Sja Ch 7 notes

Sjamaar Chapter 7 Notes PhL 8.14.15 Cutting first real new ground here after Cape Cod trip several weeks ago. Chapter 7: Differential Forms on Manifolds. 7.1 First definition In earlier sections, we learned about differential forms defined only on Rn. The question now is how you might generalize the concept of differential forms for use on a manifold M of dimension n within RN. Earlier we always had things like dxdy where these were Rn coordinates. But didn't we already do some forms work on surfaces? Stokes' theorem for example involves regions and boundaries either of which could be surfaces. I should review that a bit before continuing. On page 47 (Chap 4) we integrate over a curve c which exists in Rn . Then in Chap 5 we do ∫c α where c is a surface of dimension k within Rn. So to the extend these integration domains were manifolds, we have in fact already done some integration of forms over manifolds. So I will be looking for what is new in this Chapter. [ 2.14.16: Chap 5 integrated forms over k-cubes and k-chains, not over manifolds. ] Sja states the "problem" here as that if you are at a point on a manifold, the "usual" basis vectors I guess of RN might take you out of the manifold so those derivatives don't exist within M. But recall from the last chapter that when Sja talked about embeddings and manifolds, a point x on the manifold M was always surrounded by an open set V whose intersection with M gives a path on the surface. So if that set V exists surrounding x in all directions, then the derivatives in all directions do make sense. When we play with (Dψ), we are saying that all these derivatives exist. The existence of that cushion of space V surrounding a point x on M is of course directly connected to the notion of "embedding" and "manifold". So Sja wants us to think of an atlas of embeddings whose charts are labeled by index i, such as ψi , which cover M, and we can then think within each of these embeddings. We break our domain space into a set of probably overlapping Ui and these go to cushions Vi with ψi. Recall that a 0-form is just a function f where we usually wrote α = ΣI fI(x) dxI and for a 0-form we would write α = f(x). In the pullback chapter we wrote α = Σ fI(y) dyI (p 36) with y = φ(x) where x was a point in our flat pullback space, picture page 37. But here as in Chap 6 we want x = ψ(t) where t is in our flat space as in the toroid picture of page 68. I now have to clarify and merge the forms world with the manifolds worlds, and get a consistent notation. On page 81 we have ψi: Ui → RN t ϵ Ui ψi(Ui) = patch on M, part of a cover of M Let's assume for the moment that we are dealing with α = ΣI fI(x) dxI where x lies on M. And for the simple 0-form case we have α = f(x) where x = ψ(t) for our embedding. At a particular patch then we have αi = f(xi) = f(ψi(t)) ≡ fi(t) // new functional form of t (different function of t) Then fi(t) is the local representative of f in the tiny mapping i. Now recall that we can write the above using the pullback operator ψi* in this way fi(t) = f(ψi(t)) = ψi* f(t) = (f o ψi)(t) . So we have fi = ψi*f and we can say that that the 0-form representative fi is a pullback of f. So now we are interested in derivatives of fi(t) in the little ψi world only (why?). Right now I think Sja is talking only derivatives with respect to t components in the flat space side. Maybe fi is Ck for all i, and then we can say that f is Ck overall. What then happens at the overlap of two charts i and j? Let t in Ui and u in Uj and both these are in an overlap region. Then x = "ψ(t)" = ψi(t) = ψj(u) x = point on the manifold M f(x) = f(ψi(t)) = f(ψj(u)) f(x) = fi(t) = fj(u) So consider: [ recall that ψ-1 exists for any embedding ] ψi(t) = ψj(u) t = ψi-1(ψj(u)) = (ψi-1 o ψj)(u) (*) fi(t) = fj(u) t = f-1(fj(u)) = (fi-1 o fj)(u) From the top right result (*) we can write fi(t) = fi[(ψi-1 o ψj)(u)] . But using pullback notation, this says fi(t) = (ψi-1 o ψj)*fi(u) . But since fi(t) = fj(u) on the left we end up with fj(u) = (ψi-1 o ψj)*fi(u) for any u in Uj and so we suppress the argument to say fj = (ψi-1 o ψj)*fi which is p 81 A "consistency condition" This is a consistency condition on representatives of the 0-form f on two regions Ui and Uj which overlap and which have embedding functions ψi and ψj. Interestingly, you can interpret fi as a pullback of fj and vice versa. It seems odd that two objects on a fairly equal footing are pullbacks of each other. We can alternately write from (*) that u = ψj-1(ψi(t)) = (ψj-1 o ψi)(t) . If we let t run over all of Ui , we get this set of possible u values. {u} = (ψj-1 o ψi)(Ui) which is p 81 B where we are now assuming that u and t and x are all in a region of chart i and j overlap. In going from region Ui with t to region Uj with u, we are doing a change of coordinates. Switching from map of New York to map of New Jersey. [82] With this as introduction, we now want to talk about forms. We just showed that on manifold M we can replace f by its local representative fi on each Ui region of the flat space, so that is how we dealt with 0-forms. Now for k-forms, we try doing the same idea: represent α locally by a form αi and mimic the above fj = (ψi-1 o ψj)*fi // line below mimics this line αj = (ψi-1 o ψj)*αi which is p 82 (7.1) Now remember that the φ* type operator affects both the function and the derivative part of something like α = ΣI fI(x) dxI, so we are generalizing off our function result to get a more general form result. Now αi is the local representative of the k-form α on Ui and for ψi. And now we have αi being a pullback of αj and vice versa, where αi and αj are now representatives of k-form α on the two regions Ui and Uj which are associated with embedding functions ψi and ψj. Also, for the function we had fi(t) = f(ψi(t)) = ψi* f(t) = (f o ψi)(t) which we abbreviate to say ψi* f(t) = fi(t) . If we replace fi by αi and f by α, then we have ψi* α = αi The k-form αi representative on Ui is thus a pullback under ψi* of the full α. Now item (7.1) stated above is a "consistency condition on the k-form representatives αi and αj where regions Ui and Uj overlap. This consistency condition involves the embedding functions for these same two regions going to the Vi and Vj cushions around x. As other notation Ωk(M) = collection of k-forms on the manifold M Now suppose we have these representative forms in a region of overlap between i and j. αi = ΣI fIdtI αj = ΣJ gJdtJ What then does (7.1) shown above say? αj = (ψi-1 o ψj)*αi ΣJ gJdtJ = (ψi-1 o ψj)* ΣI fI(t)dtI Lets define φ ≡ ψi-1 o ψj Then we have (see page 15 of raw Ch 3 notes or page 42 B ) αj = (ψi-1 o ψj)*αi = φ*αi or ΣJ gJdtJ = φ* (ΣI fI(t)dtI ) = ΣI fI(φ(t))dφI = ΣI,J fI(φ(t)) det(Dφ)I,J dtJ Comparing the two sides we get gJ = ΣI fI(φ(t)) det(Dφ)I,J = ΣI φ*fI(t) det(Dφ)I,J . (**) Write this out then as gJ = ΣI (ψi-1 o ψj)*fI det[D (ψi-1 o ψj)]I,J which is page 82A So this relates the component functions gI of the form representative αj to those fI of αi. This is what Sja means by "components". I might call these things coefficient functions of the differentials in the form. So I arrive at the end of Section 7.1, I am still standing but slightly dazed. What has happened here? We deal with a manifold as being covered by Ui sets in the flatspace [ so Vi intersect M in the M space ] . In each Ui part of the problem we have a form αi. Since regions i and j overlap, we have strange consistency conditions for points in such overlap which have the form of pullbacks. The point is that we have a nice embedding ψi well defined in our little portion i of the problem. One hopes that eventually there will be some examples. Comments: According to this "first definition" of "a form on a manifold", you have a general form α on a manifold M and it is defined in terms of its pullbacks αi on each covering patch Ui. As noted above, αi = ψi*α where ψi is the transformation or embedding associated with patch Ui. This certainly seems a reasonable approach to me. We even have an expression above gJ = ΣI fI(φ(t)) det(Dφ)I,J showing how the form coefficients of αj are related to those of αi, where φ ≡ ψi-1 o ψj . 7.2 Second definition 2.15.16: In this "second definition" of "a form on a Manifold", a form is identified with an element of the dual wedge space which I call Λk in wedge doc. This of course leads to a world of alternating functions, but I am not clear how this association "helps me do something useful". This is a Grassman theory section and I just know it is going to be brutal. The wedge product is going to appear, along with alternating multi-linear functions, that whole ball of wax is coming down the pike! [page 83] 2.15.16: Thus stuff was totally strange to me on first reading. Now that I have written wedge doc, it is all I think pretty well understood. The opening half page is rather shocking. It seems to be an alternate forms universe as presented by Grassmann. The key new idea is this dxi ≡ eiT = (0,0,...1,..0) basic form dxi = a unit vector This is a totally mysterious identification of a differential distance with a vector, which is really an identification of dxi with λi which is a basis vector of the dual wedge space Λ1 = L1. The reader has to forget that dxi in some other World is a differential. Here it is just a symbol identified with eiT. Then we write a 1-form as α = Σi fi dxi = Σi fi eiT = (f1,f2....fn) = a row vector of functions (expanded on the dual basis) So now a 1-form is identified with a vector of that dual space! The following gray was written before wedge doc and I have corrected it where I can changing to ^ . I will just guess (this guess was not quite right!) for a 2-form something like this α = Σij fij dxi dxj = Σi fij [eiT ^ eiT]= a rank-2 tensor of functions, such as in my tensor doc Actually, I will refine this a bit. Recall from tensor doc that one useful tensor expansion of tensor A is this where Aijk are contravariant tensor components. But in Cartesian space up and down don't matter, so they are also covariant components. So maybe the 2-form will be α = Σi fij dxi dxj = Σij fij ei^ ej where fij are covariant rank-2 tensor components, and these ei are those of Sjamaar, not the tangent base vectors of tensor doc. This is just a conjecture right now, and I don't know about the transpose things. Any way, Sja is reporting out f = (f1,f2....fn)T as a covariant vector = covector. I think in general, any vector in any dual space is called a covector. So in this completely new approach, a 1-form is a covariant vector, or covector for short. I suppose this is not too different from the "association" we earlier discussed between 1-forms α and vector fields F. Another definition I think I have seen somewhere: (yet another meaning of the * symbol! ) (Rn)* = space of constant row vectors = the dual of Rn Side Note: If g is a vector function we can write dg = Σi(∂ig) dxi = (Dg) dx = (∂1g, ∂2g.....∂ng) = the usual Jacobian where we have used the new fact that dxi ≡ eiT. The Dual Space Concept I think this is something I already know a little about from tensor doc. Recall there from page 75: ______________________________________________________________________________ Notes on Reciprocity (Duality) 1. Suppose some set of vectors bn forms a complete basis for x-space. Can one find a set of vectors Bn that have the property Bm bn = δm,n ? // duality relation; Ba and ba are reciprocal (6.2.8) As shown below, the answer is normally "yes", and the vectors Bn are uniquely determined by the bn. One says that the set {Bn} is "dual to" the set {bn} and vice versa. If we regard bn as a basis, then Bn is the "dual basis", and Bm bn = δm,n is the "duality relation". Another terminology is that the vectors Bn are "reciprocal to" the vectors bn and vice versa. 3. One can solve for the Bn in terms of the bn. Each Bn has N components, so there are N2 unknowns. The duality relation Bm bn = δm,n is a set of N2 equations. This is basically a Cramer's Rule problem in N2 variables. Since the bn form a complete basis, one can expand Bm on the bn with some coefficients we will call w'mn ( at this point w'mn is unknown), Bm = w'mnbn . (6.2.9) Then from (6.2.8), δm,k = Bm bk = w'mn bn bk . // bn bk = ij(bn)i(bk)j . (6.2.10) Define matrix W' by, W'nk ≡ bn bk (6.2.11) and note that W'nk is symmetric. Then (6.2.10) says δm,k = w'mnW'nk or w'W' = 1 or w' = W'-1 . (6.2.12) Assuming for the moment that detW' ≠ 0, the solution is given by w' = W'-1 . The (5.6.4) "Digression" showed that (A-1)T = (AT)-1 for invertible A, so w'T = (W'-1)T = (W'T)-1 = W'-1 = w' and therefore w' is symmetric as well. Since W' is known from (6.2.11), w' = W'-1 and the Bm = w'mnbn of (6.2.9) have been found. Finally, Bm Bn = w'mibi w'njbj = w'miw'nj bi bj = w'miw'njW'ji = w'miδni = w'nm . (6.2.13) __________________________________________________________________________ So in this tensor doc context, I can regard the bi = Sja vi as some basis vectors in the vector space spanned by the bi, and then I can regard the Bi = Sja λi as a set of row vectors that span the "row space". You can think of each row vector application as a linear functional (which is a special case of a linear mapping) Bm x = number Bm : Rn → R x in Rn λm x = number λm : Rn → R In fact any row vector acts as a functional in this sense, not just the basis vectors. Fine. y x = number y : Rn → R y in Rn* x in Rn Sja wants to refer to the vector space of column vectors x as V = subset of Rn spanned by the bj = ej. Sja wants to refer to the vector space of row vectors y as V* = subset of Rn* spanned by the bj = ej. Then the space V* is "dual to" V, and V* is "the dual space". This all seems quite innocent. [page 84] 2.15.16 review OK to here Example 7.2 . Here V is a space of real functions (not just a piece of Rn) in the Stakgold sense defined on [a,b]. A functional of interest is μ[f] = ∫f dx which gives you a real number. Somehow I*f = integral value. Sja is only here giving an example of a linear functional, he says nothing any "dual space" to V. [but below I show that this particular functional is just one of a infinite number of such in V* ] 2.15.16: This little example does make an association between a dual space functional μ and something involving dx. That is a Big Issue I am trying to understand right now: why you are allowed to associate the basis function λi with the differential dxi. You could write μ[f] = ∫Σi fidxi = ∫α and then your functional is the integral of α. But Sja is identifying α itself with a functional somehow! Example 7.3. In this example, the functional is μ[x] = v x for some fixed v where now we have the "standard inner product" used in Rn. I have been using this notation all along, while Sja has been avoiding it. Comment: Recall that scalar products exist in Hilbert Spaces which are much more specific that a generic vector space. Sja wants to think in terms of the more general vector space idea, so Sja is not going to use any inner products. Now, suppose vi are the basis vectors of V of dimension n. (bi in tensor doc). Sja expands a general vector v as [ Sja often uses vi as a basis function instead of my preferred ei] v = Σj cj vj where vi is a supposed basis for V. Now define a covector λj(v) in the dual space according to λj(v) ≡ cj This λj is a linear functional on V (different functional for each j value). The functional is defined so it equals the coefficient that appears for vj in the expansion of general V vector v. So then v = Σj λj(v) vj cj = λj(v) Notice that no dot product was used. So this object λ(v) is a row vector cT of constants cj which when linearly combined with the vj gives some vector v. Using dot products I would say then that λj(v) = v vj and then v = Σj (v vj) vj . But his definition does not require the existence of an inner product, it just requires a vector space. Working again with his definition, it seems clear that λj(vj) = 1 λj(vi) = 0 i≠ j and you arrive at this just looking at the linear combination. For example, vi = Σj λj(vi) vj λj(vi) = δj,i So we then have this set of coordinate functions λi and each of these is a linear functional mapping Rn → R such as λj(v) = cj. [ Below I show the these λi really are linear. ] OK, so I think the idea is to try to avoid talking about dot products to be general to any vector space, not just a Hilbert space. 2.15.16: The above Sja notes seem confusing. Now I would write λj(v) = vj as the action of λ on a vector, it projects out the jth coordinate. But above I write λj(v) ≡ cj . I think this was a confusion on my part and Sja probably never says what I write above. Digression: Wiki and Stakgold on Dual Spaces I looked at wiki at this point. It says that the space V* is the space of all linear functionals on V. So let's look again at the above examples. When V = Rn which is a finite dimensional space, I think this is the most general linear functional you can write: v = Σj=1n cj vj = Σj λj(v) vj cj = λj(v) Now exactly what is the linear functional here? I would say that each of the λj functions is a linear functional. I show linearity below. So for each j we have λj: Rn → R. Thus, we are aware right away of a set of n different functionals on V. Below we show that these λj functionals form a basis for V* and therefore V* has dimension n. Wiki would say that the elements of V* are called covectors, so therefore each of the λj functions would be a covector. Wiki allows that λj(v) = [λj,v] = <λj,v> are other common notations. Wiki says for finite n V, you can take ei as the V basis and then ei is the V* basis where eiej = δij, so this is the same as my tensor doc idea. The dual space is the space of row vectors. Wiki goes on to talk about V being ∞ dimensional, but does not really address any space of functions. But: Suppose V instead of being Rn is some infinite dimensional function space. Consider μ[f] = ∫f dx where f lies in V. Here μ is a linear functional and we have μ: V → R. So what is the dual space V* in this case? I have displayed only one linear functional, but V* is the space of all such linear functionals. So I don't have an answer here. Notice that μ[f] = ∫fα dx for α ≠ 1 would not be a linear functional. But the form μ[f] = ∫f dx s(x) might also be a linear functional where s(x) is a weight function. Stak talks about this subject. [ Below we have Riesz who says each linear functional is associated with some element of V. So my linear functional just stated is associated with s(x) of V. You would say f s = ∫f(x) s(x) dx . ] In Stakgold, a functional is called T[x] where x lies in V. When V is a Hilbert Space, you can state the famous Riesz Representation Theorem Stak chap 2. " Any continuous linear functional which acts on any S in any Hilbert space V can be "represented" as T[x] = <x,f> (for all x in V) where f is some unique element of V " Here we have x being some vector in V, and the theorem claims that any linear functional can be expressed as the inner product of x with some other unique element of V which above is called f. Thus, the number of linear functionals would be the same as the number vectors x in V. For V = Rn both these numbers are infinite, and you can think of f as being the row vectors.  For a function space where x is a function, I might restate this theorem as " Any continuous linear functional which acts on any S in any function Hilbert space V can be "represented" as T[f] = <f,F> (for all f in V) where F is some unique function of V " . Then T[f] = ∫f(x) F(x) dx = <f,F> Then the example above happens to use the function F(x) = 1. So then my conjecture of weight functions would be correct. Note that Riesz is in the context of a Hilbert Space, not a simple vector space. Now back to Sjamaar: Lemma 7.4. The λi form a basis for V*. Now what does λi mean without an argument? It is a function so can only exist in a space of functions. [ I now refer to this as a functional.] He would have then to show that if λ were an arbitrary function, you could write λ = Σ di λi where di are some constant coefficients. If we could construct di, then maybe we have our proof. So evaluate this arbitrary function λ for one of the V-space basis vectors λ(vj) = Σ di λi(vj) = Σ di δi,j = dj . So there you have dj constructed. We then have λ = Σ λ(vj) λi and we have then shown that λi is a basis since it can generate any function λ by linear combination. So I guess this looks good, and again we have avoided mention of any inner product. If you had an inner product, you could write λj(v) = v vj but we don't want to assume an inner product, so we just have these functions λj. Fine by me. Because the set of n functions λi spans V*, we must have dim(V*) = n. So I have just shown that: Fact : The V* basis of functions {λi} is dual to the basis of vectors {vj} of V. Note that λi is a scalar function of a vector just as λj(v) = v vj would suggest were there an IP. Note that the functions λi are in fact functionals since they map to R. [ correct ] Are the λi linear functionals? Let's investigate: v = Σj λj(v) vj (αv) = Σj λj(αv) vj = α [ Σj λj(v) vj ] = Σj (α λj(v)) vj = Σj (λj(αv)) vj Since the coefficients are unique, we have λj(αv) = α λj(v) first part of showing linear. Next, v + v' = Σj [λj(v) + λj(v')] vj = Σjλj(v+v') vj λj(v+v') = λj(v) + λj(v') So YES, the functions λi are linear functions and are linear functionals. No dot product needed. Example 7.5. Let ei be the basis vectors of V, as usual. According to our fiat statement above, we know that dxi = eiT which lies in the dual space Rn* . It seems pretty reasonable that the {eiT} form a basis of Rn* and in fact eiTej = ei ej = δi,j so then we can say: " the set {dxi} form a basis in Rn*". I guess one could write v = Σ viei general vector in Rn vT = Σ vieiT general vector in Rn* So in this example you can really associate the dual space with simple row vectors, not so much abstract functions. We have just shown that Fact: The set {dxi} form a basis in Rn* where Rn is the usual space of vectors v. Comment: At this point we are claiming that the differentials dxi form a basis for the space V* which is dual to the space V which lies in Rn. In this world, we have never used the fact that dxi is a differential, it is just a symbol so far! I agree, it is interesting all in itself to come of with some kind of vector space in which the dxi are "basis vectors". Then are really this: dxi(v) = linear functional on Rn where dxi(v) = eiT v = ei v = vi Here I define as shown by the last equal sign, forget Hilbert Space. So we end up with dxi(v) = vi linear functional dxi but Sja does not write down this fact anywhere. Comment: Notice that so far, there is no meaning to something like dxidxj other than it is the product of two linear functionals, and you could then say dxidxj means dxi(v)dxj(v) = vivj. This is NOT the meaning for dxidxj that will appear below as the wedge product. For example, if we use the meaning of this paragraph, we would get dxi(v)dxi(v) = vivi whereas we know that dxidxi = 0 as a wedge. Final random item on page 84: Suppose we have L: V → W where V has basis vi and W has basis wi. Consider Lvj = some vector in W = Σi li,j wi defines some numbers li,j Again, no inner product yet. Lemma 7.6. Imagine this situation for L: V → W vi = basis in V λi = basis for V* wi = basis for W μi = basis for W* Then compute μi(Lvj) = μi( Σk lk,j wk) = Σk lk,j μi(wk) = Σk lk,j δi,k = li,j Not sure why this is interesting, but it is correct. [page 85] 2.15.16 review OK to here Definition: Sja defines the simple property of a k-multilinear function λ(v1,v2.....vk) This is λ : Vk → R, so there are k distinct vector arguments. If there are two arguments, it is bilinear = 2-multilinear. Some simple examples are given. [ these are in fact multilinear functionals since they map to R ] Comment: I think this is the obvious generalization of a linear functional of one vector variable to many vector variables in the same space V. Of course Vk = V V .. V which is a direct product space as in my tensor doc discussion. More: A multilinear function is linear on each argument separately. In my own notation here is an example where we consider four arguments so we have λ(v1,v2, v3, v4). Below I show linearity just for the third argument, but the same equations would apply an each argument separately. λ(v1,v2, αv3, v4) = α λ(v1,v2, v3, v4) // scalar rule λ(v1,v2, u + w, v4) = λ(v1,v2, u, v4) + λ(v1,v2, w, v4) // addition rule Sja likes to combine these into a single rule like this λ(v1,v2, αu + βw, v4) = αλ(v1,v2, u, v4) + βλ(v1,v2, w, v4) page 85A I find his notation more confusing than mine. PL Exercise. What could you say about λ(Mv1,Mv2, Mv3, Mv4) where M is a matrix? Perhaps write Mv1 = M [Σi(v1)iei] = Σi(v1)i Mei Now Mei is a vector, so I have written Mv1 as a linear combination of vectors. Then we can say λ(Mv1,Mv2, Mv3, Mv4) = Σi(v1)i λ(Mei,Mv2, Mv3, Mv4) Then you could do this with each argument to get λ(Mv1,Mv2, Mv3, Mv4) = Σiiii (v1)i(v2)i(v3)i(v4)i λ(Mei,Mei,Mei,Mei) There is not much more one can do with this structure. Definition: If you add the property that swapping any two arguments negates λ, then you have an alternating or asymmetric k-multilinear function. The big example here is det(c1,c2....ck) is a well-known alternating k-multilinear function where ci are the column vectors of the matrix whose determinant we are expressing. Sja has discussed this animal earlier, and there I broke it down into two properties, each of which determinants have. He gives a second simpler alternating example that is bilinear. Now Sja does a very weird thing. He defines a single function which has the following name: "λ1λ2....λk" The arguments of this function are going to be v1,v2...vk which are the basis vectors of V. And here is how the function is defined: λ1λ2....λk(v1,v2...vk) ≡ det[λi(vj)] The matrix here is this: λ1(v1) λ1(v2) λ1(v3) ..... λ1(vk) λ2(v1) λ2(v2) λ2(v3) ..... λ2(vk) ... λk(v1) λk(v2) λk(v3) ..... λk(vk) This fancy function is a k-multilinear function because that is the way determinants work. It is also alternating since that is also how they work. This function λ1λ2....λk is called "the wedge product of the λ's" function and one writes then [λ1 ˄ λ2 ˄....˄ λk](v1,v2...vk) ≡ det[λi(vj)] (*) I see the mechanical definitions, but I have no idea how this relates to anything I know about forms. Make this clearer: λ1 ˄ λ2 ˄....˄ λk = "λ1λ2....λk" = the name of a certain alternating k-multilinear functional Question: Can we identify λi(vj) = δi,j inside the determinant shown in (*)? Answer: Well, in the current context, the vi are arbitrary arguments of the multilinear function and are general vectors in V. They could be, but are in general not, the set of basis vectors. Only if we happened to evaluate the wedge thing at the basis vectors would be make such a claim. Now: AkV = the complete set of alternating k-multilinear functions. [ = my space Λkf or just Λk] This AkV is in fact a vector space, and as such, it should have some basis. Comment: Recall that V* is the collection of linear functionals on V -- the dual space. For finite V with basis vectors vi we found that this dual space is spanned by the set of functionals λi. Each of these functionals is a function of v where v lies in V. Now we generalize so that V is replaced by Vk = V V .. V. [ nope ] What is the corresponding dual space? Perhaps it is the set of multilinear functionals on Vk. I don't know how many there are right now, but write one of these as μ(v1, v2.....vk). I presume then that the set of all possible such multilinear functionals would be the dual space (Vk)*. That would seem to be the logical generalization. For reasons not yet clear, Sja wants us to consider only multilinear functionals μ which have the alternating property, and instead of (Vk)* he refers to this restricted space as AVk. So before the reasons for being interested in this sort of dual space are given, Sja wants to develop some properties of this space, and only then can he explain the motivation. Sja is not referring to either (Vk)* or AVk as a "dual space", but either seems a reasonable generalization of the dual space concept. Naively you might think that (Vk)* = V* V* .. V* and then maybe the basis vectors of (Vk)* would be λi λj ..... λk i = 1..k, j = 1..k etc I will come back to this idea later. This approach however does not work in that alternating property. But maybe this would be the normal dual space of (Vk)* . I think it is. 2.15.16: In these original notes above I am confusing the tensor and tensor dual space with the wedge and wedge dual space. I now understand how all this works. Properties of the space AVk. 1. It is a vector space. Well, (Vk)* is a vector space since it is a dual space, and if we restrict to multilinear functionals which have the alternating property, we still get a vector space. Key fact is that if you add two alternators, the sum will be alternating. 2. As a vector space, AVk must have some basis vectors and must have some dimension. [page 86] 2.15.16 review OK to here Define I as our usual increasing multi-index. Example 7.11. Here V = R3 with the usual ei basis vectors and the usual eiT dual space basis vectors. Recall from page 83 that we identify eiT = dxi . So I guess dxi is a basis vector in V*. In general, a dual space basis vector was the function λi so I guess we can identify λi = eiT = dxi for i = 1..3. // my Big Mystery, why is this allowed? Now I is an ordered multi-index with k components. The space is V = Rn so k is some integer ≤ n. Suppose I = (1,2) so we are working with k = 2. Then we can say, dxI = "dx1dx2" = dx1 ˄ dx2 = the 2-multilinear functional as shown in page 85 C We are supposed to regard "dx1dx2" as the function name in the sense of page 85 C. Then we can evaluate this functional dxI(v1,v2) for the following specific arguments: dxI(e1, e2) = det(λi(ej) = = = = 1 Note that dxi(ej) = δij as we know from earlier, hence the rightmost two = expressions above. Now Sja has defined a new object here without saying so. Namely eI ≡ (e1, e2) I = (1,2) in our example. Then we have dxI(eI) = dxI(e1, e2) so the above line becomes dxI(eI) = det(λi(ej) = = = = 1 This is what was confusing me but now seems just fine. Now he says let's take another multi-index J = (2,3) so then dxI(eJ) = dxI(e2, e3) = = = 0 Notice how you insert the vectors in the argument list of dxI into the determinant based on position. In this example, notice how we found this result dxI(eJ) = δI,J = 1 only if the two k=2 multi-indices are the same. He will now generalize this result: Lemma 7.12. The claim here is in fact that [ 2.15.16: old hat now when written using λI] dxI(eJ) = eIT(eJ) = λI(eJ) = δI,J for any two ordered multi-indices I and J of order k Proof: I agree with 86 A as a displayed determinant. Remember that all the ir indices are different, and all the js indices are different, since part of ordered multi-indices. Thus, for example, j2 can match only one index in I (at most), and vice versa. I think this means that each row in the matrix can have only a single 1 with the other row entries 0. When I = J, all those 1's are on the diagonal and det = 1. It is not obvious what happens if I ≠J so Sja discusses that in the text. I did not track the proof exactly, but he is showing that if I≠J, then either a whole row or a whole column vanishes, so det = 0. Lemma 7.13. If λ(vI) vanishes for all ordered multi-indices I of degree k, then λ ≡ 0. Question: What is the meaning of λ(vI) here? This is Sja's first use of isolated symbol λ. I guess this is meant to be some arbitrary general element of the space AVk, some λ(v1, v2....vk) k-multilinear functional of generic arguments. But, λ(vI) = λ(vi1, vi2....vik) = λ evaluated at a set of basis vectors which are in increasing order So the claim is that if your generic λ vanishes for all vI of degree k, then λ ≡ 0. Proof: Let wi be an arbitrary set of vector arguments. If we can show that λ(w1, w2....wk) = 0 for any such set of vectors, then surely λ ≡ 0. Sja expands each wi on the basis vectors vi as shown top p 87, but really he means us to do this: wj = Σi ajivi j = 1,2....k The summation index used for each wj is different and we denote it by ij. Then we get λ(w1, w2....wk) = λ( Σi a1ivi, Σi a2ivi.... Σi akivi) = ΣiΣi....Σi a1ia2i....aki λ(vi, vi ....vi) = page 87 B Now our premise of Lemma 7.13 is that λ(vI) = 0 for all increasing multi-index choices of the arguments. But swapping any pair of vector arguments negates λ, so then we really have λ({v}) = 0 for any set of basis vector arguments. If two arguments are the same, we also get 0 due to the negation fact. Thus, the premise implies that λ(vi, vi ....vi) for any values of the subscripts. Thus λ(w1, w2....wk) = 0 and then we conclude that λ ≡ 0. [ page 87] 2.15.16 review OK to here Theorem 7.14. Assume that V has basis {v1, v2....vn} and thus dimension n [ the vi are of course vectors in V ] V* has basis {λ1, λ2....λn} and thus dimension n [ the λi functionals on V ] Then the set of functions λI where we enumerates all multi-indices I forms a basis for AkV. [ fine ] Note that: λI ≡ "λiλi .....λi" = a function name My proof: Want to show that you can write λ = ΣI cI λI for any λ, so then λI is a basis. Evaluate this equation at argument vJ where J is another multi-index. Then we have λ(vJ) = ΣI cI λI(vJ) = ΣI cI δI,J // by Lemma 7.12 !! = cJ Thus for any λ we can do the expansion on cI shown and we have constructed the solution coefficients to be cI = λ(vI) and easy to show this is a unique solution. So yes, this proof is analogous to 7.4 where we showed that any simple λ could be expanded on the λi so λi was a compete basis for V*. PL Example. Suppose V is Rn. We have this fancy space AkV which is the space of alternating multilinear functionals on Vk. The basis of V* is {λi} = {eiT) = {dxi}. We have just shown that the basis for AkV is then λI = eIT = dxI where I is an ordered index having k elements, and where these are shorthands for function names like λ1λ2... = e1Te2T.... = dx1dx2... You can write these instead (for example) as λ1λ2..λk. = e1Te2T....ekT = dx1dx2...dxk . λ1˄λ2˄...˄λk = e1T˄e2T˄....˄ekT = dx1˄dx2˄...˄dxk The most general element of AkV can then be written λ = a general element of AkV = ΣI aI dxI // = ΣI aI λI where aI is a coefficient that goes with basis element dxI. We know that aI = λ(eI). [ How do we know this? Well, look at λ just above and close to get λ(eJ) = ΣI aI dxI(eJ) = ΣI aI λI(eJ) = ΣI aI δIJ = aJ . For some reason Sjammar is referring to the k-form normally called α by the name λ. Fine. Now I would write the above as α = a general element of Λk = ΣI fI dxI = ΣI fI λI = a general k-form fI = α(eI) In this notation, we are closing the k-form α or λ with a Vk vector eI to make a tensor function. Let's rewrite the above showing all arguments: λ = ΣI λ(eI) dxI ok λ(v1,v2.....vk) = ΣI λ(eI) dxI(v1,v2.....vk) ok It would take a lot of effort to write this out using proper subscripts like i1 and j1. Now replace the letter λ by the letter α in a hope to make contact with earlier book sections: α = ΣI α(eI) dxI // α(eI) = fI in notation of Chapter 2 page 17 ok α(v1,v2.....vk) = ΣI α(eI) dxI(v1,v2.....vk) ok Here all the vi and the factors of eI like e3 are elements of V. There is no x sitting in this thing yet! Pause: The above equations "look like" the differential forms we discussed in Chapter 2. But in that Chapter dx1 was a differential distance, whereas here dx1 is a basis vector e1T in the dual space to Rn. So somehow we have these parallel developments and I suspect there is going to be an isomorphism between the two Worlds. [ still looking for this Big Mystery connection on 2.15.16 ] Now in page 87 A he simply "tacks on" the x label like this: α(v1,v2.....vk) = ΣI α(eI) dxI(v1,v2.....vk) αx(v1,v2.....vk) = ΣI αx(eI) dxI(v1,v2.....vk) (*) αx = ΣI αx(eI) dxI = ΣI fI(x) dxI = a k-form of earlier chapters. This connect fI(x) = αx(eI) is explicitly stated by Sja as page 88 B. Comment: In Chapter 2 we had α = ΣI fI dxI and we always there regarded fI as fI(x), see any example there for example. So what was α in Chapter 2 has become αx here in Chapter 7. Here x is I think a point on the manifold M where we have a tangent space TxM and perhaps things like eI are specific to that tangent space and should perhaps be eI(x). In Chapter 2, x was a point in some simple Rn I think. Comment: In order to make contact with Chapter 2, he has to "tack on" the x label as shown. That is to say, in the dual space development, we have fI above being α(eI) and there is no "x label" anywhere. But if I were to write eI(x), then I would have fI = α(eI(x)) = <α | eI(x)> = fI(x) and then we have the Chapter 2 x argument. I think I am onto something here. Bell starting to ring here. This space AkV certainly looks like the space of k-forms as treated in earlier chapters. We obtain the alternating wedge product rule (which was of course intentionally inserted at the start to obtain that result). We obtain the same multi-index general form for a k-form. Comment: The LHS general function αx(v1,v2.....vk) has k-vector arguments. The coefficient αx(eI) is this same function evaluated at a particular set of k vector arguments. no confusions to this point, but yes confusion below this point!! Comment 2.15.16. Perhaps this is the gist of Sja's argument: If you blindly associate dxi with λi, and thus blindly associate a form with an element of Λk (the dual wedge space), you arrive at something that looks like the form of Chapter 2. But I still don't see how making this Λk connection "buys me anything". How does it help me? [ page 88] 2.15.16 review OK to here Page 88 A is extremely mysterious and will take me several hours to interpret. I spent two days or so on the first half of page 88 and was unable to make full sense of it. See my detailed attempts in doc "Attempt to analyze section Pullbacks re-examined on page 88". My conclusion there is this: Summary: If I am willing to simply define gJ ≡ φ*α(eJ) as a page 88 definition of gJ, I am able to show that this gJ agrees exactly with the Chapter 3 result for gJ which was gJ(x) ≡ ΣI fI[φ(x)] det(Dφ)I,J . But the whole business is very cloudy since in Chapter 3 things are functions of x, whereas in Chapter 7 page 88 things are functions of those k multilinear arguments. I am really unable to make a clean connection between the two Worlds. Maybe this will get clarified later. I gave it my best shot today. I could go off and read some other detailed source like Spivak, but that would take many more days I suspect. Maybe Sjamaar will clarify things in upcoming chapters. Here is how one can start off: Start with page 87 A and page 88 B which combined say αx(v1,v2.....vk) = ΣI αx(eI) dxI(v1,v2.....vk) But in the spirit of the x and y use in the pullback discussion of Chapter 2, replace the above with αy(v1,v2.....vk) = ΣI αy(eI) dyI(v1,v2.....vk) (**) or αy(v1,v2.....vk) = ΣI fI(y) dyI(v1,v2.....vk) fI(y) = αy(eI) Now we apply a pullback mimicking Chapter 3 φ*αy(v1,v2.....vk) = ΣI φ*[αy(eI)] [φ*dyI(v1,v2.....vk)] . (*) Now make these assumptions: (***) φ*[αy(eI)] = αφ(x)(eI) // action on the coefficient [φ*dyI(v1,v2.....vk)] = dyI( (Dφ)v1, (Dφ)v2, .,,,, (Dφ)vk) // action on the differential Inserting these assumptions into (*) we get φ*αy(v1,v2.....vk) = ΣI φ*[αy(eI)] [φ*dyI(v1,v2.....vk)] . = ΣI αφ(x)(eI) dyI( (Dφ)v1, (Dφ)v2, .,,,, (Dφ)vk) Now look back at the template (**) above and make substitutions to write the template as, αφ(x)((Dφ)v1,(Dφ)v2.....(Dφ)vk) = ΣI αφ(x)(eI) dyI((Dφ)v1,(Dφ)v2.....(Dφ)vk) Comparing the last two equations which have the same RHS, we obtain a new equation, φ*αy(v1,v2.....vk) = αφ(x)((Dφ)v1,(Dφ)v2.....(Dφ)vk) I can next make this definition: (φ*α)x(v1,v2.....vk) ≡ φ*αy(v1,v2.....vk) (****) and then the previous equation becomes (φ*α)x(v1,v2.....vk) = αφ(x)( (Dφ)v1,(Dφ)v2.....(Dφ)vk) // page 88A Thus, with the assumptions (***) and (****) made above, I have successfully derived page 88A. It purports to tell you "how to do a pullback" in our new world of multilinear functions. At this point, without specializing the argument set (v1,v2.....vk), there is nothing I can do with any of the above objects having those (Dφ)vi type arguments. If we do specialize to the case (v1,v2.....vk) = (ej1,ej2... ejk) ≡ eJ where J is an ordered index, THEN I am able in the mentioned doc to show that (φ*α)x(eJ) = ΣI fI(φ(x)) det(Dφ)I,J fI(y) = αy(eI) This is the gJ object which appeared in Chapter 3. So in Chapter 7 if we define gJ ≡ (φ*α)x(eJ) then the gJ so defined in Chapter 7 agrees with the gJ defined differently in Chapter 3. So I guess Sja is just trying to maintain some connection between the two different Worlds. [88] Forms on Manifolds 2.15.16 review OK to here Based on the above discussion, we associate each function in AVk with a k-form. Recall that the value of x was "tacked on" so we had. αx(v1,v2.....vk) = ΣI αx(eI) dxI(v1,v2.....vk) = ΣI fI(x) dxI(v1,v2.....vk) In this tack-on process, we create label called x, and this label is supposed to lie on manifold M. It is true that we can construct a vector space TxM at this point x, whose basis vectors lie on M for very small dx. But I don't see why Sja says: "map αx on the vector space TxM " "one for each x on M" I agree that you have an αx for each such x, and that there is a TxM at each x, but why "on"? It is really defined on M, not on TxM. Then he seems to say that if there is a single x = ψ(t) [ fig on page 68 of torus], then the pullback ψ* acting on a form is given by our result above with general φ replaced with this embedding function ψ, so (ψ*α)t(v1,v2.....vk) = αψ(t)((Dψ)v1,(Dψ)v2.....(Dψ)vk) x = ψ(t) // p 88 F But this is really consistent with (φ*α)y(v1,v2.....vk) = ΣI αφ(x)(eI) dyI( (Dφ)v1, (Dφ)v2, .,,,, (Dφ)vk) y = φ(x) which makes me think he may have mislabeled the earlier thing with x instead of y or t. But I know what he means. When you do a pullback on α, the above is one way to write the result. In page 88F we have t in U. It seems odd now that he goes out of his way to call ψ*α (as shown just above) a "local representative" where we have no local index like the i or j we had before. Maybe this is "the only one". Ah, but I now see that he did throw in the word "local", so he is just omitting labels like i or j. His BT book quote (Bott and Tu) makes no sense to me because I don't see how our k-form α "eats k-tuples of tangent vectors". Is he saying that the arbitrary arguments v1,v2.....vk are "tangent vectors"? Well, it is true that we have x = ψ(t) in the M context and that there are tangent vectors at the point x for space TxM. The number of such tangent vectors is the dimensionality of the surface M and I don't think this is the same as k necessarily. So the quote is lost on me. 2.15.16: Everything plays out at some point x on manifold M which has dimension k. The tangent space TxM at point x is of dimension k (k = 2 on the torus for example). I think you are supposed to regard the vector arguments v1,v2.....vk as being vectors in TxM (and not just generic arguments) and I would say that the ei are basis vectors for TxM and are thus x-dependent. So at point x when you create a tensor function by closing a form onto a set of vector arguments v1,v2.....vk , you are in fact "eating (closing onto) k-tuples of tangent vectors" and you are "spitting out numbers", in that <α | v1,v2.....vk> = scalar. The k-form α is defined at each point x, hence αx, and so "inhabits M" in that sense. In wedge doc these vector arguments are generic, there is no manifold or tangent space. It helps to read Sja's "forms on manifolds" text on page 88. I confirm all this below in the original notes. Example 7.17. M = a curve in RN. Dimensionality(M) = 1 so at a point x on M there is only one tangent vector. In this example, he assumes v lies in TxM where v is the argument of αx(v). Conjecture: Maybe you must have k = m, where m is the dimensionality of M. [correct] Then you could take those arbitrary vector arguments vi to be the basis vectors of TxM. Then the BT comment makes more sense. Then on a 2D surface, you would always be talking a 2-form and not any other kind of form. This is consistent with he earlier comment that we have (ψ*α)t(v1,v2.....vk) = αψ(t)((Dψ)v1,(Dψ)v2.....(Dψ)vk) x = ψ(t) // p 88 F and α is a multilinear map "on the vector space TxM ". OK, it is not till the lower half of page 88 that Sja even mentions M in this chapter, and so I guess we have something new here: (1) For a manifold M of dimension n, you set k = n, and you take those general arguments above (v1,v2.....vk) to be vectors in the tangent space at x, which is TxM. So you have a little "local Rn" at the point x. In Chapter 2 we talked about forms on Rn as a full Euclidian space, but now to make forms "work" on a manifold M, we have to talk about the local Rn = Rk associated with those tangent vectors. At each different x, we have a different local Rn. So now (v1,v2.....vk) are arbitrary vectors, but they are vectors in the local Rn for point x. I think that is why he is trying to say. This does sound like the fibre bundle idea again where at point x we have a vector space TxM. Now Definition 7.16 makes more sense: A k-form αx on M is a chosen element of AVk (an alternating multilinear map) whose arguments vi are vectors in the space TxM at the point x . Thus αx is defined on TxM for the point x. Now let dim(TxM) = dim(M) = n. Maybe you are allowed to talk about any k ≤ n just as we did in Chapter 2, so that then is why k and n here are not identified. [ I think only k-form with dim M = k makes sense? Well, on a 2D surface you could talk about an area or a curve, so I guess less than a k form also works. ] Example 7.17 Revisited. M = a curve in RN. Dimensionality(M) = 1 so at a point x on M there is only one tangent vector v. In this example, he assumes v lies in TxM where v is the argument of αx(v). Since n = 1 for M, the only interesting value of k is k = 1, so we would have a 1-form of some sort. There is only one argument of αx, so write αx(v). Now let us choose to have αx(v) = ± ||v|| = a real number. // this is a tensor function Notice that this is a form, not a linear functional. It is not linear. This 1-form happens to be the speed of the only tangent vector, and we have the usual direction issue. I don't think this is the ONLY αx you could invent, but this is one of interest. Somehow this form is going to be related to arc length, but in the Sja notes world it is not yet clear why that would be the case. In Buck p 320 we see L(γ) = ∫dt ||v||, so we are integrating a 1-form α = ||v|| dt where t is the parameter. Although Sja Chap 4 was dedicated to the integration of 1-forms, arc length never made an appearance there. But the pullback integration variable was in fact called t and we had ∫c α = ∫g(t)dt as the pullback. In the final section, Sja now talks about two local regions i and j which overlap. He writes Wi = ψi(Ui) Wj = ψj(Uj) where there is some overlap region WiWj on the manifold. For a point x in the overlap region we would have x = ψi(t) t in Ui x = ψj(u) u in Ui Then as in Section 7.2 we find results like ψi(t) = ψj(u) t = ψi-1 [ψj(u)] = (ψi-1 o ψj) (u) and he considers αi = ψi*α αj = ψj*α I guess we can then say α = [ψi*]-1αi α = [ψj*]-1αj so then [ψi*]-1αi = [ψj*]-1αj and then αj = ψj* [ψi*]-1αi Maybe can write this as αj = ψj* ( [ψi-1]* αi ) Then we look at page 38 regarding the "natural rule" which is φ*(ψ*α) = (ψ o φ)*α Applying this rule, we find that ψj* ( [ψi-1]* αi ) = ( ψi-1 o ψj)* (αi) and we then get the result αj = ( ψi-1 o ψj)* (αi) // page 89 B In Section 7.1 we just conjectured this to be true. Now that in Section 7.2 we have created a theory of forms on M, we have obtained this result as part of the theory. I admit that my proof here is not very good, but good enough for me now. I probably should do the exercises, but I will postpone them instead. He shows how you can "antisymmetrize" any k-multilinear function to get one in AVk. This is what happens in Spivak I recall. 2.15.16 Review. I finally reached this point at 10:30 AM, so this concludes my "review" of the hugely complicated chapter 7. This Chapter has exactly two sections which are the "two definitions": 7.1 forms on manifolds using αi = ψi*α for each patch on the manifold, αi = pullback 7.2 forms on manifolds using the monstrous dual space machinery with λi = dxi as postulate.