Measure in Sja and Tensor REVIEWED

Measure in Sjamaar and Tensor Doc PhL 4.17.16 5/16/16: This doc has a ton of practice work, much of which is now in tensor doc. See red comments for a rough index of what is happening. I have been putting off this topic and the time has come. I think it will explain why forms integration "works" and gives the classical results for line integrals and surface integrals. This will result in an addition of some sort to tensor doc where I have the nested cofactor stuff. It should tie in with my non-square matrix linear algebra stuff, and with Sjamaar's chapter 8 on "volume forms". I want to start with Tensor Doc and this issue. I will first do a rereading of tensor doc Chapter 8 while editing it live. Issue: should I replace concatenation by composition in the summary section? Tensor Doc and the Nested Cofactor Stuff -- all rewritten and installed in tensor doc!! // 5.15.15 8.1 Overview of Chapter 8 OK 8.2 The differential N-piped mapping OK, I did not get lost, seems fine. 8.3 Properties of the finite N-piped spanned by the en in x-space Getting sign conventions set in a covariant manner, simple formulas for things. OK 8.4 Back to the differential N-piped mapping: how edges, areas and volume transform (a) The Setup OK, showing all the same formulas in detail, edge area and volume (b) Edge Transformation, does not say much, exercise notation. OK (c) Area transformation, tensor density stuff, lots of results, complicated but seems OK (d) Volume Transformation, summary at the end, seems all OK (e) Covariant Magnitudes, again with a nice summary, seems all OK (f) Two Theorems : g'nn g' = cof(g'nn) and |(Πxi≠nei)| = all OK, 3 proofs! (g) Cartesian-View Magnitude Ratios all OK, cof appears lots of times. (h) Nested Cofactor Formulas OK THIS is where I want to add my new stuff, and luckily the eq num system will allow it!!!! But let's continue for continuity's sake and see where this is all leading (i) Transformation of arbitrary differential vectors, areas and volume N.B. I write dA = (dx[1]) x (dx[2]) ... x (dx[N-1]) and dA = dx[1] x dx[2]. This looks to me like a Hodge correspondence representation of wedge products used in differential forms!! This might be key to my problem of understanding why differential forms "works" for integration. Keep this in mind. (i) Transformation of arbitrary differential vectors, areas and volume (resuming from above) opening section OK Review and covariant form of the dA and dV equations Interesting. and I think OK How things look in developmental notation. Not too relevant, but OK (j) Concatenation (Composition) of Transformations OK (k) Examples of area magnitude transformation for N = 2,3,4 seems OK Example 2: Spherical Coordinates: area patches OK 8.5 Transformation of Differential Volume applied to Integration OK, a short section. 8.6 Interpretations of the Jacobian . A review section, seems OK. 8.7 Volume integration of a tensor field under linear transformations OK DONE!! This is a very difficult chapter to wade through, always has been. The nested cofactor business. I claim already the following dAn = J (Πi≠ndx'i) en (8.4.a.1)  |dAn| = |J| |en| (Πi≠ndx'i) = |J| (Πi≠ndx'i) dAn = dA'n dA'n ≡ Πi≠ndx'i Back in (5.12.14) and hopefully somewhere in Ch 7 I note that g' = J2 g = J2 1 = J2 for Chapter 8 so then I have from 3 equations above, |dAn| = |J| (Πi≠ndx'i) = (Πi≠ndx'i) = (Πi≠ndx'i) Now I want to say that dAn is a face on an n-piped, and this face itself is an n-1 piped which itself has faces with areas. I need some kind of notation to be able to talk about this stuff recursively. Let's do some tryouts: dAn, dAn = a face vector/magnitude of an n-piped, which face is of dimension n-1 dAn,i, dAn,i This would be the face-area vector and magnitude for the ith face (of dimension n-2) of the nth face (of dimension n-1) of the original n-piped (of dimension n). Fine. Now can I produce expressions for these things? [ is this the same as the ∂M boundary business?? ] Let's go all the way back to the start of Section 8.3. How should one think about this recursion? Try starting with an 3-piped in x-space and 3-cube in x'-space. I always think of the volume here as a scalar, not having a direction, because in R3 there is no room for a direction which could have to be in the 4th dimension. But as I start marching down the hierarchy, I do have "direction". I start with An = σ (-1)n-1 e1 x e2 ... x eN // en missing σ ≡ sign[det(Sab)] = sign[det(Rab)] where now on the finite n-piped one of the faces, An, is now a vector quantity. This face is the cross product of all basis vectors spanning the original N-piped in RN except for en . I would be inclined to say An,i = (sign) e1 x e2 ... x eN // en and ei missing An,i,j = (sign) e1 x e2 ... x eN // en and ei and ej missing and so on. I think I could support this idea for a finite N-cube and its hierarchy of faces going down the chain. So let fiddle first with this object An,m = (sign) e1 x e2 ... x eN // en and em missing (An,m)i = (sign) εiabc..x (e1)a(e2)b.... (eN)x // en and em missing The vectors are still each of dimension N. Maybe I define an S matrix like this S = [ e1, e2, e3 ... eN] // nothing missing N x N matrix Sn = [ e1, e2, e3 ... eN] // en missing N x (N-1) matrix Sn,m = [ e1, e2, e3 ... eN] // en and em missing N x (N-2) matrix Both of the latter would be a "tall" S matrix that is obtained from the original matrix S by crossing out one and two columns and maintaining all rows. How do I handle the signs? Well this is not simple. I read tensor Appendix A which talks about the reciprocal vector Ek and you see how the det(R) and the (-1)1-k phase factor appear here and there. If I am just doing "measure", maybe I can ignore the "signs" at least for now and try to get to something relevant. Start again dAn = dA'n dA'n ≡ Πi≠ndx'i Conjecture that for area of an area, and again one more level down, dAn = Πi≠ndx'i dAn,m = Πi≠n,m dx'i dAn,m,k = Πi≠n,m,k dx'i I can see that at each level we are dealing with a square matrix, first nxn , then mxm, and so on. I know that g' = RRT = STS in the case that g = 1 and this is an "original" N x N matrix. Now since cofactor is the same as minor for diagonal elements, and since a minor is a determinant, I claim that cof(g'nn) = det(g')with row n and column n crossed out cof[cof(g'nn)]mm = det(g')with row n,m and column n,m crossed out and so on. How about this dAi = Πi≠idx'i dAi,i = Πi≠i,i dx'i dAi,i,i = Πi≠i,i,i dx'i // obs notation ! Then maybe write in multi-index notation dAI = d (*dx'I) Just winging it here. The (*dx'I) looks a lot like the Hodge thing. And I want to rule out rows I and columns I. The further you go down in the hierarchy, the more rows and columns you rule out! dAI = (*dx'I) (*dx'I) and this looks a little like Sjamaar except seems it ought to be det(RTR). But for square R I can just put that in and write dAI = (*dx'I) (*dx'I) Note added 4.19.16: Sjamaar old says μM = n *dx which looks a little like the above! THEN looking at wedge Fact , Fact: If tall R has full rank n, then (RTR)-1 exists. For any tall R, (RRT)-1 does not exist. (10.6.c.7) If I write as shown above with RTR ordering, then det(RTR) still exists for "tall" R, since the inverse exists. so the result dAI = (*dx'I) (*dx'I) still has a meaning when R is non-square!! This is then my generalized measure formula for all levels of the hierarchy !!! Likely wrong, but seems like a good guess. I could try it out for the spherical patch example in tensor doc! I am here using my I index to "drop down from the top" in the hierarchy. Maybe better to instead use it to build up from the bottom. Start over. [ more now on the RTR business, no written up in both tensor doc Ch 8 and wedge doc App F! ] Recall from Section 10.6, R** = [u'1, u'2 ....u'n] (2.5.9) (10.6.e.5) where these are the tangent base vectors in x'-space and they each have m elements and this is a way to write the tall R matrix. Recall next the theorem that Sjamaar claims on his old page 92 which says the following: "The above set of n vectors spans the Tx'M tangent space on the manifold surface. If you make an n-piped from these vectors, it will have the following volume Vn = and then the differential volume would be dVn = Πi=1ndu'i and then this would be the proper "measure" on the surface of manifold M, and this fact has nothing to do with differential forms. This formula fits exactly into my Chapter 10. I have not derived it by the way. But I think I can make it fit with the cofactor formulas of tensor doc. But how then does this relate to ∫φ α = Σ'I Σ'J (!Syntax Error, I!Syntax Error, I ..... !Syntax Error, I) fI(φ(t)) det(RIJ(t)) dtjdtj....dtj R= (Dφ) or in older notation ∫φ α ≡ ∫φ Σ'I fI(x') dx'^I = (!Syntax Error, I!Syntax Error, I ..... !Syntax Error, I) Σ'J Σ'I fI(x') det(RIJ(x))dxjdxj....dxj How would I directly integrate a function over the surface M in n dimensions? A little piece of integrand at some point x' would be f(x') Πi=1n du'i while on the other hand ∫[something over the M surface region] = ∫Σ'I fI(x) ( Πi=1n du'i) ?? In my spherical example I have dAr = r2sinθ dθdφ ρdφ rdθ ρ = rsinθ dAθ = rsinθ drdφ ρdφ dr dAφ = rdrdθ rdθ dr (8.4.k.7) where dAr for example is a little piece of true area on the surface of a sphere. The Πi=1n du'i for the first line is played by dθdφ but of course this does not have the dimensions of area. We want Cartesian space type area, not parameter space type area. Question: What does it mean to integrate something over a surface? [ This ended up in Section 10.10 ] Example 1: Find the average temperature on a flat plate. What does that mean? Let T(x,y) be temperature. If you took 1,000,000 random temperature readings on this plate (random locations on the plate), and divided by 1,000,000, that would be the average temperature. How would you take those readings? Each reading is a random (x,y) selection. The probability of this (x,y) showing up in your random distribution is p(x,t)dxdy = dxdy/(ab) where a,b are the plate dimensions. Any AREA is just as likely. You drip a sticky micro ball on the plate from high up and it lands somewhere. So here is what we are talking about <T> = (1/ab) ∫dxdy T(x,y) = (1/A) ∫dA T(x,y) where dA is the are of a tiny patch of surface. If you switch to polar coordinates to describe points on the plate, then dA = rsinθdrdθ where rsinθ is the Jacobian. This same kind of integral would be used to compute the average density of a flat plate which has areal density ρ(x,y). And you could compute the center of mass with a similar integral. You would compute <x> = (1/A) ∫dA x T(x,y) <y> = (1/A) ∫dA y T(x,y) Example 2: Find the average normal component of a magnetic field on a flat plate. Imagine some B(x,y). the answer to this problem is <Bn> = (1/ab) ∫dxdy B(x,y) = (1/A)∫dA B(x,y) where in this problem it happens that = = constant. Each point on the surface has a well-defined normal. Comment. For both these examples, one could consider a disk instead of a square, and then you would use dA = dxdy → (rsinθ)drdθ where the Jacobian appears. You would show that such dA paper the surface perfectly. Example 1A: Find the average temperature on a sphere Example 2A: Find the average normal component of a magnetic field on sphere. It seems clear to me that I know the answer to both these questions: <T> = (1/A) ∫dA T(θ,φ) <Bn> = (1/A)∫dA B(θ,φ) where A = 4πR2 and where dA = R2sinθdθdφ and where = . I know this is the right dA because I just compute it visually to say dA = (Rdθ)(Rsinθdφ) Example 1B: Find the average temperature on an arbitrary surface. Example 2B: Find the average normal component of a magnetic field on an arbitrary surface. Again it seems clear to me that I know the answer to both these questions: <T> = (1/A) ∫S dA T(r) <Bn> = (1/A)∫S dA B(r) = (1/A)∫S dA B(r) dA = dA The meaning of these integrals is clear: dA is a local area element on the surface, and is a local normal at a point r on the surface. You just have to figure out what these things are for a given surface. Question: How do you compute the above objects if the surface is x' = F(x) ? That is to say, the "arbitrary surface" is defined by r' = x' = F(x,y) which is a mapping from some flat surface to the arbitrary surface. Obviously you need to find expressions for (r') and dA(r') for any point r' which lies on the surface. Notice that this is a very general way to define an "arbitrary surface", and we like it for that reason. As a start to finding (r') and dA(r') it would be good to go to a point r' on the surface and find some coordinates that define area and normal at that point. This certainly is where those tangent base vectors called u'i(r') come into play! For i = 1,2 these span the tangent space Tr'M, and the third i = 3 when normalized will be your normal vector. But how exactly do you do this? You don't just need base vectors, you need some "coordinates" like θ and φ on the sphere. Well, perhaps you need local coordinates at a point r' on the surface. Well, here is an arbitrary finite point ξ in the tangent plane at point r' (ξ1,ξ2) = ξ = ξ1u'1 + ξ2u'2 dξ = dξ1u'1 + dξ2u'2 = u'3 / |u'3| Now what is dA ? In general u'1 and u'2 are not perpendicular. This only happens if x' = F(x,y) defines an orthogonal coordinate system. [ the non-square R matrix examples ] Example 3. For a spherical surface for example you would have x' = x y' = y z' = ± and we can restrict our interest to the upper hemisphere with z' = + . This is already one of my wedge doc examples and I have computed that R = S = where I show one of at least three viable S matrices to go with this R matrix. I also show that R** = [u'1, u'2 ] (2.5.9) (10.6.e.5) So the I conclude for this example that u'1 = u'2 = u'3 = I need some kind of "as needed" u'3 and I could just take it to be u'3 = u'1 x u'2 . u'3 = e'1 { 0*[-y/Q] - [-x/Q]*1 } + e'2 { [-x/Q] * 0 - 1*[-y/Q]} + e'3 { 1*1 - 0*0} = e'1{x/Q} + e'2 {y/Q} + e'3 = (x/) e'1 + (y/)e'1 + e'3 = and I have entered this on the right above. At this point, I know the three vectors u'i. I don't yet know dA, but |u'3|2 = x2/Q2+ y2/Q2 + 1 = x2/(R2-x2-y2)+ y2/(R2-x2-y2) + (R2-x2-y2)/(R2-x2-y2) = [ x2+ y2 + (R2-x2-y2)] / (R2-x2-y2) = R2/(R2-x2-y2) Then I claim that = u'3 / |u'3| = u'3 / [ R/ ] = (/R) u'3 = (/R) = Not very enlightening but there is some result like this. I was hoping to get = ' , but I don't have any spherical coordinates yet. But in x'-space I do know that r' = x'1e'1 + x'2e'2 + x'3e'3 |r'|2 = x'12 + x'12 +x'32 + = x2+ y2 + (R2- x2- y2) = R2 So I then know that ' = (1/R)(x1e'1+ x2e'2 + e'3) = = so yes, I get the expected result. Now what about dA? I know that u'1 x u'2 form a finite 2-piped which has this area area = | u'1 x u'2| = | u'3| = R/ Then the differential 2-piped of interest would have this area dA = | (dξ1u'1) x (dξ2u'2)| = [R/ ] dξ1dξ2 But now what? I do know that dx' = Rdx if that can help and I also have dx = Sdx' dξ1 = dξ1u'1 dξ2 = dξ1u'2 Suppose I select dx' = dξ1u'1 Then dx(1) = Sdx' = S(dξ1u'1) = dξ1 Su'1 = dξ1 u1 yes, see (10.7.2) dx(2) = Sdx' = S(dξ1u'2) = dξ1 Su'2 = dξ1 u2 ?? Then an element of area in x-space would be just dξ1dξ2 which is just dx1dx2. So maybe dA = [R/ ] dx1dx2 = [R/ ] dxdy and maybe then that is the dA of interest! What does this look like in sphericals? x' = Rsinθcosφ y' = Rsinθsinφ z' = Rcosθ R2- x'2-y'2 = R2 ( 1 - sin2θ) = R2cos2θ = z'2 Then I am getting dA = [R/z'] dxdy = (1/cosθ)dxdy Actually this seems sort of reasonable at north pole and at equator. Now I think the upper hemisphere is mapped into by a disk in x-space, so consider the average temperature calculation <T> = (1/A) ∫S' dA' T(r) = ∫S dA T(r) / ∫S dA Look first at the denominator: ∫S' dA' = ∫S [R/ ] dxdy dimensions OK Here you see the pullback idea being used. The region S' is in x'-space, while S is in x-space. So you end up here with a purely x-space integral. You can do it in polars: x = rcosu y = rsinu dxdy = rdrdu = Then ∫S' dA' = R !Syntax Error, Idr!Syntax Error, Idu r (1/) = R * [!Syntax Error, Idr r (1 /) ] [ !Syntax Error, Idu ] = 2πR [!Syntax Error, Idr r (1 /) ] I will have Maple do this integral, Then I find that ∫S' dA' = 2πR * R = 2πR2  which is the correct result. then <T> = (1/2πR2) ∫S' dA' T(r) 2πR2<T> = ∫S' dA' T(r) = ∫S [R/ ] dxdy T(x,y,) = ∫S [R/ ] rdrdu T(rcosu,rsinu,) My hemisphere example here is complicated by the fact that I want to use polar coordinates in x-space, and that is really not the key point! Go back and review this temperature calculation: 1. For the sphere example I did this: = u'3 / |u'3| where u'3 = u'1 x u'2 dA = | (dξ1u'1) x (dξ2u'2)| = [R/ ] dξ1dξ2 = [R/ ] dx1dx2 The whole thing really does look like we are "pulling back" dA to x-space. In a more general case, things are the same but u'3 = u'1 x u'2 would have to be computed from the R matrix. The computation of <Bn> = (1/A)∫dA B(θ,φ) for a general surface involves no new feature really! It is the same dA, the same A, and you compute as shown above (I computed it and never used it, but here you would need to use it. ) Look again at the computation of A dA = | (dξ1u'1) x (dξ2u'2)| = [R/ ] dξ1dξ2 = dξ1dξ2 | u'1 x u'2| = dξ1dξ2 | u'3 | = dx dy | n | Then our temperature integral or scalar integral would be ∫S f(F(x) [ | n | dx dy ] n = u'3 = u'1 x u'2 and THIS is what Bucks are saying on page 368. It finally makes sense to me!!! It is an unnormalized normal vector that appears. Then | n | dx dy is dA in x'-space but pulled back to x-space coordinates. Now go back to R** = [u'1, u'2 ] (2.5.9) (10.6.e.5) u'3 = u'1 x u'2 (u'3)i = εijk (u'1)j (u'2)k = εijk Rj1 Rk2 where ε is the permutation tensor only! Question: I am tempted to say here that both x-space and x'-space are Cartesian so then g = g' = 1. I always think this when I draw the two spaces with a surface hanging in x'-space. But I think that means that the R matrix must then be locally a rotation. Then you would have g' = RTR and it makes sense. But then there is the confusion of having another set of coordinates u'i and ξi which are not Cartesian. In any event, coordinates like (u'2)k are relative to the e'n basis vectors in x'-space which are axis-aligned. Note that x-space and x'-space are both Cartesian in the examples I am thinking of so up/down indices are the same. Then = εijk εij'k' Rj1 Rk2 Rj'1 Rk'2 = (δjj'δkk' - δjk'δkj') Rj1 Rk2 Rj'1 Rk'2 = Σjk [ Rj1 Rk2 Rj1 Rk2 - Rj1 Rk2 Rk1 Rj2 ] and I know this brings us to page 299 of Buck and the forms of the "k thing" there. Lower all indices assuming that g = 1 and g' = 1, so then | u'3 |2 = Σjk [ Rj1 Rk2 Rj1 Rk2 - Rj1 Rk2 Rk1 Rj2 ] = Σj(Rj12) Σk(Rj22) - Σj(Rj1 Rj2) Σk(Rk1Rk2) = Σj(Rj12) Σk(Rj22) - [Σj(Rj1 Rj2)]2 and THIS is Buck p 299 B But I know from the cross product formula that there is another way to write this result | u'3 |2 = det2 + det2 + det2 and this other form is Buck p 299 E. So at least I see why these two forms are the same. But this is the same as Bucks page 368 A!!! OK, I have these three Buck puzzle pieces aligned! Definition: Following Bucks, let's define k(x') ≡ | u'3 | = | u'1 x u'2 | = either functional form above, where Rij = Rij(x') Conjecture: Suppose x'-space is R4 instead of R3. My conjecture is this: You get your three u'i from the R matrix, then you set u'4 = u'1 x u'2 x u'3 where this is the generalized cross product of tensor doc and it is the new normal vector n. And of course we have the connection to the dual vectors called Ei or ei in tensor doc, and here I guess are u'i. And of course the "volume" measure gets involved. And then there is some connection to the wedge product! So lots more pieces may fall out soon! An = σ (-1)n-1 e1 x e2 ... x eN // en missing σ ≡ sign[det(Sab)] = sign[det(Rab)] An,i = (sign) e1 x e2 ... x eN // en and ei missing An,i,j = (sign) e1 x e2 ... x eN // en and ei and ej missing dAn = Πi≠ndx'i dAn,m = Πi≠n,m dx'i dAn,m,k = Πi≠n,m,k dx'i So I have some other way to state the magnitude of An etc above? This is one formula that tensor doc is missing and I think I can fix that. Meanwhile, the issue of functionals. To me, it makes no sense to integrate a functional. Maybe a tensor function. Is there some way to have the functionals "act on something" so integration makes more sense? Consider again this stuff from above, dA' = | (dξ1u'1) x (dξ2u'2)| = [R/ ] dξ1dξ2 = dξ1dξ2 | u'1 x u'2| = dξ1dξ2 | u'3 | = dx dy | n | dx(1) = Sdx'(1) = S(dξ1u'1) = dξ1 Su'1 = dξ1 u1 yes, see (10.7.2) dx(2) = Sdx'(2) = S(dξ1u'2) = dξ1 Su'2 = dξ1 u2 ?? So write dx'(1) = dξ1u'1 dx(1) = dξ1 u1 dx'(2) = dξ1u'2 dx(2) = dξ1 u2 These are little differentials in x'-space that lie on the surface, that lie within Tx'M. They pull back as shown on the right to axis-aligned differentials in x-space. This fact then motivates changing names so dx'(1) = dx1u'1 dx(1) = dx1 u1 dx'(2) = dx2u'2 dx(2) = dx2 u2 dA' = | dx'(1) x dx'(2)| ∫S' T(x') dA' = ∫S' T(x') | dx'(1) x dx'(2)| ?? ∫S' B(x') ' dA' = ∫S' B(x') ' | dx'(1) x dx'(2)| ?? This seems to be the way dx' objects appear in the integrand for a classical surface integral! Not really what I was expecting to see. Can write as ∫S' T(x') dA' = ∫S' T(x') | u'1 x u'2| dξ1dξ2 ∫S' B(x') ' dA' = ∫S' B(x') ' | u'1 x u'2| dξ1dξ2 Now use definition above to write k(x') ≡ | u'3 | = | u'1 x u'2 | = either functional form above, where Rij = Rij(x') and then we have ∫S' T(x') dA' = ∫S' T(x') k(x') dξ1dξ2 ∫S' B(x') ' dA' = ∫S' B(x') ' k(x') dξ1dξ2 and we get a definition notion of area measure dA' = k(x') dξ1dξ2 But still such integrals don't make sense. How do you "integrate over dξ1" for example? What does make sense is this: ∫S' T(x') dA' = ∫S T(x') k(x') dx1dx2 where x' = F(x) ∫S' B(x') ' dA' = ∫S B(x') ' k(x') dx1dx2 Now a person can actually DO these integrals, over domain in x-space, deducing ' as in example above. Conclusion: I don't really have any way to write the integral directly in x'-space in some form such as ∫S' T(x') dA' = ∫S' T(x') dx'1dx'2 incorrect! ∫S' B(x') ' dA' = ∫S' B(x') ' dx'1dx'2 ∫S' T(x') dA' = ∫S' T(x') | dx'(1)(x') x dx'(2)(x')| correct ∫S' B(x') ' dA' = ∫S' B(x') ' | dx'(1)(x') x dx'(2)(x')| The last form is OK, but I am showing with dx'(1)(x') that, for example, the direction of dx'(1) changes as you move to different points on the surface S'. No one on this planet can do the above integrals unless you tell them exactly what dx'(1)(x') is. You have to quadrilate the surface S' and add up the contributions from each 2-piped. There is no English word quadulate or quadrilate, but I know what it means. So imagine mechanically doing the integral by quadrilation. You must cover the surface with a fine mesh of quads. What is the area of one such tiny quad? It has to have "extent" so I have been writing dA' = k(x') dξ1dξ2 So I evaluate k(x') at point x' on the surface. I have some small numbers dξ1dξ2 which I just mechanically read off my mesh picture. Each quad could have a different dξ1,dξ2 pair for example. there is no reason for them to have to be the same. The main idea is that they have to be very small. In a 1D Riemann integral you could similarly set each dx interval to be different, as long as they are small the sum in the limit will be good. But it would be simplest to take dξ1 to be the same for every little quad, and similarly for dξ2. Then ∫S' T(x') dA' ≈ ΣS' T(x') k(x') Δξ1Δξ2 But now we want to "parametrize" the surface with variables ξ1 and ξ2 so each point has a unique value for this pair. That is what the mapping does! So you could say ξ = F-1(x') but I usually call this x. So then we get ∫S' T(x') dA' ≈ ΣS' T(x') k(x') Δx1Δx2 → ∫S T(x'(x)) k(x'(x)) dx1dx2 and there is your temperature integral. I have given a mechanical method of getting to this resulting integral, and that mechanical process can be called "pulling back" to x-space. In effect I then have dA'(x') = k(x') dx1dx2 Question: What does the temperature integral "look most closely like" in a differential forms integral? α = Σ'I fI(x') λ'^I = Σ1≤i<i≤3 fii(x') dx'i ^ dx'i = f12(x') dx'1 ^ dx'2 + f13(x') dx'1 ^ dx'3 + f23(x) dx'2 ^ dx'3 When you pull this back you get ∫φ α = ∫[0,1]2 G(t) dt1 ^ dt2 G(t) ≡ [ f12(φ(t)) + f13(φ(t)) + f23(φ(t)) ] . = H(φ(t)) n(t) = ? = T |n| OK, someone walks up to you and gives you this problem. Given n and given T, find H such that H n = T |n| OK try writing H = Hn + H11 + H22 . Then, H n = Hn n = Hn |n| = T |n| So the solution to the problem is this: any H with Hn = T will work. In particular, H = Hn works. So can take G(t) = H(φ(t)) n(t) = T |n| where H = T . Then Hi = T ( )i = T (n)i / |n| recall |n| = = known Then the appropriate differential form for a temperature integral is this f12 = H3 = T (n)3 / |n| f13 = -H2 = - T (n)2 / |n| f23 = H1 = T (n)1 / |n| and so α = f12(x') dx'1 ^ dx'2 + f13(x') dx'1 ^ dx'3 + f23(x) dx'2 ^ dx'3 = [ T (n)3 / |n|] dx'1 ^ dx'2 + [ - T (n)2 / |n|] dx'1 ^ dx'3 + [T (n)1 / |n|] dx'2 ^ dx'3 = (T/ |n|) { (n)3 dx'1 ^ dx'2 - (n)2 dx'1 ^ dx'3 + (n)1 dx'2 ^ dx'3 } = (T/ |n|) { (n)3 *dx'3 + (n)2 *dx'2 + (n)1 *dx'1 } = (T/ |n|) { n *dx'} = T *dx' So there is the α that works for the temperature integral. Now what does this look like after pullback? ∫S' α = ∫S'T(x') (x') *dx' = ∫S G dx1dx2 = ∫S T |n| dx1 dx2 = ∫S T |n| λ1 ^ λ2 Just by the by, consider (λ1 ^ λ2) (dx1, dx2) = λ1(dx1)λ2(dx2) = (dx1)1 (dx2)2 = dx1 dx2 which is certainly interesting. Is the above wedge product really correct? The general rule from (4.4.20b) is this (λi^ λj)(vr,vs) = (1/2) [ (vr)i(vs)j - (vr)j(vs)i] Now set vr = dx1 vs = dx2 (λ1 ^ λ2) (dx1, dx2) = (1/2) [ (dx1)1(dx2)2 - (dx1)2(dx2)1] = (1/2) [ (dx1)1(dx2)2 - 0 * 0] = (1/2) dx1 dx2 So yes, it is correct except for my non-Spivak normalization. [ the model for understanding bra ket integration interp, now installed in Section 10.11 ] Now go back to the general form ∫S' α' = ∫S' Σ'I fI(x') λ'i ^ λ'i = ∫S G(x) λ1 ^ λ2 so again we have this functional = functional equation ∫S' Σ'I fI(x') λ'i ^ λ'i = ∫S G(x) λ1 ^ λ2 functional in Λ'2 functional in Λ2 These are functionals in different spaces. Not sure how to "close" with a set of vectors. Consider [λ'^I R] = ΣJ RIJ λ^J In this equation, both sides are functionals in the same space, which is x-space. How about this approach: in F(x) notation: α' = Σ'I fI(x')λ'^I = Σ'I fI(x')<e'^I | [∫S' α' ] R = F* [∫S' α' ] = ∫S F*(α') = ∫S F*( Σ'I fI(x') λ'^I) = ∫S Σ'I F*(fI(x')) F*(λ'^I) = ∫S Σ'I fI(F(x)) ΣJ RIJ λ^J = ∫S Σ'I fI(F(x)) Σ'J det(RIJ) λ^J jr = 1,2...n = ∫S Σ'I fI(F(x)) Σ'J det(RIJ) <u^J | jr = 1,2...n n > k Note: There are k factors in <u^J |, each factor of the form <ujr | where jr = 1,2...n . Try closing this now on an arbitrary set of k vectors vr1 in Rn to get [∫S' α' ] R |vr1, vr2, .... vrk> = ∫S Σ'I fI(F(x)) Σ'J det(RIJ) <u^J |vr1, vr2, .... vrk> Then use (8.3.9a), (λj ^ λj ^ .... ^ λj)(vr,vr.....vr) = (1/k!) det [ (vr)j ] and we then have [∫S' α' ] R |vr1, vr2, .... vrk> = = ∫S Σ'I fI(F(x)) Σ'J det(RIJ) (1/k!) det [ (vr)j ] = ∫S Σ'I fI(F(x)) ΣJ RIJ (1/k!) det [ (vR)J ] I don't want to lose the sum on J. NOW take a tensor function of this vectors argument: dxr1, dxr2, .... dxrk where, dxri = dxriuri // no sum on the ri so then [∫S' α' ] R | dxr1, dxr2, .... dxrk> = ∫S Σ'I fI(F(x)) Σ'J det(RIJ)<u^J |dxr1, dxr2, .... dxrk> = ∫S Σ'I fI(F(x)) Σ'J det(RIJ) dxr1dxr2....dxrk <u^J |ur1, ur2, .... urk> Next use (8.3.9b) to get = ∫S Σ'I fI(F(x)) Σ'J det(RIJ) dxr1dxr2....dxrk (1/k!) det [ δrj ] = (1/k!) ∫S Σ'I fI(F(x)) dxr1dxr2....dxrk Σ'J det(RIJ) det [ δrj ] = (1/k!) ∫S Σ'I fI(F(x)) dxr1dxr2....dxrk ΣJ RIJ det [ δrj ] = (1/k!) ∫S Σ'I fI(F(x)) dxr1dxr2....dxrk ΣJ RIJ ΣP(-1)P δrP(1)j1 δrP(2)j2 .... = (1/k!) ΣP(-1)P ∫S Σ'I fI(F(x)) dxr1dxr2....dxrk ΣJ RIJ δrP(1)j1 δrP(2)j2 .... = (1/k!) ΣP(-1)P ∫S Σ'I fI(F(x)) dxr1dxr2....dxrk Σj1j2.... Ri1j1Ri2j2.... δrP(1)j1 δrP(2)j2 .... = (1/k!) ΣP(-1)P ∫S Σ'I fI(F(x)) dxr1dxr2....dxrk Ri1rP(1)Ri2rP(2) = (1/k!) ∫S Σ'I fI(F(x)) dxr1dxr2....dxrk det(RIR) Now suppose I sum both sides on R: Σ'R [∫S' α' ] R | dxr1, dxr2, .... dxrk> = (1/k!) ∫S Σ'I fI(F(x)) Σ'Rdxr1dxr2....dxrk det(RIR) Now change names from rs to js Σ'J [∫S' α' ] R | dxj1, dxj2, .... dxjk> = (1/k!) ∫S Σ'I fI(F(x)) Σ'Jdxj1dxj2....dxjk det(RIJ) = (1/k!) ∫S Σ'I fI(F(x)) Σ'J det(RIJ) dxj1dxj2....dxjk One more time: k! Σ'J [∫S' α' ] R | dxj1, dxj2, .... dxjk> = ∫S Σ'I fI(F(x)) Σ'J det(RIJ) dxj1dxj2....dxjk At least this obscure method generates the correct right hand side which appears in the official, ∫φ α = Σ'I Σ'J (!Syntax Error, I!Syntax Error, I ..... !Syntax Error, I) fI(F(x)) det(RIJ(x)) dxjdxj....dxj R= (DF) I think in Spivak normalization I would set k! = 1. Summary: Start off with these two ingredients, <∫S' α' | R = ∫S Σ'I fI(F(x)) Σ'J det(RIJ) λ^J = ∫S Σ'I fI(F(x)) Σ'M det(RIM) <u^M | // ordered M sum = ∫S Σ'I fI(F(x)) ΣM RIM <u^M | // full M sum where we replace the usual dummy ordered multiindex J by M. The action of R acting to the left converts a functional in Λk(Rm) to a functional in Λk(Rn). The integral is treated here as a mesh sum over a region bounded by an x'-dependent boundary. F* restates the boundary using F(x) in place of x', so S' becomes S. The second ingredient is this element of the space Vk(Rn) | q > ≡ Σ'J | dxJ> = Σ'J [| dxj1> | dxj2> ..... | dxjk>] = Σ'J | dxj1, dxj2, .... dxjk> where dxi = dxiui Note that <∫S' α' | R = a dual rank-k tensor in Λk(Rn) = a bra in Λk(Rn) V*k(Rn) | q > = Σ'J | dxJ> = a rank-k tensor in Vk(Rn) = a ket in Vk(Rn) The tensor product scalar product is then <∫S' α' | R | q > = ∫S Σ'I fI(F(x)) ΣM RIM <u^M | Σ'J | dxJ> = Σ'I Σ'J ∫S fI(F(x)) ΣM RIM <u^M | dxJ> = Σ'I Σ'M Σ'J ∫S fI(F(x)) ΣM RIM dxjdxj....dxj <u^M | uJ> The last scalar product is given by <u^M | uJ> = λ^M(uJ) = (λm1 ^ λm2 ^ ... ^ λmk) (uj,uj, ...uj) = (1/k)! det [ δjm ] // non-Spivak normalization for wedge product = det [ δjm ] // Spivak normalization for wedge product = det [ δJM ] Using the Spivak normalization we then have <∫S' α' | R | q > = Σ'I Σ'J ∫S fI(F(x)) ΣM RIM det(δJM ) dxjdxj....dxj Now consider ΣM RIM det(δJM ) = ΣM RIM ΣP(-1)P δP(J)M = ΣP(-1)P ΣM RIM δMP(J) = ΣP(-1)P RIP(J) = det(RIJ) The final result is then <∫S' α' | R | q > = Σ'I Σ'J ∫S fI(F(x)) det(RIJ) dxjdxj....dxj Another intermediate result <∫S' α' | R | q > = ∫S Σ'I fI(F(x)) ΣM RIM <u^M | q > = ∫S Σ'I fI(F(x)) Σ'J det(RIJ)<u^J | q > where dummy ordered multiindex M is replaced by J. [F*(∫S' α' )] ( Σ'J(dxj1, dxj2, .... dxjk) // but function is k-multilinear, so... = Σ'J [F*(∫S' α' )](dxj1, dxj2, .... dxjk) // tensor function form = Σ'J Σ'I Σ'M∫S fI(F(x)) det(RIM) λ^M(dxj1, dxj2, .... dxjk) = Σ'J Σ'I Σ'M∫S fI(F(x)) det(RIM) dxjdxj....dxj λ^M(uj, uj, .... uj) = Σ'J Σ'I Σ'M∫S fI(F(x)) det(RIM) dxjdxj....dxj det(δMJ) = Σ'J Σ'I ∫S fI(F(x)) [ ΣMRIM det(δMJ)] dxjdxj....dxj = Σ'J Σ'I ∫S fI(F(x)) [ det(RIJ)] dxjdxj....dxj <∫S' α' | R = ∫S Σ'I fI(F(x)) Σ'J det(RIJ) dx^J First, construct this linear combination of vectors in Vk(Rn) | q > = Σ'J | dxj1, dxj2, .... dxjk> = Σ'J | dxj1> | dxj2> ..... | dxjk> = Σ'J | dxJ> Then have [∫S' α' ] R | q > = A simpler case, dx1 = dx1u1 dx2 = dx2u2 ... dxk = dxkuk <u^J | dx1, dx2, .... dxk> = // If k < n, this is an incomplete scalar product = dx1dx2....dxk (1/k!) [ δj1 δj2..δjk + signed permutations] = dx1dx2....dxk (1/k!) det(δJZ) . At this point we then have, [∫S' α' ] R = ∫S Σ'I fI(F(x)) Σ'J det(RIJ) <u^J | dx1, dx2, .... dxk> = ∫S Σ'I fI(F(x)) Σ'J det(RIJ) dx1dx2....dxk (1/k!) det(δJZ) = (1/k!) ∫S Σ'I fI(F(x)) Σ'J det(RIJ) det(δJZ) dx1dx2....dxk = (1/k!) ∫S Σ'I fI(F(x)) ΣJ RIJ det(δJZ) dx1dx2....dxk where the last two lines show alternate ways to present the J sum. Maybe at this point I will write out the det object in ΣP notation, using (8.3.9.a), = (1/k!) ∫S Σ'I fI(F(x)) ΣJ RIJ ΣP(-1)P [δP(1)jδP(2)j2...δP(k)jk ] dx1dx2....dxk = (1/k!) ΣP(-1)P ∫S Σ'I fI(F(x))dx1dx2....dxk ΣJ RIJ δP(1)jδP(2)j2...δP(k)jk ] Now imagine writing out RIJ and then you get = (1/k!) ΣP(-1)P ∫S Σ'I fI(F(x))dx1dx2....dxk RIP(Z) = (1/k!) ∫S Σ'I fI(F(x))dx1dx2....dxk ΣP(-1)P RIP(Z) = (1/k!) ∫S Σ'I fI(F(x))dx1dx2....dxk det(RIZ) The result is then, <∫S' α' ] R | dx1, dx2, .... dxk> = (1/k!)∫S Σ'I fI(F(x)) det(RIZ) dx1dx3 ...dxk But this result is too simple! It is supposed to have a sum on J. The R matrix is m x n and for this k form we are only getting data from the first k columns, I am sure that is wrong. Here is the official result, ∫φ α = Σ'I Σ'J (!Syntax Error, I!Syntax Error, I ..... !Syntax Error, I) fI(φ(t)) det(RIJ(t)) dtjdtj....dtj R= (Dφ) But I have assumed that k = n, so there is then only one term in Σ'J and maybe this too becomes ∫φ α = (!Syntax Error, I!Syntax Error, I ..... !Syntax Error, I) Σ'I fI(φ(t)) det(RIZ(t)) dt1dt2....dtk R= (Dφ) This is pretty non-standard interpretation methinks! I am saying this: ∫S' α' is really equal to <∫S' α' ] R | dx1, dx2, .... dxk> in some sense. I am off to the Maze!