How does a tensor function transform

How does a tensor function transform? PhL 11.22.15 Here is what I first wrote in App C.5: _________________________________________________________________ Under a Chapter 2 transformation from x-space to x'-space, arbitrary basis vectors ei transform to new basis vectors e'i = Rijej where R is the linearized form of a general transformation x' = F(x). We shall interpret this in bra-ket notation to say <e'i| = Rij <ej| . Then for a tensor product of the form (C.5.1) one has < e'i,e'i, .... e'i | = RijRij.....Rij < ej,ej, .... ej| . <e'I| = RIJ <eJ| (C.5.17) Closing the above "bra" equation from the right with "ket" |T> gives, < e'i,e'i, .... e'i | T> = RijRij.....Rij < ej,ej, .... ej| T> (C.5.18) According to (C.5.7) we interpret this to state, T'jj....j = RijRij.....Rij Tjj....j T'I = RIJ TJ (C.5.19) This is the usual rule for the transformation of a rank-k tensor, as shown for example in the last line of (2.1.6). One can write this as T(e'i,e'i, .... e'i) = RijRij.....Rij T(ei,ei, .... ei) or T(Rijej,Rijej, .... .Rijej) = RijRij.....Rij T(ej,ej, .... ej) . This is valid for any matrix Rab. But it is only valid if the ei form a basis. ______________________________________________________________________ What do I think of this discussion? First, what is the meaning of this: e'i = Rijej It says that each of the new vectors is an independent linear combination of the old vectors. Since the vectors ej form a basis, the e'i can be ANY vectors you want. So why not instead write vi = Rijej or in fancier notation vi = ΣjRijej You are saying: "Since the {ej} form a basis, I can write each vi as a linear combination of the ej and the coefficients of those linear combinations are the R matrix." Then consider T(vi,vi, .... vi) = T( ΣjRijej, ΣjRijej, .... ΣjRijej) = T( ΣjRijej, ΣjRijej, .... ΣjRijej) // rename dummies = ΣjRij T( ej, ΣjRijej, .... ΣjRijej) // linear in first arg = ΣjRij ΣjRijT( ej, ej, .... ΣjRijej) // linear in 2nd arg .... Σjj...j Rij Rij ....Rij T( ej, ej, ....ej) So JUST FROM k-multilinear we learn this fact T(vi,vi, .... vi) = Σjj...j Rij Rij ....Rij T( ej, ej, ....ej) = Σjj...j Rij Rij ....Rij Tjj....j Now compare this to the tensor rule with a transformation matrix R related to some x' = F(x): T'ii....i = Σjj...j Rij Rij ....Rij Tjj....j Then we get T'ii....i = T(vi,vi, .... vi) That is something NEW right now. Interpretation: Each randomly selected set of vectors {vj} defines a matrix R. That is to say: Pick a set of k vectors {vj}, then write each one as a linear combination of the basis {ei}. That action then determines all elements of the matrix Rij. Now imagine that we have some transformation x' = F(x) such that this matrix R is the differential of that transformation. THEN we get the above equation. Here the prime means the x'-space defined by x' = F(x). As you vary the set of vectors {vj}, you vary the meaning of x'-space and so you vary T'ii....i. Summary: We choose a set of vectors {vj} and that determines matrix R, and we then use that R to go from x-space to x'-space, integrating somehow to obtain a viable x' = F(x). Then with implied sums. T(vi,vi, .... vi) = Rij Rij ....Rij T( ej, ej, ....ej) which we identify with T'ii....i = Rij Rij ....Rij Tjj....j We can WRITE the equation, T(v'i,v'i, .... v'i) = Rij Rij ....Rij T( vj, vj, ....vj) but this is valid only if the {vi} form a basis for Vk. If the vi do NOT form a basis, perhaps two are equal, then there are certain vectors v'i that "cannot be reached". You can still say that v'i = ΣiRijvi but your vectors v'i will be restricted. The transformation matrix R falls below full rank and det(R) = 0 and x' = F(x) is singular. So problems occur at the singular points. Go back to App C.5. Consider this statement: You have to assume that the {vi} form a basis in order to create a dual basis {vi} Suppose the vi are linearly dependent. What happens when you try to form the vi ? I have to go look at tensor doc to find out. I did that. There I assume the bn are a basis and I write Bm = w'mnbn and the conclusion is that w' = W'-1 where W'nk ≡ bn bk . Then det(W') = det(bn bk). Suppose you can write b2 = Σi≠2 αibi Then in the matrix W'nk ≡ bn bk we have one row being a linear combination of other rows and so the det(W') = 0 and we connot create w so we cannot create the dual basis. Fine. Back to Appendix C. I have an ok to here mark there, and then it says One can say that the tensor Tii....i and the tensor function T(vi,vi, .... vi) are both representations of the same abstract tensor T/T in two different bases, |vI> and |eI>. (1) Suddenly I have added that the vi are a set of basis vectors! (2) I am stuck with this problem that the abstract tensor has two names, T and T . So I have two separate problems here to deal with, now 8 AM 11/22/15 Sun. Problem (2). If I call the tensor T, then I get T(vi,vi, .... vi) = < vi,vi, .... vi | T > I guess I would say T = T and I just use T when I write tensor functions so the reader will know when I write T that I am implying a tensor function. But I also use T when I talk about functionals with the idea that functionals are script font. So then T = T would identify a tensor T with a functional T which I think is wrong. I like this statement, <T | vi,vi, .... vi > = T(vi,vi, .... vi) because here you "see" that T is in the role of a functional. So the issue is then here: Tii....i = < ei,ei, .... ei | T > Here T ϵ Vk whereas T ϵ V*k so you CANNOT under and circumstances say T = T. What about that isomorphism between Vk and V*k ? Is there one? Did I ever mention it? Yes, see (2.11.10). So I can maybe resolve by saying T ϵ Vk T ϵ V*k T ↔ T "corresponds to" in the isomorphism Then the correct notation is to put T in a bra, and maybe T in a ket?? T → <T| ϵ V*k T → | T> ϵ Vk Then I like <T | vi,vi, .... vi > = T(vi,vi, .... vi) because the functional is on the left (the dual space object) and the Vk element is on the right. It is then very confusing to the reader to swap this around. And T is what you must display here in the bra. I then have to deal with this <T | ei,ei, .... ei > = T(ei,ei, .... ei) = Tii....i T = ΣI TIeI ϵ Vk T = ΣI TIλI ϵ V*k I guess this is all OK. There are two separate objects, but the TI coefficients appear in both.