multiplying k tensors

Multiplying k-tensors. I have having plenty of trouble getting this nailed down. Suppose you are given these two ordered expansions: T = Σi<i<....<i k! Aii....i (ei^ ei^ .....^ ei) . rank k, T ϵ Lk S = Σj<j<....<j k'! Bjj....j (ej^ ej .....^ ej) . rank k', S ϵ Lk' That is to say, you KNOW these coefficients A and B. You also know that symmetric expansions exist which have this form, 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' (4.6.47) but assume that you don't know the T and S coefficients, you only know the A and B ones. You also know that A = Alt(T) B = Alt(S) Is there a way to obtain T and S from A and B? Go back to just the case of T and A: T = Σii...i Tii...i (ei ^ ei ^ .... ^ ei) Aii...i = (1/k!) ΣP (-1)S(P) TP(i)P(i)...P(i) = [Alt(T)]ii...i OK, lets try a few things. Suppose i1 = Qj1 and so on. Then one would write AQ(j)Q(j)...Q(j) = (1/k!) ΣP (-1)S(P) TPQ(j)PQ(j)...PQ(j) = (1/k!) ΣP (-1)S(PQ)(-1)S(Q) TPQ(j)PQ(j)...PQ(j) (-1)S(Q)AQ(j)Q(j)...Q(j) = (1/k!) ΣP (-1)S(PQ)TPQ(j)PQ(j)...PQ(j) Pause. There are many different T's that would give the same A ! Any T whose antisymmetric part is A will do the trick. So why not then just take T = A ! OK, I will do that. We then have these four equations A = Σi<i<....<i k! Aii....i (ei^ ei^ .....^ ei) . rank k, T ϵ Lk B = Σj<j<....<j k'! Bjj....j (ej^ ej .....^ ej) . rank k', S ϵ Lk' and A = Σii....i Aii....i (ei^ ei .....^ ei) . rank k, T ϵ Lk B = Σjj....j Bjj....j (ej^ ej .....^ ej) . rank k', S ϵ Lk' Now "work with" just the symmetric equations. We know that AB = Σii....i[Aii....i Bii....i] (ei ei ...... ei) (4.5.23) = Σii....i[AB]ii....i (ei ei ...... ei) where we have taken T = A and S = B. So the above equation really just defines the coefficient. Now wedge the above symmetric expansion expressions to get A ^ B = Σii....i[Aii....i Bii....i] (ei^ ei ......^ ei) = Σii....i[AB]ii....i (ei^ ei ......^ ei) We now write this in ordered form as A ^ B = Σi<i<....<i (k+k')! (AB)Aii....i (ei^ ei ......^ ei) where (AB)A = Alt(AB) The final result is then A ^ B = Σi<i<....<i (k+k')! [ Alt(AB)]ii....i (ei^ ei ......^ ei) Even though A and B are already antisymmetric, we still need Alt(AB) to antisym on the cross indices. Review: We are handed these two expansions A = Σi<i<....<i k! Aii....i (ei^ ei^ .....^ ei) . rank k, T ϵ Lk B = Σj<j<....<j k'! Bjj....j (ej^ ej .....^ ej) . rank k', S ϵ Lk' where A and B are each totally antisymmetric. We then find that A ^ B = Σi<i<....<i (k+k')! [Alt(AB)]ii....i (ei^ ei ......^ ei) where [Alt(AB)]ii....i = ΣP (-1)S(P) (AB)P(i)P(i)...P(i) = ΣP (-1)S(P) AP(i)P(i)...P(i) BP(i)P(i)...P(i) At least this is an explicit expression for the coefficient in A ^ B. Perhaps some examples are now in order to test this general result. k = 1 and k' = 1 : A = Σi Ai (ei) B = Σj Bj (ej) [Alt(AB)]ii = ΣP (-1)S(P) AP(i)BP(i) = (1/2) [ AiBi - BiAi ] Then we find A ^ B = Σi<i (2)!(1/2) [ AiBi - BiAi ] (ei^ ei) = Σi<i [ AiBi - BiAi ] (ei^ ei) This agrees with (4.3.12). k = 2 and k' = 1 : We assume A is antisym A = Σi<i 2! Aii (ei^ ei) B = Σj Bj (ej) [Alt(AB)]iii = ΣP (-1)S(P) AP(i)P(i)BP(i) = (1/6) [ A12B3 - A13B2 + A31B2 - A32B1 + A23B1 -A21B3 ] = (1/6) [ A12B3 - A13B2 - A13B2 + A23B1 + A23B1 +A12B3 ] = (1/6) [ 2A12B3- 2A13B2 + 2A23B1 ] = (1/3) [ A12B3- A13B2 + A23B1 ] Then our result is A ^ B = Σi<i<i 3! [Alt(AB)]iii (ei^ ei^ ei ) = Σi<i<i (3!) (1/3) [ A12B3- A13B2 + A23B1 ] (ei^ ei^ ei ) = Σi<i<i 2 [ A12B3 - A13B2 + A23B1 ] (ei^ ei^ ei ) Question: Suppose T and S are NOT totally antisymmetric and have extra pieces., T = A + X S = B + Y Then we get T = Σii....i [Aii....i + Xii....i ] (ei^ ei .....^ ei) S = Σjj....j [Bjj....j + Yjj....j] (ej^ ej .....^ ej) ********** T = Σii...i Xii...i (ei ^ ei ^ .... ^ ei) Is this zero? Try example T = Σabc Xabc(ea ^ eb ^ ec) Xabc = (2/3)Tabc - (Tcab + Tbca )/3 Write it out. T = Σabc [(2/3)Tabc - (Tcab + Tbca )/3](ea ^ eb ^ ec) = (2/3)Σabc Tabc(ea ^ eb ^ ec) - (1/3) Σabc Tcab(ea ^ eb ^ ec) - (1/3) Σabc Tbca(ea ^ eb ^ ec) = (2/3)Σabc Tabc(ea ^ eb ^ ec) - (1/3) Σabc Tabc(ec ^ ea ^ eb) - (1/3) Σabc Tabc(eb ^ ec ^ ea) = (2/3)Σabc Tabc(ea ^ eb ^ ec) - (1/3) Σabc Tabc(ea ^ eb ^ ec) - (1/3) Σabc Tabc(ea ^ eb ^ ec) = 0 So I want some simple way to show that only Alt(T) of T Howcum I have no "projection formulas". For example, I should be saying T = Σii....i Tii....i (ei ei ..... ei) . (5.2.1) Tii....i = T (ei ei ..... ei) I had this theorem, (ab) (cd) = (ac)(bd) Is there some wedge equivalent thing? (ab) (cd) ≡ Σij(ab)ij(cd)ij (ab), (cd) ϵ VV (2.9.12) = Σijaibjcidj . (a^b) (c^d) ≡ Σij(a^b)ij(c^d)ij (a^b), (c^d) ϵ V^V (2.9.12) = (1/4)Σij[aibj - ajbi] [cidj - cjdi] = (1/4)[(ac)(bd) - (bc)(ad) - (ad)(bc) +(ac) (bd) ] = (1/2) [ (ac)(bd) - (bc)(ad)] So this would be my new wedge version of (2.9.13): Theorem: (a^b) (c^d) = [(ac)(bd) - (bc)(ad)]/2 a,b,c,d ϵ V (2.9.13) Now try to apply this M = Σab [M(e,e')]ab (ea^e'b ) (ei^e'j) M = Σab [M(e,e')]ab (ei^e'j) (ea^e'b) = Σab [M(e,e')]ab [(ei ea)(e'j e'b) - (ei e'b)(e'j ea)]/2 = Σab [M(e,e')]ab [δia δjb - δib δja ]/2 = { [M(e,e')]ij - [M(e,e')]ji} / 2 (2.10.5) ****************** Σii...i Xii...i (ei ^ ei ^ .... ^ ei) = Σii...i { Tii...i - [Alt(T)] ii...i - [Sym(T)] ii...i} (ei ^ ei ^ .... ^ ei) = Σii...i { Tii...i - [Alt(T)] ii...i} (ei ^ ei ^ .... ^ ei) = Σii...i { Tii...i - (1/k!) ΣP (-1)S(P) TP(i)P(i)...P(i)} (ei ^ ei ^ .... ^ ei) Write the second term as - Σii...i (1/k!)ΣP (-1)S(P) TP(i)P(i)...P(i)(ei ^ ei ^ .... ^ ei) ******************8