Study the pullback equation in Dirac Notation REVIEWED

Study the pullback equation in Dirac Notation PhL 2.29.16 5.17.16: I think here I am trying to start with the end in mind, the pullback defined in terms of tensor functions, and I am trying to thread backwards to some Dirac thing that makes sense. This is now all resolved to my satisfaction in Section 10, in particular (10.7.16) . The tensor function pullback equation is this: [φ*(αx)](v1,v2...vk) = αφ(t) (Rvn,Rvn...Rvn) . < φ*(xα) | v1, v2, ...vk> = < αφ(t) | Rv1, Rv2, ...Rvk> Can this be processed in some way? < αφ(t) | Rv1, Rv2, ...Rvk> = < αφ(t) | R | v1, v2, ...vk> = < RTαφ(t) | v1, v2, ...vk> = < αφ(t) | R | v1, v2, ...vk> What is the meaning of < RTαφ(t)| where R = (Dφ) = a matrix. We need to think of αφ(t) as a row vector. So write αx = Σ'I fI(x) λI = Σ'I fI(x) λ^I = Σ'I fI(x) (xλi ^ xλi....^ xλi) = Σ'I fI(x) AltI (xλi xλi... xλi) = Σ'I fI(x) AltI [ (xei)T(xei)T ... (xei)T ] = Σ'I fI(x) AltI [ <xei | <xei |.... <xei | ] Now how do you write αx in bra notation? I guess just reuse the same name and write <αx| = Σ'I fI(x) AltI [ <xei | <xei |.... <xei | ] Now each column vector xei in x-space has n components. The transpose of the above is |αx> = Σ'I fI(x) AltI [ |xei > |xei >... |xei> ] Now suppose R is a matrix with n rows and m columns so is n x m. Then RTei conforms. So apply RT to the above to get RT |αx> = Σ'I fI(x) AltI [ | RTxei > |RTxei >... |RTxei> ] Now transpose this equation to get <αx| R = Σ'I fI(x) AltI [ <RTxei | <RTxei |.... <RTxei | ] So I can start over just doing this [φ*(αx)](v1,v2...vk) = αφ(t) (Rvn,Rvn...Rvn) . pullback definition < φ*(xα) | v1, v2, ...vk> = < αφ(t) | Rv1, Rv2, ...Rvk> in Dirac = <αφ(t) | R | v1, v2, ...vk> = < αφ(t) | R | v1, v2, ...vk> = Σ'I fI(φ(t)) AltI [ <RTφ(t)ei | <RTφ(t)ei |.... <RTφ(t)ei | ] | v1, v2, ...vk> Thus I find this functional equation < φ*(xα) | = Σ'I fI(φ(t)) AltI [ <RTφ(t)ei | <RTφ(t)ei |.... <RTφ(t)ei | ] This then is the functional equivalent of the tensor function pullback definition. On the right, the vectors in the bras have m components, while the ei have n components. Presume can write as < φ*(xα) | = Σ'I fI(φ(t)) <RTφ(t)ei | ^ <RTφ(t)ei |....^ <RTφ(t)ei | (*) Now how might you write <RT φ(t)ei | = <RTei| recall λi = <ei| I answer this question in Appendix A below, and result is (RT xei) = Σj=1m Rij tej or (RT φ(t)ei) = Σj=1m Rij tej Now I can use this k times in (*) to get < φ*(xα) | = Σ'I fI(φ(t)) [ < Σj=1m Rij tej| ^ < Σj=1m Rij tej| ... ] = Σ'I fI(φ(t)) Σjj...j=1m RijRij ....Rij <tej| ^ <tej| ..... = Σ'I fI(φ(t)) ΣJ RIJ <te^J | So given the equation [φ*(αx)](v1,v2...vk) = αφ(t) (Rvn,Rvn...Rvn) I have been able to convert this to a functional statement which is < φ*(xα) | = Σ'I fI(φ(t)) ΣJ RIJ <te^J | or φ*(xα) = Σ'I fI(φ(t)) ΣJ RIJ tλJ and then you could close this with | v1, v2, ...vk> to get the above tensor statement. Appendix A. Consider the conjecture (Rei) = Σj=1m Rji ej [ This came up elsewhere. It is not meaningful except for square R. The issue is that (Rei) = e'i is in a different space than ej so cannot be a linear combination of ej for non-square R // 5.17.16.] We start with a set of n basis vectors ej of space Rn, could call these xej . This is a column vector having n components. Now suppose R is a matrix with m rows and n columns. Then Rei "conforms" and produces a vector with m components which is some vector in Rm . The basis in Rm is called tei and there are m of these basis vectors and each has m components. Since (Rei) is some vector in Rm we know we can write (R xei) = Σj=1m Aji tej where Aji are some coefficients we with to compute. To do this, we do some steps. First ten (R xei) = ten Σj=1m Aji tej = Ani But writing out the dot product ten (R xei) = Σr=1m (ten)r (R xei)r = Σr=1m (ten)r Σs=1n Rrs (xei)s = Σr=1m δnr (Σs=1n Rrs δis ) where n can only range 1 to m, i ranges 1 to n = Σr=1m δnr Rri = Rni Therefore we have shown that the coefficients are in fact the matrix elements Ani = Rni Thus I have shown that in this context of two spaces (R xei) = Σj=1m Rji tej I know it is trivial to show that the following is also true (raise and lower indices etc)\ (R xei) = Σj=1m Rji tej Finally, suppose we rename the matrix to be RT so then (RT xei) = Σj=1m (RT)ji tej Where now RT is a matrix with m rows and n columns so R is a matrix with n rows and m columns Finally taking a risky step, I use my definition of transpose to write (RT)ji = Rij // reflect in vertical line Then this last result becomes (RT xei) = Σj=1m Rij tej Here is a drawing of this situation: put drawing here!! **** it gets an eq number too!