decoding one equation of Spivak

Decoding one equation in Spivak PhL 2.19.16 It is unfortunate that Spivak is one of my main sources, yet sometimes I need an entire 30-page document like this one to "decode" just one of his equations. That is not the kind of material I want to be writing. The equation at issue here is the following on page 89 of Spivak: (1) In some sense I know he is creating a 1-form called df from a 0-form called f. But what is p? What variable(s) is f a function of? How does this equation relate to Sjamaar's idea of a flat space on the left and a manifold on the right, one instance of which is his figure on page 37? In trying to correlate with Sjamaar, it is unfortunate as well that Sjamaar uses at least three different conventions for his variables. Let's review this issue right here. I will dig a little into each chapter in the following few pages. 1. In Chapter 2, Sjamaar writes things like this: α = ΣifI(x)dxI page 17, a k-form on Rn f(x) = f(x) a 0-form df = Σi (∂f/∂xi)dxi p 20, a 1-form d(df) = 0 d2 = 0, page 22 Everything here is a function of x, so there is some underlying x-space", no pictures are drawn, it is just some Rn . Now consider α = ΣIfI(x)dxI fI = a scalar function with label I dα = ΣI dfI(x) ^ dxI // page 20 A I never realized it before, but there must be a had in there as I have just shown. We are adding another vector to the wedge product inside dxI. This is hidden in Sjamaar! You can now write dfI(x) = Σj(∂fI/∂xj) dxj // dual vector = sum of dual vectors Then we get dα = ΣI dfI(x) ^ dxI = ΣI Σj(∂fI/∂xj) dxj ^ dxI // another new result 2. In Chapter 3, Sjamaar changes things and writes now α = ΣifI(y)dyI page 36 so suddenly we are operating in some y-space instead of x-space. We then have a transformation y = φ(x) goes with Figure on page 37 With this transformation is associated a certain pullback operator φ* which he endows with these properties φ* f = f o φ f is a function property 1 So consider f(y) starting function in y-space (φ*f)(x) = f(φ(x)) = (f o φ)(x) pulling it back to x-space So we end up with (φ*f) being a function in x-space, totally clear . The second property Sjamaar provides is this dyI starting basic form in y-space (φ*(dyI)) = dφI page 37 D property 2 An example of this would be (φ*(dy1dy2)) = dφ1dφ2 His page 38 example is important to me. There he shows φ1(x), φ2(x) each is an explicit function of x dφ1(x) = (∂φ1/∂x1)(x)dx1 + (∂φ1/∂x2)(x)dx2 = explicit dφ2(x) = (∂φ2/∂x1)(x)dx1 + (∂φ2/∂x2)(x)dx2 = explicit So what he really is saying is this: (φ*(dy1^dy2)) = dφ1(x)^dφ2(x) = q(x) dx1^dx2 q(x) = explicit I know that one could write this last equations as dφ1(x) = (Dφ)11(x)dx1 + (Dφ)12(x)dx2 = explicit dφ2(x) = (Dφ)21(x)dx1 + (Dφ)22(x)dx2 = explicit In matrix notation these two equations would then be dφ(x) = (Dφ(x)) dx but beware, in general (Dφ) is not a square matrix: n cols, m rows Now if the space on the right on page 37 was Rm = R1 then we have y = φ(x) and there is only a single y variable and a single scalar φ function. Then we have just dφ(x) = (Dφ)11(x)dx1 + (Dφ)12(x)dx2 and in matrix notation dφ(x) = (Dφ(x)) dx (Dφ(x)) = a matrix consisting of a single row. (example of non-square) Sja proves the three little pullback rules shown middle of page 38. Comment: In the last equation above, notice that φ is a transformation function, it is not a function that is part of some form like fI, so this seems unrelated to our Spi clip I am studying in this doc. The φ function instead is related to the notion of doing a pullback. However: almost the very next equation in Spi says this: (2) and I would write this as df(x) = (Df(x)) dx and then this has exactly the form I show above. Spi seems to be proving his theorem 4.7 from the previous clip (1), given that dxi = λi. In Ch 3 Sja goes on to use property 1 and property 2 to pull back a general case k-form. This action is sort of buried within Sja page 40. I will outline the steps right here for a k-form α = Σ'I fI(y) dyI in this sum, each dyI has k-factors φ*α = Σ'I φ*(fI(y)) φ*(dyI) = Σ'IfI(φ(x)) dφI where I use Σ' for my ordered sums. Now I do have this equation from above, dφ(x) = (Dφ(x)) dx a non-square matrix in general, it has n cols and m rows but this dφ(x) is NOT the dφI product shown above. I am able to show elsewhere that in fact dφI = Σ'J det (Dφ(x)IJ) dxJ det is of a kxk minor of the mxn matrix Dφ where both dφI and dxI are in ordered form, and that is what makes the det appear. I do not have a problem with that. So we end up then with φ*α = Σ'I fI(φ(x)) [Σ'J det (Dφ(x)IJ) dxJ ] = Σ'J [ Σ'I fI(φ(x)) det (Dφ(x)IJ) ] dxJ = Σ'J gJ(x)dxJ gJ(x) = [ Σ'I fI(φ(x)) det (Dφ(x)IJ) ] So I am very happy with Sja Chapter 3 really. This is the pulled back form in x-space. 3. In Chapter 4, Sjamaar changes notation yet again. He now has a new version of figure page 37 left side t-space domain U = a < t < b right side x-space Rn but coordinate there is concealed, but is (x,y) in an example I am sure Sja intends to use x on the right. So his transformation is now x = c(t) // replacing y = φ(x) of Chapter 3. So in this new notation context, and considering he is doing only 1-forms so k = 1, what do my equations above look like? y = φ(x) → x = c(t) y→x x→ t α = Σ'I fI(y) dyI → α = fI(x) dxI = fi(x) dxi dxI = just one factor φ*α = Σ'J [ Σ'I fI(φ(x)) det (Dxφ(x)IJ) ] dxJ → c*α = Σ'J [ Σ'I fI(c(t)) det (Dtc(t)IJ) ] dtJ = [ Σ'I fI(c(t)) det [(Dtc(t))I1)] ] dt = [ Σi fi(c(t)) det [(Dtc(t))i1)] ] dt = [ Σi fi(c(t)) (Dtc(t))i1) ] dt // det of 1x1 matrix is matrix = [ Σi=1n fi(c(t)) ∂tci(t) ] dt // agrees Sja p 47 *(4.1) He then writes ∫c α = ∫c fi(x) dxi = ∫C f(x) dx = !Syntax Error, Ic*α = !Syntax Error, Idt [ Σi=1n fi(c(t)) ∂tci(t) ] This is the "line integral" of a function defined on a curve over that curve c. The dt integral is then the pullback of this integral where it is pulled back from x-space to t-space. This is the first time Sjamaar address any kind of integral in his pdf. This is the integral of a 1-form. 4. In Chapter 5, Sjamaar generalizes the above Chapter 4 work and the context is now y = φ(x) → x = c(t) y→x x→ t where x has n components and t has k components. The integer n is not mentioned for some reason but I am sure it is still n. Our pull back stuff is then α = Σ'I fI(y) dyI → α = Σ'I fI(x) dxI φ*α = Σ'J [ Σ'I fI(φ(x)) det (Dφ(x)IJ) ] dxJ → c*α = Σ'J [ Σ'I fI(c(t)) det (Dc(t)IJ) ] dtJ Sja then considers this thing integrated over a "unit cube" in t-space,. ∫c α = ∫c Σ'I fI(y) dyI = ∫[0,1]k c*α = ∫[0,1]k Σ'J [ Σ'I fI(c(t)) det (Dc(t)IJ) ] dtJ = Σ'J Σ'I ∫[0,1]k dtJ [fI(c(t)) det ((Dc(t))IJ) ] = ∫[0,1]k Σ'J gJ(t) dtJ Here in ∫c α the c represents a surface or manifold of dimension k. The letter c is bolded both here and in the 1-form Chapter 4 because it is a set of points in Rn , be it a curve or a surface. Note that dtJ is still a wedge product!! The pullback above has many terms in the Σ'J sum, and each of those corresponding dtJ is an ordered wedge product of k of the dti selected from your set of n. Sjamaar now wants to consider a special case where m = k, so then the sum above has only one term and dtJ is the full volume differential in the left "flat" space. Then the above becomes ∫c α = ∫[0,1]k g(t) dt1^dt2^...^dtk He then comes to the issue of Stokes' Theorem. For that theorem, he wants to replace α by dα above as the k form, so he has ∫c dα = ∫[0,1]k g(t) dt1^dt2^...^dtk = ∫[0,1]k c*(dα) = ∫[0,1]k d(c*α) p 63 A Since dα is a k-form, d(c*α) is also a k-form in the pulled back space, and that means that c*α must be a k-1 form, so it must then be true that c*α = Σi=1k gi(t) dt1^dt2^ ...dti....^dtk p 63 B where red italic dti means dti is missing. The sum is just over all the possibly allowed differentials for a k-1 form. I will now try to regurgitate the page 63 proof of Stokes theorem for k-cubes and chains. We have d(c*α) = d [ Σi=1k gi(t) dt1^dt2^ ...dti....^dtk ] = Σi=1k dgi(t) ^ dt1^dt2^ ...dti....^dtk = Σi=1k [Σj=1k(∂gi(t)/∂tj) dtj] ^ dt1^dt2^ ...dti....^dtk I have updated the Ch 2 notes above to show the extra hat here! Now if j ≠ i, the total wedge product will vanish, so we have only the j = i terms surviving and we get d(c*α) = Σi=1k (∂gi(t)/∂ti) dti ^ dt1^dt2^ ...dti....^dtk We now slide the dti into the red hole and pick up a sign (-1)i-1 = (-1)i+1 because you have to slide i-1 places. Then we have d(c*α) = Σi=1k (-1)i+1 (∂gi(t)/∂ti) (dt1^dt2^ ...^dtk) where we are back to the full volume element. Putting this into the above we get ∫c dα = ∫[0,1]k d(c*α) = Σi=1k (-1)i+1 ∫[0,1]k (∂gi(t)/∂ti) (dt1^dt2^ ...^dtk) // page 63 C This then is the pullback of ∫c dα, where dα is a k-form, to t-space. I will pause here because this is where I got into trouble and sent Sjamaar an email and I am not sure I ever cleared that matter up. If that were cleared up, I could continue to obtain Stokes' Theorem on a k-cube. But right now I am trying to stay focused on a different matter which is the dual space stuff, so let this lie for now. I will skip Chapter 6 hoping that the second manifold definition φ(x) = c won't be important for my purpose. We then come to 5. In Chapter 7, Sjamaar puts his differential forms onto manifolds, somehow. In the "first definition" this is done by treating the manifold as the union of a bunch of patches labeled by i. On each patch the form takes some value αi and we have certain consistency conditions. It is the second definition where the dual space business arises and that is my main interest. I want to nail down his notation on the various spaces here, as I have done in each of the above chapter sections. On page 88 Sja talks about φ: U→V and about φ(x). His example says α = Σ'I fI(y) dyI so I think he is exactly in the Chapter 3 (page 37 figure) framework where y = φ(x) is the transformation. Recall above that in this notation environment we had α = Σ'I fI(y) dyI φ*α = Σ'J gJ(x)dxJ where gJ(x) = [ Σ'I fI(φ(x)) det (Dφ(x)IJ) ] ________________________________________________________________________________ OK, so much for Sjamaar's various notations. I will now process page 87-88 from scratch. Whenever dxi does NOT mean dλi, I will put it in red. So dxi = λi dxi = a calculus differential What is Example 7.15 trying to show? Write λ = ΣIaI dxI dxI = λi ^ λi .... ^ λi in dual space Λk λ = Σ'I aii...i λi ^ λi .... ^ λi My language for this in wedge doc is T^ = Σii....i Tii....i (λi^ λi .....^ λi) . T^ = ΣI TI λ^I (8.4.4) T^ = Σ1≤i<i<....<i≤n Aii...i (λi ^ λi .....^ λi) . T^ = Σ'I AI λ^I (8.4.7) Now the objects T could be fields and could therefore depend on x , or they could be constants. I am not sure why Sja wants the aI to be constants in his λ = ΣIaI dxI. But let's now add the generic arguments from Vk to get our tensor functions, so then λ(v1,v2....vk) = Σ'I aii...i (λi ^ λi .... ^ λi)(v1,v2....vk) OK, if aI is constant not depending on the vI, then λ is linear and alternating. I agree. There is no mention yet of aI possibly depending on x in some x-space. Now write the above more generally as λ(vJ) = ΣIaI dxI(vJ) = ΣIaI λ^I(vJ) Then for sure λ(eJ) = ΣIaI λ^I(eJ) = ΣIaI δIJ = aJ and we get the main result of Example 7.15 which is this aJ = λ(eJ) I am very familiar with this result in wedge doc where it appears as T(ej,ej, .... ej) = <T | ej,ej, .... ej > = Σii....i Tii....i (ej)i (ej)i....(ej)i = Tjj....j . T(eJ) = TJ (2.11.f.9) I will now review how I arrived at this last result: T^ = ΣI TI λ^I // symmetric tensor expansion T^(vJ) = ΣI TI λ^I(vJ) T^(eJ) = ΣI TI λ^I(eJ) = TJ So it is important to have ΣI symmetric expansion in this proof. So in Example 7.15, you really have to assume that λ = ΣIaIdxI is a full symmetric sum, and only then does it follow that aI = λ(eI) . Since Sja uses ΣJ for both symmetric and ordered sums, things are not quite clear which he means, and now we know for thios case. Thus we can write λ = ΣIaI dxI = ΣI λ(eI) dxI Now we want to generalize so that this form has an x-dependent coefficient, so we are talking then about a form in x-space, which is NOT the starting position for p 37, but write it anyway as λx = ΣIaI(x) dxI = ΣIλx(eI) dxI or αx = ΣIfI(x) dxI = ΣIαx(eI) dxI Notice that αx is a dual space functional on the left, while αx(eI) on the right is this functional evaluated at eI. I could easily put this into Dirac notation. So I am now at the starting blocks on page 88 top. I want x to be the variable of the pull back space on left side of p 37, so lets start now with αy = ΣIfI(y) dyI = ΣIαy(eI) dyI a k-form in y-space of page 37 fig where in our new coordinates we have dyi = λi. What is this form when it is pulled back to x-space? We know from work above that αy = Σ'I fI(y) dyI (0) (φ*α)x = Σ'J gJ(x)dxJ where gJ(x) = [ Σ'I fI(φ(x)) det (Dφ(x)IJ) ] where I have now added little subscripts on the forms indicating where they are being evaluated. Let's see if we can obtain this same result from the very obscure starting point of page 88 A which says (φ*α)x(v1,v2....vk) = αy(Rv1, Rv2.....Rvk) (1) We anticipate a result of this form (φ*α)x = ΣJ gJ(x)dxJ (2) so that (φ*α)x(vI) = ΣJ gJ(x)dxJ(vI) (3) and then (φ*α)x(eI) = ΣJ gJ(x)dxJ(eI) = gI(x) (4) Thus we can anticipate a result of this form (I will repair Σ, Σ' stuff later, trying to get first cut) (φ*α)x = ΣJ gJ(x)dxJ = ΣJ (φ*α)x(eJ) dxJ (5) Now go back to (1) and evaluate at eJ and use (4) as well to say gJ(x) = (φ*α)x(ej,ej....ej) = αy(Rej, Rej.....Rej) (6) waypoint 1 Now install the αy expansion from (0) above to get gJ(x) = αy(Rej, Rej.....Rej) = Σ'I fI(y) dyI(Rej, Rej.....Rej) waypoint 2 (7) Now recall from wedge doc (λj ^ λj ^ .... ^ λi)(vi,vi....vi) = (1/k!) det[ (vi)j] (8.3.9a) = (1/k!) det (vIJ) (8) which implies that (λi ^ λi ^ .... ^ λi)(v1,v2....vk) = (1/k!) det[ vZI ] = dyI(v1,v2....vk) (9) So far then we have dyI(v1,v2....vk) = (1/k!) det[ vZI ] (10) Here vZI is a kxk matrix whose row indices are 1,2...k and whose column indices are ii, i2.. ik. A typical element of the matrix would be (v2)i . Now if we replace v2 → Rej a typical element of that matrix becomes ( Rej)i = Ris(ej)s = Risδjs = Rij . Therefore dyI(Rej, Rej.....Rej) = (1/k!) det[ RIJ] = (1/k!) det[ (Dφ)IJ] (11) We find then that gJ(x) = Σ'I fI(y) dyI(Rej, Rej.....Rej) = Σ'I fI(y) (1/k!) det[ (Dφ)IJ] (12) Now I think I have finally learned the reason for the Spivak normalization which in fact Sjamaar quietly uses on page 85 C. So I will need to work this in. In Spivak normalization we get above gJ(x) =Σ'I fI(y)det[ (Dφ)IJ] using our tensor-function starting point for a description of the pullback, (φ*α)x(v1,v2....vk) = αy(Rv1, Rv2.....Rvk) Since this gJ(x) is the same expression found in Chapter 3, it is a justified way to write the pullback in tensor function space. I could perhaps maintain my own normalization and replace the Spivak rule with this rule (φ*α)x(v1,v2....vk) = k! αy(Rv1, Rv2.....Rvk) (13) or something like that. The original rule appears here in Spivak (14) where which compare to (φ*α)x(v1,v2....vk) = αy((Dφ)v1, (Dφ)v2.....(Dφ)vk) The translation rules for Spivak to Sjamaar are: f → φ p → x f(p) → φ(x) ω(p) → αx ω(f(p)) → αy so the above Spivak equation becomes (φ*α)x(v1,v2....vk) = αy[ (Dφ)(x) v1, etc] The match is almost perfect, not sure why he has f(p) as subscript on [ (Dφ)(x) v1]y. So I am finally happy with (14) above from Spivak. Sja now transforms from y = φ(x) to x = ψ(t) on the bottom of page 88, so we have (ψ*α)t(v1,v2....vk) = αx((Dψ(t))v1, (Dψ(t))v2.....(Dψ(t))vk) // page 88 C Now you think of point x being on a manifold and you think of αx being an element of the cotangent space which is just the dual space to TxM. I still have many questions: why do I need tensor functions? why are we associating forms with the dual space functionals? clear up which sums are Σ and which are Σ' what is the connection to ordinary differentials?