yet another view on the transformation issue

Yet another view on the transformation issue 1.19.16 This topic is refusing to stabilize, each day it moves again, slithers around. I had established the idea of Transformation A where basis vectors don't move. Thus I had vi = v ei = <v | ei> = vTei v'i = v' ei = <v' | ei> = v'Tei which "looks like a scalar, being the dot product of two vectors", but which under transformation A in fact transforms as a component of a vector. So I already had the conflicting idea that ei is both a vector (in that it has 3 components) and a scalar (something that does not change under a transformation). It is a non-transforming vector under transformation A. I could transform it to get e'i = Rei. So maybe the term "non-transforming vector" is better than saying "scalar". Aside: I keep thinking that somehow writing the dot product in some different notation is going to clarify things, but I now think this idea is wrong. Expansions would be v = Σi vi ei v' = Σi v'i ei So in fact all of these objects are non-transforming vectors under transformation A: ei = |ei> eiT = <ei| ei = |ei> eiT = <ei| Now we want to talk about functionals! This is the Big Jump for me. I can write λi(v) ≡ eiT v = ei v = <ei| v > = vi rank-1 tensor function As a function, λi(v) takes scalar values, so you would say it was a scalar function. In general, if I write a function F(x) I imply a scalar-valued function, whereas F(x) implies a vector-valued function. For that reason then I don't want to write λi(v) where λ is bolded. The $64 question is this: what is meant by λi all by itself? Somehow it is this: λi ≡ eiT = ei = <ei| rank-1 tensor functional λi : V → K But when I write λi = <ei|, I have this conflict: λi is a scalar but ei is a non-transforming vector. So when it is by itself, I really have λi ≡ (ei)T = a non-transforming row vector. Being a vector, it should be bolded the way ei is bolded. It is "something that has components". Note that each ei is a vector. Thus, each λi is a different row vector, so you really should write λi = (ei)T . Then the idea that λi is a vector in the dual space V* is not so painful. It is however a non-transforming vector. So how can we both bold and not-bold the same symbol? λi ≡ eiT = a non-transforming row vector λi(v) = eiTv = ei v = vi λi(v') = eiTv' = ei v' = v'i Note added later: If you had κ(v) = vv you would not bold κ. Note added later: You could say λe in place of λi . Then λ is not bolded, but the label ei is bolded. Then you would have λe(v) = eiTv = scalar. Let's put a temp hold on the last question to ask a new one. Question: How does λi(v) "transform"? Answer #1: Since λi(v) = vi , λi(v) transforms the same way that vi transforms, which is as a component of a transforming vector. Thus, v'i = Rij vj λi(v') = Rijλj(v) Check on Answer #1: Consider λi(v') = λi(Rv) = ei Rv = (ei)k (Rv)k = (ei)k Rkjvj = δikRkjvj = Rijvj = ok λi(v') = λi(Rv) = ei Rv = (Rv)i = Rijvj = ok λi(v') = v'i = Rijvj = ok So we pass this check. Question added later: Do the above lines have anything to do with the linearity of λi ? Linearity would say for example λi(2v) = 2λi(v) λi(v + w) = λi(v) + λi(w) Well, v' = Rv is NOT a multiple of vector v. nor is it a sum of v and some other specified vector w, so the answer must be: no, the transformation rule for λi is not implied by its linearity, it is implied by it's specific form which happens to be linear. Claim: λj(v) does not transform as a vector field where v plays the role of x. Proof: Let φi(x) be a vector field. Then φ'i(x') = Rijφi(x) But the rule λi(v') = Rijλj(v) is missing the prime on the left, it does not say λ'i(v') = Rijλj(v) . Question: How does α(v) transform, where α(v) = Σiαiλi(v) ? Answer #1: [ Rejected below.] Assume that α(v') = Σiαiλi(v') where we assume that αi are constant coefficients which comprise a non-transforming vector α. With that assumption, α(v') = Σiαiλi(v') = ΣiαiRijλj(v) = ??? // using answer #1 of the previous question α(v') = Σiαiλi(v') = Σiαiv'i = α v' α(v) = Σiαiλi(v) = Σiαivi = α v α(v) = Σiαi(ei v) In the non-dual space I would say v = Σiviei = Σiviei Then under xform A, I would say the ei "stay put", but I would never say the vi formed a non-transforming vector. Transpose to get vT = ΣivieiT = Σivi<ei| = < v | Here vi are part of a transforming vector and vT is a transforming row vector which would transform I guess with RT, have to ponder that. What you see above is a general vector of V*. Suppose I rename that vector to be α. Then αT = ΣiαieiT = Σiαi<ei| = < α | = general vector in V* = Σiαi λi α = Σiαiei = Σiαiei = general vector in V I am now inclined to say that the αi are components of a transforming vector. OK, then take the above and close onto a general v in V to get α(v) = αTv = Σiαi λi(v) = Σiαi<ei| v> = < α | v > Now back up to λi(v) = < ei | v > = vi = a number = a scalar in K Theorem: (a) λi(v) is, and therefore transforms as, the ith component of a vector λi(v) = < ei | v > = vi λi(v') = < ei | v' > = v'i (b) α(v) transforms as a scalar field α'(v') = < α' | v' > = < α | v > = α(v) Here < α | is a general vector in V*, and | v > is a general vector in V . Both vectors are transforming vectors under transformation A. (c) < α | = Σiαi <ei| transforms as a vector in V* (d) | v > = Σivi |ei> transforms as a vector in V (e) OK, but what about this other object, α(v') = < α | v' > = < α | Rv > = < RTα | v > = <α" | v > = α"(v) It is just some OTHER tensor function. OK, let's try moving from V to V2 T = ΣijTij ei ej rank 2 tensor in V2 |T> = ΣijTij |ei,ej> = ΣijTij |ei> |ej> Now transpose to get TT = ΣijTij eiT ejT <T| = ΣijTij <ei,ej | = ΣijTij <ei| <ej| <T| = ΣijTij λi λj = most general rank-2 tensor in space V*2 Maybe we do α,v → A,T. So |T> = ΣijTij |ei> |ej> // general element of V2 <A| = ΣijAij λi λj // general element of V*2 If we close these together, what happens? A(T) = <A| T> = ΣijAij Σi'j'Ti'j' <ei| <ej| |ei'> |ej'> = ΣijAijTij = scalar formed by contracting these two rank-2 tensors. This is fancier than what we actually see in Spivak. Now instead of closing onto a general T, suppose we close onto this pure element of V2, |v1> |v2> ϵ V2 Then A(v1,v2) = <A| |v1> |v2> = ΣijAij (v1)i(v2)j = scalar This scalar is formed from a rank-2 tensor and two rank-2 vectors. Then A'(v'1,v'2) = A(v1,v2) and this transforms as a scalar field of two arguments! One is allowed to ask about A(v'1,v'2) = ΣijAij (v'1)i(v'2)j = <A| Rv1> |Rv2> = < RTA |v1> |v2> = (RTA)(v1,v2) = some OTHER tensor function. This is where I got on 1.19.16. Reread it on 1.20.16 and added some notes. It still seems good. In the last case, I would say that A(v1,v2) was a rank-2 tensor function. It is bilinear and it transforms as a scalar under Transformation A. The function is not a vector in bra space, or a vector in ket space, it is the closure of tensor Aij onto a pure ket in V2. Question: Why are these tensor functions of any interest to me? I always seem to forget why Spivak uses them. The only possible interest of course is in one of my "dual worlds". Chap 6 talks about the dual tensor product world. In Section 6.6 I talk about the tensor product of two tensors, and then in 6.7 I talk about the product of two tensor functions. I find this simple rule (TS)(v1,v2....vk, vk+1....vk+k') = T(v1,v2....vk) S(vk+1,vk+2....vk+k') (6.7.1) I just read Section 6.7 and it seems correct and reasonable. In bra ket this says < T | < S | |v1, v2 ....vk> |vk+1, vk+2 ....vk+k'> = < T |v1, v2 ....vk> < S |vk+1, vk+2 ....vk+k'> So why are you allowed to break the scalar product in the huge space into two multiplying pieces? It goes way back to (2.11.15) where I say (λiλj)(v1,v2) = λi(v1)λj(v2) = scalar * scalar = scalar ϵ K . (2.11.15) which says [<ei| <ej|] [|v1> |v2>] = <ei|v1><ej|v2> How does Appendix D look right now?