Attempt to analyze section Pullbacks re-examined on page 88 (CH7)

Attempt to analyze section "Pullbacks re-examined" on page 88 PhL 8.17.15 Summary: If I am willing to simply define gJ ≡ φ*α(eJ) as a page 88 definition of gJ, I am able to show that this gJ agrees exactly with the Chapter 3 result for gJ which was gJ(x) ≡ ΣI fI[φ(x)] det(Dφ)I,J . But the whole business is very cloudy since in Chapter 3 things are functions of x, whereas in Chapter 7 page 88 things are functions of those k multilinear arguments. I am really unable to make a clean connection between the two Worlds. Maybe this will get clarified later. I gave it my best shot today. I could go off and read some other detailed source like Spivak, but that would take many days I suspect. Maybe Sjamaar will clarify things in upcoming chapters. ___________________________________________________________________________________ Start with page 87 A and page 88 B which combined say αx(v1,v2.....vk) = ΣI αx(eI) dxI(v1,v2.....vk) // αx(eI) = fI(x) (1a) ____________________________________ 2.17.16: In case this idea is missed, here is a proof of the above expansion. Assume it is valid, then evaluate (1a) at this point in function space, αx(e1,e2.....ek) = ΣI αx(eI) dxI(e1,e2.....ek) = ΣI αx(eI) λI(eZ) = ΣI αx(eI) δIZ = αx(eZ) We then end up with a consistent result that αx(eZ) = αx(eZ). _______________________________________________________ But in the spirit of the x and y use in the pullback discussion of Chapter 2, replace the above with αy(v1,v2.....vk) = ΣI αy(eI) dyI(v1,v2.....vk) y = φ(x) (1b) or αy(v1,v2.....vk) = ΣI fI(y) dyI(v1,v2.....vk) (1c) which in short has the Chapter 3 pre-pullback form of α = ΣI fI(y)dyI . // each side is a functional, no tensor function arguments (1d) Evaluate (1b) at v1,v2.....vk = eJ to get αy(eJ) = ΣI αy(eI) dyI(eJ) . (2) Now do the pullback mimicking Chapter 3 φ*αy(eJ) = ΣI φ*[αy(eI)] [φ*dyI(eJ)] . (3) Recall that we used to say (in Ch 3) that φ* f(y) = f(φ(x)) for a 0-form or function. That is to say, we just replaced the argument y with y = φ(x) which was the forward transformation. And for a differential we did the replacement φ* dyI = dφI = (Dφ)dxI. This is discussed best in my Ch 3 meta notes I think. Hypothesize now in our new situation this parallel idea, φ*[αy(eI)] = αφ(x)(eI) // that is, y→φ(x) in the label of α (4) or φ*[ fI(y)] = fI(φ(x)) and [φ*dyI(eJ)] = dyI( (Dφ)ej1, (Dφ)ej2, .,,,, (Dφ)ejk) (5) Comment 2.15.16: Compare this to our Ch 3 meta notes result φ*dyI = dφI(x) = ΣJ det(RI,J) dxJ = ΣJ [det[(Dφ)I,J] dxJ // ordered sum on J In Chapter 7 the above would be interpreted as a statement about functionals, φ*dyI = dφI(x) = ΣJ det(RI,J) λJ = ΣJ [det[(Dφ)I,J] λJ and you could close this onto eK to get the following tensor function statement, [φ*dyI](eK) = dφI(x)(eK) = ΣJ det(RI,J) λJ(eK) = ΣJ [det[(Dφ)I,J] λJ(eK) = ΣJ [det[(Dφ)I,J] δJK = det[(Dφ)I,K] which would say [φ*dyI](eJ) = dφI(x)(eJ) = det[(Dφ)I,J] Now maybe we can show that det[(Dφ)I,J] =?= dyI( (Dφ)ej1, (Dφ)ej2, .,,,, (Dφ)ejk) = dyI( (Dφ)eJ ) = λI [(Dφ)eJ ] = λi1 [ (Dφ)ej1] λi2 [(Dφ)ej2] ..... = [ (Dφ)ej1]i1 [ (Dφ)ej2]i2 .... but I don't see how det is going to appear. Pause on this. End comment. [ but I do this detail below!] Based on these assumptions, we then find that φ*αy(eJ) = ΣI φ*[αy(eI)] [φ*dyI(eJ)] = ΣI αφ(x)(eI) dyI( (Dφ)ej1, (Dφ)ej2, .,,,, (Dφ)ejk) . (6) Now looking back at (1b), make substitutions in (1b) to write αφ(x)((Dφ)ej1, (Dφ)ej2..... (Dφ)ejk) = ΣI αφ(x)(eI) dxI((Dφ)ej1, (Dφ)ej2..... (Dφ)ejk) . (7) Now (6) and (7) have the same RHS so we may equate the LHS's to get φ*αy(eJ) = αφ(x)[(Dφ)ej1, (Dφ)ej2..... (Dφ)ejk] (8a) I guess you could define (φ*α )x ≡ φ*[αy] = φ*[αφ(x)] functions of k vector arguments. Then you get (φ*α )x(eJ) = αφ(x)[(Dφ)ej1, (Dφ)ej2..... (Dφ)ejk] (8a) It seems to me a weird way to label the left side, but it is certainly allowable. I guess the idea is that the object (φ*α ) is some new form, and it should be a tack-on x label like any other form, hence (φ*α )x My (8a) above is waypoint 1 which I have checked in red on page 88 as 1. Now recall, αy(v1,v2.....vk) = ΣI fI(y) dyI(v1,v2.....vk) (1c) Substitute into this v1 = (Dφ)ej1 and y = φ(x) to get αφ(x)((Dφ)ej1, (Dφ)ej2..... (Dφ)ejk) = ΣI fI(φ(x)) dyI[(Dφ)ej1, (Dφ)ej2..... (Dφ)ejk] (9) Now use this in equation (8a) to obtain (φ*α )x(eJ) = ΣI fI(φ(x)) dyI[(Dφ)ej1, (Dφ)ej2..... (Dφ)ejk] (10) This matches waypoint 2 so another red check #2 goes on page 88 near 88 A. Now go back to page 85C where we would say "dy1dy2 ...dyk"(v1,v2.....vk) = det[ dyi(vj)] dyI(v1,v2.....vk) = det[ dyI(vj)] (11a) The matrix inside the determinant here is this dyi1(v1) dyi1(v2) ... dyi1(vk) dyi2(v1) dyi2(v2) ... dyi2(vk) ... dyik(v1) dyik(v2) ... dyik(vk) (11b) Now in (11a) and (11b) replace v1 → (Dφ)ej1 and so on to get this new equation and matrix dyI((Dφ)ej1, (Dφ)ej2..... (Dφ)ejk) = det[ dyI((Dφ)ej1)] or in shorthand, dyI( (Dφ)eJ) = det [dyI((Dφ)eJ)] (12a) where the matrix is dyi1((Dφ)ej1) dyi1((Dφ)ej2) ... dyi1((Dφ)ejk) dyi2((Dφ)ej1) dyi2((Dφ)ej2) ... dyi2((Dφ)ejk) ... dyik((Dφ)ej1) dyik((Dφ)ej2) ... dyik((Dφ)ejk) (12b) But for example dyi1((Dφ)ej2) = eTi1((Dφ)ej2 = (Dφ)i1,j2 (13) Then from (12a) and (12b) we end up with dyI( (Dφ)eJ) = det(Dφ)I,J Now insert this into (10) to get (φ*α )x(eJ) = ΣI fI(φ(x)) dyI[(Dφ)ej1, (Dφ)ej2..... (Dφ)ejk] (10) = ΣI fI(φ(x)) det(Dφ)I,J = ΣI φ*fI(x) det(Dφ)I,J (14) So here is what I have shown in all the above (where y = φ(x) ) (φ*α )x(eJ) ≡ φ*[αy(eJ)] = ΣI φ*fI(x) det(Dφ)I,J (15a) This agrees with waypoint red checkmark #4. Another way to write the above is (φ*α )x(eJ) ≡ φ*[αy(eJ)] = ΣI fI(φ(x)) det(Dφ)I,J (15b) However, nowhere have I referred to this as gJ. I have simply derived it as you see it in (15). Now what was our result from Chapter 3 on pullbacks? φ*α = φ* [ ΣI fI dyI ] = ΣI φ*[fI] φ*[dyI] = ΣI fI[φ(x)] dφI = ΣI fI[φ(x)] ΣJ det(Dφ)I,J dxJ = ΣI,J fI[φ(x)] det(Dφ)I,J dxJ = ΣJ gJ(x) dxJ where gJ(x) ≡ ΣI fI[φ(x)] det(Dφ)I,J (16) This does show that gJ(x)of Chapter 3 = (φ*α )x(eJ) of page 88 (17) Now, in Chapter 3 we had, as shown above in (16) φ*α = ΣJ gJ(x) dxJ gJ(x) ≡ ΣI fI[φ(x)] det(Dφ)I,J // chapter 3 (18) What is the page 88 equation corresponding to the left equation in 18? I want to find something like this: φ*α = ΣI fI[φ(x)] ΣJ det(Dφ)I,J dxJ // Chapter 3 pullback result I do have this result with general arguments: φ*αy(v1,v2.....vk) = ΣI αφ(x)(eI) dyI( (Dφ)v1, (Dφ)v2, .,,,, (Dφ)vk) . = ΣI fI(φ(x)) dyI( (Dφ)v1, (Dφ)v2, .,,,, (Dφ)vk) (19) If these are general vectors vi , I don't know what to do next! But if I evaluate the arguments at v1 = ej1 and so on I get φ*αy(ej1,ej2.....ejk) = ΣI αφ(x)(eI) dyI( (Dφ)ej1, (Dφ)ej2, .,,,, (Dφ)ejk) . = ΣI fI(φ(x)) det(Dφ)I,J or φ*αy(eJ) = ΣI fI(φ(x)) det(Dφ)I,J Plan A. Now if we were to mult both sides by dxJ and sum over ordered index J, we would get ΣJ φ*αy(eJ)dxJ = ΣI,J fI(φ(x)) det(Dφ)I,J dxJ = φ*α of chapter 3. Plan B. I could try this det(Dφ)I,J = ΣJ det(Dφ)I,J δI,J = ΣJ det(Dφ)I,J dxJ(eI) Then my page 88 result becomes Chapter 3: φ*[dyI] = dφI = ΣJ det(Dφ)I,J dxJ Page 88: [φ*dyI(eJ)] = ΣJ det(Dφ)I,J dxJ(eI) Now they look close. But it is not really correct. Plan C. Try continuing from (18). We started above with αy(v1,v2.....vk) = ΣI fI(y) dyI(v1,v2.....vk) (1c) Had we applied the pullback to this thing directly with general arguments, we would have gotten φ*αy(v1,v2.....vk) = ΣI αφ(x)(eI) dyI( (Dφ)v1, (Dφ)v2, .,,,, (Dφ)vk) . = ΣI fI(φ(x)) dyI( (Dφ)v1, (Dφ)v2, .,,,, (Dφ)vk) (19) Can we just apply the multilinear rule to this last function? Then consider dyI( (Dφ)v1, (Dφ)v2, .,,,, (Dφ)vk) The rule is only helpful if you could express (Dφ)v1 as a linear combination of vectors. But certainly (Dφ)v1 = Σk q1k ek // since ek is a basis Now we know (Dφ)v1 = (Dφ) Σk(v1)kek = Σk(v1)k [ (Dφ)ek ] Now dive to the next level and write (Dφ)ek = Σi ski ei ejT (Dφ)ek = Σi ski ejTei = skj = (Dφ)j,k Thus we have that (Dφ)ek = Σi(Dφ)i,k ei Now go back a few steps and we have shown that (Dφ)v1 = Σk(v1)k [ (Dφ)ek ] = Σk(v1)k Σi(Dφ)i,k ei = Σi(v1)i Σk(Dφ)k,i ek // swap indices = Σk [ Σi(v1)i (Dφ)k,i ] ek = Σk [q1k ] ek q1k ≡ Σi(v1)i (Dφ)k,i So I have finally succeeded in writing (Dφ)v1 as a linear combination of vectors. Then I guess I can say dyI( (Dφ)v1, (Dφ)v2, .,,,, (Dφ)vk) = Σkq1k dyI(ek, (Dφ)v2, .,,,, (Dφ)vk) Now replace summation index k by j1 so we then have = Σj1 q1,j1 dyI(ej1, (Dφ)v2, .,,,, (Dφ)vk) Then I could do this on all the arguments and end up with = Σj1,j2...jk q1,j1 q2,j2.... qk,jk dyI(ej1, ej2, .,,,, ejk) Now things get fuzzy. I is an ordered index, but J here is not an ordered index. Give up. Plan D Sja mysteriously makes this definition gJp88 ≡ φ*α(eJ) which I interpret to mean gJp88 ≡ (φ*α )x(eJ) ≡ φ*[αy(eJ)] Given this definition, I have shown above that gJp88 = ΣI φ*fI(x) det(Dφ)I,J = exactly the same as gJch3 The problem is that the definitions of the two gJ seem unrelated: φ*α = ΣJ gJch3(x) dxJ // chapter 3 definition φ*α(eJ) = gJp88 // page 88 definition So I am not really happy with Sja's claim that "therefore our new definition agrees with the old" unless he means our new definition of gJ. He writes φ*α = ΣJ gJdxJ as page 88 C but this is not consistent with φ*α(eJ) = gJ page 88 D because then you would have to get φ*α = ΣJ φ*α(eJ) dxJ but dxJ does not appear in any of his page 88 equations! Conclusion: For now I am going to have to "let this go" because I don't have enough information. 2.15.16. Reviewed the above, good to have this detail in case I someday want to do a complete rewrite of this stuff, but not crucial at the moment.