Section 10_12 old I think

Section 10.12 Integration of Differential Forms PhL 4.13.16 10.12 Integration of 1-forms and 2-forms This section is presented in the x = φ(t) notation introduced in Section 10.9 and illustrated in Fig (10.9.3) which we replicate here, (10.9.3) The main result of Section 10.11 is this description of the integration of a k-form over a surface, ∫S' αx' = ∫S'[Σ'I fI(x') dx'i ^ dx'i ^ ... ^ dx'i ] // αx' = Σ'I fI(x') dx'^I ≡ ∫S [ Σ'J gJ(x) dxj ^ dxj ^ ... ^ dxj ] // first definition (pull back) ≡ ∫S [ Σ'J gJ(x) dxjdxj ... dxj] // second definition where gJ(x) = Σ'I fI(F(x)) det(RIJ) and x' = F(x) , R = (DF) . (10.11.7) Using the x = φ(t) notation we rewrite the above with some specialization as, ∫φ αx = ∫φ Σ'I fI(x) dxi ^ dxi ^ ... ^ dxi // αx = Σ'I fI(x) dx^I ≡ ∫[0,1]k Σ'J gJ(t) dtj ^ dtj ^ ... ^ dtj // first definition (pull back) ≡ (!Syntax Error, I!Syntax Error, I ... !Syntax Error, I) Σ'J gJ(t) dtjdtj ... dtj // second definition where gJ(t) = Σ'I fI(φ(t)) det(RIJ) and x = φ(t) , R = (Dφ) . (10.12.1) Here the pulled-back integration region formerly called S is taken to be the unit cube in k dimensions, written above as [0,1]k and referred to as a k-cube. The pre-pullback integration region formerly called S' is here called φ, with the idea that this region is φ([0,1]k). Note: A k-chain is a linear combination of k-cubes and is used by both Sjamaar (p 65) and Spivak (p 97) in their derivations of Stokes' Theorem. In fact, Spivak's entire Chapter 4 which includes his discussion of tensor products, wedge products and tensor functions is entitled Integration on Chains. We wish now to look in more detail at the integration of 1-forms and 2-forms. (a) Line Integral as example of pulling back a 1-form in Rm Consider this 1-form in x-space Rm , α = Σi fi(x) xλi = Σi fi(x) dxi . We wish to define a meaning for the integration of this 1-form α over a piece of the curve x = φ(t), ∫φ α = ∫φ Σi fi(x) dxi = integral of a 1-form over a piece of the curve φ in Rm . The transformation x = φ(t) is a mapping φ: t → Rm. Variable t is often called a "parameter". Comment: Officially it is the mapping φ which is "the curve", but one loosely refers to the image (trace) of this mapping in Rm as "the curve". The distinction is necessary because many mappings can have the same image curve, such as φ(t) and φ(t2), where the parameter is "re-speeded" (reparametrized). This picture shows the general respeeding idea : Here the same red curve is the image of two different transformations x = φ(t) and x = ψ(t) with different domain intervals, and ψ(t) = φ(f(t)) where f(t) is a monotonic respeeding function. A special case would be [a,b] = [c,d] = [0,1] to which our example φ(t) and ψ(t) = φ(t2) would apply. Curves φ and ψ are called smoothly equivalent curves and ∫φ α is the same for any two such curves (Buck p 386 Theorem 2). A similar but generalized reparametrization comment applies to integration of 2-forms and k-forms. So imagine that we have a curved line hanging in Rm space and as t varies perhaps from 0 to 1 in t-space, we move along the image curve in Rm. The problem is how to integrate a 1-form along this curve. We can define the calculational meaning of the above integral in two steps, each being a definition: First definition: ∫φ α ≡ ∫[0,1] φ*(α) = the integral in t-space of the pullback of α over the 1-cube [0,1] On the left is an integral of the 1-form α over a curve φ in Rm. On the right is an integral of a different 1-form φ*(α) (the pullback of α) over a 1-cube [0,1] in R1. Note that α lies in xΛ1(Rm) while φ*(α) lies in tΛ1(R). Since our usual 1-form pullback mapping is φ* : xΛ1(Rm) → tΛ1(Rn), we have n = 1. The "tall" m x n R-matrix for this problem is then an m x 1 matrix which is just a column vector of m elements ∂tφi , Ri1 = (D(t)φ)i1 = ∂φi(t)/∂t . // t1 ≡ t, the only coordinate in t-space We then compute the pullback of α, α = Σi fi(x) xλi = Σi fi(x) dxi = f(x) dx dx ≡ (dx1, dx2, .... dxm) φ*(α) = Σi φ*(fi(x)) φ*(dxi) // (10.9.6) item 3 = Σi fi(φ(t)) Σj=1n Rij dtj // (10.9.6) items 1 and 5 = Σi fi(φ(t)) Ri1 dt1 // n = 1 = Σi fi(φ(t)) [∂φi(t)/∂t] dt // *** = g(t) dt where g(t) ≡ Σi fi(φ(t)) [∂φi(t)/∂t] = Σi fi(φ(t)) ∂tφi(t) = f(φ(t)) (∂tφ) . The object φ*(α) = g(t) dt is a 1-form in dual t-space tΛ1(R). Using the definition given above, one then has, ∫φ α = ∫[0,1] φ*(α) = ∫[0,1] g(t) dt . Thus the integral of the 1-form α over the curve φ in x-space is defined to be equal to the integral of the 1-form g(t) dt over a 1-cube in t-space. So far no regular calculus integrals have appeared. Second definition: ∫[0,1] g(t) tλ = ∫[0,1] g(t) dt ≡ !Syntax Error, Ig(t) dt . On the left is the integral of a 1-form on a 1-cube, on the right is an ordinary calculus integral of a function over the interval [0,1] of the real axis. It is this second definition that motivates giving the dual space basis vector tλ the cosmetic name dt. If one flips the "orientation" of the integration domain, so that [0,1] becomes [1,0], the result changes sign, and of course this fact agrees with the usual notion that !Syntax Error, Ig(t) dt = - !Syntax Error, Ig(t) dt . We have then shown that ∫φ α = ∫φ f(x) dx = ∫[0,1] g(t) dt = !Syntax Error, I f(φ(t)) (∂tφ(t)) dt = !Syntax Error, Ifi(φ(t)) [∂φi(t)/∂t] dt This result appears in Sjamaar Ch 4 Eq (4.1) with φ = c and m = n. Since dx = (∂tφ(t)) dt, the above can be written concisely as ∫φ α = ∫φ f(x) dx = !Syntax Error, I f(φ(t)) dx where dx = (∂tφ(t)) dt . Normally this is written "∫φ f(x) dx " = !Syntax Error, I f(φ(t)) (∂tφ(t)) dt where the operational meaning of the object on the left is given by the calculus integral on the right. The object on the left is called "the line integral of a vector field f(x) over a curve φ". As noted earlier, this integral is independent of any smooth monotonic respeeding of φ(t). Notice that the only locations where f(x) is "sensed" in this integral are points on the curve x = φ(t). The endpoints 0 and 1 can be replaced by a and b so the unit 1-cube [0,1] becomes [a,b]. Such a change is a reparametrization as in Fig ** above. We redisplay the earlier Fig (10.9.20) to illustrate the above discussion, where βt = g(t) dt : One does not need to use "differential forms" and "dual spaces" and so on just to compute a line integral, but the above example shows how a line integral fits into the framework of differential forms and linear functionals. In dealing with 1-forms one encounters no wedge products, but see section (b) below. Example for Rm = R2: the "angle form" problem mentioned in (10.5.10). In this problem we have specific functions f1 and f2, a specific range [0,2π] for the t-space domain, and a specific curve x = φ(t) = (x1,x2). αx = Σi=12 fi(x) xλi = f1(x) dx1 + f2(x) dx2 = - (x2/r2) dx1 + (x1/r2) dx2 where r2 ≡ (x1)2 + (x2)2 x1 = φ1(t) = cos t ∂tφ1(t) = -sin t t = [0,2π] x2 = φ2(t) = sin t ∂tφ2(t) = cos t t is the polar angle of the vector x = (x1,x2) r2 = (x1)2 + (x2)2 = cos2 t + sin2 t = 1 vector x lies on the unit circle in x-space f1(φ(t)) = - (x2/r2) = - sin t f2(φ(t)) = + (x1/r2) = cos t α = - sin t dx1 + cos t dx2 φ*(α) = g(t) dt pullback of α g(t) = fi(φ(t)) [∂φi(t)/∂t] = [ f1(φ(t))∂tφ1(t) + f2(φ(t))∂tφ2(t)] = [ (-sin t)(-sin t) + (cos t)(cos t) ] = 1 so φ*(α) = 1 dt ∫φ α = ∫[0,2π] φ*(α) = ∫[0,2π] g(t) dt = ∫[0,2π] dt = !Syntax Error, Idt = 2π . So the integral of this particular 1-form α around the unit circle gives the number 2π. (b) Surface integral as example of pulling back a 2-form in Rm For the moment we set m = 3 and consider this 2-form in x-space R3 , α = Σ'I fI(x) xλ^I = Σ'I fI(x) dx^I = Σ1≤i<i≤3 fii(x) dxi ^ dxi = f12(x) dx1 ^ dx2 + f13(x) dx1 ^ dx3 + f23(x) dx2 ^ dx3 . We wish to define a meaning for the integration of this 2-form α over a piece of the surface x = φ(t), ∫φ α = ∫φ Σ'I fI(x) dxi ^ dxi = integral of a 2-form over the surface φ in R3 . The transformation x = φ(t) is a mapping φ: (t1,t2) → R3. Officially it is the mapping φ which is the "surface", but we loosely refer to the image (trace) of this mapping in R3 as "the surface". The distinction is necessary because many mappings can have the same image surface, such as φ(t1,t2) and φ(t12,t22) where the parameters are respeeded (see Comment above). So imagine that we have a 2D surface hanging in R3 and as t1 and t2 vary (each perhaps from 0 to 1 in t-space), we move around on the image surface in R3. The problem is how to integrate a 2-form over this surface. We can define the calculational meaning of the above integral in two steps, each being a definition: First definition: ∫φ α ≡ ∫[0,1]2 φ*(α) = the integral in t-space of the pullback of α over the 2-cube [0,1]2 On the left we have an integral of the 2-form α over a surface φ in R3. On the right we have an integral of a different 2-form φ*(α) (the pullback of α) over a 2-cube [0,1]2. Note that α lies in xΛ2(R3) while φ*(α) lies in tΛ2(R2). Since our usual mapping is φ* : xΛ1(Rm) → tΛ1(Rn) we have m = 3 and n = 2. The "tall" m x n R-matrix for this problem is then a 3 x 2 matrix, Rij = (Dφ)ij = ∂φi/∂tj = ∂jφi i = 1,2,3 j = 1,2 R = . We then compute the pullback of α, α = Σ1≤i<i≤3 fii(x) xλ^I = Σ1≤i<i≤3 fii(x) dxi ^ dxi φ*(α) = Σ1≤i<i≤3 φ*(fii(x)) φ*(dxi ^ dxi) // (10.9.6) item 3 = Σ1≤i<i≤3 fii(φ(t)) Σ1≤j<j≤2 det dtj ^ dtj // (10.9.16) = Σ1≤i<i≤3 fii(φ(t)) det dt1 ^ dt2 // only one term in Σ1≤j<j≤2 = Σ1≤i<i≤3 fii(φ(t)) det dt1 ^ dt2 // more compact notation // these determinants are the 2x2 minors of the matrix R shown above = Σ1≤i<i≤3 fii(φ(t)) dt1 ^ dt2 // Jacobian notation for determinants = Σ1≤i<i≤3 [ fii(φ(t)) ] dt1 ^ dt2 = Σ1≤i<i≤3 gii(t) dt1 ^ dt2 where gii(t) = fii(φ(t)) . Since in this example there are so few terms (namely three) in the sum Σ1≤i<i≤3 , we just write them out φ*(α) = [ f12(φ(t)) + f13(φ(t)) + f23(φ(t)) ] dt1 ^ dt2 = G(t) dt1 ^ dt2 where G(t) ≡ [ f12(φ(t)) + f13(φ(t)) + f23(φ(t)) ] . The pullback φ*(α) = G(t) dt1 ^ dt2 is a 2-form in dual t-space tΛ2(R2). Using the definition given above, we then have ∫φ α ≡ ∫[0,1]2 φ*(α) = ∫[0,1]2 G(t) dt1 ^ dt2 . Thus the integral of the 2-form α over the surface φ in x-space is defined to be equal to the integral of the 2-form g(t) dt1 ^ dt2 over a 2-cube in t-space. So far no regular calculus integrals have appeared. Second definition: ∫[0,1]2 g(t) tλ1 ^tλ2 = ∫[0,1]2 G(t) dt1 ^ dt2 ≡ !Syntax Error, I!Syntax Error, I G(t1,t2) dt1dt2 . and now we have an ordinary calculus double integral which evaluates to a real number. Comment on orientation Orientation of the domain surface is a tricky business in a k-form integral but can at worst cause confusion about the sign of the result. To make things more "visible", suppose [a,b] = [c,d] = [1,0]. In t-space R2, if [a,b] = (b-a)1 = 1 and [c,d] = (d-c)2= 2, then the 2-cube (unit square) integration domain [0,1]2 can be regarded as being 1 ^ 2 . Recall from the discussion near (4.3.14) that one can regard A ^ B as connected with the signed area of a parallelogram (2-piped), so A ^ B is associated with the area of the front side and B ^ A with the area of the back side, akin to the cross product situation A x B, and these two area have the same magnitude but opposite sign. So we can regard, ∫1 x 2 1 = 1 area of the front side of a unit square is 1 area unit ∫2 x 1 1 = -1 area of the back side of a unit square is -1 are unit The integral appearing above is this ∫[0,1]2 G(t) dt1 ^ dt2 = ∫1 x 2 G(t) dt1 ^ dt2 = !Syntax Error, I!Syntax Error, I G(t1,t2) dt1dt2 . Now consider ∫1 x 2 G(t) dt1 ^ dt2 = ∫1 x 2 [- G(t) dt2 ^ dt1] = – ∫1 x 2 G(t) dt2 ^ dt1 ∫2 x 1 G(t) dt1 ^ dt2 = [- ∫1 x 2] G(t) dt2 ^ dt1 = – ∫1 x 2 G(t) dt2 ^ dt1 In both these equations a minus sign is generated. In the first the minus sign arises because the 2-form called G(t) dt2 ^ dt1 is the negative of the different 2-form called G(t) dt2 ^ dt1. In the second equation the integration domain 1 ^ 2 refers to the front side of the unit square, while 2 x 1 refers to the back side of the unit square, and these domains differ by a minus sign. If both changes are made at once one gets ∫1 x 2 G(t) dt1 ^ dt2 = + ∫2 x 1 G(t) dt2 ^ dt1 If the integration domain is written simply as [0,1]2, one ends up with ∫[0,1]2 G(t) dt1 ^ dt2 = + ∫[0,1]2 G(t) dt2 ^ dt1 and this seems to be a contradiction since everyone knows that dt2 ^ dt1 = - dt1 ^ dt2 . The issue here is that the two [0,1]2 domains are not the same, they just look the same. See Sjamaar's page 64 Remark 5 where he treats the domain 2 ^ 1 as a "respeeding" (reparametrization) of the domain 1 ^ 2 which reverses the orientation of that domain. Further processing: Using the Hodge correspondence suggested in (4.3.17) we define three new function names Fi f12 ≡ F3 f23 ≡ F1 f13 ≡ - F2 with a minus in the last line since f13 is in anticyclic order compared to (4.3.17). Then ** becomes ∫φ α = !Syntax Error, I!Syntax Error, I[ F3(φ(t)) – F2(φ(t)) + F1(φ(t)) ] dt1dt2 = !Syntax Error, I!Syntax Error, I[ F1(φ(t)) – F2(φ(t)) + F3(φ(t)) ] dt1dt2 = !Syntax Error, I!Syntax Error, I[ F1(φ(t)) + F2(φ(t)) + F3(φ(t)) ] dt1dt2 = !Syntax Error, I!Syntax Error, I F(φ(t)) n(t) dt1dt2 = !Syntax Error, I!Syntax Error, I F(φ(t)) [ n(t) dt1dt2] = !Syntax Error, I!Syntax Error, I F(φ(t)) dA where n below is a vector normal to the surface in R3 at point x = φ(t), n(t) ≡ ( , , ) dA ≡ n(t) dt1dt2 // Buck p 335, 403 That n really is a normal vector can be verified by showing that n xuj = 0 for any tangent base vector xuj in the tangent space TxM, where from (E.2) (xuj)i = Rij = ∂jφi , see Buck p 336. If we forward map a tiny rectangle of area dt1dt2 at t in t-space to the surface in x-space, it maps into a tiny oriented area dA at x on the surface which has normal n and area |dA| = |n| dt1dt2 . Going back to the original 2-form α we can write α = Σ1≤i<i≤3 fii(x) dxi ^ dxi = f12(x) dx1 ^ dx2 + f13(x) dx1 ^ dx3 + f23(x) dx2 ^ dx3 = F1(x) dx2 ^ dx3 + F2(x) dx3 ^ dx1 + F3(x) dx1 ^ dx2 = F1(x) dS1 + F2(x) dS2 + F3(x) dS3 = F(x) dS // dS1 = *dx1 etc, so can say dS = *dx where in cyclic order we define the following differential area 2-forms dS1 ≡ dx2 ^ dx3 dS2 ≡ dx3 ^ dx1 dS3 ≡ dx1 ^ dx2 Then our conclusion from the above can be concisely written, ∫φ α = ∫φ F(x) dS = !Syntax Error, I!Syntax Error, I F(φ(t)) dA where dA ≡ n(t) dt1dt2 . Normally this is written "∫φ F(x) dS " = !Syntax Error, I!Syntax Error, I F(φ(t)) n(t) dt1dt2 where the operational meaning of the object on the left is given by the calculus integral on the right. The object on the left is called "the surface integral of a vector field F(x) over a surface φ". Generalization to Rm If the 2D surface lies in Rm instead of R3, the above results are easily generalized. The 2-form α is then, α = Σ1≤i<i≤m fii(x) xλ^I = Σ1≤i<i≤m fii(x) dxi ^ dxi = Σ'I fI(x) dx^I where there are terms in the ordered sum Σ'I. The pullback also has terms, being φ*(α) = Σ1≤i<i≤m fii(φ(t)) dt1 ^ dt2 = Σ1≤i<i≤m gii(t) dt1 ^ dt2 gii(t) = fii(φ(t)) = G(t) dt1 ^ dt2 G(t) = Σ1≤i<i≤m fii(φ(t)) . The pullback φ*(α) = G(t) dt1 ^ dt2 is still a 2-form in dual t-space tΛ2(R2). Then ∫φ α ≡ ∫[0,1]2 φ*(α) = ∫[0,1]2 G(t) dt1 ^ dt2 ≡ !Syntax Error, I!Syntax Error, I G(t1,t2) dt1dt2 Here G(t) is a sum of terms each of which is a 2x2 Jacobian weighted by a function fii. (c) Integration of a k-form The k-form of interest is α = Σ'I fI(x) xλ^I = Σ'I fI(x) dx^I = Σ'I fii...i(x) dxi ^ dxi ^ ...^ dxi We wish to define a meaning for the integration of this k-form α over a piece of the surface x = φ(t), ∫φ α = ∫φ Σ'I fI(x) dx^I = integral of a k-form over the surface of dimension n in Rm . Since the surface is embedded in Rm we have n ≤ m. The k-form has k ≤ n, although in our previous examples we had k = n. For example, in Example (b) one could map a line segment in R2 to a curve in Rm which would lie on the surface in Rm of that example. This is illustrated here in red, where now αx = Σi fi(x) dxi is a 1-form. If we agree to place the red line segment in t-space on the t1 axis as shown, then this situation is really no different from Example (a) where t2 can be ignored. For this reason, we limit our interest to k = n ≤ m. The transformation x = φ(t) is a mapping φ: (t1,t2,..tk) → R3. Officially it is the mapping φ which is the "surface", but we loosely refer to the image (trace) of this mapping in Rm as "the surface". The distinction is necessary because many mappings can have the same image, such as φ(t1,t2, ...) and φ(t12,t22, ...) where the parameters are respeeded (see Comment above). So imagine that we have an n-dimensional surface hanging in Rm and as t1 ... tn vary (each perhaps from 0 to 1 in t-space), we move around on the image surface in Rm. The problem is how to integrate a k-form over this surface. We can define the calculational meaning of the above integral in two steps, each being a definition: First definition: k = n ∫φ α ≡ ∫[0,1]k φ*(α) = the integral in t-space of the pullback of α over the k-cube [0,1]k On the left of this equation we have an integral of the k-form α over a surface φ in Rm. On the right we have an integral of a different 2-form φ*(α) (the pullback of α) over a k-cube [0,1]k in Rn. Note that α lies in xΛk(Rm) while φ*(α) lies in tΛk(Rn) [ n = k ] Our usual mapping is φ* : xΛk(Rm) → tΛk(Rn) with m ≥ n. The "tall" m x n R-matrix for this problem is then a general m x n matrix, Rij = (Dφ)ij = ∂φi/∂tj = ∂jφi i = 1,2,3..m j = 1,2..n We then compute the pullback of α, α = Σ1≤i<i≤..<i≤m fii...i(x) dxi ^ dxi ^ ...^ dxi = Σ'I fI(x) dx^I φ*(α) = Σ'I φ*(fI(x)) φ*(dx^I) // (10.9.6) item 3 = Σ'I fI(φ(t)) ΣJ RIJ dt^J // (10.9.6) items 1 and 4 = Σ'I fI(φ(t)) Σ'J det(RIJ(t)) dt^J // (10.9.12) R = (Dφ(t)) = Σ'J [Σ'I fI(φ(t)) det(RIJ(t)) ] dt^J = Σ'J gJ(t) dt^J = Σ'J gJ(t) dtj ^ dtj ^ ...^ dtj where gJ(t) ≡ Σ'I fI(φ(t)) det(RIJ(t)) . The pullback φ*(α) = Σ'J gJ(t) dt^J is a k-form in dual t-space tΛk(Rn). Using the definition given above, we then have ∫φ α ≡ ∫[0,1]k φ*(α) = ∫[0,1]k Σ'J gJ(t) dt^J . Thus the integral of the k-form α over the surface φ in x-space is defined to be equal to the integral of the k-form Σ'J gJ(t) dt^J over a k-cube in t-space. So far no regular calculus integrals have appeared. Second definition: ∫[0,1]k Σ'J gJ(t) tλ^J = ∫[0,1]k Σ'J gJ(t) dtj ^ dtj ^ ...^ dtj ≡ (!Syntax Error, I!Syntax Error, I ..... !Syntax Error, I) Σ'J gJ(t) dtjdtj....dtj . On the left is the integral of a k-form over a k-cube, on the right is an ordinary calculus k-fold integral of a function over the unit cube in k dimensions. It is this second definition that motivates giving the dual space basis vectors tλi the cosmetic names dti. We have then shown that ∫φ α = ∫[0,1]k Σ'J gJ(t) dtj ^ dtj ^ ...^ dtj = (!Syntax Error, I!Syntax Error, I ..... !Syntax Error, I) Σ'J Σ'I fI(φ(t)) det(RIJ(t))dtjdtj....dtj = Σ'I Σ'J (!Syntax Error, I!Syntax Error, I ..... !Syntax Error, I) fI(φ(t)) det(RIJ(t)) dtjdtj....dtj R= (Dφ) ok to here. I am not happy with the integral of a form suddenly being a real number, but it may be related to how the dα operation turns a real thing into a form, bumping up the rank 1. Note that dt1 ^ dt2 is in standard increasing order. For the integral against dt2 ^ dt1 = - dt1 ^ dt2 one gets the area of the "back side" of the surface, regarded as the negative of the area of the "front side". See Sjamaar Ch 8 and Buck Ch 6 for discussions of orientation of surfaces in various dimensions.