clarify pullback of a differential form REVIEWED

Clarify meaning of the pullback of a raw differential form PhL 2.22.16 This doc has four parts : (1) clarification of the fact that dyI≠ dφI , now well understood by me (2) my original writeup of integration of 1-forms, 2-forms and then k-forms, now in Sections 10.11-13. (3) my original comments on tangent spaces TxM now in Section 10.2 (4) my original comments on manifolds now in Section 10.2 1. A misunderstanding of Sjamaar on pulling back a "differential" This misunderstanding has been corrected. In my reading and raw notes on Sja's pullback stuff in Ch 3, I had a misunderstanding and you can see it right in those raw notes. Here was my thinking: yi = φi(x) dyi = Σj (∂φi/∂xj) dxj dyi = dφi = same thing y = φ(x) dyI = Σ'J det[ (Dφ)I,J] dxJ dyI = dφI = same thing I was doing ordinary calculus adjusted slightly for the fact that there are wedges in the forms and that things are ordered so you get the odd looking "Jacobian". Basically I was thinking as the last line being a form of a Jacobian adjusted for non-square transformations. So with this point of view, to say that φ*(dyI) = dφI is treated as a merely cosmetic thing that really just means that dyI = dφI . This is how I describe things in my raw notes for Section 3.2. Today I realize that this is completely wrong. Although you do have y = φ(x), you do not have dyI = dφI. In fact, dφI is the pullback of dyI [ objects are in different spaces! ] . That is to say φ*(dyi) = dφi = Σj (∂φi/∂xj) dxj dyi ≠ dφi φ*(dyI) = dφI = Σ'J det[ (Dφ)I,J] dxJ dyI ≠ dφI Sjamaar uses variables y and x and shows the two spaces on page 37 and that is fine. BUT, what the above two lines really say is this φ*(λi) = dφi = Σj(∂φi/∂xj) λj λi ≠ dφi φ*(λ^I) = dφI = Σ'J det[ (Dφ)I,J] λ^J λ^I ≠ dφI where the λ are the dual space basis vectors as usual. If you wanted, you could replace λi with the "cosmetic name" dxi and write the above as φ*(dxi) = dφi = Σj(∂φi/∂xj) dxj dx ≠ dφ φ*(dxI) = dφI = ΣJ det[ (Dφ)I,J] dxJ dxI ≠ dφI The first above line is what we see in Spivak page 90 2. How do a differential pullback from the tensor function point of view All is well here, last equation appears in wedge Section 10.9 as (10.8.16) item F. // 5.17.16 The tensor function rule would be this (φ*λi)(v1) = λi(Rv1) = (Rv1)i = Σj Rij(v1)j = Σj Rijλj(v1) and then (φ*λi) = Σj Rijλj and inserting the cosmetic name for λi and R = (Dφ) (φ*dxi) = Σj (Dφ)ijdxj and this agrees with Spi quote above and with my Sja derivation. __________________________________________________________________________________ In the following sections, we illustrate the use of pullbacks to define the integration of differential forms over curves and "surfaces". 1.1 Line integral example of pulling back a 1-form in Rn Consider an integral in x-space, where we integrate a 1-form over a curve x = φ(t) where t is a single scalar variable (parameter) and x is a vector in V = Rn : α = Σi fi(x) λi = a 1-form (in x-space Rn) , sum runs i = 1 to n α ϵ Λ1(Rn) ∫φ α = ∫φ Σi fi(x) λi = integral of a 1-form over the curve φ in Rn The transformation x = φ(t) is a mapping φ: t → Rn. Officially it is the mapping φ which is the "curve", but we loosely refer to the image of this mapping in Rn as "the curve". The distinction is necessary because many mappings can have the same image curve, such as φ(t) and φ(2t) where the parameter is "re-speeded". So imagine that we have a 3D curved line having in R3 and as t varies perhaps from 0 to 1 in t-space, we move along the image curve in R3. The problem is how to integrate a 1-form along this curve. The traditional notation, motivated as shown below, is to give the dual-space basis vector λi the cosmetic name dxi . Whenever we use this cosmetic name, we make it red and italic. Whereas dxi is a calculus differential along the xi axis of x-space, dxi is the vector basis functional λi. For example dxi(v) = λi(v) = vi There is no object dxi(v) where dxi is a calculus differential. Our two lines above may now be written α = Σi fi(x) dxi = a 1-form (in x-space Rn) , sum runs i = 1 to n ∫φ α = ∫φ Σi fi(x) dxi = integral of a 1-form over the curve φ in Rn First definition: We then define the meaning of this integral as follows, ∫φ α ≡ ∫[0,1] φ*α = the integral in t-space over the 1-cube [0,1] On the left we have an integral of the 1-form α over a curve φ in Rn. On the right we have an integral of a different 1-form φ*α (the pullback of α) over a 1-cube [0,1]. Note that α ϵ Λ1(Rn) while φ*α ϵ Λ1(R). We shall give a formal definition φ*α below, but for the moment we use a set of "rules" describing how it works. Therefore: ∫φ α = ∫[0,1] φ*α = ∫[0,1] φ*[Σi fi(x) λi] = ∫[0,1] Σi φ*[fi(x)] φ*[λi] // rule φ*(fα) = φ*(f)φ*(α) The rule (definition) of the pullback of a function (a 0-form) is this: φ*[fi(x)] = fi(φ(t) ) = a function of t, the only variable of t-space The rule (definition) of the pullback of a basis vector λi of the dual space (a 1-form) is this φ*[λi] = (∂φi/∂t) λ = (∂φi/∂t) dt . Since W = t-space is a one-dimensional space, the dual space W* has a single basis vector λ. We can give this dual space basis vector the cosmetic name dt if we want. If t-space were Rm instead of R1, the right side would be Σj=1m(∂φi/∂tj) λj = Σj=1m(∂φi/∂tj) dtj , but in our current example m = 1. Installing the definitions ** and ** into ** we find ∫φ α = ∫[0,1] Σi fi(φ(t)) (∂φi/∂t) dt = ∫[0,1] Σi fi(φ(t)) (∂φi/∂t) λ = ∫[0,1] g(t) λ = ∫[0,1] g(t) dt g(t) ≡ Σi fi(φ(t)) (∂φi/∂t) Thus the integral of the 1-form α over the curve φ is defined to be equal to the integral of the 1-form g(t)λ over a 1-cube. So far there are no regular calculus integrals appearing. Second definition: ∫[0,1] g(t) λ ≡ !Syntax Error, Ig(t) dt On the left is the integral of a 1-form on a 1-cube, on the right is an ordinary calculus integral of a function over the interval [0,1] of the real axis. It is this second definition that motivates giving the dual space basis vector λ the cosmetic name dt. With this name, the above equation appears as ∫[0,1] g(t) dt ≡ !Syntax Error, Ig(t) dt and one then tends to forget that this is in fact a definition, since both sides pretty much look the same. When we do a 2-form example below, the sides don't quite look so much the same. The bottom line for our example then is this: ∫φ α = !Syntax Error, I[ Σi fi(φ(t)) (∂φi/∂t) ] dt and the right side is a normal well-defined one-dimensional calculus integral. So we have shown that ∫φ f(x) dx = !Syntax Error, I[ fi(φ(t)) (∂φi/∂t) ] dt where the integrand on the left is the 1-form f(x) dx = Σi fi(x) dxi = Σi fi(x) λi where λi is a basis vector in the dual space (Rn)* . This kind of integral is called a "line integral" though of course it is an integral over the curved line φ in Rn . One does not need to use "differential forms" and "dual spaces" and so on just to compute a line integral, but the above example shows how a line integral fits into the framework of differential forms and linear functionals. In dealing with 1-forms one encounters no wedge products, but see the next example. Comment: Authors often define a curve not as x = φ(t) but as x = γ(t) [Buck] or x = c(t) [Sjamaar]. In this case all occurrences of the symbol φ above get replaced by γ or c, eg, R = (Dφ) → (Dγ) or (Dc). 4. Surface integral example of pulling back a 2-form in R3 The most general 2-form in R3 can be written α = f1 dx2^dx3 + f2 dx1^dx3+ f3 dx1^dx2 but for our example we choose a special case just to keep things very simple, α = f1 dx2^dx3 = f1 λ2 ^ λ3 . One could regard this as a special case of a 2-form in (Rn)* as long as n ≥ 3. The objects dx2 and dx3 are merely cosmetic names for the dual space vectors λ2 and λ3 which are two of the three basis vectors of the dual space (R3)*. This 2-form α is thus an element of Λ2(R3). Assume now that x = φ(t) defines a surface φ in x-space. Here φ: R2 → R3. We now integrate the 2-form α over this surface in x-space: ∫φ α = ∫φ f1 λ2 ^ λ3 = ∫φ f1 dx2^dx3 First definition: We then define the meaning of this integral as follows, ∫φ α ≡ ∫[0,1]2 φ*α = ∫[0,1]2 φ*[f1(x) λ2 ^ λ3] = ∫[0,1]2 Σi φ*[f1(x)] φ*[λ2 ^ λ3] Here the object φ*α is known as the pullback of α along φ. The rule for (definition of) the pullback of a function (a 0-form) is this: φ*[f1(x)] = f1(φ(t)) = a function of t in t-space = R2 The rule for (definition of) the pullback of a basis vector λ2 ^ λ3 of the dual space R2* (a 2-form) is this φ*[λ2 ^ λ3] = (φ*λ2) ^ (φ*λ3) // an easily proven rule for φ* = [ Σj(∂φ2/∂tj)λj ] ^ [ Σj(∂φ3/∂tj)λj ] // sums run js = 1 to 2 = Σjj (∂φ2/∂tj)(∂φ3/∂tj) λj ^ λj. From now on we shall use our compact Chapter 2 R-matrix notation R = (Dφ) Rij = (Dφ)ij = (∂φi/∂tj) where the R matrix is the differential of the transformation x = φ(t). Obviously all these objects are functions of t, such as Rij(t), but it is economical to suppress this fact below. Then the above says φ*[λ2 ^ λ3] = Σjj R2jR3j λj ^ λj = Σjj R2jR3j dtj ^ dtj // cosmetic names dts for λs [Please do not confuse the matrix element R2j with Euclidean space R2. It just happens that we use the letter R for these two unrelated objects.] At this point we must do a step that was not present in the previous example. We need to rewrite the above pullback φ*[λ2 ^ λ3] as a 2-form expanded on the true set of basis vectors of the dual space (R2)* where we recall from ** that the (n,k) basis vectors of Λk are ordered basis vector wedge products. To this end we continue the above : = Σj<j R2jR3j dtj ^ dtj + Σj<j R2jR3j dtj ^ dtj = Σj<j R2jR3j dtj ^ dtj + Σj<j R2jR3j dtj ^ dtj // j1↔j2 in 2nd sum = Σj<j R2jR3j dtj ^ dtj + Σj<j R2jR3j [ - dtj ^ dtj ] // wedge rule = Σj<j [ R2jR3j - R2jR3j] dtj ^ dtj = Σj<j det dtj ^ dtj = Σj<j det λj ^ λj . If the surface φ were in R3, this sum would have three terms 1<2 , 1<3 , 2<3. If the surface φ were in R4, this sum would have six terms 1<2 , 1<3 , 1<4, 2<3, 2<4, 3< 4 . As a shorthand, we abbreviate the above 2x2 matrix this way, = RIJ = R[2,3][j,j] I = [2,3] J = [j1,j2] Now only properly ordered dual-space basis vector products appear in our expression for φ*[λ2 ^ λ3], φ*(λ2 ^ λ3) = Σj<j det ( R[2,3][j,j]) λj ^ λj . Note that λ2 ^ λ3 is a 2-form in Λ2(R3) whereas λj ^ λj and φ*(λ2 ^ λ3) are 2-forms in Λ2(R2). In multiindex notation the above pullback of λ2 ^ λ3 "along φ" can be written, φ*(λ^I) = Σ'J det(RIJ) λ^J I = [2,3] J = [j1,j2] Installing the definitions ** and ** into ** we find ∫φ α = ∫[0,1]2 f1(φ(t)) Σj<j det λj ^ λj = ∫[0,1]2 f1(φ(t)) Σj<j det( R[2,3][j,j]) λj ^ λj = ∫[0,1]2 f1(φ(t)) Σj<j det( R[2,3][j,j]) dtj ^ dtj // cosmetic names = Σj<j ∫[0,1]2 f1(φ(t))det( R[2,3][j,j]) dtj ^ dtj = Σj<j ∫[0,1]2 gjj(t) dtj ^ dtj gjj(t) ≡ f1(φ(t)) det( R[2,3][j,j]) = Σj<j ∫[0,1]2 gjj(t) λj ^ λj . // back to true basis vector names Thus the integral of the 2-form α over the surface φ is defined to be equal to the integral of the pullback 2-form Σj<j gjj(t) dtj ^ dtk over a 2-cube. So far no regular calculus integrals appear. Second definition: Σj<j ∫[0,1]2 gjj(t) λj ^ λj ≡ Σj<j!Syntax Error, I!Syntax Error, I Σjk gjk(t) gjj(t) dtj dtj . On the left is the integral of a certain 2-form on a 2-cube, while on the right is a certain sum of ordinary calculus double integrals of a function of two variables tj and tj. It is this second definition that motivates giving the dual-space basis vector λj the cosmetic name dtj. With these names, the above equation appears as Σj<j ∫[0,1]2 gjj(t) dtj ^ dtj ≡ Σj<j!Syntax Error, I!Syntax Error, Igjj(t) dtj dtj and one then tends to forget that this is in fact a definition, since both sides pretty much look the same. But now they are not quite the same. On the left the 2 form has a wedge product dtj ^ dtk which is an element of the dual space Λ2, whereas the calculus integral on the right has a normal calculus differential product dtidtj . The bottom line for our example is this: ∫φ α = ∫φ f1 λ2 ^ λ3 = ∫φ f1 dt2 ^ dt3 = ∫φ f1 dA23 = Σj<j!Syntax Error, I!Syntax Error, If1(φ(t)) det( R[2,3][j,j]) dtj dtj // R = (Dφ) Here we have added yet another cosmetic name dA23 ≡ λ2 ^ λ3 to make it more suggestive that we are doing a surface integral where dA23 suggests a piece of area dt2 ^ dt3 . This kind of integral is officially called a "surface integral". We can now redo the entire previous example assuming a general 2-form in R3 instead of the special example we used. That general form would be (using a slightly different notation that above), α = f23(x) dx2^dx3 + f13(x) dx1^dx3 + f12(x) dx1^dx2 = Σi<i fii(x) dxi ^ dxi = Σi<i fii(x) λi ^ λi . The development goes as shown above, where we quote just a few waypoints: φ*[ λi ^ λi] = Σj<j det ( R[i,i][j,j]) λj ^ λj or φ*[λ^I] = Σ'J det(RIJ) λ^J I = [i1,i2] J = [j1,j2] ∫φ α = Σi<i ∫[0,1]2 fii(φ(t)) Σj<j det( R[i,i][j,j]) λj ^ λj = Σi<i Σj<j!Syntax Error, I!Syntax Error, Ifii(φ(t)) det( R[i,i][j,j]) dtj dtj where the last item is a sum of conventional calculus double integrals. In multiindex notation we would write the various equations above as: α = Σ'I fI(x) λ^I I = [i1,i2] general 2-form φ*[λ^I] = Σ'J det(RIJ) λ^J I = [i1,i2] J = [j1,j2] pulled back λ^I ∫φ α = Σ'I ∫[0,1]2 fI(φ(t)) Σ'J det( RIJ) λ^I R = (Dφ) integral of 2-form = Σ'I Σ'J!Syntax Error, I!Syntax Error, IfI(φ(t)) det( RIJ) dtj dtj normal integral Comment: Authors often define a 2D surface not as x = φ(t) but as x = Σ(t) [Buck] . 5. The general case: pulling back a k-form in Rn Now x = φ(t) describes a "surface" of dimension k embedded in an n dimensional space Rn, so in this case φ: Rk → Rn. If k = 2 and n = 3, this would be a regular surface in R3. If n > 3 and k = n-1, one usually calls the "surface" a hypersurface, one dimension down from the embedding space. But we allow here any values k,n as long as k < n so x = φ(t) describes a "surface" of some dimension k within Rn. The task here is to integrate a k-form over the k-dimensional surface. We ignore here technical details concerning the possible shape of that surface (as we did in the previous examples), and these details are dealt with in a discussion of "manifold" type surfaces (manifolds) which can be treated as the union of easy-to-handle "surface patches". Our interest right now is defining the integral in terms of a pullback. The most general k-form has the following form α = Σ'I fI(x) λ^I I = [i1, i2....ik] i1< i2< ... < ik where as usual Σ'I indicates a sum of (increasing) ordered terms and λ^I ≡ λi ^ λi ..... ^ λi The pullback of α is given by φ*α = Σ'I φ*[fI(x)] φ*[λ^I] = Σ'I fI(φ(t)) Σ'J det(RIJ) λ^J J = [j1, j2....jk] We showed in the previous example how this form Σ'J det(RIJ) λ^J arises, and we show it more generally in Appendix XX. Recall that the transformation differential is given by R = (Dφ) Rij = (Dφ)ij = (∂φi/∂tj) φ: Rk → Rn. x = φ(t) So in our general example, index i on φi runs 1 to n, whereas index j on tj runs 1 to k. Thus, the R matrix is in general a non-square matrix having n rows and k columns. However, the submatrix RIJ is square, being kxk, so det(RIJ) is well defined. In fact det(RIJ) is just a minor of matrix R obtained by crossing out all rows other than those indicated by I and crossing out all columns other than those indicated by J. So the k-form α ϵ Λk(Rn) is pulled back to a k-form φ*α ϵ Λk(Rk). First definition: We then define the meaning of this k-form integral over surface φ in terms of a pull back integral over the unit k-cube, ∫φ α ≡ ∫[0,1]k φ*α = ∫[0,1]k φ*[ Σ'I fI(x) λ^I] = ∫[0,1]k Σ'I φ*[fI(x)] φ*[λ^I] = ∫[0,1]k Σ'I fI(φ(t)) Σ'J det(RIJ) λ^J = ∫[0,1]k Σ'I fI(φ(t)) Σ'J det(RIJ) (λj ^ λj .... ^ λj ) = ∫[0,1]k Σ'I fI(φ(t)) Σ'J det(RIJ) (dtj ^ dtj .... ^ dtj ) // cosmetic names dts Here the integral of the k-form α over the k-dimensional "surface" φ within Rn is equal to the integral of another k-form integrated over a unit k-cube. Second definition: The above pullback integral is then set equal to the following sum of calculus integrals. = (!Syntax Error, I!Syntax Error, I ....!Syntax Error, I) Σ'I fI(φ(t)) Σ'J det(RIJ) dtjdtj ....dtj = Σ'IΣ'J (!Syntax Error, I!Syntax Error, I ....!Syntax Error, I) fI(φ(t)) det(RIJ(t)) dtjdtj ....dtj To summarize, ∫φ α = ∫φ [Σ'I fI(x) λ^I] = ∫[0,1]k φ*[Σ'I fI(x) λ^I] = Σ'IΣ'J (!Syntax Error, I!Syntax Error, I ....!Syntax Error, I) fI(φ(t)) det(RIJ(t)) dtjdtj ....dtj where there are k integrals 0 to 1. In painful detail we can write this last expression as = Σi<i...<i Σj<j...<j (!Syntax Error, I!Syntax Error, I ....!Syntax Error, I) * ir ϵ 1..n, jr ϵ 1..k fii...i(φ(t)) det [ R[i,i...i][j,j...j](t) ] dtjdtj ....dtj where ir ϵ {1,2..n} , jr ϵ {1,2..k}, and R = (Dφ). This is a completely explicit formula for computing the integral of an arbitrary k-form over a "surface" of dimension k embedded in Rn, where as noted we ignore the "manifold" issues (but see below). 6. The tangent and cotangent spaces and notational issues Let's return to the example above of integrating a 2-form over some smooth surface like a partial torus embedded in R3. The partial torus exists as a surface in x-space and we shall refer to this surface as M. At some point x on the torus one can form a coordinate system such that one coordinate is normal to the surface and the other 2 are tangent to the surface. The two tangent vectors span what is known as the tangent space at point x, and is often written TxM. Suppose we denote the two tangent vectors of the tangent space TxM by e1 and e2, and perhaps e3 is then the normal vector. The problem with this notation is that if one moves from x on M to some nearby point x' = x+dx on M, the basis vectors in general move a little bit so that e3 can remain normal and e1 and e2 remain tangent. Thus a better notation for these basis vectors is something like (ei)x. So then we can more properly say that the basis vectors (e1)x and (e2)x span the tangent space TxM. One usually thinks of these tangent vectors having their tails glued to the point x on M of interest, but it is sometimes useful to think of these vectors as being translated to the origin. One can imagine the conglomeration of all the tangent spaces TxM for all x on M as forming a tangent bundle in the sense of a fiber bundle where there exists a different vector space at each point x, but we shall not pursue that avenue. Of more immediate interest is the fact that the tangent space TxM in our example of dimension 2 has a dual space which we might call (TxM)*. This is the space of rank-2 linear functionals defined on TxM. The basis vectors in (TxM)* are the λi functionals. Recall that λi(v) = <ei|v> = vi λi(ej) = <ei|ej> = δij where λi(v) is what we call a rank-1 tensor function, and λi is a linear functional. The point here is that if the basis vectors (ei)x of the tangent space TxM move as x moves on M, then so too must the basis vectors λi move with x, so we really should something like this xλi(v) = <(ei)x | v> = xvi xλi(ej) = <(ei)x|(ej)x> = δij Thus the various λi basis vectors appearing in the above examples really should have extra labels and in fact that does help to distinguish them. For example, in our 2-form example above we had φ*[λ2 ^ λ3] = Σjj R2jR3j λj ^ λj where the 2-form λ2 ^ λ3 exists in Λ2(R3) where R3 was our x-space. On the other hand, the λj ^ λj dual vector on the right here exists in Λ2(R2) which was our t-space. So it would be better to write the above equation as φ*[xλ2 ^ xλ3] = Σjj R2jR3j tλj ^ tλj This extra notation shows first of all which space a given linear functional λr belongs to, and secondly it shows that the functional xλr varies as x varies. Instead of writing (ei)x for a basis vector in TxM, one could write ei(x). But then one wants to write λi(x) as the corresponding dual vector. This notation is rather inconvenient when one wants to write out the corresponding tensor function [λi(x)](v), but it is workable and Spivak uses it (our x he calls p). Now the space-identification benefit of writing λ2 as xλ2 is also provided by the "cosmetic" notation that xλ2 = dx2 and similarly tλj = dtj. The fact that xλ2 is a function of x (on M) is not particularly demonstrated by the notation dx2 so one must just keep this in mind. Normally in Euclidean space the calculus differential dx2 is not a function of x, but xλ2 = dx2 is a different animal. So one could revamp the examples above by adding left subscripts t and x as appropriate on all λ dual basis vectors which appear. We have seen here a sort of added "benefit" of the cosmetic notation. Recall that in the notation λi = <ei|, one refers to ei as a covector and then λi(v) = <ei|v> = ei v. The prefix "co" associates the vector with the dual space V* of V. So it is perhaps natural to refer to the space (TxM)* which is dual to the tangent space TxM as the cotangent space at point x. This term cotangent has nothing to do with the geometric cotangent of any angle (though doubtless one could concoct something). The conglomeration of cotangent spaces (TxM)* for x ϵ M is then some kind of cotangent bundle, a dual fiber bundle. 7. Comments on Manifolds We shall not present a formal theory of manifolds here, Sjamaar has a very readable discussion. Instead, we just make a few comments related to manifolds. Everyone knows that the surface of the Earth can be completely covered by a full set of maps or charts (maps of regions of water are in fact normally called charts). The complete set of charts is called an atlas. Typically a chart overlaps other abutting charts at its edges, and things have to be "consistent" in such an overlap region of two charts, meaning that both charts must present the same data in that region. For each individual chart (t-space), there is a smooth mapping to a patch on the Earth globe which is embedded in x-space. For a chart indicated by subscript i, this map might be x = φ(i)(t). One may not be able to handle integration over a manifold with a single such mapping (a single chart), but it is always possible to cover a manifold with a finite set of mappings x = φ(i)(t). By integrating carefully over each patch of the globe, pulling back the 2-form there to a 2-form on a rectangle in t-space, one can integrate a 2-form over the entire Earth. This idea of course applies to a manifold M ("surface") of dimension k within Rn. One papers the "surface" M with sufficiently small k-dimensional pieces of paper and each such patch of M can then be mapped back to a chart and the integral of a k-form on that patch is pulled back to become an integral of a k-form in the chart's t-space just as we did in the examples above. There are various technical requirements a manifold must have. For example x = φ(i)(t) must be a 1-to-1 mapping (the manifold cannot self-intersect), and the differential R(i) = (Dφ(i)) must be 1-to-1 as well. There are additional continuity requirements. Here is a simple example of a manifold (a circle in R2) being covered by two charts. Most of the circle is "covered" by the larger chart whose t-space is the red line segment at the top. But a small part of the circle shown in blue is covered by a second chart whose t-space is the short lower blue line segment. There is some overlap of the charts. So to integrate a 1-form over this circle = manifold, we could do two pullbacks of 1-forms. The circle cannot be covered by just the upper chart extended to be very long because then the bottom point on the circle will correspond to both ends of the red segment and then the mapping is not 1-to-1. There is in general an issue of manifold smoothness which is well illustrated by the following nice drawing from Spivak's book (page 110), Here a 2-form on the dark gray curved patch on the torus (in x-space) can be pulled back to a 2-form on the gray square 2-cube on the right (our t-space). The lighter gray regions of 1 higher dimension show a sort of required "elbow room" one must have around the patches in both spaces. For example, for a point x on the curved patch one needs to be able to differentiate cleanly in all directions in order to obtain the differential R = (Dφ) which appears in the pullback formulas above. Our map φ is called h in Spivak's picture, and he would like this mapping to have C∞ continuity in both directions φ and φ-1, which mapping is called a diffeomorphism because it is infinitely differentiable in both directions. Sjamaar has a somewhat lesser requirement for the mapping φ. ____________________________________________________________________________ Comment: If you don't get things into ordered form, you get the wrong answer! Let's back up in the above to this point: ∫φ α = ∫[0,1]2 f1(φ(t) ) Σj,k=12 R2jR3k λj ^ λk Rewrite as ∫φ α = ∫[0,1]2 f1(φ(t) ) Σj1,j2=12 R2j1R3j2 λj1 ^ λj2 = ∫[0,1]2 f1(φ(t) ) ΣJ R[2,3]J λ^J where ΣJ is the full symmetric sum and I = [2,3] . Then ΣJ RIJ λ^J = Σj1,j2=12 R2j1R3j2 λj1 ^ λj2 = (Σj1<j2 + Σj1>j2) R2j1R3j2 λj1 ^ λj2 = Σj1<j2 [ R2j1R3j2 - R2j2R3j1] λj1 ^ λj2 = Σj1<j2 det λj1 ^ λj2 = Σj1<j2 det RIJ λj1 ^ λj2 = Σ'J det RIJ λ^J Then we have ∫φ α = ∫[0,1]2 f1(φ(t) ) Σ'J det R[2,3]J λ^J = ∫[0,1]2 f1(φ(t) )Σ'J det R[2,3]J λj1 ^ λj2 = ∫[0,1]2 f1(φ(t)) Σ'J det R[2,3]J dtj1 ^ dtj2 NOW, after doing this, suppose we use our second definition to get ∫φ α = !Syntax Error, I!Syntax Error, I f1(φ(t)) Σ'J det [R[2,3]J] dtj1 dtj2 = calculus integral (1) Here then is my question. Is this the same result as going the other path done earlier above, ∫φ α = ∫[0,1]2 f1(φ(t) ) Σj1,j2=12 R2j1R3j2 dtj1 dtj2 (2) No, this is NOT the same result! Just compare the sums: Σ'J det [R[2,3]J] dtj1 dtj2 = Σj1<j2 [ R2j1R3j2 - R2j2R3j1] dtj1 dtj2 (1)' Σj1,j2=12 R2j1R3j2 dtj1 dtj2 (2') This is a disaster because it includes dt1dt1 which is nonsense. Even if we ignore those terms, we would have Σj1,j2=12 R2j1R3j2 dtj1 dtj2 = (Σj1<j2 + Σj1>j2)R2j1R3j2 dtj1 dtj2 = Σj1<j2R2j1R3j2 dtj1 dtj2 + Σj1<j2R2j2R3j1 dtj2 dtj1 = Σj1<j2 [ R2j1R3j2 + R2j2R3j1] dtj1 dtj2 which is a completely different result!!!