Rewrite of some Chapter 7 Sections

Rewrite of some Chapter 7 Sections PhL 4.4.15 SECTION (g) (g) Matrix Multiplication in the Standard Notation. Although there are various forms of matrix multiplication, the most standard form is that obtained when all matrices have a "down-tilt" form. Consider this example, Cab = AacBcb (7.9.g.1) where it is assumed that all three objects are down-tilt rank-2 tensors. Although these are "split level" matrices, one can see that the index c has the correct "adjacency" property to justify matrix multiplication. A second requirement is that any matrix summed index should be a genuine contraction with one index up and the other down. The above tensor transformation rule can then be written in this more compact matrix notation ( SN means Standard Notation, dt means down-tilt) : C = AB // all down-tilt [C = AB]SN,dt . (7.9.g.2) By application of suitable g tensors to (7.9.g.1) (or by lowering index a and raising index b on both sides), one gets Cab = AacBcb . Application of the Tilt Reversal Rule on index c then gives Cab = AacBcb . (7.9.g.3) Again the adjacency and contraction rules are met, so this equation can also be represented by C = AB // all up-tilt [C = AB]SN,ut (7.9.g.4) where ut means up-tilt. It has just been shown that, [C = AB]SN,dt [C = AB]SN,ut . (7.9.g.5) The conclusion is that in Standard Notation, matrix notation can be used if all matrices in the equation being represented are either all down-tilt or all up-tilt. As will be shown later in section (u), such matrix equations are all "covariant" in that both sides of the equation have the same tensor transformation property, and this is due to the fact that the matrix summation index is a contraction. One might wonder about the conversion from developmental notation (DN) to standard notation (SN) of the following matrix equation, where A and B are contravariant rank-2 tensors, [ AB = C ]DN AacBcb = Cab → AacBcb = [???]ab . (7.9.g.6) It is not hard to show entirely in developmental notation that, unless STS = 1 (meaning S is a rotation), the object Cab is not a tensor. So for general underlying F, C is not a tensor, even though A and B are tensors. This fact is reflected in the LHS of the partially translated equation, where AacBcb is seen to not be a tensor because the c index is not a contraction (it is not tilted). The rules for translation from developmental to standard notation specify what to do with R, S, and with tensors. Since C is none of these objects, we cannot really "translate" the equation [ AB = C ]DN into standard notation. However, we could define Cab ≡ AacBcb and then write [ AB = C ]DN AacBcb = Cab → AacBcb = Cab [ AB = C ]SN,up (7.9.g.7) with the understanding that Cab is not a contravariant rank-2 tensor, and we now have yet a third kind of matrix multiplication in the Standard Notation, qualified by "up" meaning all indices are up on all objects. One could of course do the same thing with covariant matrices and write [ = ]DN accb = ab → AacBcb = Cab [ AB = C ]SN,dn (7.9.g.8) The situation can be summarized in this picture: (needs repair!! ) (7.9.g.9) In standard notation, there are four different kinds of "matrix multiplication" and only the down-tilt and up-tilt equations are the same equation, and only they are covariant The moral here is that if one chooses to use matrix multiplication notation in the Standard Notation, one needs to be very clear which multiplication form one is dealing with. What about the special case [ AB = 1 ]DN where A is a contravariant tensor? In this case, it must be that B = A-1 and then one has, [ AB = 1 ]DN Aac(A-1)cb = δab → Aac(A-1)cb = (AA-1)ab = (1)ab = δab that is to say: [ A(A-1) = 1 ]SN,up (7.9.g.10) and in this case all four Standard Notation equations are valid, the other three being: [ AB = 1 ]SM,dn Aac(A-1)cb = (AA-1)ab = (1)ab = δab [ AB = 1 ]SM,ut Aac(A-1)cb = (AA-1)ab = (1)ab = δab [ AB = 1 ]SM,dt Aac(A-1)cb = (AA-1)ab = (1)ab = δab (7.9.g.11) so that [ AB = 1 ]DN [ AB = 1 ]SN,up [ AB = 1 ]SN,dn [ AB = 1 ]SN,ut [ AB = 1 ]SN,dt (7.9.g.12) Now what about the special case [ RS = 1 ]DN ? Since R is not a contravariant tensor the above logic does not apply in the sequence stated. But we have rules for translating R and S, so [ RS = 1 ]DN RacScb = δab → RacScb = δab [ RS = 1]SN,dt (7.9.g,13) and since the four SN forms are equivalent (S = R-1), we end up with the same conclusion as above, [ RS = 1 ]DN [ RS = 1 ]SN,up [ RS = 1 ]SN,dn [ RS = 1 ]SN,ut [ RS = 1 ]SN,dt (7.9.g.14) confirming what was stated in a preliminary way in item 2 above. Finally, what about the special case [ STS = 1 ]DN ? Again using the rule that Sij → Sij, [STS = 1 ]DN ScaScb = δab → ScaScb = δab ??? (7.9.g.15) The combination ScaScb does not have the proper adjacency to be matrix multiplication. In order to achieve adjacency, we must have some notion of "transpose" in the Standard Notation. That notion is the subject of the next section, and the result is going to be this: (AT)ab = Aba (AT)ab = Aba (AT)ab = Aba (AT)ab = Aba . (7.9.g.16) Therefore Sca = (ST)ac and we then have [STS = 1 ]DN ScaScb = δab → (ST)ac Scb = δab ??? (7.9.g.17) Although adjacency of the c index is now obtained, the form matches none of the Standard Notation forms for matrix multiplication defined above. Thus, no contact is being made with any of these standard forms by this translation. Since STS = 1, one must have S = (ST)-1 and therefore all four of our stated SN forms are equivalent. So we then have this interesting situation: [STS = 1 ]DN (ST)ac Scb = δab [STS = 1 ]SN,dt the other three SN forms (7.9.g.18) What this says is that [STS = 1 ]DN and [STS = 1 ]SN,dt are completely different equations. Specifically, [STS = 1 ]DN ScaScb = δab → (ST)ac Scb = δab or Sca Scb = δab (7.9.g.19) [STS = 1 ]SN,dt (ST)acScb = δab or Sca Scb = δab . (7.9.g.20) The very last equation on the right can be written (as will be seen below) Rac Rbc = δab and this is one of the standard orthogonality conditions on R and this is always true for any underlying transformation F. Therefore, [STS = 1 ]SN,dt is always true, and this means that in terms of Standard Notation down-tilt matrix multiplication, the matrix S is always real orthogonal. In contrast, in terms of Developmental Notation matrix multiplication, S is only real orthogonal if it is a rotation matrix! That is to say, in the general case: [ST = S-1 ]DN [ST = S-1 ]SN,dt (7.9.g.21) So the very meaning of the equation ST = S-1 is dependent on which notion of matrix multiplication is involved. The operator S-1 is really defined in terms of SS-1 = S-1S = 1 and this involves matrix multiplication. One final thing to note: (Sab)DN = (Sab)SN ≠ (Sab)SN = gac(Scb)SN // unless it happens that g = 1 (7.9.g.22) so clearly when one writes Sab or Rab one should be clear which notation is involved.