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Section D.12 for Covariant Tensor I will just copy paste and edit. D.12 How determinants of rank-2 tensors transform In this document we have encountered only a few determinants of tensors (like gij) and tensor-like objects (like Rij and Sij). Nevertheless, we would like to know how the determinant of a rank-2 tensor transforms under x = F(x). To this end, we first rewrite the traditional determinant formula in a covariant form. Once this is done, the conclusions come quickly. We start with this mechanical statement of the determinant of a matrix Mij det(Mij) = εab..x M1aM2b....MNc (D.12.1) where ε is the permutation tensor. This form is "mechanical" just in the sense that if you drew a picture of the matrix Mij and mechanically evaluated det(Mij), you obtain the above result. We can now restate this determinant switching from the permutation tensor εabc...x to the index-all-up Levi-Civita tensor. As stated below (D.4.1), we take εabc..x to be equal to the permutation tensor. We then have det(Mij) = εab..x M1aM2b....MNc (D.12.2) which certainly looks more "covariant" since all summed indices appear to be contracted. But with fixed upper indices 1,2...N "hanging out", it is not quite clear what is going on here. Since a through x are all contracted, and since weight(ε) = - 1, one might be tempted to say det(Mij) is a scalar density of weight -1, but this incorrect. To get a better view of things, it is useful to rewrite the above determinant as follows (proof follows) det(Mij) = (1/N!) εAB...X εab..x MAa MBb ...MXx . (D.12.3) Again, both ε's shown here are in effect permutation tensors. We shall now show that the right sides of the previous two equations are exactly the same. First write the right hand side of (D.12.3) as RHS (D.12.3) = (1/N!) εAB...X { εab..xMAa MBb ...MXx } . (D.12.4) The bracketed quantity can be expanded as (defining the bracket to be QAB..X ) , QAB..X ≡ {εab..xMAa MBb ...MXx} = MA1 MB2 ...MXN + all signed permutations . (D.12.5) Notice that Q12..N ≡ {εab..xM1a M2b ...MNx} = M11 M22 ...MNN + all signed permutations . (D.12.6) This is just a mechanical evaluation of det(Mij) and we conclude that Q12..N = det(Mij) . (D.12.7) The tensor QAB..X is totally antisymmetric as this example demonstrates QBA..X = εabc..xMBaMAbMCc ...MXx = [ - εbac..x] MAbMBaMCc ...MXx = - [ εabc..x] MAaMBbMCc ...MXx // a↔b = - QAB..X . (D.12.8) Then using theorem (D.3.1) we can write QAB..X in this form. QAB..X = K εAB..X . (D.12.9) To evaluate K, use the standard ordering 123..N to find from the above and (D.12.7), Q12..N = K ε12..N = K = det(Mij) . (D.12.10) Then, since K = det(Mij) one finds that QAB..X = det(Mij) εAB..X . (D.12.11) Therefore (D.12.4) can be written RHS (D.12.3) = (1/N!)εAB...X { εab..xMAa MBb ...MXx } = (1/N!)εAB...X {QAB..X } = (1/N!)εAB...X {det(Mij) εAB..X} = det(Mij) (1/N!) ΣAB..X (εAB...X)2 (D.12.12) The sum ΣAB..X (εAB...X)2 is a "sum of ones" and there is a one for each permutation of AB..X. All other terms are zero. There are N! total permutations including the first, so ΣAB..X (εAB...X)2 = N! (D.12.13) which agrees with the last equation of (D.10.37). Therefore, RHS (D.12.3) = det(Mij) (1/N!) ΣAB..X (εAB...X)2 = det(Mij) . (D.12.14) This concludes our too-lengthy proof that the right sides of (D.12.2) and (D.12.3) are identical. Now we start with the proven result, det(Mij) = (1/N!) εab..x εAB...X MAa MBb ...MXx . (D.12.3) But now all indices are contracted, so this is the final Eq (D.5.10) says that εABC... = g εABC... so the above can be written det(Mij) = (1/g) (1/N!) εab..x εAB...X MAa MBb ...MXx (D.12.15) Now, finally, we have a form in which all tensor indices are contracted with no loose ends. We can then use theorem (D.2.3) about the additivity of weights. Recall from (D.4.9) and (D.6.7) that both ε tensors shown in (D.12.15) have weight -1, and that the object 1/g has weight +2 from (D.1.7). Adding, we find that the object det(Mij) has weight 0. We have therefore proven: (scalar = scalar density of weight 0) Theorem: The determinant det(Mij) of a mixed rank-2 tensor Mij transforms as a scalar under the transformation x = F(x). (D.12.16) Comment: Since our Chapter 2 S matrix Sij is not a tensor, J = det(Sij) is not a scalar, and is fact not a tensor of any kind since it bridges x-space and x'-space. It is now straightforward to determine the transformation nature of the other three rank-2 tensor types, and in fact to find simple relations between the four determinants. In all these cases, we use of the up-down altering property of the g tensor of (7.4.11), the "up-tilt" or "down-tilt" version of matrix multiplication of (7.8.9), the matrix rule det(AB)=det(A)det(B), facts (7.5.20) and (7.5.21), and the weight of g and 1/g as determined in (D.1.6) and (D.1.7) : det(Mij) = det(giaMaj) = det(gij) det(Mij) = g det(Mij). -2 -2 + 0 det(Mij) = det(Miagaj) = det(Mij)det(gij) = { g det(Mij)}{g-1} = det(Mij) 0 0 det(Mij) = det(giaMaj) = det(gij) det(Mij) = g-1 det(Mij) = g-1 det(Mij) (D.12.17) +2 +2 + 0 The weight additions are shown under each line of equations. The conclusions are these: det(Mij) = det(Mij) // scalar densities of weight 0 ( = scalar) det(Mij) = g det(Mij) // scalar density of weight -2 g = det(gij) det(Mij) = g-1 det(Mij) // scalar density of weight +2 g-1 = det(gij) (D.12.18) Here we have related each of the three partner determinants to the down-tilt determinant, and have shown the weight of each type of determinant. Example: We know that gij is a rank-2 tensor from Section 5.7. Thus, we expect to find from (D.12.18) that, det(gij) = det(gij) => det(δij) = det(δij) or 1 = 1 ok, weight 0 det(gij) = g det(gij) => g = g det(δij) = g*1 = g ok, weight -2 det(gij) = g-1 det(gij) => g-1 = g-1det(δij) = g-1* 1 = g-1 ok, weight 2 (D.12.19) Equations (D.12.18) are valid only for rank-2 tensors. They are not valid for R and S. Comment: Had we used the "avoided" transpose notation shown below (7.9.2), we could claim that the first line of (D.12.18) is a statement of the fact that det(A) = det(AT).