SUPPLEMENTARY READING FOR MATH 53: DIFFERENTIALFORMS AND THE GENERAL STOKES FORMULAEDWARD FRENKEL, UC BERKELEYThe purpose of these notes is to outline briefly a general formalism which allows fora uniform treatment of various integrals studied in Chapter 16 of Stewart’s book. Moredetails can be found in the book by I. Madsen and J. Tornehave, From Calculus toCohomology, Cambridge University Press, 1997.1. What do we integrate?Let Y be an m–dimensional domain in Rn(of course, m ≤ n). What can we integrateover Y ? Let for example m = 1, i.e., Y is a curve C. We have learned how to integratethe following expressions over curves in R3:(1.1) P dx + Qdy + Rdz,where P, Q, R are functions in (x, y, z). How do we perform this integration in practice?We first parametrize our curve: x = x(t), y = y(t), z = z(t), a ≤ t ≤ b. In other words,we identify our curve C with a flat one-dimensional domain: the interval [a, b]. Thenwe writeZC(P dx + Qdy + Rdz) =Z[a,b]P∂x∂tdt + Q∂y∂tdt + r∂z∂tdt(here we use the fact that dx =∂x∂tdt, etc.). The RHS is a familiar one-dimensionalintegral. This is what the book calls the line integral of the vector field F = hP, Q, Riover C, but as we see it is more natural to think that we really integrate not F, butthe expression (1.1).Likewise, the general expression that can be integrated over a surface S in R3(thecase m = 2) is(1.2) Adydz + Bdzdx + Cdxdy,where A, B, C are functions in (x, y, z). The integral of this expression over S can becomputed as follows. We parametrize S:x = x(u, v), y = y(u, v), z = z(u, v), (u, v) ∈ D ⊂ R2,Date: December 3, 2009.12 EDWARD FRENKEL, UC BERKELEYthat is identify S with a flat domain D in R2. After that we can write(1.3)Z ZS(Adydz + Bdzdx + Cdxdy) =Z ZDA∂(y, z)∂(u, v)dudv + B∂(z, x)∂(u, v)dudv + C∂(x, y)∂(u, v)dudv.Here we use the change of variables formulas:dxdy =∂(x, y)∂(u, v)dudv, etc .,where∂(x,y)∂(u,v)denotes the familiar Jacobian.Note that, according to the last formula, we havedxdy = −dydx,because∂(x, y)∂(u, v)= −∂(y, x)∂(u, v).The right hand side of formula (1.3) is what the book calls the flux of the vectorfield hA, B, Ci (look at formula (9) of Sect. 16.7 and expand ru× rv). But this formulashows that it is more natural to think that we are integrating the expression (1.2).Finally, the most general expression that we can integrate over a solid E in R3hasthe form(1.4) f(x, y, z)dxdydz,and the corresponding integral isZ Z ZEf(x, y, z)dxdydz,which is the standard triple integral. Note that we do not need to parametrize E: unlikeC and S, it is already flat (in other words, (x, y, z) themselves are the flat coordinates).After these examples one can guess that the general object that we should integrateover an m–dimensional domain Y in Rn(with coordinates x1, . . . , xn) is the sum ofterms of the form(1.5) f(x1, . . . , xn)dxi1dxi2. . . dxim,where f (x1, . . . , xn) is a function in x1, . . . , xn.We must have exactly m factors dxjin this expression, since we are going to integratem times. Moreover, all dxij’s appearing in a product dxi1dxi2. . . dximmust be distinct.Indeed, we don’t want to integrate dxdx, say, over a surface S in R3– the result wouldbe 0.So for example for m = 2 and n = 3 we have the following possibilities:dxdy, dydx, dydz, dzdy, dxdz, dzdx.But actually dxdy is the same as −dydx, as we have seen above. Hence there are only3 truly distinct elements: dxdy, dydz, dzdx. Therefore the sum of terms of the form3(1.5) becomes formula (1.2). Other examples are given by formulas (1.1) and (1.4) –they correspond to the cases when m = 1, n = 3 and m = 3, n = 3, respectively.How to integrate the expression (1.5) over an m-dimensional domain Y in Rn? Inthe same way as in the case n = 3 (see above). First, parametrize the domain, i.e.,identify it with a flat domain Z ⊂ Rmwith coordinates z1, . . . , zm. This means writingeach xias a function xi(z1, . . . , zm). After that we can writeZYfdxi1dxi2. . . dxim=ZZf∂(xi1, xi2, . . . , xim)∂(z1, z2, . . . , zm)dz1dz2. . . dzm,where we have included the Jacobian of the corresponding change of variables. TheRHS is a standard multiple integral, which can be evaluated as an iterated integral inthe same way as we do it for m = 2 and m = 3.Terminology: the expression (1.5) is called a differential form of degree m, or anm–form. Note that a 0–form is just the same as a function. Thus, (1.1) is a 1–form,(1.2) is a 2–form, etc. Conclusion: we are integrating m–forms.2. de Rham differentialNow we introduce an operation d that manufactures an (m + 1)–form, denoted bydω, out of any m–form ω. On an m–form (1.5) it acts as follows:d (f(x1, . . . , xn)dxi1dxi2. . . dxim) =nXj=1∂f∂xjdxjdxi1dxi2. . . dxim.A general m–form is a sum of terms of the form (1.5). The result of application of d toa general form is just the sum of the applications of d to each of these terms (in otherwords, d is a linear map).This operation is called the de Rham differential, after French mathematician Georgesde Rham.Examples. (1) Let ω be a 0–form, that is a function f . Then we find:df =nXj=1∂f∂xjdxj.This is nothing but the usual differential of the function f considered as a 1–form.(2) Let ω be a 1–form (1.1) in R3. Thendω =∂P∂xdxdx +∂P∂ydydx +∂P∂zdzdx+∂Q∂xdxdy +∂Q∂ydydy +∂Q∂zdzdy+∂R∂xdxdz +∂R∂ydydz +∂R∂zdzdz.4 EDWARD FRENKEL, UC BERKELEYBut remember that dxdx = dydy = dzdz = 0, and also that dydx = −dxdy, etc.Therefore we can rewrite this formula as follows:dω =∂R∂y−∂Q∂zdydz+∂P∂z−∂R∂xdzdx+∂Q∂x−∂P∂ydxdy.We see that the functions arising here are the components of curl of the vector fieldhP, Q, Ri. Thus we obtain a very nice interpretation of curl using the de Rham differ-ential!(3) Let ω be a 2–form given by (1.2). Then we find in the same way as in example(2):dω =∂A∂x+∂B∂y+∂C∂zdxdydz.The function in the RHS is nothing but div of the vector field hA, B, Ci.An important property of d is that it is nilpotent: d2= 0 (check this!). This impliesthat if ω = dν, then dω = 0. This fact is a generalization of the following facts that wehave seen in our course:curl(~∇f) = 0, div(curl F) = 0.Moreover, the converse is also true (it is known as the Poincare lemma):If ω is an m–form well-defined on the entire Rn, such that dω = 0, then there existsan (m − 1)–form ν, such that ω =