tensor and wedge v5

The Tensor Product and the Wedge Product PhL 11.10.15 1. The Tensor Product 2 1.1 The Tensor Product as a Quotient Space 3 1.2 The Tensor Product in Category Theory 8 2. A Brief Review of Tensors in Covariant Notation 9 2.1 R, S and how tensors transform : Picture A 10 2.2 The metric tensors g and g' and the dot product 14 2.3 The basis vectors en and en 15 2.4 The basis vectors un and un 17 2.5 Expansion of vectors onto the u and e basis vectors in x-space 17 2.6 A change in notation : Picture E 19 2.7 Simple analysis of a specified basis en using Picture E 22 2.8 The Outer Product of Tensors and Use of 24 2.9 The Inner Product (Contraction) of Tensors 27 Dot products in spaces VV, VW, VVV and VWX 29 2.10 Tensor Expansions 30 (a) Rank-2 Tensor Expansion and Projection 30 (b) Rank-k Tensor Expansion and Projection 32 2.11 The dual spaces V* , V*V* and V*k 33 (a) The basis λi, general vector in V*, isomorphism V ~ V*, scalarity of α(v) 33 (b) Rank-2 tensors in V*V* 35 (c) Rank-k tensor in V*k 37 (d) Multiindex notation 39 3. Outer Products and Kronecker Products 40 3.1 Outer Products Revisited: Compatibility of Chapter 1 and Chapter 2 40 3.2 Kronecker Products 43 4. The Wedge Product of 2 vectors built on the Tensor Product 50 4.1 The tensor product of 2 vectors 50 4.2 The tensor product of 2 dual vectors 53 4.3 The wedge product of 2 vectors 55 (a) Definition of the wedge product of 2 vectors and the space L2 55 (b) How big is the space L2 compared to the space V2? 57 (c) Wedge products and determinants: the geometry connection 58 (d) Components 60 (e) Dot Products 62 4.4 The wedge product of 2 dual vectors 62 5. The Tensor Product of k vectors : the vector spaces Vk and T(V) 66 5.1 Pure elements, basis elements, and dimension of Vk 66 5.2 Tensor Expansion for a tensor in Vk ; the ordinary multiindex 67 5.3 Rules for product of k vectors 67 5.4 The Tensor Algebra T(V) 68 5.5 Comments about tensors 70 5.6 The Tensor Product of two or more tensors in T(V) 70 6. The Tensor Product of k dual vectors : the vector spaces V*k and T(V*) 73 6.1 Pure elements, basis elements, and dimension of V*k 73 6.2 Tensor Expansion for a tensor in V*k ; the ordinary multiindex and Tk(V) 74 6.3 Rules for product of k vectors 76 6.4 The Tensor Algebras T(V*) and T(V) 76 6.5 Comments about tensors 78 6.6 The Tensor Product of two or more tensors in T(V*) 78 6.7 The Tensor Product of two or more functions in T(V) 80 7. The Wedge Product of k vectors : the vector spaces Lk and L(V) 82 7.1 Definition of the wedge product of k vectors 83 7.2 Properties of the wedge product of k vectors 84 7.3 The vector space Lk and its basis 87 7.4 Tensor Expansions for a tensor in Lk 89 7.5 The alt, Alt and Sym operators 92 7.6 Expansions for the wedge product of k vectors 95 7.7 Number of elements in Lk compared with Vk. 96 7.8 Multiindex notation 97 7.9 The Exterior Algebra L(V) 98 Associativity of the Wedge Product 98 7.10 The Wedge Product of two tensors 102 (a) Wedge Product of two tensors T and S 102 (b) Commutivity Rule for the Wedge Product of two tensors T and S 104 (c) Multiindex Notation Rehash of Section 7.10 105 7.11 The Wedge Product of N tensors in L(V) 107 8. The Wedge Product of k dual vectors : the vector spaces Λk and Λ(V) 111 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 direct 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 functions 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(w3,v4) + (v2, w1+w2) – (v2,w1) – (v2,w2) = an element of F(VxW) 3(w3,v4) = 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. 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) had 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 1.4 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, as the author likes to think of mathematics. 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 used 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. We need 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). 2. A Brief 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. 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, make a change of notation, and then apply the machinery to the simple situation of a change of basis. This then provides a bulletproof way to understand expansions of vectors and other objects onto basis vectors. This helps one avoid paradoxes like the following: Paradox. Let v = Σnvnen be the expansion of a vector v onto a basis en. Since em is a vector, use this same expansion to expand v = em to get em = Σn(em)n en. This implies that (em)n = δm,n. Thus, the only possible basis vectors em allowed in the universe are axis-aligned unit vectors! (2.1) This paradox involves confusion about the meaning of a component subscript. Hopefully this paradox will motivate the reader to slog through the following presentation of covariant notation. This review is based on our Ref ** 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 Ref ** 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 Ref **. 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 Ref ** 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}. Then x'-space is a vector space V'. We ask the reader not to confuse these vector space designations with the vector V (components Va and Va) which will be our prototype vector in the vector space V, V ϵ V . Were it not for some invested history, we would change symbols to avoid this confusion. 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 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) Here we are using the Einstein convention where repeated indices are implicitly summed. That is to say, an expression XbYb means ΣbXbYb. This convention reduces clutter in equations which are already plenty cluttered with lots of indices flying around. The matrices S and R in (2.1.2) and Fig (2.1.1) are linearizations of the generally non-linear transformation x' = F(x) in the close neighborhood of a selected point x in x-space and x' = F(x) in x'-space. Therefore, S and R are in general functions of x, though we suppress this dependence. 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 . (7.5.16)' (2.1.4) Then (2.1.2) can be written V'a = RabVb Rab ≡ (∂x'a/∂xb) = ∂bx'a V'a = RabVb Rab ≡ (∂xb/∂x'a) = ∂'axb . (2.1.5) Vectors which transform under 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)' (2.1.6) 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)' (2.1.7) 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)' (2.1.8) These are just renditions of (2.1.3) with (2.1.4) ( that is to say, SR = RS = 1 ). Using these rules, one shows that the inverses of the vector transforms shown above in (2.1.5) are these Vb = RabV'a (7.6.7)' Vb = RabV'a (7.6.8)' (2.1.9) where the summation index a on R is not "abutted" against the following vector. 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 Rab ≡ (∂x'a/∂xb) = ∂bx'a dx'a = Rabdxb Rab ≡ (∂xb/∂x'a) = ∂'axb . (2.1.10) From (2.1.9) the inverses of (2.1.10) are dxb = Rabdx'a Rab ≡ (∂xb/∂x'a) = ∂'axb (2.1.11) dxb = Rabdx'a Rab ≡ (∂x'a/∂xb) = ∂bx'a . 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 (2.1.12) 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)' (2.1.13) Comments: 1. We have suppressed the fact that in general everything above is a function of x (or 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 particle) and vectors that are fields like A'μ(x') = Λμν Aν(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. 2. 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) (2.1.14) V'a,b ≡ ∂'bV'a = (Rbc(x)∂c)(Rad(x)Vd(x)) = Rbc(x) Rad(x) (∂cVd(x)) + Rbc(x)(∂cRad(x)) Vd(x) . It is this second term that causes ∂cVd to not transform as a rank-2 tensor. It only transforms as a tensor if it happens that x' = F(x) is linear (as in special relativity). This problem is dealt with by introducing the covariant derivative Vd;c as discussed in Ref *** Appendix F, see for example (F.9.5). 3. We have chosen to write (∂x'a/∂xb) as Rab as a space-saving notation. This quantity is often called "the differential" of the transformation x' = F(x) at point x. Ref ** deals only with transformations where x and x' have the same number of components, 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. 4 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. 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 δa,c so gdngup = 1. 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 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 ... ] 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' and either is called the invariant distance. As per Comment 1 of the previous section, in general everything is a function of x. For example, Va(x) = gab(x)Vb(x) V'a(x') = g'ab(x')V'b(x') g'ab(x') = Raa'(x) Rbb'(x) ga'b'(x) . 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). They 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 at some point x, while the reciprocal vectors en are "dual" to the en in that en em = δnm . In fact we have en em = g'nm en em = δnm en em = g'nm (7.18.1)' (2.3.2) For coordinate line examples, see e.g. (3.2.8)' , (3.4.3)' or (3.4.7)'. For "duality" see text above (6.2.8)'. The reciprocal base vectors are the tangent base vectors of the inverse transformation x = F-1(x'). 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 (2.3.3), 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 are obviously dependent on location in space. The metric tensor is gab = and also depends on spatial location through r. It turns out that the labels on the basis vectors are raised and lowered by g', just the way the component indices on a vector are raised and lowered: en = g'ni ei en = g'ni ei (7.18.1)' (2.3.3) These basis vectors are like any other vectors in x-space, and so they have contravariant and covariant components. 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) Assuming these components are x-space components, we look at (en)i = Rni and make these observations: (1) the n index of (en)i goes up and down with g' as shown in (2.3.3). (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 (2.2.13). 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 sum is over the component 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 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. Comment: 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 functions Rni(x) from (2.3.4). 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. We shall pursue this subject a bit more in Section 2.7 below for the special case of constant en. 2.4 The basis vectors un and un One can also define a set of "axis aligned unit vectors" in x-space as follows (un)i = δni (un)i = δni (7.18.3)' (2.4.1) which can be compared with (2.3.4). Relations involving the u basis vectors are un um = gnm un um = δnm un um = gnm (7.18.3)' (2.4.2) and un = gni ui un = gni un (7.18.3)' (2.4.3) 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 . (2.4.4) According to (2.3.3), 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 (2.4.4) is "covariant" (even though it is not a true tensor equation), so we can at once write out all four forms the dot products between the e and u basis vectors : en um = Rnm en um = Rnm en um = Rnm en um = Rnm . (2.4.5) Reminder: In general, we have en(x) and un(x) and gnm(x). Then (un(x))i = δni is valid at every point x in x-space. 2.5 Expansion of vectors onto the u and e basis vectors in x-space A vector V can be expanded onto the bases defined above as follows (implied sum on n), V = Vn un where un V = Vn V = Vn un where un V = Vn V = V'n en where en V = V'n V = V'n en where en V = V'n . (7.13.10)' (2.5.1) This reveals the interesting fact that the coefficients for expansions on the u basis vectors are the x-space components of the vector V, whereas the coefficients for expansions on the e basis vectors are the x'-space components of the vector. The projection equations on the right all arise from the duality relations of the vector pairs, en em = δnm and un um = δnm . To verify that V'n are the expansion coefficients on the third line, consider en V = en (Vk uk ) = Vk (en uk) = Vk Rnk = RnkVk = V'n (2.5.2) where we used the first line of (2.5.1), then the first line of (2.4.5) and then (2.1.2). Inserting the dot products shown in (2.5.1) into the expansions of (2.5.1) gives. V = (un V) un V = (un V) un V = (en V) en V = (en V) en . (2.5.3) We now write the above four expansions for the cases V = um, um, em, em . Please ignore the rightmost column of equations for now, they will be referenced by the next section. um = (un um) un = δnm un = um // not very interesting (um(u))n = δnm um = (un um) un = gnm un // a fact already known in (2.4.3) (um(u))n = gnm um = (en um) en = Rnm en (um(e))n = Rnm um = (en um) en = Rnm en (um(e))n = Rnm um = (un um) un = gnm un // a fact already known in (2.4.3) (um(u))n = gnm um = (un um) un = δnm un = um // not very interesting (um(u))n = δnm um = (en um) en = Rnm en (um(e))n = Rnm um = (en um) en = Rnm en (um(e))n = Rnm em = (un em) un = Rmn un (em(u))n = Rmn em = (un em) un = Rmn un (em(u))n = Rmn em = (en em) en = δnm en = em // not very interesting (em(e))n = δnm em = (en em) en = g'nm en // a fact already known in (2.3.3) (em(e))n = g'nm em = (un em) un = Rmn un (em(u))n = Rmn em = (un em) un = Rmn un (em(u))n = Rmn em = (en em) en = g'nm en // a fact already known in (2.3.3) (em(e))n = g'nm em = (en em) en = δnm en = em // not very interesting (em(e))n = δnm (2.5.4) The new information these equations provide is this: um = Rnm en em = Rmn un um = Rnm en em = Rmn un um = Rnm en em = Rmn un um = Rnm en em = Rmn un (2.5.5) In each pair, the second equation can be obtained from the first by reversing the tilt of the summation index n. These equations show that the matrix R in its various forms is the "basis change matrix" which relates the u and e basis vectors. Each u basis vector is a certain linear combination of the e basis vectors, and vice versa, and the R matrix provides the coefficients for those linear combinations. As noted in the Comment at the end of Section 2.3, one can regard the above equations regarding en to apply to an arbitrary set of basis vectors en(x), with corresponding dual en(x). 2.6 A change in notation : Picture E Due to the way expansions work, as shown in (2.5.1), we shall now cosmetically modify Picture A noted above, so that x-space becomes u-space (right side) and x'-space becomes e-space (left side). The metric tensors will then be called g(u) on the right, and g(e) on the left. (2.1.1) (2.6.1) The equations of all earlier sections of Chapter 2 can be mapped from Picture A to Picture E as follows: X → X(u) where X is any x-space object X' → X(e) where X' is any x'-space object However, the R matrix elements stay the same, since they are neither x-space nor x'-space objects. Examples: Rab → Rab g→ g(u) dx'a = Rabdxb → dx(e)a = Rabdx(u)b g' →g(e) V'a = RabVb → V(e)a = RabV(u)b V = Vn un → V = V(u)n un V = V'n en → V = V(e)n en . The four expansions of (2.5.1) are now (showing the n sum explicitly) V = Σn V(u)n un where un V = V(u)n V = Σn V(u)n un where un V = V(u)n V = Σn V(e)n en where en V = V(e)n V = Σn V(e)n en where en V = V(e)n from (2.5.1) (2.6.2) We can now make the following observations: 1. When a vector V is expanded on the un, the expansion coefficients are V(u)n . 2. When a vector V is expanded on the un, the expansion coefficients are V(u)n . 3. When a vector V is expanded on the en, the expansion coefficients are V(e)n . 4. When a vector V is expanded on the en, the expansion coefficients are V(e)n . (2.6.3) Notice that there is a distinction for example between V(u)n and V(e)n . Here the index n labels the vector component, but these two components are not the same. The number V(u)n is the component of V in the un basis, whereas V(e)n is the component of the same vector V but in the en basis. Since the bases are different, the components are different. We can apply the four equations shown on the right of (2.6.2) sequentially to the basis vectors V = um, um, em, em to obtain a set of 16 equations. The very first equation would be un um = um(u)n. One can then look up the dot product to find that un um = δnm and then one gets the result that um(u)n = δnm . The next equation is un um = um(u)n and we look up this dot product to find un um = gnm and so um(u)n = gnm. Rather than do all these calculations, since the dot products are already listed in (2.5.4), we can just read off the 16 results we want. This then produces the rightmost column of equations in (2.5.4) which we transcribe here: (um(u))n = δmn (um(u))n = g(u)mn (em(u))n = Rmn (em(u))n = Rmn (um(u))n = g(u)mn (um(u))n = δmn (em(u))n = Rmn (em(u))n = Rmn (um(e))n = Rnm (um(e))n = Rnm (em(e))n = δmn (em(e))n = g(e)mn (um(e))n = Rnm (um(e))n = Rnm (em(e))n = g(e)mn (em(e))n = δmn . (2.6.4) We have replaced g→g(u) and g'→g(e) and have made cosmetic changes such as δnm = δmn as well as g(u)mn = g(u)nm since all metric tensors are symmetric. Equations (2.6.4) give the contravariant (up) and covariant (down) components of all four basis vector types evaluated in both the u and e bases. Notice in (2.6.4) [ col 3 row 3] that (em(e))n = δmn . According to the Comment at the end Section 2.3, this equation applies for an arbitrary set of basis functions em(x). The equation is eminently reasonable. Suppose we expand em = [ Σn (em(e))n en] . We can see that we must have (em(e))n = δmn. Expanding a basis vector on its own basis yields a single term in the sum. Being true for any basis, (em(e))n = δmn must also be true for the u basis, and we see as well that (um(u))n = δmn in (2.6.4) [ col 2 row 2]. Let's now review the paradox (2.1) presented at the start of Chapter 2, but in our "improved" notation. We first restate the paradox in covariant notation: Paradox. Let v = Σnvnen be the expansion of a vector v onto a basis en. Since em is a vector, use this same expansion to expand v = em to get em = Σn(em)n en. This implies that (em)n = δmn. Thus, the only possible basis vectors em allowed in the universe are axis-aligned unit vectors! (2.6.5) Now we go through the same paradox presentation with more precise notation; Paradox? Let v = Σnv(e)nen be the expansion of a vector v onto a basis en. Since em is a vector, use this same expansion to expand v = em to get em = Σn(em(e)n) en. This implies that (em(e))n = δmn. This agrees with (2.6.4) column 3 row 3. There is no paradox. Without clear labeling, one assumes that by default the paradox statement (em)n = δmn means (em(u))n = δmn and this seems to rule out general basis vectors. In reality, (em)n = δmn means (em(e))n = δmn which is blatantly true as noted above, and this still allows us to have arbitrary basis vector components (em(u))n = Rmn as indicated by (2.6.4) col 3 row 1. (2.6.6) Conclusion: One must be careful when dealing with components of vectors and tensors to understand the space to which a vector or tensor belongs. For example, in Picture A we had a vector V'a(x') belonging to x'-space and we are now writing that as (V(e))a (x(e)) belonging to e-space using Picture E. When V is a basis vector like en (n is a label not an index), we have (en')a(x') → (en(e))a (x(e)). Now look back at (2.3.4) which says this (en)i = Rni (en)i = Rni . (7.18.1)' (2.3.4) Which components of en and en are implied here? Looking at (2.6.4) we see that the more precise equations are these, (en(u))i = Rni (en(u))i = Rni . (2.6.7) col 3 row 1 col 4 row 2 We now restate the transformation rules for rank-1 and rank-2 tensors in our (2.6.1) Picture E context: [V(e)]a = Rab[V(u)]b from (2.1.5) [M(e)]ab = Raa' Rbb' [M(u)]a'b' from (2.1.6) (2.6.8) Showing the field dependence on x adds a lot of clutter, so we keep it suppressed. For example [V(e)(x(e))]a = Rab(x(u))[V(u)(x(u))]b . (2.6.9) 2.7 Simple analysis of a specified basis en using Picture E As a first step, select g(u) = 1 so that u-space is the usual Cartesian space and the un = un are the usual orthonormal unit vectors of Cartesian space. To save space, we shall refer to g(e) as g. Suppose we are handed a set of constant-in-space basis vectors en specified by their components relative to the u-space axes. From (2.6.4), (em(u))n = Rmn (2.7.1) so we know the matrix Rab. What is the simplest way to fit this scenario into the tensor environment of Picture E ? 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 gnm can be computed from the dot product gnm = en em = (en(u))i (em(u))i = RniRmi . (2.7.2) This gnm can then be inverted to determine gnm. The reciprocal base vectors are then given by (2.3.3), en = gnm em (2.7.3) with components (en(u))i = gnm (em(u))i = gnm Rmi . (2.7.4) Since g(u) = 1, the second index on R is allowed to move up and down for free. We then rewrite the above equations as (em(u))n = Rmn // the em are the rows of matrix R gnm = RniRmi = RniRTim = (RRT)nm g = RRT en = gnm em = hnm em // where we define hnm ≡ gnm (en(u))i = hnmRmi = (hR)ni // the en are the rows of matrix (hR) (2.7.5) Exercise: You are handed these three constant vectors en in a 3-dimensional u-space, e1 = (2,-1,3) // for example, (e1(u))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" (2.7.6) Use Maple to compute g, h and then (hR). First enter the three en vectors and construct matrix R, Then compute the covariant metric tensor g = g(e) = 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 : 2.8 The Outer Product of Tensors and Use of Going back to our economical Picture A notation, consider two vectors which transform in the usual rank-1 tensor manner relative to some underlying transformation F (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 (2.1.6), 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. 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", 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 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 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". 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)(e)] ij = RiaRjb [(vw)(u)]ab // Picture E (2.6.1) or (vw)' ij = RiaRjb (vw)ab // Picture A (2.1.1) (2.8.13) As for bases, one can consider (ab)ij = aibj of (2.8.9) in several situations depending on the spaces in which the two vectors lie. Here are two examples ( the basis must of course match on the two sides ) [(ab)(e)]ij = a(e)ib(e)j [(ab)(e,u)]ij = a(e)ib(u)j . (2.8.14) As an example of the second equation, one could write [(enem)(e,u)]ij = (en(e))i (em(u))j = δni Rmj (2.8.15) where the last expression comes from (2.6.4). As a final outer product example, consider the outer product of three vectors, Mijk = aibjck a,b,c ϵ V (2.8.16) Using our naming method for outer products, this becomes (abc)ijk = aibjck (2.8.17) with this obvious extension to the outer product of any number of vectors (abc....)ijk.... = aibjck.... (2.8.18) Associativity of The outer product operator as defined here is an associate operator. Consider for example, (ABv)abcde = AabBcdve [(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 (2.1.8), such internal index contractions behave as a scalar, which is to say, behave as if they weren't there at all. A proof appears in Ref ** (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 tensor indicates 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 (2.3.3) 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 (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 is going 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. A few 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 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. Recall from (2.6.2) that for rank-1 we had the expansion V = Σa [V(e)]a ea . (2.10.1) (a) Rank-2 Tensor Expansion and Projection For a rank-2 tensor M we then write (Picture E), M = Σab [M(e)]ab eaeb . (2.10.2) As a check, take a tensor component of both sides of (2.10.2) in the ei basis : [M(e)]ij = { Σab [M(e)]ab eaeb}(e)ij = Σab [M(e)]ab (eaeb)(e)ij = Σab [M(e)]ab (ea(e))i (eb(e))j // (2.8.9) = Σab [M(e)]ab δaiδbj // (2.6.4) [ col 3 row 3] = [M(e)]ij (2.10.3) so things are consistent. 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 us a tensor expansion in VW where the basis vectors are en and e'n for V and W, 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 variables 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) Exercise: Consider these two expansions of the rank-2 tensor M, M = Σab[M(e)]ab eaeb M = Σab[M(u)]ab uaub (2.10.8) How are the coefficients in these expansions related? Method 1: Use the result in (2.5.5) that em = Rmiui. Then M = Σab[M(e)]ab eaeb = Σab[M(e)]ab (Raiui)(Rbjuj) = Σab[M(e)]abRiaRbj (ui)(uj) // next do i ↔ a and j↔ b = Σab { [M(e)]ijRaiRjb} uaub (2.10.9) Comparing with the second line of (2.10.8) one sees that [M(u)]ab = Σab RaiRjb [M(e)]ij . (2.10.10) Method 2: Consider the rule for the transformation of a rank-2 tensor as stated in (2.1.6) but converted to Picture E: [M(e)]ab = Σa'b'Raa' Rbb' [M(u)]a'b' . (2.10.11) We can invert this equation using the R matrix orthogonality rules (2.1.8), but instead we just use this rule of thumb (which can be derived from those rules): Reverse the tilt on all R matrices, then swap indices on each R. // inversion rule (2.10.12) Thus, the inversion of the above line is [M(u)]ab = Σa'b'Ra'a Rb'b [M(e)]a'b' (2.10.13) and this then agrees with (2.10.10). (b) Rank-k Tensor Expansion and Projection Defining Tii....i ≡ [T(e)] ii....i, the expansion of a general rank-k tensor T with contravariant coefficients can be written T = Σii....i Tii....i (eiei... ei) . (2.10.14)  The coefficients can be projected out according to (eiei... ei) T = Tii...i (2.10.15) with an appropriate generalization of the dot product to Vk , (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 = (v1 u1) (v2 u2) .... (vk uk) . (2.10.15) In ordinary multiindex notation, I ≡ i1, i2...ik eI ≡ ei,ei, .... ei TI ≡ Tii...i (2.10.16) the above equations can be compactly written as T = ΣI TI eI (2.10.14) (2.10.17) eI T = TI . (2.10.15) (2.10.18) 2.11 The dual spaces V* , V*V* and V*k In our treatment of the dual space objects we shall attempt to use Greek or script letters for all dual space vectors and tensors encountered. Vectors in V will continue to be represented by Latin 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. 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". (a) The basis λi, general vector in V*, isomorphism V ~ V*, scalarity of α(v) The basis λi Given n vectors {ei} which form a basis for V, one can find another set of n basis vectors {ei} in V such that, ei ej = δij . (2.3.2) (2.11.2) These are the same ei discussed in Section 2.3. Section 2.7 gave an example of computing the ei from the ei. We can then define a basis {λi) for V* as a set of n linear functionals λi such that λi(v) = ei v . λi: V → K (2.11.3) Recall that vectors like v in V have components which lie in field K, usually taken to be the reals. Notice that i is a label, not a component. Functional λi is manifestly linear since λi(kv) = kλi(v) and λi(v + v') = λi(v) + λi(v') k ϵ K . (2.11.4) Here v and v' are vectors in V. As noted earlier, the definition of linearity is often combined into a single statement, λi(k1v + k2v') = k1λi(v) + k2λi(v') k1, k2 ϵ K . (2.11.5) The vectors ei are sometimes called "covectors" or "reciprocal vectors" or "dual vectors" which form a "dual basis" to the ei. From (2.11.3) and (2.11.2) one then has λi(ej) = ei ej = δij . (2.11.6) Therefore, λi(v) = λi(Σj vj ej) = Σj vjλi(ej) = Σj vjδij = vi . λi: V → K (2.11.7) The linear functional λi applied to vector v generates the e-basis component vi of v. For this reason, the λi(v) is often referred to as "the ith coordinate function". Comments: 1. The basis functionals λi can be defined directly from λi(ej) = δij without use of the dot product. Note that (2.11.7) λi(v) = vi does not depend on the existence of a dot product. The dot product implies that the vector space V is also a Hilbert Space and one need not assume this fact, but in our applications V will always be a Hilbert Space. 2. The covector ei is associated with the basis functional λi but one should not identify λi = ei. For one thing, λi is a scalar-valued functional while ei is a vector. One sometimes sees λi(v) written as ei(v) or as e*i(v), but we use λi to emphasize the distinction between λi and ei. [ other authors?? ] 3. It is not hard to show that λi is a basis for V*. The proof relies on the fact that, since ei form a basis for V, the covectors ei also form a basis for V. 4. Some authors use notation v*i(v) in place of our λi(v) and vi in place of our ei. General vector in V* A general linear functional α in V* can be written as a linear combination of the basis functionals λi, where the coefficients αi ϵ K form a vector α in V (note various distinct meanings for symbol α ) α = Σiαiλi = general vector in V* α(v) = Σiαiλi(v) α: V → K . (2.11.8) On the right we show the corresponding function α(v). Evaluating at v = ej, α(ej) = Σiαiλi(ej) = Σiαiδij = αj // using (2.11.6) (2.11.9) so α(ei) picks off the coefficient αi appearing in the expansion (2.11.8). Isomorphism between V and V* Consider now, α(v) = Σiαiλi(v) = Σiαivi = α v = Σiαivi . (2.11.10) There is one-to-one mapping between the set of vectors α in V and the set of linear functionals α in V*. Consider for example, two functionals α1 and α2, α1(v) = α1 v for all v in V α2(v) = α2 v for all v in V (2.11.11) Then, α1 ≠ α2 α1(v) cannot equal α2(v) for all v in V, so α1 ≠ α2 α1 ≠ α2 α1 ≠ α2 , otherwise one would have α1 = α2 (2.11.12) This one-to-one mapping V ↔ V* with α ↔ α exhausts both spaces so is really a bijection which we shall call an isomorphism. For this reason, V and V* have the same dimension n. The basis vectors of V are the n basis vectors ei , while the basis vectors for V* are the n basis functionals λi. Scalar under transformations Imagine a version of Picture A (2.1.1) where we transform from x-space with basis vectors ei to some x'-space with basis vectors e'i. Under such a transformation, a dot product like α v transforms as a scalar. So, whereas v and α transform as vectors, the function α(v) transforms as a scalar field, just as in (2.2.6). That is to say, α(v) = α v = α' v' = α'(v') . (2.11.13) Although the function α(v) transforms as a scalar, the associated vector α (covector) transforms as a rank-1 tensor (vector), and it is through this isomorphic connection that we loosely refer to the function α(v) as being a rank-1 tensor. When functional α is expanded on the basis functionals λi, the coefficients αi are the rank-1 tensor components, and these coefficients are projected out by (2.11.9). (b) Rank-2 tensors in V*V* We want the object λiλj to be a bilinear functional over the space VxV such that λiλj: VxV → K . λiλj ϵ V*V* (2.11.14) The natural way to accomplish this desire is to write (λiλj)(v1,v2) = λi(v1)λj(v2) = scalar * scalar = scalar ϵ K (2.11.15) Then using (2.11.7) this can be written (λiλj)(v1,v2) = (v1)i(v2)j . (2.11.16) The function (λiλj)(v1,v2) is manifestly bilinear in its two arguments. Recall from (1.1.5) and (1.1.6) that bilinearity is the key ingredient in the tensor product of two vectors. The most general bilinear functional T in V*V* can be expanded in the following manner, analogous to (2.10.2), T = Σab Tab λaλb // Tab = [T(λ)]ab in the sense of (2.6.8) (2.11.17) where the coefficients Tab are assumed to transform as a covariant rank-2 tensor. Again, due to the isomorphism between the set of such rank-2 tensors and the set of bilinear functionals T, we refer to the functional T as a rank-2 tensor, or just a 2-tensor, in V* V*. One may evaluate the functional T at (v1,v2) using (2.11.16) to get T(v1,v2) = Σab Tab (λaλb)(v1,v2) = Σab Tab (v1)a(v2)b (2.11.18) where now the function T(v1,v2) is manifestly bilinear in its arguments. Evaluation at (e1,e2) yields T(ei,ej) = Σab Tab λa(ei) λb(ej) = ΣabTab δaiδbj = Tij . (2.11.19) This shows that the coefficients Tij of the bilinear functional T can be obtained from the function T(v1,v2) evaluated at the basis vectors (ei,ej), similar to what happened in (2.11.9). Since the rightmost expression in (2.11.18) is a fully contracted outer product of three tensors (Section 3.1), one sees that T(v1,v2) transforms as a scalar field over V x V, T'(v'1,v'2) = T(v1,v2) . (2.11.20) Once again, although T transforms as a scalar, it is referred to as a 2-tensor because its expansion coefficients form a rank-2 tensor. As a special case, consider T = α β. We find that α β = Σab (α β)ab λaλb = Σabαaβb λaλb (2.11.21) where we now use the outer product (2.8.9) for the two vectors α,β ϵ V*. Evaluation at (v1,v2) along with (2.11.16) gives (α β)(v1,v2) = Σabαaβb λaλb(v1,v2) = Σabαaβb (v1)a(v2)b = [Σaαa (v1)a][Σbαb (v1)b] = α(v1)β(v2) . (2.11.22) Once again, α β is called a 2-tensor even though (α β)(v1,v2) transforms a scalar. Evaluation at (e1,e2) with use of (2.11.9) yields (α β)(ei,ej) = α(ei)β(ej) = αiβi = (α β)ij (2.11.23) analogous to (2.11.19). (c) Rank-k tensor in V*k The equations of this subsection appear rather complicated, being written in a systematic notation. The reader is encouraged to refer back to the equations for k = 2 of the previous subsection if confusion arises. The related equation numbers of are shown in italics. As a shorthand notation we write, Vk ≡ VxVx....xV k factors // Cartesian product Vk ≡ VV....V k factors // tensor product V*k ≡ V*V*....V* k factors // tensor product . (2.11.24) Various equations above can be generalized as follows: (λiλi ... λi): Vk → K . (2.11.14) (2.11.25) (λiλi ... λi)(v1,v2....vk) = λi(v1)λi(v2) ... λi(vk) (2.11.15) (2.11.26) (λiλi ... λi)(v1,v2....vk) = (v1)i(v2)i ... (vk)i . (2.11.16) (2.11.27) The full set (eiei ... ei) for is = 1,2..n forms a basis for Vk The full set (λiλi ... λi) for is = 1,2..n forms a basis for V*k. The most general k-multilinear (rank-k) functional T in V*k can be expanded as T = Σii...i Tii...i (λiλi ... λi) (2.11.17) (2.11.28) with T(v1,v2, ...vk) = Σii...i Tii...i (λiλi ... λi)(v1,v2, ...vk) = Σii...i Tii...i λi(v1)λi(v2) ... λi(vk) = Σii...i Tii...i (v1)i(v2)i ... (vk)i . (2.11.18) (2.11.29) The function T(v1,v2, ....vk) is thus seen to be k-multilinear in its k vector arguments. As a special case, T(ej,ej, ... ej) = Σii...i Tii...i λi(ej)λi(ej) ... λi(ej) = Σii...i Tii...i δjiδji ... δji = Tjj...j (2.11.19) (2.11.30) showing how this evaluation of T(v1,v2, ....vk) projects out the rank-k tensor component Tjj....j . Since (2.11.29) shows T(v1,v2, ....vk) as a full contraction of tensors, it transforms as a scalar, T'(v'1,v'2, ....v'k) = T(v1,v2, ....vk) . (2.11.20) (2.11.31) As a special case, consider T = αj αj ... αj, which is the tensor product of k different rank-1 tensors in V*. We find that : (αj αj ... αj) = Σii...i (αj αj ... αj)ii...i (λiλi ... λi) = Σii...i (αj)i(αj)i ...(αj)i (λiλi ... λi) . (2.11.21) (2.11.32) Evaluation at (v1,v2, ....vk) along with (2.11.27) gives (αj αj ... αj)(v1,v2, ....vk) = Σii...i (αj)i(αj)i ...(αj)i (λiλi ... λi)(v1,v2, ....vk) = Σii...i (αj)i(αj)i ...(αj)i (v1)i(v2)i ... (vk)i = [Σi(αj)i(v1)i] [Σi(αj)i(v2)i] ..... [Σi(αj)i(vk)i] = αj(v1)αj(v2) ... αj(vk) . (2.11.22) (2.11.33) Finally, (αj αj ... αj)(ei,ei, .... ei) = αj(ei)αj(ei) ... αj(ei) = (αj)i(αj)i .... (αj)i = (αj αj ... αj)ii....i (2.11.23) (2.11.34) Once again, due to the isomorphism between functionals T and functions T(v1,v2, ....vk), one loosely refers to the scalar field T(v1,v2, ....vk) as a rank-k tensor, although T is the actual rank-k tensor object whose components Tabc... transform by the usual rule e.g. (2.1.6). One might say that the rank-k tensor T is "represented by" the k-multilinear function T(v1,v2, ....vk) , just as it is represented by Tabc... in the sense of tensor components. Consider these equations from above T(v1,v2, ....vk) = Σii....i Tii....i (v1)i(v2)i ... (vk)i T(ej,ej, .... ej) = Tjj....j The official tensor components in terms of Chapter 2 transformations are Tjj....j. The corresponding function T(v1,v2, ....vk) appears to be a continuation (like an analytic continuation) of the tensor components Tjj....j off the specific argument points (ej,ej, .... ej). There is probably a formal description of this structure perhaps in terms of fiber bundles. In any event, Spivak on page 75 refers to our k-multilinear function T(v1,v2, ....vk) as a k-tensor. (d) Multiindex notation The cluttered equations above can be written in a dense multiindex notation. Forms like those shown below commonly appear in the literature. Ultimately one must refer to the original equation to remove any possibly ambiguity of the notation. Italic equation numbers show the equation being abbreviated. I ≡ i1, i2...ik J ≡ j1, j2...jk // ordinary multiindices λI ≡ λiλi ... λi eJ ≡ ej,ej, .... ej vZ ≡ v1,v2....vk // Z is meant to imply labels 1,2....k (2.11.35) λI(vZ) ≡ (λiλi ... λi)(v1,v2....vk) (2.11.26) vZI ≡ (v1)i(v2)i ... (vk)i a k-multilinear function λI(vZ) = vZI a k-multilinear function (2.11.27) T = ΣI TIλI general k-tensor expansion in V*k (2.11.28) T(vZ) = ΣI TIλI(vZ) = ΣI TI vZI a k-multilinear function (2.11.29) T(eJ) = TJ project out tensor components (2.11.30) T'(v'Z) = T(vZ) T(v1,v2, ....vk) transforms as a scalar (2.11.31) αJ ≡ (αj αj ... αj) tensor product of k vectors of V* (αJ)I ≡ (αj αj ... αj)ii....i = (αj)i(αj)i .... (αj)i outer product αJ(vZ) ≡ αj(v1)αj(v2) ... αj(vk) αJ = ΣI (αJ)I λI expansion of the tensor product of k vectors (2.11.32) αJ(vz) = ΣI (αJ)I λI(vZ) = ΣI (αJ)I vZI = (αJ)(vZ) (2.11.33) αJ(eI) = (αJ)I (2.11.34) 3. Outer Products and Kronecker Products 3.1 Outer Products Revisited: 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 hung 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 α,β,γ... are arbitrary tensors, having multiindices I,J,K (for example I = {i1,i2,i3} if α 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, ( α β γ ....)IJK... = αI βJ γK...... (3.1.14) The tensor here is α β γ ...., and it is the tensor product and the outer product of the individual tensors α,β,γ.... 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) = ei Basis(X) = ei Let W and Y be vector spaces of dimension n' and m' Basis(W) = e'i Basis(Y) = e'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) 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 eie'j] = ΣijFij (ST)(eie'j) = ΣijFij (Sei)(Te'j) . (3.2.5) The action of S on a vector v (and T on w) can be written x = (Sv) = Σa[Sv]aea = Σa(ΣbSabvb)ea = Σab(Sabvb)ea y = (Tw) = Σc[Tw]ce'c = Σc(ΣdTcdwd)e'c = Σcd(Tcdwd)e'c . (3.2.6) Select v = ei and w = e'j in these last two equations to get, (Sei) = ΣabSab(ei)b ea (Te'j) = ΣcdTcd(e'j)d e'c . (3.2.7) Then the tensor product appearing in (3.2.5) can be written (Sei)(Te'j) = [ ΣabSab(ei)b ea] [ ΣcdTcd(e'j)d e'c] = Σabcd Sab(ei)bTcd(e'j)d (eae'c) (3.2.8) and so the action of the tensor product operator (ST) is given by. (ST)[ ΣijFij eie'j] = ΣijFij (Sei)(Te'j) // (3.2.3) = Σijabcd FijSab(ei)bTcd(e'j)d (eae'c) // (3.2.8) = Σac { Σijbd FijSab(ei)bTcd(e'j)d } (eae'c) // regroup = Σac Gac (eae'c) where Gac = Σijbd FijSab(ei)bTcd(e'j)d . (3.2.9) We have then shown the action of operator ST on a general element of VW : (ST) { ΣijFij eie'j } = Σac Gac (eae'c) (ST) : VW → XY where Gac = Σijbd Fij Sab (ei)b Tcd (e'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. 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. 4.1 The tensor product of 2 vectors Basics. Consider two vector spaces V and W (defined over field K) of dimension n and n'. Let {ei} = basis of V dim(V) = n v = Σi=1n vi ei = general vector in V vi ϵ K {e'i} = basis of W dim(W) = n' w = Σj=1n'wj e'j = general vector in W wj ϵ K {eie'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), (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 (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", (kv) w = v (kw) = k (vw) // k = 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 (k1w1+ k2w2) = k1(vw1) + k2(vw2) // k1,k2 = scalar (ϵ K) (k1v1 + k2v2) w = k1(v1w)+ k2(v2w) .// k1,k2 = 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) (kv) (kw) = k (vw) // wrong! (unless k = 1) As noted in Appendix B the second rule applies to a direct sum . Using the correct "rules" above, one may write v w = ( Σiviei)( Σjwjej') = Σijviwj (eie'j) (4.1.6) showing how this pure tensor product vector can be expressed in terms of the basis functions. 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), where Tij ≡ [T(e,e')]ij are the expansion coefficients, T ≡ Σij Tij eie'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 using Tij = [T(e)]ij, T ≡ Σij Tij eiej 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 (2.1.6) or [T(e)]ab = Raa' Rbb' [T(u)]a'b' Picture E . (2.6.8) Covariant expansion forms. The rank-2 tensor T can be expanded in many bases 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 eiej T ϵ VW T ≡ Σij Tij eiej T ϵ VW T ≡ Σij Tij eiej T ϵ VW T ≡ Σij Tij eiej T ϵ VW . (4.1.9) Default Notation. In the rest of this document, unless otherwise specified, expansions shall always be on the ei basis, and the label (e) appearing for example on [T(e)]ab in (2.6.8) or [ei(e)]b in (2.6.4) will be omitted. Thus, Tab = [T(e)]ab (ei)b = [ei(e)]b = δib (eiej)ab = [(eiej)(e)]ab = [ei(e)]a [eb(e)]b = (ei)a (ej)b = δia δjb . (4.1.10) Similarly, in the next section, Tij will by default mean [T(λ)]ij unless otherwise stated. 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). 4.2 The tensor product of 2 dual vectors 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. 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* {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 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, (kα) β = α (kβ) = k (α β) k 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) where Tij ≡ [T(λ,λ')]ij are coefficients in the field K. 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) is manifestly bilinear in its arguments. Setting v = ei and w = e'j , one finds that T(ei,e'j) = Tij ϵ K (2.11.19) (4.2.10) which provides an interpretation of Tij as the function T(v,w) evaluated at two basis vectors ei,e'j. This may be compared with α(ei) = αi in (4.2.3). An arbitrary rank-2 tensor T can be represented either by its usual components Tij or by the bilinear function T(v,w). More specifically, 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)(ei,e'j) = α(ei) β(e'j) = αiβj = (α β)ij (2.11.23) (4.2.14) Back to V*V*. 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). One says that the linear functionals (α β) and T in the dual space V*2 are "tensors" of rank 2. Whereas T of (4.1.8) was a rank-2 tensor in V2, this object T is a rank-2 tensor in the dual space V*2. In the abstract, it is T which is the real tensor and it is shown expanded on a particular basis {λi} which in turn is associated with a particular basis {ei} of V as discussed in Section 2.11. The space of bilinear functionals on V2 = VxV (which includes any T above) is just V*2 = V*V*. Due to the isomorphism between the functionals T and the functions T(v,w), we can also refer to V*2 as the space of all bilinear functions T(v1,v2). Fact: The vector space V*2 is equivalent to the vector space of bilinear functions on V. (4.2.15) Comment: It is possible to emphasize the parallelism between the dual-world and the non-dual world by considering, in analogy with the functions λi(v), alternate functions vi(λ). Either world is then the dual of the other world. This is the approach taken on page 2 of Benn and Tucker where v and λ are called x and X. 4.3 The wedge product of 2 vectors (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 ei 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 (projmec ref), but this notation seems uncommon. 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, (ka) ^ b = [ (ka)b - b(ka)]/2 = k [ ab - ba ]/2 = k (a ^ b) k = 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 ^ (kb) and a ^ (b + c). To summarize, we have a set of rules as follows: (ka) ^ b = k (a ^ b) (a+c) ^ b = (a ^ b) + (c ^ b) k ϵ K a ^ (kb) = k (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 of the space L2 can be written this way, T = Σij Tij ei ^ ej . Σij ≡ Σi=1n Σj=1n (4.3.5) where the Tij are the expansion coefficients. For example, if Tij = aibj this would be, T = Σij aibj ei ^ ej = (Σiaiei) ^ ( Σjbjej) = 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 Tab = Σij Tij (ei ^ ej)ab = Σij Tij(eiej - ejei)ab/2 = Σij Tij[ (eiej)ab - (ejei)ab] / 2 = Σij Tij [ eiaejb - ejaeib] / 2 = Σij Tij [ δia δjb - δja δib] / 2 = (1/2)[ Tab -Tba] Tab = - Tba (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, (ei^ ej) = [eiej - ejei]/2 n(n-1)/2 independent elements in this set (4.3.8) (ei * ej) ≡ [eiej+ ejei]/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 ei ^ ej + Σij T'ij ei ^ ej = Σij (Tij+T'ij) ei ^ ej . (4.3.9) And if (a ^ b) is an element of L2 then so is k(a ^ b) = (kα) ^ 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 (ei ^ ej) = Σi≠ j Tij (ei ^ ej) // (ei ^ ei) = 0 = Σi<j Tij (ei ^ ej) + Σi>j Tij (ei ^ ej) = Σi<j Tij (ei ^ ej) + Σj>i Tji (ej ^ ei) // i↔j in second sum = Σi<j Tij (ei ^ ej) - Σi<j Tji (ei ^ ej) // (ej ^ ei) = - (ei ^ ej) = Σi<j (Tij - Tji) (ei ^ ej) (4.3.7) = Σi<j Aij (ei ^ ej) Aij ≡ (Tij - Tji) = 2Tij 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. Below we use this terminology, T = Σij Tij (ei ^ ej) = the "symmetric expansion" of T T = Σi<j Aij (ei ^ ej) = the "ordered expansion" of T Meanwhile, the most general element of V2 can be written T = Σij Tij (ei ej) . (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. (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 (ei ^ ej) = Σi<j (aibj- ajbi) (ei ^ ej) = Σi<j det (ei^ej) 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 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 e1^e2 = det(a,b) e1^e2 = [ a1b2 - a2b1] e1^e2 . (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 we will show that for V = R3 the triple wedge product of three vectors is given by, a ^ b ^ c = det(a,b,c) (e1^ e2^ e3) *** make sure this is shown somewhere ! (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 e1^e2 + e1^e3 + e2^e3 = [a1b2 - a2b1] e1^e2 + [a3b1 - a1b3] e3^e1 + [a2b3 - a3b2] e2^e3 . (4.3.16) The coefficients are those which appear in the normal "cross product" of two contravariant vectors, a x b = [a1b2 - a2b1] e3 + [a3b1 - a1b3] e2 + [a2b3 - a3b2] e1 . We do not wish, however, to identify for example e1^e2 with e3. After all, e3 is a basis vector in V, whereas e1^e2 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") e1^e2 ↔ e3 e2^e3 ↔ e1 e3^e1 ↔ e2 // = - e1^e3 (4.3.17) in which case one can say a ^ b = [a1b2 - a2b1] e1^e2 + [a3b1 - a1b3] e3^e1 + [a2b3 - a3b2] e2^e3 ↔ a x b = [a1b2 - a2b1] e3 + [a3b1 - a1b3] e2 + [a2b3 - a3b2] e1 (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 *(e1^e2) = e3 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 stated above, a ^ b = det e1^e2 + dete1^e3 + dete1^e4 + dete2^e3 +dete2^e4 + dete3^e4 . (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, (eie'j)rs = (ei)r(e'j)s = δirδjs // VW (eiej)rs = (ei)r(ej)s = δirδjs // VV = V2 The components of (ei^ ej) are given by. (ei^ ej)rs = (1/2)[eiej - ejei]rs = (1/2)[(eiej)rs - (ejei)rs] = (1/2)[(ei)r(ej)s - (ej)r(ei)s] = (1/2) (δir δjs - δis δjr) (ei^ ej)rs = - (ei^ ej)sr = - (ej^ ei)rs . // two forms of antisymmetry (4.3.20) We now examine the pure wedge product a ^ b using the 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 (ei ^ ej)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) (ei ^ ej)rs = Σi<j det (ei ^ ej)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 // Trs = (Trs- Tsr)/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), Trs = Σij Tij (ei ^ ej)rs = Σij Tij (δir δjs - δis δjr)/2 = (Trs- Tsr)/2 = Ars/2 (4.3.28) Using the ordered double sum (4.3.10), Trs = Σi<j Aij (ei ^ ej)rs = Σi<j Aij [δirδjs - δjrδis]/2 . (4.3.29) For r = s one has [..] = 0 so Trs = 0. Otherwise, 2 Trs = 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 Trs = Ars/2 for all r,s ϵ (1,n) (4.3.31) which agrees with (4.3.30) 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 (4.3.32) with this special case (ei^ej) (cd) = [(eic)(ejd) - (ejc)(eid)]/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), (ei^ej) (c^d) = [cidj - cjdi]/2 . (4.3.35) 4.4 The wedge product of 2 dual vectors 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* x 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 which some authors call 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 : (kα) ^ β = k (α ^ β) (α+γ) ^ β = (α ^ β) + (γ ^ β) α ^ (kβ) = k (α ^ β) α ^ (β + γ) = (α ^ β) + (α ^ γ) (4.4.4) where α,β,γ are vectors in V* and k 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 . // Tij is in italics, T on the left is not (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) = 2Tij 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 = aibj 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) Components (functions) For the tensor product of two evaluated basis functionals we have, (λiλ'j)(vr,vs) = λi(vr)λ'j(vs) = (vr)i(vs)j // V*W*; r and s are vector labels (λiλj)(vr,vs) = λi(vr)λj(vs) = (vr)i(vs)j // V*V* For (λi^λj), (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)(vr,vs) = - (λi^ λj)(vs,vr) = - (λj^ λi)(vr,vs) // two forms of antisymmetry (4.4.20) We now examine the pure wedge product α ^ β using the 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 = er and vs = es this becomes, using (4.2.6), (α ^ β)(er,es) = 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 = er and vs = es gives (α ^ β)(er,es) = Σi<j (1/2) det [δirδjs - δjrδis] (4.4.26) Repeating the argument (4.3.27) this becomes (α ^ β)(er,es) = (αrβs - αsβr)/2 // = [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(er,es) = Σij Tij (1/2) [δirδjs - δjrδis] = (Trs - Tsr)/2 = Ars // see (4.4.10) (4.4.30) Using the ordered double sum (4.4.10), T(vr,vs) = Σi<j Aij (λi ^ λj)rs = Σi<j Aij [δirδjs - δjrδis]/2 (4.4.31) For r = s one has [..] = 0 so T(er,es) = 0. Otherwise, 2T(vr,vs) = 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(vr,vs) = Ars/2 for all r,s ϵ (1,n) (4.4.33) 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. We conclude that : Fact: The vector space Λ2(V) is equivalent to the vector space of antisymmetric bilinear functions on V. (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 of bilinear functions on V. (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) 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 ei ei ..... ei . (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) (ei ei ..... ei)jj...j = (ei)j (ei)j .... (ei)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 ei might be the same. This will always be the case if k > n where n ≡ dim(V). For example, k = 3 and n = 2, one such element would be e1 e1 e2 ≠ 0. 5.2 Tensor Expansion for a tensor in Vk ; the ordinary multiindex We took a preliminary look at this expansion at the end of Section 2.10. Here we fill in more details. A rank-k tensor T in Vk has this general expansion on the er basis, T = Σii....i Tii....i (ei ei ..... ei) . (5.2.1) In the notation of (2.10.2) or (2.10.14) we identify Tii....i = [T(e)]ii....i . As expected, [T]jj...j = Σii....i Tii....i (ei ei ..... ei)jj...j = Σii....i Tii....i (δij δij .... δij) = Tjj...j . (5.2.1a) Using the notion of a multiindex I (an ordinary multiindex), I ≡ {i1, i2, .....ik} // each is ranges 1,2....n (5.2.2) and a shorthand notation for the basis vectors eI ≡ ei ei ..... ei (5.2.3) the general rank-k tensor T in Vk can be expanded in the following compact restatement of (5.2.1), T = ΣI TI eI . (5.2.4) 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), (α + β)(v1v2 .....vk) = [(α+β)v1]v2 .....vk = [αv1+βv1]v2 .....vk (5.3.4) = α(v1v2 .....vk)+ β(v1v2 .....vk) α,β ϵ K and multiplication by a scalar is distributive over "vector addition", α [(v1v2 .....vk) + (v'1v'2 .....v'k)] = α (v1v2 .....vk) + α (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=1∞ 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 ei Σij Tij eiej Σijk Tijk eiejek + ...... 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 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) 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? Consider two tensors of rank k and k' expanded as in (5.2.1), T = Σii....i Tii....i (ei ei ..... ei) . rank k, T ϵ Vk (5.6.1) S = Σjj....j Sjj....j (ej ej ..... ej) rank k', S ϵ Vk' . (5.6.2) Multiplying these together with one gets, using the rules (5.3.1), TS = [Σii....iTii....i(ei ei ..... ei)][Σjj....jSjj....j(ej ej ..... ej)] = Σii....i Σjj....jTii....i Sjj....j(ei ei ..... ei) (ej ej ..... ej) (5.6.3) = Σii....ijj....jTii....i Sjj....j(ei ei ..... ei ej ej ..... ej) (5.6.4) = Σii....iii....i[Tii....i Sii....i] (ei ei ...... ei) . (5.6.5) Notice that the (2.8.22) associativity of is used going from (5.6.3) to (5.6.4). In the last step (5.6.5), we renamed the dummy jr summation indices so that j1 = ik+1 , j2 = ik+2 and so on. Rewriting the last equation, TS = Σii....i[Tii....i Sii....i] (ei ei ...... ei) . (5.6.6) Clearly the product TS is an element of Vk+k'with the following tensor components, (TS)ii....i = Tii....i Sii....i . (5.6.7) Thus the tensor on the left is the outer product of the two tensors on the right, similar to (3.1.13). Both sides of this equation of course transform in the same manner in the sense of (2.1.6). We have just shown that if T ϵ Vk and S ϵ Vk', then TS ϵ Vk+k. Thus we have strengthened the claim made in (5.4.5) that T(V) is closed under the operation : the tensor product of an Vk tensor with an Vk' tensor lies in Vk+k' which is in T(V). It is easy then to show that this is true for the tensor product of any two multivectors as defined below (5.4.6). The tensor product of three or more tensors works in the same fashion. For example, if R has rank k" then we find that TSR ϵ Vk+k'+k" with the following outer product relation, (TSR)ii....i = Tii....i Sii....iRii....i . (5.6.8) Using the ordinary multiindices of (5.2.2-4), the above equations can be considerably compacted : T = ΣI TI eI S = ΣJ SJ eJ I = {i1, i2, .. ik} J = {j1, j2, .. jk'} (5.6.9) (5.6.1) (5.6.2) eI ≡ ei ei ..... ei eJ ≡ ej ej ..... ej TI = Tii....i SJ = Sjj....j TS = ΣI,J TISJ eIeJ = ΣI (TS)I eI (TS)I = TISI' (5.6.10) (5.6.3) (5.6.6) (5.6.7) I = {i1, i2, .. ik+k'} eI ≡ ei ei ..... ei I = {i1, i2, .. ik} I' = {ik+1, ik+2, .. ik+k'} TSR = ΣI,J,K TISJRK eIeJeK = ΣI (TSR)I eI (TSR)I = TISI'RI" (5.6.11) (5.6.8) I = {i1, i2, .. ik+k'+k"} eI ≡ ei ei ..... ei I = {i1, i2, .. ik} I' = {ik+1, ik+2, .. ik+k'} I" = {ik+k'+1, ik+k'+2, .. ik+k'+k"} In the more systematic notation discussed at the end of Section 6, the tensor product of N tensors Ti of rank ki is given by T1T2...TN = ΣI (T1IT2I .... TNI) eI = ΣI (T1T2....TN)I eI (5.6.12) 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. I = I1 I2 .... IN = {i1, i2......ik+k+...k} eI = ei ei ..... ei The rank of this product tensor is then K = Σi=1N ki and the tensor is an element of VK. Example 1: The tensor product of two rank-1 tensors. TS = Σii[Ti Si] (ei ei) = Σij TiSj (eiej) = Σij (TS)ij (eiej) (TS)ab = Σij TiSj (eiej)ab = Σij TiSj δiaδjb = TaSb (5.6.13) Example 2: The tensor product of two rank-2 tensors. TS = Σiiii Tii Sii (ei ei ei ei ) (TS )abcd = TabScd (5.6.14) 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(ei ei ..... ei ej ej ..... ej) = Σii....iTii....i (κ') (ei ei ..... ei) = κ'T and ST = Σjj....jii....i Sjj....jTii....i (ej ej ..... ej ei ei ..... ei) = Σii....i (κ') Tii....i ( ei ei ..... ei) = 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 (5.6.15) 6. The Tensor Product of k dual vectors : the vector spaces V*k and T(V*) Comment: This Chapter 6 is a copy, paste and edit version of Chapter 5 -- a translation from non-dual to dual. One might think such a translation could be trivially implemented with a "translation table" which had rules like v1→ α1 and ei → λi and so on. Although this works for some equations, it does not work for others, as noted in the edited text below. There are sufficient differences between the dual and non-dual worlds that one just has to write it all out. For example, basis vectors ei have components in the non-dual world, but the basis functionals λi in the dual world don't have components, but can be evaluated at vectors of V. We proceed with this brute force translation. _________________________________________________________________________________ Our task is now to generalize the tensor product from V*2 to V*k, where V*k ≡ V*V* .... V* . // tensor product of k vector spaces, each one is V* (6.1) We are setting up for a parallel treatment in Chapter 7 where becomes ^, so certain rather obvious statements will be made here to allow for comparison later with the wedge product. comment about bold font reappearing here? 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) 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 . (6.1.2) The subscripts in (6.1.1) and the superscripts in (6.1.2) are labels, not components. The next two equations in Section 5.1 involve taking components of (5.1.1) and (5.1.2), (v1 v2 ..... vk)jj...j = (v1)j (v2)j .... (vk)j (5.1.3) (ei ei ..... ei)jj...j = (ei)j (ei)j .... (ei)j = δij δij .... δij , (5.1.4) but in the dual space functionals don't have components. The analogous equations involve evaluation of the above functionals at a set of basis vector arguments (ej, ej ....ej): (α1 α2 ..... αk)(ej, ej ....ej) = α1(ej)α2(ej) ... αk(ej) = (α1)j(α2)j ... (αk)j (6.1.3) (λi λi ..... λi)(ej, ej ....ej) = λi(ej)λi(ej)... λi(ej) = (ej)i (ej)i ...(ej)i = δji δji .... δji (6.1.4) Note that in expression α1(ej) the α1 is a functional, but in (α1)j the α1 is a vector in V which is associated with the functional α1 according to α1 = Σi(α1)iλi. 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 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, k = 3 and n = 2, one such element would be λ1 λ1 λ2 ≠ 0. We now add a new equation not present in Chapter 5. It is a restatement of (6.1.3) for arbitrary vector arguments, (λi λi ..... λi)(vj, vj ....vj) = λi(vj)λi(vj)... λi(vj) = (vj)i (vj)i ...(vj)i (6.1.6) This is a generalization of (2.11.15): (λiλj)(v1,v2) = λi(v1)λj(v2). On the second line we make use of (2.11.7): λi(v) = vi. 6.2 Tensor Expansion for a tensor in V*k ; the ordinary multiindex and Tk(V) We took a preliminary look at this expansion at the end of Section 2.10. Here we fill in more details. A rank-k tensor T in V*k has this general expansion on the λr basis, T = Σii....i Tii....i (λi λi ..... λi) . (6.2.1) In the notation of (2.10.2) or (2.10.14) we identify Tii....i = [T(λ)]ii....i . The next equation in Section 5.2 evaluates (5.2.1) to show that [T]jj...j = Tjj...j , T = Σii....i Tii....i (ei ei ..... ei) . (5.2.1) [T]jj...j = Tjj...j (5.2.1a) Again, we cannot take components of the functional T in V*k but we can evaluate it at (ej, ej ....ej) . Then, "as expected", T(ej, ej ....ej) = Σii....i Tii....i(λi λi ..... λi)(ej, ej ....ej) = Σii....i Tii....i δji δji .... δji // (6.1.4) = Tjj....j (6.2.1a) which projects out the coefficients in (6.2.1) in analogy with (5.2.1). Using the notion of a multiindex I (an ordinary multiindex), I ≡ {i1, i2, .....ik} // each is ranges 1,2....n (6.2.2) and a shorthand notation for the basis vectors λI ≡ λi λi ..... λi (6.2.3) the general rank-k tensor T in V*k can be expanded in the following compact restatement of (6.2.1), T = ΣI TI λI . (6.2.4) Again we add a new equation not present in Chapter 5 by evaluating expansion (6.2.1), T(vj, vj ....vj) = Σii....i Tii....i(λi λi ..... λi)(vj, vj ....vj) = Σii....i Tii....i λi(vj)λi(vj)... λi(vj) = Σii....i Tii....i (vj)i (vj)i .... (vj)i (6.2.5) which is similar to (6.1.6) and is a generalization of (2.11.18): T(v1,v2) = Σab Tab (v1)a(v2)b . The space Tk(V) of k-tensor functions. From (6.2.5) we see a 1-to-1 correspondence between the rank-k tensors Tii....i and the k-multilinear functions T(vj,vj ...vj). We think of Tii....i as a rank-k tensor in V*k, and we shall say that T(vj, vj ....vj) is a k-tensor in a space Tk(V) which is isomorphic to V*k. V*k is a space of k-multilinear functionals, while Tk(V) is the corresponding space of k-multilinear functions. (6.2.6) Note: It is a coincidence that we named our k-tensor functional T and that the space is called Tk(V). Another k-tensor S would also be in Tk(V). We use the name Tk(V) because that is the name used by Spivak on page 75. 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) The proof of this fact follows that of the text near (1.1.9). For example, the "0" in V*k is represented by (6.1.1) with one or more vectors being 0, since for example, α10 .....αk = α1(α2 - α2) .....αk = α1α2 .....αk - α1α2 .....αk = 0 . (6.3.3) "Vector multiplication" is distributive over scalar addition (here the "vector" is α1α2 .....αk), as one finds applying the rules (6.3.1), (a + b)(α1α2 .....αk) = [(a+b)α1]α2 .....αk = [aα1+bα1]α2 .....αk (6.3.4) = a(α1α2 .....αk)+ b(α1α2 .....αk) a,b ϵ K and multiplication by a scalar is distributive over "vector addition", a [(α1α2 .....αk) + (α'1α'2 .....α'k)] = a (α1α2 .....αk) + a (α'1α'2 .....α'k) . (6.3.5) All the above equations are meaningful for any positive integer k, regardless of the value n = dim(V*). 6.4 The Tensor Algebras T(V*) and 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 = V*0V*1V*2 .... The reader unfamiliar with direct sums will find a description in Appendix B including a simple "tall vector" method of visualizing such spaces. The Dual Tensor Algebra Normally one does not add apples and oranges, so one does not add items of the form αβ ϵ V*2 to those of the form αβγ ϵ V*3. However, as one writer notes, (dual) fruit salad is great, and so we could define a very large dual vector space of the form 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) We know this is true since T(V*) = Σk=0∞ V*k and we showed in (6.3.2) that each V*k is a vector space. For example, the "0" element in T(V*) is the direct sum of the "0" elements of all the V*k. 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 α βκ ρση = sum of 4 elements of T(V*) = an element of T(V*) s(k1 α βκ ρση) = (sk1) (sα) (sβ)κ ρ(sσ)η = element of T(V*) (6.4.4) 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 (βκ)(ρση) = βκρση = ϵ V*5 = ϵ T(V*) . // (βκ) ϵ V*2 ,(ρση) ϵ V3 (6.4.5) Here we have used the associative property (2.8.22) applied to vectors. This closure claim is stated more generally below (6.6.7). For later comparison with the corresponding wedge picture, here we have: Object lin comb is Rank(grade) Space s scalar ϵ K 0 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) Since T(V*) is closed under the operations + and , it is "an algebra" (the space V*k 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. 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 (5.2.1). A dual multivector is any linear combination of dual rank-k tensors for any 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) Due to the isomorphism noted in (6.2.6), we have this similar tensor algebra for functions, T(V) ≡ T0(V) T1(V) T2(V) T3(V) ....... = Σk=1∞ Tk(V) (6.4.8) 6.5 Comments about tensors See Section 5.5 concerning why linear combinations of permutated tensors are tensors. In Chapter 6 the tensors don't have indices so this discussion does not directly apply, but the coefficients like Tii....i in (6.2.1) are such tensors and then the comments apply to them. 6.6 The Tensor Product of two or more tensors in T(V*) The tensor algebra T(V*) shown in (6.4.1) is closed under both + and . It seems evident how one would add two tensors of T(V*) of the form (6.4.2), but how would one multiply two tensors? Consider two tensors of rank k and k' expanded as in (5.2.1), T = Σii....i Tii....i λi λi ..... λi . rank k, T ϵ V*k (6.6.1) S = Σjj....j Sjj....j λj λj ..... λj rank k', S ϵ V*k' . (6.6.2) Multiplying these together with one gets, using the rules (6.3.1), TS = [Σii...iTii....i(λi λi ..... λi)][Σjj...jSjj....j(λj λj ..... λj)] = Σii...i Σjj...jTii....i Sjj....j(λi λi ..... λi) (λj λj ..... λj) (6.6.3) = Σii...ijj...jTii....i Sjj....j(λi λi ..... λi λj λj ..... λj) (6.6.4) = Σii...iii...i[Tii....i Sii....i] (λi λi ...... λi) . (6.6.5) Notice that the (2.8.22) associativity of is used going from (6.6.3) to (6.6.4). In the last step (6.6.5), we renamed the dummy jr summation indices so that j1 = ik+1 , j2 = ik+2 and so on. Rewriting the last equation, TS = Σii....i[Tii....i Sii....i] (λi λi ...... λi) . (6.6.6) Clearly the product TS is an element of V*k+k'with the following tensor components, (TS)ii....i = Tii....i Sii....i . (6.6.7) Thus the tensor on the left is the outer product of the two tensors on the right, similar to (3.1.13). Both sides of this equation of course transform in the same manner in the sense of (2.1.6). We have just shown that if T ϵ V*k and S ϵ V*k', then TS ϵ V*k+k. Thus we have strengthened the claim made in (6.4.5) that T(V*) is closed under the operation : the tensor product of an V*k tensor with an V*k' tensor lies in V*k+k' which is in T(V*). It is easy then to show that this is true for the tensor product of any two dual multivectors as defined below (6.4.6). The tensor product of three or more tensors works in the same fashion. For example, if R has rank k" then we find that TSR ϵ V*k+k'+k" with the following outer product relation, (TSR)ii....i = Tii....i Sii....iRii....i . (6.6.8) Using the ordinary multiindices of (6.2.2-4), the above equations can be considerably compacted : T = ΣI TI λI S = ΣJ SJ λJ I = {i1, i2, .. ik} J = {j1, j2, .. jk'} (6.6.9) (6.6.1) (6.6.2) λI ≡ λi λi ..... λi λJ ≡ λj λj ..... λj TI = Tii....i SJ = Sjj....j TS = ΣI,J TISJ λIλJ = ΣI (TS)I λI (TS)I = TISI' (6.6.10) (6.6.3) (6.6.6) (6.6.7) I = {i1, i2, .. ik+k'} λI ≡ λi λi ..... λi I = {i1, i2, .. ik} I' = {ik+1, ik+2, .. ik+k'} TSR = ΣI,J,K TISJRK λIλJλK = ΣI (TSR)I λI (TSR)I = TISI'RI" (6.6.11) (5.6.8) 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"} In the more systematic notation discussed at the end of Section 6, the tensor product of N tensors Ti of rank ki is given by T1T2...TN = ΣI [(T1)I(T2)I .... (TN)I] λI = ΣI (T1T2....TN)I λI (6.6.12) 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. I = I1 I2 .... IN = {i1, i2......ik+k+...k} λI ≡ λi λi ..... λi The rank of this product tensor is then K = Σi=1N ki and the tensor is an element of V*K. Example 1: The tensor product of two rank-1 tensors. TS = Σii[TiSi] (λi λi) = Σij TiSj (λiλj) = Σij (TS)ij (λiλj) (TS)(ea,eb) = Σij (TS)ij (λiλj)(ea,eb) = Σij (TS)ij δaiδbj = (TS)ab (6.6.13) Example 2: The tensor product of two rank-2 tensors. TS = Σiiii Tii Sii (λiλiλiλi) (TS)(ea,eb,ec,ed) = TabScd = (TS)abcd (6.6.14) In both examples the basis-vector function evaluations produce the expected expansion coefficients. 6.7 The Tensor Product of two or more functions in T(V) Recall the expansion of two dual tensors shown in (6.6.4), TS = Σii...ijj...jTii....i Sjj....j (λi λi ..... λi λj λj ..... λj) . (6.6.4) Evaluate this functional at (v1,v2....vk, vk+1....vk+k') and then use (6.1.6) to get (TS)(v1,v2....vk, vk+1....vk+k') = Σii...ijj...jTii....i Sjj....j (λi λi ..... λi λj λj ..... λj)(v1,v2....vk, vk+1....vk+k') = Σii...ijj...jTii....i Sjj....j (v1)i (v2)i .. (vk)i (vk+1)j (vk+2)j .. (vk+k')j // (6.1.6) = [ Σii...iTii....i (v1)i (v2)i .. (vk)i ] * [Σjj...j Sjj....j (vk+1)j (vk+2)j .. (vk+k')j] = T(v1,v2....vk) S(vk+1,vk+2....vk+k') . // (6.1.6) twice We have then obtained the rule for multiplying two dual tensor functions, (TS)(v1,v2....vk, vk+1....vk+k') = T(v1,v2....vk) S(vk+1,vk+2....vk+k') (6.7.1) Recall comments (6.2.6) above, so Tk(V) is isomorphic to V*k and the elements of Tk(V) are k-tensors. Equation (6.7.1) combines the k-tensor function T ϵ Tk(V) and the k'-tensor function S ϵ Tk'(V) and generates a (k+k')-tensor function called (TS) where (TS) ϵ Tk+k'(V). It is useful to repeat the above development in multi-index notation: TS = ΣI,J TISJ λIλJ (TS)(vI,vI') = ΣI,J TISJ λIλJ(vI,vI') = ΣI,J TISJ λI(vI)λJ(vI') = ΣI,J TISJ (vI)I(vI')J = [ΣITI(vI)I] [ ΣJSJ(vI')J ] = T(vI) S(vI') I = {i1, i2, .. ik} , I' = {ik+1, ik+2, .. ik+k'} . (6.7.2) Then we can extend the idea easily to TSR = ΣI,J,K TISJRK λIλJλK (TSR)(vI,vI',vI") = ΣI,J,K TISJRK λIλJλK(vI,vI',vI") = ΣI,J,K TISJRK (vI)I(vI')J(vI")K = [ΣITITI(vI)I] [ ΣJSJ(vI')J ] [ΣKRK(vI")K ] = T(vI) S(vI')R(vI") I" = {ik+k'+1, ik+k'+2, .. ik+k'+k"} (6.7.3) which we can then write out longhand as (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") . (6.7.4) This shows a combination of a k-tensor and a k'-tensor and a k"-tensor to produce a resulting (k+k'+k")-tensor in the space Tk+k'+k"(V). Finally in systematic notation, 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. (6.7.5) we have T1T2...TN = ΣII...I (T1)I(T2)I ...(TN)I λIλI...λI (6.7.6) (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) so get the general result (T1T2...TN)(vI, vI ... vI) = T1(vI) T2(vI) .... TN(vI) . (6.7.7) This then is the tensor product of N ki-tensors to produce an K-tensor (T1T2...TN) where K = k1+k2 + ... +kN. (6.7.8) 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 . 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) e1^e2 where det(a,b) is the signed area of the 2-piped (parallogram) spanned by a and b. Then b ^ a = [ -det(a,b)] e1^e2 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) ( vP(j) vP(j) ..... vP(j)) = (1/k!) [ (vj vj ..... vj) + all signed permutations ] . (7.1.2) A detailed explanation of the ΣP (-1)S(P) notation is presented in Appendix 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. P(jr) means jP(r) so P acts on the jr subscripts. 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!) Σii...i εii...i (vi vi ..... vi) ir = 1 to k = (1/k!) [ (v1 v2 ..... vk) + all signed permutations ] . (7.1.3) The second line in (7.1.3) is the alternate notation shown in going from (A.2.4) to (A.2.5). Each summation index ir runs from 1 to k, and ε is the permutation tensor described in Section A.2: if any two ε indices are swapped, ε changes sign, and ε12..k = 1. Thus, εii...i = 0 if any two or more indices are the same. From our viewpoint, the (1/k!) normalization appearing in (7.1.2,3) 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 5. For other authors, the (1/k!) is replaced by 1 (give an example please). 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,s1,s2 = scalar ϵ K or v1^(s1v2 + s2v'2)^v3^.....^vk = s1(v1^v2^v3^ .....^vk) + s2(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). Our candidate expansions (7.1.2) and (7.1.3) satisfy (7.2.3) because they are k-multilinear. For example, v1^ (v2 + v'2) ^ .....^ vk = (1/k!) ΣP (-1)S(P) ( vP(1) [vP(2)+ v'P(2)] ..... vP(k)) = (1/k!) ΣP (-1)S(P) { ( vP(1) vP(2) ..... vP(k)) + ( vP(1) v'P(2) ..... vP(k)) } // (5.3.1) = (1/k!) ΣP (-1)S(P)(vP(1) vP(2) ..... vP(k)) + (1/k!) ΣP (-1)S(P)(vP(1) v'P(2) ..... vP(k)) = v1^v2^v3^ .....^vk + v1^v'2^v3^ .....^vk . Going from the first line above to the second line we have used the fact that the product is k-multilinear (also by fiat) as declared in (5.3.1). In similar fashion, our candidate expansions satisfy the scalar rule, v1^(sv2)^v3^ .....^ vk = (1/k!) ΣP (-1)S(P) ( vP(1) [svP(2)] ..... vP(k)) = (1/k!) ΣP (-1)S(P) { s( vP(1) vP(2) ..... vP(k)) } // (5.3.1) = s { (1/k!) ΣP (-1)S(P)(vP(1) vP(2) ..... vP(k)) } = s (v1^v2^v3^ .....^vk ) where again we use the fact that the product satisfies the scalar rule in (5.3.1). 3. The wedge product changes sign if any vector pair is swapped. (7.2.4) Because (7.1.2) vj^ vj^ .....^ vj = (1/k!) ΣP (-1)S(P) ( vP(j) vP(j) ..... vP(j)) (7.1.2) has the general form shown in (A.1.1) (with i→j) Fjj...j = ΣP (-1)S(P) RP(j)P(j)...P(j) (A.1.1) we know from (A.3.1) that vj^ vj^ .....^ vj is totally asymmetric in the jr indices. This means that vj^ vj^ .....^ vj changes sign if any two indices are swapped. Thus our candidate forms (7.1.2) and (7.1.3) meet the sign-change requirement imposed at the start of this section. 4. Wedge product of vectors vanishes if any two vectors are the same. Given a sign change 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 vectors 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) ( vP(j) vP(j) ..... vP(j))ii...i = (1/k!) ΣP (-1)S(P) (vP(j))i(vP(j))i ... (vP(j))i (7.2.8) where we have used the tensor outer product form (2.8.18). Since this matches the form (A.4.4), Fii...ijj...j = ΣP (-1)S(P) fP(j)i fP(j)i ...fP(j)i , (A.4.4) one can rewrite (7.2.8) with the P operators moved to the ir indices, (vj^ vj^ .....^ vj)ii...i = (1/k!) ΣP (-1)S(P) (vj)P(i)(vj)P(i) ... (vj)P(i) . (7.2.9) According to (A.4.5), then, we have Fact: (vj^ vj^ .....^ vj)ii...i is totally antisymmetric in both the labels jr and the indices ir. (7.2.10) 8. Associative Property of the wedge product. This topic is addressed below (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) Lk is a vector space (7.3.2) Fact (5.3.2) we 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), (ej ^ ej ^ .... ^ ej) . (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 (ej ^ ej ^ .... ^ ej) = (-1)S(P) (ei ^ ei ^ .... ^ ei) 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 (ei ^ ei ^ .... ^ ei) 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 (ei ^ ei ^ .... ^ ei) 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 e1, e3 and e5 are all equal to the one ordered element e1^e3^e5 with a + or - sign : e1^e3^e5 = (-1)0 e1^e3^e5 = +e1^e3^e5 135 e1^e5^e3 = (-1)1 e1^e3^e5 = - e1^e3^e5 153→135 e3^e1^e5 = (-1)1 e1^e3^e5 = - e1^e3^e5 315 →135 e3^e5^e1 = (-1)2 e1^e3^e5 = +e1^e3^e5 351→315→135 e5^e1^e3 = (-1)2 e1^e3^e5 = +e1^e3^e5 513→153→135 e5^e3^e1 = (-1)3 e1^e3^e5 = - e1^e3^e5 531→513→153→135 For k = 2 and n = 3, the 3 basis elements are e1^e2, e1^e3, e2^e3 and = 3. Components of the basis elements for Lk Now reconsider the basis vectors of the vector space Lk , (ej ^ ej ^ .... ^ ej) . (7.3.3) This wedge product can be expanded using (7.1.2), (ej ^ ej ^ .... ^ ej) = (1/k!) ΣP (-1)S(P) ( eP(j) eP(j) ..... eP(j)) . (7.3.8) The components of the above equation are (ej ^ ej ^ .... ^ ej)ii...i = (1/k!) ΣP (-1)S(P)( eP(j) eP(j) ..... eP(j))ii...i = (1/k!) ΣP (-1)S(P) (eP(j))i ( eP(j))i...( eP(j))i = (1/k!) ΣP (-1)S(P) δP(j)i δP(j)i... δP(j)i . (7.3.9) = (1/k!) ΣP (-1)S(P) δjP(i) δjP(i)...δjP(i) // from (A.4.4) (7.3.10) and from (A.4.5) we conclude that, Fact: (ej^ ej^ .....^ ej)ii...i is totally antisymmetric in both the labels jr and the indices ir. (7.3.11) We saw an example of both antisymmetries for k = 2 back in equation (4.3.20), (ei^ ej)rs = - (ei^ ej)sr = - (ej^ ei)rs . // two forms of antisymmetry (4.3.20) Either form (7.3.9) or (7.3.10) can be expressed in our usual informal notation (ej ^ ej ^ .... ^ ej)ii...i = (1/k!) [ δji δji...δji + signed permutations] (7.3.12) where these permutations can be taken to act on either the jr or the ir. As shown in (A.4.8), the above can be written as a determinant. For example, 3! (ej ^ ej ^ ej)iii = det = det ≡ det ( δjjjiii ) = det(δJI) // in multindex notation (7.3.13) 7.4 Tensor Expansions for a tensor in Lk Lk can be regarded as the vector space whose "vectors" (rank-k tensors) can be written in this ordered-sum form, T = Σ1≤i<i<....<i≤n Aii...i (ei ^ ei ^ .... ^ ei) . (7.4.1) This is so because the set (ei ^ ei ^ .... ^ ei) with 1 ≤ i1 < i2 < ..... < ik ≤ n forms a complete basis for Lk, as discussed just above. Example: If n = 3 and k = 2, then T = Σ1≤i<i≤3 Aii (ei^ ei) = A12 (e1^e2) + A13 (e1^e3) + A23 (e2^e3) . (7.4.2) On the other hand, one can also expand this same tensor T in a (redundant) symmetric sum as, T = Σii...i Tii...i (ei ^ ei ^ .... ^ ei) ir = 1,2..n (7.4.3) as in (4.3.5) for k = 2. Here Tii...i are a set of coefficients. What then is the connection between the Aii...i and the Tii...i coefficients? Start with the symmetric form (7.4.3), T = Σii...i Tii...i (ei ^ ei ^ .... ^ ei) ir = 1,2..n = Σi≠i≠...≠i Tii...i (ei ^ ei ^ .... ^ ei) . // (7.2.5) (7.4.4) Partition the summation space as follows (1 ≤ ir ≤ n), Σi≠i≠...≠i = [ Σi<i<...<i + Σi<i<...<i + many similar reorderings ] . (7.4.5) The total sum can be written in this manner, using the permutation sum notation, Σi≠i≠...≠i = ΣP ΣP(i)<P(i)<...<P(i) (7.4.6) where P are the k! permutations of the k integers {1,2,...k}. Using the form (7.4.6), the sum (7.4.4) may be rewritten as, T = ΣP ΣP(i)<P(i)<...<P(i) Tii...i (ei ^ ei ^ .... ^ ei) . (7.4.8) In (A.7.1) it is shown that, ΣP [ΣP(i)<P(i)<...<P(i)] fii...i = Σi<i<...<i [ΣP fP(i)P(i)...P(i)] . (A.7.1) Within the ΣP permutation sum, the permutation operators P have moved from the summation indices to the summand indices. One then has from (7.4.8), T = Σi<i<...<i ΣP [ TP(i)P(i)...P(i) (eP(i) ^ eP(i) ^ .... ^e P(i)) ] . (7.4.9) But we know from (7.3.5) that ( eP(i) ^ eP(i) ^ .... ^e P(i)) = (-1)S(P) (ei ^ ei ^ .... ^ ei) (7.4.10) where S(P) is the number of swaps associated with permutation P. Thus. T = Σi<i<...<i [ΣP (-1)S(P) TP(i)P(i)...P(i)] (ei ^ ei ^ .... ^ ei) (7.4.11) which we can compare with the ordered sum (7.4.1), T = Σi<i<....<i Aii...i (ei ^ ei ^ .... ^ ei) . (7.4.1) Thus, since the basis is complete, the relation between the A and T coefficients is given by Aii...i = ΣP (-1)S(P) TP(i)P(i)...P(i) i1 < i2 < ..... < ik = [ Tii...i + all signed permutations ] // k! terms (7.4.12) The Aii...i appear in the expansion (7.4.1) only for index values 1 ≤ i1 < i2 < ..... < ik ≤ n, but we can interpret (7.4.12) as defining Aii...i for all index values. Since (7.4.12) has the general form shown in (A.3.1), we conclude that Fact: Aii...i is a totally antisymmetric tensor. (7.4.13) 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 (7.4.14) Reconsider now (7.4.3) with ir → jr, T = Σjj...j Tjj...j (ej ^ ej ^ .... ^ ej) . jr = 1,2..n (7.4.4) We may compute the components of T by applying ii...i to both sides, Tii...i = Σjj...j Tjj...j (ej ^ ej ^ .... ^ ej)ii...i = (1/k!) ΣP (-1)S(P) Σjj...j Tjj...j δjP(i) δjP(i)...δjP(i) // (7.3.10) = (1/k!) ΣP (-1)S(P) TP(i)P(i)...P(i) = (1/k!) Aii...i . // (7.4.12) (7.4.15) which says T = (1/k!)A (7.4.15a) and from (7.4.13) it follows that T must be a totally antisymmetric rank-k tensor. This shows that the expansions (7.4.1) and (7.4.3) for T are capable only of describing a totally antisymmetric tensor T. Fact: The space Lk is the space of totally antisymmetric rank-k tensors T. (7.4.16) In contrast, the space Vk is the space of all rank-k tensors T, so Lk Vk. See Section 7.7 below. Recall that the pure vector wedge product v1^ v2^ .....^ vk is k-multilinear in the vi, and so is the underlying tensor product v1 v2 ..... vk. For k = 1, (7.4.15) says Ti = Σj Tj(ej)i = ΣjTjδji = Ti (7.4.17) so for a vector there is no distinction between Tk and Tk . 7.5 The alt, Alt and Sym operators One can regard the sum shown in (7.4.12) as being an operation "alt" performed on the tensor T to produce another tensor A which is totally antisymmetric. If one defines [alt(X)]ii...i ≡ ΣP (-1)S(P) XP(i)P(i)...P(i) (7.5.1) then (7.4.12) becomes Aii...i = [alt(T)]ii...i or A = alt(T) . (7.5.2) It is useful to define another version of the alt operator which has a scaling factor 1/k! , [Alt(X)]ii...i ≡ (1/k!) ΣP (-1)S(P) XP(i)P(i)...P(i) (7.5.3) so that Alt(X) = (1/k!)alt(X) . (7.5.4) It is shown in (A.3.1,2) that alt(X) and therefore Alt(X) are totally antisymmetric tensors for any tensor X. If X is already totally antisymmetric, one finds that alt(X) = k! X // since all k! terms in (7.5.1) are the same X = totally antisymmetric (7.5.5) Alt(X) = X . X = totally antisymmetric (7.5.6) We showed in (7.4.15a) that T = (1/k!) A . (7.5.7) so (7.5.2) can also be written T = Alt(T) . // (1/k!) A = (1/k!) alt(T) (7.5.8) One can define a total symmetrizing operator Sym in a manner similar to (7.5.3) but where the (-1)S(P) is omitted, [Sym(X)]ii...i ≡ (1/k!) ΣP XP(i)P(i)...P(i) (7.5.9) and then the tensor Sym(X) is totally symmetric for any tensor X. If X is already totally symmetric, then Sym(X) = X X = totally symmetric (7.5.10) Fact: The operators Sym and Alt are both projection operators, meaning that Sym2 = Sym and Alt2 = Alt when applied to any tensor X. (7.5.11) Proof: Alt(Alt(X)) = Alt(A) // where A ≡ Alt(X) = totally antisymmetric = A // by (7.5.6) = Alt(X) // since A ≡ Alt(X) Sym(Sym(X)) = Sym(S) // where S ≡ Sym(X) = totally symmetric = S // by (7.5.10) = Sym(X) // since S ≡ Sym(X) Fact: The Sym and Alt projection operators are orthogonal : Sym(Alt(X)) = 0 Alt(Sym(X)) = 0 (7.5.12) These seemingly reasonable claims are proven in Appendix A.9. We can define a third projection operator this way, Else() ≡ 1 - Alt() - Sym() // projection operator Else(X) = X - Alt(X) - Sym(X) . // applied to X (7.5.13) One can then decompose an arbitrary tensor X into three pieces, X = Alt(X) + Sym(X) + Else(X) = A + S + E (7.5.14) Then Alt(X) = Alt(Alt(X)) + Alt(Sym(X)) + Alt(Else(X)) = Alt(X) + 0 + Alt(Else(X)) Alt(Else(X)) = 0 Alt(E) = 0 . (7.5.15) Sym(X) = Sym(Alt(X)) + Sym(Sym(X)) + Sym(Else(X)) = 0 + Sym(X) + Sym(Else(X)) Sym(Else(X)) = 0 Sym(E) = 0 . (7.5.16) This verifies that the "else" piece E of a tensor has neither a totally antisymmetric nor a totally symmetric component. Examples: For a rank-2 tensor Xab one finds Aab = (Xab - Xba) Sab = (Xab + Xba) Eab = Tab - Aab - Sab = 0 (7.5.17) so the leftover else piece Eab is null. On the other hand, for a rank-3 tensor Xabc, Aabc = (Xabc - Xacb + Xcab - Xcba + Xbca - Xbac) Sabc = (Xabc + Xacb + Xcab + Xcba + Xbca + Xbac) Eabc = Xabc - Aabc - Sabc = Xabc - (Xabc - Xacb + Xcab - Xcba + Xbca - Xbac) - (Xabc + Xacb + Xcab + Xcba + Xbca + Xbac) = Xabc - (Xabc + Xcab + Xbca ) = Xabc - (Xcab + Xbca ) (7.5.18) so in this case the leftover piece Eabc is not null. Notice that Xcab = Xabc if X is either totally symmetric or totally antisymmetric Xbca = Xabc if X is either totally symmetric or totally antisymmetric and for this reason (7.5.18) shows that Eabc = 0 if X is either totally symmetric or totally antisymmetric. Fact: (vj^ vj^ .....^ vj) = Alt(vj vj ..... vj) (7.5.19) Proof: Recall from definition (7.1.2) vj^ vj^ .....^ vj = (1/k!) ΣP (-1)S(P) ( vP(j) vP(j) ..... vP(j)) . (7.1.2) Therefore (vj^ vj^ .....^ vj )ii....i = (1/k!) ΣP (-1)S(P) ( vP(j) vP(j) ..... vP(j))ii....i = (1/k!) ΣP (-1)S(P) (vP(j))i(vP(j))i......(vP(j))i // outer product = (1/k!) ΣP (-1)S(P) (vj)P(i)(vj)P(i)......(vj)P(i) // (A.6.1) = (1/k!) ΣP (-1)S(P) (vj vj ..... vj)P(i)P(i) ...P(i) // outer product = Alt(vj vj ..... vj)ii....i // (7.5.3) def of Alt and therefore (vj^ vj^ .....^ vj) = Alt(vj vj ..... vj). QED Corollary: (ej^ ej^ .....^ ej) = Alt(ej ej ..... ej) (7.5.20) 7.6 Expansions for the wedge product of k vectors The symmetric expansion is very straightforward. Let Tii...i = (v1)i (v2)i ... (vk)i . (7.6.1) Then the symmetric expansion (7.4.3) gives, T = Σii...i Tii...i (ei ^ ei ^ .... ^ ei) (7.4.3) = Σii...i (v1)i (v2)i ... (vk)i (ei ^ ei ^ .... ^ ei) (7.6.2) = [Σi(v1)iei] ^ [Σi(v2)i ei] ^ .... ^ [Σi (vk)i ei] = v1 ^ v2 ^ ... ^ vk . (7.6.3) 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 (7.4.12), Aii...i = ΣP (-1)S(P) TP(i)P(i)...P(i) // = [alt(T)]ii...i (7.5.1,2) = ΣP (-1)S(P) (v1)P(i)(v2)P(i)... (vk)P(i) (7.6.4) = ΣP (-1)S(P) (v1v2...vk)P(i)P(i)...P(i) // outer product = [alt(v1v2...vk)]ii...i // (7.5.1) (7.6.5) so that, A = alt(v1v2...vk) . (7.6.6) Then the ordered expansion (7.4.1) becomes, v1 ^ v2 ^ ... ^ vk = Σi<i<....<i Aii...i (ei ^ ei ^ .... ^ ei) = Σi<i<....<i [alt(v1v2...vk)]ii...i (ei ^ ei ^ .... ^ ei) . (7.6.7) Since (7.6.4) has the general form (A.4.4) with jr = r, we can write Aii...i = det = det . (7.6.8) As noted in (A.4.9), for k < n the second determinant above is a minor of matrix Q ≡ [v1,v2....vk] whose columns are the vectors vi. The minor is the full width of Q but only has the rows specified by i1...ik. When k = n, the minor is the full det(Q). See the k=2 example in Fig (4.3.13). We then have these three variations of the vector wedge product expansion: v1 ^ v2 ^ ... ^ vk = Σii...i (v1)i (v2)i ... (vk)i (ei ^ ei ^ .... ^ ei) (7.6.2) v1 ^ v2 ^ ... ^ vk = Σi<i<....<i [alt(v1v2...vk)]ii...i (ei ^ ei ^ .... ^ ei) (7.6.7) v1^ v2^ .....^ vk = Σi<i<....<i det (ei ^ ei ^ .... ^ ei). (7.6.9) Example: Evaluate each of these three expansions for k = 2: v1 ^ v2 = Σii(v1)i (v2)i(ei ^ ei) = [Σi(v1)iei] ^ [Σi(v2)iei] = v1 ^ v2 v1 ^ v2 = Σi<i [alt(v1v2]ii(ei ^ ei) = [(v1)i(v2)i - (v1)i(v1)i] (ei ^ ei) v1^ v2 = Σi<i det . (7.6.10) The first line is an identity while the last two lines agree with (4.3.12). 7.7 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 . (7.7.1) 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, (7.7.2) 7.8 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 (7.4.3) : T = Σii...i Tii...i (ei ^ ei ^ .... ^ ei) (7.4.3) T = ΣI TI eI where eI ≡ ei ^ ei ^ .... ^ ei TI ≡ Tii...i and I ≡ {i1, i2,.... ik} with 1 ≤ ir ≤ n = ordinary multiindex, n = dim(V) . (7.8.1) The more significant notation involves the ordered expansion (7.4.1) 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 (ei ^ ei ^ .... ^ ei) (7.4.1) T = Σ'I AI eI where eI ≡ ei ^ ei ^ .... ^ ei AI ≡ Aii...i and I ≡ {i1, i2,.... ik} with 1 ≤ i1< i2<....< ik ≤ n = ordered multiindex, n = dim(V) (7.8.2) Here are some unofficial multiindex notations for other equations developed above: T = v1 ^ v2 ^ ... ^ vk T = (^vZ) (7.6.3) Tii...i = (v1)i (v2)i ... (vk)i ≡ (vZ)I TI = (vZ)I (7.6.1) with the idea that Z = 1,2...k . Continuing on, T = Σii...i (v1)i (v2)i ... (vk)i (ei ^ ei ^ .... ^ ei) T = ΣI (vZ)I eI (7.6.2) A = alt(v1v2...vk) A = alt(vZ) (7.6.6) v1 ^ v2 ^ ... ^ vk = Σi<i<....<i [alt(v1v2...vk)]ii...i (ei ^ ei ^ .... ^ ei) . (7.6.7) (^vZ) = Σ'I alt(vZ)I eI Aii...i = det AI = det(vZI) (7.6.8) v1^ v2^ .....^ vk = Σi<i<....<i det (ei ^ ei ^ .... ^ ei). (7.6.9) (^vZ) = Σ'I det(vZI) eI 7.9 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 (7.9.1) Here L0 = the space of scalars, L1 = V the space of vectors, L2 = V ^ V V2 the space of antisymmetric rank-2 tensors (7.4.3), and so on. The most general element of the space L(V) would have the form X = s ΣiTi ei Σij Tij ei^ej Σijk Tijk ei^ej^ek + ..... or X = s ΣiTi ei Σi<j Aij ei^ej Σi<j<k Aijk ei^ej^ek + ..... (7.9.2) 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 (ei^ej)^ek 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 . (7.9.3) The product (v1^v2)^v3 = (1/2)(v1v2 - v2v1) ^ v 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. (7.9.4) What this in effect does is define an array of new objects to be the same as v1^ v2^ .....^ v . 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 . (7.9.5) 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 . (7.9.6) Once (7.9.4) is established, it is not hard to show that : Fact: The wedge product of k tensors T1^ T2^ .....^ Tk 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. (7.9.7) This fact is then exactly analogous to the similar axiomatic statement for associativity made in (2.8.22). Proof by example: Consider (in multiindex notation) three tensors T1,T2,T3 of rank k,k',k" : (T1^ T2) ^ T3 = ( (ΣIT1IeI) ^ (ΣJT2JeJ) ) ^ (ΣKT3KeK) = ΣIT1IΣJT2J { ( eI ^ eJ) ^ (ΣKT3KeK) } // rules (7.2.3) = ΣIT1IΣJT2J ΣKT3K (eI ^ eJ) ^ (eK) // rules (7.2.3) again = ΣI,J,K T1IT2JT3K (eI ^ eJ ^ eK ) // axiom (7.9.4), see detail below Compare to T1^ T2 ^ T3 = (ΣIT1IeI) ^ (ΣJT2JeJ) ^ (ΣKT3KeK) = ΣI,J,K T1IT2JT3K (eI ^ eJ ^ eK) // rules (7.2.3) Our example shows that for arbitrary tensors, (T1^ T2) ^ T3 = T1^ T2 ^ T3. Here we illuminate the key detail above: (eI ^ eJ) ^ (eK) = ( (ei^ ei^...^ ei) ^ (ej^ ej^...^ ej' ) ) ^ (ek^ ek^...^ ek" ) = ( ei^ ei^...^ei ^ ej^ ej^...^ ej' ) ^ (ek^ ek^...^ ek" ) = ei^ ei^...^ ei ^ ej^ ej^...^ ej' ^ ek^ ek^...^ ek" (eI ^ eJ ^ eK) = (ei^ ei^...^ ei) ^ (ej^ ej^...^ ej' ) ^ (ek^ ek^...^ ek" ) = ei^ ei^...^ ei ^ ej^ ej^...^ ej' ^ ek^ ek^...^ ek" In each step above the rule (7.9.4) for vectors (applied to basis vectors) is used. Having faced up to the issue of associativity, we now resume the discussion of L(V). This large space L(V) is in fact itself a vector space. (7.9.8) 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) (7.9.9) 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 (7.9.10) Here we have used the associative property (7.9.4). This closure claim is stated more generally below (7.10.6). 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) (7.9.11) 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 values of k. Note that s1( a^b) s2(c^d^e) = (s1a)^b (s2c)^d^e = (a'^b) (c'^d^e) so it is also correct to say that a k-vector is any sum of k-blades since any linear combination can be written as a sum as shown in the above example. Unlike in the tensor product world, in wedge world the above list (7.9.11) 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 (7.9.1) 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 (7.9.12) 7.10 The Wedge Product of two tensors (a) Wedge Product of two tensors T and S Consider these Eq. (7.4.1) ordered expansions of general tensors T and S of rank k and k', T = Σi<i<....<i Aii....i (ei^ ei .....^ ei) rank k, T ϵ Lk S = Σj<j<....<j Bjj....j (ej^ ej .....^ ej) rank k', S ϵ Lk' (7.10.1) where from (7.5.2), Aii...i = [alt(T)]ii...i or A = alt(T) Bjj....j = [alt(S)]jj....j or B = alt(S) . (7.10.2) The corresponding symmetric expansions (7.4.3) of T and S are given by, T = Σii....i Tii....i (ei^ ei .....^ ei) . rank k, T ϵ Lk S = Σjj....j Sjj....j (ej^ ej .....^ ej) . rank k', S ϵ Lk' . (7.10.3) We form the wedge product of these two tensors in a manner similar to the equations leading to (5.6.6) : T^S = [ Σii....iTii....i (ei^ ei .....^ ei)]^[ Σjj....j Sjj....j (ej^ ej .....^ ej)] (a) = Σii....i Σjj....jTii....i Sjj....j(ei^ ei .....^ ei) ^ (ej^ ej .....^ ej) (b) = Σii....ijj....jTii....i Sjj....j(ei^ ei .....^ ei^ ej^ ej .....^ ej) (c) = Σii....iii....i[Tii....i Sii....i] (ei^ ei ......^ ei) (d) = Σii....i[Tii....i Sii....i] (ei^ ei ......^ ei) (e) = Σii....i[TS]ii....iii....i(ei^ ei ......^ ei) (7.10.4) where, as in (5.6.7), we use in the last line the standard outer product notation, [TS]ii...iii...i = Tii...i Sii...i . (7.10.5) Notice that the (7.9.4) associativity of ^ is used going from (a) to (b). Eq. (7.10.4) shows that the product T^S is an element of Lk+k' with the following tensor components, T^S = Σii....i (T^S)ii....i(ei^ ei ......^ ei) (T^S)ii....i = [TS]ii...i. (7.10.6) Thus we have strengthened the claim made in (7.9.10) that L(V) is closed under the operation ^ : the wedge product of an Lk tensor with an Lk' tensor lies in Lk+k' which is in L(V). It is easy then to show that this is true for the wedge product of any two multivectors as defined below (7.9.11). The next step is to express (7.10.4) as an ordered sum rather than a symmetric sum. Momentarily replace TS by the symbol T,' so that (7.10.4) reads, T^S = Σii....iT ' ii...i(ei^ ei ......^ ei) . (7.10.7) This symmetric sum can be replaced by the ordered sum (7.4.1) T^S = Σi<i<....<i A' ii...i(ei^ ei ......^ ei) (7.10.8) where, according to (7.4.12) and then (7.5.2), A' ii...i = ΣP (-1)S(P) T ' P(i)P(i)...P(i) = [alt(T ' )]ii...i . // A ' = alt(T ') (7.10.9) Restoring T ' = TS gives this ordered expansion for the wedge product of tensors T and S, T ^ S = Σi<i<....<i [alt(TS)]ii...i (ei^ ei ......^ ei) . with (7.10.10) [alt(TS)]ii...i = ΣP (-1)S(P) TP(i)P(i)...P(i)S P(i)P(i)..P(i) where ΣP is over all permutations of (the subscripts of) {i1, i2, ....ik+k'}. Example 1: k = 1 and k' = 1 so that T and S are just vectors (n ≥2). The above equations reduce to T ^ S = Σi<i [alt(TS)]ii (ei^ ei) . [alt(TS)]ii = ΣP (-1)S(P) TP(i)SP(i) = TiSi - SiTi (7.10.11) so that T ^ S = Σi<i [TiSi - SiTi](ei^ ei) = Σi<j [TiSj - SiTj] (ei^ ej) // remove italics due to (7.4.16) (7.10.12) in agreement with (4.3.12). Example 2: k = 2 and k' = 1 so that T is a rank-2 tensor and S is still a vector (n ≥ 3) T ^ S = Σi<i<i3 [alt(TS)]iii (ei^ ei^ ei) . (7.10.13) From (7.10.10), [alt(TS)]iii = ΣP (-1)S(P) TP(i)P(i)SP(i) (7.10.14) [alt(TS)]abc = TabSc - TacSb + TcaSb - TcbSa + TbcSa - TbaSc // see (7.4.14) so that T ^ S = Σa<b<c {TabSc - TacSb + TcaSb - TcbSa + TbcSa - TbaSc} (ea^ eb^ ec) (7.10.15) where the summation range is 1 ≤ a < b < c ≤ n. In the case dim(V) = n = 3, there is only one term which has a = 1, b = 2 and c = 3, T ^ S = {T12S3 - T13S2 + T31S2 - T32S1 + T23S1 - T21S3} (e1^ e2^ e3) // n = 3 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 (7.10.4b) reads, T^S = Σii....ijj....jTii....i Sjj....j (ei^ ei .....^ ei^ ej^ ej .....^ ej) = Σii....iTii....i (κ') (ei^ ei .....^ ei) = κ'T and S^T = Σjj....jii....i Sjj....j Tii....i ( ej^ ej .....^ ej^ ei^ ei .....^ ei) = Σii....i (κ') Tii....i ( ei^ ei .....^ ei) = κ'T 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 (7.10.16) These special case results are seen to be the same as those for TS shown in (5.6.15). (b) Commutivity Rule for the Wedge Product of two tensors T and S Recall the expansion of T^S from (7.10.4)(b), T^S = Σii....ijj....jTii....i Sjj....j(ei^ ei .....^ ei^ej^ ej .....^ ej) (7.10.17) Swapping T↔S , 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 (ej^ ej .....^ ej^ei^ ei .....^ ei) = Σii....ijj....j Tii....i Sjj....j(ej^ ej .....^ ej^ei^ ei .....^ ei) (7.10.18) [ Note: In multiindex these equations are: T^S = ΣI,JTISJ eI^eJ and S^T = ΣI,JTISJ eJ^eI .] Equations (7.10.17) and (7.10.18) are identical except for the last factor involving the basis vectors. Consider the basis vector factor appearing in (7.10.18), (ej^ ej .....^ ej^ ei^ ei .....^ ei) . // eJ^eI To make this match the basis factor in (7.10.17), 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, (ej^ ej .....^ ej^ ei^ ei .....^ ei) = (-1)k' ei ^ (ej^ ej .....^ ej^ ei .....^ ei) = (-1)k' (-1)k' ei ^ ei ^ (ej^ ej .....^ ej .....^ ei) = etc. = = [(-1)k']k ( ei^ ei .....^ ei ^ ej^ ej .....^ ej) . // (-1)kk' eI^eJ (7.10.19) Inserting (7.10.19) into (7.10.18) and comparing with (7.10.17) we arrive at this well-known result, sometimes called "graded commutivity" since the nature of the commutivity depends on the grades (ranks) of the two tensors, S^T = (-1)kk'T^S ranks of the two tensors are k and k' . (7.10.20) The wedge product of two tensors commutes if kk' is even, and anticommutes if kk' is odd. 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) (7.10.21) 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 (7.10.20) says S^T = T^S, consistent with (7.10.16) line 1. If k=k'=0 so T = κ and S = κ', rule (7.10.20) says S^T = T^S, consistent with (7.10.16) line 3. (7.10.22) (c) Multiindex Notation Rehash of Section 7.10 For "practice" and for use in the next section, we give an abbreviated copy, paste and edit rehash of Section 7.10 above using the multiindex notation. Whenever there is a confusion, one must write things out in detail. Equation numbers from above are shown in italics. __________________________________________________________________________________ Consider these Eq. (7.4.1) ordered expansions of general tensors T and S of rank k and k', T = Σ'I AI eI I = {i1, i2, .. ik} eI ≡ (ei^ ei .....^ ei) rank k S = Σ'J BJ eJ J = {j1, j2, .. jk'} eJ ≡ (ej^ ej .....^ ej) rank k' (7.10.1) where from (7.5.2), AI = [alt(T)]I BJ = [alt(S)]J . (7.10.2) The corresponding symmetric expansions (7.4.3) of T and S are given by, T = ΣI TI eI // symmetric expansions rank k, T ϵ Lk S = ΣJ SJ eJ . rank k', S ϵ Lk' (7.10.3) We form the wedge product of these two tensors in a manner similar to the equations leading to (5.6.6), T^S = [ ΣITI eI] ^ [ ΣJ SJ eJ] (a) = ΣI ΣJTI SJ (eI^eJ) (b) = ΣI,JTISJ (eI^eJ) I = {i1, i2, .. ik} (c) = ΣI,I'[TI SI'] (eI^eI') I ≡ {i1...ik+k'}, I' ≡ {ik+1...ik+k'} so I I' = I (d) = ΣI [TI SI'] eI eI ≡ (ei^ ei ......^ ei) (e) = ΣI (TS)I eI // symmetric expansion of T^S (7.10.4) where, as in (5.6.7), we use in the last line the standard outer product notation, (TS)I = TI SI' . (7.10.5) Eq. (7.10.4) shows that the product T^S is an element of Lk+k', T^S = ΣI (T^S)I eI with (T^S)I = (TS)I . (7.10.6) The next step is to express (7.10.4) as an ordered sum rather than a symmetric sum. Momentarily replace TS by the symbol T,' so that (7.10.4) reads, T^S = ΣIT 'I eI . (7.10.7) This symmetric sum can be replaced by the ordered sum (7.4.1) T^S = Σ'I A' I eI (7.10.8) where, according to (7.4.12) and then (7.5.2), A' I = ΣP (-1)S(P) T ' P(I) = [alt(T ' )]I . // A ' = alt(T ') (7.10.9) Restoring T ' = TS gives this ordered expansion for the wedge product of tensors T and S, T ^ S = Σ'I [alt(TS)]I eI. (7.10.10) [alt(TS)]I = ΣP (-1)S(P) TP(I)S P(I') where ΣP is over all (k+k')! permutations of (the subscripts of) {i1, i2, ....ik+k'}. Commutivity Rule for the Wedge Product of two tensors T and S Recall the expansion of T^S from (7.10.4)(b), T^S = ΣI,J TISJ (eI ^ eJ) (7.10.17) Swapping T↔S , T↔S, k↔k' and I ↔ J gives the following form for the wedge product S^T , S^T = ΣJ,I SJTI (eJ ^ eI) = ΣI,J TISJ (eJ ^ eI) . (7.10.18) Equations (7.10.17) and (7.10.18) are identical except for the last factor involving the basis vectors. (eJ ^ eI) = [(-1)k']k (eI ^ eJ) // slide eJ left through eI (7.10.19) S^T = (-1)kk'T^S ranks of the two tensors are k and k' . (7.10.20) ________________________________________________________________________________ 7.11 The Wedge Product of N tensors in L(V) First, we mimic the multiindex development just above to obtain the wedge product of three tensors: T = ΣI TI eI rank k I = {i1, i2, .. ik} S = ΣJ SJ eJ rank k' I' ≡ {ik+1...ik+k'} R = ΣK RK eK rank k" I" = {ik+k'+1, ik+k'+2, .. ik+k'+k"} (7.11.1) T^S^R = (ΣITIeI) ^ (ΣJSJeJ) ^ (ΣKRKeK) = ΣI,J,K TISJRK (eI^eJ^eK) . // symmetric expansion = ΣI,I',I" TISI'RI" (eI^eI'^eI") = ΣI(TSR)I eI (TSR)I = TISI'RI" (7.11.2) = Σ'I [alt(TSR)]IeI // ordered expansion, eI ≡ (ei^ ei ......^ ei) where [alt(TSR)]I = ΣP (-1)S(P) TP(I)SP(I')RP(I") (7.11.3) Sample reordering rule: S^T^R = ΣI,J,K TISJRK (eJ^eI^eK) = ΣI,J,K TISJRK [ (-1)kk'(eI^eJ^eK)] = (-1)kk' T^S^R (7.11.4) Systematic Tensor Products To develop a more systematic approach, consider the first three tensors in a product sequence, 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} . (7.11.5) Define the following "cumulative ranks", κ1 = k1 // cumulative ranks, as in κ2 = k1+ k2 κ3 = k1+ k2 + k3 ... κN = k1 + k2 + ... + kN = Σi=1N ki . (7.11.6) Then rewrite (7.11.5), 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} ... TN = tensor of rank kN IN = {iκ+1,iκ+2.....iκ} . (7.11.7) Define, T1P(I) ≡ T1P(i)T1P(i)...T1P(i) T2P(I) ≡ T2P(i)T2P(i) ....T2P(i) T3P(I) ≡ T3P(i)T3P(i) ....T3P(i) ... TNP(I) ≡ TNP(i)TNP(i) ....TNP(i) . (7.11.8) We can now write out the product of any number of tensors. In each case we show the symmetric expansion first, then the ordered expansion. Wedge Product of 2 Tensors (7.11.9) T1^T2 = ΣII T1IT2I (eI^eI) = ΣI (T1T2)IeI T1^T2 = Σ'I [alt(T1T2)]I eI eI ≡ ei^ ei .....^ ei where [alt(T1T2)]I = ΣP (-1)S(P) T1P(I)T2P(I) Wedge Product of 3 Tensors (7.11.10) T1^T2^T3 = ΣIII T1IT2IT3I (eI^eI^eI) = ΣI (T1T2T3)I eI T1^T2^T3 = Σ'I [alt(T1T2T3)]I eI eI ≡ ei^ ei .....^ ei where [alt(T1T2T3)]I = ΣP (-1)S(P) T1P(I)T2P(I)T3P(I) Wedge Product of N Tensors (7.11.11) T1^T2^...^TN = ΣII...I T1IT2I....TNI (eI^eI ... ^eI) = ΣI (T1T2 ....TN)I eI T1^T2^...^TN = Σ'I [alt(T1T2...TN)]I eI eI ≡ ei^ ei .....^ ei where [alt(T1T2...TN)]I = ΣP (-1)S(P) T1P(I)T2P(I)...TNP(I) . Sign Rule for swapping two tensors Swapping two tensors in a tensor product results in either + or - the same tensor, as shown for example in (7.11.4) Consider an example where we have a wedge product of 9 tensors. The eI basis function groups are eI ^ eI ^ eI ^ eI ^ eI ^ eI ^ eI ^ eI^ eI which goes with T1 ^ T2 ^ T3 ^ T4 ^ T5 ^ T6 ^ T7 ^ T8 ^ T9 . 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 eI ^ eI ^ eI ^ eI ^ eI ^ eI ^ eI ^ eI^ eI = eI ^ eI ^ eI ^ eI ^ eI ^ eI ^ eI ^ eI^ eI (-1)kk = eI ^ eI ^ eI ^ eI ^ eI ^ eI ^ eI ^ eI^ eI (-1)(k+k)k = eI ^ eI ^ eI ^ eI ^ eI ^ eI ^ eI ^ eI^ eI (-1)(k+k+k)k = eI ^ eI ^ eI ^ eI ^ eI ^ eI ^ eI ^ eI^ eI (-1)(k+k+k+k)k Now with this as a starting point, we slide eI to the right, one group at a time, eI ^ eI ^ eI ^ eI ^ eI ^ eI ^ eI ^ eI^ eI = eI ^ eI ^ eI ^ eI ^ eI ^ eI ^ eI ^ eI^ eI (-1)kk = eI ^ eI ^ eI ^ eI ^ eI ^ eI ^ eI ^ eI^ eI (-1)k(k+k) = eI ^ eI ^ eI ^ eI ^ eI ^ eI ^ eI ^ eI^ eI (-1)k(k+k+k) and now we have successfully swapped eI ↔ eI 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 . (7.11.12) 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 . (7.11.13) If the sum of the ranks of the two swapped tensor is even, in effect m = krks . Example 1: T1 ^ T2 ^ T3 = (-1)m T2 ^ T1 ^ T3 r = 1 s = 2 m = (0)(k1+k2) + k1k2 = k1k2 (-1)m = (-1)kk (7.11.16) 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 (7.11.17) 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 swapping two of these tensors produces a minus sign phase = (-1)m where m ≈ krks = 1*1 = 1 in agreement with (7.2.4). (7.11.18) 8. The Wedge Product of k dual vectors : the vector spaces Λk and Λ(V)