,2,1) + f(2,3,1) - f(2,1,3) ] // (A.2.2) s(1,2,3) =(1/6) [ f(1,2,3) + f(1,3,2) + f(3,1,2) + f(3,2,1) + f(2,3,1) + f(2,1,3) ] // (A.3.2) e(1,2,3) = f(1,2,3) - (1/6) [ f(1,2,3) - f(1,3,2) + f(3,1,2) - f(3,2,1) + f(2,3,1) - f(2,1,3) ] - (1/6) [ f(1,2,3) + f(1,3,2) + f(3,1,2) + f(3,2,1) + f(2,3,1) + f(2,1,3) ] = f(1,2,3) - (1/3) [ f(1,2,3) + f(3,1,2) + f(2,3,1) ] = (2/3) f(1,2,3) - (1/3) [ f(3,1,2) + f(2,3,1) ] (A.4.7) so in this case the leftover piece e(1,2,3) is not null. In the case that f is either totally antisymmetric or totally symmetric, we know from (A.2.18) and (A.3.18) that all cyclic permutations of f are the same (for k = odd). In these cases, we can see explicitly from (A.4.7) that e(1,2,3) = 0, as expected. A.5 Application to Tensors We now restate the "generic" results of Sections A.2, A.3 and A.4 for this special case: f(1,2...k) = Tii...i . // a "tensor" T ϵ Vk . (A.5.1) Here T is any rank-k tensor (either in the weak or strong sense mentioned in ***). This f seems perhaps an odd looking "function", but one can consider it to be just an evaluation of this more respectable mapping, f(a,b,c,...q) = Tiii...i a,b,c... ϵ {1,2...k} f: {1,2...k}k → Vk . (A.5.2) This technical mapping issue is not important because we are just regarding Tii...i as a "carrier" of the labels 1,2,3..k, from the point of view of doing permutations. The actual indices like i1 could be arbitrary objects (labeled pancakes) as far as the permutation theorems are concerned, but in our applications we have in mind that i1 is an integer in the range 1,2....n where n = dim(V) and n is unrelated to the tensor rank k. In Section A.8 we shall instead apply our results to "tensor functions", f(1,2,3...k) = T(vi, vi, ....vi) Again, from a permutation point of view, T(vi, vi, ....vi) is just a carrier of the labels 1,2...k. The permutation theorems don't care whether or not vi happens to be a vector in V labeled by i1, or even whether or not vi happens to be an argument of a function T. Here then are some Section A.2,A.3,A.4 results translated according to f(1,2,3...k) = Tii...i . For some of the translations, we show the actual equation from above, then its translation. For others we just state the translated result. In all the results below, one can always specialize to the case i1,i2...ik → 1,2,...k. The resulting equations are then as if our mapping were f(1,2...k) = T12...k. Note then that iP(r) → i(r) in a superscript. (a) Alt Equations (translated from Section A.2 above) The basic Alt definition of (A.2.1) g(1,2...k) = [Alt(f)](1,2...k) ≡ (1/k!) ΣP (-1)S(P)f( P(1),P(2)...P(k) ) (A.2.1) becomes, Gii...i = [Alt(F)]ii...i = (1/k!) ΣP (-1)S(P) Fii...i or G = Alt(F) . // definition of Alt acting on a tensor (A.5.3a) We can define the object AltI[Fii...i] in the following obvious manner AltI[Fii...i] ≡ (1/k!) ΣP (-1)S(P) Fii...i = [Alt(F)]ii...i (A.5.3b) In multiindex notation : [Alt(F)]I = AltI[FI] = (1/k!) ΣP (-1)S(P) F P(I) (A.5.3c) Examples: Gii = [Alt(F)]ii = (1/2) [ Fii - Fii ] = AltI [Fii] Giii = [Alt(F)]iii = (1/6) [ Fiii - Fiii + the other four terms ] (A.5.4) In practice, we might more easily write Gabc = [Alt(F)]abc = (1/6) (Fabc - Facb + Fcab - Fcba + Fbca - Fbac) but when it comes time to prove permutation-related theorems, we use indices like iii . Continuing on, R[1,2....k] = [R(1),R(2).....R(k)] of (A.2.3) becomes, with R→ P, P Tii...i = Tii...i (A.2.3) (A.5.5) Fact: Any permutation P is a linear operator, so P( ΣiaiTiii...i) = Σiai(PTiii...i) (A.2.5) (A.5.6) Definition: A tensor Tii...i is totally antisymmetric if it changes sign when any two superscripts are swapped. (A.2.9) (A.5.7) Example: Tii...i = - Tii...i or Tabc = -Tbac Fact: Tii...i totally antisymmetric P Tii...i = (-1)S(P) Tii...i , where P is any permutation of [1,2..k]. (A.2.10) (A.5.8) Fact: The function Tii...i ≡ [Alt(F)]ii...i is totally antisymmetric in its indices. (A.2.11) (A.5.9) Fact: Alt is a linear operator, so Alt (ΣjajTjii...i) = Σjaj [Alt(Tj)]ii...i. (A.2.12) (A.5.10) Fact: Alt is a projection operator, so Alt(Alt(T)) = Alt(T) . (A.2.13) (A.5.11) Fact: If T is a totally antisymmetric rank-k tensor, then Alt(T) = T . (A.2.16) (A.5.12) (b) Sym Equations (translated from Section A.3 above) The basic Sym definition of (A.3.1) g(1,2...k) = [Sym (f)](1,2...k) ≡ (1/k!) ΣP f( P(1),P(2)...P(k) ) (A.3.1) becomes, Gii...i = [Sym (F)]ii...i = (1/k!) ΣP Fii...i (A.5.12) G = Sym (F) . // definition of Sym acting on a tensor Examples: Gii = [Sym(F)]ii = (1/2) [ Fii + Fii ] Giii = [Sym(F)]iii = (1/6) [ Fiii + Fiii + the other four terms ] (A.5.13) Gabc = [Sym(F)]abc = (1/6) (Fabc + Facb + Fcab+ Fcba + Fbca + Fbac) Definition: A tensor Tii...i is totally symmetric if it is unchanged when any two superscripts are swapped. (A.3.9) (A.5.14) Example: Tii...i = Tii...i or Tabc = Tbac = Tacb Fact: Tii...i totally symmetric P Tii...i = Tii...i, where P is any permutation of [1,2..k]. (A.3.10) (A.5.15) Fact: The function Tii...i ≡ [Sym(F)]ii...i is totally symmetric in its indices. (A.2.11) (A.5.16) Fact: Sym is a linear operator, so Sym (ΣjajTjii...i) = Σjaj [Sym(Tj)]ii...i. (A.3.12) (A.5.17) Fact: Sym is a projection operator, so Sym (Sym (T)) = Sym (T) . (A.3.13) (A.5.18) Fact: If T is a totally symmetric rank-k tensor, then Sym(T) = T . (A.3.16) (A.5.19) (c) Alt/Sym and Other Equations (translated from Section A.4 above) Fact: The projection operators Alt and Sym are orthogonal, so Alt(Sym(f)) = Sym(Alt(f)) = 0 . (A.4.1) (A.5.20) Fact: A tensor Tii...i can be decomposed in the following manner: (A.5.21) Tii...i = Aii...i + Sii...i + Eii...i (A.4.3) Alt(A) = A Sym(A) = 0 Alt(E) = 0 (A.4.4) Alt(S) = 0 Sym(S) = S Sym(E) = 0 (A.4.5) where A is totally antisymmetric, S is totally symmetric, and E is whatever is left over. Definition: Tensor product: (TS)ii...i ≡ Tii...i Sii...i, where the ranks of tensors T,S are k,k'. (A.2.19) (A.5.22) A.6 The permutation tensor ε The permutation tensor ε of rank k is written εii...i where each subscript must be an element of {1,2...k}. The values of the tensor are these: ε12..k = +1 εii...i changes sign if any two indices are swapped . Therefore, if two or more indices are the same, εii...i = 0 . εii...i ≡ εii...i (A.6.1) The tensor εii...i has kk components, but only k! of those components are non-zero. One arrives at k! by allowing k values for i1, then only (k-1) values for i2, and so on. The tensor εii...i is totally antisymmetric by (A.5.7) since any index swap causes a minus sign. Fact: Apart from scale, the εii...i tensor is the only totally antisymmetric tensor one can construct. (A.6.2) Proof: From the definition of εii...i, we see that if Aii...i is a arbitrary totally antisymmetric tensor, then one can write Aii...i = [ A12...k] εii...i . (A.6.3) Here εii...i does the bookkeeping for swaps of index pairs. The scale factor is then A12...k . Use of the ε tensor We noted already that all our permutation results can be specialized to ir → r. For example, [Alt(T)]ii...i = (1/k!) ΣP (-1)S(P) Tii...i (A.5.3) then becomes [Alt(T)]12...k = (1/k!) ΣP (-1)S(P) TP(1)P(2)...P(k) . (A.6.4) Now we make the following claim, ΣP (-1)S(P) TP(1)P(2)...P(k) = Σii...i εii...i Tii...i, ir = 1,2...k . (A.6.5) Each of the ir sums runs from 1 to k. Notice that each side has k! non-vanishing terms in its sum. Suppose P[1,2,3...k] = [i1, i2, i3 ...ik]. Then we claim that the parity of the permutation is given by (-1)S(P) = εii...i . (A.6.6) To see why this is so, start off with the identity permutation P = 1 which has (-1)S(P) = (-1)0 = 1. In this case P[1,2...k] = [1,2...k] and conveniently ε123...k = 1, so both sides of ** agree. Now swap 1↔2 and then the left side is (-1)1 = -1 and the right side is ε213...k = - ε123...k = - 1 and again both sides agree. Now swap 2↔3. The left side is (-1)2 and the right side is - ε132...k = ε123...k = 1, and again both sides agree. In this way one can exhaust all permutations P and the equation is always true. On the left side of (A.6.5) the permutations are enumerated by P, while on the right they are enumerated by i1, i2, i3 ...ik which is restricted by the ε tensor to be a permutation of 1,2,3...k. Basically the notation on each side of (A.6.5) is describing the same instructions for forming the sum. Example with k = 3 (A.6.7) ΣP (-1)S(P) TP(1)P(2)P(3) = T123 - T213 + T231 - T321 + T312 - T132 . The only simple way to form this sum is to keep doing swaps. We show in red the pair that will be swapped to make the next term on the right. Compare then to Σiii3 εiii Tiii . To enumerate the terms, we use the 3! = 6 non-zero values of εiii in the same order as above Σiii3 εiii Tiii = ε123 T123 + ε213 T213 + ε231 T231 + ε321T321 + ε312 T312 + ε132 T132 = (1) T123 + (-1) T213 + (1) T231 + (-1) T321 + (1) T312 + (-1)T132 = T123 - T213 + T231 - T321 + T312 - T132 Here the signs of the ε factors alternate as shown because each one is obtained by an index pair swap on the preceding term. Application Consider, Tii...i = (vi vi ..... vi) . We can specialize this to say T12...k = (v1 v2 ..... v3) and then apply P using (A.2.4) with R→P to get TP(1)P(2)...P(k) = (vP(1) vP(2) ..... vP(k)) . Then (A.6.5) ΣP (-1)S(P) TP(1)P(2)...P(k) = Σii...i εii...i Tii...i, ir = 1,2...k (A.6.5) becomes ΣP (-1)S(P) (vP(1) vP(2) ..... vP(k)) = Σii...i εii...i (vi vi ..... vi), ir = 1,2...k (A.6.8) A.7 The wedge-product-of-vectors Alt equation Here we consider a new application for our generic function f[1,2...k], namely, f[1,2,....k] = (vj vj ..... vj) . (A.7.1) Here the js label the generic objects vj and is some generic operator. Then, consider the generic Alt definition, [Alt(f)](1,2...k) ≡ (1/k!) ΣP (-1)S(P)f( P(1),P(2)...P(k) ) . (A.2.1) Formally speaking, the left side would have to be written something like this, [Alt(f)](1,2...k) = [Alt((vj vj ..... vj))](1,2...k) = Alt(vj vj ..... vj) . (A.7.2) On the first line the asterisks are place holders which will get the arguments in the argument list. This lets us make a formal association f → (vj vj ..... vj) for a function without arguments. On the right side of (A.2.1) we have f( P(1),P(2)...P(k) ) = (vj vj ..... vj) . (A.7.3) We end up then with this statement Alt(vj vj ..... vj) = (1/k!) ΣP (-1)S(P) (vj vj ..... vj) . (A.7.4) If it happens that means the tensor product, and if the vj happen to be vectors in V, then the above expression happens to be our definition (7.1.2) for the wedge product of k vectors: (vj ^ vj ^ .....^ vj) = Alt(vj vj ..... vj) . (A.7.5) As noted earlier, we can always specialize replacing jr → r. Then f[1,2,....k] = (v1 v2 ..... vk) . Alt(v1 v2 ..... vk) = (1/k!) ΣP (-1)S(P) (vP(1) vP(2) ..... vP(k)) (A.7.6) (v1 ^ v2 ^ .....^ vk) = Alt(v1 v2 ..... vk) . (A.7.7) A.8 Application to Tensor Functions We now restate the "generic" results of Sections A.2, A.3 and A.4 for this special case: f(1,2...k) = T(vi,vi....vi) // a "tensor function" T ϵ V*k . (A.8.1) We apologize for copy, paste and edit, but things really are exactly parallel to the tensor discussion above. Here T is any rank-k tensor function. This f seems perhaps an odd looking "function", but one can consider it to be just an evaluation of this more respectable mapping, f(a,b,c,...q) = T(vi,vi....vi) a,b,c... ϵ {1,2...k} f: {1,2...k}k → Vf*k . (A.8.2) This technical mapping issue is not important because we are just regarding T(vi,vi....vi) as a "carrier" of the labels 1,2,3..k, from the point of view of doing permutations. The actual indices like i1 could be arbitrary objects (labeled pancakes) as far as the permutation theorems are concerned, but in our applications we have in mind that i1 is an integer in the range 1,2....n where n = dim(V) and n is unrelated to the tensor rank k. Here then are some Section A.2,A.3,A.4 results translated according to f(1,2,3...k) = T(vi,vi....vi) . For some of the translations, we show the actual equation from above, then its translation. For others we just state the translated result. In all the results below, one can always specialize to the case i1,i2...ik → 1,2,...k. The resulting equations are then as if our mapping were f(1,2...k) = T(v1,v2....vk). Note then that iP(r) → i(r) in a subscript. (a) Alt Equations (translated from Section A.2 above) The basic Alt definition of (A.2.1) g(1,2...k) = [Alt(f)](1,2...k) ≡ (1/k!) ΣP (-1)S(P)f( P(1),P(2)...P(k) ) (A.2.1) becomes, G(vi,vi....vi) = [Alt(F)](vi,vi....vi) = (1/k!) ΣP (-1)S(P)F(vi,vi....vi) (A.8.3) G = Alt(F) // definition of Alt acting on a tensor Examples: G(vi,vi) = [Alt(F)](vi,vi) = (1/2)[ F(vi,vi) - F(vi,vi) ] (A.8.4) G(vi,vi,vi) = [Alt(F)](vi,vi,vi) = (1/6)[ F(vi,vi,vi) - F(vi,vi,vi+ the other four terms ] In practice, we might more easily write G(va,vb,vc) = [Alt(F)](va,vb,vc) = (1/6) [ F(va,vb,vc) - F(vb,va,vc) + the other four terms ] but when it comes time to prove permutation-related theorems, we use subscripts like i1. Continuing on, R[1,2....k] = [R(1),R(2).....R(k)] of (A.2.3) becomes, with R→ P, P T(vi,vi....vi) = T(vi,vi....vi) . (A.2.3) (A.8.5) Fact: Any permutation P is a linear operator, so P( ΣrarTr(vi,vi....vi)) = Σrar(PTr(vi,vi....vi)) (A.2.5) (A.8.6) Definition: A tensor function T(vi,vi....vi) is totally antisymmetric if it changes sign when any two arguments are swapped. (A.2.9) (A.8.7) Example: T(vi,vi....vi) = - T(vi,vi....vi) or T(va,vb....vq) = - T(vb,va....vq) Fact: T(vi,vi....vi) totally antisymmetric P T(vi,vi....vi) = (-1)S(P) T(vi,vi....vi) , where P is any permutation of [1,2..k]. (A.2.10) (A.8.8) Fact: The function T(vi,vi....vi) ≡ [Alt(T)](vi,vi....vi) is totally antisymmetric in its labels. (A.2.11) (A.8.9) Fact: Alt is a linear operator, so Alt (ΣjajTj(vi,vi....vi)) = Σjaj [Alt(Tj)](vi,vi....vi). (A.2.12) (A.8.10) Fact: Alt is a projection operator, so Alt(Alt(T)) = Alt(T) . (A.2.13) (A.8.11) Fact: If T is a totally antisymmetric rank-k tensor, then Alt(T) = T . (A.2.16) (A.8.12) (b) Sym Equations (translated from Section A.3 above) The basic Sym definition of (A.3.1) g(1,2...k) = [Sym (f)](1,2...k) ≡ (1/k!) ΣP f( P(1),P(2)...P(k) ) (A.3.1) becomes, G(vi,vi....vi) = [Sym(F)](vi,vi....vi) = (1/k!) ΣP F(vi,vi....vi) (A.8.12) G = Sym(F) // definition of Alt acting on a tensor Examples: G(vi,vi) = [Sym(F)](vi,vi) = (1/2)[ F(vi,vi) + F(vi,vi) ] (A.8.13) G(vi,vi,vi) = [Sym(F)](vi,vi,vi) = (1/6)[F(vi,vi,vi)+F(vi,vi,vi+ the other four terms ] In practice, we might more easily write G(va,vb,vc) = [Sym(F)](va,vb,vc) = (1/6) [ F(va,vb,vc) + F(vb,va,vc) + the other four terms ] but when it comes time to prove permutation-related theorems, we use subscripts like i1. Definition: A tensor function T(vi,vi....vi) is totally symmetric if it is unchanged when any two arguments are swapped. (A.3.9) (A.8.14) Example: T(vi,vi....vi) = T(vi,vi....vi) or T(va,vb....vq) = T(vb,va....vq) Fact: T(vi,vi....vi) totally symmetric P T(vi,vi....vi) = T(vi,vi....vi), where P is any permutation of [1,2..k]. (A.3.10) (A.8.15) Fact: The function T(vi,vi....vi) ≡ [Sym(F)]T(vi,vi....vi) is totally symmetric in its labels. (A.2.11) (A.8.16) Fact: Sym is a linear operator, so Sym(ΣjajTj(vi,vi....vi)) = Σjaj [Sym(Tj)](vi,vi....vi). (A.3.12) (A.8.17) Fact: Sym is a projection operator, so Sym (Sym (T)) = Sym (T) . (A.3.13) (A.8.18) Fact: If T is a totally symmetric rank-k tensor, then Sym(T) = T . (A.2.16) (A.8.19) (c) Alt/Sym and Other Equations (translated from Section A.4,A.6 and A.7 above) Fact: The projection operators Alt and Sym are orthogonal, so Alt(Sym(T)) = Sym(Alt(T)) = 0 . (A.4.1) (A.8.20) Fact: A tensor function T(vi,vi....vi) can be decomposed in the following manner: (A.8.21) T(vi,vi....vi) = A(vi,vi....vi) + S(vi,vi....vi) + E(vi,vi....vi) (A.4.3) Alt(A) = A Sym(A) = 0 Alt(E) = 0 (A.4.4) Alt(S) = 0 Sym(S) = S Sym(E) = 0 (A.4.5) where A is totally antisymmetric, S is totally symmetric, and E is whatever is left over. Definition: Tensor product: (TS)(vi,vi....vi) ≡ T(vi,vi....vi) S(vi,vi....vi), where the ranks of tensors T,S are k,k'. (A.2.19) (A.8.22) The following are based on Section A.6 and concern use of the ε tensor with tensor functions. If A is totally antisymmetric, then A(vi,vi....vi) = [ A(v1,v2....vk)] εii...i . (A.6.3) (A.8.22) The tensor function [Alt(T)](v1,v2....vk) can be expressed as, ΣP (-1)S(P) T(vP(1),vP(2)....vP(k)) = Σii...i εii...i T(vi,vi....vi) (A.6.5) (A.8.23) Let T(vi,vi....vi) = (αi αi ..... αi)(vi,vi....vi) so T(v1,v2....vk) = (α1 α2 ..... αk)(v1,v2....vk) Then (A.8.23) gives this way to write (αj ^ αj ^ .....^ αj) : ΣP (-1)S(P) (αP(1) αP(2) ..... αP(k)) (A.6.8) (A.8.24) = Σii...i εii...i (αi αi ..... αi), ir = 1,2...k The following is based on Section A.7. (αj ^ αj ^ .....^ αj) = Alt(αj αj ..... αj) (A.7.5) (A.8.25) (d) Alt/Sym when there are two sets of indices It is not uncommon to encounter objects like the following (Xjj...j)ii...i XJI where the jr are labels and the ir are tensor component indices. An example would be the components of a wedge product of k vectors (vj^ vj^ .....^ vj)ii...i . In this situation, we have to clarify which of the two sets of indices is being acted upon by the Alt operator. We might do this as follows, using AltI and AltJ , AltI[(Xjj...j)ii...i] = (1/k!) ΣP (-1)S(P) (Xjj...j)ii...i AltJ[(Xjj...j)ii...i] = (1/k!) ΣP (-1)S(P) (Xjj...j)ii...i In general, the above two objects are different. Now recall Fact (A.5.12) from above, Fact: If T is a totally antisymmetric rank-k tensor, then Alt(T) = T . (A.2.16) (A.5.12) If it happens that (Xjj...j)ii...i is totally antisymmetric in the ir , then AltI[(Xjj...j)ii...i] = (Xjj...j)ii...i If it happens that (Xjj...j)ii...i is totally antisymmetric in the jr , then AltJ[(Xjj...j)ii...i] = (Xjj...j)ii...i If it happens that (Xjj...j)ii...i is totally antisymmetric separately in the ir and the jr, then we have AltI[(Xjj...j)ii...i] = AltJ[(Xjj...j)ii...i] since both are equal to (Xjj...j)ii...i Recalling Fact (A.5.19) Fact: If T is a totally symmetric rank-k tensor, then Sym(T) = T . (A.3.16) (A.5.19) we conclude a similar fact for SymI and SymJ . We then summarize these results Fact: If (Xjj...j)ii...i is totally antisymmetric in both sets of indices, then AltI[(Xjj...j)ii...i] = AltJ[(Xjj...j)ii...i] (A.8.26) Fact: If (Xjj...j)ii...i is totally symmetric in both sets of indices, then SymI[(Xjj...j)ii...i] = SymJ[(Xjj...j)ii...i] (A.8.27) Example: According to (7.2.9) the object (vj^ vj^ .....^ vj)ii...i is totally antisymmetric in both sets of indices. Therefore, AltI (vj^ vj^ .....^ vj)ii...i = AltJ (vj^ vj^ .....^ vj)ii...i (A.8.28) Another case of interest is when (Xjj...j)ii...i is an outer product of identical rank-2 tensors, (Xjj...j)ii...i = Tji Tji ..... Tji . Then AltI[(Xjj...j)ii...i] = (1/k!) ΣP(-1)P Tji Tji ..... Tji = (1/k!) det(Tj*i ) . According to (A.1.19) we can move the P(..) operators from the i subscripts to the j subscripts to get = (1/k!) ΣP(-1)P Tji Tji ..... Tji = AltJ[(Xjj...j)ii...i] . We are just swapping the rows and columns in the determinant shown above. Thus, Fact: AltI [ Tji Tji ..... Tji ] = AltJ [ Tji Tji ..... Tji ] = (1/k!) det(Tj*i ) (A.8.29) Either form gives the same expression (1/k!) [Tji Tji ..... Tji + all signed permutations]. In multiindex notation, we rewrite this last Fact as Fact: AltI [TJI] = AltJ [TJI] = (1/k!) det(TJI) if TJI has factored form (A.8.30) The above is of course true for TIJ or any other index positions. If TJI has the factored form shown above, then Fact: TJI = TP(J)P(I) where P = any permutation of the index subscripts (A.8.31) Proof: Reordering the index subscripts this way just reorders the terms in the product of the factors. For example if TJI = Tji Tji then if P[1,2] = [1,2] then TP(J)P(I) = TjiTji. Fact : ΣJ det(TIJ) xJ = ΣJ k! TIJ x^J if T has factored form. (A.8.32) Here xJ = xj xj .... xj , x^J = xj ^ xj ....^ xj and each xj is a vector labeled by j . That is to say, xj is not the jth component of vector x. Proof: LHS = ΣJ det(TIJ) xJ = ΣJ [k! AltI(TIJ)] xJ // (A.8.30) = ΣJ k! [ (1/k!) ΣP (-1)P TP(I)J] xJ // (A.1.3) for Alt = ΣP (-1)P [ΣJ TP(I)J xJ] // reorder = ΣP (-1)P [ΣJ TP(I)P(J) xP(J)] // (A.1.20) that ΣJ fJ = ΣJ fP(J) = ΣP (-1)P ΣJTIJ xP(J) // (A.8.31) = k! ΣJTIJ [ (1/k!) ΣP (-1)P xP(J)] // reorder = k! ΣJTIJ AltJ(xJ) // (A.3.1) for Alt = k! ΣJTIJ x^J = RHS // (7.4.3) █ Fact : ΣI TI u^I = Σ'I k! AltI(TI) u^I for any tensor TI (A.8.33) ΣI TI u^I = Σ'I k! AltI(TI) u^I for any tensor TI This theorem (first line) says that if the symmetric sum ΣI is replaced by the ordered sum Σ'I , then the coefficients TI get replaced by k! AltI(TI) . Example: Tii gets replaced by 2! AltI(Tii) = 2! ( (1/2!)[Tii -Tii] ) = [Tii -Tii] . Then ΣI TI u^I = Σii Tii ui^ ui = Σi<i [Tii -Tii] ui^ ui Proof: This theorem (first line) was proved in (7.4.4) through (7.4.16). Here we just review that proof using multindex notation: T^ = ΣI TI u^I u^I = ui^ ui^ .....^ ui (7.4.4) T^ = Σ'I AI u^I . (7.4.7) T^ = Σi≠i≠...≠i TI u^I (7.4.9) T^ = ΣP Σi<i<...<i TI u^I (7.4.12) T^ = Σi<i<...<i ΣP TP(I) u^P(I) // using (A.9.1) below (7.4.13) u^P(I) = (-1)S(P)u^I (7.4.14) T^ = Σi<i<...<i [ΣP (-1)S(P) TP(I)] u^I = Σ'I [Alt(T)] u^I (7.4.15) AI = k! [Alt(T)]I by comparing (7.4.7) and (7.4.15) (7.4.16) The theorem goes through with the I-tilt the other way, which is the second line of (A.8.33). Alternatively, we can take the first result and apply the "tilt reversal rule" (2.9.1) to get the second line from the first line. █ Fact : ΣI TI x^I = Σ'I k! AltI(TI) x^I for any tensor TI (A.8.34) ΣI TI x^I = Σ'I k! AltI(TI) x^I for any tensor TI Proof: In the proof of the previous Fact, the basis vectors ui played a placeholder role and the same proof works with any set of vectors xi where x^I is the wedge product of those vectors. █ Fact : ΣI TIJ x^I = Σ'I k! AltI(TIJ) x^I for any tensor TIJ (A.8.35) ΣI TJI x^I = Σ'I k! AltI(TJI) x^I for any tensor TJI Proof: The first line is the first line of the previous Fact where a bystander J multiindex has been added. The same idea for the second line. Fact : ΣI TIJ x^I = Σ'I det(TIJ) x^I if TIJ has factored form (A.8.36) ΣI TJI x^I = Σ'I det(TJI) x^I if TJI has factored form Proof: ΣI TIJ x^I = Σ'I k! AltI(TIJ) x^I // (A.8.35) = Σ'I det(TIJ) x^I // (A.8.30) ΣI TJI x^I = Σ'I k! AltI(TJI) x^I // (A.8.35) = Σ'I det(TJI) x^I // (A.8.30) █ A.9 The Ordered Sum Theorem The ordered sum theorem states that, (ΣP [ΣP(i)<P(i)<...<P(i)]) fii...i = Σi<i<...<i [ΣP fP(i)P(i)...P(i)] . (A.9.1) Rather than present a formal proof, we look at the two simplest cases and the general case is then obvious. k = 2: First, consider Q ≡ Σi≠i fii = [Σi<i + Σi>i] fii = [Σi<i + Σi<i] fii= (ΣP[ΣP(i)<P(i)]) fii . (A.9.2) On the other hand, Q = [Σi<i + Σi<i] fii = Σi<i fii + Σi<i fii = Σi<i fii + Σi<i fii //dummy swap i1↔i2 in 2nd term = Σi<i [ fii + fii] = Σi<i [ΣP fP(i)P(i)] . (A.9.3) Thus we have proven the Theorem for k = 2, (ΣP [ΣP(i)P(i)]) fii = Σi<i [ΣP fP(i)P(i)] (A.9.4) k = 3: First, consider Q ≡ Σi≠i≠i fiii = (Σi<i<i + Σi<i<i + Σi<i<i + Σi<i<i + Σi<i<i + Σi<i<i) fiii = (ΣP [ΣP(i)<P(i)<P(i)]) fiii . (A.9.5) On the other hand we can rename the summation indices in all but the first sum to get Q = (Σi<i<i + Σi<i<i + Σi<i<i + Σi<i<i + Σi<i<i + Σi<i<i) fiii as is 2↔3 1↔3 = Σi<i<i fiii + Σi<i<i fiii + Σi<i<i fiii + 3 more sums = Σi<i<i [ fiii + fiii + fiii + 3 more terms ] = Σi<i<i [ΣP fP(i)P(i)P(i)]. (A.9.6) Thus we have proven the Theorem for k =3, (ΣP [ΣP(i)<P(i)<P(i)]) fiii = Σi<i<i [ΣP fP(i)P(i)P(i)] . (A.9.7) The argument for a k-fold sum proceeds in the same manner, and we end up with (ΣP [ΣP(i)<P(i)<...<P(i)]) fii...i = Σi<i<...<i [ΣP fP(i)P(i)...P(i)] . (A.9.1) A.10 The Alt Operator in Dirac bra-ket notation We won't make use of what follows, but it seems a reasonable way to incorporate operators like Alt and Sym into the Dirac bra-ket notation. In the above discussion, when we say g = Alt(f), what we mean is that, g(1,2...k) = Alt(f)(1,2...k) = (1/k!) ΣP(-1)P f(P(1),P(2)....P(k) ) . (A.10.1) The stripped down statement g = Alt(f) is somewhat vague because there are no visible labels for Alt to act upon. We can give g = Alt f a more concrete meaning using the Dirac bra-ket notation. Write, g(1,2...k) = <g | 1,2...k> <g | = functional (A.10.2) Alt(f)(1,2...k) = < Alt f |1,2...k > < Alt f | = functional (A.10.3) Here |1,2...k > is a ket in some generic permutation space Gk . Next we define Alt |1,2...k> ≡ (1/k!) ΣP(-1)P |P(1),P(2)...P(k)> (A.10.4) < Alt f | 1,2...k> ≡ <f | Alt 1,2...k> . (A.10.5) In (A.10.5) we are pretending that Alt is a symmetric (self-adjoint) operator AltT = Alt which can be swung for free from the bra space to the ket space. It is just notation. Then we can interpret the stripped statement g = Alt(f) to mean < g | = < Alt f | . (A.10.6) When both sides are closed with the ket |1,2,...k> we get the intended result, < g | 1,2,...k> = < Alt f |1,2...k> = <f | Alt 1,2...k> = <f | [(1/k!) ΣP(-1)P |P(1),P(2)...P(k)>] = (1/k!) ΣP(-1)P <f | P(1),P(2)...P(k)> (A.10.7) which then says g(1,2..k) = (1/k!) ΣP(-1)P f(P(1),P(2)....P(k) ) . (A.10.1) (A.10.8) In our two applications we have tensor: | 1,2...k> = | ei, ei ...ei> <T| 1,2...k> = <T|ei, ei ...ei> = Tii...i (A.10.9) tensor function: | 1,2...k> = | vi, vi ...vi> <T| 1,2...k> = <T|vi, vi ...vi> = T(vi, vi ...vi) . (A.10.10) Example: Prove that Alt(Alt(f)) = Alt(f) in Dirac notation. That is, show < Alt Alt f | = <Alt f |. Proof: Let | i1,i2...ik> be an arbitrary permutation of | 1,2...k> ( an arbitrary ket in Gk ) < Alt Alt f | i1,i2...ik> = < Alt f | Alt i1,i2...ik> = < Alt f | (1/k!) ΣP(-1)P iP(1),iP(2)...iP(k)> = <f | Alt (1/k!) ΣP(-1)P iP(1),iP(2)...iP(k)> = (1/k!) ΣP(-1)P <f | Alt iP(1),iP(2)...iP(k)> = (1/k!) ΣP(-1)P <f | (1/k!) ΣQ (-1)Q iQP(1),iQP(2)...iQP(k)> = (1/k!)2 ΣP <f | ΣQ (-1)QP iQP(1),iQP(2)...iQP(k)> = (1/k!)2 ΣP <f | ΣQ (-1)Q iQ(1),iQ(2)...iQ(k)> // rearrangement theorem (A.1.3) = (1/k!)2 (k!) <f | ΣQ (-1)Q iQ(1),iQ(2)...iQ(k)> // ΣP(1) = k! = <f | Alt i1,i2...ik> = <Alt f | i1,i2...ik> (A.10.11) Since this is true for all kets in Gk we conclude that < Alt Alt f | = <Alt f |. QED Appendix B: Direct Sum of Vector Spaces There are nine short numbered sections below. Here are the headings: 1. Axioms for 2. Direct Sum Space VW 3. Basis for VW 4. Z = VW is a vector space 5. vw does not commute 6. Visualization of the Direct Sum 7. Extension to multiple products 8. Application to tensor products 9. Direct sum of operators (matrices): Block Diagonal Form 1. Axioms for Let vi ϵ V and wi ϵ W where V and W are vector spaces, and α ϵ K is a scalar. The direct sum operator can be defined by these rules (axioms), v1w1 + v2w2 + ... + vkwk = (v1+v2 + ...+ vk) (w1+w2 + ... + wk) (B.1) (αv)(αw) = α(vw) . (B.2) In slightly more concise notation (B.1) can be written Σi=1k (viwi) = (Σi=1kvi) (Σi=1kwi). For k = 2 (B.1) becomes, v1w1 + v2w2 = (v1+v2) (w1+w2) . (B.3) Since V and W are vector spaces, each has a 0 element and we can write v10 + 0w2 = (0+v1) (0+w2) = v1w2 // (B.3) with w1 = 0 and v2 = 0 (B.4) (αv)0 = α(v0) // w = 0 0(αw) = α(0w) // v = 0 (B.5) 2. Direct Sum Space VW Define space Z by Z ≡ VW (B.6) and let zi ≡ viwi ϵ Z. (B.7) One might write : (V,W) → VW : (v,w) ↦ (vw) Lemma: Given some zi, we can find vi and wi such that zi = viwi . (B.8) Proof: When we say Z ≡ VW, we mean these spaces are the same, so there is a 1-to-1 correspondence between elements of zi ϵ Z and elements viwi ϵ VW. 3. Basis for VW Let us assume that: ei form a basis of dimension n for V e'j form a basis of dimension n' for W Fact: A basis for Z can be written as {e10, e20 ........en0, 0e'1, 0e'2, ....0e'n'} (B.9) Proof: Let vi be components of vector v, and wi the components of vector w. Consider: { v1(e10) + v2(e20) + ... + vn(en0)} + {w1(0e'1) + w2(0e'2) + ... + wn'(0e'n')} ={ (v1e1)0 + (v2e2)0 + ...+ (vnen)0} + {0(w1e'1) + 0(w2e'2) + ... + 0(wn'e'n')} // (B.5) = { (v1e1+ v2e2 + ...+ vnen )(0+0..+0)} + {(0+0+..+0) (w1e'1+ w2e'2 + ...+ wne'n )} // (B.1) = (v1e1+ v2e2 + ...+ vnen ) 0 +0 (w1e'1+ w2e'2 + ...+ wne'n ) = (v1e1+ v2e2 + ...+ vnen ) (w1e'1+ w2e'2 + ...+ wne'n ) // (B.4) = v w This z = v w is an arbitrary element of Z, and we have therefore shown that an arbitrary element of Z can be expanded on the basis shown in (B.9) and that no smaller basis will do the job. QED Fact: If dim(V) = n and dim(W) = n'. then dim(VW) = n + n' (B.10) Proof: Count the basis elements shown in (B.9). Compare this Fact with that shown in (4.1.1) : Fact: If dim(V) = n and dim(W) = n', then dim(VW) = n * n' . (4.1.1) 4. Z = VW is a vector space Fact: If V and W are vector spaces, then Z = VW is a vector space. (B.11) Proof: We just run down the required axioms listed for example on the wiki vector space page. The conclusion one reaches is that the vector space properties are "induced" from V and W into Z. The fact that + is commutative within V and W causes + to be commutative within Z : z1 + z2 = v1w1 + v2w2 = (v1+v2) (w1+w2) = (v2+v1) (w2+w1) = v2w2 + v1w1 = z2 + z1 . Addition in Z is associative because it is associative in V and W: (z1+ z2) + z3 = ( v1w1 + v2w2) + v3w3 = (v1+v2)(w1+w2) + v3w3 = (v1+v2+v3) (w1+w2+w3) = v1w1 + (v2+v3)(w2+w3) = v1w1 + ( v2w2 + v3w3) = z1 + (z2 + z3) . The zero element in Z is 0 = 00 since vw + 0 = vw + 00 = (v+0)(w+0) = vw . The additive inverse of z = vw is -z = (-v)(-w) since z + (-z) = vw + (-v)(-w) = (v-v)(w-w) = 00 = 0 . For scalars a,b we have a(bz) = (ab)z compatibility since a(bz) = a(b[vw]) = a[ (bv)(bw) ] = (abv)(abw) = (ab)(vw) = (ab)z . Identity for scalar multiplication requires that 1(z) = z : 1(z) = 1(vw) = (1v)(1w) = vw = z . Distributive requirement #1: a(z1+z2) = az1+ az2 (a = scalar) a(z1+z2) = az3 = a(v3w3) = (av3)(aw3) = (av1+av2)(aw1+aw2) = (av1)(aw1) + (av2)(aw2) = a(v1w1) + a(v2w2) = az1+ az2 Distributive requirement #2 : (a+b)z = az + bz (a,b = scalars) (a+b)z = (a+b)(vw) = [(a+b)v][(a+b)w] = [av+bv][aw+bw] = (av)(aw) + (bv)(bw) = a(vw) + b(vw) = az + bz QED 5. vw does not commute Fact: vw ≠ wv unless V = W and v = w. (B.12) Proof: V≠W: If V≠W, the object wv makes no sense since it would require w ϵ V and v ϵ W. V=W: vw - wv = vw + (-w)(-v) = (v-w)(w-v) ≠ 0 unless v = w. Compare (B.12) to the Fact stated in and below (4.1.1), Fact: vw ≠ wv unless V = W and v = w. (4.1.1) However: There is certainly an isomorphism between VW and WV. Writing VW ~ WV one could certainly then say that vw ~ wv . The same could be said for the operator. 6. Visualization of the Direct Sum Consider this example v = ϵ R2 w = ϵ R3 z = vw = = ϵ R5 . (B.13) Here we visualize the direct sum vector z as a tall column vector which is the stacking of the two smaller column vectors v and w. In the tall column vector, v and w each occupy a private region. Here then are the rules (B.3) and (B.2) : v1w1 + v2w2 = + = = (v1+v2) (w1+w2) (B.14) (αv1)(αv2) = = α = α(v1v2) . (B.15) The fact (B.10) that dim(VW) = dim(V) + dim(W) is demonstrated by 5 = 2+3. The fact (B.12) that vw ≠ wv is demonstrated since (a,b,s,t,u)T ≠ (s,t,u,a,b)T. 7. Extension to multiple products The axioms for a triple direct sum are these, v1w1x1 + v2w2x2 + ... + vkwkxk = (v1+v2 + ...+ vk) (w1+w2 + ... + wk) (x1+x2 + ... + xk) (B.16) (αv)(αw)(αx) = α(vwx) (B.17) and from this one can imagine an arbitrary number of involved in a direct sum. One can derive these two equations from (B.1) and (B.2) by assuming associativity and then grouping things for example as v1w1x1 + v2w2x2 + ... + vkwkxk = (v1w1)x1 + (v2w2)x2 + ... + (vkwk)xk = [(v1w1) + (v2w2) + ... + (vkwk)] (x1 + x2 + ... + xk) = [ (v1+v2+ .. +vk) (w1+w2+ .. +wk)] (x1 + x2 + ... + xk) = (v1+v2 + ...+ vk) (w1+w2 + ... + wk) (x1+x2 + ... + xk) and (αv)(αw)(αx) =[(αv)(αw)] (αx) = [α(vw)] (αx) = α [ (vw x] = α[vwx] . One can define Z ≡ VWX zi ≡ viwix1 ϵ Z (B.6)' : (V,W,X) → VWX : (v,w,x) ↦ (vwx) We leave it to the reader to prove the following extended claims: Lemma: Given some zi, we can find vi, wi and xi such that zi = viwixi . (B.8)' Fact: If dim(V) = n, dim(W) = n' and dim(X) = n". then dim(VWX) = n + n' + n". (B.10)' Fact: If V.W and X are vector spaces, then Z = VWX is a vector space. (B.11)' The extension of the "tall vector" visualization to the triple sum seems fairly obvious where one ends up stacking three vectors to make a single tall vector. 8. Application to tensor products Define the vector product space Vk as in (5.1). If T = V2V3 one can write t = Σij Tij uiuj Σijk Tijk uiujuk ϵ T // t = vw If T = V1V2V3 one can write t = ΣiTiui Σij Tij uiuj Σijk Tijk uiujuk ϵ T // t = vwx and in this manner we eventually arrive at (5.4.3) for T(V) T(V) ≡ V0 V V2 V3 ....... (5.4.2) t = s ΣiTi ui Σij Tij uiuj Σijk Tijk uiujuk ... ϵ T(V), s ϵ K (5.4.3) For the space V*k the objects being direct-summed are functionals instead of tensors, but the formalism is exactly the same, τ = s ΣiTi λi Σij Tij λiλj Σijk Tijk λiλjλk + ...... s ϵ K (6.4.3) 9. Direct sum of operators (matrices): Block Diagonal Form Let vector spaces V and W have dimension n and n'. Let S, S1 and S2 be n x n matrices which we can regard as linear operators in V. Let T, T1 and T2 be n' x n' matrices which we can regard as linear operators in W. Then in the direct product space VW we can write these operator equations, (S1+S2)(T1+T2) = S1T1 + S2T2 α(ST) = αS αT (S1T1)(S2T2) = (S1S2)(T1T2) (B.18) and the action of operator S T of VW on a vector of VW is given by (S T) (vw) = (Sv)(Tw) . (B.19) Just as we visualized the direct sum of two vectors in (B.13), it is helpful to visualize the above three matrix equations graphically: (B.20) The direct sum operator ST is represented as an (n+n')x(n+n') matrix that is in "block diagonal form" where the matrix outside the blocks is filled with zeros. A triple direct sum has this visualization, (B.21) In all cases, the entire area outside the diagonal blocks is set to 0. The rules for such operators are, (S1 + S2)(T1 + T2)(R1 + R2) = S1T1R1 + S2T2R2 α(STR) = αS αT αR (S1T1R1)(S2T2R2) = (S1S2)(T1T2) (R1R2) (STR) (vwx) = (Sv)(Tw)(Rx) . (B.22) Appendix C: Theorems on Pre-Symmetrization The Rearrangement Theorem (A.1.3) is used to prove three other theorems (One, Two and Three) where we have attempted to abstract as much as possible the "permutational nature" of the objects involved by using a generic permutation space with elements |1,2...k>. Then in Section C.4 the theorems are summarized and are generalized to apply to arbitrary tensor products. Finally, the generic theorems are applied to tensors and tensor functions. The reader uninterested in the theorem details would do well to skip right to Section C.4. It is assumed that the reader is familiar with App. A.1 and the first part of App. A.2, C.1 Theorem One Consider the following set of k+k' integers, {1,2....k, k+1,k+2....k+k'} = {1,2....k+k' } . (C.1.1) Partition this list into a low and high group by defining z ≡ 1,2....k Z = k+1,k+2....k+k' (C.1.2) Then {1,2....k+k'} = {z,Z}. (C.1.3) Now let Q be a permutation of the lower integers [1,2...k] = z. There are k! possible permutations, so we know that ΣQ (1) = k! . (C.1.4) We can extend the meaning of Q so it applies to the entire set of integers {1,2....k+k'} merely by stating that this extended Q' does not alter the higher integers. Then Q'(z) = Q(z) = z' = some permutation of the lower integers (C.1.5a) Q'(Z) = Z // since Q has no effect on the higher integers (C.1.5b) Q'(z, Z) = {Q'(z), Q'(Z)} = {Q(z), Z} . *** (C.1.5c) Now imagine we have a function f of the lower integers and a function F of the higher ones, f[z] = f[1,2....k] F[Z] = F[k+1,k+2....k+k'] . (C.1.6) Here are two applications we shall consider later on, f[z] = f[1,2....k] = Tii...i = components of a rank-k tensor f[z] = f[1,2....k] = T(vi,vi, .... vi) = a rank-k tensor function . (C.1.7) Now let P be a general permutation of [1,2....k+k'] = [z,Z]. P(z,Z) = [P(z), P(Z)] . (C.1.8) Notice that QP and PQ are undefined since P and Q operate in different spaces, but Q'P and PQ' are both defined since both permutations Q' and P operate in the space of [1,2....k+k']. Recall now the meaning of S(Q) as the number of swaps required to go from z to Q(z) . This is the same as the number of swaps required to go from [z,Z] to Q'[z,Z] = [Q(z),Z]. Therefore S(Q) = S(Q') (C.1.9) We shall now prove the following theorem : Theorem One ΣP (-1)σS(P) f[P(z)] F[P(Z)] = ΣP (-1)σS(P) f^[P(z)] F[P(Z)] (C.1.10) where f^[z] ≡ (1/k!) ΣQ (-1)σS(Q) f[Q(z)] σ = 1 or 0 The purpose of σ is to state the theorem with and without the (-1)S(P) factor. We shall use the notation f^ in most of this section to apply for both values of σ, but at the end, we shall distinguish these two cases by writing: f^[z] ≡ (1/k!) ΣQ (-1)S(Q) f[Q(z)] // = Alt(f), see (A.2.1) fs[z] ≡ (1/k!) ΣQ f[Q(z)] . // = Sym(f), see (A.3.1) (C.1.11) At the end of this section we will show that the above Theorem One with σ = 1 and σ = 0 is equivalent to the statements: Alt(fF) = Alt(f^F) f^ = Alt(f) σ = 1 Sym(fF) = Sym(fsF) fs = Sym(f) σ = 0 (C.1.12) Proof of Theorem One: Our first task is to process the second line of (C.1.10), f^[z] = (1/k!) ΣQ (-1)σS(Q) f[Q(z)] = (1/k!) ΣQ (-1)σS(Q') f[Q'(z)] . // (C.1.9) and (C.1.5a) (C.1.13) Apply permutation P to the above equation and use (A.2.8) to get P f^[z] = f^[P(z)] = (1/k!) ΣQ (-1)σS(Q') f [PQ'(z)] . (C.1.14) Then, RHS (C.1.10) = ΣP (-1)σS(P) f^[P(z)] F[P(Z)] = ΣP (-1)σS(P) { (1/k!) ΣQ (-1)σS(Q') f[PQ'(z)] } F[P(Z)] // (C.1.14) for f^[P(z)] = (1/k!) ΣQ ΣP (-1)σS(PQ') f[PQ'(z)]} F[P(Z)] // reorder and use (A.1.10) = (1/k!) ΣQ ΣP(-1)σS(PQ') f[PQ'(z)]} F[PQ'(Z)] // Q'(Z) = Z from (C.1.5b) = (1/k!) ΣQ ΣP(-1)σS(P) f[P(z)]} F[P(Z)] // rearrangement theorem (A.1.3) = ΣP(-1)σS(P) f[P(z)]} F[P(Z)] { (1/k!) ΣQ (1)} // reorder = ΣP(-1)σS(P) f[P(z)]} F[P(Z)] { 1 } // ΣQ (1) = k! from (C.1.4) = LHS (C.1.10) QED (C.1.15) Recall now definitions of the generic Alt and Sym operators, [Alt(f)](1,2...k) ≡ (1/k!) ΣP (-1)S(P)f( P(1),P(2)...P(k) ) (A.2.1) (C.1.16) [Sym(f)](1,2...k) ≡ (1/k!) ΣP f( P(1),P(2)...P(k) ) (A.3.1) (C.1.17) Using (C.1.16) and (C.1.17) , the second line of (C.1.10) can be restated f^ = Alt(f) σ = 1 totally antisymmetric fs = Sym(f) σ = 0 totally symmetric . (C.1.18) Recall next the definition of a tensor product in our generic function space, (fg)(1,2,....k+k') ≡ f(1,2...k) g(k+1,k+2....k+k') . (A.2.19) (C.1.19) Then we can write (fF)(1,2,....k+k') = f(1,2..k)F(k+1,k+2...k+k') = f(z) F(Z) (f^F)(1,2,....k+k') = f^(1,2..k)F(k+1,k+2...k+k') = f^(z) F(Z) . (C.1.20) Theorem One (with σ = 1) can then be stated in this manner, ΣP (-1)S(P)(fF)(P(1),P(2),...P(k+k')) = ΣP (-1)S(P)(f^F)(P(1),P(2),...P(k+k')) (C.1.10)σ=1 Add a factor 1/(k+k')! to both sides and use the Alt definition (C.1.15) with k→ k+k' to get, [Alt(fF)](1,2...k+k') = [Alt(f^F)](1,2...k+k') or Alt(fF) = Alt(f^F) f^ = Alt(f) . (C.1.21) Taking σ = 0 in (C.1.10) gives ΣP (fF)(P(1),P(2),...P(k+k')) = ΣP (f^F)(P(1),P(2),...P(k+k')) (C.1.10)σ=0 Use this with the Sym definition (C.1.16) with k→ k+k' to get Sym(fF) = Sym(fsF) fs = Sym(f) (C.1.22) C.2 Theorem Two This section is a copy, paste and edit version of Section C.1. Equations that are the same have italicized equation numbers. Consider the following set of k+k' integers, {1,2....k, k+1,k+2....k+k'} = {1,2....k+k' } (C.1.1) Partition this list into a low and high half by defining z ≡ 1,2....k Z = k+1,k+2....k+k' (C.1.2) Then {1,2....k+k'} = {z,Z} (C.1.3) Now let R be a permutation of the upper integers {k+1,k+2....k+k'} = Z. There are k'! possible permutations, so we know that ΣR (1) = k'! (C.2.4) We can extend the meaning of R so it applies to the entire set of integers {1,2....k+k'} merely by stating that this extended R' does not alter the lower integers. Then R'(Z) = R(Z) = Z'' = some permutation of the upper integers (C.2.5a) R'(z) = z // since R has no effect on the lower integers (C.2.5b) R'(z, Z) = {R'(z), R'(Z)} = {z, R(Z)} *** (C.2.5c) Now imagine we have a function f of the lower integers and a function F of the higher ones, f[z] = f[1,2....k] F[Z] = F[k+1,k+2....k+k'] (C.1.6) We use [] in place of () merely to improve clarity below. Think of the integers as generic labels on the functions. Here are two applications we shall consider later on, F[Z] = F [k+1,k+2....k+k'] = Sii...i = components of a rank-k' tensor F[Z] = F [k+1,k+2....k+k'] = S(vi,vi, .... vi) = a rank-k' tensor function (C.2.7) Now let P be a general permutation of {1,2....k+k'} = {z,Z}. P(z,Z) = {P(z), P(Z)} (C.1.8) Notice that RP and PR are undefined since P and R operate in different spaces, but R'P and PR' are both defined since both permutations R' and P operate in the space of {1,2....k+k'}. Recall now the meaning of S(R) as the number of swaps required to go from Z to R(Z) . This is the same as the number of swaps required to go from {z,Z} to R'{z,Z} = {z,R(Z)}. Therefore S(R) = S(R') (C.2.9) We shall now prove the following theorem : Theorem Two ΣP (-1)σS(P) f[P(z)] F[P(Z)] = ΣP (-1)σS(P) f[P(z)] F^[P(Z)] (C.2.10) where F^[Z] ≡ (1/k'!) ΣR (-1)σS(R) F[R(Z)] σ = 1 or 0 The purpose of σ is to state the theorem with and without the (-1)S(P) factor. We shall use the notation F^ in most of this section to apply for both values of σ, but at the end, we shall distinguish these two cases by writing: F^[z] ≡ (1/k!) ΣQ (-1)S(Q) F[Q(z)] // = Alt(F), see (A.2.1) Fs[z] ≡ (1/k!) ΣQ F[Q(z)] // = Sym(F), see (A.3.1) (C.1.11) At the end of this section we will show that the above Theorem Two with σ = 1 and σ = 0 is equivalent to the statements: Alt(fF) = Alt(fF^) F^ = Alt(F) σ = 1 Sym(fF) = Sym(fFs) Fs = Sym(F) σ = 0 (C.2.12) Proof of Theorem Two: Our first task is to process the second line of (C.2.10), F^[Z] = (1/k'!) ΣR (-1)σS(R) F[R(Z)] = (1/k'!) ΣR (-1)σS(R') F[R'(Z)] / (C.2.9) and (C.2.5a) (C.2.13) Apply permutation P to the above equation and use (A.2.8) to get P F^[Z] = F^[P(Z)] = (1/k'!) ΣR (-1)σS(R') F [PR'(Z)] (C.2.14) Then, RHS (C.2.10) = ΣP (-1)σS(P) f[P(z)] F^[P(Z)] = ΣP (-1)σS(P) f[P(z)]{(1/k'!) ΣR (-1)σS(R') F [PR'(Z)]} // (C.2.14) for F^[P(Z)] = (1/k'!)ΣR ΣP(-1)σS(PR') f[P(z)] F [PR'(Z)] // reorder and (A.1.10) = (1/k'!)ΣR ΣP(-1)σS(PR') f[PR'(z)] F [PR'(Z)] // R'(z) = z from (C.2.5b) = (1/k'!)ΣR ΣP(-1)σS(P) f[P(z)] F [P(Z)] // rearrangement theorem (A.1.3) = ΣP(-1)σS(P) f[P(z)] F [P(Z)] { (1/k'!) ΣR (1) } // reorder = ΣP(-1)σS(P) f[P(z)] F [P(Z)] {1 } // ΣR (1) = k'! from (C.2.4) = LHS (C.2.10) QED (C.2.15) Using (C.1.15) and (C.1.16) , the second line of (C.2.10) can be restated F^ = Alt(F) σ = 1 totally antisymmetric Fs = Sym(F) σ = 0 totally symmetric . (C.2.18) Following the same arguments used the end of Section C.1, one obtains the following equivalent restatement of Theorem Two (just move the subscript from f to F) Alt(fF) = Alt(fF^) F^ = Alt(F) (C.1.21) Sym(fF) = Sym(fFs) Fs = Sym(F) (C.1.22) Alternate Proof of Theorem 2 An alternate proof of Theorem Two is two start with Theorem One and just make these changes z ↔ Z f↔F k↔k' Q→R Here is Theorem One ΣP (-1)σS(P) f[P(z)] F[P(Z)] = ΣP (-1)σS(P) f^[P(z)] F[P(Z)] where f^[z] ≡ (1/k!) ΣQ (-1)σS(Q) f[Q(z)] σ = 1 or 0 (C.1.10) and here is Theorem One with the above changes applied, ΣP (-1)σS(P) F[P(Z)] f[P(z)] = ΣP (-1)σS(P) F^[P(Z)] f[P(z)] where F^[z] ≡ (1/k'!) ΣR (-1)σS(R) F[R(Z)] σ = 1 or 0 . (C.1.10)swap This is the same as Theorem Two which we quote from above, ΣP (-1)σS(P) f[P(z)] F[P(Z)] = ΣP (-1)σS(P) f[P(z)] F^[P(Z)] where F^[Z] ≡ (1/k'!) ΣR (-1)σS(R) F[R(Z)] σ = 1 or 0 (C.2.10) We went ahead with the detailed proof for two reasons. First, the swap proof might not be convincing to the reader. Second, the detailed proof provides steps which are crucial to proving Theorem Three below. C.3 Theorem Three Now both functions have a ^ subscript : ΣP (-1)σS(P) f[P(z)] F[P(Z)] = ΣP (-1)σS(P) f^[P(z)] F^[P(Z)] (C.3.1) where f^[z] ≡ (1/k!) ΣQ (-1)σS(Q) f[Q(z)] F^[Z] ≡ (1/k'!) ΣR (-1)σS(R) F[R(Z)] σ = 1 or 0 The purpose of σ is to state the theorem with and without the (-1)S(P) factor. At the end of this section we will show that the above Theorem Three with σ = 1 and σ = 0 is equivalent to the statements, Alt(fF) = Alt(f^F^) . f^ = Alt(f) F^ = Alt(F) (C.3.5) Sym(fF) = Sym(fsFs) . fs = Sym(f) Fs = Sym(F) (C.3.6) This theorem will involve both R and Q, as well as R' and Q' from earlier sections. Note that R'Q' = Q'R' (C.3.3) because Q' acts only on the lower integers in (1,2...k+k') while R' acts only on the upper integers. Proof of Theorem Three: Recall these results from previous sections, f^[P(z)] = (1/k!) ΣQ (-1)σS(Q') f [PQ'(z)] (C.1.13) F^[P(Z)] = (1/k'!) ΣR (-1)σS(R') F [PR'(Z)] (C.2.13) Then, RHS (C.3.1) = ΣP (-1)σS(P) f^[P(z)] F^[P(Z)] = ΣP (-1)σS(P){ (1/k!) ΣQ (-1)σS(Q') f [PQ'(z)]}{ (1/k'!) ΣR (-1)σS(R') F [PR'(Z)]} = (1/k!)(1/k'!)ΣQΣR ΣP (-1)σS(PR'Q') f [PQ'(z)] F [PR'(Z)] // reorder and (A.1.10) = (1/k!)(1/k'!)ΣQΣR ΣP (-1)σS(PR'Q') f [PQ'R'(z)] F [PR'Q'(Z)] // Q'(Z) = Z from (C.1.5b) // R'(z) = z from (C.2.5b) = (1/k!)(1/k'!)ΣQΣR ΣP (-1)σS(PR'Q') f [PR'Q'(z)] F [PR'Q'(Z)] // (C.3.3) R'Q' = Q'R' = (1/k!)(1/k'!)ΣQΣR ΣP (-1)σS(P[R'Q']) f [P[R'Q'](z)] F [P[R'Q'](Z)] = (1/k!)(1/k'!)ΣQΣR ΣP (-1)σS(P) f [P(z)] F [P(Z)] // rearrangement theorem (A.1.3) = ΣP (-1)σS(P) f [P(z)] F [P(Z)] { (1/k!)ΣQ(1) }{ (1/k'!)ΣR(1) } // reorder = ΣP (-1)σS(P) f [P(z)] F [P(Z)] { 1 }{ 1 } // (C.1.4) and (C.2.4) = LHS (C.3.1) QED (C.3.4) The endgame steps of Section C.1 are identical here with the change F→F^, giving Alt(fF) = Alt(f^F^) . f^ = Alt(f) F^ = Alt(F) (C.3.5) Sym(fF) = Sym(fsFs) . fs = Sym(f) Fs = Sym(F) (C.3.6) C.4 Summary and Generalization Summary of the Three Theorems Theorems One, Two and Three have shown that, in our generic function space, Alt[TS] = Alt[T^S] = Alt[TS^] = Alt[T^S^] where T^ = Alt(T) S^ = Alt(S) (C.4.1) Sym[TS] = Sym[TsS] = Sym[TSs] = Sym[TsSs] where Ts = Sym(T) Ss = Sym(S) . (C.4.2) One can of course rewrite these statements as Alt[TS] = Alt[Alt(T)S] = Alt[TAlt(S)] = Alt[Alt(T)Alt(S)] (C.4.3) Sym[TS] = Sym[Sym(T)S] = Sym[TSym(S)] = Sym[Sym(T)Sym(S)] . (C.4.4) Intuitively these equations are easily interpreted: If one is going to totally antisymmetrize a tensor product, the act of pre-antisymmetrizing one or more of the tensors makes no difference. So adding any ^ subscripts to objects inside an Alt makes no difference. If one is going to totally symmetrize a tensor product, the act of pre-symmetrizing one or more of the tensors makes no difference. So adding any s subscripts to objects inside an Alt makes no difference. Various "theorems" can be generated by "adding hats" to the insides of an Alt expression. Example: Consider. Alt[ABC] = Alt[(AB)C] = Alt[(AB)^C] = Alt[Alt(AB)C] Alt[ABC] = Alt[A(BC)] = Alt[A(BC)^] = Alt[AAlt(BC)] (C.4.5) Therefore Alt[Alt(AB)C] = Alt[ABC] = Alt[AAlt(BC)] (C.4.6) Replacing A,B,C with the obscure names ω,η,θ gives Alt[Alt(ω η) θ] = Alt[ω η θ] = Alt[ω Alt(η θ)] . (C.4.7) This may be compared with Spivak page 80 from which we quote, (C.4.8) Generalization of the three theorems The theorems derived above can be generalized in the following manner. Suppose for example we have a set of integers 1,2,3......(k1+k2+...+kN) = 1,2,3....κN. Instead of partitioning Z into 2 groups Z = (z,z') as done above, we partition the integers into N groups Z = (z1, z2....zN) as follows: κ1 = k1 // "cumulative ranks", as in (7.11.6) κ2 = k1+ k2 κ3 = k1+ k2 + k3 ... κN = k1 + k2 + ... + kN = Σi=1N ki . (7.11.6) Z = (z1, z2....zN) (C.4.9) z1 = 1,2,3...κ1 // the partitions z2 = κ1+1,κ2+2,...κ2 z2 = κ2+1,κ2+2,...κ3 ... zN = κN-1, κN-1 + 1, ...κN . And instead of functions f and F, we have functions f1, f2....fN. Whereas for N = 2 we had 22-1 = 3 theorems, for general N there will be 2N - 1 theorems. If we define L ≡ ΣP (-1)σS(P)f1[P(z1)]f2[P(z2)] ... fN[P(zN)] // Left side of theorems (C.4.10) then here are those theorems: ( exercise for the reader: use induction or brute force ) 1. L = ΣP (-1)σS(P)(f1)^[P(z1)]f2[P(z2)] ... fN[P(zN)] 2. L = ΣP (-1)σS(P)f1[P(z1)](f2)^[P(z2)] ... fN[P(zN)] 3. L = ΣP (-1)σS(P)(f1)^[P(z1)](f2)^[P(z2)] ... fN[P(zN)] 4. L = ΣP(-1)σS(P)f1[P(z1)]f2[P(z2)](f3)^[P(z3)] ...fN[P(zN)] ...... (2N-1). L = ΣP(-1)σS(P)(f1)^[P(z1)](f2)^[P(z2)](f3)^[P(z3)] ...(fN)^[P(zN)] (C.4.11) Translated to Alt/Sym, notation, we then find for the case N = 3 Alt[TSR] = Alt[T^SR] = Alt[TS^R] = Alt[TSR^] = Alt[T^S^R] = Alt[T^SR^] = Alt[TS^R^] = Alt[T^S^R^] (C.4.12) Sym[TSR] = Sym[TsSR] = Sym[TSsR] = Sym[TSRs] = Sym[TsSsR] = Sym[TsSRs] = Sym[TSsRs] = Sym[TsSsRs] (C.4.16) One can write these using X^ = Alt(X) and Xs = Sym(X) to obtain nested equations as we did earlier. In general one can write Alt[(T1)(T2) ...... (TN)] = Alt[(T1)a(T2)a ...... (TN)a] (C.4.17) where each ai can independently be a blank or can be a ^. This then is the ultimate statement that arbitrary pre-antisymmetrizing of one or more tensors in a totally antisymmetric product makes no difference. Similarly, Sym[(T1)(T2) ...... (TN)] = Sym[(T1)a(T2)a ...... (TN)a] (C.4.18) where each ai can independently be a blank, (Ti) , or can be an s, (Ti)s. This then is the ultimate statement that arbitrary pre-symmetrizing of one or more tensors in a totally symmetric product makes no difference. Application to Tensors and Tensor Functions All the work done above in Appendix C has been "generic", meaning the various operations are with respect to generic permutation functions like f(1,2...k). The work can be applied to tensors or tensor functions according to these simple translation rules f[1,2....k] = Tii...i = components of a rank-k tensor f[1,2....k] = T(vi,vi, .... vi) = a rank-k tensor function (C.1.7) (C.4.19) For example, consider our result (C.4.1) above that Alt(TS) = Alt(T^S) (C.4.1) (C.4.20) In the generic space this equation means [Alt(TS)](1,2...k+k')= [Alt(T^S)](1,2...k+k') . (C.4.21) Translated from the generic space to the tensor space, one gets [Alt(TS)]ii...i = [Alt(Alt(T)S)]ii...i (C.4.22) where for example [Alt(T)]ii...i = (1/k!) ΣP (-1)S(P) Tii...i (C.4.23) Translated from the generic space to the tensor function space, one gets instead, [Alt(TS)](v1, v2.....vk+k') = [Alt(Alt(T)S)](v1, v2.....vk+k') (C.4.24) where for example [Alt(T)](v1, v2....vk) = (1/k!) ΣP (-1)S(P) T(vP(1), vP(2)....vP(k)) (C.4.25) Here we follow our convention of putting dual-space tensor names into script/italic font. Appendix D: A Unified View of Tensors and Tensor Functions In this section multiindex notations are shown in red to the right. D.1 Tensor functions in Dirac notation The vector space V has dimension n, and k ≤ n. The vector space is real, so <a|b> = <b|a>. In the bra-ket notation (Paul Dirac, 1939), a rank-k tensor functional T is represented by the bra <T| which is an element of the dual space V*k. Meanwhile, elements of the space Vk are written as kets which are a tensor product of smaller kets, | vi,vi, .... vi > = |vi> |vi> ..... |vi> . | vI> (D.1.1) Here the ir are labels, not components. Each vi is a vector in V having n components (vi)j . The tensor function T(vi,vi, .... vi) is then represented by the application of the functional <T| to vectors in Vk so that, <T | vi,vi, .... vi > = T(vi,vi, .... vi) . T(vI) = <T | vI > (D.1.2) Due to the tensor product (of vector spaces) construction of the "ket" shown in (D.1.1), the function shown in (D.1.2) is manifestly k-multilinear. The bra-ket notation represents an inner product (scalar product) so the spaces here are Hilbert spaces, not just vector spaces. The covariant tensor Tii....i we claim is this (each label ir ranges from 1 to n ), Tii....i = <T | ui,ui, .... ui > . TI = <T | uI > (D.1.5) The n vectors |ui> for i=1 to n form a basis for V, and the n*k kets | ui,ui, .... ui > form a basis for Vk. From this one would conclude from (D.1.2) that Tii....i = T(ui,ui, .... ui) ir = 1 to n TI = T(uI) (D.1.6) in agreement with our established fact (6.2.5). The contravariant form is then Tii....i = T(ui,ui, .... ui) . TI = T(uI) (D.1.7) Let us now assume that the n vectors |vi> for i=1 to n form some alternative basis for V, and then the n*k kets | vi,vi, .... vi > form an alternative basis for Vk. The dual basis is {vi} where vi vj = δij as in (2.11.2) for the ui basis and its dual uj. Looking at our two equations from above, <T | vi,vi, .... vi > = T(vi,vi, .... vi) ir = 1 to n (D.1.2) <T | ui,ui, .... ui > = Tii....i ir = 1 to n (D.1.5) one can say that the tensor Tii....i and the tensor function T(vi,vi, .... vi) are both representations of the same abstract tensor <T| in two different Vk bases, |vI> and |uI>. Both bases have dimension n*k. Recall T = ΣI TIuI ϵ Vk = a tensor |T> = ΣI TI |uI > T = ΣI TIλI ϵ V*k = a tensor functional <T| = ΣI TI <uI| T ~ T by the isomorphism V*k ~ Vk [see circa (2.11.12)] . (D.1.8) Notice that for the basis {vi}, <vI|vJ> = < vi|vj>< vi|vj> .... < vi|vj> = (vi vj)(vi vj) .... (vi vj) = δijδij...δij // see (2.3.2) for basis {vr} with dual basis {vr} = δIJ . // orthonormal basis in the multiindex notation (D.1.9) This result applies as well to the basis |uI>, so <uI|uJ> = <vI|vJ> = δIJ . (D.1.10) D.2 Basis change matrix The basis-change transformation matrix between the |vI> and |eI> bases is given by, MIJ ≡ < ui,ui, .... ui | vj,vj, .... vj > MIJ = <uI|vJ> (D.2.1) = < ui|vj>< ui|vj> .... < ui|vj> // see (2.9.17) = (ui vj)(ui vj) .... (ui vj) = λi(vj)λi(vj) .... λi(vj) // see (2.11.3) = (vj)i (vj)i ...(vj)i // see (2.11.7) = (vJ)I . // using a multiindex notation shown below (7.8.2) (D.2.2) There are n*k values for I and n*k values for J, so matrix M has dimension nk x nk. Entirely in multiindex notation, MIJ = <uI|vJ> = (vJ)I // mixed, see (2.1.6) line 2 or (D.2.3) MIJ = <uI|vJ> = (vJ)I . // pure covariant, see (2.1.6) line 4 The transpose is then, (MT)JI = MIJ = <uI|vJ> = <vJ| uI> = (vJ)I // Hilbert Space is real (MT)JI = MIJ = <uI|vJ> = <vJ| uI> = (vJ)I . (D.2.4) In the bra-ket notation completeness of an orthonormal basis is expressed this way: 1 = ΣJ |uJ><uJ| = ΣJ |uJ><uJ| = ΣJ |vJ><vJ| = ΣJ |vJ><vJ| . (D.2.5) Proof: (example) Consider a general Vk tensor T : (1) |T> = 1|T> = ΣJ |uJ><uJ| T> = ΣJ TJ |uJ> // so basis |uJ> must be complete (2) |uI> = 1|uI> = ΣJ |uJ><uJ|uI> = ΣJ |uJ>δJI = |uI> // why orthonormal is needed Therefore the up-tilt basis-change matrix M is real orthogonal, meaning MMT = 1 or MT = M-1 : (MMT)IK = ΣJ MIJ(MT)JK = ΣJ <uI|vJ><vJ| uK> = <uI| (ΣJ|vJ><vJ| )uK> = <uI | 1 | uK> = <uI | uK> = uI uK = δIK // see (2.11.2) or (D.2.6) MMT = 1 . // real orthogonal in the multi-index sense In quantum mechanics with complex Vk, one gets instead MM† = 1 (M is unitary) . The connection then between the tensors and tensor functions is given by, TI = <uI| T> = <uI| 1 | T> = <uI| ΣJ |vJ><vJ| T> = ΣJ <uI|vJ><vJ| T> = ΣJ <uI|vJ><T| vJ> = ΣJ MIJ T(vJ) . // MIJ = (vJ)I (D.2.7) Going the other direction, T(vI) = <vI | T > = <vI | 1 | T > = <vI | ΣJ |uJ><uJ| T > = ΣJ <vI|uJ> <uJ|T > = ΣJ (MT)IJ TJ . // (MT)IJ = (vI)J (D.2.8) Example of (D.2.7): Tii = Σj,j=1n (vj)i(vj)i T(vj,vj) ir = 1 to n or Tij = Σa,b=1n (va)i(vb)j T(va,vb) . // n2 terms in the sum (D.2.7a) Example of (D.2.8): T(vi,vi) = Σj,j=1n (vi)j (vi)j Tjj ir = 1 to n or T(vi,vj) = Σa,b=1n (vi)a (vj)bTab . // n2 terms in the sum (D.2.8a) Comment: These examples can be compared to a simple quantum mechanics case. Let |x> be a basis vector describing a 1D particle at location x (coordinate representation), and let |p> be a basis vector describing a plane-wave particle having momentum p (momentum representation). Then it turns out that the basis change matrix is <x|p> = ψp(x) = C eipx where C is a normalization constant. So the basis change "matrix" (continuous matrix subscripts p and x) is a function of p, just as the basis change matrix in (D.2.8a) is a function of vi and vj. D.3 Transformations of tensors and tensor functions In this section we write vectors in bold font. Consider two sets of n vectors vi and v'i where vi form a basis for V. One can then write, v'i = Qijvj i = 1,2...n implied sum on j (D.3.1) where Qij is a matrix describing the linear combinations of the vi that make up the v'i. Since the tensor function T(vi,vi, .... vi) is k-multilinear, one can certainly write T(v'i,v'i, .... v'i) = QijQij .... Qij T(vj,vj, .... vj) (D.3.2) or just showing the ket part, | v'i,v'i, .... v'i > = QijQij .... Qij | vj,vj, .... vj > . (D.3.3) Equation (D.3.2) vaguely resembles the Chapter 2 transformation of a covariant tensor field, T'ii...i (x') = RijRij .... Rij Tjj...j (x) (D.3.4) where x' = F(x) and dx'  = R dx. // R is the differential of F. The resemblance is perhaps closer if we restrict x' = F(x) to be a linear transformation, so then x' = R x or x'i = Rijxj . (D.3.5) The resemblance between (D.3.2) and (D.3.4) we claim is really superficial and misleading, which is the main reason for bringing it up. We just make a few comments on this matter. The linearized transformations of Chapter 2 like v'i = Rijvj for a vector vi are component transformations. The j on vj is a component index, and v'i = Rijvj (v' = Rv) is an instruction for creating a new vector v' by linearly combining the components of v. Transformation (D.3.5) is such a component transformation. In contrast, the transformation (D.3.1) that v'i = Qijvj is not a component transformation. It constructs n new vectors v'i by linearly combining the n vectors vi. The j on vj is a label, not a component index. In (D.3.4), the left-side object T'ii...i(x') has a prime on T. It is a tensor different from Tjj...j (x), and this would be true even if there were no x dependence of the field. In (D.3.2), the left-side object T(v'i,v'i, .... v'i) has no prime, it is the same T as on the right. In fact, as was shown in (2.11.f.8), the object T(v1, v2, ...vk) under any Chapter 2 component transformation transforms as a scalar, so there are no Rij or Qij matrices involved, T'(v'1, v'2, ...v'k) = T(v1, v2, ...vk) . T'(v'Z) = T(vZ) (2.11.f.8) where (v'r)i = Rij (vr)j r = 1...k implied sum on j i = 1...n T(v1, v2, ...vk) is a scalar because it is the scalar product of a rank-k tensor functional T = <T| with a rank-k tensor | v1, v2, ...vk >, just as <a | b> = a b is a scalar. (D.3.2) is nothing more than a statement that the tensor function T(v1, v2, ...vk) is k-multilinear. D.4 Tensor Functions and Quantum Mechanics Eq. (D.1.2) defining a tensor function as a bra-ket combination <T | vi,vi, .... vi > = T(vi,vi, .... vi) (D.1.2) has the following quantum mechanics incarnation, which was the original use Dirac intended for his bra-ket notation, <ψ| r1, r2...rk> = ψ(r1, r2...rk) . (D.1.3) Here object <ψ| plays the role of the abstract tensor <T|, and the generic arguments vi become the physical positions ri of k particles. The object ψ is a functional in V*k which gets applied to |r1,r2....rk> = |r1>|r2>....|rk> and the resulting function ψ(r1,r2...rk) is called a "wavefunction" which describes the "probability amplitude" that the k particles are near spatial locations r1, r2, ....rk. The probability that the k particles are near these spatial locations is given by |ψ(r1,r2...rk)|2dnr1dnr2 ...dnrk. "Near" means that ri lies somewhere in the range ri to ri+dnri , It happens that in quantum mechanics literature it is the ket that is the functional in V*k and the bra which is the element of Vk. So in a physics text one always sees equations like, ψ(r1, r2...rk) = <r1, r2...rk| ψ>. (D.1.4) This is a long-standing convention difference between the physics and math worlds. When talking about a functional f applied to a vector x, it seems natural to have f(x) = <f | x>, which is the math convention. The physics person writes <x|ψ> = ψ(x) and says that the state vector |ψ> is being projected onto the coordinate representation basis element <x|. Usually ψ is not called a "functional". A ket is thought of as a vector v, while the bra is a transpose vector vT and then <v1|v2> = v1Tv2 in a matrix notation sense, so here it seems logical to put "the vector", whether v, v2 or ψ, on the right. Our functional T maps elements of Vk to the real numbers, and <a|b> = <b|a> = a b, so one can "for free" switch the role of which is the functional, and which is the ket acted upon by the functional. In quantum mechanics the functional maps to complex numbers, and <a|b> = a* b where * is complex conjugation. Then <b|a> = b* a = (a b*)* = a* b = <a|b>*. And <v1|v2> = v1T* v2 = v1†v2 . It is a crucial element of quantum mechanics that the space Vk is complex and not real. In the math world, one usually sees instead <b|a> = b a* . If the k particles are electrons or other half-integral spin particles which are in a "symmetric spin state", then the wavefunction (D.1.4) must be replaced by [Alt(ψ)](r1, r2...rk) in order to make it be totally antisymmetric in the coordinates ri, as required by "Fermi statistics" for half-integral spin particles. We mention this just to show that the Alt operator and the permutation group in general have important applications in quantum mechanics. Appendix E: Chapter 10 with x' = F(x) changed to x = φ(t) The material here is just for completeness and is intended only for perusal. It shows how the development of Chapter 10 appears for x = φ(t) in place of x' = F(x). In some ways, the x = φ(t) results concerning differential forms are simpler that those expressed in the x' = F(x) notation. The less pleasant aspect is that tensors (including metric tensors), basis vectors, and their spaces need an extra x or t label to distinguish the two spaces (now t-space and x-space), whereas in the x' = F(x) approach this distinction is accomplished by a prime versus no prime. We do use part of this notation in Section 10.9 since it brings our results into a more standard form for comparison with other sources. Translation Table → x-space → t-space x'-space → x-space F → φ general transformation name x' = F(x) → x = φ(t) general transformation equation R,S → R,S differential matrices (no change in name) F* → φ* pullback function V → tV vector in t-space V' → xV vector in x-space e → te tangent base vectors in t-space u → tu axis-aligned basis vectors in t-space g → tg metric tensor in t-space u' → xu tangent base vectors in x-space e' → xe axis-aligned basis vectors in x-space g' → xg metric tensor in x-space Λ'k → xΛk dual space to Rm Λk → tΛk dual space to Rn λ'i = dx'i → xλi = dxi basis vector in dual space to Rm λi = dxi → tλi = dti basis vector in dual space to Rm (E.1) In this new notation, the "kinematics package" of (10.6.a.1) with adjustment (10.6.d.1) for "tall" R appears as (a) x = φ(t) xform Rij ≡ (∂xi/∂tj) = ∂j(t)xi R = (Dφ) xV = R tV vector Sij ≡ (∂ti/∂xj) = ∂j(x)ti (b) xei with (xei)j = δij axis-aligned basis vectors in x-space (i = 1..m) tei tei = S xei tangent base vectors in x-space (i = 1..n) (c) tui with (tui)j = δij axis-aligned basis vectors in t-space (i = 1..n) xui xui= R tui tangent base vectors in t-space (i = 1..n) (xui)j = Rjk (tui)k (d) x1 = | xei> <xei| = | xei> <xei| = | xui> <xui| = | xui> <xui| completeness in x-space t1 = | tei> <tei| = | tei> <tei| = | tui> <tui| = | tui> <tui| completeness in t-space (e) (tuj)i = tui tuj = <tui | tuj > = tgij = xui xuj = <xui | xuj > (tej)i = tui tej = <tui | tej > = Sij = Rji (xej)i = xei xej = <xei | xej > = xgij = tei tej = <tei | tej > (xuj)i = xei xuj = <xei | xuj > = Rij = Sji (f) tei = xgij tej xei = xgij xej tui = tgij tuj xui = tgij xuj tei = xgij tej xei = xgij xej tui = tgij tuj xui = tgij xuj (g) <tej | S | xei> = <xei | R | tej> = xgij <tej | S | xui> = <xui | R | tej> = Sij = Rji <tuj | S | xei> = <xei | R | tuj> = Rij = Sji <tuj | S | xui> = <xui | R | tuj> = tgij . (h) S = RT Sij = (RT)ij = Rji R = ST Rij = (ST)ij = Sji (i) SR = 1 SST = RTR = 1 (10.6.a.1) (E.2) The uniqueness table of (10.6.d.2) becomes the following Metric tensors tgij, tgij unique xgij unique, since xgij = RiaRjb tgab xgij not unique, since xgij = RiaRjb tgab = SaiSbj tgab and Sij not unique Transformation matrices Rij = Sji unique (tall R matrix from x' = F(x)) Rij = Sji unique since Rij = tgjaRia and both tgja and Ria are unique Rji = Sij not unique, see (10.6.c.3) Rij = Sji not unique, since Rij = xgia Raj and xgia not unique Axis-aligned basis vectors (tuj)i unique since (tuj)i = tgji (xej)i unique since (xej)i = xgij (= δij) (tuj)i unique since (tuj)i = tgji (xej)i not unique since (xej)i = xgij (tuj)i unique since (tuj)i = tgji (xej)i unique since (xej)i = xgij (tuj)i unique since (tuj)i = tgji (xej)i unique since (xej)i = xgij Tangent base vectors (tej)i not unique since (tej)i = Rji (xuj)i unique since (xuj)i = Rij (tej)i not unique since (tej)i = Rji (xuj)i not unique since (xuj)i = Rij (tej)i unique since (tej)i = Rji (xuj)i unique since (xuj)i = Rij (tej)i unique since (tej)i = Rji (xuj)i not unique since (xuj)i = Rij (10.6.d.2) (E.3) Other parts of the development translate as follows: Basis Vectors {tui} i = 1,2...n basis for t-space , axis-aligned (tui)j = δij components of these basis vectors in t-space . (10.6.e.1) (E.4) xui = (10.6.e.4) (E.5) R** = [xu1, xu2 ....xun] R has full rank n basis for TxM complete (10.6.e.5) (E.6) Non-Dual Pull Backs xui = R tui |xui> = R |tui> i = 1,2..n push forward (10.7.1) (E.7) tui = S xui |tui> = S |xui> i = 1,2..n pull back (10.7.2) (E.8) xui = R tui |xui> = R |tui> i = 1,2..n push forward tui = RT xui |tui> = RT |xui> i = 1,2..n pull back (10.7.4) (E.9) xei = R tei |xei> = R |tei> i = 1,2..n push forward tei = RT xei |tei> = RT |xei> i = 1,2..n pull back (10.7.5) (E.10) Dual Pull Backs (xui)T= (tui)T RT <xui| = <tui|RT i = 1,2..n push forward (tu)T = (xu)T R <tui| = <xu|R i = 1,2..n pull back (10.7.6) (E.11) (xei)T= (tei)T RT <xei| = <tei|RT i = 1,2..n push forward (tei)T = (xei)T R <tei| = <xei|R i = 1,2..n pull back (10.7.7) (E.12) <xei| φ* = <xei| R = Rij<tuj| pullback operator φ* = R <xui| φ* = <xui| R = <tui| (10.7.9) (E.13) (10.7.11) (E.14) Further translations of significant equations appear in Section 10.9 References Spivak Sjamaar Benn and Tucker Dirac Conrad (Chapter 9 and above 7.1.4) Shankar Messiah Denker both Birkhoff and MacLane books Lang Algebra book Roman book that contains a tensor product chapter. Suter Loring Tu's book Buck book