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Scratch Edit for Section 7.9 PhL 11.19.15 this has been installed, do not edit 7.9 The Wedge Product of two or more tensors in L(V) 1 (a) Wedge Product of two tensors T^ and S^ 1 (b) Special cases of the wedge product T^^ S^. 3 (c) Commutivity Rule for the Wedge Product of two tensors T^ and S^ 4 (d) Wedge Product of three or more tensors 5 (e) Commutativity Rule for product of N tensors 7 7.9 The Wedge Product of two or more tensors in L(V) (a) Wedge Product of two tensors T^ and S^ Here we shall mimic the developmental approach used in Section 5.6 for the tensor product. As before, we quietly "break in" the multiindex notation. The symmetric expansions (7.4.4) of T^ and S^ are given by, T^ = Σii....i Tii....i (ei^ ei .....^ ei) . rank k, T^ ϵ Lk (7.9.a.1) ΣITIe^I S^ = Σjj....j Sjj....j (ej^ ej .....^ ej) . rank k', S^ ϵ Lk' . (7.9.a.2) ΣJSJe^J We form the wedge product of these two tensors in a manner similar to (5.6.3) : T^^S^= [Σii....iTii....i (ei^ ei .....^ ei)]^[ Σjj....j Sjj....j (ej^ ej .....^ ej)] [ ΣITIe^I] ^ [ΣJTJe^J] (a) = Σii....i Σjj....jTii....i Sjj....j(ei^ ei .....^ ei) ^ (ej^ ej .....^ ej) ΣI,JTISJ(e^I) ^ (e^J) (b) = Σii....ijj....jTii....i Sjj....j(ei^ ei .....^ ei^ ej^ ej .....^ ej) ΣI,JTISJ(e^I ^ e^J) (c) = Σii....iii....i[Tii....i Sii....i] (ei^ ei ......^ ei) ΣI,I'TISI'(e^I ^ e^I') (d) = Σii....iii....i[Tii....i Sii....i] (ei^ ei ......^ ei) ΣI,I'[TS]I,I'(e^I ^ e^I') (e) = Σii....i[TS]ii...i(ei^ ei ......^ ei) (7.9.a.3) ΣI (TS)I e^I Comparing lines one sees that I ≡ i1, i2...ik I' ≡ ik+1, ik+2, ....ik+k' I ≡ I, I' = i1,i2...ik+k' e^I ≡ (ei^ ei ....^ ei) e^I' ≡ (ei^...^ei) e^I ≡ (ei^ ei ....^ ei) (7.9.a.4) Notice that the (2.8.22) vector associativity of ^ is used going from (a) to (b). The conclusion is that T^^ S^ = ΣI (TS)I e^I I ≡ I, I' = i1,i2...ik+k', e^I ≡ (ei^ ei ....^ ei) . (7.9.a.5) Since the e^I are basis vectors in Lk+k', we have shown that: T^ ϵ Lk and S^ ϵ Lk' T^^ S^ ϵ Lk+k' L(V) . (7.9.a.6) Thus we have strengthened the claim made in (7.8.10) that L(V) is closed under the operation ^. Recall now from (7.3.8) the relationship between e^I and eI, (ei ^ ei ^ .... ^ ei) = Alt(ei ei .... ei) e^I = Alt(eI) (7.3.8) and also the Chapter 5 expansion of the tensor product TS, TS = ΣI (TS)I eI I ≡ I, I' = i1,i2...ik+k', eI ≡ (ei ei .... ei) . (5.6.5) Applying Alt to this last equation gives Alt( TS) = ΣI (TS)IAlt(eI) // Alt is linear = ΣI (TS)I e^I // (7.3.8) above = T^^ S^ so T^^ S^ = Alt(TS) (7.9.a.7) with components [T^^ S^]I = [Alt(TS)]I = ΣP(-1)S(P) (TS)P(I) // (A.5.3) = ΣP(-1)S(P) TP(I)SP(I') . (7.9.a.8) This last line is an explicit instruction for computing the components of the tensor T^^ S^ . We have added this new notation, TP(I) ≡ Tii...i for I = i1, i2...ik (7.9.a.9) Example: Let S and T both be rank-2 tensors so k = k' = 2 . Then [T^^ S^]I = [T^^ S^]iiii = (1/4!) ΣP(-1)S(P)TiiSii = (1/24) [ TiiSii - TiiSii + TiiSii - TiiSii + 20 more terms ] (7.9.a.10) Here as elsewhere we show in red the indices to be swapped to make the next term. From (7.9.c.6) below, T^^ S^ = (-1)2*2 S^^ T^ = S^^ T^. (7.9.a.11) (b) Special cases of the wedge product T^^ S^. Assume T^ and S^ have rank k and k'. If S = κ' ϵ K = a scalar, then rank(S) = k' = 0 and (7.9.4b) reads, T^^ S^ = Σii....ijj....jTii....i Sjj....j (ei^ ei .....^ ei^ ej^ ej .....^ ej) = Σii....iTii....i (κ') (ei^ ei .....^ ei) = κ'T (7.9.b.1) 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 (7.9.b.2) so we find that T^S = S^T = κ'T . If T = κ and S = κ', the result above would be T^^S^ = κκ' and S^^T^ = κ'κ and so T^^S^ = S^^T^ = κκ'. Thus, T^^S^ = κ^S^ = S^^T^ = S^^κ = κS^ if T^ = κ ϵ V0 T^^S^ = T^^κ' = S^^T^ = κ'^T^ = κ'T^ if S^ = κ' ϵ V0 T^^S^ = κ^κ' = S^^T^ = κ'^κ = κκ' if T^,S^ = κ,κ' ϵ V0 (7.9.b.3) These special case results are seen to be the same as those for TS shown in (5.6.15). When T is rank-0 or rank-1 we can write T^ = T according to (7.4.17), but we continue to use T^ (c) Commutivity Rule for the Wedge Product of two tensors T^ and S^ Recall the expansion of T^^ S^ from (7.9.a.3) item (b), T^^ S^ = Σii....ijj....jTii....i Sjj....j(ei^ ei .....^ ei^ej^ ej .....^ ej) ΣI,J TI SJ ( e^I ^ e^J) (7.9.c.1) Swapping T↔S, k↔k' and i ↔ j gives the following form for the wedge product S^^T^ , S^^T^ = Σjj....jii....iSjj....j Tii....i (ej^ ej .....^ ej^ei^ ei .....^ ei) ΣJ,I SJTI (e^J ^ e^I) = Σii....ijj....j Tii....i Sjj....j(ej^ ej .....^ ej^ei^ ei .....^ ei) ΣI,J TISJ (e^J ^ e^I) (7.9.c.2) Equations (7.9.c.1) and (7.9.c.2) are identical except for the last factor involving the basis vectors. Consider the basis vector factor appearing in (7.9.c.2), (e^J ^ e^I) = (ej^ ej .....^ ej^ ei^ ei .....^ ei) . (7.9.c.3) To make this match the basis factor in (7.9.c.1), we have to slide all the red basis vectors to the left through all the black basis vectors. Each time a red passes through a black, we pick up a minus sign due to the rule (7.2.4). Thus, (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) (7.9.c.4) Therefore, (e^J ^ e^I) = (-1)kk' (e^I ^ e^J) . (7.9.c.5) Inserting this result into (7.9.c.2) gives S^^ T^ = (-1)kk'T^^ S^ ranks of the two tensors are k and k' . (7.9.c.6) Since the commutivity sign is a function of the ranks (grades) of the tensors, this statement is often referred to as "graded commutivity". The wedge product of two tensors commutes if kk' is even, and anticommutes if kk' is odd. Using (7.9.a.7) the above becomes. Alt(ST) = (-1)kk'Alt(TS) . (7.9.c.7) Example: If k = k' = 1, (-1)kk' = -1 and we recover the simple rule for vectors, S^^T^ = - T^^S^ // S and T are rank-1 tensors (vectors) (7.9.c.8) as first stated in (4.3.2). One must keep in mind that the result S^^T^ = - T^^S^ is not valid for arbitrary tensors S^ and T^. Examples: If k = 0 so T = κ, rule (7.9.c.6) says S^^T^ = T^^S^, consistent with (7.9.b.3) line 1. If k=k'=0 so T = κ and S = κ', rule (7.9.c.6) again says S^^T^ = T^^S^, consistent with (7.9.b.3) line 3. (7.9.c.9) (d) Wedge Product of three or more tensors Mimicking (5.6.7) we write T^^S^^R^ = [ΣITIe^I]^[ΣJ SJe^J]^[ΣK SKe^K] (a) = ΣI,J,K TISJRK (e^I) ^ (e^J) ^ (e^K) (b) = ΣI,J,K TISJRK (e^I ^ e^J ^ e^K ) // associative of ^ used here (d) = ΣI,I',I" TISI'RI" (e^I ^ e^I' ^ e^I") // rename multiindices J→I',K→I" I ≡ i1, i2...ik I' ≡ ik+1, ik+2, ....ik+k' I" ≡ ik+k'+1, ik+k'+2, ....ik+k'+k" e^I ≡ (ei^....^ ei) e^I' ≡ (ei^...^ei) e^I" ≡ (ei^...^ei) (e) = ΣI (TSR)I e^I e^I ≡ (ei^....^ ei) I ≡ I, I',I" = i1,i2...ik+k'+k" (7.9.d.1) The outer product form is TISI'RI" = (TSR)I,I',I" = (TSR)I . The conclusion is this: T^^S^^R^ = ΣI (TSR)Ie^I I ≡ I, I',I" = i1,i2...ik+k'+k" , e^I ≡ (ei^....^ ei) (7.9.d.2) Since the e^I are basis vectors in Lk+k'+k", we have shown that: T^ ϵ Lk and S^ ϵ Lk' and R^ ϵ Lk" T^^S^^R^ ϵ Lk+k'+k" L(V) . (7.9.d.3) Recalling the Chapter 5 result, TSR = ΣI (TSR)IeI I ≡ I, I',I" = i1,i2...ik+k'+k" eI ≡ (ei.... ei) (5.6.8) and (7.3.8) that e^I = Alt(eI) we find, Alt(TSR) = ΣI (TSR)I Alt(eI) // Alt is linear = ΣI (TSR)Ie^I = T^^S^^R^ so T^^S^^R^ = Alt(TSR) (7.9.d.4) and then [T^^ S^^R^]I = [Alt(TSR)]I = ΣP(-1)S(P) (TSR)P(I) // (A.5.3) = ΣP(-1)S(P) TP(I)SP(I') RP(I") (7.9.d.5) which gives implicit instructions for how to compute the components of T^^S^^R^ . Using the systematic notation outlined in (5.6.11) and (5.6.12), and generalizing the above development for the wedge product of three tensors, we find the following expansion for the wedge product of N tensors of L(V), (T1)^^(T2)^^...^(TN)^ = ΣI (T1IT2I .... TNI) e^I = ΣI (T1T2....TN)I e^I (7.9.d.6) where e^I = ei^ ei .....^ ei = ei^ ei .....^ eκ and (T1T2....TN)I = T1IT2I .... TNI The rank of this product tensor is then κ = Σi=1N ki and the tensor is an element of Lκ L(V). Recall (5.6.13), T1T2...TN = ΣI (T1IT2I .... TNI) eI = ΣI (T1T2....TN)I eI . (5.6.13) Applying Alt to both sides shows that again with e^I = Alt(eI) shows that (T1)^^(T2)^^...^(TN)^ = Alt(T1T2...TN) . (7.9.d.7) Components of this tensor are computed as follows: [(T1)^^(T2)^^...^(TN)^]I = [Alt(T1T2...TN)]I = ΣP(-1)S(P) (T1T2...TN)P(I) // (A.5.3) = ΣP(-1)S(P) T1P(I)T2P(I) ....TNP(I) (7.9.d.8) where T1P(I) ≡ T1ii...i for I1 = i1, i2...ik T2P(I) ≡ T2ii...i for I2 = i1, i2...ik (e) Commutativity Rule for product of N tensors To reduce clutter, in this section we abbreviate e^I by eI and (Ti)^ by Ti. Consider an example where we have a wedge product of 9 tensors. The eI basis function groups are eI ^ eI ^ eI ^ eI ^ eI ^ eI ^ eI ^ eI^ eI (7.9.e.1) which goes with T1 ^ T2 ^ T3 ^ T4 ^ T5 ^ T6 ^ T7 ^ T8 ^ T9 . (7.9.e.2) The sign caused by swapping T3 ↔ T7 will be the same as the sign swapping eI ↔eI in the basis function. We do it one step at a time, first sliding the group eI to the left using (7.9.c.5), 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 (7.9.e.3) 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) (7.9.e.4) 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 . (7.9.e.5) Based on this result, we claim that : Fact: In a product of tensors T1^T2^T3.... of rank k1, k2, k3 ... , if two tensors are swapped Tr ↔ Ts (with r < s), the resulting tensor incurs the following sign relative to the starting tensor, sign = (-1)m where m = (kr+1+kr+2 ...+ks-1)(kr+ks) + krks . (7.9.e.6) 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 (7.9.e.7) Example 2: T1 ^ T2 ^ T3 = (-1)m T3 ^ T2 ^ T1 r = 1 s = 3 m = (k2)(k1+k3) + k1k3 = k1k2 + k1k3 + k2k3 (-1)m = (-1)kk+kk+kk (7.9.e.8) 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 = - 1 where m ≈ krks = 1*1 = 1 in agreement with the basic vector swap rule (7.2.4). (7.9.e.9)