save old Appendix A section f

Appendix A (f) PhL 11.19.11 I am yanking this section from Appendix A and will store it here probably never to be used. What I show here is this: If you think of the ε tensor as something that does not transform, it is just the bookkeeping object, THEN if you think about a cross product of contravariant vectors, the resulting object is a vector ONLY if transformation F is a rotation. I later show in Appendix D that if ε is the real Levi-Civita tensor, THEN the cross product is in fact a covariant vector of weight -1 and everything makes sense. So here is the yanked section: ____________________________________________________________________________ (j) Transformation of a cross product and Ek contravariance Recall the cross product of N-1 vectors in N dimensional space, Q = B x C x D x ... x X / N-1 factors, N-2 crosses If B,C,D... are all contravariant vectors ( for example B'i = RijBj ) then one can show the following interesting fact Q'a = Maa' Qa' where Maa' = cof(Raa') which says Qa transforms according to a linear transformation M, but that linear transformation is not R, and therefore Qa does not transform as a contravariant vector. The above rule for Q transformation can be derived from this expression involving the cofactor of a matrix element εabcd...x Rbb' Rcc' ... Rxx' = εa'b'c'..x' cof(Raa')  which in turn is based on this ε expansion of the cofactor of a matrix element cof(Aqp) = εiii...[i→q]...i Ai1 Ai2 ...... Aip.......... AiN = (-1)p-1 εqiii...i...i Ai1 Ai2 ...... Aip.......... AiN This in turn can be verified by multiplying both sides by Aqp and summing on q Σq Aqp cof(Aqp) = Σq Aqp εiii...[i→q]...i Ai1 Ai2 ...... Aip.......... AiN = Σi Aip εiii...[i→i]...i Ai1 Ai2 ...... Aip.......... AiN = εiii.....i Ai1 Ai2 ...... Aip.......... AiN = det(A) In the special case that RRT = 1 (=> det(R) ≡ σ = ±1) one can show that cof(Raa') = σ Raa' so that Q'a = σ Raa Qa' Finally, if R is a true rotation so det(R) = σ = 1, then Qa is in fact a contravariant vector Q'a = Raa Qa' given that Q is the direct product of contravariant vectors. Replacing R by ST, one arrives at exactly the same conclusion stated in terms of covariant vectors. The cross product notation was used to obtain this Picture B expression for Ek, Ek ≡ det(R) (-1)k-1 e1 x e2 x ......x eN // ek missing; N > 2 We have just shown that e1 x e2 x ......x eN in the formula above is not a contravariant vector under a general transformation F, but instead transforms according to [e'1 x e'2 x ......x e'N ]i = Mij[e1 x e2 x ......x eN ]j Mij = cof(Rij) Therefore, the cross product e1 x e2 x ......x eN, though a vector, is not a tensorial vector (see Section 2 (k)). Meanwhile, the quantity det(R) has no tensor properties at all. Like R, it has one foot in x-space and the other in x'-space. So det(R), while being a scalar, is not a tensorial scalar. When these two oddball objects are combined as shown above, the resulting object Ek comes out being a contravariant vector! This fact is much more obvious from the other expression for Ek Ek = g'ki ei since a linear combination of contravariant vectors transforms as a contravariant vector. In the special case that x' = F(x) is an N-dimensional rotation, then e1 x e2 x ......x eN is a contravariant vector, and det(R) = 1 is a tensorial scalar, and then Ek is a contravariant vector just from the cross product representation above.