Chapter 7 notation change

Trying out notation change in Chapter PhL 11.18.15 1.15.16 What notation change is that? I think the new notation is to use T^ to represent a totally antisymmetric tensor, so in this new notation T^ = Alt(T) would work for any T. 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 (parallelogram ) 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 equivalent 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) = Alt(vj vj ..... vj) 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 multiindex notation (7.3.13) 7.4 Tensor Expansions for a tensor in Lk Recall now the tensor expansion for a most-general tensor T in Vk is, T = Σii....i Tii....i (ei ei ..... ei) . Tϵ Vk (5.2.1) (7.4.1) Consider then the similar-looking most-general object in Lk, Σii....i Tii....i (ei^ ei^ .....^ ei) = Σii....i Tii....i Alt (ei ei ..... ei) // (7.3.8) = Alt[ Σii....i Tii....i (ei ei ..... ei) ] // (A.5.10), Alt is linear = Alt(T) ≡ T^ (7.4.2) where we define this notation, T^ ≡ Alt(T). (7.4.3) From (7.4.2) we then have the following fully general element of Lk, T^ = Σii....i Tii....i (ei^ ei^ .....^ ei) (7.4.4) We refer to this type of expansion as a symmetric expansion, and we know it is redundant since the symmetric sum includes each true basis vector k! times. According to (A.5.9), we know from (7.4.3) that Fact: T^ii....i is a totally antisymmetric tensor. (7.4.5) Therefore, Fact: The space Lk is the space of all totally antisymmetric rank-k tensors T^. (7.4.6) In contrast, the space Vk is the space of all rank-k tensors T, so Lk Vk. See Section 7.7 below. Since the set (ei ^ ei ^ .... ^ ei) with 1 ≤ i1 < i2 < ..... < ik ≤ n forms a complete basis for Lk, as discussed below (7.3.3), it must be possible to express T^ in the following manner T^ = Σ1≤i<i<....<i≤n Aii...i (ei ^ ei ^ .... ^ ei) . (7.4.7) 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) . (7.4.8) What then is the connection between the Aii...i and the Tii...i coefficients? Start with the symmetric form (7.4.4), T^ = Σii...i Tii...i (ei ^ ei ^ .... ^ ei) ir = 1,2..n = Σi≠i≠...≠i Tii...i (ei ^ ei ^ .... ^ ei) . // (7.2.5) (7.4.9) Partition the summation space as follows (1 ≤ ir ≤ n), Σi≠i≠...≠i = [ Σi<i<...<i + Σi<i<...<i + many similar reorderings ] . (7.4.10) The total sum can be written in this manner, using the permutation sum notation, Σi≠i≠...≠i = ΣP ΣP(i)<P(i)<...<P(i) (7.4.11) where P are the k! permutations of the k integers {1,2,...k}. Using the form (7.4.11), the sum (7.4.9) may be rewritten as, T^ = ΣP ΣP(i)<P(i)<...<P(i) Tii...i (ei ^ ei ^ .... ^ ei) . (7.4.12) 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 can be moved from the summation indices to the summand indices. One then has from (7.4.12), T^ = Σi<i<...<i ΣP [ TP(i)P(i)...P(i) (eP(i) ^ eP(i) ^ .... ^e P(i)) ] . (7.4.13) But we know from (7.3.5) that ( eP(i) ^ eP(i) ^ .... ^e P(i)) = (-1)S(P) (ei ^ ei ^ .... ^ ei) (7.4.14) 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) (7.4.15) which we can compare with the ordered sum (7.4.7), T^ = Σi<i<....<i Aii...i (ei ^ ei ^ .... ^ ei) . (7.4.5) Thus, since the basis is complete, the relation between the A and T coefficients is given by Aii...i = ΣP (-1)S(P) TP(i)P(i)...P(i) i1 < i2 < ..... < ik = [ Tii...i + all signed permutations ] // k! terms = k! [Alt(T)]ii...i . // (A.5.3) def of Alt or A = k!Alt(T) = k! T^. // (7.4.3) (7.4.16) The Aii...i appear in the expansion (7.4.7) only for index values 1 ≤ i1 < i2 < ..... < ik ≤ n, but we can interpret (7.4.16) as defining Aii...i for all index values. Since A = k! T^ , (7,4,5) show that Fact: Aii...i and T^ii...i are both totally antisymmetric tensors. (7.4.17) Comment: Tii...i and Aii...i are both rank-k tensors, see (5.5.3). Examples: (relating the A and T coefficients) Aab = Tab - Tba // as in (4.3.10) k = 2 Aabc = Tabc - Tacb + Tcab - Tcba + Tbca - Tbac k = 3 (7.4.18) For k = 1, (7.4.9) says (T^)j = Σi Ti(ei)j = ΣiTiδij = Tj (7.4.17) so for a vector there is no distinction between (T^)j and Tj (and in fact V1 = L1). . 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)