Archived Ch 6 retired 1_23_16

Archived Ch 6 removed from main doc v6 PhL 1.23.16 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. Analogous equations are these : (α1 α2 ..... αk)(vj,vj....vj) = α1(vj)α2(vj) ..... αk(vj) // (A.8.22) (6.1.3a) (α1 α2 ..... αk)(ej,ej....ej) = α1(ej)α2(ej) ..... αk(ej) = (α1)j(α2)j ...(αk)j // (2.11.9) (6.1.3b) (λi λi ..... λi)(vj,vj....vj) = λi(vj)λi(vj) ..... λi(vj) = (vj)i(vj)i ....(vj)i // (2.11.7) (6.1.4a) (λi λi ..... λi)(ej,ej....ej) = λi(ej)λi(ej) ..... λi(ej) [ = (ej)i(ej)i ....(ej)i ] = δijδij ....δij // (2.11.6) (6.1.4b) 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 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 . 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) Section 5.2 evaluates its equation (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) We cannot take components of the functional expansion (6.2.1), but we can evaluate it at (vj,vj ...vj), T(vj, vj ....vj) = Σii....i Tii....i(λi λi ..... λi)(vj, vj ....vj) = Σii....i Tii....i (vj)i(vj)i ....(vj)i . // (6.1.4a) (6.2.5) A special case then is T(ej, ej ....ej) = Σii....i Tii....i δji δji .... δji // (6.1.4b) = Tjj....j (6.2.6) which projects out the coefficients in (6.2.1) in analogy with (5.2.1). In multiindex notation, fact (6.2.6) appears in (C.5.6) as the statement TI = T(eI) = <eI| T>. In Section C.5 the object T(vj,vj ...vj) = T(vI) = <vI | T > and the object Tii....i = TI = <eI | T > are treated as two representations of the abstract tensor T projected onto different bases. In this | T > context, or just looking at (6.2.1a), one sees that there is a 1-to-1 correspondence between the rank-k tensors Tii...i and the k-multilinear functions T(vj,vj ...vj). We shall refer to Tii...i as a "tensor" (as in Section 2) and T(vj,vj ...vj) as a "tensor function". We shall then say that the tensor T lies in Vk, T lies in V*k, and T(vj,vj ...vj) lies in V*kf (subscript f for function). The two spaces V*k and V*kf are isomorphic. Then T is a "tensor functional". Different authors have various names for these three objects. Some refer to them all just as "tensors" which is a reasonable stance. Spivak for example (p 75) refers to our space V*kf as Tk(V), and to a tensor function as a k-tensor. His tensor functions are indicated by non-script letters like S and T, while we use script letters S and T to distinguish them from our Chapter 2 tensors S and T. (6.2.7) 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*). Since V*k is a vector space, so is the isomorphic space of tensor functions V*kf . 6.4 The Tensor Algebras T(V*) and Tf(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) All objects listed in the left column are tensor functionals, but we just call them tensors above and below. 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 (6.2.1). 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) Equation (6.4.1) gives a direct sum decomposition of T(V*), the space of tensor functionals. Due to the isomorphism noted in (6.2.7), we have this similar tensor algebra for tensor functions, Tf(V*) ≡ V*0f V*f V*2f V*3f ....... = Σk=1∞ V*kf . (6.4.8) 6.5 Comments about tensors Section 5.5 shows that linear combinations of permutated tensors are still tensors. In the dual tensor world, one would say the following T'(vi, vj) = RiaRjb T(va, vb) 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)