New Chapter 6 on dual with k vectors

Chapter 6 Attempt PhL 11.9.15 This has been installed on 11/9/15 around 7:45 PM. 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. 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 (5.1.1) and the superscripts in (5.1.2) 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. 6.2 Tensor Expansion for a tensor in V*k ; 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 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. 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 which projects out the coefficients in (6.2.1). 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) 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 Algebra T(V*) Brief Digression: The Direct Sum of Vector Spaces See Section 5.4 for this digression. The 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 . // from (5.4.1) (6.4.2) 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 (6.4.3) with all coefficients in a field K. Fact: This large space T(V*) is in fact itself a vector space. (6.4.4) 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. 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) + (sβ) κ + ρ(sσ)η = element of T(V*) . (6.4.5) 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 (6.4.6) 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*) (6.4.7) 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 + ... = ∞ (6.4.8) 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 (6.4.2) is closed under both + and . It seems evident how one would add two tensors of T(V*) of the form (6.4.3), 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 (6.4.6) 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 (6.4.7). 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 (λi λi λi λi ) (TS)(ea,eb,ec,ed) = TabScd = (TS)abcd (6.6.14) In both examples the basis-vector function evaluations produce the expected expansion coefficients.