Section 7.12 rough draft v3

Rough Draft for Section 7.10 (brand new) PhL 11.20.15 7.10 Mixed tensor/wedge products of tensors in T(V) and L(V) Recall the expansions of rank-k tensor in T in Vk and T^ in Lk , T = Σii....i Tii....i (ei ei ..... ei) T ϵ Vk (5.6.1) T^ = Σii...i Tii...i (ei ^ ei ^ .... ^ ei) T^ ϵ Lk Vk (7.4.3) T^ = Alt(T) (7.10.1) Fact: In general T^ ≠ T but there are three exceptions: (7.10.2) 1. If T is a scalar k ϵ K (rank 0 tensor), then by definition we set T^ = T = k so then T^ = T and L0 = V0. 2. If T is a vector (rank 1 tensor), then the above equations read, T = Σi Ti (ei) T ϵ V1 T^ = Σi Ti (ei) T^ ϵ L1 V1 In this case T^ = T and of course L1 = V1 . 3. If T happens to be a totally antisymmetric tensor, one knows from (7.5.6) that Alt(T) = T. In this case T^ = Alt(T) = T. Double Products Consider these four tensor products all of which are elements of Vk+k', TS = ΣI(TS)IeI = ΣITISI'IeI T^S = ΣI(T^S)IeI = ΣIT^ISI'IeI TS^ = ΣI(TS^)IeI = ΣITIS^I'IeI T^S^ = ΣI(T^S^)IeI = ΣIT^IS^I'IeI . Since T^ = Alt(T), in general the coefficients T^I will differ from the coefficients TI, and the same for S, so in general the four tensors shown above are different. So Fact: T^S^, TS^, T^S, TS all exist, but in general are different (7.10.9) The exceptions to "are different" are those listed in (7.10.2). For example if T is a scalar or vector or is totally antisymmetric, then T^S = TS and T^S^ = TS^ . For the wedge product, we know from (7.10.4d) that T^^ S^ = ΣI (TS)I e^I Since the wedge product is defined only on tensors in L(V) T(V), certain objects are in general undefined, Fact: T^^ S^ always defined T ^ S^ , T^^ S, T ^ S in general are all undefined (7.10.11) The exceptions to "are all undefined" are cases where S^ = S and/or T^ = T, as outlined in (7.10.2). Triple Products Rule 1: Suppose in (7.10.9) and (7.10.11) we make the replacements T = (AB) and S = C. Then, (AB)^C^, (AB)C^, (AB)^C, (AB)C all exist, but in general are different (AB)^^C^ always defined (AB)^C^ , (AB)^^C, (AB)^C in general are all undefined (7.10.12) Exceptions: The above general classification must be modified in these special cases: (7.10.13) 1. If A,B = κ,κ' (scalars), then AB is scalar κκ' by (5.6.15) so by (7.10.7) item 1 (AB)^ = (AB) . 2. If A = κ (scalar) and B=vector, AB is vector κB by (5.6.15) so by (7.10.7) item 2 (AB)^ = (AB). The conclusion also applies to B = κ (scalar) and A=vector. 3. C is a scalar or a vector or is totally antisymmetric, since then C^ = C. Example: Suppose A,B,C are vectors a,b,c, so special case 3 above applies. Then (ab)^ = Alt(ab) = (ab- ba)/2 = a^b (7.10.14) and the classification (7.10.12) becomes, (a^b)c, (ab)c exist but are different: (a^b)c = (abc - bac)/2, (ab)c= abc (a^b)^c = a^b^c defined (ab)^c undefined since (ab) is not in L(V) (7.10.15) Rule 2: Suppose in (7.10.9) and (7.10.11) we instead make the replacements T = (A^B) and S = C. Then, (A^B)^C^, (A^B)C^, (A^B)^C, (A^B)C all exist, but in general are different (A^B)^^C^ always defined (A^B)^C^ , (A^B)^^C, (A^B)^C in general are all undefined (7.10.16) Exceptions: The above general classification must be modified in these special cases: (7.10.17) 1. If A,B = κ,κ' (scalars), A^B is scalar κκ' by (7.10.16) so by (7.10.7) item 1 (A^B)^ = (A^B) . 2. If A = κ (scalar) and B=vector, A^B is vector κB by (7.10.16) so by (7.10.7) item 2 (A^B)^ = (A^B). The conclusion also applies to B = κ (scalar) and A=vector. 3. C is a scalar or a vector or is totally antisymmetric, since then C^ = C. Example: Let A,B,C be vectors a,b,c, so special case 3 above applies. We then have (a^b)^ = (a^b) so (a^b) c exists and equals (1/2)[ab - ba]c = (abc - bac)/2 (a^b)^c = a^b^c defined (7.10.18) Quad Products In (7.10.12) replace C by (CD) to get (AB)^(CD)^, (AB)(CD)^, (AB)^(CD), (AB)(CD) exist but differ (AB)^^(CD)^ always defined (AB)^(CD)^ , (AB)^^(CD), (AB)^(CD) in general are all undefined (7.10.19) There are special cases as in (7.10.13) 1 and 2 for which (AB)^ = (AB) and/or (CD)^ = (CD). Example: Suppose A,B,C,D = a,b,c,d are all vectors. Then (ab)^ = (a^b) and (cd)^ = (c^d) and the classification (7.10.15) becomes, (a^b)(c^d), (ab)(c^d), (a^b)(cd), (ab)(cd) all exist, but are different (a^b)^(c^d) defined (ab)^(c^d) , (a^b)^(cd), (ab)^(cd) all undefined (7.10.20) General Products Here are some sample cases, where we ignore special cases. There are no special cases for example if rank ki ≥ 1 for all tensors. (ei ei ... ei) (ej ej ... ej) exists, is associative by (2.8.22) (ei^ ei ... ^ ei) ^ (ej^ ej ... ^ ej) exists, ^ is associative by (7.9.4) (ei ei ... ei) (ej^ ej ... ^ ej) exists, expand right factor with (7.1.2) (ei ei ... ei) ^ (ej^ ej ... ^ ej) does not exist, left factor not in L(V) ( T ) ^ ( S^ ) does not exist, as already shown in (7.10.11) [T1 T2 T3 T4] [ (T5)^^ (T6)^)^ (T7)^] exists, can expand right factor in products [T1 T2 T3 T4] ^ [ (T5)^^ (T6)^)^ (T7)^] does not exist, left factor not in L(V) T1 T2 T3 ... TN exists, see (5.6.12) T1 (T2)^ T3 ... TN exists, use expansion (7.10.6) for (T2)^ T1 (T2T3) ... TN exists, is associative by (2.8.22) T1 (T2^T3) ... TN exists, (T2^T3) can be expanded in all basis T1 (T2^T3)^ ... TN exists, (T2^T3)^ can be expanded in all basis (T1)^^ (T2)^^ (T3)^^...^ (TN)^ exists, see (7.11.11) (T1)^^ T2 ^ (T3)^^...^ (TN)^ exists only if T2 is totally antisymmetric so T2 = (T2)^ (T1)^^ (T2T3) ^...^ (TN)^ does not exist, (T2T3) not in L(V) (T1)^^ (T2T3)^ ^...^ (TN)^ exists (7.10.21) Theorem: (Spivak) Alt( Alt(A^ B^) C^) = Alt(A^ B^ C^) or [ (A^ B^)^ C^]^ = (A^ B^ C^)^ I think I need new theorems from Appendix C which right now I have only in dual notation.