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delete this file after a few days. This is the Title PhL 3.26.05 Note that page numbering is turned on in this template. Question: When we say 7. The w matrix. Suppose the bm are a complete set of basis vectors in x-space. It must then be possible to express the dual vectors with some w'nm coefficients as Bn = wnmbm. [ In standard notation. bn = wnm bm ]. Why is it that wnm must be symmetric. Is there not some direct manner? In other words, what is the solution for those dual vectors Bn bm = δn,m ? This then says wnkbk bm = δn,m Let's call bk bm = Wkm. It would normally be 'km. Clearly Wkm is symmetric. Then we have wnk Wkm = δn,m We know the Wkm because we know g and we know the bk . How do we invert? This says wW = 1 => w = W-1 Is the inverse of a symmetric matrix symmetric? (A-1)ab = [cof(AT)]ab /det(A) = cof(ATab) /det(A) = cof(Aba) /det(A) (A-1)Tab = (A-1)ba = cof(Aab) /det(A) = cof(ATab) /det(A) = (A-1)ab Rewrite start of Notes on Duality. 1. Suppose the vectors bn form a complete basis for x-space. Can one find a set of vectors Bn that has the property Bm bn = δm,n? As shown below, the answer is normally "yes", and the vectors Bn are uniquely determined by the bn. One says that the set {Bn} is "dual to" the set {bn} and vice versa. If we regard bn as a basis, then Bn is the "dual basis", and Bm bn = δm,n is the "duality relation". Another terminology is that the vectors Bn are "reciprocal to" the vectors bn . 2. Above it was shown that En em = δn,m so the En and the en vectors are dual to each other. The en are the tangent base vectors, and the En are "reciprocal to" the en which is why we call them the reciprocal base vectors. 3. Solving for the Bn in terms of the bn. Each Bn has N components, so there are N2 unknowns. The duality relation Bm bn = δm,n is a set of N2 equations. This is basically a Cramer's Rule problem in N2 variables. Since the bn form a complete basis, one can expand Bm = wmnbn . Then δm,k = Bm bk = wmn bn bk // bn bk = ij(bn)i(bk)j Define Wnk ≡ bn bk and note that Wnk is symmetric. Then δm,k = wmnWnk or wW = 1. Assuming for the moment that detW ≠ 0, the solution is given by w = W-1 . The Section k (e) "Digression" showed that (A-1)T = (AT)-1 for invertible A, so wT = (W-1)T = (WT)-1 = W-1 = w and therefore w is symmetric. Since w is known, the Bm = wmnbn have been found. 4. In the case that bn = en and Bn = En, one finds that Wnk = en ek = 'nk , the covariant metric tensor. Then w = W-1 must be the contravariant metric tensor wmn = g'mn. Then Bm = wmnbn says En = g'mn en which agrees with our original definition of the En. Moreover, det(W) = det( 'ab) = g' of Section 5 (k), so as long as g' ≠0, one has detW ≠ 0. 5. For the general bn case, if one imagines that the bn are the tangent base vectors for some transformation Fb with some Rb , then the issue of detW ≠ 0 boils down to g' ≠ 0 for that transformation. In the case that x-space is Cartesian, we know that J2 = det(')/ det() = g'/g = g', and the assumed invertibility of transformation F means that J ≠ 0 everywhere, so g' ≠ 0 and then detW ≠ 0. In all cases considered in this document, detW ≠ 0. 6. If the bm are true tensorial vectors, then the Bm = wmnbn will be as well and then Bn bm is a tensorial scalar. Therefore if Bn bm = δn,m in x-space, then so also B'n b'm = δn,m in x'-space, where bm' = Rbm and B'n = RBn. For example, E'n e'm = δn,m in x'-space where em' = Rem and E'n = REn . 7. In section (e) we shall encounter another dual pair Un um = U'n u'm = δn,m which is associated with the inverse transformation x = F-1(x'). 8. One major significance of the equation Bn bm = δn,m is that it allows the following expansions: V = Σn kn Bn where km = V bm V = Σn cn bn where cm = V Bm so that for example bm V = bm [Σn kn Bn] = Σn kn bm Bn = Σn kn δm,n = km. These expansions are explored in section (f) below for the two dual sets En, en and Un, un.