scraps started 1_25_16

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 (8.4.4) 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) ************ 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) ******* 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) ************* In the previous section, the n vectors vi can be regarded as linear combinations of the n vectors ei, vj = Σr Mrjer or |vj> = ΣrMrj |er> . r = 1,2...k (D.3.1) ei vj = Σr Mrjei er = Mij = <ei|vj> = (vj)i  as in (D.2.3) Then, since we showed below (D.1.2) that T is k-multilinear, one finds T(vi,vi, .... vi) = T( ΣjMjiej, ΣjMjiej, .... ΣjMjiej) T(vI) = T(MJIeJ) = Σjj...j MjiMji .... Mji T(ej,ej, .... ej) . = MJI T(eJ) (D.3.2) But T(ej,ej, .... ej) = Tjj....j from (D.1.6) so, now with implied sums on the jr, T(vi,vi, .... vi) = MjiMji .... MjiTjj....j . T(vI) = MJITJ (D.3.3) We can regard the coefficient matrix R as being the differential of the linear transformation x' = F(x) = Rx which maps x-space to x'-space as shown in Fig (2.1.1). STOP! Comparing (D.3.3) to (D.2.8a) we see that we have simply rewritten things using Rij = (vi)j so we know exactly what this Rij transformation is. So Rij = (vi)j and then (Rx)i = Rijxj = (vi)jxj. Of course this R thing is linear, so we can identify x' = F(x) = Rx if we want. But what is the purpose of doing all this?? For a rank-k tensor we know then that the transformation rule is, similar to the fourth line of (2.1.6), T'ii....i = RijRij .... Rij Tjj....j . T'I= RIJTJ (D.3.4) Comparison of (D.3.4) with (D.3.3) provides this interpretation of T'ii....i, T'ii....i = T(vi,vi, .... vi) . T'I = T(vI) (D.3.5) Since we earlier assumed that the vectors vi form a basis for V, we can define arbitrary vectors v'i as linear combinations v'i = ΣjRijvj where the coefficients now define a new matrix R. We take this to be the differential of x' = F(x) = Rx. Equation (D.3.2) then becomes (or just use k-multilinearity), T(v'i,v'i, .... v'i) = RijRij .... Rij T(vj,vj, .... vj) . T(v'I) = RIJ T(vJ) (D.3.6) Therefore (implied sums 1 to k for all ir ), T '(vi,vi, .... vi) ≡ T(Rijvj,Rijvj, .... Rijvj) va' = Rabvb x' = Rx = T(v'i,v'i, .... v'i) = RijRij .... Rij T(vj,vj, .... vj) . (D.3.7) We can interpret this as "the transformation rule for a rank-k tensor function" which is akin to "the transformation rule for a rank-k tensor", T'ii....i = RijRij .... Rij Tjj....j va = Rabeb x' = Rx . (D.3.8) In multiindex notation we now rewrite and compare transformation equations (D.3.7), (D.3.2) and (D.3.8). We have reordered the terms a bit, and all three underlying transformations are x' = F(x) = Rx . The first two equations are transformations for tensor functions, the third is for a tensor T: T '(vI) = RIJ T (vJ) = T(RIJvJ) = T(v'I) va' = Rabvb (D.3.7) T '(eI) = RIJ T (eJ) = T(RIJeJ) = T(vI) va = Rabeb (D.3.2) T'I = RIJ TJ va = Rabeb (D.3.8) (D.3.9) Comments: 1. A rank-k tensor field of multiple arguments transforms as T'I(x',y', ...) = RIJ TJ(x,y,...) x' = Fx, y' = Fy, ... see (2.1.15) (d) (D.3.10) 2. The components of vectors ei, vi and v'i have not been mentioned (so far) in this appendix. D.4 When is a scalar not a scalar? For fun, we present this as a sort of mystery, but the mystery is soon cleared up. We start with a k = 1 example. Recall equation (2.11.10) concerning a rank-1 functional α, α(v) = Σiαiλi(v) = Σiαivi = α v = Σiαivi . (2.11.10) (D.4.1) If we move from x-space to x'-space as described in Fig. (2.1.1), we expect to find that α(v) = α v = α' v' = α'(v') (D.4.2) so we would describe α(v) as a "scalar field" (rank-0), similar to (2.1.15) (e). On the other hand, if we set T = α in the discussion of Section D.3 just above, we find that α(v'i) = Rij α(vj) (D.4.3) which says that α(v) transforms as a tensor function of rank-1 (vector). So is α a scalar or is it a vector? The same issue of course arises at the k = 2 level. Recall now this equation, T(v1,v2) = Σab Tab (v1)a(v2)b . (2.11.18) (D.4.4) We claim that T(v1,v2) is scalar field of two arguments, as in the example (2.1.15) item (f). This is so because it is a contraction of a rank-2 tensor Tab against two vectors with upper a and b indices. For the doubtful, here is a direct derivation of this fact: (v1')a = Rar(v1)r (v2')b = Rbs(v2)s T'ab = RadRbeTde (D.4.5) T'(v'1,v'2) = Σab T'ab (v1')a(v2')b = [RadRbeTde] [ Rar(v1)r ][ Rbs(v2)s] // insert the above, all implied sums = RadRbeRarRbs Tde(v1)r(v2)s = (RadRar)(RbeRbs) Tde(v1)r(v2)s = δdrδesTde(v1)r(v2)s // (2.1.8) item 1, twice = Tde(v1)d(v2)e = T(v1,v2) . (D.4.6) On the other hand, from the discussion of Section D.3 we find that T(v'i,v'i) = RijRijT(vj,vj) or T(v'a,v'b) = RaiRbjT(vi,vj) . (D.4.7) This says that T(vi,vj) transforms as a rank-2 tensor function. So is this T a scalar, or is it a rank-2 tensor? Doubtless the reader has spotted the explanation for these paradoxes. Recall the strong-sense of the word "tensor" as discussed below (4.1.8): a true tensor is a tensor with respect to some underlying transformation. In our k = 1 example above, in (D.4.1) we use this transformation, (v'i)a = Rab (vi)b . (D.4.8) With respect to this component transformation (and a similar one for αi), α(v) does indeed transform as a scalar field. But in (D.4.3) the transformation is v'i = Rij vj (D.4.9) which is a completely different vector transformation which does not involve components (recall that the j on vj is a label, not a component). With respect to this transformation, α(v) does indeed transform as a vector function. For k = 2 one has the same situation. The component transformations are shown in (D.4.5) which lead to the function T(v1,v2) transforming as a scalar field. But it is the vector transformation v'i = Rij vj which is used to get (D.4.7), and with respect to this different transformation, T(v1,v2) transforms as a rank-2 tensor function. **************