the transpose in std notation

The transpose in tensor Standard Notation PhL 2.9.12 Theorem 1: If A is a true rank-2 tensor, then (AT)ab = Aba Proof: Since the pure contravariant tensor Aab is a matrix, we certainly could write (std notation) (AT)ab = Aba If we now apply the metric tensor ga'a to both sides, sum on a', and replace a' with a, we get (AT)ab = Aba QED. Similarly we could have raised the b index to get (AT)ab = Aba. And finally we can do both these operations at the same time. The conclusion is: (AT)ab = Aba (AT)ab = Aba (AT)ab = Aba (AT)ab = Aba For the mixed tensors, notice that transposing is the same as reflecting the indices through the vertical index axis. We do get a "left to right" switch in all cases, and each index keeps its tensorial sense (up or down) Theorem 2: Theorem 1 applies to object R even though it is not a true rank-2 tensor. Proof: By the same argument above, since object Rab is a "matrix", and we can certainly say (RT)ab = Rba (*) We know that index b on Rba is a g' index and a is a g index. This is also true of (RT)ab. Therefore, if we apply ga'a to both sides, sum on a, then replace a' by a, we get (RT)ab = Rba On the other hand, if we start again with (*), we can apply g'b'b to both sides etc to get (RT)ab = Rba Finally, we can apply both ga'a and g'b'b at the same time to get (RT)ab = Rba Thus we obtain the same four results as in Theorem 1, even though R is not a tensor. The proof was slightly different because we used g in some places and g' in other places. Theorem 3: The same conclusions apply to object S. The only difference from the R case is a swapping of the roles of g and g'. Theorem 4 on Matrix Multiplication in Standard Notation. A combination of two rank-2 tensors with a single summed index can be regarded as a matrix multiplication in Standard Notation only if the following two conditions are met: (a) the sum index must be on the right side of the first factor and on the left side of the second factor. (b) the sum index must be up on one factor and down on the other, ie, it must be a contraction. Examples: Cac = AabBbc = (AB)ac and C = AB Cac = AabBbc = (AB)ac and C = AB Cac = AabBbc does NOT imply C = AB Cac = AabBcb is not in the right form, but Cac = AabBcb = AabBTbc and C = ABT Special Case: ( assumes g = 1) dev not:: g'ac = RabRcb = Rab(R)Tbc => g' = RRT std not: g'ac = RabRcb is not in the right form, but g'ac = RabRcb = Rab(R)Tbc and g' = RRT This last case does not fit with the other examples because (1) R is not a rank-2 tensor, and (2) the condition (b) is violated, the sum index is down in both factors. We know, however, from our developmental notation work that in g' = RRT the object g' on the LHS is a true rank-2 tensor even though neither of the R's is true tensor. Therefore the combination RRT is also a true rank-2 tensor. More generally we would say in dev notation that g' = RgRT and g,g' comprise a true rank-2 tensor. So we claim that the claim of this Special Case is in fact valid despite violating Theorem 4 in two ways. We then find in the standard notation the same matrix equation g' = RRT that we find in the devel notation. But in the standard notation, we have to "know" that both summation indices are down. In all these examples (including the Special Case), once you establish a matrix/tensor equation with no indices showing, you can then apply any indices you want to get a valid equation. For example, the special case has: g' = RRT => g'ac = Rab(R)Tbc = RabRcb g'ac = Rab(R)Tbc = RabRcb g'ac = Rab(R)Tbc = RabRcb g'ac = Rab(R)Tbc = RabRcb // = the translation above Here is another special case: Another Special Case: ( assumes = 1) dev not:: 'ac = SbaSbc = (ST)abSbc => ' = STS std not: g'ac = SbaSbc is not in the right form, but g'ac = SbaSbc = (ST)abSbc and g' = STS We then find in the standard notation the same matrix equation ' = STS that we find in the devel notation. But in the standard notation, we have to "know" that both summation indices are up. Theorem 5. Here then is how we need to present the translation from dev to std notation for various objects: Pure rank-2 tensors (including say gab and g'ab etc) Aab → Aab ab → Aab I never had mixed tensors in the standard notation, so I don't have to provide translations for them. The R object: Rij → Rij (RT)ij = Rji → Rji = (RT)ij Rewrite these two lines: Rij → Rij (RT)ij → (RT)ij Note well the second line! This is where I made my Big Mistake on 2.9.12. By the exact same argument, we find that Sij → Sij (ST)ij → (ST)ij I will soon work this stuff into tensor doc. Theorem 6 Concerning How the Inverses Translate The correct translation for inverses is the same as for non-inverses. That is to say (R-1)ij → (R-1)ij (S-1)ij → (S-1)ij so this situation is quite different from that of the transpose. Proof for the R line above. We know that in dev notation, (R-1)ij = Sij → Sij (S-1)ij = Rij → Rij In the standard notation space, we would define (S-1) as the object which makes this be true: (S-1)ik Skc = δic Now the derivative's chain rule tells us that (∂x'i/∂xk) (∂xk/∂x'c) = δic or Rik Skc = δic Therefore it must be true that (S-1)ik = Rik But Rik is the translation of (S-1)ik . Therefore the correct translation must be (S-1)ik → (S-1)ik which is the same as the rule for translating S itself, Sik → Sik A sample translation: Rij(R-1)jk = (RR-1)ik = δik → Rij(R-1)jk = (RR-1)ik = δik In the Std Not world, it is true that R = S-1 just as in the Dev Not world. Theorem 7: In standard notation, we have ST = R and S = RT , even though this is not true in the developmental notation. None of these four objects is a tensor. Proof: (ST)ab = Sba according to Theorem 3 = Rab according to Section 7 *** Therefore (ST)ab = Rab We can then apply appropriate g and/or g' to both sides to get the other forms (ST)ab = Rab (ST)ab = Rab (ST)ab = Rab and in general then we have ST = R Let's take this particular incarnation of the above (ST)ab = Rab Now recall our rules for forward translation Rij → Rij (RT)ij → (RT)ij Sij → Sij (ST)ij → (ST)ij We don't have translation rule involving the shape Rab , so we have to do this Rab = g'aa'Ra'b' gb'b We know how to back-translate the RHS here: 'aa'Ra'b'gb'b → g'aa'Ra'b' gb'b = Rab Therefore if we start with (ST)ab = Rab and back-translate the LHS it becomes (ST)ab . Then our back-translated equation is this (ST)ab = 'aa'Ra'b'gb'b or ST = ' R g = ( gT RT 'T )T = (g RT ')T Now ST = ' R g apply RT to both sides from the left RT ST = RT ' R g Now LHS = (SR)T = 1T= 1 so have 1 = RT ' R g Now apply from the right = RT ' R And this is one of my equations in Section 5. Thus, when we take this Std Notation equation ST = R and back translate it to dev notation, it becomes = RT ' R Theorem 8. ST = S-1 in the Standard Notation world, so S is unitary (but only in the Std world!) Proof: From the end of Theorem 6 we claim that R = S-1 But from Theorem 7 we have R = ST Therefore ST = S-1 This seems amazing to me. Let's now try a back translation and see where it leads. Start with this incarnation, (ST)ij = (S-1)ij (*) Here is our table of translations Rij → Rij (RT)ij → (RT)ij Sij → Sij (ST)ij → (ST)ij (R-1)ij → (R-1)ij (S-1)ij → (S-1)ij = Rij The LHS of (*) back translates into (ST)ij = Sji in the dev not world. The RHS is not in our table, but we know that (S-1)ij = Rij = g'ii'Ri'j'gj'j →back→ 'ii'Ri'j'gj'j = ('Rg)ij Thus our back-translated equation is Sji = ('Rg)ij = STij or 'Rg = ST So right multiply by RT to get 'Rg RT = 1 => Rg RT = g' = a standard result.