Integration on Manifolds

Transcription

1 7 Integration on anifolds Page 399 The integral of an n-form on an n-manifold is defined by piecing together integrals over sets in R n using a partition of unity subordinate to an atlas. The change of variables theorem guarantees that the integral is well defined, independent of the choice of atlas and partition of unity. Two basic theorems of integral calculus, the change of variables theorem and Stokes theorem, are discussed in detail along with some applications. 7.1 The Definition of the Integral The aim of this section is to define the integral of an n-form on an oriented n-manifold and prove a few of its basic properties. We begin with a summary of the basic results in R n. Integration on R n. Suppose f : R n R is continuous and has compact support. Then fdx 1...dx n is defined to be the Riemann integral over any rectangle containing the support of f Definition. Let U R n be open and ω Ω n (U) have compact support. If, relative to the standard basis of R n, ω(x) = 1 n! ω i 1...i n (x)dx i1 dx in = ω 1...n (x)dx 1 dx n, where the components of ω are given by ω i1...i n (x) =ω(x)(e i1,...,e in ), then we define ω = ω 1...n (x)dx 1 dx n. U R n

2 4 7. Integration on anifolds Recall that if ζ is any integrable function and f : R n R n is any diffeomorphism, the change of variables theorem states that ζ f is integrable and ζ(x 1,...,x n )dx 1 dx n R n = J Ω f(x 1,...,x n ) (ζ f)(x 1,...,x n )dx 1 dx n, (7.1.1) R n where Ω = dx 1 dx n is the standard volume form on R n and J Ω f is the Jacobian determinant of f relative to Ω. This change of variables theorem can be rephrased in terms of pull backs in the following form Theorem (Change of Variables in R n ). Let U and V be open subsets of R n and suppose f : U V is an orientation-preserving diffeomorphism. If ω Ω n (V ) has compact support, then f ω Ω n (U) has compact support as well and f ω = ω (7.1.2) U Proof. If ω = ω 1...n dx 1 dx n, then f ω =(ω 1...n f)(j Ω f)ω, where the n-form Ω = dx 1 dx n is the standard volume form on R n. As discussed in 6.5, J Ω f>. Since f is a diffeomorphism, the support of f ω is f 1 (supp ω), which is compact. Then from equation (7.1.1), f ω = (ω 1...n f)(j Ω f)dx 1 dx n U R n = ω 1...n dx 1 dx n = ω. R n V V Integration on a anifold. Suppose that (U, ϕ) is a chart on a manifold and ω Ω n () has compact support. If supp(ω) U, we may form ω U, which has the same support. Then ϕ (ω U) has compact support, so we may state the following Definition. Let be an orientable n-manifold with orientation [Ω]. Suppose ω Ω n () has compact support C U, where (U, ϕ) is a positively oriented chart. Then we define ω = (ϕ) ϕ (ω U) Proposition. Suppose ω Ω n () has compact support C U V, where (U, ϕ), and (V,ψ) are two positively oriented charts on the oriented manifold. Then ω = ω. (ϕ) (ψ) Proof. By Theorem 7.1.2, ϕ (ω U) = (ψ ϕ 1 ) ϕ (ω U). Hence ϕ (ω U) = ψ (ω U). (Recall that for diffeomorphisms, we have f =(f 1 ) and (f g) = f g.)

3 7.1 The Definition of the Integral 41 Thus, we may define U ω = ω, where (U, ϕ) is any positively oriented chart containing the compact (ϕ) support of ω. ore generally, we can define ω where ω has compact support not necessarily lying in a single chart as follows Definition. Let be an oriented manifold and A an atlas of positively oriented charts. Let P = {(U α,ϕ α,g α )} be a partition of unity subordinate to A. Define ω α = g α ω (so ω α has compact support in some U i ) and let ω = ω α. (7.1.3) P α Proposition. (i) The sum (7.1.3) contains only a finite number of nonzero terms. (ii) For any other atlas of positively oriented charts and subordinate partition of unity Q we have ω = ω. The common value is denoted ω, and is called the integral of ω Ωn (). P Proof. For any m, there is a neighborhood U such that only a finite number of g α are nonzero on U. By compactness of supp ω, a finite number of such neighborhoods cover the support of ω. Hence only a finite number of g α are nonzero on the union of these U. For (ii), let P = {(U α,ϕ α,g α )} and Q = {(V β,ψ β,h β )} be two partitions of unity with positively oriented charts. Then the functions {g α h β } satisfy g α h β (m) = except for a finite number of indices (α, β), and Σ α Σ β g α h β (m) = 1, for all m. Since Σ β h β = 1, we get ω = g α ω = h β g α ω = g α h β ω = ω. P α β α α β Q Global Change of Variables. This result can now be formulated very elegantly as follows Theorem (Change of Variables Theorem). Suppose and N are oriented n-manifolds and f : N is an orientation-preserving diffeomorphism. If ω Ω n (N) has compact support, then f ω has compact support and ω = f ω. (7.1.4) N Proof. First, note that supp(f ω)=f 1 (supp(ω)), which is compact. To prove equation (7.1.4), let {(U i,ϕ i )} be an atlas of positively oriented charts of and let P = {g i } be a subordinate partition of unity. Then {(f(u i ),ϕ i f 1 )} is an atlas of positively oriented charts of N and Q = {g i f 1 } is a partition of unity subordinate to the covering {f(u i )}. By Proposition 7.1.6, f ω = g i f ω = ϕ i (g i f ω) i i R n = ϕ i (f 1 ) (g i f 1 )ω i R n = (ϕ i f 1 ) (g i f 1 )ω i R n = ω. N Q

4 42 7. Integration on anifolds This result is summarized by the following commutative diagram: Ω n () f f N R Ω n (N) Definition. Let (,µ) be a volume manifold. Suppose f F() and f has compact support. Then we call fµ the integral of f with respect to µ. The reader can check that since the Riemann integral is R-linear, so is the integral just defined. easures (Optional). The next theorem will show that the integral defined by equation (7.1.4) can be obtained in a unique way from a measure on. (The reader unfamiliar with measure theory can find the necessary background in Royden [1968]; this result will not be essential for future sections.) The integral we have described can clearly be extended to all continuous functions with compact support. Then we have the following Theorem (Riesz Representation Theorem). Let (,µ) be a volume manifold. Let β denote the collection of Borel sets of, the σ-algebra generated by the open (or closed, or compact) subsets of. Then there is a unique measure m µ on β such that for every continuous function of compact support fdm µ = fµ. (7.1.5) Proof. Existence of such a measure m µ is proved in books on measure theory, for example Royden [1968]. For uniqueness, it is enough to consider bounded open sets (by the Hahn extension theorem). Thus, let U be open in, and let C U be its characteristic function. We construct a sequence of C functions of compact support ϕ n such that ϕ n C U, pointwise. Hence from the monotone convergence theorem, ϕ n µ = ϕ n dm µ C U dm µ = m µ (U). Thus, m µ is unique. The space L p (,µ), p R, consists of all measurable functions f such that f p is integrable. For p 1, the norm ( ) 1/p f p = f p dm µ makes L p (,µ) into a Banach space (functions that differ only on a set of measure zero are identified). The use of these spaces in studying objects on itself is discussed in 7.4. The next propositions give an indication of some of the ideas. If F : N is a measurable mapping and m is a measure on, then F m is the measure on N defined by F m (A) = m (F 1 (A)). If F is bijective, we set F (m N )=(F 1 ) m N.Iff : R is an integrable function, then fm is the measure on defined by (fm )(A) = fdm for every measurable set A in. A

5 7.1 The Definition of the Integral Proposition. Suppose and N are orientable n-manifolds with volume forms µ and µ N and corresponding measures m and m N.LetF be an orientation preserving C 1 diffeomorphism of onto N. Then F m N = ( J (µ,µ N )F ) m. (7.1.6) Proof. Let f be any C function with compact support on. By Theorem 7.1.7, fdm N = fµ N = F (fµ N )= (f F ) ( J (µ,µ N )F ) µ N N = (f F ) ( J (µ,µ N )F ) dm. As in the proof of Theorem 7.1.9, this relation holds for f chosen to be the characteristic function of F (A). That is, ( m N (F (A)) = J(µ,µ N )F ) dm. A Jacobians and Divergence. In preparation for the next result, we notice that on a volume manifold (,µ), the flow F t of any vector field X is orientation preserving for each t R (regard this as a statement on the domain of the flow, if the vector field is not complete). Indeed, since F t is a diffeomorphism, J µ (F t ) is nowhere zero; since it is continuous in t and equals one at t =, it is positive for all t Proposition. Let be an orientable manifold with volume form µ and corresponding measure m µ.letx be a (possibly time-dependent) C 1 vector field on with flow F t. The following are equivalent (if the flow of X is not complete, the statements involving it are understood to hold on its domain): (i) div µ X =; (ii) J µ F t =1for all t R; (iii) F t m µ = m µ for all t R; (iv) Ft µ = µ for all t R; (v) fdm µ = (f F t)dm µ for all f L 1 (,µ) and all t R. Proof. Statement (i) is equivalent to (ii) by Corollary Statement (ii) is equivalent to (iii) by equation (7.1.6) and to (iv) by definition. We shall prove that (ii) is equivalent to (v). If J µ F t = 1 for all t R and f is continuous with compact support, then (f F t )µ = (f F t )(Ft µ) = Ft (fµ)= fµ. Hence, by uniqueness in Theorem 7.1.9, we have fdm µ = (f F t)dm µ for all integrable f, and so (ii) implies (v). Conversely, if (f F t )dm µ = fdm µ then taking f to be continuous with compact support, we see that (f F t ) µ = fµ = Ft (fµ)= (f F t )Ft µ = Thus, for every integrable f, Hence J µ F t = 1, and so (v) implies (ii). (f F t )dm µ = (f F t )(J µ F t )dm µ. (f F t )(J µ F t )µ.

6 44 7. Integration on anifolds Transport Theorem. The following result is central to continuum mechanics (see Example below and 8.2 for applications) Theorem (Transport Theorem). Let (,µ) be a volume manifold and X a vector field on with flow F t.forf F( R) and letting f t (m) =f(m, t), we have ( ) d f f t µ = dt F t(u) F t(u) t + div µ(f t X) µ (7.1.7) for any open set U in. Proof. By the flow characterization of Lie derivatives and Proposition , we have ( ) d f dt F t (f t µ)=ft t µ + Ft X (f t µ) ( ) f = Ft t µ + Ft [( X f t )µ + f t (div µ X)µ] [( ) ] f = Ft t + div µ(f t X) µ. Thus, by the change of variables formula, d f t µ = d [( ) ] f Ft (f t µ)= Ft dt F t(u) dt U U t + div µ(f t X) µ ( ) f = t + div µ(f t X) µ. F t(u) Example. Let ρ(x, t) be the density of an ideal fluid moving in a compact region of R 3 with smooth boundary. One of the basic assumptions of fluid dynamics is conservation of mass: the mass of the fluid in the open set U remains unchanged during the motion described by a flow F t. This means that d ρ(x, t) d 3 x = (7.1.8) dt F t(u) for all open sets U. By the transport theorem, equation (7.1.8) is equivalent to the equation of continuity ρ + div(ρu) =, (7.1.9) t here u represents the velocity of the fluid particles. We shall return to this example in 8.2. Recurrence. As another application of Proposition , we prove the following Theorem (Poincaré Recurrence Theorem). Let (,µ) be a volume manifold, m µ the corresponding measure, and X a time-independent divergence-free vector field with flow F t. Suppose A is a measurable set in such that m µ (A) <, F t (x) exists for all t R if x A, and F t (A) A. Then for each measurable subset B of A and T, there exists S T such that B F S (B). Therefore, a trajectory starting in B returns infinitely often to B. Proof. By Proposition , the sets B,F T (B),F 2T (B),... all have the same finite measure. Since m µ (A) <, they cannot all be disjoint, so there exist integers k>l> satisfying F kt (B) F lt (B). Since F kt =(F T ) k (as X is time-independent), we get F (k l)t (B) B.

7 7.1 The Definition of the Integral 45 The Poincaré recurrence theorem is one of the forerunners of ergodic theory, a topic that will be discussed briefly in 7.4. A related result is the following Theorem (Schwarzschild Capture Theorem). Let (,µ) be a volume manifold, X a time-independent divergence-free vector field with flow F t, and A a measurable subset of with finite measure. Assume that for every x A, the trajectory t F t (x) exists for all t R. Then for almost all x A (relative to m µ ) the following are equivalent: (i) F t (x) A for all t ; (ii) F t (x) A for all t. Proof. Let A 1 = t F t(a) be the set of points in A which have their future trajectory completely in A. Similarly, consider A 2 = t F t(a). By Proposition , for any τ, which shows, by letting τ, that µ(a 1 )=µ(f τ (A 1 )) = µ F τ (A) t τ ( ) µ(a 1 )=µ F t (A) = µ(a 1 A 2 ). Reasoning similarly for A 2, we get µ(a 1 )=µ(a 1 A 2 )=µ(a 2 ), so that Let t R µ(a 1 \(A 1 A 2 )) = µ(a 2 \(A 1 A 2 )) =. S =(A 1 (A 1 A 2 )) (A 2 (A 1 A 2 )); then m µ (S) =ands A. oreover, we have A 1 S = A 1 A 2 = A 2 S, which proves the desired equivalence. So far only integration on orientable manifolds has been discussed. A similar procedure can be carried out to define the integral of a one-density (see 6.5) on any manifold, orientable or not. The only changes needed in the foregoing definitions and propositions are to replace the Jacobians with their absolute values and to use the definition of divergence with respect to a given density as discussed in 6.5. All definitions and propositions go through with these modifications. Vector Valued Forms. If F is a finite-dimensional vector space, F -valued one-forms and one-densities can also be integrated in the following way. If ω = l i=1 ωi f i, where f 1,...,f n is an ordered basis of F, then we set l ( ) ω = ω i f i F. i=1 It is easy to see that this definition is independent of the chosen basis of F and that all the basic properties of the integral remain unchanged. On the other hand, the integral of vector-bundle-valued n-forms on is not defined unless additional special structures (such as triviality of the bundle) are used. In particular, integration of vector or general tensor fields is not defined.

8 46 7. Integration on anifolds Exercises Let be an n-manifold and µ a volume form on. IfX is a vector field on with flow F t show that d dt (J µ(f t )) = J µ (F t )(div µ X F t ). Hint: Compute (d/dt)ft µ using the Lie derivative formula Prove the following generalization of the transport theorem ( ) d ωt ω t = dt t + Xω t, F t(v ) F t(v ) where ω t is a time dependent k-form on and V is a k-dimensional submanifold of (i) Let ϕ : S 1 S 1 be the map defined by ϕ(e iθ )=e 2iθ, where θ [, 2π]. Let, by abuse of notion, dθ denote the standard volume of S 1. Show that the following identity holds: ϕ (dθ) =2 dθ. S 1 S 1 (ii) Let ϕ : N be a smooth surjective map. Then ϕ called a k-fold covering map if every n N has an open neighborhood V such that ϕ 1 (V )=U 1 U k, are disjoint open sets each of which is diffeomorphic by ϕ to V. Generalize (i) in the following way. If ω Ω n (N) is a volume form, show that ϕ ω = k ω. N Define the integration of Banach space valued n-forms on an n-manifold. Show that if the Banach space is R l, you recover the coordinate definition given at the end of this section. If E, F are Banach spaces and A L(E, F), define A L(Ω(,E), Ω(,F)) by (A α)(m) =A(α(m)). Show that ( ) ( ) A = A on Ω n (,E) Let and N be oriented manifolds and endow N with the product orientation. Let p : N and p N : N N be the projections. If α Ω dim () and β Ω dim N (N) have compact support show that α β := (p α) (p N β) has compact support and is a (dim + dim N)-form on N. Prove Fubini s Theorem ( )( ) α β = α β. N (Fiber Integral). Let ϕ : N be a surjective submersion, where dim = m and dim N = n. The map ϕ is said to be orientable if there exists η Ω p (), where p = m n, such that for each y N, j y η is a volume form on ϕ 1 (y), where j y : ϕ 1 (y) is the inclusion. An orientation of ϕ is an equivalence class of p-forms under the relation: η 1 η 2 iff there exists f F(), f> such that η 2 = fη 1. (i) If ϕ : N is a vector bundle, show that orientability of ϕ is equivalent to orientability of the vector bundle as defined in Exercise N

9 7.1 The Definition of the Integral 47 (ii) If ϕ is oriented by η and N by ω, show that ϕ ω η is a volume form on. The orientation on defined by this volume is called the local product orientation of (compare with Exercise (vi)). (iii) Let Ω k ϕ() :={ α Ω k () ϕ 1 (K) supp(α) is compact, for any compact set K N }, the fiber-compactly supported k-forms on. Show that Ω k ϕ() isanf()-submodule of Ω k (), and is invariant under the interior product, exterior differential, and Lie derivative. (iv) If α Ω k+p ϕ (), k and y N, define a p-form α y on ϕ 1 (y), with values in T y N T y N (k times) by [α y (x)(u 1,...,u p )](T x ϕ(v 1 ),...,T x ϕ(v k )) = α(x)(v 1,...,v k,u 1,...,u p ), where ϕ(x) =y, x, v 1,...,v k T x, and u 1,...,u p ker(t x ϕ)=t x (ϕ 1 (y)). Assume ϕ is oriented. Define the fiber integral :Ω k+p ϕ () Ω k (N) by ( α fib fib ) (x) = ϕ 1 (y) α y, if ϕ(x) =y; the right-hand side is understood as the integral of a vector-valued p-form on the oriented p-manifold ϕ 1 (y). (a) Prove that α is a smooth k-form on N. fib Hint: Use charts in which ϕ is a projection and apply the theorem of smoothness of the integral with respect to parameters. (b) Show that if ϕ : N is a locally trivial fiber bundle with N paracompact, is surjective. fib (v) Let β Ω l (N) have compact support and let α Ω k+p ϕ (). Show that ϕ β α Ω k+l+p ϕ () and that (ϕ β α) =β α. fib Hint: Let E = Ty N Ty N (k times) and let F be the wedge product l + k times. Define A L(E,F) bya(γ) =β(y) γ and show that (ϕ β α) y = A (α y ) using the notation of Exercise Then apply to this identity and use Exercise fib (vi) Assume N is paracompact and oriented, ϕ is oriented, and endow with the local product orientation. Prove the following iterated integration (Fubini-type) formula = by following the three steps below. N fib fib

10 48 7. Integration on anifolds Step 1: Using a partition of unity, reduce to the case = N P where ϕ : N is the projection and,n,p are Euclidean spaces. Step 2: Use (v) and Exercise to show that for β Ω n (N) and γ Ω p (P ) with compact support, (β γ) = (β γ) N fib Step 3: Since,N, and P are ranges of coordinate patches, show that any ω Ω m () with compact support is of the form β γ. (vii) Let ϕ : N and ϕ : N be oriented surjective submersions and let f :, and f : N N be smooth maps satisfying f ϕ = ϕ f. Show that fib f = f fib, where fib denotes the fiber integral of ϕ. (viii) Let ϕ : N be an oriented surjective submersion and assume X X() and Y X(N) are ϕ-related. Prove that fib i X = i Y, fib fib d = d, fib fib X = Y. fib (For more information on the fiber integral see Bourbaki [1971] and Greub et al..) Let ϕ : N be a smooth orientation preserving map, where and N are volume manifolds of dimension m and n respectively. For α Ω k () with compact support, define the linear functional ϕ α :Ω m k R by (ϕ α)(β) = ϕ β α for all β Ω m k (N); that is, ϕ α is a distributional k-form on N. Ifm<k,set ϕ α =. If there is a γ Ω n m+k (N) satisfying (ϕ α)(β) = β γ, identify ϕ α with γ and say ϕ α is of form-type. Prove the following statements. (i) If ϕ is a diffeomorphism, then ϕ α is the usual push-forward. (ii) If α is a volume form, this definition corresponds to that for the push-forward of measures. (iii) If ϕ is an oriented surjective submersion, show that ϕ α = α, as defined in Exercise 7.1-6(iv). fib Hint: Prove the identity ( ) ϕ β α = β α N fib using Exercise 7.1-6(v) and (vi) Let (,µ) be a paracompact n-dimensional volume manifold. (i) If (N,ν) is another paracompact n-dimensional volume manifold and f : N is an orientation reversing diffeomorphism, show that N ω = f ω for any ω Ω n (N) with compact support. Hint: Use the proof of Theorem

11 7.1 The Definition of the Integral 49 (ii) If η Ω n () has compact support and denotes the manifold endowed with the orientation [ µ], show that η = η and Hint: IfA = {(U i,ϕ i )} is an oriented atlas for (,[µ]), then is an oriented atlas for (,[ µ]). A = {(U i,ϕ i ψ i )}, ψ i (x 1,...,x n )=( x 1,x 2,...,x n ) Let ω n be the standard volume form on S n. Show that using the following steps. (i) Let R n+1 be the annulus Sn ω n = 2m+1 π m, if n =2m, m 1 (2m 1)!! Sn ω n = 2πm+1, if n =2m +1,m m! { x R n+1 <a< x <b< } and let f :]a, b[ S n A be the diffeomorphism f(t, s) =ts. Use Exercise (ii) to show that for x R n+1, f ( e x 2 Ω n+1 ) = t n e t2 (dt ω n ) where Ω n+1 = e 1 e n+1 for {e 1,...,e n+1 } the standard basis of R n+1, and where dt ω n denotes the product volume form on ]a, b[ S n. (ii) Deduce the equality e x 2 R n+1 Ω n+1 = b a t n e t2 dt ω n. S n (iii) Let a and b to deduce the equality ( + n+1 t n d t2 dt ω n = e du) u2. S n Prove that e u2 du = π, t 2m e t2 dt = (2m 1)!! π 2 m+1, and t 2m+1 e t2 = m! 2, to deduce the required formula for ω S n n.

12 41 7. Integration on anifolds 7.2 Stokes Theorem Stokes theorem states that if α is an (n 1)-form on an orientable n-manifold, then the integral of dα over equals the integral of α over, the boundary of. As we shall see in the next section, the classical theorems of Gauss, Green, and Stokes are special cases of this result. Before stating Stokes theorem formally, we need to discuss manifolds with boundary and their orientations Definition. Let E be a Banach space and λ E.Let E λ = { x E λ(x) }, called a half-space of E, and let U E λ be an open set (in the topology induced on E λ from E). Call Int U = U {x E λ(x) > } the interior of U and U = U ker λ the boundary of U. IfE = R n and λ is the projection on the jth factor, then E λ is denoted by R n j and is called positive jth half-space. Rn n is also denoted by R n +. We have U =IntU U, IntU is open in U, U is closed in U (not in E), and U Int U =. The situation is shown in Figure Note that U is not the topological boundary of U in E, but it is the topological boundary of U intersected with that of E λ. This inconsistent use of the notation U is temporary. x n int U R n + U Figure Open sets in a half-space A manifold with boundary will be obtained by piecing together sets of the type shown in the figure. To carry this out, we need a notion of local smoothness to be used for overlap maps of charts Definition. Let E and F be Banach spaces, λ E, µ F, U be an open set in E λ, and V be an open set in F µ. A map f : U V is called smooth if for each point x U there are open neighborhoods U 1 of x in E and V 1 of f(x) in F and a smooth map f 1 : U 1 V 1 such that f U U 1 = f 1 U U 1. We define Df(x) =Df 1 (x). The map f is a diffeomorphism if there is a smooth map g : V U which is an inverse of f. (In this case Df(x) is an isomorphism of E with F.) We must prove that this definition of Df is independent of the choice of f 1, that is, we have to show that if ϕ : W E is a smooth map with W open in E such that ϕ (W E λ ) =, then Dϕ(x) = for all x W E λ.ifx Int(W E λ ), this fact is obvious. If x (W E λ ), choose a sequence x n Int(W E λ ) such that x n x; but then = Dϕ(x n ) Dϕ(x) and hence Dϕ(x) =, which proves our claim Lemma. Let U E λ be open, ϕ : U F µ be a smooth map, and assume that for some x Int U, ϕ(x ) F µ. Then Dϕ(x )(E) F µ =kerµ. Proof. The quotient F/ ker µ is isomorphic to R, so that fixing f with µ(f) >, the element [f] F/ ker µ forms a basis. Therefore [f] determines the isomorphism T f : F/ ker µ R given by T f ([y]) = t, where t R

13 7.2 Stokes Theorem 411 is the unique number for which t[f] =[y]. This isomorphism in turn defines the isomorphism S f :kerµ R F given by S f (y, t) =y + tf which induces diffeomorphisms (in the sense of Definition 7.2.2) of ker µ [, [ with F µ and of ker µ ], ] with { y F µ(y) }. Denote by p : F R the linear map given by S 1 f followed by the projection ker µ R R, so that y F µ (respectively, ker µ, { y F µ(y) }) if and only if p(y) (respectively, =, ). Notice that the relation ϕ(x + tx) =ϕ(x )+Dϕ(x ) tx + o(tx), where lim t o(tx)/t =, together with the hypothesis (p ϕ)(x) for all x U, implies that (p ϕ)(x + tx)=+(p Dϕ)(x ) tx + p(o(tx)), whence for t> ( ) o(tx) (p Dϕ)(x ) x + p. t Letting t, we get (p Dϕ)(x ) x for all x E. Similarly, for t< and letting t, we get (p Dϕ)(x ) x for all x E. The conclusion is (Dϕ)(x )(E) ker µ. Intuitively, this says that if ϕ preserves the condition λ(x) and maps an interior point to the boundary, then the derivative must be zero in the normal direction. The reader may also wish to prove Lemma from the implicit mapping theorem. Now we carry this idea one step further Lemma. Let U be open in E λ, V be open in F µ, and f : U V be a diffeomorphism. Then f restricts to diffeomorphisms Int f :IntU Int V and f : U V. Proof. Assume first that U =, that is, that U ker λ =. We shall show that V = and hence we take Int f = f. If V, there exists x U such that f(x) V and hence by definition of smoothness there are open neighborhoods U 1 U and V 1 F, such that x U 1 and f(x) V 1, and smooth maps f 1 : U 1 V 1, g 1 : V 1 U 1 such that f U 1 = f 1, g 1 V V 1 = f 1 V V 1. Let x n U 1, x n x, y n V 1 \ V, and y n = f(x n ). We have and similarly Df(x) Dg 1 (f(x)) = lim (Df(g 1(y n )) Dg 1 (y n )) y n f(x) = lim D(f g 1)(y n )=Id F y n f(x) Dg 1 (f(x)) Df(x) =Id E so that Df(x) 1 exists and equals Dg 1 (f(x)). But by Lemma 7.2.3, Df(x)(E) ker µ, which is impossible, Df(x) being an isomorphism. Assume that U. If we assume V =, then, working with f 1 instead of f, the above argument leads to a contradiction. Hence V. Let x Int U so that x has a neighborhood U 1 U, U 1 U =, and hence U 1 =. Thus, by the preceding argument, f(u 1 )=, and f(u 1 )isopeninv\ V. This shows that f(int U) Int V. Similarly, working with f 1, we conclude that f(int U) Int V and hence f :IntU Int V is a diffeomorphism. But then f( U) = V and f U : U V is a diffeomorphism as well.

14 Integration on anifolds Now we are ready to define a manifold with boundary Definition. A manifold with boundary is a set together with an atlas of charts with boundary on ; charts with boundary are pairs (U, ϕ) where U and ϕ(u) E λ for some λ E and an atlas on is a family of charts with boundary satisfying A1 and A2 of Definition 3.1.1, with smoothness of overlap maps ϕ ji understood in the sense of Definition See Figure If E = R n, is called an n-manifold with boundary. Define Int = U ϕ 1 (Int(ϕ(U))) and = U ϕ 1 ( (ϕ(u))) called, respectively, the interior and boundary of. The definition of Int and makes sense in view of Lemma Note that 1. Int is a manifold (with atlas obtained from (U, ϕ) by replacing ϕ(u) E λ by the set Int ϕ(u) E); 2. is a manifold (with atlas obtained from (U, ϕ) by replacing ϕ(u) E λ by ϕ(u) E =kerλ); 3. is the topological boundary of Int in (although Int is not the topological interior of ). Summarizing, we have proved the following Proposition. If is a manifold with boundary, then its interior Int and its boundary are smooth manifolds without boundary. oreover, if f : N is a diffeomorphism, N being another manifold with boundary, then f induces, by restriction, two diffeomorphisms Int f :Int Int N and f : N. If n = dim, then dim(int ) =n and dim( ) =n 1. x n ϕ i ϕ j x n ϕji R n_ 1 R n_ 1 Figure Boundary charts To integrate a differential n-form over an n-manifold, must be oriented. If Int is oriented, we want to choose an orientation on compatible with it. In the classical Stokes theorem for surfaces, it is crucial that the boundary curve be oriented, as in Figure The tangent bundle to a manifold with boundary is defined in the same way as for manifolds without boundary. Recall that any tangent vector in T x has the form [dc(t)/dt] t=τ, where c :[a, b] is a C 1 curve, a<b, and τ [a, b]. If x, we consider curves c :[a, b] such that c(b) =x. If ϕ : U U E λ is a chart at m, then [d(ϕ c)(t)/dt] t=b in general points out of U, as in Figure

15 7.2 Stokes Theorem 413 Figure Orientation for surfaces Therefore, T x is isomorphic to the model space E of even if x (see Figure 7.2.5). It is because of this result that tangent vectors are derivatives of C 1 -curves defined on closed intervals. Had we defined tangent vectors as derivatives of C 1 -curves defined on open intervals, T y for y would be isomorphic to ker λ and not to E. In Figure 7.2.4, E = R n, λ is the projection onto the n-th factor, and c(t) is defined on a closed interval whereas the C 1 -curve d(t) is defined on an open interval. x y T y T x Figure Tangent spaces at the boundary. Having defined the tangent bundle, all of our previous constructions including tensor fields and exterior forms as well as operations on them such as the Lie derivative, interior product, and exterior derivative carry over directly to manifolds with boundary. One word of caution though: the fundamental relation between Lie derivatives and flows still holds if one is careful to take into account that a vector field on has integral curves which could run into the boundary in finite time and with finite velocity. (If the vector field is tangent to, this will not happen.) ϕ x n ϕ(u) n x =(,...,, 1) R n 1 Figure Oriented boundary charts

16 Integration on anifolds Next, we turn to the problem of orientation. As for manifolds without boundary a volume form on an n-manifold with boundary is a nowhere vanishing n-form on. Fix an orientation on R n. Then a chart (U, ϕ) is called positively oriented if T u ϕ : T u R n is orientation preserving for all u U. If is paracompact this latter condition is equivalent to orientability of (the proof is as in Proposition 6.5.2). Therefore, for paracompact manifolds, an orientation on is just a smooth choice of orientations of all the tangent spaces, smooth meaning that for all the charts of a certain atlas, called the oriented charts, the maps D(ϕ j ϕ 1 i )(x) :R n R n are orientation preserving. The reader may wonder why for finite dimensional manifolds we did not choose a standard half-space, like x n to define the charts at the boundary. Had we done that, the very definition of an oriented chart would be in jeopardy. For example, consider =[, 1] and agree that all charts must have range in R + = { x R x }. Then an example of an orientation reversing chart at x =1isϕ(x) =1 x; in fact, every chart at x = 1 would be orientation reversing. However, if we admit any half-space of R, so charts can be also in R = {x R x }, then a positively oriented chart at 1 is ϕ(x) =x 1. See Figure x x R + 1 x 1 x x x 1 R_ Figure Boundary charts for [, 1] Once oriented charts and atlases are defined, the theory of integration for oriented paracompact manifolds with boundary proceeds as in 7.1. Finally we define the boundary orientation of. Ateveryx, the linear space T x ( ) has codimension one in T x so that there are (in a chart on intersecting ) exactly two kinds of vectors not in ker λ: those for which their representatives v satisfy λ(v) > orλ(v) <, that is, the inward and outward pointing vectors. By Lemma 7.2.4, a change of chart does not affect the property of a vector being outward or inward (see Figure 7.2.4). If dim = n, these considerations enable us to define the induced orientation of in the following way Definition. Let be an oriented n-manifold with boundary, x and ϕ : U R n λ a positively oriented chart, where λ (R n ). A basis {v 1,...,v n 1 } of T x ( ) is called positively oriented if { (Tx ϕ) 1 } (n),v 1,...,v n 1 is positively oriented in the orientation of, where n is any outward pointing vector to R n λ at ϕ(x). For example, we could choose for n the outward pointing vector to R n λ and perpendicular to ker λ. If λ : R n R is the projection on the n-th factor, then (T x ϕ) 1 (n) = / x n and the situation is illustrated in Figure Theorem (Stokes Theorem). Let be an oriented smooth paracompact n-manifold with boundary and α Ω n 1 () have compact support. Let i : be the inclusion map so that i α Ω n 1 ( ). Then i α = dα (7.2.1a) or for short, α = dα (7.2.1b)

17 If =, the left hand side of equation (7.2.1a) or (7.2.1b) is set equal to zero. 7.2 Stokes Theorem 415 Proof. Since integration was constructed with partitions of unity subordinate to an atlas and both sides of the equation to be proved are linear in α, we may assume without loss of generality that α is a form on U R n + with compact support. Write α = n ( 1) i 1 α i dx 1 (dx i )ˆ dx n, (7.2.2) i=1 where ˆ above a term means that it is deleted. Then n α i dα = x i dx1 dx n, (7.2.3) and thus U dα = i=1 n Rn α i i=1 x i dx1 dx n. (7.2.4) There are two cases: U = and U. If U =, we have α =. The integration of the ith term U in the sum occurring in equation (7.2.4) is ( α i ) x i dxi dx 1 (dx i )ˆ dx n (no sum) (7.2.5) R n 1 R and + ( αi / x i )dx = since α i has a compact support. Thus, the expression in equation (7.2.4) is zero as desired. If U, then we can do the same trick for each term except the last, which is, by the fundamental theorem of calculus, ( α n ) R n 1 x n dxn dx 1 dx n 1 = α n (x 1,...,x n 1, )dx 1 dx n 1. R n 1 (7.2.6) since α n has compact support. Thus, dα = U α n (x 1,...,x n 1, )dx 1 dx n 1 R n 1 (7.2.7) On the other hand, α = α = ( 1) n 1 α n (x 1,...,x n 1, )dx 1 dx n 1. (7.2.8) U R n + R n + But R n 1 = R n + and the usual orientation on R n 1 is not the boundary orientation. The outward unit normal is e n = (,...,, 1) and hence the boundary orientation has the sign of the ordered basis { e n,e 1,...,e n 1 }, which is ( 1) n. Thus equation (7.2.8) becomes α = ( 1) n 1 α n (x 1,...,x n 1, )dx 1 dx n 1 U R n + =( 1) 2n 1 R n 1 α n (x 1,...,x n 1, )dx 1 dx n 1. (7.2.9) Since ( 1) 2n 1 = 1, combining equations (7.2.7) and (7.2.9), we get the desired result.

18 Integration on anifolds This basic theorem reduces to the usual theorems of Green, Stokes, and Gauss in R 2 and R 3, as we shall see in the next section. For forms with less smoothness or without compact support, the best results are somewhat subtle. See Gaffney [1954], orrey [1966], Yau [1976], Karp [1981] and the remarks at the end of Supplement 7.2B. Next we draw some important consequences from Stokes theorem Theorem (Gauss Theorem). Let be an oriented paracompact n-manifold with boundary and X a vector field on with compact support. Let µ be a volume form on. Then (div X)µ = i X µ. (7.2.1) Proof. Recall that (div X)µ = X µ = di X µ + i X dµ = di X µ. The result is thus a consequence of Stokes theorem. If carries a Riemannian metric, there is a unique outward pointing unit normal n along, and and carry corresponding uniquely determined volume forms µ and µ. Then Gauss theorem reads as follows Corollary. (div X) dµ = X,n dµ, where X,n is the inner product of X and n is the outward unit normal. Proof. Let µ denote the volume element on induced by the Riemannian volume element µ Ω n (); that is, for any positively oriented basis v 1,...,v n 1 T x ( ), and charts chosen so that n = / x n at the point x, ( µ (x)(v 1,...,v n 1 )=µ (x) ) x n,v 1,...,v n 1. Since ( (i X µ )(x)(v 1,...,v n 1 )=µ (x) X i (x)v i + X n (x) ) x n,v 1,...,v n 1 = X n (x)µ (x)(v 1,...,v n 1 ) and X n = X,n, the corollary follows by Gauss theorem Corollary. If X is divergence-free on a compact boundaryless manifold with a volume element µ, then X as an operator is skew-symmetric; that is, for f and g F(), Proof. X[f]gµ = fx[g]µ. Since X is divergence free, X (hµ) =( X h)µ for any h F(). Thus, X[f]gµ + fx[g]µ = X (fg)µ = X (fgµ). Integration and the use of Stokes theorem gives the result.

19 7.2 Stokes Theorem Corollary. If is compact without boundary X X(), α Ω k (), and β Ω n k (), then X α β = α X β. Proof. Since α β Ω n (), the formula follows by integrating both sides of the relation di X (α β) = X (α β) = X α β + α X β and using Stokes theorem Corollary. If is a compact orientable, boundaryless n-dimensional pseudo-riemannian manifold with a metric g of index Ind(g), then d and δ are adjoints, that is, dα, β µ = dα β = α δβ = α, δβ µ for α Ω k () and β Ω k+1 (). Proof. Recall from Definition that δβ =( 1) nk+1+ind(g) d β, so that dα β α δβ = dα β +( 1) nk+ind(g) α d β = dα β +( 1) nk+ind(g)+k(n k)+ind(g) α d β = dα β +( 1) k α d β = d(α β) since k 2 + k is an even number for any integer k. Integrating both sides of the equation and using Stokes theorem gives the result. The same identity dα, β µ = α, δβ µ holds for noncompact manifolds, possibly with boundary, provided either α or β has compact support in Int(). Supplement 7.2A Stokes Theorem for Nonorientable anifolds Let be a nonorientable paracompact n-manifold with a smooth boundary and inclusion map i :. We would like to give meaning to the formula dρ = i ρ. in Stokes theorem. Clearly, both sides makes sense if dρ and i (ρ) are defined in such a way that they are densities on and, respectively. Here d should be some operator analogous to the exterior differential,

20 Integration on anifolds and ρ should be a section of some bundle over analogous to n 1 (). Denote the as yet unknown bundle analogous to k () by k t () and its space of sections Ωk τ (). Then we desire an operator d :Ω k τ () Ω k+1 τ (), k =,...,n, and desire n τ () to be isomorphic to (). To guess what k τ () might be, let us first discuss n τ (). The key difference between an n-form ω and a density ρ is their transformation property under a linear map A : T m T m as follows: ω(m)(a(v 1 ),...,A(v n )) = (det A) ω(m)(v 1,...,v n ) ρ(m)(a(v 1 ),...,A(v n )) = det A ρ(m)(v 1,...,v n ) for m and v 1,...,v n T m.ifv 1,...,v n is a basis, then det(a) > ifapreserves the orientation given by v 1,...,v n and det(a) < ifa reverses this orientation. Thus ρ can be thought of as an object behaving like an n-form at every m once an orientation of T m is given; that is, ρ should be thought of as an n-form with values in some line bundle (a bundle with one-dimensional fibers) associated with the concept of orientation. This definition would then generalize to any k; k τ () will be line-bundle-valued k-forms on. We shall now construct this line bundle. At every point of there are two orientations. Using them, we construct the oriented double covering (see Proposition 6.5.7). Since is not a line bundle, some other construction is in order. At every m, a line is desired such that the positive half-line should correspond to one orientation of T m and the negative half-line to the other. The fact that must be taken into account is that multiplication by a negative number switches these two half-lines. To incorporate this idea, identify (m, [µ],a) with (m, [ µ], a) where m, a R, and [µ] is an orientation of T m. Thus, define the orientation line bundle σ() ={ (m, [µ],a) m, a R, and [µ] is an orientation of T m ) }/ where is the equivalence relation (m, [µ],a) (m, [ µ], a). Denote by m, [µ],a the elements of σ(). It can be checked that the map π : σ() defined by π( m, [µ],a ) =m is a line bundle with bundle charts given by ψ : π 1 (U) ϕ(u) R, ψ( m, [µ],a ) =(ϕ(m),ɛa), where ϕ : U R n is a chart for at m, and ɛ =+1ifT m ϕ :(T m,[µ]) (R n, [ω]) is orientation preserving and 1 if it is orientation reversing, [ω] being a fixed orientation of R n. The change of chart map of the line bundle σ() is given by (x, a) U R R n R ( (ϕ j ϕ i ) 1 (x), sign(det D(ϕ j ϕ 1 i )(x)) ) U R. If is paracompact, then σ() is an orientable vector bundle (see Exercise ). If in addition is also connected, then is orientable if and only if σ() is trivial line bundle; the proof is similar to that of Proposition Definition. A twisted k-form on is a σ()-valued k-form on. The bundle of twisted k-forms is denoted by k τ () and sections of this bundle are denoted Ωk τ () or Γ ( k τ ()). Locally, a section ρ Ω k τ () can be written as ρ = αξ where α Ω k (U) and ξ is an orientation of U regarded as a locally constant section of σ() over U. The operators d :Ω k τ () Ω k+1 τ () and i X :Ω k τ () Ω k 1 τ (), where X X(), are defined to be the unique operators such that if ρ = αξ in the neighborhood U, then dρ =(dα)ξ and i X ρ =(i X α)ξ. One has X = i X d + d i X. Note that if is orientable, k τ () coincides with k (). Next we show that the line bundles () and n τ () are isomorphic. If λ () m and v 1,...,v n T m, define ϕ(λ) :T m T m σ() m

21 7.2 Stokes Theorem 419 by setting Φ(λ)(v 1,...,v n )= m, [σ(v 1,...,v n )],λ(v 1,...,v n ), if {v 1,...,v n } is a basis of T m, and setting it equal to, if {v 1,...,v n } are linearly dependent, where [σ(v 1,...,v n )] denotes the orientation of T m given by the ordered basis {v 1,...,v n }.Φ(λ) is skew symmetric and homogeneous with respect to scalar multiplication since if {v 1,...,v n } is a basis and a R, we have and Φ(λ)(v 2,v 1,v 3,...,v n )= m, [σ(v 2,v 1,v 3,...,v n )],λ(v 2,v 1,v 3,...,v n ) = m, [ σ(v 1,v 2,...,v n )],λ(v 1,...,v n ) = m, [σ(v 1,...,v n )], λ(v 1,...,v n ) = Φ(λ)(v 1,...,v n ) Φ(λ)(av 1,v 2,...,v n )= m, [σ(av 1,v 2,...,v n )],λ(av 1,...,v n ) = m, [(sign a)σ(v 1,...,v n )], a λ(v 1,...,v n ) = m, [σ(v 1,...,v n )],aλ(v 1,...,v n ) = aφ(λ)(v 1,...,v n ). The proof of additivity is more complicated. Let v 1,v 1,v 2,...,v n T m.ifboth{v 1,...,v n } and {v 1,v 2,...,v n } are linearly dependent, then so are {v 1 +v 1,v 2,...,v n } and the additivity property of Φ(λ) is trivially verified. So assume that {v 1,...,v n } is a basis of T m and write v 1 = a 1 v a n v n. Therefore oreover, if (i) a 1 >, then (ii) a 1 =, then and (iii) 1 <a 1 <, then (iv) a 1 = 1, then and λ(v 1,v 2,...,v n )= a 1 λ(v 1,...,v n ), and λ(v 1 + v 1,v 2,...,v n )= 1+a 1 λ(v 1,...,v n ). [σ(v 1,...,v n )] = [σ(v 1,v 2,...,v n )] = [σ(v 1 + v 1,v 2,...,v n )]; [σ(v 1,...,v n )] = [σ(v 1 + v 1,v 2,...,v n )] Φ(λ)(v 1,v 2,...,v n )=; [σ(v 1,...,v n )] = [ σ(v 1,v 2,...,v n )] = [σ(v 1 + v 1,v 2,...,v n )]; [σ(v 1,...,v n )] = [ σ(v 1,v 2,...,v n )] Φ(λ)(v 1 + v 1,v 2,...,v n )=;

22 42 7. Integration on anifolds (v) a 1 < 1, then [σ(v 1,...,v n )] = [ σ(v 1,v 2,...,v n )] = [ σ(v 1 + v 1,v 2,...,v n )]. Additivity is now checked in all five cases separately. For example, in case (iii) we have Φ(λ)(v 1 + v 1,v 2,...,v n )= m, [σ(v 1 + v 1,v 2,...,v n )],λ(v 1 + v 1,...,v n ) = m, [σ(v 1,...,v n )], (1 + a 1 )λ(v 1,...,v n ) = m, [σ(v 1,...,v n )],λ(v 1,...,v n ) + m, [ σ(v 1,v 2,...,v n )], a 1 λ(v 1,...,v n )] =Φ(λ)(v 1,...,v n )+Φ(λ)(v 1,v 2,...,v n ). Thus Φ has values in n τ (). The map Φ is clearly linear and injective and thus is an isomorphism of () with n τ (). Denote also by Φ the induced isomorphism of Ω() with Ωn τ (). The integral of ρ Ω n τ () is defined to be the integral of the density Φ 1 (ρ) over. In local coordinates the expression for Φ is Φ(a dx 1 dx n )=(adx 1 dx n )ξ n, where ξ n is the basis element of the space sections of σ(u) given by ξ n (u)(v 1,...,v n )= u, [σ(v 1,...,v n )], sign(det(v i j )), where (vi j ) are the components of the vector v j in the coordinates (x 1,...,x n )ofu. Therefore and Φ 1 ((adx 1 dx n )ξ) = aξ ξ n dx 1 dx n U (adx 1 dx n )bξ n = U ab dx 1 dx n for any smooth functions a, b : U R. Finally, for the formulation of Stokes Theorem, if i : is the inclusion and ρ Ω n 1 τ (), the induced twisted (n 1)-form i ρ on is defined by setting (i ρ)(m)(v 1,...,v n 1 )= m, [sign[µ n ]σ( / x n,v 1,...,v n 1 )], ρ (m)(v 1,...,v n 1 ), if v 1,...,v n 1 are linearly independent and setting it equal to zero, if v 1,...,v n 1 are linearly dependent, where (x 1,...,x n ) is a coordinate system at m with described by x n = and ρ(m)(v 1,...,v n 1 )= m, sign[µ m ],ρ (m)(v 1,...,v n 1 ) with ρ (m) skew symmetric; moreover sign[µ m ] = +1 (respectively, 1) if [µ m ] and [σ( / x n,v 1,...,v n 1 )] define the same (respectively, opposite) orientation of T m.if = U, where U is open in R n + and ρ = αaξ Ω n 1 τ (U), then In particular, if ζ = i ρ =( 1) n i (aα)ξ n 1. n ζ i dx 1 (dx i )ˆ dx n, i=1

23 7.2 Stokes Theorem 421 we have (i ρ)(x 1,...,x n 1 )=( 1) n a(x 1,...,x n 1, )( 1) n 1 α n (x 1,...,x n 1, ) dx 1 (dx i )ˆ dx n ξ n 1 = a(x 1,...,x n 1, )α n (x 1,...,x 1, ) dx 1 (dx i )ˆ dx n ξ n 1. With this observation, the proof of Stokes theorem gives the following Theorem (Nonorientable Stokes Theorem). Let be a paracompact nonorientable n-manifold with smooth boundary and ρ Ω n 1 τ (), a twisted (n 1)-form with compact support. Then dρ = The same statement holds for vector-valued twisted (n 1)-forms and all corollaries go through replacing everywhere (n 1)-forms with twisted (n 1)-forms. For example, we have the following Theorem (Nonorientable Gauss Theorem). Let be a nonorientable Riemannian n-manifold with associated density µ. Then for X X() with compact support div(x)µ = (X n)µ where n is the outward unit normal of, µ is the induced Riemannian density of and X µ = (div X)µ. i ρ. For a concrete situation in R 3 involving these ideas, see Exercise Supplement 7.2B Stokes Theorem on anifolds with Piecewise Smooth Boundary The statement of Stokes theorem we have given does not apply when is, say a cube or a cone, since these sets do not have a smooth boundary. If the singular portion of the boundary (the four vertices and 12 edges in case of the cube, the vertex and the base circle in case of the cone), is of Lebesgue measure zero (within the boundary) it should not contribue to the boundary integral and we can hope that Stokes theorem still holds. This supplement discusses such a version of Stokes theorem inspired by Holmann and Rummler [1972]. (See Lang [1972] for an alternative approach.) First we shall give the definition of a manifold with piecewise smooth boundary. A glance at the definition of a manifold with boundary makes it clear that one could define a manifold with corners, by choosing charts that make regions near the boundary diffeomorphic to open subsets of a finite intersection of positive closed half-spaces. Unfortunately, singular points on the boundary such as the vertex of a cone need not be of this type. Thus, instead of trying to classify the singular points up to diffeomorphism and then make a formal intrinsic definition, it is simpler to consider manifolds already embedded in a bigger manifold. Then we can impose a condition on the boundary to insure the validity of Stokes theorem.

24 Integration on anifolds Definition. Definition Let U R n 1 be open and f : U R be continuous. A point p on the graph of f, Γ f = { (x, f(x)) x U }, is called regular if there is an open neighborhood V of p such that V Γ f is an (n 1) dimensional smooth submanifold of V.Letρ f denote the set of regular points. Any point in σ f =Γ f \ρ f is called singular. The mapping f is called piecewise smooth if ρ f is Lebesgue measurable, π(σ f ) has measure zero in U (where π : U R U is the projection) and f π(σ f ) is locally Hölder; that is, for each compact set K π(σ f ) there are constants c(k) >, <α(k) 1 such that for all x, y K. f(x) f(y) c(k) x y α(k) Note that ρ f is open in Γ f and that Int(Γ f ) ρ f, where Γ f = { (x, y) U R y f(x) }, is a manifold with boundary ρ f.thusρ f has positive orientation induced from the standard orientation of R n. This will be called the positive orientation of Γ f. We are now ready to define manifolds with piecewise smooth boundary Definition. Let be an n-manifold. A closed subset N of is said to be a manifold with piecewise smooth boundary if for every p N there exists a chart (U, ϕ) of at p, ϕ(u) =U U R n 1 R, and a piecewise smooth mapping f : U R such that ϕ(bd(n) U) =Γ f ϕ(u) and ϕ(n U) =Γ f ϕ(u). See Figure σ n = singular points N U U ρ n = regular points ϕ Γ f ϕ(u) Γ f ϕ(u) U Figure Singular boundary charts It is readily verified that the condition on N is chart independent, using the fact that the composition of a piecewise smooth map with a diffeomorphism is still piecewise smooth. Thus, regular and singular points of bd(n) make intrinsic sense and are defined in terms of an arbitrary chart satisfying the conditions of the preceding definition. Let ρ N and σ N denote the regular and singular part of the boundary bd(n) ofn in. To formulate Stokes theorem, we define η, for η an n-form (respectively, density) on with compact N support. This is done as usual via a partition of unity; ρ N and σ N play no role since they have Lebesgue measure zero in every chart: ρ N because it is an (n 1)-manifold and σ N by definition. It is not so simple to define bd(n) ζ for ζ Ωn 1 () (respectively, a density). First a lemma is needed.

25 7.2 Stokes Theorem Lemma. Let ζ Ω n 1 (U R), where U is open in R n 1, supp(ζ) is compact and f : U R is a piecewise smooth mapping. Then there is a smooth bounded function a : ρ f R, such that i ζ = aλ where i : ρ f U R is the inclusion and λ Ω n 1 (ρ f ) is the boundary volume form induced by the canonical volume form of U R R n on Int(Γ f ) ρ f. Proof. The existence of the function a on ρ f follows since Ω n 1 (ρ f ) is one-dimensional with a basis element λ. We prove that a is bounded. Let p ρ f and v 1,...,v n 1 T p (ρ f ) be an orthonormal basis with respect to the Riemannian metric on ρ f induced from the standard metric of R n, and denote by n the outward unit normal. Then a(p)=a(p)(dx 1 dx n )(p)(n, v 1,...,v n 1 ) = a(p)λ(p)(v 1,...,v n 1 )=ζ(p)(v 1,...,v n 1 ). Let v i = v j i x j. p Since x 1,..., p x n and n, v 1,...,v n 1 p are orthonormal bases of T p (U R), we must have v j i 1 for all i, j. Hence if then ζ = n ζ i dx 1 (dx i )ˆ dx n, i=1 a(p) = ζ(p)(v 1,...,v n 1 ) n = ζ i (p) ( 1) i (sign σ)v σ(1) i=1 σ S n 1 n = ζ i (p) (n 1)! i=1 1 v σ(n 1) n 1 which is bounded, since ζ has compact support. In view of this lemma and the fact that σ f has measure zero, we can define ζ = Γ f i ζ = ρ f aλ. ρ f Now we can define, via a partition of unity, the integral of η Ω n 1 () (or a twisted (n 1)-form) by η = η. bd(n) ρ N Theorem (Piecewise Smooth Stokes Theorem). Let be a paracompact n-manifold and N a closed submanifold of with piecewise smooth boundary. If (i) is orientable and ω Ω n 1 () has compact support, or