Matrix Notation and dev vs std notation

Matrix Notation and dev vs std notation (keep) PhL 4.9.12 Here I ponder a few issues and I think they are now all resolved and nailed down in tensor doc Section 7 (i) item 7. [ Things nailed down often come un-nailed, I have found.] Problem 1. Consider this equation in dev notation, and assume that g ≠ 1 : AB = C ie AabBbc = Cac If A and B are contravariant tensors and we are not sure yet what C is, how do we convert such an equation to standard notation? We know that Aab→ Aab and Bab→ Bab but we don't have any idea what to do with C because we don't know it is a tensor. So we don't know how to convert such an equation to std notation!! [ I have dealt with this question in new Section 7 (i) item 7. ] Problem 2. Using only dev notation, show that if we assume Cac is a tensor, we get a contradiction. [ a useful exercise, I think the flow below is all OK, and the conclusion is true. ] If C is a tensor, then it should be true that C'ij = Rii'Rjj'Ci'j'. and this is also true for A and B. Therefore we find that C'ij = RiaRjcCac = RiaRjc AabBbc now we know that, since S = R-1, Aab = Saa'Sbb'A'a'b' Bab = Saa'Sbb'B'a'b' Bbc = Sba'Scb'B'a'b' Bbc = SbnScmB'nm Then we have C'ij = RiaRjc Saa'Sbb'A'a'b' SbnScmB'nm = RiaRjc Saa'Sbb' Sbn Scm A'a'b'B'nm Now try to group the R and S factors optimally, = Ria Saa' Rjc Scm Sbb' Sbn A'a'b'B'nm = δia' δjm Sbb' Sbn A'a'b'B'nm = Sbb' Sbn A'ib'B'nj But if C were a tensor, then our equation would be a tensor equation and we would have A'abB'bc = C'ac so that C'ij = A'ib'B'b'j Then the following would have to be true : Sbb' Sbn A'ib'B'nj = A'ib'B'b'j and since must be true for any tensor A, then this must be true Sbb' Sbn B'nj = B'b'j But this is only true if Sbb' Sbn = δnb' or STnb Sbb' = δnb' or STS = 1 Only if this last fact were true would it be true to say that C is a tensor. But in general this is NOT true, so in general C is not a tensor, despite the fact that A and B are tensors. Problem 3. If we write the dev nota equation AB = 1, is 1 a tensor? [In Cartesian coordinates g = δ, and then g is a tensor and g' = RRg. But one would never make the claim that δ' = RRδ where [δ']ab = δa,b because this is just not true unless R is a rotation. So my answer here is that the object [1]ab = δa,b is not a tensor if by that you mean [1]'ab = δa,b. It might be good to get this fact into tensor doc somewhere. ] The equation is AabBbc = δac so the question is: is δac a tensor? If it were, we would write δ'ac = Raa'Rcc'δa'c' = Raa'Rca' = (RRT)ac Only in the case RRT = 1 do we get what we want. If g ≠ 1, then RRT is some weird object, it is not even g' . So in the sense that we were thinking 1' = 1 in both spaces, then in general 1 is not a tensor. That is rather shocking. So AB =1 has the same overall problem that AB = C has in going from dev to std nota. [ I have dealt with the following issue in new Section 7 (i) item 7. ] Fact: Standard notation is a set of rules for S, R and for tensors! It has no rules for oddball objects. What about the equation RS = 1? RabSbc = δac RabSbc = δac This special matrix equation translates OK, but it is a special case!! What about STS = 1? We can try this SabScb = δac → SabScb = δac I think this translation is OK as well. The result is also RbcRbc = δac . This is a condition on the matrix S written in either notation. The equation STS = 1 is not a true tensor equation and the objects in it are not even tensors.