Role of the Multilinear Functions

Role of the Multilinear functions PhL 8.24.15 Last Update: 8.25.15 I have finished reading Sja and am now looking back at various things. I think back-and-fill and cross correlate and digest and crystallize and congeal and consolidate is going to take a series of essays on my part to get this stuff understood. So this doc is today's essay. The "Grassmann section" as I might call it (perhaps incorrectly) is a theory section called "7.2 Second Definition" on page 82 of Ch 7. Sja talks first about dual spaces and their covectors. Recall that a dual space V* is the set of all linear maps over vector space V. In our work on forms, we have some underlying space of interest, Rn which has coordinates xi or perhaps a vector v with components vi. This might by the usual (x,y) space in R2. In Sja page 84, if V has basis {vi}, then V* has basis functionals λi. He shows that if you just think about the meaning of space V* which contains things like f(v1,v2...) = Σici(vi)vi then (c1,c2...) is the covector for this particular functional in V* and he shows that ci(vi) = λi(vj) = δi,j or λi(vj) = (eTi)j = (eTj)i and this is a "coordinate function". These λi functionals form the dual basis to V. Notice that we can say that λi V* where λi is a functional and V* is a space of functionals. We don't need to show the arguments. Notice also that λi(v) = λi(Σj cjvj) = Σj cj λi(vj) = Σj cj δi,j = ci λ(v) = cT so you think of λi as a functional on the space V* for any vector v in V. Question. As you see, one can write either λi(v) = ci or λ(v) = c . We claim that λi is one of a set of basis functionals in the functional space V*. But we need ALL of these functionals in order to generate the vector cT which we thought was a vector in the dual space to V. So which is it? Is λi a covector in V*, or is cT a covector in V* ? I have no idea! Resolution: I went off and wrote another doc "the meaning of lamba-i.doc" [ where is this doc? wedge stuff ] to clear up this confusion. Here are some of the things I learned. Sjamaar's presentation is not really clear, symbols are under-labeled, no mention of qi objects. The correct solution is that your take a generic basis vi of V and you construct the set of dual basis vectors qi as instructed in tensor doc. This set is the reciprocal basis, and is really a basis. Then you know that qi vi = δi,j . You then define λi(v) ≡ qiTv . This λi is obviously a linear functional on V, since v is an arbitrary element of V. And we then have λi(vj) ≡ δi,j. There are really two different "dual spaces". You can think of either vi or qi as a basis for V, and then you can think of qiT as a basis for what I call VT which is the corresponding "row vector space". Sometimes people regard this space VT as the "dual space of V". But the dual space of Sjamaar is a slightly different dual space. It is the space V* of linear functionals which act on V. Our λi of λi(v) ≡ qiTv is a particular linear functional. I show in the doc (and Sjamaar shows) that the set of λi form a basis for functionals in V*. Since λi(v) ≡ qiTv, we can associate λi as a basis vector in V* with qiT being a basis vector in VT. A general functional could be written as λ(v) = qTv so for each qT in VT you end up with a specific functional λ in V*. So the size of the two spaces V* and VT is the same. Of course VT and V also have the same size. Thus dim(V*) = dim(V) as Sjamaar shows. To summarize, there are two dual spaces. One dual space has vectors which are row vectors like qT and I call that VT. The other dual space has elements which are scalar functionals like λ (not vectors). You could if you wanted combined all the scalar functionals λi into a vector λ, but this is never done and is just not a useful way to think of things, but I do it in a few places below regarding determinants. So far, nothing mysterious, but the reader wonders what is the relevance to the above λi discussion to differential forms? There is no answer at this point. Alternating functions and AkV Sja then goes on the define AkV as the space of alternating linear functionals on V having k vector arguments. Within this space, he then draws attention to a certain type of element which has this form "λ1λ2....λk"(v1,v2...vk) ≡ det[λi(vj)] = det [ λ(v1), λ(v2).....λ(vk) ] λ = (λ1,λ2....λk)T where on the right I show the column vectors of the associated matrix. This is just a particular example. Here is the matrix whose determinant is taken, λ1(v1) λ1(v2) λ1(v3) ..... λ1(vk) λ2(v1) λ2(v2) λ2(v3) ..... λ2(vk) ... λk(v1) λk(v2) λk(v3) ..... λk(vk) Here is another AkV element: "λ2λ1....λk"(v1,v2...vk) ≡ det[λi(vj)] = det [ λ(v1), λ(v2).....λ(vk) ] now λ = (λ2,λ1....λk)T = – "λ1λ2....λk"(v1,v2...vk) This is a different AkV element. In terms of AkV, the determinant guarantees that "λ1λ2....λk"(v1,v2...vk) is an "alternating function" of the vector arguments vi. Swapping two arguments swaps the two corresponding columns of the matrix and thus negates the function. But that same determinant ALSO causes the function "λiλi .....λi" to have this same alternating property if you switch any two indices! Such a swap causes a swap of two rows of the matrix! This fact then shows that if you consider a function like "λ1λ1....λk" where two indices are the same, that function is 0 ! I claim then that you could write "λiλi .....λi"(v1,v2...vk) = εii .....i "λ1λ2....λk"(v1,v2...vk) In this family of functions there are k*k..*k = kk members, but only k*(k-1)....*1 = k! are nonzero. So all the nonvanishing functions are trivially related to the first function "λ1λ2....λk". Now we could use the wedge symbol by definition to write (λ1 ˄ λ2 ˄.... ˄λk)(v1,v2...vk) ≡ "λ1λ2....λk"(v1,v2...vk) = det[λi(vj)] This is another way to write the NAME of this AkV element. It is still a function of all the arguments. I would say that this was NOT true: (λ1 ˄ λ2 ˄.... ˄λk)(v1,v2...vk) = (λ1(v1) ˄ λ2(v2) ˄.... ˄λk(vk)) wrong! because that gives the impression that the result is a function only of λ1(v1) when in fact it is a function of all the objects λ1(vi) . Now: If you realize that the object "λiλi .....λi" is a functional in AkV, you don't have to show its arguments because you know, since it is a k-multilinear functional, you can always display the arguments, but the object "λiλi .....λi" is the name of a functional in AkV . In analogy, you can think of f(x) as a function of x, but the function is f and you don't have to always say f(x) to talk about function f. Nevertheless, obviously if you evaluate the functional "λiλi .....λi" at some point in Vk, you get some particular real number, and different evaluation points will have different real numbers. With this in mind, we can simply say (λ1 ˄ λ2 ˄.... ˄λk) AkV The reader now pauses again to ask: all the above is quite clear and fine, but what does it have to do with differential forms? Well, suppose we simply attach a new name to λi as follows "dxi" ≡ λi and we don't at this point associate dxi with a "differential". Then we have (dx1 ˄ dx2 ˄.... ˄dxk) AkV Consider then the following object α(v1,v2...vk) ≡ f12..k (dx1 ˄ dx2 ˄.... ˄dxk) (v1,v2...vk) where f12..k is a real number which is a coefficient. I think if would be incorrect to require f12..k to also be an alternating function f12..k(v1,v2...vk). Right now I think f12..k is just a constant real number which does not depend on (v1,v2...vk). Without showing arguments α ≡ f12..k (dx1 ˄ dx2 ˄.... ˄dxk) This is of course an element of AkV because any linear combination of (dx1 ˄ dx2 ˄.... ˄dxk) is in AkV. This expression "looks like" a particular differential form of the type we studied in Buck, though there were no wedge symbols. If we now incorporate the technology of the "increasing multi index I" with all that has been said above, we can write this as the most general possible "differential form α" of order k: α = ΣI fI dxI dxI = dxi ˄ dxi ..... ˄ dxi with i1 < i2 ....< ik Again we can show arguments if we want α(v1,v2...vk) = ΣI fI dxI(v1,v2...vk) α(v1,v2...vk) = ΣI fI "dxi ˄ dxi ..... ˄ dxi"(v1,v2...vk) But the arguments are not needed if you are just talking about functionals in AkV . So we go back to α = ΣI fI dxI = most general element of AkV At this point, the reader must admit that in terms of these functional labels dxi = λi the resulting α certainly "looks like" a differential form like f(x,y)dxdy which you could integrate over a piece of R2. But "looks like" is not very precise. These dxI have nothing whatsoever to do with calculus differentials! Let's compare these two animals, β = f(x1,x2)dx1dx2 // a calculus integrand α(v1,v2) = f [dx1 ˄ dx2](v1,v2) // a differential form where recall that (v1,v2) is some set of basis vectors in R2. Let's rewrite as α(r1,r2) = f [dx1 ˄ dx2](r1,r2) r1, r2 in R2 Well, not very clear is it! But now let's change the scenario. Go back to α(v1,v2) = f [dx1 ˄ dx2](v1,v2) where (v1,v2) is some basis in R2. Comment: I had the passing idea that you could regard arguments (v1,v2) as being two arbitrary vectors in the tangent space TxM, but I don't think that is right. They are just arbitrary arguments which any AVk functional must have. Now let's just assume that for a manifold M, the above "form" varies depending on the point x on M you choose. We now just "tag" things with x in this arbitrary manner αx(v1,v2) = fx [dx1 ˄ dx2]x(v1,v2) where all three items are then functions of x, as indicated by this label. This step seems quite arbitrary. We can write it this way αx(v1,v2) = f(x) [dx1 ˄ dx2]x(v1,v2) so at least we have f(x) in there, looking more like calculus. If we omit these generic arguments we have αx = f(x) [dx1 ˄ dx2]x which compare to the above calculus thing. I think here we have to assume that our manifold M is just the plane, so then [dx1 ˄ dx2]x if differentials might be the same at all points. Then x = (x1,x2) and we get αx = f(x1,x2) [dx1 ˄ dx2] or αx(v1,v2) = f(x1,x2) [dx1 ˄ dx2](v1,v2) This now seems quite close now to our calculus result β = f(x1,x2)dx1dx2. Remark: The object α is a theoretical object in AkV, whereas β is some real-world calculus thing which I can integrate to get a number. Sjamaar has provided the reader with many technical tools, but has failed to make this connection. The question is this: what really does the function "λ1λ2"(v1,v2), which we can write as [λ1 ˄ λ2](v1,v2), have to do with the calculus area patch dx1dx2 ? In his differential forms Chapter 2, we get lots of good tools. We learn what "d" and "*" do to a differential form. We learn that d2α = 0. We are told about closed and exact forms. But the English language explanation of "what is really going on here" is completely missing, His discussion of forms is divorced from the physical world. Pullbacks at least relate integrals over curves to flat-space integrals But as noted above, I have lots of trouble just with a simple 2D flat space integral. Still, I will comment on the pullback stuff here. The pullback chapters 3 and 4 are perhaps closest to real world integrals. ∫c α = !Syntax Error, Ig(t) dt. It seems that only after you do a "pullback" do you get something that can be integrated in the real world. Here he pulls back the abstract integral of the abstract form α over a general curve c to get a physical integral. So you never integrate actually "on the curve", you always integrate in the "flat" pullback space. He jumped widely from 1-forms to k-forms at the end of Ch 2 and maybe we should have had a visit to the world of 2-forms as done in Buck. Let's try to specialize Ch 2 ending to 2-forms. First, here is the "general theory" α = ΣI fI(x) dxI φ*α = ΣI φ*(fI(x)) φ*(dxI) = ΣI fI(φ(x)) ΣJ det(Dφ)I,J dxJ = ΣJ [ ΣI fI(φ(x))det(Dφ)I,J] dxJ = ΣJ gJ(x) dxJ If you wanted to do an integral, you would write ∫M α = ∫U φ*α = ∫U ΣJ gJ(x) dxJ Here M is the curved "surface" while U is the "flat space" over which we know how to integrate [but I don't yet really have the flat-space integral form-to-integral connection] . In the above notation, this flat space has coordinates xJ. The mapping φ(x) is what takes you from the flat integrating space to the general surface M. Now Sjamaar is IMHO guilty of doing a bad choice of variable names going from one chapter to the next, which added lots of confusion to me. A calculus book that used y = f(x) in one chapter and x = f(y) in another chapter would also be confusing. So let's now rewrite the above "general theory" where the flat space U has coordinate t and then t = φ(x) is the mapping from M to U. Note how in Ch 3 he instead on page 37 uses y = φ(x). So I will make a better picture: α φ*α x t t = φ(x) In this picture the two arrows are oppositely directed. The domain U is "flat" and we know how to integrate something over that domain. Now rewrite the above "general theory" α = ΣI fI(x) dxI = defined for the right side picture φ*α = ΣI φ*(fI(x)) φ*(dxI) = ΣI fI(φ(t)) ΣJ det(Dφ)I,J dtJ = ΣJ [ ΣI fI(φ(t))det(Dφ)I,J] dtJ = ΣJ gJ(t) dtJ Then here is how we do an integral ∫M α = ∫U φ*α = ∫U ΣJ gJ(t) dtJ gJ(t) = ΣI fI(φ(t))det(Dφ)I,J Here the dtJ integral is a flat space one we know how to do! And gJ(t) is just some scalar function. Now let's tone this down from the general case to k = 2 α = f12(x) dx1˄dx2 = defined for the right side picture φ*α = φ*(f12(x)) φ*(dx1˄dx2) = [f12(φ(t)) det(Dφ)12,12] (dt1˄dt2) = gJ(t) (dt1˄dt2) I assume that the wedge product survives here. There is always the orientation sign question which Sja worries about a lot, but let's say we identify the front side of U with the outside of M. Then I claim ∫U gJ(t) (dt1˄dt2) = !Syntax Error, I!Syntax Error, I gJ(t1,t2) dt1dt2 = !Syntax Error, I[ !Syntax Error, I gJ(t1,t2) dt1] dt2 as in the Sja Remark 5.3. I know that the above is the "rule", but I don't at all see how we can replace the theoretical wedge product dt1˄dt2 with some real-world differentials dt1dt2 multiplied together. At this point, I don't really associate (dt1˄dt2) with a rectangular patch of area! So there is work still to be done, the picture is incomplete. Consider then for general vector arguments that (λ1˄λ2) (v1,v2) = det Now specialize this to the case that v1 and v2 are a basis for R2 (possibly non-orthogonal). Then we can say (λ1˄λ2) (v1,v2) = det = det = 1 How exactly should I interpret this result? Note in passing that (λ1˄λ2) (v1,v1) = det = det = 0 So I continue to be puzzled by the claim that (λ1˄λ2) (v1,v2) = 1 for any pair of basis vectors (v1,v2) which would translate into, for example, (dt1˄dt2) (v1,v2) = 1 for any pair of basis vectors (v1,v2) Comment: By not using ˄ symbols, both Sjamaar and Buck are able to masquerade various activities which would I think be clearer with the symbols. So I end this essay for now being still confused about the relation between (λ1˄λ2)(v1,v2) of the Grassmann world and dt1dt2 being the area of a differential area patch on a flat surface. Those (v1,v2) arguments seem to play no role in anything. Maybe there is a homomorphism between the Grassmann forms world and the calculus world, and if so, someone should state that fact. So I have now to look at other sources once again. I have squeezed all I can out of Buck and Sjamaar. Arapura. He starts off doing what seems to be normal calculus stuff, but then suddenly on page 8 bottom he shows his first ˄ symbol. He digresses to discuss u ˄ v . He refers to this as a 2-vector which we are supposed to think in terms of a real physical n-piped which is spanned by u and v . This object has an area, it has a plane in which it lies (normal to that plane would be u x v ), and the boundary has two circulation senses. This discussion makes no connection to any AV2 2-multilinear functionals. This is fine, but then suddenly he writes, where dx seems to be a scalar and not a vector. So now all that u ˄ v vector stuff seems not to relate. So right at this point, Arapura has dropped the ball at least for what I want to know. Here is an interesting piece of his work: This seems to suggest that dx is our ordinary differential dx and we then maintain the wedge between differentials. There is no mention of "dxdy"(v1,v2) as an alternating function in AV2. Next comes this also interesting result: Probably this can be written as a pullback thing, but the point I make is that on the right we have dudv of ordinary calculus, there is no wedge there. Like Sjamaar, he defines the thing on the left to be what is on the right. Here comes the Catch 22: which seems to say you can drop the wedge for a flat space integral. More on this theme So again you can just drop the wedge when you feel like it! Here is more Again this suggests the for a flat space integral like in R3 you can drop the wedges. Arapura only references Spivak and Flanders, like Sjamaar. Donu Arapura is at Purdue, and this pdf is listed on his list of "expository materials". He is a smilin' guy, So OK, he is not going to provide the connection. However, his stuff lends credence to the idea that "in flat space you can drop the wedge symbols".