covariant transpose update of tensor doc

Rewrite of tensor doc sections using the covariant transpose. 3.16.16. This task is complete as of 8 AM 3.17.16, but I will keep this doc just for the record since it shows most of the edits I did. Commentary in red, installed sections in blue. I am right now searching for all mentions of the transpose of a matrix in tensor doc. The first hit is section 1 below 1 ********************************************************************************* The following was installed as an overwrite on 3/16/16 into tensor doc, so do not edit here!!! Transpose Matrices and Rotations. It is possible to define the "covariant transpose" MT (italic T) of a matrix M in Standard Notation, and things work out as follows, where M is any rank-2 tensor, (MT)ab = Mba (RT)ab = Rba (ST)ab = Sba (MT)ab = Mba (RT)ab = Rba (ST)ab = Sba (MT)ab = Mba (RT)ab = Rba (ST)ab = Sba (MT)ab = Mba (RT)ab = Rba (ST)ab = Sba . // covariant transpose (7.9.3) The general rule is that, in each equation, the indices are reflected in a vertical line separating them. The notation is "covariant" in that one can raise and lower indices at will in each equation and thereby create a new valid equation. The drawback of this notation is that, when the indices are tilted, MT in general differs from what we might call the "matrix transpose" MT of a matrix M. When one transposes a matrix, one normally means to swap the rows and columns, and that means to swap the indices. One would then write (MT)ab = Mba (RT)ab = Rba (ST)ab = Sba (MT)ab = Mba (RT)ab = Rba (ST)ab = Sba (MT)ab = Mba (RT)ab = Rba (ST)ab = Sba (MT)ab = Mba (RT)ab = Rba (ST)ab = Sba . // matrix transpose (7.9.4) The middle two lines (tilted indices) are very non-covariant in form since for example (MT)ab = Mba has the up and down sense of index a and b not even matching. The use of the MT matrix transpose is in theorems like det(M) = det(MT) where rows and columns are swapped without changing the determinant. Comments: 1. The issue here is that one might have (MT)ab = Mba ≠ (MT)ab = Mba . If M is an x-space tensor and if the x-space metric tensor is g = 1, then up and down index positions don't matter and our problem goes away, (MT)ab = Mba = (MT)ab = Mba so then MT = MT. (7.9.5) 2. For R and S: As shown in (7.5.9), g' acts on the first index of R while g acts on the second index. Conversely as shown in (7.5.10), g acts on the first index of S while g' acts on the second index. So up and down indices are the same for R and S only if both g = 1 and g' = 1. Only in this special case, which implies a (local) rotation for x' = F(x), does one have RT = RT and ST = ST . This situation arises in Appendix E where we have x" = FM(x) and R and S (there called M and N) are rotations. (7.9.6) 3. As it turns out, regardless of g and g', one has det(MT) = det(MT) = det(M) (7.9.7) for any index positions on the matrices. This is shown in (D.12.20). 4. As shown in the next few paragraphs, in either up-tilt or down-tilt notations one can write, RRT = RTR = 1 SST = STS = 1 RS = SR = 1 RT = R-1 = S ST = S-1 = R . (7.9.8) Thus, both the R matrix and S matrix are "covariant real-orthogonal" for any underlying transformations x' = F(x), even when R and S are not rotation matrices. Since the matrix elements are real, R and S are also "covariant unitary". We shall generally avoid using the covariant transpose notation since it leads to results which can be confusing although correct, but there will be times when we use it. To restate the potential confusion, in developmental notation RRT = 1 (or RT = R-1) implies that R is a "rotation", where we include in this term the possibility of axis reflections. But in Standard Notation, RRT = 1 is valid for any matrix R associated with transformation x' = F(x), rotations and non-rotations alike. To see why this is so, just write out orthogonality rule #4 from (7.6.4) : Rab Rcb = δac  Rab (RT)bc = δac  RRT = 1 (up-tilt) . (7.9.9) To distinguish these situations, we have to use a somewhat unpleasant notation such as, [RRT = 1]DN R is a "rotation" DN = Developmental Notation [RRT = 1]SN,ut for any R associated with F SN,ut = Standard Notation, up-tilt (7.9.10) because the matrices like Rab in (7.9.4) have indices of the "up tilt" configuration. How then does one recognize a rotation matrix directly in Standard Notation? We translate using the rule Rik→ Rik shown in (7.5.2), [RRT= 1]DN RikRjk = δi,j → RikRjk = δi,j . A simpler alternate form for recognizing a rotation can be obtained as follows, RikRjk= δi,j // now apply Sai to both sides SaiRikRjk= Saiδi,j (SR)akRjk = Saj // but SR = 1 δakRjk = Saj = Rja // right side from (7.5.13) Rja = Rja // if R is a rotation Therefore, in Standard Notation a "rotation" can be identified in either of these two ways, RikRjk = δi,j or Rja = Rja // "rotation" . (7.9.11) Instead of trying to use this covariant transpose T superscript in the standard notation, we shall normally get rid of T in the developmental notation and then do our translation to standard notation. For example we start in developmental notation, M' = R M RT  M'ad = RabMbc(RT)cd = RabMbcRdc = RabRdcMbc (7.9.12) Then we make the conversion using (7.5.2) Rik → Rik and (7.4.1) Mab → Mab : M'ad = RabRdcMbc → (M')ad = RabRdcMbc (7.9.13) and this then is the Standard Notation rule for the way a contravariant rank-2 tensor transforms, as was shown in the first line of (7.5.8). Determinant of a Matrix. In developmental notation one writes det(A) = εabc... A1aA2bA3c..... = εabc... Aa1 Ab2Ac3..... (7.9.14) where the Aij are components of the contravariant rank-2 tensor A and where ε is the permutation tensor discussed above in Section 7.7. We have argued above that the notion of a rank-2 tensor being a matrix in Standard Notation is only viable for mixed rank-2 tensor of either the down-tilt or up-tilt variety. Thus, the matrix determinants of interest in Standard Notation would be these: det(Adt) = det(A**) = εabc... A1a A2bA3c..... = εabc... Aa1 Ab2Ac3..... (7.9.15) det(Aut) = det(A**) = εabc... A1a A2bA3c..... = εabc... Aa1 Ab2Ac3..... (7.9.16) Determinants of a rank-2 tensor A rarely appear in this document, but they do appear for the non-tensor objects R and S as shown in (7.5.17), as needed for the Jacobian in (5.12.6). For more information on determinants of rank-2 tensors see Section D.12. end of 1 ****************************************** Continuing my search in tensor doc for transpose stuff. Found nothing in the rest of Chapter 7. Found nothing in the remaining Chapters, so will start scanning appendices. (paging through). nothing in : A, B, C, thru D.11 Section D.12 talks about determinants. 2*************************************************************** The following was installed as an overwrite on 3/17/16 into tensor doc, so do not edit here!!! 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. Comment: The objects below like det(Mij) and det(Mij) of course have no dependence on the "indices" i and j other than on the up/down position of these indices. Elsewhere we write these objects as det(M**) and det(M**) where the "wildcards" just show the nature of the matrix elements (down-tilt or all-down). Here, for technical reasons to be seen below, we maintain the notations det(Mij) and det(Mij). We start with this mechanical statement of the determinant of a matrix Mij det(Mij) = εab..x M1aM2b....MNx (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....MNx (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 that is 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, QAB..X ≡ {εab..xMAa MBb ...MXx} = MA1 MB2 ...MXN + all signed permutations . (D.12.5) meaning all signed permutations of the upper indices. 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) 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 -2 0 +2 0 det(Mij) = det(giaMaj) = det(gij) det(Mij) = g-1 det(Mij) = g-1 det(Mij) . (D.12.17) +2 +2 0 +2 0 The weights 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. Go back now to the first line of (D.12.18), det(Mij) = det(Mij) = det([MT]ji) where MT is the "covariant transpose" of M as shown in (7.9.3). This shows that det(M**) = det([MT]**) or det(Mut) = det(MTut) . // ut = up-tilt Since det(M) = det(MT) is valid for any matrix regardless of index position (MT is the "matrix transpose" of M), we find that det(MTut) = det(MTut) = det(Mut) . A similar argument beginning with det(Mij) = det(Mij) = det([MT]ji) det(Mdt) = det(MTdt) shows that det(MTdt) = det(MTdt) = det(Mdt) . // dt = down-tilt Since MT = MT for both-up or both-down index positions, we arrive at this interesting generalization of the traditional determinant theorem det(M) = det(MT) : Fact: det(M) = det(MT) = det(MT) for all four possible index positions (D.12.20) where T is the matrix transpose and T is the covariant transpose (7.9.3). The scalar density weights of these determinants are given above and do depend on the index positions. end of 2 ************************************************************** Continuing my search in tensor doc for transpose stuff. Found nothing in the rest of App D. nothing in : started into E In E.5 there is discussion of transposes. It is all regular T type matrix transpose stuff, no edits needed. We are OK until a section starting at (E.7.6) which I will edit here 3 **************************************************************** The following was installed as an overwrite on 3/17/16 into tensor doc, so do not edit here!!! Bases are related by a transformation. Consider again, [A(b)]nm = <bn | A | bm > = (bn)T A bm = [bn]i Aij [bm]j = <bn|ui><ui|A|uj><uj|bm> . (E.7.6) We lower index m on both sides (using w'ab as noted above) and reverse the j tilt to get [A(b)]nm = <bn | A | bm > = (bn)T A bm = [bn]i Aij [bm]j = <bn|ui><ui|A|uj><uj|bm> . (E.7.7) One could then define the following tensor-like object, Bni ≡ [bn]i . (E.7.8) The first index on B is raised and lowered by w', while the second is raised and lowered by g, so this object is a bit like R and S in its non-tensor nature. Lowering n and raising i then gives Bni = [bn]i = (BT)in , (E.7.9) where we use the covariant transpose of a tilted matrix described in (7.9.3). One then has [A(b)]nm = Bni Aij(BT)jm . (E.7.10) Since all the matrices are tilted the same way and summed indices are contractions, this is one of the "legal" Standard Notation matrix multiplication forms like (7.8.6) and we then write, A(b) = BABT or more precisely [A(b) = BABT ]SN,dt (E.7.11) where SN,dt means Standard Notation, down-tilt, as described below (7.8.6). The matrix equation A(b) = BABT shows that the [A(b)]nm are related to the Aij by a "congruence transformation" with a matrix Bni = [bn]i whose rows are the basis vectors bn. When bm = um , matrix B is the identity matrix, and when bm = em one has Bni = [en]i = Rni, so that B = R in this case. In Standard Notation the general R matrix is "covariant real orthogonal", [RRT = 1]SN,dt and [RT = R-1]SN,dt (see (7.9.3) and following text) so in fact one has for the bm = em basis, A(e) = B A BT = R A RT = R A R-1 = R A S . (E.7.12) Specifically in this case, [A(e)]nm = RniAijSjm = RniRmjAij = A'nm . (E.7.13) which is the expected result looking at the second line of (7.5.8). Comment: Recall that in Developmental Notation one has RRT = 1 and RT = R-1 only when R is a rotation, whereas in Standard Notation one has RRT = 1 and RT = R-1 for any R matrix, see (7.9.3). We used this fact above to show that a basis change from basis un to bn on operator A can be thought of as a congruence transformation by a matrix B whose rows are the vectors bn. This is a standard concept in linear algebra where one operates in Cartesian x-space. end 3 *********************************************** We are then OK through (E.8.21) where we have to do another clip 4 ************************************* The following was installed as an overwrite on 3/17/16 into tensor doc, so do not edit here!!! One can compare the last line of (E.8.20) to the first line of the generic (7.5.8) showing how a rank-2 tensor transforms. The second line of (7.5.8) then tells us the manner in which the mixed tensor Aab transforms under FM : A" ab = Maa' Mbb' Aa'b' = Maa'Aa'b'Mbb' . (E.8.21) If we use the Standard Notation covariant transpose shown in (7.9.3) [RM = M], then Mbb' = (MT)b'b and the above becomes A" ab = Maa'Aa'b' (MT)b'b = [ MAMT]ab from which we conclude that A" = MAMT // that is to say, [ A" = MAMT]SN,dt (E.8.22) Comment: This matrix result is a special case of the generic result (E.7.11) that A(b) = BABT when bn = n, but it is not obvious how this works out so here are some details. In order to compute Bni ≡ [bn]i in (E.7.8) one has to know bn. One sees from (7.18.6) that bn = w'ni bi where w'nm = bn bm = n m = (h'nh'm)-1 en em = (h'nh'm)-1g'nm  w'nm = (h'nh'm)g'nm . // since wnm and wnm are inverses (E.8.23) Therefore, bn = w'nm bm = (h'nh'm)g'nm m = h'ng'nm em = h'n en  Bni ≡ [bn]i = h'n (en)i (E.8.24) But (E.8.12) says Mni = h'n(en)i so we have shown that Bni = Mni so BABT = MAMT. end 4 ************************************* We then get to (E.9.17) where we may have some problems. OK, I cleaned this up in-line yesterday 3/16/16 and it is done, the idea being to explain why M A MT = M A MT . Are there more covariant transpose issues beyond (E.9.17) in tensor doc? Scan: rest of E OK, F, G, H, I, J, K done There is a transpose in (K.2.5). But it says it is a rotation, so no issue. I will just ignore all the T superscripts here since in the Lai world. This concludes my large-scale edits of Tensor Doc with regard to the covariant transpose. I had to edit all the sections shown above, and it is now all done, cross off the errata list.