Section 7.12 rough draft

Rough Draft for Section 7.12 (brand new) PhL 1.14.15 7.12 Mixed tensor/wedge products of tensors in T(V) and L(V) We now must distinguish between a tensor T in the tensor product world (Chapter 5) and a tensor T^ in the wedge product world (Chapter 7). We therefore write (ir= 1,2..n), T = Σii....i Tii....i (ei ei ..... ei) T ϵ Vk (5.6.1) (7.12.1) T^ = Σii...i T^ii...i (ei ^ ei ^ .... ^ ei) T^ ϵ Lk Vk (7.4.3) (7.12.2)  Suppose the two sets of coefficients are the same, so Tii....i = T^ii...i or T = T^ . (7.12.3) Then from (7.5.8) we have T^ = Alt(T^) = Alt(T) T^ ϵ Lk Vk , T ϵ Vk assumes T = T^ . (7.12.4) which says T^ is just an antisymmetrized version of T. Consider now our result (7.10.6) for the wedge product of two tensors T^ ϵ Lk and S^ ϵ Lk', (T^^S^) = Σii....i (T^S^)ii...i(ei^ ei ......^ ei) = Σii....i Tii...i Sii...i(ei^ ei ......^ ei) (7.12.5) Since the result lies in Lk+k' we can write (T^^S^) = (T^^S^)^ where the extra ^ reminds us that this product lies in the tensor algebra L(V). Since Lk+k' Vk+k', the notation (T^^S^) also makes sense, where now the extra reminds us that (T^^S^) lies in T(V). So (T^^S^)^ = (T^^S^) = (T^^S^). (7.12.6) We wish now to explore the various other possibilities for subscripts in the expression (Ta^Sb)c where a,b,c ϵ {,^}. Is there any meaning to (T^S^)? To find out, we insert the expansions *** to get T^S^ = { Σii....i Tii....i (ei ei ..... ei)} ^ {Σjj...j S^jj...j (ej ^ ej ^ .... ^ ej)} = Σii....iΣjj...jTii....i S^jj...j * (ei ei ..... ei) ^ (ej ^ ej ^ .... ^ ej) . (7.12.7) We have a problem. The last line shows a wedge product of an element of Vk with an element of Lk'. In the special case that k = k' = 1 this is a wedge product of two objects in V1 = L1 which is well defined, T^S^ = ( TS^ - S^T)/2 k = k' = 1 . (7.12.8) In all higher cases, nothing presented in this document gives any meaning to the wedge product shown in (7.12.7). For example, these objects are undefined, where a,b,c,d are vectors, a^(bc) (ab)^c (ab)^(cd) (7.12.9) [ Note that a(b^c) and a(b^c)d are defined, but are not of the form T^S^ .] We conclude that T^S^ has no meaning unless k = k' = 1. The same argument shows that T^^S and T^S have no meaning. We have then proven: Fact 1: Consider the eight objects (Ta^Sb)c with a,b,c ϵ {,^} where k,k' are the rank of T,S: (7.12.10) (1) If k = k' = 1, then all 8 objects are the same and one can write any as (T^S) = (TS-ST)/2 . (2) Otherwise, (Ta^Sb)c is defined only if a = b = ^ (2 of 8 objects are defined), In this case one has (T^^S^)^ = (T^^S^) = (T^^S^) = Σii....i T^ii...i S^ii...i(ei^ ei ......^ ei) . (7.10.4d) (3) These 6 objects are therefore undefined: (T^S^)^ , (T^^S)^ , (T^S)^ , (T^S^) , (T^^S) , (T^S) Now what about (TaSb)c? Since T,T^,S,S^ all lie in Vk, (TaSb) lies in Vk+k' for any a and b. Since (TaSb)^ does not lie in Lk+k', these objects have no meaning. Even (T^S^)^ has no meaning because, even though T^ and S^ are totally antisymmetric, the result (T^S^) is not totally antisymmetric and therefore does not lie in Lk+k'. We have then proven: Fact 2: Consider the eight objects (TaSb)c with a,b,c ϵ {,^} where k,k' are the rank of T,S: (7.12.11) The four objects with c = are meaningful, while those with c = ^ are meaningless, (TS) , (T^S), (TS^), (T^S^) , are all defined (TS)^ , (T^S)^ , (TS^)^, (T^S^)^ are all meaningless We can extend this analysis to the general case: Fact 3: Consider the 2N+1 objects [(T1)a^(T2)a^ .....^(TN)a]b with ai,b ϵ {,^} and rank(Ti) = ki. (7.12.12) (1) If all ki= 1, then all 2N+1 objects are the same and the wedge-of-vectors expansion (7.1.2) applies. (2) Otherwise, [(T1)a^(T2)a^ .....^(TN)a]b is defined only if all ai= ^ (2 of 2N+1 objects are defined): [(T1)^^(T2)^^ .....^(TN)^]^ = [(T1)^^(T2)^^ .....^(TN)^] = [(T1)^^(T2)^^ .....^(TN)^] and the resulting wedge product is given by (7.11.11). If one or more of the ai = , the object is undefined. Here are a few undefined objects [(T1)^(T2)^^ .....^(TN)^]^ , [(T1)^(T2)^^ .....^(TN)^] , [(T1)^(T2)^ .....^(TN)^] Fact 4: Consider the 2N+1 objects [(T1)a(T2)a .....(TN)a]b with ai,b ϵ {,^} and rank(Ti) = ki. (7.12.13) The 2N objects with b = are meaningful, while those with b = ^ are meaningless. Examples: [(T1)^(T2) .....(TN)^] well defined [(T1)^(T2) .....(TN)^]^ meaningless What about something like [(T1)a^(T2)a^(T3)a(T4)a^(T5)a]b ? (7.12.14) First of all, we have no rule that a mixture of and ^ products is associative, we only know that each separately is associative, as in (2.8.22) for and (7.9.7) for ^ . So consider then [(T1)a^(T1)a^[(T1)a(T1)a]^(T1)a]b . (7.12.15) According to Fact 2 (7.12.11) only object [(T1)a(T1)a] = [(T1)a(T1)a] is well defined and is some tensor in the space Vk+k, call it R. According to Fact 3 (7.12.12) the resulting [(T1)a^(T1)a^ R ^(T1)a]b is undefined, so we conclude that (7.12.15) is undefined. Consider next, [ [(T1)a^(T2)a^(T3)a] [(T4)a^(T5)a] ]b ** (7.12.16) Here [(T4)a^(T5)a] is some B in Vk+k provided a4 = a5 = ^, otherwise it is undefined. The other factor [(T1)a^(T2)a^(T3)a] is some A in Vk+k+k provided a1 = a2 = a3 = ^, otherwise it is undefined. So assume all five ai = ^. We then have [A B ]b and this is defined if b = . So of all the 26 versions of (7.12.16), only this one is defined, [ [(T1)^^(T2)^^(T3)^] [(T4)^^(T5)^] ] = [ A B ] We have no general rule for determining the validity of a general tensor product, but we have provided tools that allow a case-by-case analysis. Note that if all operators are or all operators are ^, there will be some defined versions of the multiple tensor product. ************************************************************ What can I say about Alt(TS) ? The Alt formula says: Let R = (TS) = exists according to Fact 2. Then for ANY tensor X, we know that, [alt(X)]ii...i ≡ ΣP (-1)S(P) XP(i)P(i)...P(i) So I guess that [alt(Rθ)]ii...i = ΣP (-1)S(P) RP(i)P(i)...P(i) = R^ii...i = (TS)^ii...i = meaningless OK, start over, try R = (T^S^) = exists according to Fact 2 Then [alt(Rθ)]ii...i = ΣP (-1)S(P) (T^S^)P(i)P(i)...P(i) = R^ = (T^S^)^ But I claim this is meaningless as I have defined it above. I could define this differently so that the extra ^ means that you start with (T^S^) and then you antisymmetrize THAT to get [(T^S^)]^ . I could try this convention: subscript is the same as no subscript. Then I would say with the new meaning, Alt(T^S^) = (T^S^)^ ϵ Lk+k' At least it has meaning. Then maybe Spivak is saying this: Alt[Alt(T^S^)R^] = Alt[T^S^R^] ?? I would then have to show that Alt[(T^S^)^ R^] = Alt[T^S^ R^] ?? I see the intuition on this thing, but maybe it is something I have to do in Appendix C in lots of detail. I can at least say this on the right Alt[(T^S^)^ R^] = Alt[(T^S^) R^] ?? Define A = (T^S^) A^ = (T^S^)^ Then I have to show that Alt[A^ R^] = Alt[A R^] But this is my (C.5.12). Go the other way. Start with (C.5.12) Alt(TaSb) = Alt(TcSd) a,b,c,d ϵ {,^} (C.5.12) Select this form: Alt(T^S^) = Alt(TS^) ** Now set T = (A^ B^) so then T^ = (A^ B^)^ = DOES make sense Then write ** as Alt( (A^ B^)^S^) = Alt( (A^ B^)S^) Now suppress the on the right, and use Alt on the left to get Alt( Alt(A^ B^)S^) = Alt( (A^ B^)S^) Now use associative on the right and we are there: Alt( Alt(A^ B^) S^) = Alt( (A^ B^ S^) because this then matches Well THAT certainly took a while. What about Spivak's equation: or Alt[Alt(TS)R] = Alt[TSR] I think that = [Alt(TS) (TS)^