Note 16. Stokes theorem Differential Geometry, 2005

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1 Note 16. Stokes theorem ifferentil Geometry, 2005 Stokes theorem is the centrl result in the theory of integrtion on mnifolds. It gives the reltion between exterior differentition (see Note 14) nd integrtion (see Note 15) of differentil forms, nd it generlizes the fundmentl theorem of clculus. Let be n m-dimensionl oriented smooth mnifold with boundry. The boundry (which, of course, my be empty) is equipped with the induced orienttion, see ilnor, top of p. 27. Stokes theorem. Let ω be compctly supported m 1-form on. Then ω. Here the integrnd ω on the right-hnd side is understood to be the restriction of ω to, nd the right-hnd side is interpreted s 0 if the boundry is empty. The result is quite remrkble, becuse it shows tht by knowing ω only on the boundry of, we cn predict property of it on the interior, nmely the totl integrl of its derivtive. Proof. We my ssume tht ω is supported inside σ(int() H m ), where U is generlized rectngle nd σ: U chrt. For if the result hs been estblished in this generlity, we cn pply it to ϕ α ω for ech element ϕ α in suitble prtition of unity (see Note 15), We my lso ssume tht ω = f dx 1 dx j 1 dx j+1 dx m for some j between 1 nd m, becuse in generl ω will be sum of such forms. Since df = m i=1 x i dx i (see Note 13, Lemm 2), it then follows tht j 1 ( 1) dx 1 dx m, where the sign ppers becuse dx j hs been moved pst the 1-forms from dx 1 up to dx j 1. Assume first tht j < m. We will prove tht then ω = 0. By definition ( 1) j 1 dx 1 dx m = ( 1) j 1 H m σ dv. Notice tht by definition σ = 1 (f σ) j,

2 2 where u 1,..., u m denote the coordintes in U (to be distinguished from the coordinte functions x i : σ(u) u i ). In the integrl over the rectngle H m, we cn freely interchnge the order of integrtions over u 1,..., u m. Let us tke the integrl over u j first (innermost), nd let us denote its limits by nd b (these re the limits for the u j vrible in the rectngle ). Now by the fundmentl theorem of clculus b (f σ) j du j = f(σ(u 1,..., u j 1, b, u j+1,..., u m )) f(σ(u 1,..., u j 1,, u j+1,..., u m )). However, since ω is supported in σ(int() H m ), it follows tht these vlues re zero for ll u 1,...u j 1, u j+1,...u m, nd hence 0 s climed. On the other hnd, since x m = 0 long the boundry σ(u H m ) = σ(u H m ), it follows immeditely tht dx m, nd hence lso ω, restricts to zero on. Therefore lso ω = 0. Next we ssume j = m. We will prove tht then ω = ( 1) H m f(σ(u 1,..., u m 1, 0)) du 1...du m 1. m Following the preceding computtion of dω, we tke the integrl over u j = u m first. This time, however, the lower limit is replced by the vlue 0 of x m on the boundry, nd the previous conclusion fils, tht f vnishes here. Insted we obtin the vlue f(σ(u 1,..., u m 1, 0)), with minus in front becuse it is the lower limit in the integrl. Recll tht there ws fctor ( 1) j 1 = ( 1) m 1 in front. Performing the integrl over the other vribles s well, we thus obtin the desired expression ( 1) m H m f(σ(u 1,..., u m 1, 0)) du 1...du m 1 for dω. On the other hnd, the integrl ω cn be computed by mens of the restricted chrt σ U H m on. However, we hve to keep trck of the orienttion of this chrt. Since the tngent vector dσ(e m ) points outwrds, the orienttion of the bsis dσ(e 1 ),..., dσ(e m 1 ) for T p is the sme s the orienttion of the bsis dσ(e m ), dσ(e 1 ),..., dσ(e m 1 ) for T p (see ilnor p. 27). The orienttion of the ltter is ( 1) m, becuse σ is positive chrt. It follows tht ω = ( 1) H m f(σ(u 1,..., u m 1, 0)) du 1...du m 1, m

3 3 s climed. The definition of ω by mens of prtition of unity is lmost impossible to pply, when it comes to explicit computtions, becuse it is very difficult to write down explicit prtitions of unity. The following proposition, of which we omit the proof, cn be used insted. The ide is to chop up the support of ω in finitely mny pieces nd tret ech of them seprtely. Proposition 1. Let be smooth oriented mnifold, nd let ω A m () be compctly supported. Suppose 1,..., n R m re domins of integrtion, nd g i : i smooth mps, i = 1,..., n, such tht ) supp ω g 1 ( 1 ) g n ( n ), b) g 1 int(1 ),..., g n int(n ) re disjoint chrts on. Then ω = n i=1 i gi ω. For exmple, the unit sphere S 2 is covered in this fshion by single mp g: S 2 of sphericl coordintes g(u, v) = (cos u cosv, cos u sinv, sinu) with = [ π/2, π/2] [ π, π], nd thus we cn compute the integrl of 2-form over S 2 by mens of its pull-bck by sphericl coordintes. Exmples. 1. Let = [, b], then m = 1 nd ω is 0-form, tht is, function f C ([, b]). The boundry consists of the two points nd b, oriented by 1 nd +1, respectively (see ilnor, p. 27). Hence nd b df = b ω = f(b) f() f (x) dx (the interprettion of the integrl of 0-form over 0-dimensionl mnifold is tht it is sum, weighted by the orienttions ±1). We now see tht in this cse Stokes theorem reduces to the fundmentl theorem of clculus. 2. Let = R 2 be n elementry domin, the boundry of which is simple closed smooth regulr curve. Then m = 2, nd ω is 1-form on. Write ω = f(x, y)dx + g(x, y)dy, then df dx + dg dy = ( y + g ) dx dy x nd hence y + g x da.

4 4 On the other hnd, the integrl ω over the boundry cn be computed s follows. Assume γ: [0, T] is the boundry curve, with endpoints γ(0) = γ(t) (nd no other self intersections). The proposition bove cn be pplied with 1 = [0, T] nd g 1 = γ. Write γ(t) = (x(t), y(t)), then by definition nd ω is the line integrl ω = g1 ω = (f γ) dx + (g γ) dy, γ f(x, y)dx + g(x, y)dy. We thus see tht in this cse Stokes theorem reduces to the clssicl Green s formul (which is equivlent with the divergence theorem for the plne): y + g x da = f(x, y)dx + g(x, y)dy. γ The orienttion of γ is determined s follows. If n nd t re the outwrd norml vector nd the positive unit tngent vector, respectively, in given point of γ, then (n,t) should be positively oriented, tht is, t = ˆn, which is exctly the stndrd counter clockwise orienttion of closed curve in R Let = R 3 be domin of integrtion, the boundry of which is compct smooth surfce S, nd let ω be 2-form on : Then nd hence ω = f(x, y, z) dy dz + g(x, y, z) dz dx + h(x, y, z) dx dy. ( x + g y + h ) dx dy dz z div(f, g, h) dv, where div(f, g, h) = x + g y + h z is the divergence of the vector field (f, g, h). On the other hnd, the integrl ω over the boundry = S cn be computed s follows. Let 1,..., n nd g i : i S be s in Proposition 1 for the mnifold S. Then S ω = n i=1 i gi ω. Let σ(u, v) = g i(u, v) be one of the functions g i. Then σ ω = (f σ) d(y σ) d(z σ)+(g σ) d(z σ) d(x σ)+(f σ) d(x σ) d(y σ). Furthermore d(y σ) d(z σ) = ( σ 2 du + σ 2 v dv) ( σ 3 du + σ 3 v dv) = ( σ 2 σ 3 v σ 2 v σ 3 ) du dv

5 5 nd similrly d(z σ) d(x σ) = ( σ 3 d(x σ) d(y σ) = ( σ 1 σ 1 v σ 1 σ 3 ) du dv v σ 2 v σ 2 σ 1 ) du dv. v Notice tht the three expressions in front of du dv re exctly the coordintes of the norml vector σ u σ v. Thus we see tht σ ω = (f σ, g σ, h σ) (σ u σ v ) du dv. Let N(u, v) denote the outwrd unit norml vector in σ(u, v), then σ u σ v = σ u σ v N nd we see tht σ ω is the surfce integrl of the function (f, g, h) N over the imge of σ = g i (recll tht the surfce integrl of function includes fctor σ u σ v = (EG F 2 ) 1/2 ). Summing over i = 1,..., n we finlly obtin the surfce integrl ω = (f, g, h) N da. S S We conclude tht in this cse Stokes theorem reduces to the divergence theorem (sometimes lso clled the Guss theorem): div(f, g, h) dv = (f, g, h) N da. 4. Let be n oriented compct smooth surfce in R 3 with boundry, nd let ω be 1-form on. Assume ω is the restriction of smooth 1-form f(x, y, z) dx + g(x, y, z) dy + h(x, y, z) dz defined on n open neighborhood of in R 3. A similr nlysis s in the previous exmple shows tht in this cse Stokes theorem reduces to the originl theorem of Stokes, which is (curlf N) da = f dx + g dy + h dz. where curl F = ( h y g z, z h x, g x y ). The preceding exmples show tht in ddition to the fundmentl theorem of clculus, Stokes theorem lso generlizes the three min theorems of clssicl vector clculus.