INTEGRATION ON MANIFOLS and GAUSS-GREEN THEOREM 1. Schwarz s paradox. Recall that for curves one defines length via polygonal approximation by line segments: a continuous curve γ : [a, b] R n is rectifiable if the supremum of the lengths L P [γ] of the polygonal approximations defined by partitions P of [a, b] is finite, and that supremum is the length: n 1 L P [γ] = γ(t i+1 ) γ(t i ), P = {t 0 = a, t 1,..., t n 1, t n = b}, L[γ] = sup L P [γ]. P P i=1 It is natural to attempt to define, say, surface area (for a surface M R 3 ) analogously, via polyhedral approximation (where a polyhedron is made up of flat triangles with all three vertices on the surface). However an example due to Hermann Schwarz (1843-1921) shows, surprisingly, that this doesn t work: even a cylinder in R 3 admits inscribed triangular polyhedra with arbitrarily large area! Example. Consider a cylinder S R 3 parametrized by: φ : [0, 2π) [0, 1] R 3, φ(x, y) = (cos x, sin x, y). Subdivide the domain into m intervals in the x factor, n intervals in the y factor, and then each of the mn rectangles into four triangles, with a common vertex at the center. This defines a triangulation of the domain, and the images of its vertices under φ define a polyhedral surface S mn inscribed in S (where the facets are triangles with vertices on S). It can be shown that the area of S mn is: A mn = 2n sin π 2n + [1 4 + 4m2 n 4 (n sin π 2n )4 ] 1/2.2n sin π n. Thus, if m = n, lim A nn = 2π, as expected. But if m = n 3, we find {A mn } n 1 is unbounded! So although the area of S is finite, the supremum sup{area(p ); P a polyhedral surface inscribed in S} isn t. Two useful solutions to the problem of surface measure are: an extrinsic one, the notion of n-dimensional Hausdorff measure of subsets of R m (m n); and a definition of surface measure for n-dimensional surfaces (manifolds) in R m, m > n. Both reduce to arc length in the one-dimensional case, and Hausdorff measure is more general. We ll only use the second one (described in sections 8.1-8.3 of [Fleming].) 1
2. The linear case. We consider a set A E contained in an n-dimensional subspace E R m, and state reasonable conditions its ndimensional measure µ E should satisfy. (i) If E = R n {0} R n R m n R m (the subspace obtained by setting the last m n coordinates equal to zero), then it is natural to require µ E (A) = m n (A), the Lebesgue measure of A as a (measurable) subset of R n. (ii) If U O(m) is an orthogonal transformation, then µ U(E) (U(A)) = µ E (A). These two conditions define µ E uniquely. The reason is any T L(R n ; R m ) admits a decomposition: T = U S, S Sym + (R n ), U O(m), the embedding of R n into R m = R n R m n as E 0 = R n {0}, obtained by setting the last m n coordinates to zero. Then if A R n we have T (A) = U(A 0 ) E = U(E 0 ), A 0 = (S(A)) E 0, and using (i) and (ii): µ E (T (A)) = µ E0 (A 0 ) = m n (S(A)) = J S m n (A), where J S = det S, the Jacobian of S L(R n ). Note that, since U U = I m : det(t T ) = det(s S) = det(s S) = (det S) 2, using = I n. Thus J S = (det T T ) 1/2 (note T T Sym + (R n )), and we have: µ E (T (A)) = J T m n (A), J T := (det T T ) 1/2, E = T (R n ) R m. In particular, given an n-dimensional subspace E R m, we can always find U O(m) so that E = U(E 0 ), and then if A E, A = U(A 0 ) where A 0 E 0 R n is m n -measurable we have: µ E (A) = m n (A 0 ). Another expression in the linear case is obtained by recalling that the m- dimensional volume of the parallelepiped P (m) R m spanned by m linearly independent vectors v 1,..., v m is: m(p (m) ) = det[v 1... v m ], the absolute determinant of the m m matrix with columns v i R m. Now if P (n) E is the parallelepiped spanned by n linearly independent vectors {v 1,..., v n } in R m, where E is the subspace spanned by the {v i }, 2
find U O(m) so that U(E 0 ) = E, and let P (n) 0 E 0 be the parallelepiped spanned by w i = U 1 v i. Then: µ E (P (n) ) = Leb n (P (n) 0 ) = Leb m (P (n) 0 [0, 1] m n ) = det m [w 1... w n e n+1... e m ] = (det n W W ) 1/2, (W W ) ij = w i w j = (det n V V ) 1/2, (V V ) ij = v i v j = w i w j, since U O(m). In summary, we have: µ E (P (v 1,..., v n )) = det n (v i v j ), E = span{v 1,..., v n }, dim(e) = n. 3. Measure and integration on manifolds.[fleming, 8.3] Recall that, by definition, a C k manifold (k 1) M of dimension n in R m (m > n) is a closed subset of R m which admits a covering by relatively open sets U = {U = Û M}, Û open in Rm, each of which is the image of a coordinate chart φ : U 0 R m injective immersion of class C k, U 0 R n open, so that φ is a homeomorphism onto its image φ(u 0 ) = U. Motivated by the linear case, we define the Jacobian of φ by: J φ (x) = det n (dφ(x) dφ(x)) 1/2, x U 0. An alternative expression is obtained by considering the parallelepiped in T φ(x) M (the tangent space at φ(x)) spanned by the basis {v i = i φ} n i=1 of the tangent space: J φ (x) = det n ( i φ j φ)(x). A subset A U is µ M -measurable if B = φ 1 (A) U 0 is Leb n -measurable, and then we set: µ M (A) = J φ (x)d n x, and for f C(U): A fdµ M = B B (f φ)(x)j φ (x)d n x. 3
If A = φ(b) = ψ(c) is contained in the intersection φ(u 0 ) ψ(v 0 ) of two coordinate patches, we need to check the consistency relation: J φ (x)d n x = J ψ (y)d n y. B This follows easily from the relation (for a C k diffeomorphism F : U 1 V 1, U 1 U 0, V! V 0 open): C φ U1 = ψ F J φ (x) = J ψ (F (x))j F (x), J F (x) = det n df (x), combined with the change of variables formula. To extend this to general subsets of M, we need partitions of unity. Let U = {U i } i 1 be a locally finite, countable covering of M by open sets U i M. A partition of unity subordinate to U is a set of smooth functions φ i : M [0, 1] with compact support, satisfying: (i) {x M; φ i (x) > 0} = U i, i; (ii) i=1 φ i(x) 1 on M. Note this sum is finite at each x M, given (i) and the assumption that the covering U is locally finite. Existence. A submanifold of R n is paracompact: any open covering V admits a countable, locally finite refinement {U i } (any U i is contained in some open set of V). And any countable, locally finite open covering admits a partition of unity subordinate to it. (See [Fleming, prop 8.6] for the compact case.) Now given A M say A is µ M -measurable if A U is µ M -measurable, for each coordinate patch U M. Let U = {U i } i 1 be a countable, locally finite open covering of M by coordinate patches, and let {φ i } i 1 be a smooth partition of unity associated with the covering. Then for A M µ M - measurable, let: µ M (A) = φ i dµ M. i A U i And for f C(M) and A M measurable, define: A fdµ M = i A U i fφ i dµ M. It is not hard to show that the result is independent both of the covering {U i } i 1 and the partition of unity {φ i } (see [Fleming]). 4
4. The case of graphs. Let U R n be open, u C 1 (U). The graph of u is an n-dimensional C 1 manifold in R n+1 : Γ = {(x, x n+1 ) U R; x n+1 = u(x)}. We need a single coordinate chart in this case, for instance the graph chart: φ : U R n+1, φ(x) = (x, u(x)) R n R. To compute the Jacobian of this chart we consider the basis of the tangent space T φ(x) Γ: i φ(x) = (e i, i u(x)), i = 1,..., n. Thus: We claim this equals: J φ (x) = det n (δ ij + i u j u) 1/2. J φ (x) = 1 + u 2. It clearly suffices to prove the following algebraic lemma. Lemma. Let v R n. Then det n (δ ij + v i v j ) = 1 + v 2. Proof. We may assume v 0. Consider the linear transformation of R n (the rank one map with range v, kernel v ): v v L(R n ), (v v )[x] = (v x)v. The matrix of the linear map T = I n + v v in the standard basis is clearly δ ij + v i v j. Let {f 1 = v/ v, f 2,..., f n } be an orthonormal basis of R n. As is well known, any x R n admits the decomposition: Thus x = (f 1 x)f 1 +... + (f n x)f n, or I n = f 1 f 1 +... + f n f n. I n + v v = I n + v 2 f 1 f 1 = (1 + v 2 )f 1 f 1 +... + f n f n. Since the matrix of this linear transformation in the basis {f i } is diagonal, with one entry equal to 1 + v 2, the others equal to 1, the claim is proved. We conclude that, for f C(Γ) and A = φ(b) Γ measurable: fdµ Γ = f(x, u(x)) 1 + u(x) 2 d n x. A 5. Gauss-Green theorem. B 5
5.1 Subgraph domains. Given an open cube U = ( 1, 1) n R n and a C 1 function u : U (a, b), consider the subgraph domain A Q = U (a, b) in R n+1 : A = {y = (x, x n+1 ) Q; a < x n+1 < u(x)}. Part of the boundary of A is the graph Γ of u over U, with outward unit normal ν, with components: i u ν i =, i = 1,..., n; ν n+1 = 1 + u 2 1 1 + u 2. In fact it is easy to see that A = ( Q Ā) Γ. Proposition. Let U be an open neighborhood of Ā in R n+1 and suppose f C 1 (U) vanishes on Q Ā. Then, for each i = 1,..., n + 1: ( i f)(y)d n+1 y = fν i dµ Γ. A Proof. i=n+1. By Fubini s theorem and the FTC in one variable: = U A ( n+1 f)(y)d n+1 y = U [ u(x) [f(x, u(x)) f(x, a)]d n x = a U Γ ( n+1 f)(x, x n+1 )dx n+1 ]d n x f(x, u(x))d n x = Γ fν n+1 dµ Γ. i=1,..., n. For simplicity we let i = 1 and set x = (x 1, z) R R n 1, for x R n (forget z if n = 1). Then let F (x) = u(x) Note that F (x) = 0 if x U. We have: a f(x, x n+1 )dx n+1 Thus 1 F (x) = ( 1 u)(x)f(x, u(x)) + A ( 1 f)(y)d n+1 y = U u(x) a ( 1 f)(x, x n+1 )dx n+1. 1 F (x)d n x ( 1 u)(x)f(x, u(x))d n x U 6
= [ ( 1,1) n 1 1 1 ( 1 F )(x 1, z)dx 1 ]d n 1 z U ( 1 u)(x)f(x, u(x))d n x where the first term vanishes, since F (1, z) = F ( 1, z) = 0 z, given that ( 1, z), (1, z) are in U; while the second term clearly equals Γ fν 1dµ Γ. This concludes the proof. 5.2 C k domains. A connected open set R n+1 is a C k domain (k 1) if for each y we may find an open rectangle Q y = U y I y (U y R n an open cube, I y R a bounded open interval) and a C k function u : U y I y so that: Q y = graph(u); Q y = {(x, x n+1 ) Q y ; x n+1 < u(x)}. Thus Q y is a subgraph domain as in 5.1. In particular, the boundary is a C k n-dimensional manifold in R n+1 (a C k hypersurface), not necessarily connected. Thus it admits an n-dimensional volume dµ, as well as a unit outward normal vector field ν defined at points of (of class C k 1 ), given in each local graph chart (as above) by the expression in 5.1. Gauss-Green theorem. Let R n+1 be a bounded C 1 domain, U an open neighborhood of its closure, f C 1 (U). Then for i = 1,..., n+1 we have: ( i f)dx = fν i dµ. Proof. Let {Q y } y be an open covering of the compact set defined as follows: (i) for y, we let Q y be an open rectangle as in the definition of C k domain (above.) (ii) for y (an open set) we let Q y be any open cube with center y, and contained in. Let {Q y1,..., Q yn } be a finite subcovering, and consider a (finite) partition of unity subordinate to it: φ j C c (U), φ j 1, {y; φ j (y) > 0} = U yj. j=1 There are two cases to consider: (a) if Q yj is of type (i) above, the G-G theorem in the subgraph domain Q yi (for the function fφ j ) gives: i (fφ j )dy = Q yj fφ j ν i dµ, Q yi i = 1,..., n + 1. 7
(Note fφ j 0 on Q yj ). Hence: i (fφ j )dy = fφ j ν i dµ, i = 1,..., n + 1. If Q yj is of type (ii), both sides of the equality are easily seen to vanish: the right-hand side since Q yj = : fφ j ν i = fφ j ν i dµ = 0, i = 1,..., n + 1; Q yi the left-hand side by Fubini s theorem and the FTC in one variable, since Q yj = Q yj and fφ j 0 on Q yj : i (fφ j ) = i (fφ j )dy = 0, Q yj i = 1,..., n + 1. Adding over j = 1,..., N, we find (for each fixed i = 1,..., n + 1: i fdy = j=1 This concludes the proof. i (fφ j )dy = j=1 fφ j dµ = fν i dµ. ivergence Theorem. Let X = (a 1,..., a n+1 ) be a C 1 vector field in a neighborhood U of, for a bounded C 1 domain R n+1. By definition, div(x) = n+1 i=1 ia i. (So each a i C 1 (U)). We have: div(x)dx = X νdµ. This follows immediately from the Gauss-Green theorem. Conversely: Exercise 1. Prove the Gauss-Green theorem, assuming the ivergence Theorem. The G-G theorem also leads to a simple proof of Stokes theorem for line integrals of 1-forms in planar C 1 bounded domains: if R 2 is a bounded C 1 domain and ω = a 1 dx 1 + a 2 dx 2 Ω 1 U is a C1 1-form in a neighborhood U of, we have: ω = dω (= ( 1 a 2 2 a 1 )d 2 x). 8
Orientation and measure on the boundary. To make sense of the line integral on, we need an orientation of the boundary. The usual one is defined using the inner unit normal N = ν. A parametrization γ(t) of a boundary component Γ is positive if the frame {T, N}, T = γ (t)/ γ (t), is a positive frame in R 2 (i.e., obtained from the standard frame {e 1, e 2 } by a rotation.) Thus we see that if Γ is parametrized by arc length, we have: ν 1 (s) = N 1 (s) = γ 2(s), ν 2 (s) = N 2 (s) = γ 1(s). From Topology, we know is a disjoint union of simple closed C 1 curves: = Γ 0 Γ 1... Γ N where Γ 0 is the boundary of the unbounded component of the complement c. Informally, Γ 0 is oriented counterclockwise, the other Γ i clockwise. Parametrizing each Γ i (with length L i ) by arc length, we have: = i=0 L i 0 ω = i=0 Li 0 [a 1 (s)γ 1(s) + a 2 (s)γ 2(s)]ds [ a 1 (s)ν 2 (s) + a 2 (s)ν 1 (s)]ds = i=0 Li 0 (a 2, a 1 ) νds. Exercise 2. Show that, for a C 1 simple closed curve Γ in R 2, the 1- dimensional volume dµ Γ equals arc length: for any f C(Γ): Γ fdµ Γ = L 0 f(γ(s))ds, where γ(s) is the arc length parametrization of Γ. Hint: Recall Γ can be locally parametrized by graph charts, and compare the expressions for dµ Γ and ds in graph parameters. while: Thus, with X the vector field X = (a 2, a 1 ): ω = i=0 Li 0 X(γ i (s)) ν(γ i (s)ds = dω = 9 X νdµ Γi = Γ i i=1 div(x)d 2 x. X νdµ
So we see that Stokes theorem is equivalent to the divergence theorem for planar C 1 domains. 6. Weak derivatives and Sobolev spaces. efinition. Let R n be open (not necessarily bounded) and let f L 1 loc () (that is, integrable in each compact subset of.) We say a function v L 1 loc () is the weak partial derivative of f in direction e i if, for every φ Cc 1 () ( test function ) we have: v(x)φ(x)dx = f(x) i φ(x)dx. Remark: It is easy to see that a function v with this property, if it exists, is unique a.e. We adopt the notation v = i f for the weak derivative of a locally integrable function (when it exists). Note that if f C 1 () this holds for v = i f, as a consequence of the Gauss-Green theorem (applied to f in any bounded C 1 domain A with Ā and A support(φ)). Thus if f C 1 () we may take i f = i f. Example 1. The function f(x) = x in = ( 1, 1) R has the weak derivative: 1 f(x) = 1, x 0; 1 f(x) = 1, x > 0. Example 2. The function g(x) = 1, x > 0; g(x) = 0, x 0 in = ( 1, 1) does not have a weak derivative in ( 1, 1). If it did, that weak derivative would have to vanish a.e. in ( 1, 1), while 1 1 gφ (x)dx = φ(1) φ(0) = φ(0) does not vanish for arbitrary φ C 1 c ( 1, 1). efinition. Let 1 p. If f L p () admits weak derivatives i f in all directions e i, and they are in L p (), we say f is in the Sobolev space W 1,p (). This space is endowed with the norm: f W 1,p = [ f p dx + i f p dx] 1/p. i Remark: Any f L p () is automatically in L 1 loc (). To see this, let K be compact and estimate fχ Kdx using Hölder s inequality. 10
Proposition. Endowed with this norm, W 1,p () is a Banach space. Proof. Let (f n ) n 1 be a Cauchy sequence in W 1,p (). (1 p < ) Then each ( i f n ) as well as (f n ) itself, is a Cauchy sequence in L p (), therefore convergent: we have f L p () and v i L p () so that: f n f in L p, i f n v i in L p, i. We have (with q = These imply: p p 1 ): ( i f n )φdx = f n i φdx φ C 1 c (), n 1, i. (f n f) i φdx f n f p i φ q 0; ( i f n v i )φdx i f n v i p i φ q 0. v i φdx = f i φdx so v i = i f, as we wished to show. φ C 1 c (), n 1, i, Approximations in W 1,p. We aim to show that kernel smoothing gives an approximation of any f W 1,p () by smooth functions in, in W 1,p norm. Recall that, given a kernel function k C (R n, [0, 1]) supported in the unit ball (k even, with unit integral) we define, for f L 1 loc () and 0 < h < d(x, ): f h (x) = 1 h n k( x y R n h )f(y)dy. (Extend f to R n by zero.) The function f h C (R n ), and we showed that f h f in L p, as h 0. Lemma. Let f W 1,p, 1 p <. Then for each x and each h > 0 such that h < d(x, ), we have: Proof. i f h (x) = 1 h n i f h (x) = ( i f) h (x) for i = 1,..., n. ( xi k)( x y h )f(y)dy = 1 h n ( yi k)( x y h )f(y)dy 11
= 1 h n k( x y h ) if(y)dy = ( i f) h (x) The third equality uses the definition of weak derivative. Note that, for x fixed, the support of y k( y x h ) is the closed ball of center x, radius h, which is a compact subset of. Remark. There is an important subtlety here. For f L p (), we may extend f by zero to R n, so the smoothing f h is defined in all of R n, for all h > 0. (And then we show f h f in L p (R n ), and a fortiori in L p ()). For the weak derivatives i f L p (), we can t extend by zero and expect the extensions continue to be weak derivatives! So ( i f) h (x) is only defined for those x whose distance to is greater than h. Or, given (that is, open and bounded, with closure contained in ), ( i f) h (x) is defined for all x, and all h > 0 sufficiently small (h < d(, )). And then ( i f) h i f in L p ( ). Proposition. Let f W 1,p (), R n open. Then f h f in L p () and i f h i f in L p ( ) (as h 0), for each. In particular, for each and h > 0 sufficiently small, the functions f h C ( ) converge to f in W 1,p ( ), as h 0 +. Proof. We know ( i f) h is defined in (for h suff. small) and converges to i f in L p ( ). On the other hand the lemma shows i f h = ( i f) h (in, for h suff. small). With a further technical device (partitions of unity associated to a locally finite covering) we can prove density of smooth functions in W 1,p (). Proposition. C () W 1,p () is dense in W 1,p (). (For R n open, not nec. bounded). Proof. Let n = {x ; x < n, d(x, ) > 1/n}, n 1. Then n and we have an exhaustion of : n n+1, n 1 n =, and an associated locally finite open covering U = {U n } n 1 of : U n = n+1 \ n 1, n 2; U 1 = 2. Let (φ n ) n 1 be a partition of unity subordinate to U. Let f W 1,p () and ɛ > 0 be given. Since φ n f has compact support in n+2, we may find h n > 0 so that: φ n f (φ n f) hn W 1,p () < ɛ 2 n. 12
Then let: g = n 1(φ n f) hn C (), (since the sum is locally finite.) Since f = n φ nf, it follows that: g f W 1,p () n (φ n f) hn φ n f W 1,p () < ɛ. This concludes the proof. Remark. In contrast with L p (), it is not true we may approximate any f W 1,p () by smooth functions of compact support f n C c () (in W 1,p norm). For example, if is bounded, nonzero constants (which are trivially in W 1,p ()) do not admit such an approximation. 13