Sja Ch 1-5 notes

Notes on Sjamaar PDF PhL 6.29.15 Had to stop on this subject on 7.20.15 due to Cape Cod. After a gap of 20 days, on 8.10.15 I read through these notes but only up to the start of Chap 3. The Chap 3 notes are a bit confused, so I then switched over to the meta doc and read it all the way through. On 8.12.15 my review is up to the start of Chap 6. Chapter 1: Introduction 3 1.1 Manifolds 3 1.2 Equations 5 1.3 Parametrizations. // several spellings are allowed 6 1.4 Configuration Spaces 6 Chapter 2: Differential Forms on Rn 7 2.1 Elementary Properties of differential forms on Rn 8 2.2 The Exterior Derivative [20] 10 2.3 Closed and Exact Forms [22] 13 2.4 The Hodge star operator [23] 14 2.5 div, grad and curl [24] 16 Chapter 3: Pulling Back Forms 19 3.1 Determinants 20 3.2 Pulling back forms 23 Chapter 4: Integration of 1-forms 32 4.1 Definition of the integral and its elementary properties [47] 32 4.2 Integration of exact 1-forms [49] 36 4.3 The global angle function and the winding number [51] 39 Chapter 5: Integration and Stokes' theorem 39 5.1 Integration of forms over chains 40 5.2 The boundary of a chain 45 5.3 Cycles and Boundaries 47 5.4 Stokes's Theorem 50 Chapter 6: Manifolds // see separate doc 58 I will do mark up right in his PDF. // Had to give up on that, something wrong with PDF. I will mark up a hard copy instead. Here is data on the other PDF I have which is 153 pages. Recently minted PhD, 1990! (that is to say, a mere 25 years ago). http://www.math.cornell.edu/m/People/Faculty/sjamaar Does he have updated notes on this site? http://www.math.cornell.edu/~sjamaar/ Here is his recent teaching activity so he is still doing his "manifolds and forms" classes. His notes on the site are the same 8.26.06 notes which I already have, so I know at least I am current. He does not seem to have notes on his other classes. Preface. Notes are for undergrads! Two more advanced books are Spivak and Flanders [on 2.11.16 I found Flanders as a djvu, and have long had Spivak.] . Says he has intended since 2006 to update these notes, but as of 2015 he has not done so. Maybe they are good and stable! [ he did an update July 2015, but I continue to work with the older notes since I have invested heavily in them with markup. ] I notice that like my early tensor doc, this 153 page PDF has no equation or figure numbers. Chapter 1: Introduction 1.1 Manifolds He writes column and row vectors exactly as I do! En is called Rn. A manifold will get defined gradually, but the first fact is that a manifold is a subset of Rn. Recall that "set" is more general than "subspace" and I think this is a theme of manifold theory-- to keep things as general as possible. A 1D manifold is a smooth non-intersecting curve. It can exist in En for n ≥ 2 I would say. A trivial curve can exist in E1, it would be just a piece of the x axis. [2] He shows curves which are closed, infinite, open at one end and closed at the other, and a circle has a missing point and can be opened topologically as shown. All these examples are manifolds. I think one idea is that each point on the curve has to have a unique tangent, and this cannot be true for an intersection. The bow tie is treated as self-intersecting so is not a manifold. [3] The spiral shown I think is the curve r = k|θ| in polar coordinates. I just made my own For negative θ, since r < 0 it gets drawn as r>0 with θ+π. For only positive θ you get, S shows how you can "pull the upper spiral apart" by adding a z coordinate to the curve so you then have a curve in 3D space. I just did this: The curve lies on a cone which has shape (x2 + y2)1/2 = z which in our case is r = z or θ = θ. [4] Examples of a 2D manifold are shown. First we see cases where they lie in a plane R2. One example is an open square with no boundary shown in dashes. Another example is closed and has a boundary and even has holes. The boundary here is finite. The closed square has non-smooth corners so is not a manifold. More interesting 2D manifolds are then shown in R3 . [5] Just as I would do if I were writing this and knew that the author knows, I would get into the Mobius strip. It is a manifold (though not orientable). He shows you would "make" a Mobius band (as he calls it) from a square piece of paper. The second picture shows how you would make a Klein bottle by first making a cylinder, then twisting it to get the bottle. The surface in this case has to pass through itself and therefore self-intersects, and thus is not a manifold. See https://en.wikipedia.org/wiki/Real_projective_plane for some pictures. That site also deals with the three objects that S treats. They seem to give some plots of a "projective plane" object. Claim is made that a Klein bottle in R4 can have no intersections. I have not followed every detail of this section which I am sure is dear to S. 1.2 Equations [7] You can usually describe a manifold by one or more equations. If there are fewer equations that there are unknowns, you get some kind of "surface" which could be a manifold. His symbol for surface is φ (bolded) whereas Bucks always use Σ. You can write φ(x) = c to represent a set of equations where the ci are all constants. Equations in general are non-linear. If the equations are all linear, then you really have a matrix equation Φ x = c (generally Φ not square) and the manifold you get in this case is called an affine subspace and S teaches this in his linear algebra course. I think the solution of Φ x = c is a surface that is planar within En of some dimension ≤ n-1 depending on the rank of Φ. These planar "flats" generally don't pass through the origin, so they are not official linear subspaces (linear subspace must include the origin). It is said that an affine subspace in En is the set of all linear subspaces of En which are then translated an arbitrary distance away from the origin. [???] Example 1.1 [7] Φ x = c where En = E3 and there are m = 2 equations. The equations are a unit circle in the plane, and a tilted plane z = -y. You can extrude the circle into a cylinder (as I have done in Lagrange) and then you see that the intersection is a tilted plane with a cylinder, which turns out to be an ellipse. This ellipse is in fact a manifold of dimension 1, meaning a curve, but we have not really defined dimension yet in detail. Aside: In Lines I claim that an ellipse in 3D looks like: Suppose I try that here. then I get A2cos2(ωt-a) + B2cos2(ωt-b) = 1 and Bcos(ωt-b) + Ccos(ωt-c) = 0 The second equation says B2cos2(ωt-b) = C2cos2(ωt-c) which suggests b = c and B = ±C, so put that equation aside for a while and look at the first equation. I know this is a rotated 2D ellipse, but the details elude me, so let it ride. [8] Example 1.2. Defines ||x|| as norm and x y as scalar product in Rn. In 3D, a sphere is a 2D surface which has the name S2. So a sphere in Rn is then called Sn-1. The equation is just ||x|| = r which is one equation in n unknowns. Another example of an equation defining a manifold. If you call this manifold M, then you have M = {x in Rn | ||x|| = r} // set notation The tangent space concept Consider the sphere situation, and pick a point x on the sphere. I know there is a tangent plane which has some basis vectors (two) all of which are perp to the normal at x. This plane does not pass through the origin and can be described by plane(x) = {y in Rn | (y-x)x = 0} = all points y such that the vector from x to that point is perp to the normal vector. For spherical our surface, you can also say (since x x = r2) plane(x) = {y in Rn | yx = r2} . This plane is an affine subspace. If you were to translate this plane to pass through the origin, it would be a genuine subspace. The official "tangent space" for a point x is taken to be the actual tangent plane translated in this way, so then one has tangent space at point x on manifold Sn-1 = {y in Rn | yx =0} = TxM Here we see a famous notation: TxM means "the tangent space at point x in M" // contains origin. Now suppose M = circle in E2. The tangent space would be a line through the origin, so in this case one would have dim(TxM ) = dim(M) = 1. But suppose we had M = a circle in E3. In this case I think TxM would be any plane in E3 which is tangent to the circle at point x on the circle. In this case, dim(TxM) = n-1 of M has dimension M. Maybe it is true that dim(M) + dim(TxM) = n ?? In any event, TxM is a true linear subspace in En. Passing note that a closed ball is a locus described by an inequality ≤ rather than an equation. 1.3 Parametrizations // several spellings are allowed [9] Describe points on a circle by parameter θ, or points on a sphere by θ,φ (two parameters). S points out that typically you cannot arrange things so that every point is "covered only once" on the surface or manifold which is being described by the parameters. I am quite used to this problem. 1.4 Configuration Spaces S is getting into some physics now. If you have a double pendulum of two connected math rods when any "state" of the system can be described by θ1,φ1, θ2,φ2 .The space spanned by these 4 parameters is called the configuration space for this system, and I suspect that space is a 4D manifold. Here we have 4 "degrees of freedom". [10]. On this page he does the spherical pendulum θ,φ (Example 1.3) and then the double spherical pendulum (Example 1.4). This is really the kinematics of the system, whether or not there is gravity does not matter. He wants you to think of θ1,φ1, θ2,φ2 = (θ1,φ1) x (θ2,φ2) = Cartesian product. [11] A Rigid body has 6 degrees of freedom which are translation and Euler angles. Example 1.6 is more detailed. Quadrilateral in E2 with four fixed rods with universal joints! How many degrees of freedom are there, ignoring overall translations in E2 ? It must be ≤ 4 since you could define 4 angles. Looking at page 11, if we keep edge AB fixed, you only have only one angle θ for BC as his drawing shows. You would then have 2 translation degrees of freedom for point A, and then one rotational degree φ to make edge AB. So I think the total is 4. If this were a rigid 2D object, you would only have a total of 3 degrees of freedom: 2 translation and one rotation. Imagine that edge b is very small relative to other edges. As C goes around in the circle shown, point D would do only a partial circle bounded by some angles. So D oscillates back and forth on a circular arc, and then C is the "wheel" and D is the "pump", like an oil well pump. Since rods can cross, there is then a separate arc locus below. Now what is "the manifold" in this elaborate example? We have the two angles for configuration space, ignoring the other 3 degrees. If this were a mapping, there are two different functions being mapped, a 1:2 mapping. Each mapping has a circular domain, the ranges do not overlap. So S wants to think of there being two circles on the left, and they are disjoint in some sense that is not quite clear to me and which he does not clarify. [12] Example 1.7 has a+b = c+d and I think this creates this situation where now the two mappings meet at one point. He somehow gets this to be two circles touching at one point. I am afraid he has just lost me on these drawings. Well, think of two circles as two Riemann sheets that touch at that point. The actual configuration space would look something like this and this shows all legal states of the system in angle parameter space. Chapter 2: Differential Forms on Rn 2.1 Elementary Properties of differential forms on Rn [17] Notice the word "on" in the title. He credits differential forms to Cartan in 1900, now sure that happened to Grassmann. Differential form: α = ΣI fI(x) dxI on Rn // multi-index, see my Buck notes. He uses Buck notation without the wedge symbols so has dx1dx6 for example. For a k-form, there are k factors of dxi so there are then k indices i1, i2....ik which must all be different. You should always say what space Rn you have in mind for your differential form. The single function fI(x) might only be defined on some open subset U of Rn. ( not clear why open is required). Example: U = Rn - the z axis = Rn - {(x,y,z) | x2+y2 ≠ 0}. // all cylinders except the line Then fi(x) = ln(x2+y2)z is defined on U but not on Rn. Note the traditional use of Greek letters for differential forms! [18] We think of the dx1dx2....dxk as describing a little differential cube within Rn and this product is the volume. S then just states that swapping any pair of dxi negates the volume, and thus dx2dx2 = 0. No justification is given at this point for this Minus Sign Rule. S uses a hat notation to mark something that is missing, and I once used a slash notation for the same thing. If n = k, then the k-form is the full volume form. [19] S then does the ordered multi-index idea where there are then (n,k) unique terms, accounting for all the swap rules. I derive this in the Buck notes for Ch 7. Definition: Ωk(U) = collection of all k-forms on U in Rn. This set is a vector space because: (1) you can add two k-forms; (2) you can multiply a k-form by a constant; (3) there is an additive 0 and a multiplicative 1; (4) a few more basic axioms are met. α = 0 thus exists as a differential form. I guess this is a 0-form with f = 0. And β = 1 with f = 1 would also exist. Dim( Ωk(U)) = ∞ because there are an infinite number of smooth coefficient functions you could choose. [p 19] Suppose all fI are constants by edict. This gives you some subset of Ωk(U). You can still add these items and multiply them by a constant and there is still 0 and 1, so this is a linear subspace of Ωk(U). Ignoring the constants, think of this space spanned by the (n,k) different dxI. Thus, the dimension of this nameless subspace of Ωk(U) is (n,k). Note Added 2.11.16. Let's call the above subspace Ωk(U)c. Elements are like 2 dx + 3 dy where the fI functions are constants. Why is this subspace a vector space? Closed under addition. Has zero element which might be 0 dx for example. Scalar mult cleanly defined. So OK. Now what is meant by a basis for this subspace? If n = 2 and k = 1 then basis is {dx,dy} because can construct any element of Ω1(U)c as lincomb of these elements. If n=3 and k = 2, then basis is {dxdy,dxdz,dydz}, same idea. In general, for a given n and k, there will be (n,k) elements in ths basis which have the form dxI. What is the dimensionality of this subspace Ωk(U)c ? It equals the number of elements in the basis, so the dimension of this space is (n,k). Now go back to the larger vector space Ωk(U). What is a basis here if n = 2 and k = 1? dx, xdx, y2dx , ...... these are all elements of Ωk(U) In the previous case, you could have said that the space has to include dx, 2dx, 3dx, 7.1 dx, ..... but all these elements are "covered" by the one element α dx where α is a scalar, meaning a real number. in the Ωk(U) case the items list above cannot be "covered" by a single element f(x,y)dx because in general f(x,y) is not a number. You cannot reach y2dx by α dx where α is a scalar. So in my example, you have to include all possible elements in the basis having the form f(x,y)dx, g(x,y)dy and each group is an infinite set since there are an infinite number of functions f(x,y). So the idea that you "span" the vector space Ωk(U)c with the dxI does not apply to Ωk(U). Note Added 2.11.16. Above I just noted that the dimension of Ωk(U)c is (n,k). In wedge doc I note that the dimension of Lk (non-dual wedge product space) is (n,k), as is the dimension of Λk for the corresponding dual wedge space. The basis elements are e^I and λ^I for these two spaces. At this point in Sja, I have no reason to connect either of my wedge doc spaces with the dx^I objects which are the basis elements of Ωk(U)c . [p 19] Definition: Exterior product of two differential forms. Assume one is a k-form and the other an m-form. The result is a k+m form. α = k-form = ΣI fI dxI β = m-form = ΣJ gJ dxJ αβ = (ΣI fI dxI)( ΣJ gJ dxJ) = ΣI,J fI gJ dxI dxJ // exterior product of two diff forms // page 19 B where note the ordering of the dxi stuff! As S points out, lots of terms will likely vanish in such an exterior product. Example: Consider κ ≡ αp = a differential form since products are allowed, where α is a k-form. The degree of this k form is then kp, meaning the number of factors of dxi. If kp exceeds n, then you must have two the same and the form vanishes. Now, when is kp > n ? k = 1 p > n k = 2 p > n ... k = k p > n In any of these cases, if p > n, the form vanishes. Thus: αp = 0 if p = n+1 regardless of k as long as k > 0 αn+1 = 0 regardless of k as long as k > 0 page 19 C Notes added 2.11.16. Here I will construct some text to be later added to wedge doc which involves the two ideas presented just above. The first affected area is Chapter 7 which does non-dual wedge k. I talk there about the wedge product of 3 or more tensors and this is the general result, (T1)^^(T2)^^...^(TN)^ = ΣI (T1IT2I .... TNI) e^I = ΣI (T1T2....TN)I e^I (7.9.d.6) where e^I = ei^ ei .....^ ei = ei^ ei .....^ ei and (T1T2....TN)I = T1IT2I .... TNI . Now suppose all these tensors are the same and are just T of rank k. Then this becomes, (T^)N ≡ T^^T^^...^T^ = ΣI (TITI .... TI) e^I = ΣI (TT....T)I e^I (7.9.d.6) where e^I = ei^ ei .....^ ei = ei^ ei .....^ ei , κ = Nk and (TT....T)I = TITI .... TI . How many distinct ei are there? Only n if dim(V) = n. So e^I = 0 if κ > n which is to say if Nk>n which is to say if N > (n/k). Thus we seen to arrive at this new rule; (T^)N = 0 if N > (n/k) Now as a special case, suppose T^ is a rank-1 tensor (vector) so k = 1. We then conclude that (V^)N ≡ V ^ V...^ V = 0 if N > n. For example, (V^)n+1 = 0 and this is then the above result αn+1 = 0 . However, this will turn out to be in the dual space, so let's repeat the above discussion in that case. (T1)^^(T2)^^...^(TN)^ = ΣI (T1)I(T2)I .... (TN)I λ^I = ΣI (T1T2....TN)I λ^I (8.9.d.6) where λ^I = λi^ λi .....^ λi = λi^ λi .....^ λκ and (T1T2....TN)I = (T1)I(T2)I .... (TN)I . Now suppose all these tensors are the same and are just T of rank k. Then this becomes, (T^)N ≡ T^^T^^...^T^ = ΣI (T)I(T)I .... (T)I λ^I = ΣI (TT....T)I λ^I (8.9.d.6) where λ^I = λi^ λi .....^ λi = λi^ λi .....^ λκ and (TT....T)I = (T)I(T)I .... (T)I . How many distinct λi are there? Only n if dim(V*) = n. So λ^I = 0 if κ > n which is to say if Nk>n which is to say if N > (n/k). Thus we seen to arrive at this new rule; (T^)N= 0 if N > (n/k) Now as a special case, suppose T^ is a rank-1 tensor (vector) so k = 1. We then conclude that (α)N ≡ α ^ α...^ α = 0 if N > n. For example, (α)n+1 = 0 and this is then the above result αn+1 = 0 . Theorem 2.1. βα = (-1)kmαβ α = k-form β = m-form We know that αβ = ΣI,J fI gJ dxI dxJ βα = ΣI,J fI gJ dxJ dxI So the question is: what happens when you slide all of dxI through dxJ ? Slice first dxi through and pick up (-1)k since need to do k swaps Then slide second dxi through, get another factor (-1)k Do this m times and you then have (-1)km QED. He shows this in p 19 D. Fine. Note that βα = ± αβ, so you almost have the two forms commuting, but you might pick up a sign. This relation is then called "graded" commutivity. Corollary 2.2: α2 = (-1)k*k α2 for α = any k-form so that α2 = 0 for k = odd. Notice that S is sneaking in some numbers that are like equation numbers, but instead are the numbers of theorems (propositions) and corollaries [ I will add this idea as well to wedge doc *** ] 2.2 The Exterior Derivative [20] This phrase refers to applying operator "d" onto a differential form "α" of any order. Only the coefficient functions are "hit" by the d operator, the dxi are constants. S starts on page 20 applying d to a 0-form to get a 1-form. The d operator does this to any coefficient: d(f) = (∂if) dxi and this adds a dxi factor raising the order of the form you acted upon by one. Question added 2.11.16. Why are the dxi treated as constants? You could also ask in wedge doc why ei and λi might be treated as constants. This entire Sja Chapter 2 has "Euclidean space" in its title, which I call Cartesian space. In that case, I would argue that yes, ei and ei are constants in space and so λi is a constant and eventually when we set λi = dxi we also have dxi = constants in space. So that is my temporary answer to this question. Then all the math of this section is justified. Acting on a general 1-form in Rn is treated bottom of page 20. Here is the result dα = d( dfidxi) = d(dfi)dxi = [ (∂jfi) dxj]dxi = Σi<j (∂jfi) dxjdxi + Σj<i (∂jfi) dxjdxi = Σi<j (∂jfi) dxjdxi (as was) + Σi<j (∂ifj) dxidxj (swap dummy sum indices) = Σi<j [ (∂ifj) - (∂jfi)] dxidxj . (2.2) Here you see the order increase by 1, and you see a very "curl like" structure formed in the bracket, but is true only for R3. In that case, (∂ifj) - (∂jfi) = εijkεkbc∂bfc = εijk[curl(f)]k = εijk[ x f ]k So for R3 I guess we have this result dα = Σi<j [ εijk[ x f ]k ] dxidxj (2.2a) I think in tensor doc I already claim that the form shown in (2.2) is the generalization of the curl. Yes, except it is all covariant derivatives. I can see that we will have to redo all this work in the covariant world at some point! [21] Now try a general 2-form in Rn : It is important to start out right!!! Here is α: α = Σi<j fijdxidxj dα = Σi<j [dfij]dxidxj = Σi<j [Σk(∂kfij)dxk]dxidxj = Σi<j Σk (∂kfij) dxkdxidxj Here Σk is over all k initially, but to get a non-zero term it cannot equal i or j. There are three possible positions k can therefore take relative to i and j: Σi<j Σk = Σk<i<j + Σi<k<j + Σi<j<k where I don't bother to show the extremal limits 1 and n for Rn. So write the three terms dα = [ Σi<j<k + Σi<k<j + Σk<i<j ] (∂kfij) dxkdxidxj = Σi<j<k(∂kfij) dxkdxidxj + Σi<k<j (∂kfij) dxkdxidxj + Σk<i<j (∂kfij) dxkdxidxj . Now the plan is to do dummy index swaps to get the last terms to look like the first. The second term requires only a single swap k↔j to give Σi<k<j (∂kfij) dxkdxidxj = Σi<j<k (∂jfik) dxjdxidxk . But the last term requires three swaps which in effect do a backwards cyclic, so we get kij→ijk Σk<i<j (∂kfij) dxkdxidxj = Σi<j<k (∂ifjk) dxidxjdxk . Now all three terms are Σi<j<k and we then have dα = Σi<j<k { (∂kfij) dxkdxidxj + (∂jfik) dxjdxidxk + (∂ifjk) dxidxjdxk } The next step is to get the dxi into ijk order in each term, so dα = Σi<j<k {+ (∂kfij) dxidxjdxk - (∂jfik) dxidxjdxk + (∂ifjk) dxidxjdxk } = Σi<j<k { (∂kfij) + (∂ifjk) - (∂jfik) } dxidxjdxk // agrees with (2.4) The object in {..} is then some higher tensor form object. Some Rules 2.5 1. d(Aα + Bβ) = A(dα) + B(dβ) α,β are any diff forms of any order, A,B constants 2. d(αβ) = (dα) β + (-1)k α (dβ) α = k-form β = l-form Proof of Rule 1: d(Aα + Bβ) = d(A ΣI fI dxI + B ΣJ gJ dxJ) = A ΣI (dfI) dxI + B ΣJ (dgJ) dxJ = A(dα) + B(dβ) QED Proof of Rule 2 d(αβ) = d[ (ΣI fI dxI)(ΣJ gJ dxJ) ] = ΣI ΣJ d(fIgJ) dxI dxJ = ΣI ΣJ [ (dfI)gJ + fI(dgJ)] dxI dxJ = ΣI ΣJ (dfI)gJ dxI dxJ + ΣI ΣJ fI(dgJ) dxI dxJ Now comes a key point. The things (dfI) and (dgJ) each contain a dxi factor, so you cannot slide things around willy nilly. So note that in the second term, (dgJ) dxI = (sign) dxI(dgJ) where the sign arises from sliding the single dxi through all the dxI guys. But the dxI goes with α which is a k-form, so there are k factors we have to slide dxs through, so you really do get (-1)k = sign. So = ΣI ΣJ (dfI)gJ dxI dxJ + (-1)kΣI ΣJ dxI fI(dgJ) dxJ = [ΣI(dfI)dxI ][ΣJ gJdxJ] + (-1)k[ΣI fI dxI][ ΣJ(dgJ) dxJ] = (dα) β + (-1)kα (dβ) . QED Note that for a 0-form (function) you get the "usual" expected result d(αβ) = (dα)β + α(dβ). Remember that βα = (-1)kl αβ from 2.1-- forms do not trivially "commute". So page 21 was chock full of stuff. [22] Rule 2.6 that d2α = 0 Proof: α = ΣI fIdxI dα = ΣI (dfI)dxI = ΣIΣk (∂kfI) dxkdxI d(dα) = ΣIΣk [d(∂kfI)]dxkdxI = ΣIΣkΣj [ ∂jkfI]dxjdxkdxI = 0 by symmetry! His proof is a little different: α = ΣI fIdxI dα = ΣI (dfI)dxI = ΣIΣk (∂kfI) dxkdxI = ΣI { Σk (∂kfI)dxk}dxI Then d(dα) = ΣI d{ Σk (∂kfI)dxk} dxI Notice that {...} is a 1-form. Recall from earlier that when α is a 1-form,. dα' = Σi<j [ (∂if'j) - (∂jf'i)] dxidxj where α' = Σk f'kdxk . So now let α' = Σk (∂kfI)dxk so that the role of f'k is played by (∂kfI) . Then d[Σk (∂kfI)dxk] = Σi<j [ (∂i[∂jfI]) - (∂j[∂ifI)] dxidxj = Σi<j [0] dxidxj = 0 and therefore above earlier we have d(dα) = ΣI d{ Σk (∂kfI)dxk} dxI = ΣI 0 dxI = 0 OK, but I think my proof is a lot simpler and does not need to draw on other theorems. Comment: How interpret d2α = 0? For a function in regular calculus you would get df = ∂idxi d2f = ∂ijdxidxj ≠ 0 So the whole reason d2α = 0 in "form world" is that asymmetry dxidxj = - dxjdxi and this is really a property of the wedge product which Sjamaar is suppressing throughout his doc so far. You have to keep in mind that in a form the object dxdy is not what it appears to be. 2.3 Closed and Exact Forms [22] If dα = 0, the form α is closed. If dα = β, the form β is exact. It is like an "exact differential" in regular calculus. β exact β closed: dβ = d(dα) = d2α = 0 so β is closed // β not closed β not exact ** If α = 0-form, α = f is closed if all fi = 0, so f is a constant. Only constant 0-forms are closed. If β is closed, it may be exact, but it depends on topology! Coming later. When is a 1-form α exact? Must find g such that dg = α. Gives system of n 1st order PDE's in (2.5): ∂ig = fi for i = 1.2..n . (2.5) If α is not closed, there can be no solution by **. Now recall that for a general 1-form in Rn, dα = Σi<j [ (∂ifj) - (∂jfi)] dxidxj . (2.2) To make sure α is closed before you test for exactness, you need to show (∂ifj) = (∂jfi) which is p 22C. Or you just show directly that dα = 0. So don't waste time trying to solve (2.5) until you show (∂ifj) = (∂jfi). Example 2.9. We are given a messy 1-form α in R3 We find that dα = 0, so closed, so can maybe find g with α = dg. S shows in detail how you solve the PDE system (2.5) ["integrating the system"], and the solution g is stated where dg = α. I remember this kind of PDE solution method I think from Sneddon. [23] Theorem without proof: Every closed 1-form α on Rn is exact! You are supposed to show this by mimicking the example just done. Exercise 2.7 addresses this proof. Note that α must be defined on all of Rn for this theorem to be true. Example 2.10: The Angle Form (a 1-form) on R2 with origin removed. They (with me) show that dα = 0 so this angle form is in fact closed, but we will later see it is not exact. This does not contradict the above theorem because the angle form is not defined on all of R2. Question: What is the integrability condition and the PDE system for an arbitrary 2-form? α = ΣI fI dxI ordered multi-index = Σi<j fijdxidxj . specific We know from (2.4) above that dα = = Σi<j<k { (∂kfij) + (∂ifjk) - (∂jfik) } dxidxjdxk so the integrability condition in this case is (∂kfij) + (∂ifjk) - (∂jfik) = 0 . If we want α = dg, where g is a 1-form g = Σk gkdxk then we have dg = Σi<j [ (∂igj) - (∂jgi)] dxidxj // from (2.2) = Σi<j fijdxidxj . So the system of equations you need to solve to find g is given by (∂igj) - (∂jgi) = fij . // agrees with S (2.6). At this point, the reader has no feel whatsoever for the significance of a form being closed and being exact. S is just getting the reader used to the machinery. 2.4 The Hodge star operator [23] The first concept here is this: dxI = some multi-index set of dxi in Rn (*dxI) = (sign) dxIc = the ones that are missing in dxI within Rn (c means complement) What is the sign? Well, when you write the combination dxI dxIc you will get ALL the dxi . When you adjust this list to put them into standard order 1,2,3... you may get a minus sign if it takes an odd number of swaps. With this sign adjustment, you always get dxI(*dxI) = dx1....dxn = the positive volume element. Now using this idea, we define α = ΣI fI dxI *α = ΣI fI (*dxI) this then is the "Hodge starred α" // Hodge circa 1930 If α is a k-form, then *α is in (n-k) form [ sometimes called dual form] . Lots of examples are given. They note that there are (n,k) k-forms α on Rn and there are (n,n-k) (n-k)-forms = *α forms = (n,k). The connection is not yet stated. Exercise 2.12: Show that **α = (-1)kn+kα if α is a k-form. α = ΣI fI dxI k-form *α = ΣI fI (*dxI) *(*α) = ΣI fI *(*dxI) Lemma: Show that *(*dxI) = (-1)kn+k dxI . If we can show this Lemma, then we continue the previous proof *(*α) = ΣI fI *(*dxI) = ΣI fI (-1)kn+k dxI = (-1)kn+k ΣI fIdxI = (-1)kn+k α . Proof of Lemma: *(*dxI) = (-1)kn+k dxI Do an example first dxI = dx2dx4 in R6 *dxI = S dx1dx3dx5dx6 dxI(*dxI) = dx2dx4 S dx1dx3dx5dx6 = S (-1)2dx1dx2dx4dx3dx5dx6 = S(-1)3 dV Think of this in terms of permutations. We have dxI = 24 *dxI = S 1356 k = 2 n = 6 dxI(*dxI) = S 241356 = S (-1)2 124356 = S(-1)2 (-1)1 123456 OK, now go back to *dxI = S dx1dx3dx5dx6 *(*dxI) = S S' dx2dx4 Now S' is set by this requirement *(*dxI) (*dxI) = 123456 So we then have S S' 24 S 1356 = 123456 The sign S cancels out and we are left with S' 241356 = 123456 and as before, S' = (-1)3 = -1. In this example then we have *(*dxI) = S S' dx2dx4 = SS' dXI and SS' = +1. The rule was (-1)kn+k with k = 2 and n = 6 so we get the right answer. But I don't have much insight on how to do the general case. But I later proved this in a separate doc "proof of a lemma v 2.doc". 2.5 div, grad and curl [24] You can write any 1-form as α = Fdx on Rn. Thus, there is a direct 1-to-1 connection between α and F. Probably there is a simple connection between a rank-2 tensor and a 2 form in that α = Fij dxidxj and F would be antisymmetric. In this way, I guess every tensor is related to a form. This is definitely a new appendix for tensor doc at some point and maybe a new whole chapter! I suspect all the forms are tensorial scalars. Note added 2.11.15 re last comment above. When I write T = ΣijTij ei ej, under a Type A transformation the basis vectors do not move, and this becomes T' = ΣijT'ij ei ej where the T'ij is the usual thing, so "this is how a tensor transforms". In the dual space T = ΣijTij λi λj and again the basis vectors do not move under a transformation, so I would then say T' = ΣijT'ij λi λj . I think the above is just the same when you replace with ^, so T = ΣijTij ei ^ ej, and then T' = ΣijT'ij ei ^ ej and you are here just dealing with tensors which are in a certain subspace of the original space. So when we eventually write T = ΣijTij dxi dxj, I would say T' = ΣijT'ij dxi dxj . Thus, my statement above that "I suspect all forms are tensorial scalars" is incorrect. However, all tensor functions are scalars, but these don't exist at this point in Sjamaar World. Let f be a 0-form and then f ≡ F is a directly related vector field. Then consider α = fdx = df. But I don't see the big deal here. If f is a 0-form, then yes, grad f appears in df. Write α = f 0-form α ↔ f dα = df 1-form dα ↔ f // namely, dα = fdx I think I now better understand Buck's ↔ symbol. It means there is a 1-to-1 mapping between the items on the two sides. Now comes a longer deal. Write α = F dx 1-form (*α) = F (*dx) where (*dx)1 = dx2dx3....dxn , for example n-1 form Later we will claim that dx and (*dx) are related to orthogonal subspaces in Rn, the way dx1 and dx2 are in R2. This idea is just mentioned at this point. Now consider d(*α) = d(ΣkFk (*dx)k = Σkd(Fk) (*dx)k = ΣkΣj∂jFk dxj(*dx)k . n-form Now consider dxj(*dx)k . The combination (*dx)k is missing dxk . If j = k, we then have dxk(*dx)k which according to p 24 B is + dx1....dxn. = dV If j≠ k, we then have dxj(*dx)k = 0 because two dxs will appear together, killing off the form. Thus, d(*α) = ΣkΣj∂jFk dxj(*dx)k = ΣkΣj∂jFk δj,kV = Σk (∂kFk)V = (div F) V . n-form This diagonal effect brings out the divergence! Now do one more step *[d(*α)] = *[ (div F) V] = (div F)(*V) = (div F)(1) = div F . 0-form So what does that really tell me? I can write (div F) in terms of the 1-form associated with F. : α = F dx 1-form α ↔ F *[d(*α)] = (div F) 0-form *[d(*α)] = div F Finally, let's repeat the above, but do things in a different order α = F dx = same starting 1-form as above dα = Σi<j [ (∂iFj) - (∂jFi)] dxidxj // our usual formula (2.2) *(dα) = Σi<j [ (∂iFj) - (∂jFi)] *(dxidxj) // agrees with p 27A He writes *(dxidxj) = (-1)i+j+1 dx1dx2.....dxi....dxj.....dxn // red ones are missing To verify that sign, consider (dxidxj) (-1)i+j+1 dx1dx2.....dxi....dxj.....dxn = (-1)i+j+1 [ dxidxjdx1dx2.....dxi....dxj.....dxn] = (-1)i+j+1[ (-1)j-1 dxidx1dx2.....dxi....dxj.....dxn ] // did not pass through dxi since missing = (-1)i+j+1[ (-1)j-1 (-1)idx1dx2.....dxi....dxj.....dxn ] = dx1dx2.....dxn = dV So OK, go back to *(dα) = Σi<j (-1)i+j+1 [ (∂iFj) - (∂jFi)] dx1dx2.....dxi....dxj.....dxn // red missing This result is a little hard to interpret in general Rn , is just a fact. Now what happens if you are in R3 ? Then write out the terms : i=1 j=2 (+) [ (∂1F2) - (∂2F1)] dx3 i=1 j=3 (-) [ (∂1F3) - (∂3F1)] dx2 i=2 j=3 (+) [ (∂2F3) - (∂3F2) dx1 Then *(dα) = [ (∂1F2) - (∂2F1)] dx3 + [ (∂2F3) - (∂3F2)] dx1 + [ (∂3F1) - (∂1F3)] dx2 = [ (∂2F3) - (∂3F2)] dx1 + [ (∂3F1) - (∂1F3)] dx2 + [ (∂1F2) - (∂2F1)] dx3 = [curl F]1 dx1 + [curl F]2 dx2 + [curl F]3 dx3 = [curl F] dx Note that α = 1-form, dα = 2-form, *(dα) = n-2 = 1-form again for R3. Therefore we have this correspondence idea: α = F dx α ↔ F one-to-one *(dα) = [curl F] dx *(dα) ↔ curl F How about the Laplacian from Ex 2.14. Start again with a 0-form α = f 0-form α ↔ f dα = df = ∂if dxi 1-form *(dα) = ∂if (*dxi) = (-1)i ∂if dx1dx2.....dxi....dxn n-1 form d[*(dα)] = (-1)i ∂j∂if dxjdx1dx2.....dxi....dxn n-form Implied sum on i and j. If j ≠ i, then dxj appears twice so get 0. If j = i, we just slide dxj into position and kill the sign d[*(dα)] = ∂i2f dxjdx1dx2.....dxi....dxn n-form Then finally *d[*(dα)] = ∂i2f = 2f // scalar Laplacian. 0-form OK, lets recap these amazing little results: α = f 0-form in Rn α ↔ f PL1 dα = df 1-form in Rn dα ↔ f // namely, dα = fdx α = f 0-form in Rn α ↔ f *d[*(dα)] = ∂i2f = 2f 0-form on Rn *d[*(dα)] ↔ 2f PL2 α = F dx 1-form on Rn α ↔ F *[d(*α)] = (div F) 0-form on Rn *[d(*α)] = div F PL3 α = F dx 1-form in R3 α ↔ F one-to-one *(dα) = [curl F] dx 1-form in R3 *(dα) ↔ curl F PL4 Now imagine this correspondence idea α = f ↔ f α = 0-form Rn dα ↔ f *d[*(dα)] ↔2f α = F dx ↔ F α = 1-form R3 only *[d(*α)] ↔ div F *(dα) ↔ curl F If you are willing to "work in the form universe", you can construct fancy differential operators from the symbols α, d and * and they certainly looks simple. At this point I did Exercise 2.12 (**α) and part of Exercise 2.18 (Maxwell). I am ready to continue on. See separate docs for those exercises "Sjamaar Ex 2_18 on Maxwells v2.doc". I do Exercise 2.6 in the Chapter 4.1 notes below, to show that a closed 1-form on Rn is exact. Recall that exact closed, so this is going the other way. The key idea is that the 1-form is closed on ALL of Rn. Chapter 3: Pulling Back Forms 3.1 Determinants This is a long digression which I can only assume he will make use of later on. Note added 2.11.16. I think the multilinear idea is very simple. The determinant of a square matrix having columns v1, v2...vk is k-multilinear in those arguments. So f(v1, v2 .. vk) = det(v1, v2 .. vk) is a k-multilinear function of its k vector arguments. It is also a totally antisymmetric function of the arguments. When I first read this stuff here, I did not understand the significance of these two properties. In wedge doc I show that (λj ^ λj ^ .... ^ λi)(vi,vi....vi) = (1/k!) det[ (vi)j] (8.3.9a) = (1/k!) det (vi,vi....vi) so this shows that the degree-k basis function shown here, closed to be a tensor function, is in fact both k-multilinear and totally antisymmetric and in this case this follows directly from the properties of the determinant. Later still I show that T^(vj,vj....vj) ≡ [Alt(T)] (vj,vj....vj) where things are then less obvious. Author reviews classic definitions of detA and comes up with his own: Three axioms which are the multilinear rule, the minus sign on swap columns, and det(ei) = 1. The multi rule needs a clearer statement: column j det( a1, a2....... [cai + c'a'i],.... an) What is confusing here is that the column marked as column j is a linear combination of any two columns of A which two are denoted by ai and a'i. The rule is then det( a1, a2....... [cai + c'a'i],.... an) = c det( a1, a2....... ai.... an) + c' det( a1, a2.......a'i,.... an) col j col j I might have presented this differently. Consider determinant det( a1, a2...... aj .....an) . Now let aα and aβ be any two columns of the matrix. Then make this replacement aj = cαaα + cβaβ . The result is det of a new matrix Q = det( a1, a2...... cαaα + cβaβ, .....an) The axiom is then this det( a1, a2...... cαaα + cβaβ, .....an) = cαdet( a1, a2...... aα .....an) + cβdet( a1, a2...... aβ .....an) . If we take cα = 1 and aα = aj, then we are adding a multiple of a column to a column, so the above rule is more general. How about breaking this down. First, here is what happens if you replace aj with a sum of any two columns, det( a1, a2...... aα + aβ, .....an) = det( a1, a2...... aα .....an) + det( a1, a2...... aβ .....an) #1 Then another rule would be det( a1, a2...... c aj .....an) = c det( a1, a2...... aj .....an) #2 Then his rule would follow as follows det( a1, a2...... cαaα + cβaβ, .....an) = det( a1, a2...... cαaα .....an) + det( a1, a2...... cβaβ .....an) = cαdet( a1, a2...... aα .....an) + cβdet( a1, a2...... aβ .....an) Sj (i) So I think he is just trying to keep his axiom count low by combining things into a single axiom. I. Add multiple of column to a different column: det( a1, a2...... aj + cβaβ, .....an) = det(A) + cβdet( a1, a2...... aβ, .....an) = 1 det(A) . mult = 1 The detA does not change because two columns are the same and swap rule says zero. II. Mult column by constant det( a1, a2...... c aj .....an) = c det( a1, a2...... aj .....an) = c det(A) mult = c III. Swap columns det(A) → (-1) detA mult = -1 So this is the content of his 3.2 Lemma page 32. 3.4 Det determined by the three axioms is unique. Fine, I skip it in these notes. 3.5 Comment that you can normalize differently if you want. Then det' I = c for that different third axiom. 3.6 This is my expansion with a permutation sum. I write it this way In his notation, he has a1 = σ(1) a2 = σ(2) ... an = σ(n) so he has which I like to write as a = A z0 det(M) = Σa sign(a) Mz0 a where A is a matrix, I can think of this as a mapping Pa in which 1 → a1 2→ a2 ... which you could write as a1 = Pa(1) a2 = Pa(2) and so on. He refers to the mapping Pa as σ. It's just another way to do it, I like my way better. [34] Comparison Me Sjamaar Parity(a) = (-1)Sa sign(σ) = (-1)l(σ) parity of permutation a relative to z0 swap inversion Sa l(σ) number of swaps, length of σ He then does a proof of his Theorem 3.6 which I skip since I have my own world on this. [35] 3.7 Theorem is a collection of standard det rules. Only the triangular rule is not in my Appendix B of Lagrange. He then claims that | det(A) | is a "volume change factor" where A maps the n-cube into an n-piped. It is true that the entire interior of the n-cube maps into the interior of the n-piped. But interior→interior does not convince me that the volume changes by this factor! I had to write a whole Appendix B in tensor doc to prove this fact, where volume is not defined by this process. I think he has pulled a fast one here. This concludes the long determinant digression, and we now get on with the main chapter topic: 3.2 Pulling back forms [36] Sja is now going to treat a subject I am quite interested in. What happens to a form under a transformation of the variables? So start with a form α = ΣIfI(y) dyI p 36 A and do this transformation on it y = φ(x) p 36 B // x = φ-1(y) if it exists Note: x-space has dimension En but y space has dimension Em as shown p 36B, so in general an inverse is not going to exist. So the number of variables in the two spaces is not generally the same. Thus, you cannot talk about dyI and dxI as if they had the same sets of dyi and dxi factors! Now define the action of operator φ* on a scalar function in this manner φ φ* fI(x) ≡ [fI * φ] fI(x) = fI(φ(x)) = fI(y) x → y ______________________________________________________________________________ Digression on the words Concatenation and Composition and use of the circle operator Comment: for a function, we are just taking f(y) → F(x) = f(φ(x)) so you could say that the new functional form F = f * φ which for me is a concatenation. Buck and Ahlfors don't seem to use this word and I could find no mention of the subject. Wiki https://en.wikipedia.org/wiki/Function_composition speaks of "composition": So they put things in "my" normal ordering which is that (gf)(x) = g(f(x)) so that the functions are in the same order on both sides. They use the circle, and even state "g circle f" above. I see from the web that my word "concatenation" is never used for this meaning! (I think I used the word in bipolar doc and elsewhere). I look in my doc notes and see concat used in reference to combining matrices. I see things like this where again you could say this is matrices. Conclusion: Concatenation is a good word for composing two linear transformations with matrices, but the correct word in general is composition, not concatenation. I stand corrected and will fix up docs at some point perhaps. But here is somebody using my word: Now, regarding composition, I note above that wiki uses (gf)(x) = g(f(x)) so the functions stay in the same order. That is the same for person above using the concat word. All sites use the circle thing. So where did I see the * symbol used? Some sources say (fg)(x) = f(g(x)) so there is no operator. Now on page 36 we get this ( A courier new letter o works well ) φ*f = f o φ = f(φ(x)) So Sja is using the standard f o φ ordering, but for some reason he wants to use φ* as if it were an operator which changes the function of f from whatever it is to φ(whatever it is). The operator φ* acts on whatever is to its right. OK, this has no name. ____________________________________________________________________ Now he wants to do a similar thing to dyI : φ* dyI = φ* (dyadyb....) ≡ (dφadφb...) y = φ(x) What exactly is this saying? Since ya = φa(x) you would think that dyI = dyadyb.... = dφadφb... without any special operator The intention is really this dyI = dyadyb.... = dφa(x) dφb(x)... [ wrong! ] so he wants to emphasize the idea using the φ* operator [wrong]. So his meaning is simply this φ* dyI = φ* (dyadyb....) ≡ dφa(x) dφb(x)... Nothing more and nothing less. Recall that φ* fI = fI(φ(x)) and this is a true function composition, whereas φ* dyI is just a cosmetic notation IMHO. Now look at the action of φ* on a form α(y) = ΣI fI(y) dyI y = φ(x) Here I have made explicit that α(y) is defined on y-space, including the yi coordinates and the dyi differentials. I could simply replace everywhere y = φ(x) to get α(φ(x)) = ΣI fI(φ(x)) dφI(x) y = φ(x) and then I could account for the "change of functional form by writing α'(x) = ΣI fI'(x) dφI(x) dφi(x) = Σj=1m(∂φi/∂xj) dxj i = 1,2..n Of course if α was a k-form, there would be only four dφi factors. He will talk more on this below. So OK, instead of using primes to show a new functional form, just write things like this α(y) = ΣI fI(y) dyI α(φ(x)) = ΣI fI(φ(x)) dφI(x) (α o φ)(x) = ΣI (fIo φ)(x) (dyI o φ)(x) Here the idea is (dyI o φ)(x) = dyI(φ(x)) = dφI(x) . It is a little stretched, but Sja is free to define things the way he wants. Then you translate the above bo (φ*α)(x) = ΣI (φ*fI)(x) dφI(x) You can then think of α'(x) = (φ*α)(x) so the operator φ* is in effect creating a function having your new functional form of x. I am happy. You can then just say α' = φ*α where φ* acts right on the form itself. You are starting with a form α(y) = ΣI fI(y) dyI defined in y-space and using y = φ(x) and all the fancy notation above you end up with (φ*α)(x) = ΣI (φ*fI)(x) dφI(x) being a form in x-space. I might have named the regions X and Y instead of U and V. Then we have φ: X → Y φ(x) = y a forward map going left x-space to right y-space φ*: Ωk(Y) → Ωk(X) φ*α = ΣI (φ*fI)(x) dφI(x) = β(x) α(y) → β(x) Words: You have a forward mapping y = φ(x) which takes points in x-space to points in y-space. You have a backward mapping β(x) = φ* α(y) which takes a k-form defined on y-space and gives you a k-form defined on x-space. You are pulling the form α back from y-space to x-space where it is β(x). The picture on page 37 is then just fine by me. 3.9 Example. Suppose we have some 2-form like α = f(y) dy1dy2 in R2. Then the pullback is doing to be φ*α = (φ*f ) φ*(dy1dy2). So here n = 2 and k = 2. The equation y = φ(x) is square in that m = n = 2. the purpose of this example is merely to compute the quantity φ*(dy1dy2) since he has not done that yet. We have y1(x) = x13x2 y2(x) = log(x1+x2) You then just use the chain rule to find that φ*(dy1dy2) = dφ1dφ2 = F(x1,x2) dx1dx2 as shown p 38 A In general terms we would get dφ1 = (∂φ1/∂xi)dxi dφ2 = (∂φ2/∂xj)dxj dφ1dφ2 = Σij [ (∂φ1/∂xi)(∂φ2/∂xj) ] dxidxj = Σij [ (∂iφ1)(∂jφ2) ] dxidxj Now while we are here, consider Σij fij dxidxj = Σi<j fij dxidxj + Σj<i fij dxidxj = Σi<j fij dxidxj + Σi<j fji dxjdxi = Σi<j ( fij – fji) dxidxj So in our example we get dφ1dφ2 = Σij [ (∂iφ1)(∂jφ2) ] dxidxj = Σi<j [ (∂iφ1)(∂jφ2) – (∂jφ1)(∂iφ2)] dxidxj and now you see a Jacobian-like object appearing. More generally dφadφb = Σi<j [ (∂iφa)(∂jφb) – (∂jφa)(∂iφb)] dxidxj = Σi<j dxidxj Even if y = φ(x) is not "square", the above line makes sense. There may be no overall Jacobian due to non-square, but this little 2x2 Jacobian chunk will always exist. Proposition 3.10 Here we have three "rules" for the φ* operator called (i), (ii) and (iii). (i) linear: φ*(aα+bβ) = aφ*α + bφ*β My proof: show that (aα+bβ) o φ = a (α o φ ) + b (β o φ ) or show that (aα+bβ)(φ(x)) = aα(φ(x)) + bβ(φ(x)) Well just write φ(x) as y and then show that (aα+bβ)(y) = aα(y) + bβ(y) Sja never really talked about "adding two forms" so maybe I will address that right here: adding two forms (must be of same degree) α + β = ΣIfIdxI + ΣIgIdxI = ΣI(fI+ gI)dxI = ΣIhIdxI = γ (some new form) cα = c ΣIfIdxI = ΣI(cfI)dxI = δ (some new form) So given this, there is nothing more to show. (ii) multiplicative: φ*(αβ) = (φ*α)(φ*β) Proof: show (αβ)(φ(x)) = α(φ(x)) β(φ(x)) show (αβ)(y) = α(y) β(y) The thing on the left is an "exterior product" as defined on page 19 B. so (αβ) = ΣI,JfI gJ dyI dyJ = obvious (ii) natural: φ*(ψ*α) = (ψ o φ)*α Proof: LHS = φ*(ψ*α) = φ*(α o ψ) = (α o ψ) o φ RHS = (ψ o φ)*α = α o (ψ o φ) But I am sure the circle operator is associative. Wiki BUT, this is for functions, and we need it for form α with ψ and φ being functions. Then LHS = φ*(ψ*α) = φ*(ψ*ΣIfI dyI) = φ* [ ΣI(ψ*fI) (ψ*dyI)] = φ* [ ΣI(fI(ψ(x))dψI] = ΣI(fI(φ(ψ(x))) φ*(dψI) So you can see that some WORK is needed in this proof, and so I just go with his work as shown. I am not sure about that (B.6) chain rule. Maybe look at that now: (B.6) Chain Rule: = Σj φ*() Show that = Σj φ*() = Σj // seems OK I have scribbled on the left facing page the last steps of his proof, let's not dwell on it now. Theorem 3.11. φ*d = dφ* these operators commute. I have read the proof twice, and it uses the same (B.6) above. How about a naive proof φ*(dα) = φ* [ΣI (dfI) dyi] = [ΣI φ*(dfI) φ*(dyi) ] = ΣI φ*(dfI) dφi d(φ*α) = d [ΣI (φ*fI) φ*(dyi)] = d [ΣI (φ*fI)dφi ] = ΣI d(φ*fI)dφi At this point we must use the rule for functions, then continue = ΣI φ*d(fI)dφi = QED His proof of the "rule for functions" is straightforward, using his B.6. But I will try a naive on that now: φ*(df) = (df)(φ(x)) = [Σj (∂f/∂yj)dyj ](φ(x)) = Σj (∂f/∂yj)dφj d(φ*f) = d(f(φ(x)) = Σi ( ∂/∂xi) f(φ(x)) dxi = Σij( ∂f/∂yj)( ∂φi/∂xi) dxi = Σj( ∂f/∂yj) dφj OK, good enough. [40] The Determinant Connection. I am happy here down through line E which I will rewrite this way dφadφb....dφk = ΣABC.... RaARbB ....RkK dxAdxB.....dxK k factors k factors Now how would I convert this to a determinant? The sums on A,B,C are full range 1 to n. The only contributions in the sum are when all the dxi are different! Let's try a simple example first dφadφb = ΣABRaARbBdxAdxB = ΣA<B (RaARbB - RaBRbA)dxAdxB You MUST get the thing into an ordered index sum in order to reveal the determinant! Let's try it for the case k = 3: dφadφbdφc = ΣABCRaARbBRcCdxAdxBdxC = ΣA<B<C[RaARbBRcC + 5 signed permutations] dxAdxBdxC So here is the general question: When you write the thing as an ordered sum A<B<C...<K, what exactly is the factor which multiplies dxAdxB.....dxK ? The claim is that this factor is a determinant. OK, consider this even more general situation ΣABC TABC = ΣA<B<C TABC + ΣP ΣP(A<B<C) TABC where TABC is assumed to vanish when any two indices are the same. What the above says is that you can break up the original triple sum into 6 different summation sectors. The first has A<B<C. The remaining sums have P(A<B<C) ≡ A'<B'<C' where (A',B',C') = P(A,B,C) and we sum over all remaining permutations. We might as well include the first term and write ΣABC TABC = ΣP ΣP(A<B<C) TABC where we now include the identity permutation. The indices on TABC are always the same, they do not feel the permutation. Now if (A'B'C') = P(ABC), then (ABC) = P-1(A'B'C'). But stop, let's try to get this into the language of my (Lagrange doc) Appendix B which has some very powerful tools! A ≡ , a vector A' = PA for example A' = = P = PA I am changing notation relative to App B to be closer to what fits this current need. Now A and A' are vectors, whereas P is a permutation matrix. The set of all matrices P which does all shuffles forms a group (permutation group of k objects). Now consider: ΣABC TABC = ΣP ΣP(A<B<C) TABC = ΣP ΣA'<B'<C' TABC Now for each term in the sum, we want to rename variables so that (A'B'C') → (ABC) I think (will show later) that this means ΣP ΣA'<B'<C' TABC = ΣA<B<C ΣPTP(ABC) For the moment, assume this is correct and let's see where it leads when TABC = RaARbBRcCdxAdxBdxC TP(ABC) = P2(RaARbBRcC) P(dxAdxBdxC) = P2(RaARbBRcC) (-1)S dxAdxBdxC where S is the number of swaps need to rearrange P(ABC) back to order (ABC). We then have ΣABC TABC = ΣA<B<C ΣPTP(ABC) = ΣA<B<C [ ΣP (-1)S P2(RaARbBRcC)] dxAdxBdxC = ΣA<B<C det(Rabc,ABC) dxAdxBdxC = dφadφbdφc where det(R) in this case is the determinant of a 3x3 submatrix of the full R matrix. The row labels are given by a,b,c while the column labels are given by A,B,C. Each ordered term in the sum thus involves different columns A,B,C of the full R matrix (but the same rows). Now using ordered notation, you might say, dφadφbdφc = dφI = ΣA<B<C det(Rabc,ABC) dxAdxBdxC = Σ J det(RI,J) dxJ and this explains the notation appearing in Sja p 41 (3.3). Now my R matrix elements are RaA = ∂φa/∂xA = ∂Aφa ≡ (Dφ)A,a RaARbB = (Dφ)A,a(Dφ)B,b If we define an ordered multi index I = AB and J = ab we might abbreviate RI,J = (Dφ)I,J So then my result above is dφI = Σ J det(RI,J) dxJ = Σ J det[(Dφ)I,J] dxJ // matches (3.3) where everything is ordered!! What happens if k = n, the max value? Then I think the entire R matrix is involved, and we get dφ1dφ2.....dφn = dVφ = det(R) dx1dx2...dxn = det(Dφ) dx1dx2...dxn This is because in the sum ΣA<B<C there is only one term plus those permutations, so ΣJ goes away. The only ordered term is 1<2<3..<n . Comments: 1. This entire discussion involves forms like dφadφbdφc and dxAdxBdxC. These are not the same as products of differentials. One example is that there is no sign swap rule for a product of differentials. 2. The maximal case (only) does seem to align with the usual integration Jacobian rule. The author first addresses integration in the next chapter. Right now we have no idea how to do integration. 3. In general, the det is only of a square portion of the full R matrix. 4. These smaller determinants only show up if you use ordered summation, or (which is the same thing) ordered multi-indices. 5. The reason the author had that long opening section on determinants was to get these little form differentials into a determinant form! Note that in φ* the asterisk is a superscript. The Hodge operator is a main line * character. Seems unfortunate to use the same symbol for both. Chapter 4: Integration of 1-forms 4.1 Definition of the integral and its elementary properties [47] [47] We now have to make a translation of Chapter 3 concepts to this particular application: Chapter 3 Chapter 4 xi in x-space = Rn and U (p37) t in t-space = R1 and [a,b] (p47) left yi in y-space = Rm and V (p37) x in Rm and U right y = φ(x) x = c(t) α α φ*α c*α The idea is that "the space on the right" holds the trace of a curve c(t) in Rm while "the space on the left" is just R1 with t in [a,b]. This is of course the parameter of the curve in Buck language. Now how do our big gun results apply here? α = Σifi(x)dxi = 1-form associated with space on the right c*α = Σifi(c(t))dci = Σifi(c(t))(∂tci) dt ≡ g(t) dt // page 42 A dxJ → dt where g(t) ≡ Σifi(c(t))(∂tci) This is the pullback, now associated with the space on the left which is just t-space. Now Sja defines the integral of a 1-form in this manner ∫c α ≡ ∫[a,b] c*α = !Syntax Error, I g(t) dt = !Syntax Error, I Σifi(c(t))(∂tci) dt In other words, the integral over the curve c in Rm of the form α is defined as the integral of the pullback c*α in R1, the space on the left. This certainly is an integral we know how to do. [ Bucks make this same definition in their book, see p 376 (7-7); of course notation is different.] Example 4.1. (the angle form). Here we have α = Σifi(x)dxi = f1(x) dx + f2(y) dy = -(y/r2)dx + (x/r2)dy r2 ≡ x2+ y2 and x = c(t) : x = cost [a,b] = [0,2π] curve is circle radius 1 in Rm = R2 y = sint r = 1 If t is time, then point (x,y) moves around the unit circle with ω = 1 radians/sec. g(t) = Σifi(c(t))(∂tci) = f1(cost,sint)∂t(cost) + f2(cost,sint)∂t(sint) = -(y/r2) [-sint] + (x/r2)[cost] = ysint + xcost = sin2t + cos2t = 1 c*α = g(t) dt = dt //extremely simple example! Then ∫c α = !Syntax Error, I g(t) dt = !Syntax Error, I 1 dt = 2π We might interpret this as the "length" of the circumference of the unit circle, but Sja says nothing yet about such length. Respeeding a Curve. [ bottom p 47] We had above that x = c(t) and now we reparametrize by saying t = p(s) so x = c(t) = c(p(s) = (c o p)(s) interval is [,] in s-space Require p'(s) ≠ 0 on this interval so the motion does not reverse itself! Compare: c : [a,b] in R1 t-space→ curve c in U within Rm (c o p) : [,] in R1 s-space → curve c in U within Rm If p'(s) > 0 then orientation of the curve is preserved, otherwise it is reversed. Example 4.2. Let t = 2s in previous example so then x = c(2s) : x = cos(2s) [a,b] = [0,π] curve is circle radius 1 in Rm = R2 y = sin(2s) r = 1 If s is the new time, then point rotates around circle at ω = 2 radians/sec and one loop uses s = [0,π]. Suppose t = 2πs2 so s runs on [0,1]. Then x = cos(2πs2) = cos(ωs) so ω = 2πs. Then the point has an angular velocity which starts at 0 and linearly increases to 2π at the end point. Theorem 4.3. Apart from a possible orientation minus sign, reparametrization does not alter the integral of a 1 form over a curve! Proof: ∫c α ≡ ∫[a,b] c*α ∫c o p α ≡ ∫[,] (c o p)* α Now we get to use our earlier painful "naturality" theorem from p 38 which says [ t = p(s) ] (c o p)* α = p*(c*α) = p*(g(t)dt) = g(p(s)) dp // recall fI(φ(x))dφI = (p*g) dp // really (p*g(t)) = g(p(s)) = (p*g)(∂p/∂s) ds Then we have ∫c o p α ≡ ∫[,] (c o p)* α = ∫[,] (p*g)(∂p/∂s) ds = ∫[,] g(p(s)(∂p/∂s) ds Recall that ∫c α ≡ ∫[a,b] c*α = !Syntax Error, I g(t) dt If we now make a normal calculus substitution t = p(s) where dt = dp = (dp/ds)ds we get ∫c α = !Syntax Error, Ig(p(s)) | (dp/ds) |ds = sign(dp/ds) ∫[,] g(p(s)(∂p/∂s) ds Thus we have shown that ∫c o p α = sign(p'(s)) ∫c α Sub-example. Consider !Syntax Error, I t dt = (1/2) t = ± 2s dt = ±2ds but dt = J ds for integration J = |±2| = +2 !Syntax Error, I t dt = !Syntax Error, I[ ±2s] [ +2ds ] = ± 4 !Syntax Error, Is ds . There are two cases to consider: + sign: = + 4 !Syntax Error, Is ds = 4 * (1/2)22 = 8 - sign = - 4 [!Syntax Error, Is ds ] = -4[ - !Syntax Error, Is ds ] = 4 (1/2) ( 0 - (-2)2) = - 8 So the key fact is that the Jacobian for integration has the |....| sitting there. In this example, the integral over the right triangle is positive because s is positive in the entire range. The integral over the left triangle is the sum of rectangle positive widths times the function which is negative there, so the triangle on the left integrates to -8. So reparametrization does not alter the integral of a 1-form over a curve, apart from minus sign if you have reversal. A fact familiar to me from Buck. [ And on 2.12.16 I firmed that idea up by adding notes going through the above proof without the need for the formal "naturality" rule. ] PL Corollary: If you are asked to compute the 1-form integral ∫c o p α = ∫[,] (c o p)* α = ∫[,] g(p(s)) (∂p/∂s) ds you might save time by instead computing the exact same number using ∫c α ≡ ∫[a,b] c*α = !Syntax Error, I g(t) dt where g(t) ≡ Σifi(c(t))(∂tci) where t runs through [a..b] and s runs through [...]. We have s = p(t) so = p(a) and = p(b). Interpretation of ∫c α . Recall α = Fdx is associated (one to one!) with vector F. If dx is a physical distance in physical space, and if F is the force on a particle, then α is the work done on that particle as it moves dx. For example if F and dx are aligned, you are doing positive work on the particle, you are accelerating it or maybe increasing its potential energy. So our integral is then specifically "the total work done on the particle" as it traverses the curve in the presence of this force field F . [49] Interesting conclusion: If particle doubles its speed along the path, the field still does exactly the same work on the particle. This is true for any variation of the speed during motion, as long as it does not go backwards. Suppose F is such that you can write F = -φ where φ is then a potential. α = F dx dα = (dF) dx = dFi dxi = (∂jFi)dxjdxi = - ∂ijφ dxjdxi = 0 So in this case dα = 0 so α is closed. Moreover, dφ = (∂iφ) dxi = - Fidxi = - Fdx = -α, so we find that in this case α = -dφ, so α is also exact. Consistent with "every exact form is closed". I now repeat the above paragraph with the other sign: Suppose F is such that you can write F = φ where φ is then a potential. α = F dx dα = (dF) dx = dFi dxi = (∂jFi)dxjdxi = ∂ijφ dxjdxi = 0 So in this case dα = 0 so α is closed. Moreover, dφ = (∂iφ) dxi = Fidxi = Fdx = α, so we find that in this case α = dφ, so α is also exact. Consistent with "every exact form is closed". 4.2 Integration of exact 1-forms [49] Theorem 4.4. Suppose α = dg so α is exact. Then consider c*α = c*(dg) = d(c*g) // from earlier theorem Now g must be a 0-form since α = dg is assumed to be a 1-form, so g is then a function. But what is it a function of? Well, α and g are both associated with the space of the curve which is Rm . If g is a 0-form on this space, then it is g(x). But then c*g = g(c(t)) = (g o c)(t) = (c*g)(t), lots of ways to write it. So yes, I agree that you could define h(t) ≡ (c*g)(t). Then dh = d(c*g) = c*α from above. Then we have ∫c α ≡ ∫[a,b] c*α =∫[a,b] d(c*g) = ∫[a,b] dh = h(b) - h(a) = g(c(b)) - g(c(a)) QED Obviously if c(a) = c(b) which would be the case for a closed curve, we get ∫c α = 0, So we have shown that the integral of an exact 1-form around a closed curve vanishes. This all applies when α = dg is exact, and that in turn applies when F = - φ and then the work done going around closed path is 0 if the force can be written in this manner (a conservative force). Note added: In my integral theorems doc I have this theorem, ∫c α = !Syntax Error, Iφ dl = φ(B) - φ(A) (9) MM 4-36 p 156 This is the case just described where F = +φ so we identify α = dφ which is then exact. Theorem 4.5. Three ways to say the same thing, with fairly long proofs. (i) α is exact, meaning there exists g such that α = dg (ii) ∫cα = 0 for any closed curve c (iii) ∫c α depends only on the endpoints of the curve c. I think I could roll my own proofs, but I read through the first two proofs, both simple. The third has many steps. Part of the third proof is this idea: Let cx be any curve from xo to x and define g(x) = ∫cx α // integral of a 1-form is a 0-form or function. This turns out to be an appropriate g such that dg = α and, if shown true, this concludes the proof that (iii)(i). Naively that certain seems reasonable. You have constructed an explicit g(x) that works! As I show elsewhere, what you need to prove is that ∂ig = fi where α = Σifi(x) dxi. Sja gives a very long proof that ∂ig = fi and therefore dg = α. I skip the details shown by my pencil mark on pages 50,51. Example 4.6. We have shown that for α = -(y/r)dx + (x/r)dy that dα = 0 so α is closed. If it were also exact, we could write α = dg for some function g. In that case, ∫c α = ∫c dg = g(a) - g(b) and then we would have that ∫c α = 0 for any closed path. But by direct calculation we found that this integral = 2π when path c is a circle (a closed path), so it must be that α is not exact, even though it is closed. Exercise 2.6. Theorem: Any closed 1-form on (all of) Rn is exact. Given α being a closed 1-form (meaning dα = 0), you want to show that α = dg where then g would be a 0-form, that is to say, g is a function. This is demonstrated on page 27 Exercise 2.6 which I am now reading. You assume that α = Σifidxi and g is expressed in terms of the fi in this fancy way: [ proof by construction of the solution ] The proof that α = dg requires showing that Σifidxi = dg = Σi∂ig dxi which is to say fi = ∂ig. Note that ∂1g(x) = f1 (from first term) + no other terms since none contain x1 Next want to show that ∂2g(x) = f2. First, look at the contribution to ∂2g from the first term in the above expansion of g(x), ∂2[!Syntax Error, If1(t,x2,x3....) dt = !Syntax Error, I∂2[f1(t,x2,x3....)] // assuming this is allowed Now pause to look at dα = 0 which we know is true (closed). This says d[Σifidxi] = 0 or Σi(dfi)dxi = 0. Write this as Σij(∂jfi)dxjdxi = 0 or Σi<j[(∂jfi) - (∂ifj)] dxjdxi = 0. // as in p 20 (2.2) I suppose this says that ∂jfi = ∂ifj and in particular ∂2f1 = ∂1f2 Thus my first term contribution above can also be written as !Syntax Error, I∂1[f2(t,x2,x3....)] dt = [ f2(x1,x2,x3.....) - f2(0,x2,x3.....)] Therefore, including the contribution of the second term I am getting ∂2g(x) = [f2(x1,x2,x3.....) - f2(0,x2,x3.....)] + f2(0,x2,x3....) = f2(x1,x2,x3.....) So OK, I accept that this tricky sum of integrals shown in Ex 2.6 does do ∂ig = fi and therefore we have shown that dg = α. Now where does the requirement that α be defined on all of Rn come in ? Well, in order to say ∂jfi(x) = ∂ifj(x) [ α is closed at x ] ... Suppose this is only true for x in some subset of Rn. If that subset has a closed boundary somewhere (or is fully closed), then ∂jfi(x) is not going to be defined at the boundary so the proof fails at any point on a boundary. More specifically: It seems that we are requiring ∂jfi(x) = ∂ifj(x) to be true in the following hypercube: [0,x1] x [0,x2] x....x [0,xn] and this includes the origin point! Our thing has to also be true at the origin. But couldn't you fix up this problem by changing the 0 lower endpoint to .01, say? I would have to change all 0's to .01's in the various integrals and their endpoints. But then all xi have to be > .01 for all the integrals to make sense. Then g(x) is only defined for xi > .01 so g(x) not defined at the origin. But the form as shown is not defined for xi < 0 ? But I suppose each integral is in fact defined. So I guess I am hazy on this theorem's proof? Again, assume expression shown is valid for all xi. Then we have g(x) for all x. But now back to the issue that ∂jfi(x) = ∂ifj(x) has then to be true at the origin if things are OK for all Rn. But as we shall see, the angle form α is not OK at the origin because the fi blow up there and so do ∂jfi. So the punctured plane is missing the origin, so g(x) won't be defined at the origin. Again, suppose in E2 we try to restrict to U = a square centered on the origin, upper right corner of the square at (x,y) = (a,a). Then we just restrict |xi| < a, say, so we have an open square. I think for this kind of special subset of R2, we might be OK in Ex 2.6. For every point in U we have ∂jfi(x) = ∂ifj(x), and for every point we presume that fi is finite (continuous). Then excluding the origin means g(x) not defined at the origin, and then you can't really say that α = dg in the punctured square since g does not exist at the origin. So, although the angle form is a 1-form, since it is only defined in the plane punctured at the origin and not over all R2 or a square centered at the origin, we conclude that the proof that α = dg fails, and it is likely that the angle form is NOT exact. We know it is not exact since the integral around a closed curve (circle) gives 2π. So I see why the Exercise 2.6 proof fails for this case. PL Theorem: Any 1-form defined on an open cube centered at the origin is exact within that cube. 2.13.16. The above notes are pretty good on this Ex 2.6 thing. My trademark "splainin'". __________________________________________________________________________ Pause to do some comparisons between Sjamaar and Buck. I added some notes in Buck Ch 7 and in doing so finally learned that Buck's ω* is the pullback written without showing the name of the function φ. Sjamaar Buck α ω a generic differential form c γ a curve y = φ(x) x = φ(u) a transformation φ*α ω* a pullback Just one other item, then I will end this pause: ∫γ ω = ∫γ [ A(r)dx + B(r)dy ] = !Syntax Error, I [ A(γ(t))γ'x(t) + B(γ(t))γ'y(t)] dt // Buck ∫c α ≡ ∫[a,b] c*α = !Syntax Error, I g(t) dt = !Syntax Error, I Σifi(c(t))(∂tci) dt // Sjamaar ________________________________________________________________________ 4.3 The global angle function and the winding number [51] Now back to bottom of page 51. We find that for the angle form, dα = dθ where we regard this as an angle in physical space. Not only do we have a problem at the origin, we also have a problem where θ jumps by 2π. If we define θ = 0 on the positive real axis in the usual manner, there is a discontinuity at 2π. We really needs some Riemann sheets or something to get winding numbers and understand this disc. [52-53] I think what Sjamaar is doing here is forming a curve which wraps around the origin a few times in terms of pieces which do not do so. For each piece we don't have a problem, then we just take the final curve to be the union of all those pieces. [ similar I think to the manifold concept ] Well look at p 52 A where he defines θ(t) = θ(0) + !Syntax Error, I c*α = the pull back of α relative to c' The mapping starts out on t in [0,2π) say, but then we just extend it to [0, 42π) or whatever. The pullback integral is perfectly well defined for any range of t. For t in (0,2π) we have θ = θ but then we have provided an extension of θ beyond this range by means of θ. Note that if you draw a vector from origin to point c on the curve c(t), the direction of that vector is given by (x,y) and if you make that a unit vector, by (ξ,η) on page 51. So as you move on the curve, we can regard this little unit vector = (ξ,η) as a "dial" which measures the angle θ . The curve on page 53 could wind around the origin 50 times, and that dial would go around 50 or so times. The pictures are quite clear on this issue. Corollary: If curve is closed after winding around the origin k times, then Δθ = k(2π) and we can call k the winding number of the closed curve c. Recall that such a curve is not a manifold because it self-intersects. The curve on the left below exceeds 2π but is still a manifold and can be mapped with an atlas, but the curve on the right with winding number 2 intersects itself so is not a manifold. OK, I think I am now happy with Chapter 4. I guess I should update the meta notes for Ch 4. Chapter 5: Integration and Stokes' theorem I argue for Stokes's theorem but both forms are OK. Sja never mentions Gauss. [ well he says Gauβ ] 5.1 Integration of forms over chains In the pullback chapter 3 we had y = φ(x) as the transformation going from left to right. In the 1-form int chapter 4 we had x = c(t) as the transformation going from left to right. In this Stokes's chapter 5 we have x = c(t) with t a vector from left to right. Although for a k-form we are not dealing with a curve, the letter c is still used. Bucks used x = Σ(u,v) for their 2-form surface work. But Sja is going to keep it fully general in this chapter for k-forms, with examples of various k. In other words, Sjamaar beefs up his Ch 4 notation for 1-forms to apply to k-forms. A bit odd that he did it this way. So as an example, consider x = c(t1,t2). This is a surface say in R3 and there is no reason such a surface cannot penetrate itself, as shown Buck p 331. In the curve chapter, our left-space domain was [a,b] in R1. For a k-form, this left-space domain becomes a "rectangular block R" as described in p 57 A and shown there in direct product notation as well. No problem for me. Now let's translate the pullback from Chapter 3. One ingredient there was this (from meta notes) dφadφbdφc....dφq = ΣA,B,C...Q Rabc...q,ABC...Q dxAdxBdxC...dxQ // non-ordered sum k factors k factors dφI = ΣJ det(RI,J) dxJ = ΣJ [det[(Dφ)I,J] dxJ // ordered sum on J, ΣA<B<C...<Q α = ΣI fI(y) dyI y = φ(x) φ*α = ΣI fI(φ(x)) dyI = ΣI fI(φ(x)) dφI(x) = ΣI fI(φ(x)) [ ΣJ det[(Dφ)I,J] dxJ = ΣJ { ΣI fI(φ(x)) det[(Dφ)I,J] } dxJ = ΣJ {gJ(x) } dxJ = ΣJ gJ(x) dxJ gJ(x) ≡ ΣI fI(φ(x)) det[(Dφ)I,J] // p 42 A and B ≡ β I will now translate all the above into Chapter 5 notation: Now let's translate the pullback from Chapter 3. Space on the left has t, space on the right has x. We had x on the left and y on the right. So we want to do x → t and y → x and φ → c: dcadcbdcc....dcq = ΣA,B,C...Q Rabc...q,ABC...Q dtAdtBdtC...dtQ // non-ordered sum k factors k factors dcI = ΣJ det(RI,J) dtJ = ΣJ [det[(Dc)I,J] dtJ // ordered sum on J, ΣA<B<C...<Q α = ΣI fI(x) dxI x = c(t) c*α = ΣI fI(c(t)) dxI = ΣI fI(c(t)) dcI(t) = ΣI fI(c(t)) [ ΣJ det[(Dc)I,J] dtJ = ΣJ { ΣI fI(c(t)) det[(Dc)I,J] } dtJ = ΣJ {gJ(t) } dtJ = ΣJ gJ(t) dtJ gJ(t) ≡ ΣI fI(c(t)) det[(Dc)I,J] // p 42 A and B ≡ β Spaces here and previous: Notice how Sjamaar wantonly redefines m,n and V and U. page 37: U Rn on the left (x) V Rm on the right (y) general pullback chap 4: [a,b] R1 on the left (t) U Rn on the right (x) pullback of 1-forms chap 5: [0,1]k or R Rk on the left (t) U Rn on the right (x) pullback of k-forms Now the notation shown above ΣJ gJ(t) dtJ is general in that it would allow something like dt1dt3 as a differential included in the sum. But in our current context, we want ALL the dt factors in the form dt1dt2.....dtk. So we can further simplify the above equations for this special case: (first the general rule, then the special case) dcadcbdcc....dcq = ΣA,B,C...Q Rabc...q,ABC...Q dtAdtBdtC...dtQ // non-ordered sum k factors k factors dc1dc2dc3....dck = ΣA,B,C...Q R123...k,ABC...Q dtAdtBdtC...dtQ // non-ordered sum k factors k factors = ΣA,B,C...Q R123...k,ABC...Q [ PABC..Q dt1dt2dt3...dtk ] // parity operator = [ ΣA,B,C...Q PABC..QR123...k,ABC...Q] dt1dt2dt3...dtk = [det(RI,I)] dt1dt2dt3...dtk Remember that in general k < n, so RI,I is a the upper left square kxk submatrix of the full matrix R. We indicate the idea that this is submatrix by writing RI,I where I indicates that we only involve the first k coordinates of the space Rn. In Rabc...q,ABC...Q we suppress indices i for k < i < n. Continuing: dcI = det(RI,I) dtI = det[(Dc)I,I] dtI dcI ≡ dc1dc2....dck α = fI(x) dxI c*α = fI(c(t)) dcI = fI(c(t)) det[(Dc)I,I]dtI = fI(c(t)) det[(Dc)I,I] dtI = gI(t) dtI gI(x) ≡ fI(c(x)) det[(Dc)I,I] // p 42 A and B ≡ β Now again since we know that I means the entire set (the volume differential) we write again dc1dc2dc3....dck = ΣA,B,C...Q R123...k,ABC...Q PABC..Q dt1dt2dt3...dtk k factors k factors dc = det(R) dt = det[(Dc)] dt // ordered sum on J α = f(x) dx dc ≡ Πi=1k dck dx ≡ Πi=1k dxk dt ≡ Πi=1k dtk c*α = f(c(t)) dc = f(c(t)) det[(Dc)]dt = f(c(t)) det(Dc) dt = g(t) dt = g(t) dt g(t) ≡ f(c(x)) det(Dc) // p 42 A and B ≡ β So finally we arrive at p 57 B c*α = g(t) dt = g(t) dt1dt2dt3...dtk ∫c α ≡ ∫R c*α = ∫R g(t) dt = ∫∫∫...∫ g(t) dt1dt2dt3...dtk p 57 B where as just noted g(t) ≡ f(c(t)) det(Dc) . Now back up just a little. We used to have y = φ(x) and now we have x = c(t) as our underlying mapping. The object c is the mapping (just as Bucks say the curve is the mapping, not the image of the mapping). In the current case, what is the image? It is some kind of k-dimensional surface within Rm (page 37). Normally one takes m = k+1, but I don't think that would be required. For k = 1, the image could be a self-intersecting curve. For k = 2 it could be a self intersecting 2D surface as shown Buck p *. For k = 3, it could be some self-intersecting 3D surface displayable in R4. What is the situation when k = 0? Chapter 4 was k = 1, so maybe this was not addressed. 1. the differential form α must be a 0-form if k = 0, so α is a function of x. We have α = f(x) 2. The only point in t space is t = 0 . This space is R0. 2. The mapping c does this : c: (t=0) → image c(0) in Rm c(0) = x So I guess we imagine that the domain has only the point 0 and this maps to point x in Rm 3. c*α = g(t) which is a function (pullback does not change degree of a form) ∫c α ≡ ∫R c*α = ∫R g(t) δ(t=0)dt = g(0) below p 57 B 4. I guess we set det(Dc) = 1 and write g(t) ≡ f(c(t)) det(Dc) and g(0) = f(c(0)) * 1 = f(x) and I dimly arrive at result p 57 C. They call it a definition: ∫c α ≡ f(x) when α = f(x) a 0-form k = 0 Reparametrization We start with x = c(t) = some surface in the range of c: Rk → Rm. I will mimic my own proof from earlier in these notes for the case k = 1, but now k = k. We now have t = p(s) as the reparam: (c o p)* α = p*(c*α) = p*(g(t)dt) = g(p(s)) dp = (p*g) dp = (p*g)dp1dp2....dpk = (p*g) det(Dp) ds1ds2.....dsn = (p*g) det(Dp) dt = g(p(s)) det(Dp) ds // agrees with p 58 B We assume that t = p(s) is one-to-one and onto (bijective) and that det(Dp) ≠ 0. Then ∫c o p α ≡ ∫ (c o p)* α = ∫ g(p(s)) det(Dp) ds1ds2.....dsn // agrees p 58 C What is ? It is the mapping of the t-cube by t = p(s) . If p were linear, might be a block, but in general it is just some unknown shape, the image of the original block under p-1: t → s . But now we just do a multiple change of variables and we can write ∫ g(p(s)) det(Dp) ds1ds2.....dsn = sign[det(Dp)] ∫R g(t) dt But we earlier had from 57 B, ∫c α ≡ ∫R c*α = ∫R g(t) dt Therefore I have shown right here that ∫c o p α = sign[det(Dp)] ∫c α // agrees with p 58 E This says that any reparametrization leaves the k-form integral unchanged except for a possible sign! This seems to me to be a fairly powerful result. [ Thus we have generalized the 1-form result of Ch 4.] Comment: Recall from Buck that in respeeding we went from [a,b] to [α,β] as the interval change with I think with f = f(t). We didn't use a separate name for the function f(t). In Sjamaar page 48 we went from p: [ , ] to [a,b] under t = p(s) so went from t = [ , ] to s = [a,b]. Now in k dimensions on page 57 we have t = p(s) and this respeeding (reparam) takes t = to s = R where p: to R. So Sja is trying to use the same kind of notation, I appreciate that concession. So the reparam t = p(s) maps one cube into the other cube in a bijective manner probably with some extra rules. In the Buck 1D case we had conditions at the endpoints, so perhaps here there are conditions on the cube boundaries. [58] Remark. In (5.1) Sja identifies dt1 ˄ dt2 with normal calculus dt1dt2 with no comment. His reconciliation of the minus sign issues is OK, but I won't be happy until I understand the wedge product. [ I suspect Sja would say the wedges are present in page 57 B, I will deal with this later. ] About k-cubes and k-chains [59] Definition of a k-cube: It is a mapping c: [0,1]k → U . Seems to me it is whatever you get in this mapping, and the image is NOT a cube. It is the image of a cube. So seems an odd name. Certain mappings would in fact map the source cube into a block in U space. [ the word "cube" refers to the shape of the domain in the mapping, generalization of an "interval" in R1 ] Example: A 0-cube is c:0 → U and we decided ∫c α = ∫c f = f(x) . The point 0 has to map to some point x, so the claim is that 0-cube = {x} . Again, this seems to be the image of a 0-cube mapping. Example: A 1-cube is c: [0,1] → U and this gives some curve in U. The 1-cube image is not a line segment. Example: A 2-cube is c:[0,1]2 → U and this gives some surface say in R3. Again the image of the 2-cube is not a block in R3. Definition of a k-chain: It is a mapping formed as a linear combination of k-cube mappings. This is clearly shown on page 59 C. Somewhere we must have shown that ∫c+d α = ∫c α + ∫d α // probably page 38 i It then follows that (mappings are linear) ∫c α = ∫a1c1+a2c2+... α = Σi ai ∫ci α which is p 59 D This last line is "an integration over a k-chain" which is based "on the integration over a k-cube". Note: a k-cube is a surface of dimension m-1 in Rm let's say, so we are integrating over a sum of surfaces, or maybe it is the union of those surfaces, when integrating over a k-chain. Each surface is of the same dimension and is a k-cube. Example: A 0-chain would be Σiai{xi}. He wants us to think of this the way we think of ρ(x) = Σi aiδ(x-xi) ai = charges at points xi Example: A 1-chain would be c = Σiaici. Perhaps this is a sum of curves in the image space, each with some weight ai. For a 1-chain remember that a 1-cube image is a curve in space, it is not a cube. For a 0-chain replace curve with point. Idea: Maybe we are going to "chain" some curves together to get a Stokesian boundary or something like that. Things are a little hazy to me at this point in terms of those charge analogies. 2.13.16. Alright, for k = 2 you could string together a bunch of patches on a donut to cover the donut. Each of these patches could from a mapping on [0,1]2 so each donut's curved patch is a 2-cube and then you cover the entire donut with a chain of these 2-cubes. I think that is the Manifold idea starting here. 5.2 The boundary of a chain The boundary of a curve is stated as ∂c = {c(1)} - {c(0)}. What does this mean? Is this the difference between two sets? It seems to be one set containing 1 point minus a second set containing a different point. Sja notation is hazy, he never uses the word set. Since this is a sum of points, it is a 0-chain, so the ∂ operator lowered you one level of chain space. For a 2-cube (a finite surface patch) we get this ∂c = c1+c2- c3-c3 = a chain of 1-cubes. In this little sum, c is a 2-cube while each of the ci is a 1-cube. ∂ lowers the degree by one. The boundary is the sum or union of the four curved segments shown (a 1-chain of 1-cubes), but they have signs based on the direction for each segment. I recall something from Buck on this but don't digress now. In the above case we have c1(t) = c(t,0) c2(t) = c(1,t) c3(t) = c(t,1) c4(t) = c(0,t) Again each c here is a 1-cube! Note that 0,1 specify location on the four sided patch but in [0,1]2. For a 3-cube boundary we might have 2k = 6 faces as on a normal cube. Perhaps c1(a,b) = c(a,b,0) c2(a,b) = c(a,b,1) // an opposite side pair c3(a,b) = c(0,a,b) c4(a,b) = c(1,a,b) // an opposite side pair c5(a,b) = c(a,0,b) c6(a,b) = c(a,1,b) // an opposite side pair We can rewrite this as c1(t1,t2) = c(t1,t2,0) c2(t1,t2) = c(t1,t2,1) // an opposite side pair c3(t1,t2) = c(0,t1,t2) c4(t1,t2) = c(1,t1,t2) // an opposite side pair c5(t1,t2) = c(t1,0,t2) c6(t1,t2) = c(t1,1,t2) // an opposite side pair And if we now define t = (t1,t2....tk-1) being on [0,1]k-1 then [ note the notation here!! ] c3,0(t) = c(t1,t2,0) c3,1(t) = c(t1,t2,1) // an opposite side pair c1,0(t) = c(0,t1,t2) c1,1(t) = c(1,t1,t2) // an opposite side pair c2,0(t) = c(t1,0,t2) c2,1(t) = c(t1,1,t2) // an opposite side pair or reordering c1,ρ(t) = c(ρ,t1,t2) t = t1,t2 c2,ρ(t) = c(t1,ρ,t2) c3,ρ(t) = c(t1,t2,ρ) where we have a pairwise labeling scheme. This is the meaning of page 60 C. The first index tells you the number of the slot where the 0 or 1 is located. Notice that these boundary faces are each defined on [0,1]2 so whereas the 3-cube c is a map of [0,1]3, the boundary ∂c is a 2-cube mapping [0,1]2. At this point the reader can look at a separate doc "detail work on k-cube faces.doc". It is 9 pages of fine detail, so I want to keep it out of this huge notes file. Here is a summary of these side notes: 1. A clarification the meaning of a k-cube 1 2. Here is the Rule for creating faces: 1 3. Face of a Face 2 4. Sign of a Face 4 5. Proof that ∂(∂c) = 0. 8 So this takes us through the end of Section 5.2, and I proved the fact that ∂2 = 0 in full detail. I accept the formula p 60D for the boundary of a k-cube. I see how it works for a 2-cube, but I have no geometric interpretation for the signs of the 6 faces of a cube, although pairs always have opposite sign, and the three + faces are adjacent, so in a huge pile of cubes you will get cancellation of all internal faces in some sense. 5.3 Cycles and Boundaries Suppose you have a 2-cube where c(t1,t2) normally describes a patch, but suppose this thing does not depend on t2. Then c(t1,t2) = c(t1) = a 1-cube = a curve. This would be a degenerate 2-cube. A degenerate 1-cube means c(t1) = x0, some constant, so your curve becomes just a point. You are losing a whole dimension in such "first order degeneracy". For a degenerate curve c(t1) = k if you integrate a form α over that curve, you really have no integration at all, so the integral is 0. Sja comments at this point that no work is done on a motionless particle, and that is what he means. What happens if you do this integral ∫c α when α is a k-form, and where c is a degenerate k-cube? Mixing lots of concepts together here: k-forms, k-cubes (chains) and integration. To evaluate this integral, I would write it as a pullback to the unit cube in the "left" space to get ∫c α = ∫[0,1]k c*α α = ΣI fI(y) dyI y = φ(x) is y = c(x) c*α = ΣI fI(c(x)) dcI = ΣI fI(c(x)) det(Dc) dxI So then ∫c α = ∫[0,1]k c*α = ∫[0,1]k ΣI fI(c(x)) det(Dc) dxI (Dc)ij = ∂jci But if c does not depend on one of the coordinates, then ∂jci for that j and all i, which is a whole column of the matrix (Dc), so det(Dc) = 0 and then we have ∫c α = 0. The argument Sja gives for this proof requires the introduction of a new function g and we then have c* = f*g* and it just seems way complicated to me. The point to me is that if you have a degenerate c, then det(Dc) = 0 and you are done! Just above I have proved [61] Lemma 5.5: " The integral of a k-form α over a degenerate k-cube is zero. " This is then of course true for a degenerate k-chain which is a linear combination of degenerate k-cubes. Suppose you had a k-cube c such that ∂c is a degenerate (k-1)-cube. In this situation, one says that the original k-cube c is closed. One also says it is "a cycle". The reasons for these words are not clear yet. Definition: A k-cube c is closed or a cycle if the (k-1)-cube ∂c is degenerate. Now suppose a k-cube c can be written as c = ∂b + c' where c' is degenerate. If this is the case, then c is said to be a boundary for the (k+1)-cube b. [ A little like a form being exact, but an extra term ]. Definition. If a k-cube c can be written as c = ∂b + c' where c' is a degenerate k-cube, then c is the boundary of the (k+1)-cube b. Example 5.6. Let c1 and c2 be two curves which form a closed loop. These curves are 1-cubes. Perhaps if c = c1+c2 then ∂c is degenerate since there is then no boundary. I would say ∂c = 0. I guess such a c is regarded as a degenerate 1-cube and therefore you would say that c = c1+c2 was closed or was a cycle using the above definitions. Now the words closed and cycle make sense. 2.13.16: If a 1-cube is non-degenerate, then it's boundary is a set of points each of which has dimension 0. For example, a semicircle curve has a boundary consisting of the two endpoints. If a 1-cube is degenerate, then the boundary is lower than the official dimension, and in this case instead of having a set of points, you have no points at all in the boundary! You might call that dimension = -1, so we dropped a dimension. If you could have a circle be a 1-cube, we know it has no boundary so ∂c = 0 and dim(∂c) = -1, so to speak. In any event, this would be an example of a degenerate 1-cube because the boundary has lower than the expected dimension. A half spherical shell is a 2-cube and it has a circular 1D boundary, so it would be a normal non-degenerate 2-cube. But if a full spherical shell could be a 2-cube, then since it has no 1D boundary at all, it is degenerate. I think that a circle can only be a 1-chain and a spherical shell a 2-chain, but I will hedge on that for now. [62] Lemma 5.7: The boundary of a degenerate k-cube is a degenerate (k-1)-cube. Proof: Let c be a degenerate k-cube. Focus just on argument ti and assume c no depend on ti. This means that ci,0 = ci,1. Now what is ∂c for our c? The full formula is this ∂c = Σj=1k (-1)j(cj,0 - cj,1). The term in this sum with j = i then vanishes so we then have ∂c = Σj≠i (-1)j(cj,0 - cj,1). // this is page 62 D (*) Since c does not depend on ti, it seems that cj,ρ also does not depend on i. Remember [c(t1,t2,t3.. tj-1,tj,tj+1..tk)]j,ρ → c(t1,t2,t3.. tj-1,ρ,tj+1..tk) → c(t1,t2,t3.. tj-1,ρ,tj..tk-1) If ti is to the left of j, we could just put it in and find that if c does not depend on ti, neither does cj,ρ. But suppose ti is to the right of j (like the tk is). Then this ti gets renumbered to be ti-1. In either case, we find that cj,ρ is degenerate because its dependence on at least one variable is missing. Thus from (*) above we conclude that ∂c is degenerate. QED. Corollary: Every boundary is a cycle Proof: If ∂c is a "boundary" of c, then we can write c = ∂b + c' by the boundary definition above, where c' is degenerate. Now consider ∂c = ∂2b + ∂c' = ∂c'. Since c' is degenerate, so is ∂c' by the above lemma 5.7. Therefore ∂c is degenerate. By the definition above, c is then a cycle (it is closed). QED Comment: In the formula c = ∂b + c' in 2D. This issue has come up somewhere before. Suppose the ellipse including interior plus whisker sticking out (line segment) is patch b. Then c = ∂b is the red perimeter of the ellipse plus the boundary points of the whisker (I show one in red). But the whisker boundary points are just points, so c' = those points = a degenerate curve. The points are loci of one less dimension and would contribute nothing to a line integral of something around the boundary. So the point is that in this case, the ellipse c is regarded as the boundary of the entire k-chain and the fact that you have c = ∂b + something else, does not matter. [62] 5.9 Unit Circle Example. This took me a while to decipher. Consider this 2-cube b b(t1,t2) = (1-t2)c(t1) where c(t1) is a unit circle parametrized by t1, meaning c(t1) = (sin(2πt),cos(2πt)) = (x,y). In this example, we shall show that ∂b = c - c' where c' = (0,0) = a degenerate 1-cube and this provides an example of the idea that c = ∂b + c' Details: looking at b above, you can see that as you vary the two parameters, you sort of fill out the area of a disk of radius 1, so this is certainly a good candidate for a 2-cube with boundary being a circle. So let's use our formula of page 60 D to compute ∂b : ∂b = Σi=12 (-1)i [ bi,0 - bi,1] // a 2-cube is a sum of 1-cubes = a sum of curves = - [ b1,0 - b1,1] + [ b2,0 - b2,1] = - b1,0 + b1,1 + b2,0 - b2,1 = four terms We then go off and compute these four terms: Be careful with the argument renaming! b(t1,t2) = (1-t2)c(t1) b1,0 = b(0,t1) = (1-t1)c(0) = (1-t1)(1,0) = (1-t1, 0) this curve is a hor line segment. b1,1 = b(1,t1) = (1-t1)c(1) = (1-t1)(1,0) = (1-t1, 0) b2,0 = b(t1,0) = (1-0)c(t1) = c(t1) b2,1 = b(t1,1) = (1-1)c(t1) = (0,0) Adding up the four terms we then get ∂b = - b1,0 + b1,1 + b2,0 - b2,1 = - (1-t1, 0) + (1-t1, 0) + c(t1) - (0,0) = c(t1) - (0,0) and therefore c(t1) = ∂b+ (0,0) In this case, the extra piece c' = (0,0) is a degenerate 1-curve. Here is a picture that might explain what is happening here (this picture is apparently implied by the explicit form of b stated above) The four terms of ∂b are the four pieces of this boundary. The two line segments (1-t1, 0) cancel, and you end up with the circle plus the point at the origin which is the (0,0). 5.4 Stokes's Theorem Recall Theorem 4.4 from page 49, where α is an exact 1-form and c is a curve and g is a 0-form = function: ∫c α = g(c(b)) - g(c(a)) where α = dg (*) which I derived in this manner, noting that c*α = c*(dg) = d(c*g) = dh where h(t) = c*g = g(c(t)), ∫c α ≡ ∫[a,b] c*α = ∫[a,b] dh = h(b) - h(a) = g(c(b)) - g(c(a)) QED Notice that the proof here builds on the notions of: the pullback to be the definition of the integral over a form and the definition of φ* in this context the theorem that d(φ*α) = φ*(dα) the general theory of differential forms This (*) is the "fundamental theorem of calculus in Rn (FTOC)" written in the language of 1-forms. Rewrite then as ∫c dg = g(c(b)) - g(c(a)) where c is a curve with endpoints c(a) and c(b). Note that c:R1→Rn so in general the curve lies in Rn. A special case is for n = 1 and then the curve is like y = f(x) in regular 1D calculus, here y = c(x). Here we are integrating an exact 1-form α = dg. Now consider our earlier result from page 57 where we integrate a 0-form over a 0-cube c where c = x = just a point. ∫c f = f(x) Using this notion, we can write g(c(b)) = ∫c g = ∫{c(b)} g. Then our FTOC reads ∫c dg =∫{c(b)} g - ∫{c(a)} g = ∫{c(b)}-{c(a)} g Now we know from the bottom of page 59 or from the general formula p 60 D that {c(b)}-{c(a)} = ∂c Therefore our FTOC appears as (replace g with symbol α) ∫c dα = ∫∂c α where α is a 0-form and c is a 1-cube, ∂c = 0-cube It turns out that this theorem stated exactly this way is also valid for α = k-form and c = (k+1)-cube and then ∂c = k-cube. This more general theorem is the famous Stokes's Theorem which then will encompass all my "integral theorems". For the proof we will instead say α = (k-1)-form c = k-cube ( c-chain, etc etc) dα = k-form ∂c = (k-1)-cube Side Question: Why isn't the above FTOC on my list of "integral theorems"? I think it only applies to the case that you have α = Fdx where F is a conservative force and only in that case is α a perfect differential. [ I had trouble with this proof because you need to show the wedge product symbols in some places and not in others. I will add those symbols where I think they go: ] Proof: We know that ∫c dα = ∫[0,1]k c*(dα) = ∫[0,1]k d(c*α) // pullback and theorem on d and * We know that α is a (k-1) form, so the pulled back c*α is also a (k-1)-form since the degree of forms does not change when you do the pullback. Therefore we can write c*α = (k-1)-form = Σi gi(t) dt1^dt2^ ...dti....^dtk dti (red) is missing page 63 B This is the most general form you can have for a (k-1)-form and gi are the coefficient functions. More detail on this: c*α = c*[ΣI fI dxI] = ΣI c*(fI) c*(dxI) = ΣI fI(c(x))dcI = ΣI fI(c(x))(Dc)dtI = ΣI gI dtI p 42A,B Note that the tj are all in proper order. Now apply d: d(c*α) = Σi (dgi) dt1^dt2^ ...dti....^dtk = Σi Σj [(∂jgi)dtj] dt1^dt2^ ...dti....^dtk In the Σj only j = i will contribute since otherwise you get dtjdtj appearing which is 0. So, d(c*α) = Σi (∂igi)dti^ dt1^dt2 ...dti....^dtk = Σi (-1)i+1 (∂igi) dt1^dt2 .....^dtk where we pick up the sign putting dti into is correct location. Aside: Proof that correct sign is (-1)i+1 : Consider: dt1dt2 ..dti...dtk = (-1)s dti dt1.....dti...dtk Suppose i = 1. this then reads dt1dt2 ..dti...dtk = (-1)s dt1dt2 ..dti...dtk s = even so s = (-1)i+1 is correct Suppose i = 2. Then dt1dt2 ..dti...dtk = (-1)s dt2 dt1dt3......dtk and here need s = odd Then we have ∫c dα = ∫[0,1]k d(c*α) = ∫[0,1]k {Σi (-1)i+1 (∂igi) dt1^dt2^ .....^dtk} = Σi (-1)i+1 ∫[0,1]k (∂igi) dt1^dt2^ .....^dtk page 63 C This is the point where we have things in standard order and we want to actually do the integration. If we do that right here, we then have ∫c dα = Σi (-1)i+1 ∫[0,1]k (∂igi) dt1dt2 .....dtk // no ^ We would then continue this way ∫[0,1]k (∂igi) dt1dt2 .....dtk = ∫[0,1]k (∂igi) dti dt1dt2 ..dti...dtk // no ^ page 63 D1 So this is the only way I can obtain page 63 D1. He slips dti to the left without a sign because we have already collapsed out the ^ symbols and we are just working on the multivariable integration. At this point we then have ∫c dα = Σi (-1)i+1 ∫[0,1]k (∂igi) dti dt1dt2 ..dti...dtk = Σi (-1)i+1∫[0,1]k-1 [ ∫[0,1] (∂igi) dti ] dt1dt2 ...dti....dtk . Ignoring other coordinates we find ∫[0,1] (∂igi) dti = !Syntax Error, I (digi(ti))dti = !Syntax Error, Idt = gi(ti=1) - gi(ti= 0) = gi(t1,t2,....ti-1,1,ti+1... tk) - gi(t1,t2,....ti-1,0,ti+1... tk) We then have ∫c dα = Σi(-1)i+1 ∫[0,1]k-1[gi(t1,t2,....ti-1,1,ti+1... tk) - gi(t1,t2,....ti-1,0,ti+1... tk)] dt1dt2 ...dti....dtk Now at this point we can reinsert the ^ symbols, without changing any ordering ∫c dα = Σi(-1)i+1 ∫[0,1]k-1[gi(t1,t2,....ti-1,1,ti+1... tk) - gi(t1,t2,....ti-1,0,ti+1... tk)] dt1^dt2^ ...dti....^dtk // dti is missing Now a notation detail we have to work on. Let x = c(t) . If we fix ti at value ρ, we can define a new mapping x = d(t1,t2....ti....tk; ti = ρ) mapping of only k-1 variables I could then define the mapping ci,ρ as ci,ρ ≡ d Now go back to c*α = (k-1)-form = Σi gi(t1,t2,t3 ..ti ...tk) dt1^dt2^ ...dti....^dtk Then how about d*α = (k-1)-form = Σi gi(t1,t2,t3 ..ρ ...tk) dt1^dt2^ ...dti....^dtk So nothing illegal here. But then with the above definition we get [ci,ρ]* α = Σi gi(t1,t2,t3 ..ρ ...tk) dt1^dt2^ ...dti....^dtk = Σi [gi]i,ρ dt1^dt2^ ...dti....^dtk (*) where I have NOT made any variable name adjustments in [gi]i,ρ . So OK, then ignoring that little adjustment issue, we have ∫c dα = Σi (-1)i+1∫[0,1]k-1[gi(t1,t2,....ti-1,1,ti+1... tk) - gi(t1,t2,....ti-1,0,ti+1... tk)] dt1^dt2^ ...dti....^dtk = Σi(-1)i+1 ∫[0,1]k-1[ [gi]i,1 - [gi]i,0] dt1^dt2^ ...dti....^dtk = Σi(-1)i+1 ∫[0,1]k-1[ (ci,1)* α - (ci,0)* α ] // using (*) above = Σi(-1)i ∫[0,1]k-1[ (ci,0)* α - (ci,1)* α ] = Σi Σρ (-1)i+ρ∫[0,1]k-1[ (ci,ρ)* α ] Now use the pullback definition to write this as = Σi Σρ (-1)i+ρ∫c(i,ρ) α Now do the sum acting on the integration domain and use formula p 60 D and you get = ∫∂c α and this then concludes the proof. Summary of this proof of the general Stokes's Theorem Step 1: ∫c dα = ∫[0,1]k c*(dα) = ∫[0,1]k d(c*α) pullback form φ*d = dφ* p 39 Step 2: c*α = (k-1)-form = Σi gi(t) dt1^dt2^ ...dti....^dtk d(c*α) = Σi (dgi) dt1^dt2^ ...dti....^dtk = Σi Σj [(∂jgi)dtj] dt1^dt2^ ...dti....^dtk = Σi (-1)i+1 (∂igi) dt1^dt2 .....^dtk // slide dti into place Step 3: ∫c dα = ∫[0,1]k d(c*α) = Σi (-1)i+1 ∫[0,1]k (∂igi) dt1^dt2^ .....^dtk Step 4: ∫c dα = ∫[0,1]k d(c*α) = Σi (-1)i+1 ∫[0,1]k (∂igi) dt1dt2 ....dtk // no ^ Step 5: ∫c dα = Σi (-1)i+1∫[0,1]k-1 [ ∫[0,1] (∂igi) dti ] dt1dt2 ...dti....dtk Step 6: ∫c dα = Σi (-1)i+1∫[0,1]k-1 [ [gi]i,1 - [gi]i,0] dt1dt2 ...dti....dtk = Σi Σρ (-1)i+ρ Σρ∫[0,1]k-1 [gi]i,ρ dt1dt2 ...dti....dtk = Σi Σρ (-1)i+ρ Σρ∫[0,1]k-1 [gi]i,ρ dt1^dt2^ ...dti....^dtk Step 7: [ci,ρ]* α = Σi [gi]i,ρ dt1^dt2^ ...dti....^dtk Step 8: ∫c dα = Σi Σρ (-1)i+ρ Σρ∫[0,1]k-1 [ci,ρ]* α = Σi Σρ (-1)i+ρ Σρ∫c(i,ρ) α = ∫∂c α // using page 60 D for ∂c expression I had to add the wedge stuff to make it work right and to agree with Sja's waypoints. Spivak has a similar proof on page 102 that I have not studied. It is not clear, but Spivak also seems to play with removing the ^ operators in a judicious fashion. After a long search I finally landed this Flanders book. But it is like all the rest, there is no definition of the wedge product, he just starts using it. [ downloaded again on about 2.11.16 ]. The last Sja comment concerns the angle form and the circle once again. Given the puncture, we know we cannot write c = ∂b + c' because when we did that, we had to have c' = {0} which is not present in the punctured domain U. We know that ∫cα = 2π. We know that the angle form is closed but not exact, and we showed earlier that dα = 0 (closed). But then if c = ∂b + c' were true and if you could apply Stokes, it would say 2π = ∫c α = ∫∂b α = ∫b dα = 0 = ∫b 0 = 0 and we would get a contradiction. Stokes Exercise 5.6. This is the big payoff for the general theorem ∫c dα = ∫∂c α . Recall this table from earlier: α = f 0-form in Rn α ↔ f PL1 dα = df 1-form in Rn dα ↔ f // namely, dα = fdx α = f 0-form in Rn α ↔ f *d[*(dα)] = ∂i2f = 2f 0-form on Rn *d[*(dα)] ↔ 2f PL2 α = F dx 1-form on Rn α ↔ F *[d(*α)] = (div F) 0-form on Rn *[d(*α)] = div F PL3 α = F dx 1-form in R3 α ↔ F one-to-one *(dα) = [curl F] dx 1-form in R3 *(dα) ↔ curl F PL4 (i) Let dα = fdx = a 1-form. Integrate this over a curve. Then ∫c dα = ∫c fdx = ∫c grad f dx = ∫∂c α = ∫∂c f = f(final end) = f(initial end) = f(c(1))-f(c(0)) This is for any curve parametrized over [0,1]. This appears as my "other theorem" in integral theorems doc: !Syntax Error, Iφ dl = φ(B) - φ(A) (9) MM 4-36 p 156 (ii) Green's Formula: Let α = fdx+gdy. Then dα = df dx + dg dy = ∂yf dydx + ∂xgdxdy = (∂xg-∂yf)dxdy So α = fdx+gdy 1-form dα = (∂xg-∂yf)dxdy 2-form Then let c = 2-cube or 2-chain. ∫c dα = ∫c (∂xg-∂yf) dxdy = ∫∂c α = ∫∂c (fdx+gdy) so ∂c = a curve This is usually called "Green's Theorem" and appears as a one-liner in the Stokes part of my doc. This can be interpreted as Stokes for A = f +g as I show there. (iii) The divergence theorem (Gauss's theorem, or Gauβ' formula says Sja) α = F [*dx] dα = dF [*dx] = dFi [*dx]i = (∂jFi) dxj [*dx]i = (∂iFi) dV So Stokes says: ∫c dα = ∫c [div F] dV = ∫∂c α = ∫∂c F [*dx] If x is in Rn, then c is a region of Rn and and ∂c is a boundary which is a surface of degree n-1. The object [*dx] for n = 3 would be *dx1 = dx2dx3 so that dx1*dx = dx1 dx2dx3 = +dV. Write [*dx] = dA if you want to think of this as a patch of area of dimension n-1. But you see that the version here is clearer that my version ∫V dV A = ∫S dSA which is hazy about the meaning of dA except when n = 3. (iv) Stokes's theorem for n = 3 α = Fdx = 1-form dα = ∂jFidxjdxi = Σij,j≠i ∂jFidxjdxi = Σi<j(∂jFi-∂iFj)dxjdxi = Σi<j(∂iFj-∂jFi)dxidxj = [curl F]k (*dxk) = curl F (*dx) = 2-form Then if c is a 2-form meaning a surface (open) and ∂c is the boundary of that surface. ∫c dα = ∫c curl F (*dx) = ∫∂c Fdx Compare to integral theorems doc, where again (*dx) = dA = dS = area element. ∫S dS ( x A) = dl A (7) Schaum 22.60 This concludes (finally) my voyage through Sjamaar Chapter 5. No meta notes yet on Ch 5. Comment on [*dx] in an area integral. F dA = Fi dAi = F1dA1 + F2dA2 + F3dA3 = F1dx2dx3 + F2dx3dx1 + F3dx1dx2 = F1[*dx1] + F2[*dx2] + F3[*dx3] = F [*dx] So this is not so strange after all. Exercise 5.5. Consider the famous equations x = rsinθcosφ y = rsinθsinφ z = r cosθ Now consider a block such that r = [0,a], θ = [0,π], φ = [0,2π]. This is close enough to a cube. Call this block B. Then the above equations can be written as a set S = f(B) . As r,θ,φ scan over the block, we get a certain set S in R3. This set is "the sphere". So roughly speaking, this is a mapping from the unit cube to a sphere. We could redo it this way x = u1sin(πu2)cos(2πu3) y = u1sin(πu2)sin(2πu3) z = u1cos(πu2) Then this map x = F(u) is a mapping from [0,1]3 a 3D sphere. See Fig 1.7 of tensor doc for the picture of the block. I suppose I could have add a picture of a sphere on the RHS. So, I know how to map from a unit cube to a sphere. I new it all along really. Exercise 5.6 to show now the general Stokes' theorem generates all the famous integral theorems: ∫c fdx = f(c(1))-f(c(0)) "other theorem" ∫c (∂xg-∂yf) dxdy =∫∂c (fdx+gdy) ∂c = a curve "Green's theorem" ∫c [div F] dV = ∫∂c F [*dx] divergence theorem (Gauss's theorem) ∫c curl F (*dx) = ∫∂c Fdx Stokes's theorem