Math 52H: Multilinear algebra, differential forms and Stokes theorem. Yakov Eliashberg

Transcription

1 Math 52H: Multilinear algebra, differential forms and Stokes theorem Yakov Eliashberg March 202

2 2

3 Contents I Multilinear Algebra 7 Linear and multilinear functions 9. Dual space Canonical isomorphism between (V ) and V The map A Multilinear functions Tensor and exterior products 7 2. Tensor product Spaces of multilinear functions Symmetric and skew-symmetric tensors Symmetrization and anti-symmetrization Exterior product Spaces of symmetric and skew-symmetric tensors Operator A on spaces of tensors Orientation and Volume Orientation Orthogonal transformations Determinant and Volume Volume and Gram matrix

4 4 Dualities Duality between k-forms and (n k)-forms on a n-dimensional Euclidean space V Euclidean structure on the space of exterior forms Contraction Complex vector spaces Complex numbers Complex vector space Complex linear maps II Calculus of differential forms 57 6 Vector fields and differential forms Differential and gradient of a smooth function Gradient vector field Vector fields Differential forms Coordinate description of differential forms Smooth maps and their differentials Operator f Coordinate description of the operator f Examples Pfaffian equations Exterior differential Properties of the operator d Curvilinear coordinate systems More about vector fields Case n = 3. Summary of isomorphisms Gradient, curl and divergence of a vector field Complex-valued differential k-forms

5 8 Preliminaries Elements of topology in an Euclidean space Partition of unity and cut-off functions Integration of differential forms and functions One-dimensional Riemann integral for functions and differential -forms Integration of differential -forms along curves Integrals of closed and exact differential -forms Integration of functions over domains in high-dimensional spaces Fubini s Theorem Integration of n-forms over domains in n-dimensional space Manifolds and submanifolds Manifolds Gluing construction Examples of manifolds Submanifolds of an n-dimensional vector space Submanifolds with boundary Tangent spaces and differential Vector bundles and their homomorphisms Orientation Integration of differential k-forms over k-dimensional submanifolds III Stokes theorem and its applications 53 0 Stokes theorem Statement of Stokes theorem Proof of Stokes theorem Integration of functions over submanifolds Work and Flux Integral formulas of vector analysis

6 0.6 Expressing div and curl in curvilinear coordinates Applications of Stokes formula 75. Integration of closed and exact forms Approximation of continuous functions by smooth ones Homotopy Winding and linking numbers Properties of k-forms on k-dimensional manifolds

7 Part I Multilinear Algebra 7

8

9 Chapter Linear and multilinear functions. Dual space Let V be a finite-dimensional real vector space. The set of all linear functions on V will be denoted by V. Proposition.. V is a vector space of the same dimension as V. Proof. One can add linear functions and multiply them by real numbers: (l + l 2 )(x) = l (x) + l 2 (x) (λl)(x) = λl(x) for l, l, l 2 V, x V, λ R It is straightforward to check that all axioms of a vector space are satisfied for V. Let us now check that dim V = dim V. Choose a basis v... v n of V. For any x V let x. x n be its coordinates in the basis v... v n. Notice that each coordinate x,..., x n can be viewed as a linear function on V. Indeed, ) the coordinates of the sum of two vectors are equal to the sum of the corresponding coordinates; 9

10 2) when a vector is multiplied by a number, its coordinates are being multiplied by the same number. Thus x,..., x n are vectors from the space V. Let us show now that they form a basis of V. Indeed, any linear function l V can be written in the form l(x) = a x a n x n which means that l is a linear combination of x... x n with coefficients a,..., a n. Thus x,..., x n generate V. On the other hand, if a x a n x n is the 0-function, then all the coefficients must be equal to 0; i.e. functions x,..., x n are linearly independent. Hence x,..., x n form a basis of V and therefore dim V = n = dim V. The space V is called dual to V and the basis x,..., x n dual to v... v n. Exercise.2. Prove the converse: given any basis l,..., l n of V we can construct a dual basis w,..., w n of V so that the functions l,..., l n serve as coordinate functions for this basis. Recall that vector spaces of the same dimension are isomorphic. For instance, if we fix bases in both spaces, we can map vectors of the first basis into the corresponding vectors of the second basis, and extend this map by linearity to an isomorphism between the spaces. In particular, sending a basis S = {v,..., v n } of a space V into the dual basis x,..., x n of the dual space V we can establish an isomorphism i S : V V. However, this isomorphism is not canonical, i.e. it depends on the choice of the basis v,..., v n. If V is a Euclidean space, i.e. a space with a scalar product x, y, then this allows us to define another isomorphism V V, different from the one described above. This isomorphism associates with a vector v V a linear function l v (x) = v, x. We will denote the corresponding map V V by D. Thus we have D(v) = l v for any vector v V. Exercise.3. Prove that D : V V is an isomorphism. Show that D = i S for any orthonormal basis S. The isomorphism D is independent of a choice of an orthonormal basis, but is still not completely canonical: it depends on a choice of a scalar product. It is sometimes customary to denote dual bases in V and V by the same letters but using lower indices for V and upper indices for V, e.g. v,..., v n and v,..., v n. However, in these notes we do not follow this convention. 0

11 Remark.4. The definition of the dual space V also works in the infinite-dimensional case. Exercise.5. Show that both maps i S and D are injective in the infinite case as well. However, neither one is surjective if V is infinite-dimensional..2 Canonical isomorphism between (V ) and V The space (V ), dual to the dual space V, is canonically isomorphic in the finite-dimensional case to V. The word canonically means that the isomorphism is god-given, i.e. it is independent of any additional choices. When we write f(x) we usually mean that the function f is fixed but the argument x can vary. However, we can also take the opposite point of view, that x is fixed but f can vary. Hence, we can view the point x as a function on the vector space of functions. If x V and f V then the above argument allows us to consider vectors of the space V as linear functions on the dual space V. Thus we can define a map I : V V by the formula x I(x) (V ), where I(x)(l) = l(x) for any l V. Exercise.6. Prove that if V is finite-dimensional then I is an isomorphism. What can go wrong in the infinite-dimensional case?.3 The map A Given a map A : V W one can define a dual map A : W V as follows. For any linear function l W we define the function A (l) V by the formula A (l)(x) = l(a(x)), x V. In other words, A (l) = l A. 2 2 In fact, the above formula makes sense in much more general situation. Given any map Φ : X Y between two sets X and Y the formula Φ (h) = h Φ defines a map Φ : F (Y ) F (X) between the spaces of functions on Y and X. Notice that this map goes in the direction opposite to the direction of the map Φ.

12 Given bases B v = {v,..., v n } and B w = {w,..., w k } in the vector spaces V and W one can associate with the map A a matrix A = M BvBw (A). Its columns are coordinates of the vectors A(v j ), j =,..., n, in the basis B w. Dual spaces V and W have dual bases X = {x,... x n } and Y = {y,..., y k } which consist of coordinate functions corresponding to the basis B v and B w. Let us denote by A the matrix of the dual map A with respect to the bases Y and X, i. e. A = M Y X (A ). Proposition.7. The matrices A and A are transpose to each other, i.e. A = A T. Proof. By the definition of the matrix of a linear map we should take vectors of the basis Y = {y,..., y k }, apply to them the map A, expand the images in the basis X = {x,... x n } and write the components of these vectors as columns of the matrix A. Set ỹ i = A (y i ), i =,..., k. For any vector u = n u j v j V, we have ỹ i (u) = y i (A(u)). The coordinates of the vector A(u) in the j=i basis w,..., w k may be obtained by multiplying the matrix A by the column n ỹ i (u) = y i (A(u)) = a ij u j. But we also have n n a ij x j (u) = a ij u j. j= j= Hence, the linear function ỹ i V has an expansion space V. Hence the i-th column of the matrix A equals has the form a a k A =... = AT. a n a kn 2 j= n j= u. u n. Hence, a ij x j in the basis X = {x,... x n } of the a i. a in, so that the whole matrix A

13 Exercise.8. Given a linear map A : V W with a matrix A, find A (y i ). Answer. The map A sends the coordinate function y i on W to the function its expression in coordinates x i. n j= a ij x j on V, i.e. to Proposition.9. Consider linear maps U A V B W. Then (B A) = A B. Proof. For any linear function l W we have (B A) (l)(x) = l(b(a(x)) = A (B (l))(x) for any x U. Exercise.0. Suppose that V is a Euclidean space and A is a linear map V V. Prove that for any two vectors X, Y V we have A(X), Y = X, D A D(Y ). (.3.) Solution. By definition of the operator D we have X, D (Z) = Z(X) for any vector Z V. Applying this to Z = A D(Y ) we see that the right-hand send of (.3.) is equal to to A D(Y )(X). On the other hand, the left-hand side can be rewritten as D(Y )(A(X)). But A D(Y )(X) = D(Y )(A(X)). Let us recall that if V is an Euclidean space, then operator B : V V is called adjoint to A : V V if for any two vectors X, Y V one has A(X), Y = X, B(Y ). The adjoint operator always exist and unique. It is denoted by A. Clearly, (A ) = A. In any orthonormal basis the matrices of an operator and its adjoint are transpose to each other. An 3

14 operator A : V V is called self-adjoint if A = A, or equivalently, if for any two vectors X, Y V one has A(X), Y = X, A(Y ). The statement of Exercise.0 can be otherwise expressed by saying that the adjoint operator A is equal to D A D : V V. In particular, an operator A : V V is self-adjoint if and only if A D = D A. Remark.. As it follows from Proposition.7 and Exercise.0 the matrix of a self-adjoint operator in any orthonormal basis is symmetric. This is not true in an arbitrrary basis..4 Multilinear functions A function l(x, X 2,..., X k ) of k vector arguments X,..., X k V is called k-linear (or multilinear) if it is linear with respect to each argument when all other arguments are fixed. We say bilinear instead of 2-linear. Multilinear functions are also called tensors. Sometimes, one may also say a k-linear form, or simply k-form instead of a k-linear functions. However, we will reserve the term k-form for a skew-symmetric tensors which will be defined in Section 2.3 below. If one fixes a basis v... v n in the space V then with each bilinear function f(x, Y ) one can associate a square n n matrix as follows. Set a ij = f(v i, v j ). Then A = (a ij ) i,j=,...,n is called the matrix of the function f in the basis v,..., v n. For any 2 vectors X = n x i v i, Y = n y j v j we have n n n n f(x, Y ) = f x i v i, y j v j = x i y j f(v i, v j ) = a ij x i y j = X T AY. i= j= i,j= i,j= Exercise.2. How does the matrix of a bilinear function depend on the choice of a basis? Answer. The matrices A and à of the bilinear form f(x, y) in the bases v,..., v n and ṽ,..., ṽ n are related by the formula à = CT AC, where C is the matrix of transition from the basis v... v n 4

15 to the basis ṽ... ṽ n, i.e the matrix whose columns are the coordinates of the basis ṽ... ṽ n in the basis v... v n. Similarly, with a k-linear function f(x,..., X k ) on V and a basis v,..., v n one can associate a k-dimensional matrix A = {a i i 2...i k ; i,..., i k n}, where a i i 2...i k = f(v i,..., v ik ). If X i = n x ij v j, i =,..., k, then j= see Proposition 2. below. n f(x,..., X k ) = a i i 2...i k x i x 2i2... x kik, i,i 2,...i k = 5

16 6

17 Chapter 2 Tensor and exterior products 2. Tensor product Given a k-linear function φ and a l-linear function ψ, one can form a (k + l)-linear function, which will be denoted by φ ψ and called the tensor product of the functions φ and ψ. By definition φ ψ(x,..., X k, X k+,..., X k+l ) := φ(x,..., X k ) ψ(x k+,..., X k+l ). For instance, the tensor product two linear functions l and l 2 is a bilinear function l l 2 defined by the formula l l 2 (U, V ) = l (U)l 2 (V ). Let v... v n be a basis in V and x,..., x n a dual basis in V, i.e. x,..., x n are coordinates of a vector with respect to the basis v,..., v n. The tensor product x i x j is a bilinear function x i x j (Y, Z) = y i z j. Thus a bilinear function f with a matrix A can be written as a linear combination of the functions x i x j as follows: n f = a ij x i x j, i,j= where a ij is the matrix of the form f in the basis v... v n. Similarly any k-linear function f with a k-dimensional matrix A = {a i i 2...i k } can be written (see 2. below) as a linear combination of functions x i x i2 x ik, i, i 2,..., i k n. 7

18 Namely, we have n f = a i i 2...i k x i x i2 x ik. i,i 2,...i k = 2.2 Spaces of multilinear functions All k-linear functions, or k-tensors, on a given n-dimensional vector space V themselves form a vector space, which will be denoted by V k. The space V is, of course, just the dual space V. Proposition 2.. Let v,... v n be a basis of V, and x,..., x k be the dual basis of V formed by coordinate functions with respect to the basis V. Then n k k-linear functions x i x ik, i,..., i k n, form a basis of the space V k. Proof. Take a k-linear function F from V k and evaluate it on vectors v i,..., v ik : F (v i,..., v ik ) = a i...i k. We claim that we have F = a i...i k x i x ik. i,...,i k n Indeed, the functions on the both sides of this equality being evaluated on any set of k basic vectors v i,..., v ik, give the same value a i...i k. The same argument shows that if a i...i k x i i,...,i k n x ik = 0, then all coefficients a i...i k should be equal to 0. Hence the functions x i x ik, i,..., i k n, are linearly independent, and therefore form a basis of the space V k Similar to the case of spaces of linear functions, a linear map A : V W induces a linear map A : W k V k, which sends a k-linear function F W k to a k-linear function A (F ) V k, defined by the formula A (F )(X,..., X k ) = F (A(X ),..., A(X k )) for any vectors X,... X k V. Exercise 2.2. Suppose V is provided with a basis v,..., v n and x i x ik, i,..., i k n, is the corresponding basis of the space V k. Suppose that the map A : V V has a matrix A = (a ij ) in the basis v,..., v n. Find the matrix of the map A : V k V k in the basis x i x ik. 8

19 2.3 Symmetric and skew-symmetric tensors A multilinear function (tensor) is called symmetric if it remains unchanged under the transposition of any two of its arguments: f(x,..., X i,..., X j,..., X k ) = f(x,..., X j,..., X i,..., X k ) Equivalently, one can say that a k-tensor f is symmetric if f(x i,..., X ik ) = f(x,..., X k ) for any permutation i,..., i k of indices,..., k. Exercise 2.3. Show that a bilinear function f(x, Y ) is symmetric if and only if its matrix (in any basis) is symmetric. Notice that the tensor product of two symmetric tensors usually is not symmetric. Example 2.4. Any linear function is (trivially) symmetric. However, the tensor product of two functions l l 2 is not a symmetric bilinear function unless l = l 2. On the other hand, the function l l 2 + l 2 l, is symmetric. A tensor is called skew-symmetric (or anti-symmetric) if it changes its sign when one transposes any two of its arguments: f(x,..., X i,..., X j,..., X k ) = f(x,..., X j,..., X i,..., X k ). Equivalently, one can say that a k-tensor f is anti-symmetric if f(x i,..., X ik ) = ( ) inv(i...i k ) f(x,..., X k ) for any permutation i,..., i k of indices,..., k, where inv(i... i k ) is the number of inversions in the permutation i,..., i k. Recall that two indices i k, i l form an inversion if k < l but i k > i l. The matrix A of a bilinear skew-symmetric function is skew-symmetric, i.e. A T = A. 9

20 Example 2.5. The determinant det(x,..., X n ) (considered as a function of columns X,..., X n of a matrix) is a skew-symmetric n-linear function. Exercise 2.6. Prove that any n-linear skew-symmetric function on R n is proportional to the determinant. Linear functions are trivially anti-symmetric (as well as symmetric). As in the symmetric case, the tensor product of two skew-symmetric functions is not skewsymmetric. We will define below in Section 2.5 a new product, called an exterior product of skewsymmetric functions, which will again be a skew-symmetric function. 2.4 Symmetrization and anti-symmetrization The following constructions allow us to create symmetric or anti-symmetric tensors from arbitrary tensors. Let f(x,..., X k ) be a k-tensor. Set f sym (X,..., X k ) := f(x i,..., X ik ) (i...i k ) and f asym (X,..., X k ) := ( ) inv(i,...,i k ) f(x i,..., X ik ) (i...i k ) where the sums are taken over all permutations i,..., i k of indices,..., k. The tensors f sym and f asym are called, respectively, symmetrization and anti-symmetrization of the tensor f. It is now easy to see that Proposition 2.7. The function f sym is symmetric. The function f asym is skew-symmetric. If f is symmetric then f sym = k!f and f asym = 0. Similarly, if f is anti-symmetric then f asym = k!f, f sym = 0. Exercise 2.8. Let x,..., x n be coordinate function in R n. Find (x x 2... x n ) asym. Answer. The determinant. 20

21 2.5 Exterior product For our purposes skew-symmetric functions will be more important. Thus we will concentrate on studying operations on them. Skew-symmetric k-linear functions are also called exterior k-forms. Let φ be an exterior k-form and ψ an exterior l-form. We define an exterior (k + l)-form ψ ψ, the exterior product of φ and ψ, as In other words, φ ψ := k!l! (φ ψ)asym. φ ψ(x,..., X k, X k+,..., X k+l ) = k!l! where the sum is taken over all permutations of indices,..., k + l. i,...i k+l ( ) inv(i,...,ik+l) φ(x i..., X ik )ψ(x ik+,..., X ik+l ), Note that because the anti-symmetrization of an anti-symmetric k-tensor amounts to its multiplication by k!, we can also write φ ψ(x,..., X k+l ) = i <...<i k,i k+ <...<i k+l ( ) inv(i,...,ik+l) φ(x i,..., X ik )ψ(x ik+,..., X ik+l ), where the sum is taken over all permutations i,..., i k+l of indices,..., k + l. Exercise 2.9. The exterior product operation has the following properties: For any exterior k-form φ and exterior l-form ψ we have φ ψ = ( ) kl ψ φ. Exterior product is linear with respect to each factor: (φ + φ 2 ) ψ = φ ψ + φ 2 ψ (λφ) ψ = λ(φ ψ) for k-forms φ, φ, φ 2, l-form ψ and a real number λ. Exterior product is associative: (φ ψ) ω = φ (ψ ω). 2

22 First two properties are fairly obvious. To prove associativity one can check that both sides of the equality (φ ψ) ω = φ (ψ ω) are equal to k!l!m! (φ ψ ω)asym. In particular, if φ, ψ and ω are -forms, i.e. if k = l = m = then ψ φ ω = (φ ψ ω) asym. This formula can be generalized for computing the exterior product of any number of -forms: φ φ k = (φ φ k ) asym. (2.5.) Example 2.0. x x 2 = x x 2 x 2 x. For 2 vectors, U = u., V = v., we have u n v n x x 2 (U, V ) = u v 2 u 2 v = u v u 2 v 2. For 3 vectors U, V, W we have x x 2 x 3 (U, V, W ) = = x x 2 (U, V )x 3 (W ) + x x 2 (V, W )x 3 (U) + x x 2 (W, U)x 3 (V ) = u v w (u v 2 u 2 v )w 3 + (v w 2 v 2 w )u 3 + (w u 2 w 2 u )v 3 = u 2 v 2 w 2 u 3 v 3 w 3. The last equality is just the expansion formula of the determinant according to the last row. Proposition 2.. Any exterior 2-form f can be written as f = a ij x i x j i<j n 22

23 Proof. We had seen above that any bilinear form can be written as f = ij a i,j x i x j. If f is skew-symmetric then the matrix A = (a ij ) is skew-symmetric, i.e. a ii = 0, a ij = a ji for i j. Thus, f = a ij (x i x j x j x i ) = a ij x i x j. i<j n i<j n Exercise 2.2. Prove that any exterior k-form f can be written as f = a i...i k x i x i2... x ik. i,<...<i k n The following proposition can be proven by induction over k, similar to what has been done in Example 2.0 for the case k = 3. Proposition 2.3. For any k -forms l,..., l k and k vectors X,..., X k we have l (X )... l (X k ) l l k (X,..., X k ) = (2.5.2) l k (X )... l k (X k ) Corollary 2.4. The -forms l,..., l k are linearly dependent as vectors of V if and only if l... l k = 0. In particular, l... l k = 0 if k > n = dimv. Proof. If l,..., l k are dependent then for any vectors X,..., X k V the rows of the determinant in the equation (2.3) are linearly dependent. Therefore, this determinant is equal to 0, and hence l... l k = 0. In particular, when k > n then the forms l,..., l k are dependent (because dimv = dimv = n). On the other hand, if l,..., l k are linearly independent, then the vectors l,..., l k V can be completed to form a basis l,..., l k, l k+,..., l n of V. According to Exercise.2 there exists a basis w,..., w n of V that is dual to the basis l,..., l n of V. In other words, l,..., l n can be viewed as coordinate functions with respect to the basis w,..., w n. In particular, we have l i (w j ) = 0 if i j and l i (w i ) = for all i, j =,..., n. Hence we have l (w )... l (w k )... 0 l l k (w,..., w k ) = = =, l k (w )... l k (w k )

24 i.e. l l k 0. Proposition 2.3 can be also deduced from formula (2.5.). Corollary 2.4 and Exercise 2.2 imply that there are no non-zero k-forms on an n-dimensional space for k > n. 2.6 Spaces of symmetric and skew-symmetric tensors As was mentioned above, k-tensors on a vector space V form a vector space under the operation of addition of functions and multiplication by real numbers. We denoted this space by V k. Symmetric and skew-symmetric tensors form subspaces of this space V k, which we denote, respectively, by S k (V ) and Λ k (V ). In particular, we have V = S (V ) = Λ (V ). Exercise 2.5. What is the dimension of the spaces S k (V ) and Λ k (V )? Answer. dim Λ k (V ) = n k = n! k!(n k)! dim S k (V ) = (n + k )!. k!(n )! The basis of Λ k (V ) is formed by exterior k-forms x i x i2... x ik, i < i 2 <... < i k n. 2.7 Operator A on spaces of tensors For any linear operator A : V W we introduced above in Section.3 the notion of a dual linear operator A : W V. Namely A (l) = l A for any element l V, which is just a linear function on V. In this section we extend this construction to k-tensors for k, i.e. we will define a map A : W k V k. 24

25 Given a k-tensor φ W k and k vectors X,..., X k V we define A (φ)(x,..., X k ) = φ (A(X ),..., A(X k )). Note that if φ is symmetric, or anti-symmetric, so is A (φ). Hence, the map A also induces the maps S k (W ) S k (V ) and Λ k (W ) Λ k (V ). We will keep the same notation A for both of these maps as well. Proposition 2.6. Let A : V W be a linear map. Then. A (φ ψ) = A (φ) A (ψ) for any φ W k, ψ W l ; 2. A (φ asym ) = (A (φ)) asym, A (φ sym ) = (A (φ)) sym ; 3. A (φ ψ) = A (φ) A (ψ) for any φ Λ k (W ), ψ Λ l (W ). If B : W U is another linear map then (B A) = A B. Proof.. Take any k + l vectors X,..., X k+l V. Then by definition of the operator A we have A (φ ψ)(x,..., X k+l ) =φ ψ(a(x ),..., A(X n+k ) = φ(a(x ),..., A(X k )ψ(a(x k+ ),..., A(X n+k )) = A (φ)(x,..., X k )A (ψ)(x k+,..., X k+n ) = A (φ) A (ψ)(x,..., X k+l ). 2. Given k vectors X,..., X k V we get A (φ asym )(X,..., X k ) = φ asym (A(X ),..., A(X k )) = (i...i k ) (i...i k ) ( ) inv(i,...,i k ) A (φ)(x i,..., X ik ) = (A (φ)) asym (X,..., X k ), ( ) inv(i,...,i k ) φ(a(x i ),..., A(X ik )) = where the sum is taken over all permutations i,..., i k of indices,..., k. Similarly one proves that A (φ sym ) = (A (φ)) sym. 3. A (φ ψ) = k!l! A ((φ ψ) asym )) = k!l! (A (φ ψ)) asym = A (φ) A (ψ). The last statement of Proposition 2.6 is straightforward and its proof is left to the reader. 25

26 Let us now discuss how to compute A (φ) in coordinates. Let us fix bases v,..., v m and w,..., w n in spaces V and W. Let x,..., x m and y,..., y n be coordinates and a... a m A =... a n... a nm be the matrix of a linear map A : V W in these bases. Note that the map A in these coordinates is given by n linear coordinate functions: y = l (x,..., x m ) = a x + a 2 x a m x m y 2 = l 2 (x,..., x m ) = a 2 x + a 22 x a 2m x m... y n = l n (x,..., x k ) = a n x + a n2 x a nm x n We have already computed in Section.3 that A (y k ) = l k = m a kj x j, k =,..., n. Indeed, the coefficients of the function l k = A (y k ) form the k-th column of the transpose matrix A T. Hence, using Proposition 2.6 we compute: j= A (y j y jk ) = l j l jk and A (y j y jk ) = l j l jk. Consider now the case when V = W, n = m, and we use the same basis v,..., v n in the source and target spaces. Proposition 2.7. A (x x n ) = det A x x n. Note that the determinant det A is independent of the choice of the basis. Indeed, the matrix of a linear map changes to a similar matrix C AC in a different basis, and det C AC = det A. Hence, we can write det A instead of det A, i.e. attribute the determinant to the linear operator A rather than to its matrix A. 26

27 Proof. We have n n A (x x n ) = l l n = a i x i a nin x in = n a i... a nin x i x in. i,...,i n= i = i n= Note that the in the latter sum all terms with repeating indices vanish, and hence we can replace this sum by a sum over all permutations of indices,..., n. Thus, we can continue A (x x n ) = a i... a nin x i x in = i,...,i n ( ) inv(i,...,in) a i... a nin x x n = det A x x n. i,...,i n Exercise 2.8. Apply the equality A (x x k x k+ x n ) = A (x x k ) A (x k+ x n ) for a map A : R n R n to deduce the formula for expansion of a determinant according to its first k rows: det A = i < <i k, j < <j n k ; i m j l ( ) inv(i,...,jn k) a,i... a,ik a k+,j... a k+,jn k a k,i... a k,ik a n,j... a n,jn k 27

28 28

29 Chapter 3 Orientation and Volume 3. Orientation We say that two bases v,..., v k and w,..., w k of a vector space V define the same orientation of V if the matrix of transition from one of these bases to the other has a positive determinant. Clearly, if we have 3 bases, and the first and the second define the same orientation, and the second and the third define the same orientation then the first and the third also define the same orientation. Thus, one can subdivide the set of all bases of V into the two classes. All bases in each of these classes define the same orientation; two bases chosen from different classes define opposite orientation of the space. To choose an orientation of the space simply means to choose one of these two classes of bases. There is no way to say which orientation is positive or which is negative it is a question of convention. For instance, the so-called counter-clockwise orientation of the plane depends from which side we look at the plane. The positive orientation of our physical 3-space is a physical, not mathematical, notion. Suppose we are given two oriented spaces V, W of the same dimension. An invertible linear map (= isomorphism) A : V W is called orientation preserving if it maps a basis which defines the given orientation of V to a basis which defines the given orientation of W. Any non-zero exterior n-form η on V induces an orientation of the space V. Indeed, the preferred set of bases is characterized by the property η(v,..., v n ) > 0. 29

30 3.2 Orthogonal transformations Let V be a Euclidean vector space. Recall that a linear operator U : V V is called orthogonal if it preserves the scalar product, i.e. if U(X), U(Y ) = X, Y, (3.2.) for any vectors X, Y V. Recall that we have U(X), U(Y ) = X, U (U(Y )), where U : V V is the adjoint operator to U, see Section.3 above. Hence, the orthogonality of an operator U is equivalent to the identity U U = Id, or U = U. Here we denoted by Id the identity operator, i.e. Id(X) = X for any X V. Let us recall, see Exercise.0, that the adjoint operator U is related to the dual operator U : V V by the formula U = D U D. Hence, for an orthogonal operator U, we have D U D = U, i.e. U D = D U. (3.2.2) Let v,..., v n be an orthonormal basis in V and U be the matrix of U in this basis. The matrix of the adjoint operator in an orthonormal basis is the transpose of the matrix of this operator. Hence, the equation U U = Id translates into the equation U T U = E, or equivalently UU T = E, or U = U T for its matrix. Matrices, which satisfy this equation are called orthogonal. If we write u... u n U = , u n... u nn then the equation U T U = E can be rewritten as, if k = j, u ki u ji =. i 0, if k j, 30

31 Similarly, the equation UU T = E can be rewritten as, if k = j, u ik u ij =. i 0, if k j, The above identities mean that columns (and rows) of an orthogonal matrix U form an orthonormal basis of R n with respect to the dot-product. In particular, we have = det(u T U) = det(u T ) det U = (det U) 2, and hence det U = ±. In other words, the determinant of any orthogonal matrix is equal ±. We can also say that the determinant of an orthogonal operator is equal to ± because the determinant of the matrix of an operator is independent of the choice of a basis. Orthogonal transformations with det = preserve the orientation of the space, while those with det = reverse it. Composition of two orthogonal transformations, or the inverse of an orthogonal transformation is again an orthogonal transformation. The set of all orthogonal transformations of an n-dimensional Euclidean space is denoted by O(n). Orientation preserving orthogonal transformations sometimes called special, and the set of special orthogonal transformations is denoted by SO(n). For instance O() consists of two elements and SO() of one: O() = {, }, SO() = {}. SO(2) consists of rotations of the plane, while O(2) consists of rotations an reflections with respect to lines. 3.3 Determinant and Volume We begin by recalling some facts from Linear Algebra. Let V be an n-dimensional Euclidean space with an inner product,. Given a linear subspace L V and a point x V, the projection proj L (x) is a vector y L which is uniquely characterized by the property x y L, i.e. x y, z = 0 for any z L. The length x proj L (x) is called the distance from x to L; we denote it by dist(x, L). Let U,..., U k V be linearly independent vectors. The k-dimensional parallelepiped spanned by vectors U,..., U k is, by definition, the set { k } P (U,..., U k ) = λ j U j ; 0 λ,..., λ k Span(U,..., U k ). 3

32 Given a k-dimensional parallelepiped P = P (U,..., U k ) we will define its k-dimensional volume by the formula VolP = U dist(u 2, Span(U ))dist(u 3, Span(U, U 2 ))... dist(u k, Span(U,..., U k )). (3.3.) Of course we can write dist(u, 0) instead of U. This definition agrees with the definition of the area of a parallelogram, or the volume of a 3-dimensional parallelepiped in the elementary geometry. Proposition 3.. Let v,..., v n be an orthonormal basis in V. Given n vectors U,..., U n let us denote by U the matrix whose columns are coordinates of these vectors in the basis v,..., v n : u... u n U :=. u n... u nn Then Vol P (U,..., U n ) = det U. Proof. If the vectors U,..., U n are linearly dependent then Vol P (U,..., U n ) = det U = 0. Suppose now that the vectors U,..., U n are linearly independent, i.e. form a basis. Consider first the case where this basis is orthonormal. Then the matrix U is orthogonal. i.e. UU T = E, and hence det U = ±. But in this case Vol P (U,..., U n ) =, and hence Vol P (U,..., U n ) = det U. Now let the basis U,..., U n be arbitrary. Let us apply to it the Gram-Schmidt orthonormalization process. Recall that this process consists of the following steps. First, we normalize the vector U, then subtract from U 2 its projection to Span(U ), Next, we normalize the new vector U 2, then subtract from U 3 its projection to Span(U, U 2 ), and so on. At the end of this process we obtain an orthonormal basis. It remains to notice that each of these steps affected Vol P (U,..., U n ) and det U in a similar way. Indeed, when we multiplied the vectors by a positive number, both the volume and the determinant were multiplied by the same number. When we subtracted from a vector U k its projection to Span(U,..., U k ), this affected neither the volume nor the determinant. Corollary Let x,..., x n be a Cartesian coordinate system. Then Vol P (U,..., U n ) = x... x n (U,..., U n ). i.e. a coordinate system with respect to an orthonormal basis 32

33 2. Let A : V V be a linear map. Then Vol P (A(U ),..., A(U n )) = det A Vol P (U,..., U n ). Proof.. According to 2.3, x... x n (U,..., U n ) = det U. 2. x... x n (A(U ),..., A(U n )) = A (x x n )(U,..., U n ) = det Ax... x n (U,..., U n ). In view of Proposition 3. and the first part of Corollary 3.2 the value x... x n (U,..., U n ) = det U is called sometimes the signed volume of the parallelepiped P (U,..., U n ). It is positive when the basis U,..., U n defines the given orientation of the space V, and it is negative otherwise. Note that x... x k (U,..., U k ) for 0 k n is the signed k-dimensional volume of the orthogonal projection of the parallelepiped P (U,..., U k ) to the coordinate subspace {x k+ = = x n = 0}. For instance, let ω be the 2-form x x 2 + x 3 x 4 on R 4. Then for any two vectors U, U 2 R 4 the value ω(u, U 2 ) is the sum of signed areas of projections of the parallelogram P (U, U 2 ) to the coordinate planes spanned by the two first and two last basic vectors. 3.4 Volume and Gram matrix In this section we will compute the Vol k P (v,..., v k ) in the case when the number k of vectors is less than the dimension n of the space. Let V be an Euclidean space. Given vectors v,..., v k V we can form a k k-matrix v, v... v, v k G(v,..., v k ) = , (3.4.) v k, v... v k, v k which is called the Gram matrix of vectors v,..., v k. Suppose we are given Cartesian coordinate system in V and let us form a matrix C whose columns are coordinates of vectors v,..., v k. Thus the matrix C has n rows and k columns. Then G(v,..., v k ) = C T C, 33

34 because in Cartesian coordinates the scalar product looks like the dot-product. We also point out that if k = n and vectors v,..., v n form a basis of V, then G(v,..., v k ) is just the matrix of the bilinear function X, Y in the basis v,..., v n. It is important to point out that while the matrix C depends on the choice of the basis, the matrix G does not. Proposition 3.3. Given any k vectors v,..., v k in an Euclidean space V the volume Vol k P (v,..., v k ) can be computed by the formula Vol k P (v,..., v k ) 2 = det G(v,..., v k ) = det C T C, (3.4.2) where G(v,..., v k ) is the Gram matrix and C is the matrix whose columns are coordinates of vectors v,..., v k in some orthonormal basis. Proof. Suppose first that k = n. Then according to Proposition 3. we have Vol k P (v,..., v k ) = det C. But det C T C = det C 2, and the claim follows. Let us denote vectors of our orthonormal basis by w,..., w n. Consider now the case when Span(v,..., v k ) Span(w,..., w k ). (3.4.3) In this case the elements in the j-th row of the matrix C are zero if j > k. Hence, if we denote by C the square k k matrix formed by the first k rows of the matrix C, then C T C = C T C and thus det C T C = det C T C. But det CT C = Volk P (v,..., v k ) in view of our above argument in the equi-dimensional case applied to the subspace Span(w,..., w k ) V, and hence Vol 2 k P (v,..., v k ) = det C T C = det G(v,..., v k ). But neither Vol k P (v,..., v k ), nor the Gram matrix G(v,..., v k ) depends on the choice of an orthonormal basis. On the other hand, using Gram-Schmidt process one can always find an orthonormal basis which satisfies condition (3.4.3). Remark 3.4. Note that det G(v,..., v k ) 0 and det G(v,..., v k ) = 0 if an and only if the vectors v,..., v k are linearly dependent. 34

35 Chapter 4 Dualities 4. Duality between k-forms and (n k)-forms on a n-dimensional Euclidean space V Let V be an n-dimensional vector space. As we have seen above, the space Λ k (V ) of k-forms, and the space Λ n k (V n! ) of (n k)-forms have the same dimension k!(n k)! ; these spaces are therefore isomorphic. Suppose that V is an oriented Euclidean space, i.e. it is supplied with an orientation and an inner product,. It turns out that in this case there is a canonical way to establish this isomorphism which will be denoted by : Λ k (V ) Λ n k (V ). Definition 4.. Let α be a k-form. Then given any vectors U,..., U n k, the value α(u,..., U n k ) can be computed as follows. If U,..., U n k are linearly dependent then α(u,..., U n k ) = 0. Otherwise, let S denote the orthogonal complement to the space S = Span(U,..., U n k ). Choose a basis Z,..., Z k of S such that Vol k (Z,..., Z k ) = Vol n k (U,..., U n k ) and the basis Z,..., Z k, U,..., U n k defines the given orientation of the space V. Then α(u,..., U n k ) = α(z,..., Z k ). (4..) 35

36 Let us first show that Lemma 4.2. α is a (n k)-form, i.e. α is skew-symmetric and multilinear. Proof. To verify that α is skew-symmetric we note that for any i < j n q the bases Z, Z 2,..., Z k, U,..., U i,..., U j,..., U n k and Z, Z 2,..., Z k, U,..., U j,..., U i,..., U n k define the same orientation of the space V, and hence α(u,..., U j,..., U i,..., U n k ) = α( Z, Z 2,..., Z k ) = α(z, Z 2,..., Z k ) = α(u,..., U i,..., U j,..., U n k ). Hence, in order to check the multi-linearity it is sufficient to prove the linearity of α with respect to the first argument only. It is also clear that α(λu,..., U n k ) = λα(u,..., U n k ). (4..2) Indeed, multiplication by λ 0 does not change the span of the vectors U,..., U n q, and hence if α(u,..., U n k ) = α(z,..., Z k ) then α(λu,..., U n k ) = α(λz,..., Z k ) = λα(z,..., Z k ). Thus it remains to check that α(u + Ũ, U 2,..., U n k ) = α(u, U 2,..., U n k ) + α(ũ, U 2,..., U n k )). Let us denote L := Span(U 2,..., U n k ) and observe that proj L (U + Ũ) = proj L (U ) + proj L (Ũ). Denote N := U proj L (U ) and Ñ := Ũ proj L (Ũ). The vectors N and Ñ are normal components of U and Ũ with respect to the subspace L, and the vector N + Ñ is the normal component of U + Ũ with respect to L. Hence, we have α(u,..., U n k ) = α(n,..., U n k ), α(ũ,..., U n k ) = α(ñ,..., U n k), and α(u + Ũ,..., U n k ) = α(n + Ñ,..., U n k). Indeed, in each of these three cases, 36

37 - vectors on both side of the equality span the same space; - the parallelepiped which they generate have the same volume, and - the orientation which these vectors define together with a basis of the complementary space remains unchanged. Hence, it is sufficient to prove that α(n + Ñ, U 2,..., U n k ) = α(n, U 2,..., U n k ) + α(ñ, U 2,..., U n k ). (4..3) If the vectors N and Ñ are linearly dependent, i.e. one of them is a multiple of the other, then (4..3) follows from (4..2). Suppose now that N and Ñ are linearly independent. Let L denote the orthogonal complement of L = Span(U 2,..., U n k ). Then dim L = k + and we have N, Ñ L. Let us denote by M the plane in L spanned by the vectors N and Ñ, and by M its orthogonal complement in L. Note that dim M = k. Choose any orientation of M so that we can talk about counter-clockwise rotation of this plane. Let Y, Ỹ M be vectors obtained by rotating N and Ñ in M counter-clockwise by the angle π 2. Then Y + Ỹ can be obtained by rotating N + Ñ in M counter-clockwise by the same angle π 2. Let us choose in M a basis Z 2,..., Z k such that Vol k P (Z 2,..., Z k ) = Vol n k P (U 2,..., U n k ). Note that the orthogonal complements to Span(N, U 2,..., U n k ), Span(Ñ, U 2,..., U n k ), and to Span(N +Ñ, U 2,..., U n k ) in V coincide, respectively, with the orthogonal complements to the the vectors N, Ñ and to N + Ñ in L. In other words, we have Next, we observe that (Span(N, U 2,..., U n k )) V = Span(Y, Z 2,..., Z k ), ( ) Span(Ñ, U 2,..., U n k ) ( Span(N + Ñ, U 2,..., U n k ) V = Span(Ỹ, Z 2,..., Z k ) and ) V = Span(Y + Ỹ, Z 2,..., Z k ). Vol n k P (N, U 2,..., U n k ) = Vol k P (Y, Z 2,..., Z k ), 37

38 Vol n k P (Ñ, U 2,..., U n k ) = Vol k P (Ỹ, Z 2,..., Z k ) and Vol n k P (N + Ñ, U 2,..., U n k ) = Vol k P (Y + Ỹ, Z 2,..., Z k ). Consider the following 3 bases of V : Y, Z 2,..., Z k, N, U 2,..., U n k, Ỹ, Z 2,..., Z k, Ñ, U 2,..., U n k, Y + Ỹ, Z 2,..., Z k, N + Ñ, U 2,..., U n k, and observe that all three of them define the same a priori given orientation of V. Thus, by definition of the operator we have: α(n + Ñ, U 2,..., U n k ) = α(y + Ỹ, Z 2,..., Z k ) = α(y, Z 2,..., Z k ) + α(ỹ, Z 2,..., Z k ) = α(n, U 2,..., U n k ) + α(ñ, U 2,..., U n k ). This completes the proof that α is an (n k)-form. Thus the map α α defines a map : Λ k (V ) Λ n k (V ). Clearly, this map is linear. In order to check that is an isomorphism let us choose an orthonormal basis in V and consider the coordinates x,..., x n V corresponding to that basis. Let us recall that the forms x i x i2 x ik, i < i 2 < < i k n, form a basis of the space Λ k (V ). Lemma 4.3. x i x i2 x ik = ( ) inv(i,...,i k,j,...,j n k ) x j x j2 x jn k, (4..4) where j < < j n k is the set of indices, complementary to i,..., i k. In other words, i,..., i k, j,..., j n k is a permutation of indices,..., n. Proof. Evaluating (x i x ik ) on basic vectors v j,..., v jn k, j < < j n q n, we get 0 unless all the indices j,..., j n k are all different from i,..., i k, while in the latter case we get (x i x ik )(v j,..., v jn k ) = ( ) inv(i,...,i k,j,...,j n k ). 38

39 Hence, x i x i2 x ik = ( ) inv(i,...,i k,j,...,j n k ) x j x j2 x jn k. Thus establishes a to correspondence between the bases of the spaces Λ k (V ) and the space Λ n k (V ), and hence it is an isomorphism. Note that by linearity for any form α = a i...i k x i x ik ) we have i < <i k n Examples. α = a i...i k (x i x ik ). i < <i k n. C = Cx x n ; in other words the isomorphism acts on constants (= 0-forms) by multiplying them by the volume form. 2. In R 3 we have x = x 2 x 3, x 2 = x x 3 = x 3 x, x 3 = x x 2, (x x 2 ) = x 3, (x 3 x ) = x 2, (x 2 x 3 ) = x. 3. More generally, given a -form l = a x + + a n x n we have l = a x 2 x n a 2 x x 3 x n + + ( ) n a n x x n. In particular for n = 3 we have (a x + a 2 x 2 + a 3 x 3 ) = a x 2 x 3 + a 2 x 3 x + a 3 x x 2. Proposition = ( ) k(n k) Id, i.e. ( ω) = ( ) k(n k) ω for any k-form ω. In particular, if dimension n = dim V is odd then 2 = Id. If n is even and ω is a k-form then ( ω) = ω if k is even, and ( ω) = ω if k is odd. 39

40 Proof. It is sufficient to verify the equality ( ω) = ( ) k(n k) ω for the case when ω is a basic form, i.e. ω = x i x ik, i < < i k n. We have (x i x ik ) = ( ) inv(i,...,i k,j,...,j n k ) x j x j2 x jn k and (x j x j2 x jn k ) = ( ) inv(j,...,j n k,i,...,i k ) x i x ik. But the permutations i... i k j... j n k and j... j n k i... i k differ by k(n k) transpositions of pairs of its elements. Hence, we get ( ) inv(i,...,i k,j,...,j n k ) = ( ) k(n k) ( ) inv(j,...,j n k,i,...,i k ), and, therefore, ( (x i x ik ) ) = ( ( ) inv(i,...,i k,j,...,j n k ) x j x j2 x jn k ) = ( ) inv(i,...,i k,j,...,j n k ) (x j x j2 x jn k ) = ( ) inv(i,...,i k,j,...,j n k )+inv(j,...,j n k,i,...,i k ) x i x ik = ( ) k(n k) x i x ik. Exercise 4.5. (a) For any special orthogonal operator A the operators A and commute, i.e. A = A. (b) Let A be an orthogonal matrix of order n with det A =. Prove that for any k {,..., n} the absolute value of each k-minor M of A is equal to the absolute value of its complementary minor of order (n k). (Hint: Apply (a) to the form x i x ik ). 40

41 (c) Let V be an oriented 3-dimensional Euclidean space. Prove that for any two vectors X, Y V, their cross-product can be written in the form X Y = D ( (D(X) D(Y ))). 4.2 Euclidean structure on the space of exterior forms Suppose that the space V is oriented and Euclidean, i.e. it is endowed with an inner product, and an orientation. Given two forms α, β Λ k (V ), k = 0,..., n, let us define α, β = (α β). Note that α β is an n-form for every k, and hence, α, β is a 0-form, i.e. a real number. Proposition The operation, defines an inner product on Λ k (V ) for each k = 0,..., n. 2. If A : V V is a special orthogonal operator then the operator A : Λ k (V ) Λ k (V ) is orthogonal with respect to the inner product,. Proof.. We need to check that α, β is a symmetric bilinear function on Λ k (V ) and α, α > 0 unless α = 0. Bilinearity is straightforward. Hence, it is sufficient to verify the remaining properties for basic vectors α = x i x ik, β = x j x jk, where i < < i k n, j < < j k n. Here (x,..., x n ) is any Cartersian coordinates in V which define its given orientation. Note that α, β = 0 = β, α unless i m = j m for all m =,..., k, and in the the latter case we have α = β. Furthermore, we have α, α = (α α) = (x x n ) = > The inner product, is defined only in terms of the Euclidean structure and the orientation of V. Hence, for any special orthogonal operator A (which preserves these structures) the induced operator A : Λ k (V ) Λ k (V ) preserves the inner product,. Note that we also proved that the basis of k-forms x i x ik, i < < i k n, is orthonormal with respect to the scalar product,. Hence, we get 4

42 Corollary 4.7. Suppose that a k-form α can be written in Cartesian coordinates as α = a i...i k x i x ik. i < <i k n Then α 2 = α, α = a 2 i...i k. i < <i k n Exercise 4.8. Show that if α, β are -forms on an Euclidean space V. Then α, β = D (α), D (β), i.e the scalar product, on V is the push-forward by D of the scalar product, on V. Corollary 4.9. Let V be a Euclidean n-dimensional space. Choose an orthonormal basis e,..., e n in V. Then for any vectors Z = (z,..., z n ),..., Z k = (z k,..., z nk ) V we have (Vol k P (Z,..., Z k )) 2 = Zi 2,...,i k, (4.2.) i < <i k n where z i... z i k Z i,...,i k = z ik... z ik k Proof. Consider linear functions l j = D(Z j ) = n z ij x i V, j =,..., k. Then i= n n l l k = z i jx i z ik jx ik = i = i,...,i k z i... z ik x i x ik = i k = i < <i k n Z i,...,i k x i... x ik. (4.2.2) In particular, if one has Z,... Z k Span(e,..., e k ) then Z...k = Vol P (Z,..., Z k ) and hence l l k = Z...k x x k = Vol P (Z,..., Z k )x x k, which yields the claim in this case. 42

43 In the general case, according to Proposition 4.7 we have l l k 2 = Zi 2,...,i k, (4.2.3) i < <i k n which coincides with the right-hand side of (4.2.). Thus it remains to check to that l l k = Vol k P (Z,..., Z k ). Given any orthogonal transformation A : V V we have, according to Proposition 4.6, the equality l l k = A l A l k. (4.2.4) We also note that any orthogonal transformation B : V V preserves k-dimensional volume of all k-dimensional parallelepipeds: Vol k P (Z,..., Z k ) = Vol k P (B(Z ),..., B(Z k )). (4.2.5) On the other hand, there exists an orthogonal transformation A : V V such that A (Z ),..., A (Z k ) Span(e,..., e k ). Denote Z j := A (Z j ), j =,..., k. Then, according to (3.2.2) we have lj := D( Z j ) = D(A (Z j )) = A (D(Z j )) = A l j. As was pointed out above we then have Vol k P ( Z,..., Z k ) = l l k = A l A l k, (4.2.6) and hence, the claim follows from (4.2.4) and (4.2.5) applied to B = A. We recall that an alternative formula for computing Vol k P (Z,..., Z k ) was given earlier in Proposition Contraction Let V be a vector space and φ Λ k (V ) a k-form. Define a (k )-form ψ = v φ by the formula ψ(x,..., X k ) = φ(v, X,..., X k ) for any vectors X,..., X k V. We say that the form ψ is obtained by a contraction of φ with the vector v. Sometimes, this operation is called also an interior product of φ with v and denoted by i(v)φ instead of v ψ. In these notes we will not use this notation. 43

44 Proposition 4.0. Contraction is a bilinear operation, i.e. (v + v 2 ) φ = v φ + v 2 φ (λv) φ = λ(v φ) v (φ + φ 2 ) = v φ + v φ 2 v (λφ) = λ(v φ). Here v, v, v 2 V ; φ, φ, φ 2 Λ k (V ); λ R. The proof is straightforward. Let φ be a non-zero n-form. Then we have Proposition 4.. The map : V Λ n (V ), defined by the formula (v) = v φ is an isomorphism between the vector spaces V and Λ n (V ). Proof. Take a basis v,..., v n. Let x,..., x n V be the dual basis, i.e. the corresponding coordinate system. Then φ = ax... x n, where a 0. To simplify the notation let us assume that a =, so that φ = x... x n. Let us compute the images v i v i φ = φ, i =,..., k of the basic vectors. Let us write n a j x x j x j+ x n. Then v i φ(v,..., v l, v l+,..., v n ) n = a j (x x j x j+ x n (v,..., v l, v l+,..., v n )) = a l, (4.3.) 44

45 but on the other hand, v i φ(v,..., v l, v l+,..., v n ) = φ(v i, v,..., v l, v l+,..., v n ) ( ) i, l = i ; = ( ) i φ(v,..., v l, v i, v l+,..., v n ) = 0, otherwise (4.3.2) Thus, v i φ = ( ) i x x i x i+ x n. Hence, the map sends a basis of V into a basis of Λ n (V ), and therefore it is an isomorphism. Take a vector v = n a j v j. Then we have v (x x n ) = n ( ) i a i x x i x i+ x n. (4.3.3) This formula can be interpreted as the formula of expansion of a determinant according to the first column (or the first row). Indeed, for any vectors U,..., U n we have v φ(u,..., U n ) = det(v, U,..., U n ) = a u,... u,n a n u n,... u n,n, where u,i. u n,i are coordinates of the vector U i V in the basis v,..., v n. On the other hand, n v φ(u,..., U n ) = ( ) i a i x x i x i+ x n (U,..., U n ) u 2,... u 2,n u = a 3,... u 3,n + + ( ) n a n u n,... u n,n u,... u,n u 2,... u 2,n u n,... u n,n. (4.3.4) 45

46 Suppose that dim V = 3. Then the formula (4.3.3) can be rewritten as v (x x 2 x 3 ) = a x 2 x 3 + a 2 x 3 x + a 3 x x 2, where a a 2 a 3 are coordinates of the vector V. Let us describe the geometric meaning of the operation. Set ω = v (x x 2 x 3 ). Then ω(u, U 2 ) is the volume of the parallelogram defined by the vectors U, U 2 and v. Let ν be the unit normal vector to the plane L(U, U 2 ) V. Then we have ω(u, U 2 ) = Area P (U, U 2 ) v, ν. If we interpret v as the velocity of a fluid flow in the space V then ω(u, U 2 ) is just an amount of fluid flown through the parallelogram Π generated by vectors U and U 2 for the unit time. It is called the flux of v through the parallelogram Π. Let us return back to the case dim V = n. Exercise 4.2. Let α = x i x ik, i < < i k n and v = (a,..., a n ). Show that k v α = ( ) j+ a ij x i... x ij x ij+... x ik. j= The next proposition establishes a relation between the isomorphisms, and D. Proposition 4.3. Let V be a Euclidean space, and x,..., x n be coordinates in an orthonormal basis. Then for any vector v V we have Dv = v (x x n ). Proof. Let v = (a,..., a n ). Then Dv = a x + + a n x n and Dv = a x 2 x n a 2 x x 3 x n x + + ( ) n a n x x n. But according to Proposition 4. the (n )-form v (x x n is defined by the same formula. 46

47 We finish this section by the proposition which shows how the contraction operation interacts with the exterior product. Proposition 4.4. Let α, β be forms of order k and l, respectively, and v a vector. Then v (α β) = (v α) β + ( ) k α (v β), provided that α β 0. Exercise 4.5. Show that the statement may be wrong if α β = 0. Proof. Note that given any indices k,... k m (not necessarily ordered) we have e i x k x km = 0 if i / {k,..., k m } and e i x k x km = ( ) J x k... i x km, where J = J(i; k,..., k m ) is the number of variables ahead of x i. By linearity it is sufficient to consider the case when v, α, β are basic vector and forms, i.e. v = e i, α = x i x ik, β = x j... x jl. If v α 0 then v α = ( ) J x i... i x ik and if v β 0 then v β = ( ) J x j... i x jl, where J = J(i; i,..., i k ), J = J(i; j,..., j l ). Note that v (α β) 0 only if e i enters exactly one of the products α or β. If it enters α then v (α β) = ( ) J x i... i x ik x j x jl = (v α) β, while α (v β) = 0. Similarly, if it enters β then v (α β) = ( ) J +m x i x ik x j... i x jl = ( ) k α (v β), while v α β = 0. 47

48 48

49 Chapter 5 Complex vector spaces 5. Complex numbers The space R 2 can be endowed with an associative and commutative multiplication operation. This operation is uniquely determined by three properties: it is a bilinear operation; the vector (, 0) is the unit; the vector (0, ) satisfies (0, ) 2 = (0, ). The vector (0, ) is usually denoted by i, and we will simply write instead of the vector (, 0). Hence, any point (a, b) R 2 can be written as a + bi, where a, b R, and the product of a + bi and c + di is given by the formula (a + bi)(c + di) = ac bd + (ad + bc)i. The plane R 2 endowed with this multiplication is denoted by C and called the set of complex numbers. The real line generated by is called the real axis, the line generated by i is called the imaginary axis. The set of real numbers R can be viewed as embedded into C as the real axis. Given a complex number z = x + iy, the numbers x and y are called its real and imaginary parts, respectively, and denoted by Rez and Imz, so that z = Rez + iimz. 49

50 For any non-zero complex number z = a + bi there exists an inverse z such that z z =. Indeed, we can set z = a a 2 + b 2 b a 2 + b 2 i. The commutativity, associativity and existence of the inverse is easy to check, but it should not be taken for granted: it is impossible to define a similar operation any R n for n > 2. Given z = a + bi C its conjugate is defined as z = a bi. The conjugation operation z z is the reflection of C with respect to the real axis R C. Note that Rez = 2 (z + z), Imz = (z z). 2i Let us introduce the polar coordinates (r, φ) in R 2 = C. Then a complex number z = x + yi can be written as r cos φ + ir sin φ = r(cos φ + i sin φ). This form of writing a complex number is called, sometimes, ttrigonometric. The number r = x 2 + y 2 is called the modulus of z and denoted by z and φ is called the argument of φ and denoted by arg z. Note that the argument is defined only mod 2π. The value of the argument in [0, 2π) is sometimes called the principal value of the argument. When z is real than its modulus z is just the absolute value. We also not that z = z z. An important role plays the triangle inequality z z 2 z + z 2 z + z 2. Exponential function of a complex variable Recall that the exponential function e x has a Taylor expansion e x = 0 x n n! = + x + x2 2 + x We then define for a complex the exponential function by the same formula e z := + z + z2 2! + + zn n! One can check that this power series absolutely converging for all z and satisfies the formula e z +z+2 = e z e z 2. 50

51 Figure 5.: Leonhard Euler ( ) In particular, we have e iy = + iy y2 2! iy3 3! + y4 4! (5..) = ( ) k y2k 2k! + i ( ) k y 2k+ (2k + )!. (5..2) k=0 k=0 But ( ) k y2k 2k! = cos y and ( ) k y2k+ k=0 k=0 (2k+)! = sin y, and hence we get Euler s formula e iy = cos y + i sin y, and furthermore, e x+iy = e x e iy = e x (cos y + i sin y), i.e. e x+iy = e x, arg(e z ) = y. In particular, any complex number z = r(cos φ + i sin φ) can be rewritten in the form z = re iφ. This is called the exponential form of the complex number z. Note that ( e iφ) n = e inφ, and hence if z = re iφ then z n = r n e inφ = r n (cos nφ + i sin nφ). Note that the operation z iz is the rotation of C counterclockwise by the angle π 2. More generally a multiplication operation z zw, where w = ρe iθ is the composition of a rotation by the angle θ and a radial dilatation (homothety) in ρ times. 5

52 Exercise 5... Compute n cos kθ and n sin kθ. 2. Compute + ( n 4) + ( n 8) + ( n 2) Complex vector space In a real vector space one knows how to multiply a vector by a real number. In a complex vector space there is defined an operation of multiplication by a complex number. Example is the space C n whose vectors are n-tuples z = (z,..., z n ) of complex numbers, and multiplication by any complex number λ = α + iβ is defined component-wise: λ(z,..., z n ) = (λz,..., λz n ). Complex vector space can be viewed as an upgrade of a real vector space, or better to say as a real vector space with an additional structure. In order to make a real vector space V into a complex vector space, one just needs to define how to multiply a vector by i. This operation must be a linear map J : V V which should satisfy the condition J 2 = Id, i.e J (J (v)) = i(iv) = v. Example 5.2. Consider R 2n with coordinates (x, y,..., x n, y n ). Consider a 2n 2n-matrix J = Then J 2 = I. Consider a linear operator J : R 2n R 2n with this matrix, i.e. J (Z) = JZ for any vector Z R 2n which we view as a column-vector. Then J 2 = Id, and hence we can define on R 2n a complex structure (i.e.the multiplication by i by the formula iz = J (Z), Z R 2n. This complex vector space is canonically isomorphic to C n, where we identify the real vector (x, y,..., x n, y n ) R 2n with a complex vector (z = x + iy,..., z n x n + iy n ). 52

53 On the other hand, any complex vector space can be viewed as a real vector space. In order to do that we just need to forget how to multiply by i. This procedure is called the realification of a complex vector space. For example, the realification of C n is R 2n. Sometimes to emphasize the realification operation we will denote the realification of a complex vector space V by V R. As the sets these to objects coincide. Given a complex vector space V we can define linear combinations λ i v i, where λ i C are complex numbers, and thus similarly to the real case talk about really dependent, really independent vectors. Given vectors v,... v n V we define its complex span Span C (V,..., v n ) by the formula { n } Span C (v,..., v n ) = λ i v i, λ i C.. A basis of a complex vector space is a system of complex linear independent vectors v,..., v n such that Span C (v,..., v n ) = V. The number of vectors in a complex basis is called the complex dimension of V and denoted dim C V. For instance dim C n = n. On the other hand, its realification R 2n has real dimension 2n. In particular, C is a complex vector space of dimension, and therefore it is called a complex line rather than a plane. Exercise 5.3. Let v,..., v n be a complex basis of a complex vector space V. Find the real basis of its realification V R. Answer. v, iv, v 2, iv 2,..., v n, iv n. There is another important operation which associates with a real vector space V of real dimension n a complex vector space V C of complex dimension n. It is done in a way similar to how we made complex numbers out of real numbers. As a real vector space the space V C is just the direct sum V V = {(v, w); v, w V }. This is a real space of dimension 2n. We then make V V into a complex vector space by defining the multiplication by i by the formula: i(v, w) = ( w, v). We will write vector (v, 0) simply by v and (0, v) = i(v, 0) by iv. Hence, every vector of V C can be written as v + iw, where v, w V. If v,..., v n is a real basis of V, then the same vectors form a complex basis of V C. 53

54 5.3 Complex linear maps Complex linear maps and their realifications Given two complex vector spaces V, W a map A : V W is called complex linear (or C-linear) if A(X + Y ) = A(X) + A(Y ) and A(λX) = λa(x) for any vectors X, Y V and any complex number λ C. Thus complex linearity is stronger condition than the real linearity. The difference is in the additional requirement that A(iX) = ia(x). In other words, the operator A must commute with the operation of multiplication by i. Any linear map A : C C is a multiplication by a complex number a = c + id. If we view C as R 2 and right the real matrix of this map in the standard basis, i we get the matrix c d. d c Indeed, A() = a = c + di and A(i) = ai = d + ci, so the first column of the matrix is equal to c and the second one is equal to d. d c If we have bases v,..., v n of V and w,... w m of W then one can associate with A an m n complex matrix A by the same rule as in the real case. Recall (see Exercise 5.3) that vectors v, v = iv, v 2, v 2 = iv 2,..., v n, v n = iv n and w, w = iw, w 2, w 2 = iw 2,... w m, w m = iw m ) form real bases of the realifications V R and W R of the spaces V and W. if a... a n A =... a m... a mn is the complex matrix of A then the real matrix A R of the map A is the real basis has order 2n 2n and is obtained from A by replacing each complex element a kl = c kl + id kl by a 2 2 matrix c kl d kl d kl c kl. Exercise 5.4. Prove that det A R = det A 2. 54

55 Complexification of real linear maps Given a real linear map A : V W one can define a complex linear map A C : V C W C by the formula A C (v + iw) = A(v) + ia(w).. If A is the matrix of A in a basis v,..., v n then A C has the same matrix in the same basis viewed as a complex basis of V C. The operator A C is called the complexification of the operator A. In particular, one can consider C-linear functions V C on a complex vector space V. Complex coordinates z,..., z n in a complex basis are examples of C-linear functions, and any other C-linear function on V has a form c z +... c n z n, where c,..., c n C are complex numbers. Complex-valued R-linear functions It is sometimes useful to consider also C-valued R-linear functions on a complex vector space V, i.e. R-linear maps V C (i.e. a linear map V R R 2 ). Such a C-valued function has the form λ = α + iβ, where α, β are usual real linear functions. For instance the function z on C is a C-valued R-linear function which is not C-linear. If z,..., z n are complex coordinates on a complex vector space V then any R-linear complexvalued function can be written as n a i z i + b i z i, where a i, b i C are complex numbers. We can furthermore consider complex-valued tensors and, in particular complex-valued exterior forms. A C-valued k-form λ can be written as α + iβ where α and β are usual R-valued k-forms. n For instance, we can consider on C n the 2-form ω = i 2 z k z k. It can be rewritten as ω = i 2 n (x k + iy k ) (x k iy k ) = n x k y k. 55

56 56

57 Part II Calculus of differential forms 57

58

59 Chapter 6 Vector fields and differential forms 6. Differential and gradient of a smooth function Given a vector space V we will denote by V x the vector space V with the origin translated to the point x V. One can think of V x as that tangent space to V at the point x. Though the parallel transport allows one to identify spaces V and V x it will be important for us to think about them as different spaces. Let f : U R be a function on a domain U V in a vector space V. The function f is called differentiable at a point x U if there exists a linear function l : V x R such that f(x + h) f(x) = l(h) + o( h ) for any sufficiently small vector h, where the notation o(t) stands for any function such that o(t) t 0. The linear function l is called the differential of the function f at the point x and t 0 is denoted by d x f. In other words, f is differentiable at x U if for any h V x there exists a limit f(x + th) f(x) l(h) = lim, t 0 t and the limit l(h) linearly depends on h. The value l(h) = d x f(h) is called the directional derivative of f at the point x in the direction h. The function f is called differentiable on the whole domain U if it is differentiable at each point of U. Simply speaking, the differentiability of a function means that at a small scale near a point x the function behaves approximately like a linear function, the differential of the function at the 59

60 point x. However this linear function varies from point to point, and we call the family {d x f} x U of all these linear functions the differential of the function f, and denote it by df (without a reference to a particular point x). Let us summarize the above discussion. Let f : U R be a differentiable function. Then for each point x U there exists a linear function d x f : V x R, the differential of f at the point x defined by the formula f(x + th) f(x) d x f(h) = lim, x U, h V x. t 0 t If v,..., v n are vectors of a basis of V, parallel transported to the point x, then we have d x f(v i ) = f x i (x), x U, i =,..., n, where x,..., x n are coordinates with respect to the chosen basis v,..., v n. Notice that if f is a linear function, f(x) = a x + + a n x n, then for each x V we have d x f(h) = a h + + a n h n, h = (h,..., h n ) V x. Thus the differential of a linear function f at any point x V coincides with this function, parallel transported to the space V x. This observation, in particular, can be applied to linear coordinate functions x,..., x n with respect to a chosen basis of V. In Section 6.6 below we will define the differential for maps f : U W, where W is a vector space and not just the real line R. 6.2 Gradient vector field If V is an Euclidean space, i.e. a vector space with an inner product,, then there exists a canonical isomorphism D : V V, defined by the formula D(v)(x) = v, x for v, x V. Of course, D defines an isomorphism V x V x for each x V. Set 60

61 f(x) = D (d x f). The vector f(x) is called the gradient of the function f at the point x U. We will also use the notation gradf(x). By definition we have f(x), h = d x f(h) for any vector h V. If h = then d x f(h) = f(x) cos ϕ, where ϕ is the angle between the vectors f(x) and h. In particular, the directional derivative d x f(h) has its maximal value when ϕ = 0. Thus the direction of the gradient is the direction of the maximal growth of the function and the length of the gradient equals this maximal value. As in the case of a differential, the gradient varies from point to point, and the family of vectors { f(x)} x U is called the gradient vector field f. We discuss the general notion of a vector field in the next section. 6.3 Vector fields A vector field v on a domain U V is a function which associates to each point x U a vector v(x) V x, i.e. a vector originated at the point x. A gradient vector field f of a function f provides us with an example of a vector field, but as we shall see, gradient vector fields form only a small very special class of vector fields. Let v be a vector field on a domain U V. If we fix a basis in V, and parallel transport this basis to all spaces V x, x V, then for any point x V the vector v(x) V x is described by its coordinates (v (x), v 2 (x),..., v n (x)). Thus to define a vector field on U is the same as to define n functions v,..., v n on U, i.e. to define a map (v,..., v n ) : U R n. Thus, if a basis of V is fixed, then the difference between the maps U R n and vector fields on U is just a matter of geometric interpretation. When we speak about a vector field v we view v(x) as a vector in V x, i.e. originated at the point x U. When we speak about a map v : U R n we view v(x) as a point of the space V, or as a vector, with its origin at 0 V. 6

62 Vector fields naturally arise in a context of Physics, Mechanics, Hydrodynamics, etc. as force, velocity and other physical fields. There is another very important interpretation of vector fields as first order differential operators. Let C (U) denote the vector space of infinitely differentiable functions on a domain U V. Let v be a vector field on V. Let us associate with v a linear operator given by the formula D v : C (U) C (U), D v (f) = df(v), f C (U). In other words, we compute at any point x U the directional derivative of f in the direction of the vector v(x). Clearly, the operator D v is linear: D v (af +bg) = ad v (f)+bd v (g) for any functions f, g C (U) and any real numbers a, b R. It also satisfies the Leibniz rule: D v (fg) = D v (f)g + fd v (g). In view of the above correspondence between vector fields and first order differential operatopts, it is sometimes convenient just to view a vector field as a differential operator. Hence, when it will not be confusing we may drop the notation D v and just directly apply the vector v to a function f. Let v,..., v n be a basis of V, and x,..., x n be the coordinate functions in this basis. We would like to introduce the notation for the vector field obtained from vectors v,..., v n by parallel transporting them to all points of the domain U. To motivate the notation which we are going to introduce, let us temporarily denote these vector fields by v,..., v n. Observe that D vi (f) = f x i, i =,..., n. Thus the operator D vi is just the operator x i of taking i-th partial derivative. Hence, viewing the vector field v i as a differential operator we will just use the notation of v i. Given any vector field v with coordinate functions a, a 2,..., a n : U R we have n D v (f)(x) = a i (x) f (x), for any f C (U), x i and hence we can write v = constants. n i= i= x i instead a i x i. Note that the coefficients a i here are functions and not 62

63 Suppose that V,, is a Euclidean vector space. Choose a (not necessarily orthonormal) basis v,..., v n. Let us find the coordinate description of the gradient vector field f, i.e. find the coefficients a j in the expansion f(x) = n a i (x) x i. By definition we have f(x), h = d x f(h) = n f x j (x)h j (6.3.) for any vector h V x with coordinates (h,..., h n ) in the basis v,..., v n parallel transported to V x. Let us denote g ij = v i, v j. Thus G = (g ij ) is a symmetric n n matrix, which is called the Gram matrix of the basis v,..., v n. Then the equation (6.3.) can be rewritten as n g ji a i h j = i,j= n f x j (x)h j. Because h j are arbitrarily numbers it implies that the coefficients with h j in the right and left sides should coincide for all j =,..., n. Hence we get the following system of linear equations: or in matrix form and thus i.e. n i= g ij a i = f x j (x), j =,..., n, (6.3.2) f a x (x) G. =., f a n x n (x) f a x (x). = G., (6.3.3) f a n x n (x) f = n i,j= where we denote by g ij the entries of the inverse matrix G = (g ij ) ij f g (x), (6.3.4) x i x j If the basis v,..., v n is orthonormal then G is the unit matrix, and thus in this case f = n f x j (x) x j, (6.3.5) 63

64 i.e. f has coordinates ( f x,..., f x n ). However, simple expression (6.3.5) for the gradient holds only in the orthonormal basis. In the general case one has a more complicated expression (6.3.4). 6.4 Differential forms Similarly to vector fields, we can consider fields of exterior forms, i.e. functions on U V which associate to each point x U a k-form from Λ k (V x ). These fields of exterior k-forms are called differential k-forms. Thus the relation between k-forms and differential k-forms is exactly the same as the relation between vectors and vector-fields. For instance, a differential -form α associates with each point x U a linear function α(x) on the space V x. Sometimes we will write α x instead of α(x) to leave space for the arguments of the function α(x). Example 6... Let f : V R be a smooth function. Then the differential df is a differential -form. Indeed, with each point x V it associates a linear function d x f on the space V x. As we shall see, most differential -form are not differentials of functions (just as most vector fields are not gradient vector fields). 2. A differential 0-form f on U associates with each point x U a 0-form on V x, i. e. a number f(x) R. Thus differential 0-forms on U are just functions U R. 6.5 Coordinate description of differential forms Let x,..., x n be coordinate linear functions on V, which form the basis of V dual to a chosen basis v,..., v n of V. For each i =,..., n the differential dx i defines a linear function on each space V x, x V. Namely, if h = (h,..., h n ) V x then dx i (h) = h i. Indeed d x x i (h) = lim t 0 x i + th i x i t = h i, independently of the base point x V. Thus differentials dx,..., dx n form a basis of the space V x for each x V. In particular, any differential -form α on v can be written as 64

65 α = f dx f n dx n, where f,..., f n are functions on V. In particular, df = f x dx f x n dx n. (6.5.) Let us point out that this simple expression of the differential of a function holds in an arbitrary coordinate system, while an analogous simple expression (6.3.5) for the gradient vector field is valid only in the case of Cartesian coordinates. This reflects the fact that while the notion of differential is intrinsic and independent of any extra choices, one needs to have a background inner product to define the gradient. Similarly, any differential 2-form w on a 3-dimensional space can be written as ω = b (x)dx 2 dx 3 + b 2 (x)dx 3 dx + b 3 (x)dx dx 2 where b, b 2, and b 3 are functions on V. Any differential 3-form Ω on a 3-dimensional space V has the form Ω = c(x)dx dx 2 dx 3 for a function c on V. More generally, any differential k-form α can be expressed as α = a i...i k dx i dx ik for some functions a i...i k on V. i <i 2 <i k n 6.6 Smooth maps and their differentials Let V, W be two vector spaces of arbitary (not, necessarily, equal) dimensions and U V be an open domain in V. Recall that a map f : U W is called differentiable if for each x U there exists a linear map l : V x W f(x) 65

66 such that l(h) = lim t 0 f(x + th) f(x) t for any h V x. In other words, f(x + th) f(x) = tl(h) + o(t), where o(t) t 0. t 0 The map l is denoted by d x f and is called the differential of the map f at the point x U. Thus, d x f is a linear map V x W f(x). When W = R then we get the notion of the differential of a function, which was introduced earlier in Section 6.. Let us pick bases in V and W and let (x,..., x k ) and (y,..., y n ) be the corresponding coordinate functions. Then each of the spaces V x and W y, x V, y W inherits a basis obtained by parallel transport of the bases of V and W. In terms of these bases, the differential d x f is given by the Jacobi matrix f x... f x k f n x... f n x k In what follows we will consider only sufficiently smooth maps, i.e. we assume that all maps and their coordinate functions are differentiable as many times as we need it. 6.7 Operator f Let U be a domain in a vector space V and f : U W a smooth map. Then the differential df defines a linear map d x f : V x W f(x) for each x V. 66

67 Let ω be a differential k-form on W. Thus ω defines an exterior k-form on the space W y for each y W. Let us define the differential k-form f ω on U by the formula (f ω) Vx = (d x f) (ω Wf(x) ). Here the notation ω Wy space W y. stands for the exterior k-form defined by the differential form ω on the In other words, for any k vectors, H,..., H k V x we have f ω(h,..., H k ) = ω(d x f(h ),..., d x f(h k )). We say that the differential form f ω is induced from ω by the map f, or that f ω is the pull-back of ω by f. Example 6.2. Let Ω = h(x)dx dx n. Then formula (3.2) implies f Ω = h f det Dfdx dx n. Here f f x... x n det Df = f n f x... n is the determinant of the Jacobian matrix of f = (f,..., f n ). Similarly to Proposition.9 we get x n Proposition 6.3. Given 2 maps and a differential k form ω on U 3 we have U f U2 g U3 (g f) (ω) = f (g ω). 67

68 6.8 Coordinate description of the operator f Consider first the linear case. Let A be a linear map V W and ω Λ p (W ). Let us fix coordinate systems x,..., x k in V and y,..., y n in W. If A is the matrix of the map A then we already have seen in Section 2.7 that A y j = l j (x,..., x k ) = a j x + a j2 x a jk x k, j =,..., n, and that for any exterior k-form ω = A i,...,i p y i... y ip i <...<i p n we have A ω = A i...i p l i... l ip. i,<...<i p n Now consider the non-linear situation. Let ω be a differential p-form on W. Thus it can be written in the form for some functions A i...i p on W. ω = A i...i p (y)dy i... dy ip Let U be a domain in V and f : U W a smooth map. Proposition 6.4. f ω = A i...,i p (f(x))df i... df ip, where f,..., f n are coordinate functions of the map f. Proof. For each point x U we have, by definition, f ω Vx = (d x f) (ω Wfx ) But the coordinate functions of the linear map d x f are just the differentials d x f i of the coordinate functions of the map f. Hence the desired formula follows from the linear case proven in the previous proposition. 68

69 6.9 Examples. Consider the domain U = {r > 0, 0 ϕ < 2π} on the plane V = R 2 with cartesian coordinates (r, ϕ). Let W = R 2 be another copy of R 2 with cartesian coordinates (x, y). Consider a map P : V W given by the formula P (r, ϕ) = (r cos ϕ, r sin ϕ). This map introduces (r, ϕ) as polar coordinates on the plane W. Set ω = dx dy. It is called the area form on W. Then P ω = d(r cos ϕ) d(r sin ϕ) = (cos ϕdr + rd(cos ϕ)) (sin ϕdr + rd(sin ϕ) = (cos ϕdr r sin ϕdϕ) (sin ϕdr + r cos ϕdϕ) = cos ϕ sin ϕdr dr r sin 2 ϕdϕ dr + r cos 2 ϕdr dϕ r 2 sin ϕ cos ϕdϕ dϕ = r cos 2 ϕdr dϕ + r sin 2 dr dϕ = rdr dϕ. 2. Let f : R 2 R be a smooth function and the map F : R 2 R 3 be given by the formula F (x, y) = (x, y, f(x, y)) Let ω = P (x, y, z)dy dz + Q(x, y, z)dz dx + R(x, y, z)dx dy be a differential 2-form on R 3. Then F ω = P (x, y, f(x, y))dy df + + Q(x, y, f(x, y))df dx + R(x, y, f(x, y))dx dy = P (x, y, f(x, y))dy (f x dx + f y dy) + + Q(x, y, f(x, y))(f x dx + f y dy) dx + + R(x, y, f(x, y))dx dy = = (R(x, y, f(x, y)) P (x, y, f(x, y))f x Q(x, y, f(x, y))f y )dx dy 69

70 Figure 6.: Johann Friedrich Pfaff ( ) where f x, f y are partial derivatives of f. 3. If p > k then the pull-back f ω of a p-form ω on U to a k-dimensional space V is equal to Pfaffian equations Given a non-zero linear function l on an n-dimensional vector space V the equation l = 0 defines a hyperplane, i.e. an (n )-dimensional subspace of V. x U. Suppose we are given a differential -form λ on a domain U V. Suppose that λ x 0 for each Then the equation λ = 0 (6.0.) defines a hyperplane field ξ on U, i.e. a family of of hyperplanes ξ x = {λ x = 0} V x, x U. The equation of this type is called Pfaffian in honor of a German mathematician Johann Friedrich Pfaff ( ). Example 6.5. Let V = R 3 with coordinates (x, y, z). Let λ = dz. Then ξ = {dz = 0} is the horizontal plane field which is equal to Span( x, y ). 2. Let λ = dz ydx. Then the plane field dz ydx is shown on Fig This plane is non-integrable in the following sense. There are no surfaces in R 3 tangent to ξ. This plane field is called a contact 70

71 z y x Figure 6.2: Contact structure structure. It plays an important role in symplectic and contact geometry, which is, in turn, the geometric language for Mechanics and Geometric Optics. 7

72 72

73 Chapter 7 Exterior differential Let Ω k (U) be the space of all differential k-forms on U. We will define a map d : Ω k (U) Ω k+ (U) which is called the exterior differential. In the current form it was introduced by Élie Cartan. We first define it in coordinates and then prove that the result is independent of the choice of the coordinate system. Let us fix a coordinate system x,..., x n in V U. As a reminder, a differential k-form w Ω k (U) has the form w = a i...,i k dx i... dx ik where a i...i k are i <...<i k functions on the domain U. Define dw := da i...i k dx i... dx ik. i <...<i k Examples.. Let w Ω (U), i.e. w = n i= a i dx i. Then dw = n da i dx i = i= n n i= j= a i dx j dx i = x j i<j n ( aj a ) i dx i dx j. x i x j For instance, when n = 2 we have d(a dx + a 2 dx 2 ) = ( a2 a ) dx dx 2 x x 2 73

74 Figure 7.: Élie Cartan (869-95) For n = 3, we get d(a dx +a 2 dx 2 +a 3 dx 3 ) = 2. Let n = 3 and w Ω 2 (U). Then ( a3 a ) ( 2 a dx 2 dx 3 + a ) ( 3 a2 dx 3 dx + a ) dx dx 2. x 2 x 3 x 3 x x x 2 and w = a dx 2 dx 3 + a 2 dx 3 dx + a 3 dx dx 2 dw = da dx 2 dx 3 + da 2 dx 3 dx + da 3 dx dx 2 ( a = + a 2 + a ) 3 dx dx 2 dx 3 x x 2 x 3 3. For 0-forms, i.e. functions the exterior differential coincides with the usual differential of a function. 7. Properties of the operator d Proposition 7.. For any 2 forms, α Ω k (U), β Ω l (U) we have d(α β) = dα β + ( ) k α dβ. 74

75 Proof. We have α = β = α β = = i <...<i k a i...i k dx i... dx ik j <...<j l b ji...j l dx j... dx jl a i...i k dx i... dx ik b j...j l dx j... dx jl i <...<i k j i <...<j l a b dx... dx dx... dx i...ik j...jl il ik jl jl i <...<i k j <...<j l d(α β) = = (b da + a db ) dx... dx dx... dx j...jl i...ik i...ik j...jl i ik j jl i <...<i k j <...<j l b da dx... dx dx... dx j...jl i...ik i ik jl jl i <...<i k j <...<j l + = a db dx... dx dx... dx i...ik j...jl i ik j jl i <...<i k j <...<j l ( ) da i...i k dx i... dx ik bj...j l dx j... dx jl i <...<i k + ( ) k a i...i k dx i... dx ik db j...j l dx j... dx jl i <...<i k j i <...<j l = dα β + ( ) k α dβ. Notice that the sign ( ) k appeared because we had to make k transposition to move db j...j l to its place. 75

76 Proposition 7.2. For any differential k-form w we have ddw = 0. Proof. Let w = a i...i k dx i... dx ik. Then we have i <...<i k dw = da i...i k dx i... dx ik. i <...<i k Applying Proposition 7. we get ddw = dda i...i k dx i... dx ik da i...i k ddx i... dx ik +... i <...<i k + ( ) k da i...i k dx i... ddx ik. But ddf = 0 for any function f as was shown above. Hence all terms in this sum are equal to 0, i.e. ddw = 0. Definition. A k-form ω is called closed if dω = 0. It is called exact if there exists a (k )-form θ such that dθ = ω. The form θ is called the primitive of the form ω. The previous theorem can be reformulated as follows: Corollary 7.3. Every exact form is closed. The converse is not true in general. For instance, take a differential -form ω = xdy ydx x 2 + y 2 on the punctured plane U = R 2 \ 0 (i.e the plane R 2 with the deleted origin). It is easy to calculate that dω = 0, i.e ω is closed. On the other hand it is not exact. Indeed, let us write down this form in polar coordinates (r, ϕ). We have Hence, x = r cos ϕ, y = r sin ϕ. ω = (r cos ϕ(sin ϕdr + r cos ϕdϕ) r sin ϕ(cosϕdr r sin ϕdϕ)) = dϕ. r2 76

77 If there were a function H on U such that dh = ω, then we would have to have H = ϕ + const, but this is impossible because the polar coordinate ϕ is not a continuous univalent function on U. Hence ω is not exact. However, a closed form is exact if it is defined on the whole vector space V. Proposition 7.4. Operators f and d commute, i.e. for any differential k-form w Ω k (W ), and a smooth map f : U W we have df w = f dw. Proof. Suppose first that k = 0, i.e. w is a function ϕ : W R. Then f ϕ = ϕ f. Then d(ϕ f) = f dϕ. Indeed, for any point x U and a vector X V x we have d(ϕ f)(x) = dϕ(d x f(x)) (chain rule) But dϕ(d x f(x)) = f (dϕ(x)). Consider now the case of arbitrary k-form w, w = a i...i k dx i... dx ik. i <...<i k Then f w = a i...i k fdf i... df ik i <...<i k where f,..., f n are coordinate functions of the map f. Using the previous theorem and taking into account that d(df i ) = 0, we get d(f w) = d(a i...i k f) df i... df ik. i <...<i k On the other hand dw = da i...i k dx i... dx ik i <...<i k 77

78 and therefore f dw = f (da i...i k ) df i... df ik. i <...<i k But according to what is proven above, we have Thus, f da i...i k = d(a i...i k f) f dw = i <...<i k d(a i...i k f) df i... df ik = df w The above theorem shows, in particular, that the definition of the exterior differential is independent of the choice of the coordinate. Moreover, one can even use non-linear (curvilinear) coordinate systems, like polar coordinates on the plane. Remark 7.5. We will show later (see Lemma 0.2) that one can give another equivalent definition of the operator d without using any coordinates at all. 7.2 Curvilinear coordinate systems A (non-linear) coordinate system on a domain U in an n-dimensional space V is a smooth map f = (f,..., f n ) : U R n such that. For each point x U the differentials d x f,..., d x f n (V x ) are linearly independent. 2. f is injective, i.e. f(x) f(y) for x y. Thus a coordinate map f associates n coordinates y = f (x),..., y n = f n (x) with each point x U. The inverse map f : U U is called the parameterization. Here U = f(u) R n is the image of U under the map f. If one already has another set of coordinates x... x n on U, then the coordinate map f expresses new coordinates y... y n through the old one, while the parametrization map expresses the old coordinate through the new one. Thus the statement 78

79 g dw = dg w applied to the parametrization map g just tells us that the formula for the exterior differential is the same in the new coordinates and in the old one. Consider a space R n with coordinates (u,..., u n ). The j-th coordinate line is given by equations u i = c i, i =,..., n; i j. Given a domain U R n consider a parameterization map g : U U V. The images g{u i = c i, i j}) U of coordinates lines {u i = c i, i j} U are called coordinate lines in U with respect to the curvilinear coordinate system (u,..., u n ). For instance, coordinate lines for polar coordinates in R 2 are concentric circles and rays, while coordinate lines for spherical coordinates in R 3 are rays from the origin, and latitudes and meridians on concentric spheres. 7.3 More about vector fields Similarly to the case of linear coordinates, given any curvilinear coordinate system (u,..., u n ) in U, one denotes by u,..., the vector fields which correspond to the partial derivatives with respect to the coordinates u,..., u n. In other words, the vector field u i u n is tangent to the u i -coordinate lines and represents the the velocity vector of the curves u = const,..., u i = const i, u i+ = const i+,..., u n = const n, parameterized by the coordinate u i. For instance, for spherical coordinates (r, θ, ϕ) in R 3 the vector fields r, θ and are mutually orthogonal. We also have r =. However the length of vector fields ϕ ϕ and θ vary. When r and θ are fixed and ϕ varies, then the corresponding point (r, θ, ϕ) is moving along a meridian of radius r with a constant angular speed. Hence, ϕ = r. 79

80 When r and ϕ are fixed and θ varies, then the point (r, ϕ, θ) is moving along a latitude of radius r sin ϕ with a constant angular speed. Hence, = r sin ϕ. θ The chain rule allows us to express the vector fields x,..., x n. Indeed, for any function f : U R we have f u i = n j= f x j x j u i, u,..., u n through the vector fields and, therefore, u i = n j= x j, u i x j For instance, suppose we are given spherical coordinates (r, ϕ, θ) in R 3. The spherical coordinates are related to the cartesian coordinates (x, y, z) by the formulas x = r sin ϕ cos θ, y = r sin ϕ sin θ, z = r cos ϕ. Hence we derive the following expression of the vector fields r, ϕ, θ x, y, z : r ϕ θ = sin ϕ cos θ x + sin ϕ sin θ y + cos ϕ z, = r cos ϕ cos θ x + r cos ϕ sin θ y r sin ϕ z, = r sin ϕ sin θ x + r sin ϕ cos θ y. through the vector fields 7.4 Case n = 3. Summary of isomorphisms Let U be a domain in the 3-dimensional space V. We will consider 5 spaces associated with U. Ω 0 (U) = C (U) the space of 0-forms, i.e. the space of smooth functions; Ω k (U) for k =, 2, 3 the spaces of differential k-forms on U; 80

81 Vect(U) the space of vector fields on U. Let us fix a volume form w Ω 3 (U) that is any nowhere vanishing differential 3-form. In coordinates w can be written as w = f(x)dx dx 2 dx 3 where the function f : U R is never equal to 0. The choice of the form w allows us to define the following isomorphisms.. Λ w : C (U) Ω 3 (U), Λ w (h) = hw for any function h C (U). 2. w : Vect(U) Ω 2 (U) w(v) = v w. Sometimes we will omit the subscript w an write just Λ and. Our third isomorphism depends on a choice of a scalar product <, > in V. Let us fix a scalar product. This enables us to define an isomorphism D = D <,> : Vect(U) Ω (U) which associates with a vector field v on U a differential -form D(v) = v,.. Let us write down the coordinate expressions for all these isomorphisms. Fix a cartesian coordinate system (x, x 2, x 3 ) in V so that the scalar product x, y in these coordinates equals x y + x 2 y 2 + x 3 y 3. Suppose also that w = dx dx 2 dx 3. Then Λ(h) = hdx dx 2 dx 3. (v) = v dx 2 dx 3 + v 2 dx 3 dx + v 3 dx dx 2 where v, v 2, v 3 are coordinate functions of the vector field v. D(v) = v dx + v 2 dx 2 + v 3 dx 3. If V is an oriented Euclidean space then one also has isomorphisms : Ω k (V ) Ω 3 k (V ), k = 0,, 2, 3. 8

82 If w is the volume form on V for which the unit cube has volume and which define the given orientation of V (equivalently, if w = x x 2 x 3 for any Cartesian positive coordinate system on V ), then w(v) = D(v), and Λ w = : Ω 0 (V ) Ω 3 (V ). 7.5 Gradient, curl and divergence of a vector field The above isomorphism, combined with the operation of exterior differentiation, allows us to define the following operations on the vector fields. First recall that for a function f C (U), gradf = D (df). Now let v Vect (U) be a vector field. Then its divergence div v is the function defined by the formula div v = Λ (d( v)) In other words, we take the 2-form v w (w is the volume form) and compute its exterior differential d(v w). The result is a 3-form, and, therefore is proportional to the volume form w, i.e. d(v w) = hw. This proportionality coefficient (which is a function; it varies from point to point) is simply the divergence: div v = h. Given a vector field v, its curl is as another vector field curl v defined by the formula curl v := d(dv) = D d(dv). If one fixes a cartesian coordinate system in V such that w = dx dx 2 dx 3 and x, y = x y + x 2 y 2 + x 3 y 3 then we get the following formulas grad f = ( f, f, f ) x x 2 x 3 div v = v x + v 2 x 2 + v 3 x 3 curl v = ( v3 v 2, v v 3, v 2 v ) x 2 x 3 x 3 x x x 2 82

83 where v = (v, v 2, v 3 ). We will discuss the geometric meaning of these operations later in Section Complex-valued differential k-forms One can consider complex-valued differential k-forms. A C-valued differential -form is a field of C-valued k-forms, or simply it is an expression α + iβ, where α, β are usual real-valued k-forms. All operations on complex valued k-forms ( exterior multiplication, pull-back and exterior differential) are defined in a natural way: (α + iβ ) (α 2 + iβ 2 ) = α α 2 β β 2 + i(α β 2 + β α 2 ), f (α + iβ) = f α + if β, d(α + iβ) = dα + idβ. We will be in particular interested in complex valued on C. Note that a complex-valued function (or 0-form) is on a domain U C is just a map f = u+iv : U C. Its differential df is the same as the differntial of this map, but it also can be viewed as a C-valued differential -form df = du+idv. Example 7.6. dz = dx + idy, d z = dx idy, zdz = (x + iy)(dx + idy) = xdx ydy + i(xdy + ydx), dz d z = (dx + idy) (dx idy) = 2idx dy. Exercise 7.7. Prove that d(z n ) = nz n dz for any integer n 0. Solution. Let us do the computation in polar coordinates. Then z n = r n e inφ and assuming that n 0 we have d(z n ) = nr n e inφ dr + inr n e inφ dφ = nr n e inφ (dr + idφ). On the other hand, ( ) nz n dz = nr n e i(n )φ d(re iφ ) = nr n e i(n )φ e iφ dr + ie iφ dφ = nr n e inφ (dr + idφ). Comparing the two expressions we conclude that d(z n ) = nz n dz. 83

84 It follows that the -form dz z n On the other hand the form dz z and hence d ( ) dz z = 0. On the other hand, z = is exact on C \ 0 for n >. Indeed, ( dz z n = d ( n)z n ). is closed on C \ 0 but not exact. Indeed, dz z = d(reiφ ) re iφ = eiφ dr + ire iφ dφ re iφ = dr r + idφ, dz z 2π = 0 idφ = 2πi 0, and hence dz z computed over the unit circle oriented counter-clockwise. is not exact. Here the integral is 84

85 Chapter 8 Preliminaries 8. Elements of topology in an Euclidean space We recall in this section some basic topological notions in a finite-dimensional Euclidean space and elements of the theory of continuous functions. The proofs of most statements are straightforward and we omit them. Let V be a Euclidean space. Notation B r (p) := {x V ; x p < r}, D r (p) := {x V ; x p r} and S r (p) := {x V ; x p = r} stand for open, closed balls and the sphere of radius r centered at a point p V. Open and closed sets A set U V is called open if for any x U there exists ɛ > 0 such that B ɛ (x) U. A set A V is called closed if its complement V \ A is open. Equivalently, Lemma 8.. The set A is closed if and only if for any sequence x n A, n =, 2,... which converges to a point a V, the limit point a belongs to A. In general, points which appear as limits of sequences of points x n A are called limit points of A. There are only two subsets of V which are simultaneously open and closed: V and. Lemma For any family U λ, λ Λ of open sets the union λ Λ U λ is open. 85

86 2. For any family A λ, λ Λ of closed sets the intersection A λ is closed. 3. The union n A i of a finite family of closed sets is closed. λ Λ 4. The intersection n U i of a finite family of open sets is open. By a neighborhood of a point a V we understand any open set U p. Given a subset X V - a subset Y X is called relatively open in Y if there exists an open set U V such that Y = X U. - a subset Y X is called relatively closed in Y if there exists a closed set A V such that Y = X A. One also call relatively and open and closed subsets of X just open and closed in X. Given any subset X V a point a V is called interior point for A if there is a neighborhood U p such that U A; boundary point if it is not an interior point neither for A nor for its complement V \ A. We emphasize that a boundary point of A may or may not belong to A. Equivalently, a point a V is a boundary point of A if it is a limit point both for A and V \ A. The set of all interior points of A is called the interior of A and denoted Int A. The set of all boundary points of A is called the boundary of A and denoted A. The union of all limit points of A is called the closure of A and denoted A. Lemma We have A = A A, Int A = A \ A. 2. A is equal to the intersection of all closed sets containg A 3. Int U is the union of all open sets contained in A. 86

87 Everywhere and nowhere dense sets A closed set A is called nowhere dense if Int A =. For instance any finite set is nowhere dense. Any linear subspace L V is nowhere dense in V if dim L < dim V. Here is a more interesting example of a nowhere dense set. Fix some number ɛ 3. For any interval = [a, b] we denote by ɛ the open interval centered at the point c = a+b 2, the middle point of, of the total length equal to ɛ(b a). We denote by C( ) := \ ɛ. Thus C( ) consists of two disjoint smaller closed intervals. Let I = [0, ]. Take C(I) = I I 2. Take again C(I ) C(I 2 ) then again apply the operation C to four new closed intervals. Continue the process, and take the intersection of all sets arising on all steps of this construction. The resulted closed set K ɛ I is nowhere dense. It is called a Cantor set. A subset B A is called everywhere dense in A if B A. For instance the the set Q I of rational points in the interval I = [0, ] is everywhere dense in I. Compactness and connectedness A set A V is called compact if one of the following equivalent conditions is satisfied: COMP. A is closed and bounded. COMP2. from any infinite sequence of points x n A one can choose a subsequence x nk converging to a point a A. COMP3. from any family U λ, λ Λ of open sets covering A, i.e. finitely many sets U λ,..., U λk which cover A, i.e. k U λk A. λ Λ U λ A, one can choose The equivalence of these definitions is a combination of theorems of Bolzano-Weierstrass and Émile Borel. A set A is called path-connected if for any two points a 0, a A there is a continuous path γ : [0, ] A such that γ(0) = a 0 and γ() = a. A set A is called connected if one cannot present A as a union A = A A 2 such that A A 2 =, A, A 2 and both A and A 2 are simultaneously relatively closed and open in A. 87

88 Figure 8.: Bernard Bolzano (78-848) Figure 8.2: Karl Weierstrass (85-897) 88

89 Figure 8.3: Émile Borel (87-956) Lemma 8.4. Any path-connected set is connected. If a connected set A is open then it is path connected. Proof. Suppose that A is disconnected. Then it can be presented as a union A = A 0 A of two non-empty relatively open (and hence relatively closed) subsets. Consider the function φ : A R defined by the formula 0, x A 0, φ(x) =, x A. We claim that the function φ is is continuous. Indeed, For each i = 0, and any point a A i there exists ɛ > 0 such that B ɛ (x) A A i. Hence the function φ is constant on B ɛ (x) A, and hence continuous at the point x. Now take points x 0 A 0 and x A and connect them by a path γ : [0, ] A (this path exists because A is path-connected). Consider the function ψ := φ γ : [0, ] R. This function is continuous (as a composition of two continuous maps). Furthermore, ψ(0) = 0, ψ() =. Hence, by an intermediate value theorem of Cauchy the function ψ must take all values in the interval [0, ]. But this is a contradiction because by construction the function ψ takes no other values except 0 and. Lemma 8.5. Any open connected subset U R n is path connected. Proof. Take any point a U. Denote by C a the set of all points in U which can be connected with a by a path. We need to prove that C a = U. 89

90 First, we note that C a is open. Indeed, if b C a then using openness of U we can find ɛ > 0 such the ball B ɛ (b) U. Any point of c B ɛ (b) can be connected by a straight interval I bc B ɛ (b) with b, and hence it can be connected by a path with a, i.e. c C a. Thus B ɛ (b) C, and hence C a is open. Similarly we prove that the complement U \ C a is open. Indeed, take b / C a. As above, there exists an open ball B ɛ (b) U. Then B ɛ (b) U \ C a. Indeed, if it were possible to connect a point c B ɛ (b) with a by a path, then the same would be true for b, because b and c are connected by the interval I bc. Thus, we have U = C a (U \ C a ), both sets C a and U \ C a are open and C a is non-empty. Hence, U \ C a is to be empty in view of connectedness of U. Thus, C a = U, i.e. U is path-connected. Exercise 8.6. Prove that any non-empty connected (= path-connected) open subset of R is equal to an interval (a, b) (we allow here a = and b = ). If one drops the condition of openness, then one needs to add a closed and semi-closed intervals and a point. Remark 8.7. One of the corollaries of this exercise is that in R any connected set is path-connected. Solution. Let A R be a non-empty connected subset. Let a < b be two points of A. Suppose that a point c (a, b) does not belong to A. Then we can write A = A 0 A, where A 0 = A (, 0), A = A (0, ). Both sets A 0 and A are relatively open and non-empty, which contradicts connectedness of A. Hence if two points a and b, a < b, are in A, then the whole interval [a, b] is also contained in A. Denote m := inf A and M := sup A (we assume that m = if A is unbounded from below and M = + if A is unbounded from above). Then the above argument shows that the open interval (m, M) is contained in A. Thus, there could be 5 cases: m, M / A; in this case A = (m, M); m A, M / A; in this case A = [m, M); m / A, M A; in this case A = (m, M]; m, M A and m < M; in this case A = [m, M]; m, M A and m = M; in this case A consists of one point. 90

91 Connected and path-connected components Lemma 8.8. Let A λ, λ Λ be any family of connected (resp. path-connected) subsets of a vector space V. Suppose A λ. Then A λ is also connected (resp. path-connected) λ Λ λ Λ Proof. Pick a point a A λ. Consider first the case when A λ are path connected. Pick a point λ Λ a A λ. Then a can be connected by path with all points in A λ for any points in λ Λ. Hence, λ Λ all points of A λ and A λ can be connected with each other for any λ, λ Λ. Suppose now that A λ are connected. Denote A := A λ. Suppose A can be presented as a union A = U U of disjoint relatively open subsets, where we denoted by U the set which contains the point a A λ. Then for each λ Λ the intersections U λ := U A λ and U λ := U A λ are λ Λ relatively open in A λ. We have A λ = U λ U λ. By assumption, U λ a, and hence U λ. Hence, connectedness of A λ implies that U λ =. But then U = =, and therefore A is connected. λ Λ U λ λ Λ Given any set A V and a point a A the connected component (resp. path-connected component C a A of the point a A is the union of all connected (resp. path-connected) subsets of A which contains the point a. Due to Lemma 8.8 the (path-)connected component C a is itself (path-)connected, and hence it is the biggest (path-)connected subset of A which contains the point a. The path-connected component of a can be equivalently defined as the set of all points of A one can connect with a by a path in A. Note that (path-)connected components of different points either coincide or do not intersect, and hence the set A can be presented as a disjoint union of (path-)-connected components. Lemma 8.5 shows that for open sets in a vector space V the notions of connected and pathconnected components coincide, and due to Exercise 8.6 the same is true for any subsets in R. In particular, any open set U R can be presented as a union of disjoint open intervals, which are its connected (= path-connected) components. Note that the number of these intervals can be infinite, but always countable. Continuous maps and functions Let V, W be two Euclidean spaces and A is a subset of V. A map f : A W is called continuous if one of the three equivalent properties hold: 9

92 . For any ɛ > 0 and any point x A there exists δ > 0 such that f(b δ (x) A) B ɛ (f(x)). 2. If for a sequence x n A there exists lim x n = x A then the sequence f(x n ) W converges to f(x). 3. For any open set U W the pre-image f (U) is relatively open in A. 4. For any closed set B W the pre-image f (B) is relatively closed in A. Let us verify equivalence of 3 and 4. For any open set U W its complement B = W \ U is closed and we have f (U) = A \ f (B). Hence, if f (U) is relatively open, i.e. f (U) = U A for an open set U V, then f (B) = A (V \ U ), i.e. f (B) is relatively closed. The converse is similar. Let us deduce from 3. The ball B ɛ (f(x)) is open. Hence f (B ɛ (f(x))) is relatively open in A. Hence, there exists δ > 0 such that B δ (x) A f (B ɛ (f(x))), i.e. f(b δ (x) A) B ɛ (f(x)). We leave the converse and the equivalence of definition 2 to the reader. Remark 8.9. Consider a map f : A W and denote B := f(a). Then definition 3 can be equivalently stated as follows: 3. For any set U B relatively open in B its pre-image f (U) is relatively open in A. Definition 4 can be reformulated in a similar way. Indeed, we have U = U A for an open set U W, while f (U) = f (U ). The following theorem summarize properties of continuous maps. Theorem 8.0. Let f : A W be a continuous map. Then. if A is compact then f(a) is compact; 2. if A is connected then f(a) is connected; 3. if A is path connected then f(a) is path-connected. 92

93 Figure 8.4: George Cantor (845-98) Proof.. Take any infinite sequence y n f(a). Then there exist points x n A such that y n = f(x n ), n =,.... Then there exists a converging subsequence x nk a A. Then by continuity lim k f(x nk ) = f(a) f(a), i.e. f(a) is compact. 2. Suppose that f(a) can be presented as a union B B 2 of simultaneously relatively open and closed disjoint non-empty sets. Then f (B ), f (B 2 ) A are simultaneously relatively open and closed in A, disjoint and non-empty. We also have f (B ) f (B 2 ) = f (B B 2 ) = f (f(a)) = A. Hence A is disconnected which is a contradiction. 3. Take any two points y 0, y f(a). Then there exist x 0, x A such that f(x 0 ) = y 0, f(x ) = y. But A is path-connected. Hence the points x 0, x can be connected by a path γ : [0, ] A. Then the path f γ : [0, ] f(a) connects y 0 and y, i.e. f(a) is path-connected. Note that in the case W = R Theorem 8.0. is just the Weierstrass theorem: a continuos function on a compact set is bounded and achieves its maximal and minimal values. We finish this section by a theorem of George Cantor about uniform continuity. Theorem 8.. Let A be compact and f : A W a continuous map. Then for any ɛ > 0 there exists δ > 0 such that for any x A we have f(b δ (x)) B ɛ (f(x)). Proof. Choose ɛ > 0. By continuity of f for every point x A there exists δ(x) > 0 such that f(b δ(x) (x)) B ɛ 4 (f(x)). 93

94 We need to prove that inf x A B δ(x) (x), and hence f(b δ(x) 2 δ(x) > 0. Note that for any point in y B δ(x) 2 (y)) B ɛ (f(y)). By compactness, from the covering (x) we have B δ(x) (y) 2 (x) we δ(x can choose a finite number of balls B δ(x j ) (x j ), j =,..., N which still cover A. Then δ = min j ) 2 k 2 satisfy the condition of the theorem, i.e. f(b δ (x)) B ɛ (f(x)) for any x A. x A B δ(x) 2 Remark on smooth maps, vector fields and differential forms on closed subsets We will often need to consider smooth maps, functions, vectors fields, differential forms, etc. defined on a closed subset A of a vector space V. We will always mean by that the these objects are defined on some open neighborhood U A. It will be unimportant for us how these objects are extended to U but to make sense of differentiability we need to assume that. In fact, one can define what differentiability means without any extension. Moreover, a theorem of Hassler Whitney asserts that any function smooth on a closed subset A V can be extended to a smooth function to a neighborhood U A. 8.2 Partition of unity and cut-off functions Let us recall that the support of a function θ is the closure of the set of points where it is not equal to 0. We denote the support by Supp(θ). We say that θ is supported in an open set U if Supp(θ) U. Lemma 8.2. There exists a C function ρ : R [0, ) with the following properties: ρ(x) 0, x ; ρ(x) = ρ( x); ρ(x) > 0 for x <. Proof. There are a lot of functions with this property. For instance, one can be constructed as follows. Take the function e x 2, x > 0 h(x) = 0, x 0. (8.2.) 94

95 The function e x 2 has the property that all its derivatives at 0 are equal to 0, and hence the function h is C -smooth. Then the function ρ(x) := h( + x)h( x) has the required properties. Lemma 8.3. Existence of cut-off functions Let C V be compact set and U C its open neighborhood. Then there exists a C -smooth function σ C,U : V [0, ) with its support in U which is equal to on C Proof. Let us fix a Euclidean structure in V and a Cartesian coordinate system. Thus we can identify V with R n with the standard dot-product. Given a point a V and δ > 0 let us denote by ψ a,δ the bump function on V defined by ( ) x a 2 ψ a,δ (x) := ρ, (8.2.2) where ρ : R [0, ) is the function constructed in Lemma 8.2. Note that ψ a,δ (x) is a C -function with Supp (ψ a,δ ) = D δ := B δ (a) and such that ψ a,δ (x) > 0 for x B δ (a). Let us denote by U ɛ (C) the ɛ-neighborhood of C, i.e. δ 2 U ɛ (C) = {x V ; y C, y x < ɛ.} There exists ɛ > 0 such that U ɛ (C) U. Using compactness of C we can find finitely many points z,..., z N C such that the balls B ɛ (z ),..., B ɛ (z N ) U cover U ɛ (C), i.e. U ɛ (C) N B 2 2 ɛ (z j ). Consider a function σ := N ψ zi, ɛ : V R. 2 The function ψ is positive on U ɛ 2 (C) and has Supp(ψ ) U. The complement E = V \ U ɛ (C) is a closed but unbounded set. Take a large R > 0 such that 2 B R (0) U. Then E R = D R (0) \ U ɛ (C) is compact. Choose finitely many points x,..., x 2 M E R such that M B ɛ (x i) E 4 R. Notice that M B ɛ (x i) C =. Denote 4 σ 2 := M ψ xi, ɛ. 4 Then the function σ 2 is positive on V R and vanishes on C. Note that the function σ +σ 2 is positive on B R (0) and it coincides with σ on C. Finally, define the function σ C,U by the formula σ C,U := σ σ + σ 2 95

96 on B ( R)(0) and extend it to the whole space V as equal to 0 outside the ball B R (0). Then σ C,U = on C and Supp(σ C,U ) U, as required.. Let C V be a compact set. Consider its finite covering by open sets U,..., U N, i.e. N U j C. We say that a finite sequence θ,..., θ K of C -functions defined on some open neighborhood U of C in V forms a partition of unity over C subordinated to the covering {U j } j=,...,n if K θ j (x) = for all x C; Each function θ j, j =,..., K is supported in one of the sets U i, i =,..., K. Lemma 8.4. For any compact set C and its open covering {U j } j=,...,n there exists a partition of unity over C subordinated to this covering. Proof. In view of compactness of there exists ɛ > 0 and finitely many balls B ɛ (z j ) centered at points z j C, j =,..., K, such that K B ɛ (z j ) C and each of these balls is contained in one of the open sets U j, j =,..., N. Consider the functions ψ zj,ɛ defined in (8.2.2). We have K ψ zj,ɛ > 0 on some neighborhood U C. Let σ C,U be the cut-off function constructed in Lemma 8.3. For j =,..., K we define ψ zj,ɛ(x)σ C,U (x), if x U, KP ψ θ j (x) = zj,ɛ(x). 0, otherwise Each of the functions is supported in one of the open sets U j, j =,..., N, and we have for every x C K θ j (x) = K ψ zj,ɛ(x) K ψ zj,ɛ(x) =. 96

97 Chapter 9 Integration of differential forms and functions 9. One-dimensional Riemann integral for functions and differential -forms A partition P of an interval [a, b] is a finite sequence a = t 0 < t < < t N = b. We will denote by T j, j = 0,..., N the vector t j+ t j R tj and by j the interval [t j, t j+ ]. The length t j+ t j = T j of the interval j will be denoted by δ j. The number max δ j is called the j=,...,n fineness or the size of the partition P and will be denoted by δ(p). Let us first recall the definition of (Riemann) integral of a function of one variable. Given a function f : [a, b] R we will form a lower and upper integral sums corresponding to the partition P: N L(f; P) = 0 N U(f; P) = The function is called Riemann integrable if 0 ( inf [t j,t j+ ] f)(t j+ t j ), ( sup [t j,t j+ ] f)(t j+ t j ), (9..) sup L(f; P) = inf U(f; P), P P 97

98 Figure 9.: Bernhard Riemann ( ) and in this case this number is called the (Riemann) integral of the function f over the interval [a, b]. The integrability of f can be equivalently reformulated as follows. Let us choose a set C = {c,..., c N }, c j j, and consider an integral sum I(f; P, C) = N 0 f(c j )(t j+ t j ), c j j. (9..2) Then the function f is integrable if there exists a limit lim I(f, P, C). In this case this limit is δ(p) 0 equal to the integral of f over the interval [a, b]. Let us emphasize that if we already know that the function is integrable, then to compute the integral one can choose any sequence of integral sum, provided that their fineness goes to 0. In particular, sometimes it is convenient to choose c j = t j, and in this case we will write I(f; P) instead of I(f; P, C). The integral has different notations. It can be denoted sometimes by notation for this integral is b a [a,b] f, but the most common f(x)dx. This notation hints that we are integrating here the differential form f(x)dx rather than a function f. Indeed, given a differential form α = f(x)dx we have f(c j )(t j+ t j ) = α cj (T j ), and hence I(α; P, C) = I(f; P, C) = N 0 α cj (T j ), c j j. (9..3) Here we parallel transported the vector T j from the point t j to the point c j [t j, t j+]. 98

99 We say that a differential -form α is integrable if there exists a limit lim I(α, P, C), which is δ(p) 0 called in this case the integral of the differential -form α over the oriented interval [a, b] and will be denoted by α, or simply [a,b] N b α. By definition, we say that α = α. This agrees with the a [a,b] [a,b] a b α cj ( T j ), and with the standard calculus rule f(x)dx = f(x)dx. a definition α = lim δ(p) 0 b [a,b] Let us recall that a map φ : [a, b] [c, d] is called a diffeomorphism if it is smooth and has a smooth inverse map φ : [c, d] [a, b]. This is equivalent to one of the following: φ(a) = c; φ(b) = d and φ > 0 everywhere on [a, b]. In this case we say that φ preserves orientation. φ(a) = d; φ(b) = c and φ orientation. < 0 everywhere on [a, b]. In this case we say that φ reverses Theorem 9.. Let φ : [a, b] [c, d] be a diffeomorphism. Then if a -form α = f(x)dx is integrable over [c, d] then its pull-back f α is integrable over [a, b], and we have φ α = α, (9..4) [a,b] [c,d] if φ preserves the orientation and φ α = α = α, [a,b] [c,d] [c,d] if φ reverses the orientation. Remark 9.2. We will show later a stronger result: φ α = α [a,b] [c,d] for any φ : [a, b] [c, d] with φ(a) = c, φ(b) = d, which is not necessarily a diffeomorphism. Proof. We consider only the orientation preserving case, and leave the orientation reversing one to the reader. Choose any partition P = {a = t 0 < < t N < t N = b} of the interval [a, b] and 99

100 choose any set C = {c 0,..., c N } such that c j j. Then the points t j = φ(t j ) [c, d], j = 0,... N form a partition of [c, d]. Denote this partition by P, and denote j := [ t j, t j+ ] [c, d], c j = φ(c j ), C = φ(c) = { c 0,..., c N }. Then we have N I(φ α, P, C) = 0 N 0 φ α cj (T j ) = N 0 α ecj (dφ(t j )) = α ecj (φ (c j )δ j ). (9..5) Recall that according to the mean value theorem there exists a point d j j, such that T j = t j+ t j = φ(t j+ ) φ(t j ) = φ (d j )(t j+ t j ). Note also that the function φ is uniformly continuous, i.e. for any ɛ > 0 there exists δ > 0 such that for any t, t [a, b] such that t t < δ we have φ (t) φ (t ) < ɛ. Besides, the function φ is bounded above and below by some positive constants: m < φ < M. Hence mδ j < δ j < Mδ j for all j =,..., N. Hence, if δ(p) < δ then we have N I(φ ( α, P, C) I(α; P, C) = α ecj (φ (c j ) φ ) (d j ))δ j = 0 ɛ N f( c j ) δ j = ɛ I(α; P, C). (9..6) m m When δ(p) 0 we have δ(p) = 0, and hence by assumption I(α; P, C) that I(φ α, P, C) I(α; P, C) 0, and thus φ α is integrable over [a, b] and b a φ α = d lim δ(p) 0 I(φ α, P, C) = lim I(α, P, C) = α. δ( P) 0 e c d α, but this implies If we write α = f(x)dx, then φ α = f(φ(t))φ (t)dt and the formula (9.) takes a familiar form of the change of variables formula from the -variable calculus: d c f(x)dx = b a f(φ(t))φ (t)dt. c 00

101 9.2 Integration of differential -forms along curves Curves as paths A path, or parametrically given curve in a domain U in a vector space V is a map γ : [a, b] U. We will assume in what follows that all considered paths are differentiable. Given a differential -form α in U we define the integral of α over γ by the formula α = γ α. γ [a,b] Example 9.3. Consider the form α = dz ydx + xdy on R 3. Let γ : [0, 2π] R 3 be a helix given by parametric equations x = R cos t, y = R sin t, z = Ct. Then γ 2π 2π α = (Cdt + R 2 (sin 2 tdt + cos 2 tdt)) = (C + R 2 )dt = 2π(C + R 2 ). 0 0 Note that α = 0 when C = R 2. One can observe that in this case the curve γ is tangent to the γ plane field ξ given by the Pfaffian equation α = 0. Proposition 9.4. Let a path γ be obtained from γ : [a, b] U by a reparameterization, i.e. γ = γ φ, where φ : [c, d] [a, b] is an orientation preserving diffeomorphism. Then eγ α = γ α. Indeed, applying Theorem 9. we get eγ α = d c γ α = d c b φ (γ α) a γ α = γ α. A vector γ (t) V γ(t) is called the velocity vector of the path γ. Curves as -dimensional submanifolds A subset Γ U is called a -dimensional submanifold of U if for any point x Γ there is a neighborhood U x U and a diffeomorphism Φ x : U x Ω x R n, such that Φ x (x) = 0 R n and Φ x (Γ U x ) either coincides with {x 2 =... x n = 0} Ω x, or with {x 2 =... x n = 0, x 0} Ω x. In 0

102 the latter case the point x is called a boundary point of Γ. In the former case it is called an interior point of Γ. A -dimensional submanifold is called closed if it is compact and has no boundary. An example of a closed -dimensional manifold is the circle S = {x 2 + y 2 = } R 2. WARNING. The word closed is used here in a different sense than when one speaks about closed subsets. For instance, a circle in R 2 is both, a closed subset and a closed -dimensional submanifold, while a closed interval is a closed subset but not a closed submanifold: it has 2 boundary points. An open interval in R (or any R n ) is a submanifold without boundary but it is not closed because it is not compact. A line in a vector space is a -dimensional submanifold which is a closed subset of the ambient vector space. However, it is not compact, and hence not a closed submanifold. Proposition Suppose that a path γ : [a, b] U is an embedding. This means that γ (t) 0 for all t [a, b] and γ(t) γ(t ) if t t. 2 Then Γ = γ([a, b]) is -dimensional compact submanifold with boundary. 2. Suppose Γ U is given by equations F = 0,..., F n = 0 where F,..., F n : U R are smooth functions such that for each point x Γ the differential d x F,..., d x F n are linearly independent. Then Γ is a -dimensional submanifold of U. 3. Any compact connected -dimensional submanifold Γ U can be parameterized either by an embedding γ : [a, b] Γ U if it has non-empty boundary, or by an embedding γ : S Γ U if it is closed. Proof.. Take a point c [a, b]. By assumption γ (c) 0. Let us choose an affine coordinate system (y,..., y n ) in V centered at the point C = γ(c) such that the vector γ (c) V C coincide with the first basic vector. In these coordinates the map gamma can be written as (γ,..., γ n ) where γ (c) =, γ j (c) = 0 for j > and γ j(c) = 0 for all j =,..., n. By the inverse function theorem the function γ is a diffeomorphism of a neighborhood of c onto a neighborhood of 0 in R (if c is one of the end points of the interval, then it is a diffeomorphism onto the corresponding one-sided neighborhood of 0). Let σ be the inverse function defined on the interval equal to ( δ, δ), [0, δ) and ( δ, 0], respectively, depending on whether c is an interior point, c = a or c = b]. 2 If only the former property is satisfied that γ is called an immersion. 02

103 so that γ (σ(u)) = u for any u. Denote = σ( ) [a, b]. Then γ( ) U can be given by the equations: y 2 = γ 2 (u) := γ 2 (σ(y )),..., y n = γ n (u) := γ n (σ(y )); y. Let us denote Denote θ = θ(δ) := max max γ j(u). j=2,...,n u P δ := { y δ, y j θ(δ)}. We have γ( ) δ. We will show now that for a sufficiently small δ we have γ([a, b]) P δ = γ( ). For every point t [a, b] \ Int denote d(t) = γ(t) γ(c). Recall now the condition that γ(t) γ(t ) for t t. Hence d(t) > 0 for all t [a, b] \ Int. The function d(t) is continuous and hence achieve the minimum value on the compact set [a, b] \ Int. Denote d := min d(t) > 0. Chose t [a,b]\int δ < min(d, δ) and such that θ(δ ) = max max γ j(u) < d. Let = { u δ }. Then j=2,...,n u c <δ γ([a, b]) P δ = {y 2 = γ 2 (u),..., y n = γ n (u); y }. 2. Take a point c Γ. The linear independent -forms d c F,..., d c F n V c by a -form l V c can be completed to a basis of V c. We can choose an affine coordinate system in V with c as its origin and such that the function x n coincides with l. Then the Jacobian matrix of the functions F,..., F n, x n is non-degenerate at c = 0, and hence by the inverse function theorem the map F = (F,..., F n, x n ) : V R n is invertible in the neighborhood of c = 0, and hence these functions can be chosen as new curvilinear coordinates y = F,..., y n = F n, y n = x n near the point c = 0. In these coordinates the curve Γ is given near c by the equations y = = y n = See.Exercise??. In the case where Γ is closed we will usually parameterize it by a path γ : [a, b] Γ U with γ(a) = γ(b). For instance, we parameterize the circle S = {x 2 + y 2 = } R 2 by a path [0, 2π] (cos t, sin t). Such γ, of course, cannot be an embedding, but we will require that γ (t) 0 and that for t t we have γ(t) γ(t ) unless one of these points is a and the other one is b. We will refer to -dimensional submanifolds simply as curves, respectively closed, with boundary etc. 03

104 Given a curve Γ its tangent line at a point x Γ is a subspace of V x generated by the velocity vector γ (t) for any local parameterization γ : [a, b] Γ with γ(t) = x. If Γ is given implicitly, as in 9.5.2, then the tangent line is defined in V x by the system of linear equations d x F = 0,..., d x F n = 0. Orientation of a curve Γ is the continuously depending on points orientation of all its tangent lines. If the curve is given as a path γ : [a, b] Γ U such that γ (t) 0 for all t [a, b] than it is canonically oriented. Indeed, the orientation of its tangent line l x at a point x = γ(t) Γ is defined by the velocity vector γ (t) l x. It turns out that one can define an integral of a differential form α over an oriented compact curve directly without referring to its parameterization. For simplicity we will restrict our discussion to the case when the form α is continuous. Let Γ be a compact connected oriented curve. A partition of Γ is a sequence of points P = {z 0, z,..., z N } ordered according to the orientation of the curve and such that the boundary points of the curve (if they exist) are included into this sequence. If Γ is closed we assume that z N = z 0. The fineness δ(p) of P is by definition is a Euclidean space). max dist(z j, z j+ ) (we assume here that V j=0,...,n Definition 9.6. Let α be a differential -form and Γ a compact connected oriented curve. Let P = {z 0,..., z N } be its partition. Then we define where I(α, P) = N 0 Γ α = α zj (Z j ), Z j = z j+ z j V zj. lim I(α, P), δ(p) 0 When Γ is a closed submanifold then one sometimes uses the notation α instead of α. Γ Γ Proposition 9.7. If one chooses a parameterization γ : [a, b] Γ which respects the given orientation of Γ then b α = γ α = α. γ a Γ Proof. Indeed, let P = {t 0,..., t N } be a partition of [a, b] such that γ(t j ) = z j, j = 0,..., N. 04

105 where U j = d tj γ(t j )) V zj I(γ α, P) N = γ α tj (T j ) = N α zj (U j ), is a tangent vector to Γ at the point z j. Let us evaluate the difference U j Z j. Choosing some Cartesian coordinates in V we denote by γ,..., γ n the coordinate functions of the path γ. Then using the mean value theorem for each of the coordinate functions we get γ i (t j+ ) γ i (t j ) = γ i (ci j )δ j for some c i j j, i =,..., n; j = 0,..., N. Thus Z j = γ(t j+ ) γ(t j ) = (γ (c j),..., γ i(c n j ))δ j. On the other hand, U j = d tj γ(t j )) V zj = γ (t j )δ j. Hence, Z j U j = δ j n (γ i (ci j ) γ i (t j)) 2. Note that if δ(p) 0 then we also have δ( P) 0, and hence using smoothness of the path γ we conclude that for any ɛ > 0 there exists δ > 0 such that Z j U j < ɛδ j for all j =,..., N. Thus N α zj ( T N j ) α zj (Z j ) δ(p) 0, and therefore b a γ α = lim I(γ α, P) = lim I(α, P) = δ( P) 0 e δ(p) 0 Γ α. 9.3 Integrals of closed and exact differential -forms Theorem 9.8. Let α = df be an exact -form in a domain U V. Then for any path γ : [a, b] U which connects points A = γ(a) and B = γ(b) we have α = f(b) f(a). γ In particular, if γ is a loop then γ α = 0. 05

106 Similarly for an oriented curve Γ U with boundary Γ = B A we have α = f(b) f(a). b b Γ Proof. We have df = γ df = d(f γ) = f(γ(b)) f(γ(a)) = f(b) f(a). γ a a It turns out that closed forms are locally exact. A domain U V is called star-shaped with respect to a point a V if with any point x U it contains the whole interval I a,x connecting a and x, i.e. I a,x = {a + t(x a); t [0, ]}. In particular, any convex domain is star-shaped. Proposition 9.9. Let α be a closed -form in a star-shaped domain U V. Then it is exact. Proof. Define a function F : U R by the formula F (x) = α, x U, I a,x where the intervals I a,x are oriented from 0 to x. We claim that df = α. Let us identify V with the R n choosing a as the origin a = 0. Then α can be written as α = n P k (x)dx k, and I 0,x can be parameterized by Hence, F (x) = t tx, t [0, ]. α = I 0,x Differentiating the integral over x j as parameters, we get But dα = 0 implies that P k x j F = x j 0 n k= F = x j = (tp j (tx)) 0 0 n k= 0 P k tx k (tx)dt + x j n P k (tx)x k dt. (9.3.) 0 P j (tx)dt. = P j x k, and using this we can further write P j tx k (tx)dt + x k 0 0 P j (tx)dt + P j (tx)dt = P j (tx)dt = P j (tx) t dp j(tx) dt + P j (tx)dt dt 0

107 Thus df = n j= F x j dx j = n P j (x)dx = α j= 9.4 Integration of functions over domains in high-dimensional spaces Riemann integral over a domain in R n. In this section we will discuss integration of bounded functions over bounded sets in a vector space V. We will fix a basis e,..., e n and the corresponding coordinate system x,..., x n in the space and thus will identify V with R n. Let η denote the volume form x... x n. As it will be clear below, the definition of an integral will not depend on the choice of a coordinate system but only on the background volume form, or rather its absolute value because the orientation of V will be irrelevant. We will need a special class of parallelepipeds in V, namely those which are generated by vectors proportional to basic vectors, or in other words, parallelepipeds with edges parallel to the coordinate axes. We will also allow these parallelepipeds to be parallel transported anywhere in the space. Let us denote P (a, b ; a 2, b 2 ;... ; a n, b n ) := {a i x i b i ; i =,..., n} R n. We will refer to P (a, b ; a 2, b 2 ;... ; a n, b n ) as a special parallelepiped, or rectangle. Let us fix one rectangle P := P (a, b ; a 2, b 2 ;... ; a n, b n ). Following the same scheme as we used in the -dimensional case, we define a partition P of P as a product of partitions a = t 0 < < t N = b,..., a n = t n 0 < < tn N n = b n, of intervals [a, b ],..., [a n, b n ]. For simplicity of notation we will always assume that each of the coordinate intervals is partitioned into the same number of intervals, i.e. N = = N n = N. This defines a partition of P into N n smaller rectangles P j = {t j x t j +,..., tn j n x n t n j }, where j = (j n+,..., j n ) and each index j k takes values between 0 and N. Let us define n Vol(P j ) := (t k j k + t k j k ). (9.4.) k= This agrees with the definition of the volume of a parallelepiped which we introduced earlier (see formula (3.3.) in Section 3.3). We will also denote δ j := max k=,...,n (tk j k + tk j k ) and δ(p) := max(δ j ). j 07

108 Let us fix a point c j P j and denote by C the set of all such c j. Given a function f : P R we form an integral sum I(f; P, C) = j f(c j )Vol(P j ) (9.4.2) where the sum is taken over all elements of the partition. If there exists a limit lim I(f; P, C) σ(p) 0 then the function f : P R is called integrable (in the sense of Riemann) over P, and this limit is called the integral of f over P. There exist several different notations for this integral: f, fdv, P P fdvol, etc. In the particular case of n = 2 one often uses notation fda, or f da. Sometime, the P P P functions we integrate may depend on a parameter, and in these cases it is important to indicate with respect to which variable we integrate. Hence, one also uses the notation like f(x, y)dx n, where P the index n refers to the dimension of the space over which we integrate. One also use the notation... f(x,... x n )dx... dx n, which is reminiscent both of the integral P f(x,... x n )dx } {{ } P dx n which will be defined later in Section 9.6 and the notation... f(x,... x n )dx... dx n for a n a n interated integral which will be discussed in Section 9.5. Alternatively and equivalently the integrability can be defined via upper and lower integral sum, similar to the -dimensional case. Namely, we define b n b U(f; P) = j M j (f)vol(p j ), L(f; P) = j m j (f)vol(p j ), where M j (f) = sup f, m j (f) = inf f, and say that the function f is integrable over P if inf U(f; P) = P j P j P U(f; P) and sup L(f, P) are sometimes called upper and lower integrals, respec- P f and f. Thus a function f : P R is integrable iff f = f. P P P sup L(f, P). P Note that inf P tively, and denoted by P Let us list some properties of Riemann integrable functions and integrals. Proposition 9.0. Let f, g : P R be integrable functions. Then. af + bg, where a, b R, is integrable and P af + bg = a P f + b P g; 2. If f g then P f P g; 08

109 3. h = max(f, g) is integrable; in particular the functions f + := max(f, 0) and f := max( f, 0) and f = f + + f are integrable; 4. fg is integrable. Proof. Parts and 2 are straightforward and we leave them to the reader as an exercise. Let us check properties 3 and Take any partition P of P. Note that M j (h) m j (h) max(m j (f) m j (f), M j (g) m j (g)). (9.4.3) Indeed, we have M j (h) = max(m j (f), M j (g)) and m j (h) max(m j (f), m j (g)). Suppose for determinacy that max(m j (f), M j (g)) = M j (f). We also have m j (h) m j (f). Thus M j (h) m j (h) M j (f) m j (f) max(m j (f) m j (f), M j (g) m j (g)). Then using (9.4.3) we have U(h; P) L(h; P) = j (M j (h) m j (h))vol(p j ) max (M j (f) m j (f), M j (g) m j (g)) Vol(P j ) = j max (U(f; P) L(f; P), U(f; P) L(f; P)). By assumption the right-hand side can be made arbitrarily small for an appropriate choice of the partition P, and hence h is integrable. 4. We have f = f + f, g = g + g and fg = f + g + + f g f + g f g +. Hence, using and 3 we can assume that the functions f, g are non-negative. Let us recall that the functions f, g are by assumption bounded, i.e. there exists a constant C > 0 such that f, g C. We also have 09

110 M j (fg) M j (f)m j (g) and m j (fg) m j (f)m j (g). Hence U(fg; P) L(fg; P) = j (M j (fg) m j (fg)) Vol(P j ) (M j (f)m j (g) m j (f)m j (g))vol(p j ) = j (M j (f)m j (g) m j (f)m j (g) + m j (f)m j (g) m j (f)m j (g)) Vol(P j ) j ((M j (f) m j (f))m j (g) + m j (f)(m j (g) m j (g))) Vol(P j ) j C(U(f; P) L(f; P) + U(g; P) L(g; P)). By assumption the right-hand side can be made arbitrarily small for an appropriate choice of the partition P, and hence fg is integrable. Consider now a bounded subset K R n and choose a rectangle P K. Given any function f : K R one can always extend it to P as equal to 0. A function f : K R is called integrable over K if this trivial extension f is integrable over P, and we define fdv := fdv. When this K will not be confusing we will usually keep the notation f for the above extension. Volume We further define the volume Vol(K) = dv = χ K dv, K P provided that this integral exists. In this case we call the set K measurable in the sense of Riemann, or just measurable. 3 Here χ K is the characteristic or indicator function of K, i.e. the function which is equal to on K and 0 elsewhere. In the 2-dimensional case the volume is called the area, and in the -dimensional case the length. 3 There exists a more general and more common notion of measurability in the sense of Lebesgue. Any Riemann measurable set is also measurable in the sense of Lebesgue, but not the other way around. Historically an attribution of this notion to Riemann is incorrect. It was defined by Camille Jordan and Giuseppe Peano before Riemann integral was introduced. What we call in these notes volume is also known by the name Jordan content. 0

111 Remark 9.. For any bounded set A there is defined a lower and upper volumes, Vol(A) = χ A dv Vol(A) = χ A dv. The set is measurable iff Vol(A) = Vol(A). If Vol(A) = 0 then Vol(A) = 0, and hence A is measurable and Vol(A) = 0. Exercise 9.2. Prove that for the rectangles this definition of the volume coincides with the one given by the formula (9.4.). The next proposition lists some properties of the volume. Proposition Volume is monotone, i.e. if A, B P are measurable and A B then Vol(A) Vol(B). 2. If sets A, B P are measurable then A B, A \ B and A B are measurable as well and we have Vol(A B) = Vol(A) + Vol(B) Vol(A B). 3. If A can be covered by a measurable set of arbitrarily small total volume then Vol(A) = 0. Conversely, if Vol(A) = 0 then for any ɛ > 0 there exists a δ > 0 such that for any partition P with δ(p) < δ the elements of the partition which intersect A have arbitrarily small total volume. 4. A is measurable iff Vol ( A) = 0. Proof. The first statement is obvious. To prove the second one, we observe that χ A B = max(χ A, χ B ), χ A B = χ A χ B, max(χ A, χ B ) = χ A + χ B χ A χ B, χ A\B = χ A χ A B and then apply Proposition 9.0. To prove we first observe that if a set B is measurable and VolB < ɛ then then for a sufficiently fine partition P we have U(χ B ; P) < VolB + ɛ < 2ɛ. Since A B then χ A χ B, and therefore U(χ A, P) U(χ B, P) < 2ɛ. Thus, inf P U(χ A, P) = 0 and therefore A is measurable and Vol(A) = 0. Conversely, if Vol(A) = 0 then for any ɛ > 0 for a sufficiently fine partition P we have U(χ A ; P) < ɛ. But U(χ A ; P) is equal to the sum of volumes of elements of the partition which have non-empty intersection with A.

112 Finally, let us prove Consider any partition P of P and form lower and upper integral sums for χ A. Denote M j := M j (χ A ) and m j = m j (χ A ). Then all numbers M j, m j are equal to either 0 or. We have M j = m j = if P j A; M j = m j = 0 if P j A = and M j =, m j = 0 if P j has non-empty intersection with both A and P \ A. In particular, B(P) := P j A. Hence, we have j;m j m j = U(χ A ; P) L(χ A ; P) = j (M j m j )Vol(P j ) = VolB(P). Suppose that A is measurable. Then there exists a partition such that U(χ A ; P) L(χ A ; P) < ɛ, and hence A is can be covered by the set B(P) of volume < ɛ. Thus applying part 3 we conclude that Vol( A) = 0. Conversely, we had seen below that if Vol( A) = 0 then there exists a partition such that the total volume of the elements intersecting A is < ɛ. Hence, for this partition we have L(χ A ; P) U(χ A ; P) < ɛ, which implies the integrability of χ A, and hence measurability of A. Corollary 9.4. If a bounded set A V is measurable then its interior Int A and its closure A are also measurable and we have in this case VolA = Vol Int A = VolA. Proof.. We have A A and (Int A) A. Therefore, Vol A = Vol Int A = 0, and therefore the sets A and Int A are measurable. Also Int A A = A and Int A A =. Hence, the additivity of the volume implies that VolA = Vol Int A+Vol A = Vol Int A. On the other hand, Int A A A. and hence the monotonicity of the volume implies that Vol Int A Vol A Vol A. Hence, Vol A = Vol Int A = Vol A. Exercise 9.5. If Int A or A are measurable then this does not imply that A is measurable. For instance, if A is the set of rational points in interval I = [0, ] R then Int A = and A = I. However, show that A is not Riemann measurable. 2. A set A is called nowhere dense if Int A =. Prove that if A is nowhere dense then either VolA = 0, or A is not measurable in the sense of Riemann. Find an example of a non-measurable nowhere dense set. 2

113 Volume and smooth maps We will further need the following lemma. Let us recall that given a compact set C V, we say that a map f : C W is smooth if it extends to a smooth map defined on an open neighborhood U C. Here V, W are vector spaces. Lemma 9.6. Let A V be a compact set of volume 0 and f : V W a C -smooth map, where dim W dim V. Then Volf(A) = 0. Proof. The C -smoothness of f and compactness of A imply that there exists a constant K such that d x f(h) K h for any x A and h V x. In particular, the image d x f(p ) of every rectangle P of size δ in V x, x A, is contained in a cube of size Kδ in W f(x). C -smoothness of f also implies that for any ɛ > 0 there exists such δ > 0 that if x A and h δ then f(x+h) f(x) d x f(h) ɛ h. This implies that if we view P as a subset of V, rather than V x, then the image f(p ) is contained in a cube in W of size 2Kδ if δ is small enough, and if P A. Let us denote dimensions of V and W by n and m, respectively. By assumption A can be covered by N cubes of size δ such that the total volume of these cubes is equal to Nδ n ɛ. Hence, f(a) can be covered by N cubes of size 2Kδ of total volume N(2Kδ) m = Nδ n (2K) m δ m n = ɛ(2k) m δ m n ɛ 0 0, because m n. Corollary 9.7. Let A V be a compact domain and f : A W a C -smooth map. Suppose that n = dim V < m = dim W. Then Vol(f(A)) = 0. Indeed, f can be extended to a smooth map defined on a neighborhood of A 0 in V R (e.g. as independent of the new coordinate t R). But Vol n+ (A 0) = 0 and m n +. Hence, the required statement follows from Lemma 9.6. Remark 9.8. The statement of Corollary 9.7 is wrong for continuous maps. For instance, there exists a continuous map h : [0, ] R 2 such that h([0, ]) is the square {0 x, x }. (This is a famous Peano curve passing through every point of the square.) Corollary 9.7 is a simplest special case of Sard s theorem which asserts that the set of critical values of a sufficiently smooth map has volume 0. More precisely, 3

114 Proposition 9.9. (A. Sard, 942) Given a C k -smooth map f : A W (where A is a compact subset of V, dim V = n, dim W = m) let us denote by Σ(f) : {x C; rank d x f < m}. Then if k max(n m +, ) then Vol m (f(σ(f)) = 0. If m > n then Σ(f) = A, and hence the statement is equivalent to Corollary 9.7. Proof. We prove the proposition only for the case m = n. The proof in this case is similar to the proof of Corollary 9.6. C -smoothness of f and compactness of Σ(f) imply that there exists a constant K such that d x f(h) K h (9.4.4) for any x A and h V x. C -smoothness of f also implies that for any ɛ > 0 there exists such δ > 0 that if x A and h δ then f(x + h) f(x) d x f(h) ɛ h. (9.4.5) Take a partition P of a rectangle P A by N n smaller rectangles of equal size. Let B be the union of the rectangles intersecting Σ(f). For any such rectangle P j B choose a point c j Σ(f). Viewing P j as a subset of V cj we can take its image P j = d cj (P bj ) W f(cj ). Then the parallelepiped P j is contained in a subspace L W f(cj ) of dimension r = rank(d cj f) < m. In view of (9.4) it also contained in a ball of radius K r N centered at the point f(c j). On the other hand, the inequality (9.4.5) implies that if N is large enough then for any point u f(p j ) there is a point in ũ P j such u ũ ɛ N. This means that f(p j) is contained in a parallelepiped centered at c j and generated by r orthogonal vectors of length C N of length C 2ɛ N parallel to the subspace L and n r > 0 vectors which are orthogonal to L, where C, C 2 are some positive constants. The volume of this parallelepiped is equal to C rcn r 2 ɛ n r N n = C 3ɛ n r N n, and hence Vol(f(P j ) < C 3ɛ n r N. The set B contains no more that N n cubes P n j, and hence Vol(f(Σ(f)) Volf(C) N n C 3ɛ n r N n = C 3 ɛ n r ɛ

115 Properties which hold almost everywhere We say that some property holds almost everywhere (we will abbreviate a.e.) if it holds in the complement of a set of volume 0. For instance, we say that a bounded function f : P R is almost everywhere continuous (or a.e. continuous) if it is continuous in the complement of a set A P of volume 0. For instance, a characteristic function of any measurable set is a.e. continuous. Indeed, it is constant away from the set A which according to Proposition has volume 0. Proposition Suppose that the bounded functions f, g : P R coincide a.e. Then if f is integrable, then so is g and we have P f = P g. Proof. Denote A = {x P : f(x) g(x)}. By our assumption, VolA = 0. Hence, for any ɛ there exists a δ > 0 such that for every partition P with δ(p) δ the union B δ of all rectangles of the partition which have non-empty intersection with A has volume < ɛ. The functions f, g are bounded, i.e. there exists C > 0 such C f(x), g(x) C for all x P. Due to integrability of f we can choose δ small enough so that U(f, P) L(f, P) ɛ when δ(p) δ. Then we have U(g, P) U(f, P) = sup g sup f) J: P J B P J P J δ 2CVolB δ 2Cɛ. Similarly, L(g, P) L(f, P) 2Cɛ, and hence U(g, P) L(g, P) U(g, P) U(f, P) + U(f, P) L(f, P) + L(f, P) L(g, P) ɛ + 4Cɛ δ 0 0, and hence g is integrable and g = P lim U(g, P) = lim δ(p) 0 U(f, P) = δ(p) 0 P f. Proposition Suppose that a function f : P R is a.e. continuous. Then f is integrable. 2. Let A V be compact and measurable, f : U W a C -smooth map defined on a neighborhood U A. Suppose that dim W = dim V. Then f(a) is measurable. 5

116 Proof.. Let us begin with a Warning. One could think that in view of Proposition 9.20 it is sufficient to consider only the case when the function f is continuous. However, this is not the case, because for a given a.e. continuos function one cannot, in general, find a continuos function g which coincides with f a.e. Let us proceed with the proof. Given a partition P we denote by J A the set of multi-indices j such that Int P j A, and by J A the complementary set of multi-indices, i.e. for each j J A we have P j A =. Let us denote C := P j. According to Proposition for any ɛ > 0 there exists j J A a partition P such that Vol(C) = Vol(P j ) < ɛ. By assumption the function f is continuous j J A over a compact set B = P j, and hence it is uniformly continuous over it. Thus there exists j J A δ > 0 such that f(x) f(x ) < ɛ provided that x, x B and x x < δ. Thus we can further subdivide our partition, so that for the new finer partition P we have δ(p ) < δ. By assumption the function f is bounded, i.e. there exists a constant K > 0 such that M j (f) m j (f) < K for all indices j. Then we have U(f; P ) L(f, P ) = j (M j (f) m j (f))vol(p j ) = (M j (f) m j (f))vol(p j ) + (M j (f) m j (f))vol(p j ) < j;p j B j;p j C ɛvolb + KVolC < ɛ(volp + K). Hence inf U(f; P) = sup L(f; P), i.e. the function f is integrable. P P 2. If x is an interior point of A and detdf(x) 0 then the inverse function theorem implies that f(x) Int f(a). Denote C = {x A; det Df(x) = 0}. Hence, f(a) f( A) f(c). But Vol( A) = 0 because A is measurable and Vol f(c) = 0 by Sard s theorem 9.9. Therefore, Vol f(a) = 0 and thus f(a) is measurable. Orthogonal invariance of the volume and volume of a parallelepiped The following lemma provides a way of computing the volume via packing by balls rather then cubes. An admissible set of balls in A is any finite set of disjoint balls B,..., B K A 6

117 Lemma Let A be a measurable set. Then VolA is the supremum of the total volume of admissible sets of balls in A. Here the supremum is taken over all admissible sets of balls in A. Proof. Let us denote this supremum by β. The monotonicity of volume implies that β VolA. Suppose that β < V ola. Let us denote by µ n the volume of an n-dimensional ball of radius (we will compute this number later on). This ball is contained in a cube of volume 2 n. It follows then that the ratio of the volume of any ball to the volume of the cube to which it is inscribed is equal to µn 2. Choose an ɛ < µn n 2 (VolA β). Then there exists a finite set of disjoint balls B n,..., B K A such that Vol( K B j ) > β ɛ. The volume of the complement C = A \ K B j satisfies ( K ) VolC = VolA Vol B j > VolA β. Hence there exists a partition P of P by cubes such that the total volume of cubes Q,..., Q L contained in C is > VolA β. Let us inscribe in each of the cubes Q j a ball B j. Then B,..., B K, B,..., B L is an admissible set of balls in A. Indeed, all these balls are disjoint and contained in A. The total volume of this admissible set is equal to K VolB j + L Vol B i β ɛ + µ n (VolA β) > β, 2n in view of our choice of ɛ, but this contradicts to our assumption β < VolA. Hence, we have β = VolA. Lemma Let A V be any measurable set in a Euclidean space V. Then for any linear orthogonal transformation F : V V the set F (A) is also measurable and we have Vol(F (A)) = Vol(A). Proof. First note that if VolA = 0 then the claim follows from Lemma 9.6. Indeed, an orthogonal transformation is, of course a smooth map. Let now A be an arbitrary measurable set. Note that F (A) = F ( A). Measurability of A implies Vol( A) = 0. Hence, as we just have explained, Vol( F (A)) = Vol(F ( A)) = 0, and hence F (A) is measurable. According to Lemma 9.22 the volume of a measurable set can be computed as a supremum of the total volume of disjoint inscribed balls. But the orthogonal transformation F moves disjoint balls to disjoint balls of the same size, and hence VolA = VolF (A). 7

118 Next proposition shows that the volume of a parallelepiped can be computed by formula (3.3.) from Section 3.3. Proposition Let v,..., v n V be linearly independent vectors. Then Vol P (v,..., v n ) = x x n (v..., v n ). (9.4.6) Proof. The formula (9.4.6) holds for rectangles, i.e. when v j = c j e j for some non-zero numbers c j, j =,... n. Using Lemma 9.23 we conclude that it also holds for any orthogonal basis. Indeed, any such basis can be moved by an orthogonal transformation to a basis of the above form c j e j, j =,... n. Lemma 9.23 ensures that the volume does not change under the orthogonal transformation, while Proposition 2.7 implies the same about x x n (v..., v n ). The Gram-Schmidt orthogonalization process shows that one can pass from any basis to an orthogonal basis by a sequence of following elementary operations: - reordering of basic vectors, and shears, i.e. an addition to the last vector a linear combination of the other ones: n v,..., v n, v n v,..., v n, v n + λ j v j. Note that the reordering of vectors v,..., v n changes neither VolP (v,..., v n ), nor the absolute value x x n (v..., v n ). On the other hand, a shear does not change x x n (v..., v n ). It remains to be shown that a shear does not change the volume of a parallelepiped. We will consider here only the case n = 2 and will leave to the reader the extension of the argument to the general case. Let v, v 2 be two orthogonal vectors in R 2. We can assume that v = (a, 0), v 2 = (0, b) for a, b > 0, because we already proved the invariance of volume under orthogonal transformations. Let v 2 = v 2 + λv = (a, b), where a = a + λb. Let us partition the rectangle P = P (v, v 2 ) into N 2 smaller rectangles P i,j, i, j = 0,..., N, of equal size. We number the rectangles in such a way 8

119 Figure 9.2: Guido Fubini ( ) that the first index corresponds to the first coordinate, so that the rectangles P 00,..., P N,0 form the lower layer, P 0,..., P N, the second layer, etc. Let us now shift the rectangles in k-th layer horizontally by the vector ( kλb N, 0). Then the total volume of the rectangles, denoted P ij remains the same, while when N the volume of part of the parallelogram P (v, v 2 ) that is not covered by rectangles P i,j, i, j = 0,..., N converges to Fubini s Theorem Let us consider R n as a direct product of R k and R n k for some k =,..., n. We will denote coordinates in R k by x = (x,..., x k ) and coordinates in R n k by y = (y,..., y n k ), so the coordinates in R n are denoted by (x,..., x k, y,..., y n k ). Given rectangles P R k and P 2 R n k their product P = P P 2 is a rectangle in R n. The following theorem provides us with a basic tool for computing multiple integrals. Theorem 9.25 (Guido Fubini). Suppose that a function f : P R is integrable over P. Given a point x P let us define a function f x : P 2 R by the formula f x (y) = f(x, y), y P 2. Then fdv n = f x dv n k dv k = f x dv n k dv k. P P P 2 P P 2 9

120 In particular, if the function f x is integrable for all (or almost all) x P then one has fdv n = f x dv n k dv k. P P P 2 Here by writing dv k, dv n k and dv n we emphasize the integration with respect to the k-, (n k)- and n-dimensional volumes, respectively. Proof. Choose any partition P of P and P 2 of P 2. We will denote elements of the partition P by P j and elements of the partition P 2 by P2 i. Then products of P j,i = P j P 2 i form a partition P of P = P P 2. Let us denote I(x) := f x, I(x) := P 2 Let us show that Indeed, we have P 2 f x, x P. L(f, P) L(I, P ) U(I, P ) U(f, P). (9.5.) L(f, P) = j m j,i (f)vol n P j,i. i Here the first sum is taken over all multi-indices j of the partition P, and the second sum is taken over all multi-indices i of the partition P 2. On the other hand, L(I, P ) = inf f x dv n k Vol k P j j x P j. P 2 Note that for every x P j we have f x dv n k L(f x ; P 2 ) = m i (f x )Vol n k (P2) i i i P 2 and hence Therefore, L(I, P ) j inf f x dv n k x P j P 2 m i,j (f)vol n k (P2)Vol i k (P j ) = i j m i,j (f)vol n k (P2). i i m i,j (f)vol n k (P i 2), m j,i (f)vol n (P j,i ) = L(f, P). i 20

121 Similarly, one can check that U(I, P ) U(f, P). Together with an obvious inequality L(I, P ) U(I, P ) this completes the proof of (9.5.). Thus we have max(u(i, P ) L(I, P ), U(I, P ) L(I, P )) U(I, P ) L(I, P ) U(f, P) L(f, P). By assumption for appropriate choices of partitions, the right-hand side can be made < ɛ for any a priori given ɛ > 0. This implies the integrability of the function I(x) and I(x) over P. But then we can write and P I(x)dV n k = lim δ(p ) 0 L(I; P ) I(x)dV n k = lim U(I; P ). δ(p ) 0 P We also have lim L(f; P) = lim U(f; P) = fdv n. δ(p) 0 δ(p) 0 P Hence, the inequality (9.5.) implies that P fdv n = P P 2 f x dv n k dv k = P f x dv n k dv k. P 2 Corollary Suppose f : P R is a continuous function. Then f = f x = f y. P P 2 P P Thus if we switch back to the notation x,..., x n for coordinates in R n, and if P = {a x b,..., a n x n b n } then we can write b n b f =... f(x,..., x n )dx... dx n. (9.5.2) a P a n P 2 The integral in the right-hand side of (9.5.2) is called an iterated integral. Note that the order of integration is irrelevant there. In particular, for continuous functions one can change the order of integration in the iterated integrals. 2

122 9.6 Integration of n-forms over domains in n-dimensional space Differential forms are much better suited to be integrated than functions. For integrating a function, one needs a measure. To integrate a differential form, one needs nothing except an orientation of the domain of integration. Let us start with the integration of a n-form over a domain in a n-dimensional space. Let ω be a n-form on a domain U V, dim V = n. Let us fix now an orientation of the space V. Pick any coordinate system (x... x n ) that agrees with the chosen orientation. We proceed similar to the way we defined an integral of a function. Let us fix a rectangle P = P (a, b ; a 2, b 2 ;... ; a n, b n ) = {a i x i b i ; i =,..., n}. Choose its partition P by N n smaller rectangles P j = {t j n x t j,..., n+ tn j x t n j + }, where j = (j,..., j n ) and each index j k takes values between 0 and N. Let us fix a point c j P j and denote by C the set of all such c j. We also denote by t j the point with coordinates t j,..., t n j n and by T j,m V cj, m =,..., n the vector t j+m t j, parallel-transported to the point c j. Here we use the notation j + m for the multi-index j,..., j m, j m +, j m+,..., j n. Thus the vector T j,m is parallel to the m-th basic vector and has the length t jm+ t jm. Given a differential n-form α on P we form an integral sum I(α; P, C) = j α(t j, T j 2,..., T j,n), (9.6.) where the sum is taken over all elements of the partition. We call an n-form α integrable if there exists a limit lim I(α; P, C) which we denote by α and call the integral of α over P. Note δ(p) 0 P that if α = f(x)dx dx n then the integral sum I(α, P, C) from (9.6.) coincides with the integral sum I(f; P, C) from (9.4.2) for the function f. Thus the integrability of α is the same as integrability of f and we have P f(x)dx dx n = P fdv. (9.6.2) Note, however, that the equality (9.6.2) holds only if the coordinate system (x,..., x n ) defines the given orientation of the space V. The integral f(x)dx dx n changes its sign with a change P of the orientation while the integral P fdv is not sensitive to the orientation of the space V. 22

123 It is not clear from the above definition whether the integral of a differential form depends on our choice of the coordinate system. It turns out that it does not, as the following theorem, which is the main result of this section, shows. Moreover, we will see that one even can use arbitrarty curvilinear coordinates. In what follows we use the convention introduced at the end of Section 8.. Namely by a diffeomorphism between two closed subsets of vector spaces we mean a diffeomorphism between their neighborhoods. Theorem Let A, B R n be two measurable compact subsets. Let f : A B be an orientation preserving diffeomorphism. Let η be a differential n-form defined on B. Then if η is integrable over B then f α is integrable over A and we have f η = A B η. (9.6.3) For an orientation reversing diffeomorphism f we have f η = η. A B Let α = g(x)dx dx n. Then f α = g f det Dfdx dx n, and hence the formula (9.6.3) can be rewritten as g(x,..., x n )dx dx n = g f det Df dx dx n. P P Here f f x... x n det Df = f n f x... n is the determinant of the Jacobian matrix of f = (f,..., f n ). Hence, in view of formula (9.6.2) we get the following change of variables formula for multiple integrals of functions. Corollary [Change of variables in a multiple integral] Let g : B R be an integrable function and f : A B a diffeomorphism. Then the function g f is also integrable and gdv = g f det Df dv. (9.6.4) B A 23 x n

124 Figure 9.3: Image of a cube under a diffeomorphism and its linearization We begin the proof with the following special case of Theorem Proposition The statement of 9.27 holds when η = dx dx n and the set A is the unit cube I = I n. We will use below the following notation. For any set A V and any positive number λ > 0 we denote by λa the set {λx, x A}. For any linear operator F : V W between two Euclidean spaces V and W we define its norm F by the formula F = max F (v) = max v = v V,v 0 F (v). v Equivalently, we can define F as follows. The linear map F maps the unit sphere in the space V onto an ellipsoid in the space W. Then F is the biggest semi-axis of this ellipsoid. Let us begin by observing the following geometric fact: 24

125 Lemma Let I = { x i 2, i =,..., n} Rn be the unit cube centered at 0 and F : R n R n a non-degenerate linear map. Take any ɛ (0, ) and set σ = z I we have see Fig. 9.6 ɛ. Then for any boundary point F B σ (F (z)) ( + ɛ)f (I) \ ( ɛ)f (I), (9.6.5) Proof. Inclusion (9.6.5) can be rewritten as F (B σ (F (z)) ( + ɛ)i \ ( ɛ)i. But the set F (B σ (F (z)) is an ellipsoid centered at z whose greatest semi-axis is equal to σ F. Hence, if σ F ɛ then F (B σ (F (z)) ( + ɛ)i \ ( ɛ)i. We will also need Lemma 9.3. Let U R n is an open set, f : U R n such that f(0) = 0. Suppose that f is differentiable at 0 and its differential F = d 0 f : R n R n at 0 is non-degenerate. Then for any ɛ (0, ) there exists δ > 0 such that see Fig ( ɛ)f (δi) f(δi) ( + ɛ)f (δi), (9.6.6) Proof. First, we note that inclusion (9.6.5) implies, using the linearity of F, that for any δ > 0 we have B δσ (F (z)) ( + ɛ)f (δi) \ ( ɛ)f (δi), (9.6.7) where z (δi) and, as in Lemma 9.30, we assume that ɛ = σ F. According to the definition of differentiability we have f(h) = F (h) + o( h ). Denote σ := σ n = ɛ n F. There exists ρ > 0 such that if h ρ then f(h) F (h) σ h σρ. 25

126 Denote δ := ρ n. Then δi B ρ (0), and hence f(z) F (z) σρ for any z δi. In particular, for any point z (δi) we have f(z) B eσρ (F (z)) = B neσδ(f (z)) = B σδ (F (z)), and therefore in view of (9.6.7) f( (δi)) ( + ɛ)f (δi) \ ( ɛ)f (δi). But this is equivalent to inclusion (9.6.6). Lemma Let Let F : R n R n be a non-degenerate orientation preserving linear map and P = P (v,..., v n ) be a parallelepiped, and η = dx dx n. Then η = F η. Here we F (P ) P assume that the orientation of P and F (P ) are given by the orientation of R n. Proof. We have F (P ) η = F (P )dx... dx n F η = det F η, and hence F η = det F η = det F VolP. P P = VolF (P ) = det F VolP. On the other hand, Proof of Proposition We have f η = (det Df)dx dx n, and hence the form f η is integrable because f is C -smooth, and hence det Df is continuous. Choose a partition P of the cube I by N n small cubes I K, K =,..., N n, of the same size N. Let c K I K be the center of the cube I K. Then I f η = N n K= I K f η. Note that in view of the uniform continuity of the function det Df, for any ɛ > 0 the number N can be chosen so large that det Df(x) det Df(x ) < ɛ for any two points x, x I K and any K =,..., N n. Let η K be the form det Df(c k )dx dx n on I K. Then f η η K det Df(x) det Df(c K ) dx dx n ɛvol(i K ) = I K I K I K ɛ N n. Thus I f η N n K= I K η K ɛ. (9.6.8) 26

127 Next, let us analyze the integral η K. Denote F K := d ck (f). We can assume without loss of I K generality that c K = 0, and f(c K ) = 0, and hence F K can be viewed just as a linear map R n R n. Using Lemma 9.3 we have for a sufficiently large N ( ɛ)f K (I K ) f(i K ) ( + ɛ)f K (I K ). Again in view of compactness of I the number N can be chosen the same for all cubes I K. Hence Note that Volf(I K ) = N n ( ɛ) n K= ( ɛ) n Vol(F K (I K )) Volf(I K ) ( + ɛ) n Vol(F K (I K )). (9.6.9) f(i K ) Vol(F K (I K )) η, and hence summing up inequality (9.6.9) over K we get N n K= Volf(I K ) = N n K= f(i K ) η = f(i) N n η ( + ɛ) n Note that η K = FK η and by Lemma 9.32 we have η K = FK η = I K I K Hence, it follows from (9.6.0) that N n ( ɛ) n K= I K η K f(i) N n η ( + ɛ) n K= K= F K (I K ) Vol(F K (I K )). (9.6.0) η = Vol(F K (I K ). I K η K. (9.6.) Recall that from (9.6.8) we have I f η ɛ N n K= I K I f η + ɛ. Combining with (9.6.) we get ( ɛ) n f η ɛ η ( + ɛ) n f η + ɛ. (9.6.2) I f(i) I Passing to the limit when ɛ 0 we get f η η f η, (9.6.3) i.e. I I f(i) I f η = η. f(i) 27

128 Proof of Theorem Let us recall that the diffeomorphism f is defined as a diffeomorphism between open neighborhoods U A and U B. We also assume that the form η is extended to U as equal to 0 outside B. The form η can be written as hdx dx n. Let us take a partition P of a rectangular containg U by cubes I j of the same size δ. Consider forms η + j := M j (h)dx dx n and η j := m j (h)dx dx n on I j, where m j (h) = inf Ij h, M j (h) = sup Ij (h). Let η ± be the form on U equal to η ± j on each cube I j. The assumption of integrability of η over B guarantees that for any ɛ > 0 if δ is chosen small enough we have B η+ B η ɛ. The forms f η ± are a.e. continuous, and hence integrable over A and we have f η f η f η +. Hence, if we prove A A A that f η ± = η ± then this will imply that η is integrable and η = η. A B A B On the other hand, η ± = η ± j and f η ± = f η ± j, where B j = f (I j ). But according B j I j A j B j to Proposition 9.29 we have f η ± j = η ± j, and hence f η ± = η ±. B j I j A B 9.7 Manifolds and submanifolds 9.7. Manifolds Manifolds of dimension n are spaces which are locally look like open subsets of R n but globally could be much more complicated. We give a precise definition below. Let U, U R n be open sets. A map f : U U is called a homeomorpism if it is continuous one-to-one map which has a continuous inverse f : U U. A map f : U U is called a C k -diffeomorpism, k =,...,, if it is C k -smooth, one-to-one map which has a C k -smooth inverse f : U U. Usually we will omit the reference to the class of smoothness, and just call f a diffeomorphism, unless it will be important to emphasize the class of smoothness. A set M is called an n-dimensional C k -smooth (resp. topological) manifold if there exist subsets U λ X, λ Λ, where Λ is a finite or countable set of indices, and for every λ Λ a map Φ λ : U λ R n such that M. M = λ Λ U λ. M2. The image G λ = Φ λ (U λ ) is an open set in R n. 28

129 M3. The map Φ λ viewed as a map U λ G λ is one-to-one. M4. For any two sets U λ, U µ, λ, µ Λ the images Φ λ (U λ U µ ), Ψ µ (U λ U µ ) R n are open and the map h λµ := Φ µ Φ λ is a C k -diffeomorphism (resp. homeomorphism). : Φ λ(u λ U µ ) Φ µ (U λ U µ ) R n Sets U λ are called coordinate neighborhoods and maps Φ λ : U λ R n are called coordinate maps. The pairs (U λ, Φ λ ) are also called local coordinate charts. The maps h λµ are called transiton maps between different coordinate charts. The inverse maps Ψ λ = Φ λ : G λ U λ are called (local) parameterization maps. An atlas is a collection A = {U λ, Φ λ } λ Λ of all coordinate charts. One says that two atlases A = {U λ, Φ λ } λ Λ and A = {U γ, Φ γ} γ Γ on the same manifold X are equivalent, or that they define the same smooth structure on X if their union A A = {(U λ, Φ λ ), (U γ, Φ γ)} λ Λ,γ Γ is again an atlas on X. In other words, two atlases define the same smooth structure if transition maps from local coordinates in one of the atlases to the local coordinates in the other one are given by smooth functions. A subset G M is called open if for every λ Λ the image Φ λ (G U λ ) R n is open. In particular, coordinate charts U λ themselves are open, and we can equivalently say that a set G is open if its intersection with every coordinate chart is open. By a neighborhood of a point a M we will mean any open subset U M such that a U. Given two smooth manifolds M and M of dimension m and n then a map f : M M is called continuous if if for every point a M there exist local coordinate charts (U λ, Φ λ ) in M and (Ũλ, Φ λ ) in M λ, such that a U λ, f(u λ ) Ũλ and the composition map G λ = Φ λ (U λ ) Ψ λ f Φλ e U λ Ũ λ R n is continuous. Similarly, for k =,..., a map f : M M is called C k -smooth if for every point a M there exist local coordinate charts (U λ, Φ λ ) in M and (Ũλ, Φ λ ) in M λ, such that a U λ, f(u λ ) Ũλ and the composition map G λ = Φ λ (U λ ) Ψ λ f Φλ e U λ Ũ λ R n 29

130 is C k -smooth. In other words, a map is continuous or smooth, if it is continuous or smooth when expressed in local coordinates. A map f : M N is called a diffeomorphism if it is smooth, one-to-one, and the inverse map is also smooth. One-to-one continuous maps with continuous inverses are called homeomorphisms. Note that in view of the chain rule the C k -smoothness is independent of the choice of local coordinate charts (U λ, Φ λ ) and (Ũλ, Φ λ ). Note that for C k -smooth manifolds one can talk only about C l -smooth maps for l k. For topological manifolds one can talk only about continuous maps. If one replaces condition M2 in the definition of a manifold by M2 b. The image G λ = Ψ λ (U λ ) is either an open set in R n or an intersection of an open set in R n with R n + = {x 0} then one gets a definition of a manifold with boundary. A slightly awkward nuance in the above definition is that a manifold with boundary is not a manifold! It would be, probably, less confusing to write this as a word manifold-with-boundary, but of course nobody does that. The points of a manifold M with boundary which are mapped by coordinate maps Ψ λ to points in R n = R n + are called the boundary points of M. The set of boundary points is called the boundary of M and denoted by M. It is itself a manifold of dimension n. Note that any (interior) point a of an n-dimensional manifold M has a neighborhood B diffeomorphic to an open ball B (0) R n, while any boundary point has a neighborhood diffeomorphic to a semi-ball B (0) {x 0} R n. Exercise Prove that a boundary point does not have a neighborhood diffeomorphic to an open ball. In other words, the notion of boundary and interior point of a manifold with boundary are well defined. Next we want to introduce a notion of compactness for subsets in a manifold. Let us recall that for subsets in a Euclidean vector space we introduced three equivalent definition of compactness, see Section 8.. The first definition, COMP is unapplicable because we cannot talk about bounded sets in a manifold. However, definitions COMP2 and COMP3 make perfect sense in an arbitrary 30

131 manifold. For instance, we can say that a subset A M is compact if from any infinite sequence of points in A one can choose a subsequence converging to a point in A. A compact manifold (without boundary) is called closed. Note that the word closed is used here in a different sense than a closed set. For instance, a closed interval is not a closed manifold because it has a boundary. An open interval or a real line R is not a closed manifold because it is not compact. On the other hand, a circle, or a sphere S n of any dimension n is a closed manifold. The notions of connected and path connected subsets of a manifold are defined in the same way as in an Euclidean space Gluing construction The construction which is described in this section is called gluing or quotient construction. It provides a rich source of examples of manifolds. We discuss here only very special cases of this construction. a) Let M be a manifold and U, U its two open disjoint subsets. Let us moreover assume that each point x M has a neighborhood which does not intersect at least one of the sets U and U. 4 Consider a diffeomorphism f : U U. Let us denote by M/{f(x) x} the set obtained from M by identifying each point x U with its image f(x) U. In other words, a point of M/{f(x) x} is either a point from x M \(U U ), or a pair of points (x, f(x)), where x U. Note that there exists a canonical projection π : M M/{f(x) x}. Namely π(x) = x if x / U U, π(x) = (x, f(x)) if x U and π(x) = (f (x), x) if x U. By our assumption each point x M has a coordinate neighborhood G x x such that f(g x U) G x =. In particular, the projection π Gx : G x G x = π(g x ) is one-to-one. We will declare by definition that G x is a coordinate neighborhood of π(x) M/{f(x) x} and define a coordinate map Φ : Gx R n by the formula Φ = Φ π. It is not difficult to check that this construction define a structure of an n-dimensional manifold on the set M/{f(x) x}. We will call the resulted manifold the quotient manifold of M, or say that M/{f(x) x} is obtained from M by gluing U with U with the diffeomorphism f. 4 Here is an example when this condition is not satisfied: M = (0, 2), U = (0, ), U = (, 2). In this case any neighborhood of the point intersect both sets, U and U. 3

132 Figure 9.4: Felix Hausdorff ( ) Though the described above gluing construction always produce a manifold, the result could be quite pathological, if no additional care is taken. Here is an example of such pathology. Example Let M = I I be the union of two disjoint open intervals I = (0, 2) and I = (3, 5). Then M is a -dimensional manifold. Denote U := (0, ) I, U := (3, 4) I. Consider a diffeomorphism f : U U given by the formula f(t) = t + 3, t U. Let M = M/{f(x) x} be the corresponding quotient manifold. In other words, M is the result of gluing the intervals I and I along their open sub-intervals U and U. Note that the points I and 4 I are not identified, but ɛ, 4 ɛ are identified for an arbitrary small ɛ > 0. This means that any neighborhood of and any neighborhood of 4 have non-empty intersection. In order to avoid such pathological examples one usually (but not always) requires that manifolds satisfy an additional axiom, called Hausdorff property: M5. Any two distinct points x, y M have non-intersecting neighborhoods U x, G y. In what follows we always assume that the manifolds satisfy the Hausdorff property M5. Let us make the following general remark about diffeomorphisms f : (a, b) (c, d) between two open intervals. Such diffeomorphism is simply a differentiable function whose derivative never vanishes and whose range is equal to the interval (c, d). If derivative is positive then the diffeomorphism is orientation preserving, and it is orientation reversing otherwise. The function f always 32

133 extends to a continuous (but necessarily differentiable function f : [a, b] [c, d] such that f(a) = c and f(b) = d in the orientation preserving case, and f(a) = d f(b) = c in the orientation reversing case. Lemma Given a, b, a b (0, ) such that a < b and a < b consider an orientation preserving diffeomorphisms f : (0, a) (0, a ) and (b, ) (b ). Then for any ã (0, a) and b (b, ) there exists a diffeomorphism F : (0, ) (0, ) which coincides with f on (0, ã) and coincides with g on ( b, ). Proof. Choose real numbers c, c, d, d such that a < c < c < d < d < b. Consider a cut-off C - function θ : (0, ) (0, ) which is equal to on (0, ã] [ c, d] [ b, ) and equal to 0 on [a, c] [d, b]. For positive numbers ɛ > 0 and C > 0 (which we will choose later) consider a function h ɛ,c on (0, ) defined by the formula ( ɛ)θ(x)f (x) + ɛ, x (0, a); ɛ, x [a, c] [d, b]; h ɛ,c (x) = ( ɛ)cθ(x) + ɛ, x (c, d); ( ɛ)θ(x)g (x) + ɛ, x (b, ). Note that h ɛ,c (x) = f (x) on (0, ã], h ɛ,c (x) = g (x) on [ b, ) and equal to C on [ c, d]. Define the function F ɛ,c : (0, ) (0, ) by the formula F ɛ,c (x) = x 0 h(u)du. Note that the derivative F ɛ,c is positive, and hence the function F ɛ,c is strictly increasing. It coincides with f on (0, ã] and coincides up to a constant with g on ( b, ). Note that when ɛ and C are small we have F ɛ,c ( b) < b < g( b), and lim F ɛ,c( b) =. Hence, by continuity one can C choose ɛ, C > 0 in such a way that F ɛ,c ( b) = g( b). Then the function F = F ɛ,c is a diffeomorphism (0, ) (0, ) with the required properties. Lemma Suppose that a -dimensional manifold M (which satisfies the Hausdorff axiom M5) is covered by two coordinate charts, M = U U, with coordinate maps Φ : U (0, ), Φ : U 33

134 (0, ) such that Φ(U U ) = (0, a) (b, ), Φ (U U ) = (0, a ) (b ) for some a, a, b, b (0, ) with a < b, a < b. Then M is diffeomorphic to the circle S. Proof. Denote by Ψ and Ψ the parameterization maps Φ and (Φ ), and set G := Φ(U U ) and G := Φ (U U ). Let h = Φ Ψ : G G be the transition diffeomorphism. There could be two cases: h((0, a)) = (0, a ), h((b, )) = (b, ) and h((0, a)) = (b, ), h((b, )) = (0, a). We will analyze the first case. The second one is similar. Let h be the continuous extension of h to [0, a] [b, ]. We claim that h(0) = a, h(a) = 0, h(b) = and h() = b. Indeed, assuming otherwise we come to a contradiction with the Hausdorff property M5. Indeed, suppose h(a) = a. Note the points A := Ψ(a), A := Ψ (a ) M are disjoint. On the other hand, for any neighborhood Ω A its image Φ(Ω) I contains an interval (a ɛ, a), and similarly for any neighborhood Ω A its image Φ (Ω) I contains an interval (a ɛ, a ). for a sufficiently small ɛ. But h((a ɛ, a)) = (a ɛ, a ) for some ɛ > 0 and hence Ω Ψ ((a, a ɛ )) = Ψ h((a, a ɛ)) = Ψ Φ Φ((a, a ɛ)) = Φ((a, a ɛ)) Ω, i.e. Ω Ω. In other words, any neighborhoods of the distinct points A, A M intersect, which violates axiom M5. Similarly we can check that h(b) = b. Now take the unit circle S R 2 and consider the polar coordinate φ on S. Let us define a map g : U S by the formula φ = πφ (x). Thus g is a diffeomorphism of U onto an arc of S given in polar coordinates by π < φ < 0. The points A = Ψ (a ) and B = Ψ (b ) are mapped to points with polar coordinates φ = πa and φ = πb. On the intersection U U we can describe the map g in terms of the coordinate in U. Thus we get a map f := g Ψ : (0, a) (b, ) S. We have f(0) = g (A ), f(a) = g (0), f() = g (B ), f(b) = g (). Here we denoted by f the continuous extension of f to [0, a] [b, ]. Thus f((0, a)) = { πa < φ < 0} and f((b, ) = {π < φ < 3π πb }. Note that the diffeomorphism f is orientation preserving assuming that the circle is oriented counter-clockwise. Using Lemma 9.35 we can find a diffeomorphism F from (0, ) to the arc { πa < φ < 3π πb } S which coincides with f on (0, ã) ( b, ) for any ã (0, a) and b (b, ). Denote ã := h(ã), b = h( b). Notice that the neighborhoods U and Ũ = Ψ ((ã, b)) cover M. Hence, the required diffeomorphism F : M S we can define by the 34

135 formula g(x), x F (x) = Ũ ; F Φ(x), x U. Similarly (and even simpler), one can prove Lemma Suppose that a -dimensional manifold M (which satisfies the Hausdorff axiom M5) is covered by two coordinate charts, M = U U, with coordinate maps Φ : U (0, ), Φ : U (0, ) such that U U is connected. Then M is diffeomorphic to the open interval (0, ). Theorem Any (Hausdorff) connected closed -dimensional manifold is diffeomorphic to the circle S. Exercise Show that the statement of the above theorem is not true without the axiom M5, i.e. the assumption that the manifold has the Hausdorff property. Proof. Let M be a connected closed -dimensional manifold. Each point x M has a coordinate neighborhood U x diffeomorphic to an open interval. All open intervals are diffeomorphic, and hence we can assume that each neighborhood G x is parameterized by the interval I = (0, ). Let Ψ x : I G x be the corresponding parameterization map. We have U x = M, and due to compactness of M we can choose finitely many U x,..., U xk such that k x M U xi i= = M. We can further assume that none of these neighborhoods is completrely contained inside another one. Denote U := U x, Ψ := Ψ x. Note that U k. Indeed, if this were the case then due to connectedness of M we would have k U xi 2 U xk 2 = and hence M = U, but this is impossible because M is compact. Thus, there exists i = 2,..., k such that U xi U. We set U 2 := U xi, Ψ 2 = Ψ xi. Consider open sets G,2 := Ψ (U U 2 ), G 2, = Ψ 2 (U U 2 ) I. The transition map h,2 := Ψ 2 Ψ G,2 : G,2 G 2, is a diffeomorphism. Let us show that the set G,2 (and hence G 2, ) cannot contain more that two connected components. Indeed, in that case one of the components of G,2 has to be a subinterval I = (a, b) I = (0, ) where 0 < a < b <. Denote I := h,2 (I ). Then at least of of the boundary values of the transition diffeomorphism h,2 I, say h,2 (a), which is one of the end points of I, has to 35

136 be an interior point c I = (0, ). We will assume for determinacy that I = (c, d) I. But this contradicts the Hausdorff property M5. The argument repeats a similar argument in the proof of Lemma Indeed, note that Ψ (a) Ψ 2 (c). Indeed, Ψ (a) belongs to U \ U 2 and Ψ 2 (c) is in U 2 \ U. Take any neighborhood Ω Ψ (a) in M. Then Ψ (Ω) is an open subset of I which contains the point a. Hence Ψ ((a, a + ɛ)) Ω, and similarly, for any neighborhood Ω Ψ 2 (c) in M we have Ψ 2 ((c, c+ɛ)) Ω for a sufficiently small ɛ > 0. But Ψ ((a, a+ɛ)) = Ψ 2 (h,2 ((a, a ɛ ))) = Ψ 2 (c, c+ɛ ), where c + ɛ = h,2 (a + ɛ). Hence Ω Ω, i.e. any two neighborhoods of two distict points Ψ (a) and Ψ 2 (c) have a non-empty intersection, which violates the Hausdorff axiom M5. If G,2 (0, ) consists of two components then the above argument shows that each of these components must be adjacent to one of the ends of the interval I, and the same is true about the components of the set G 2, I. Hence, we can apply Lemma 9.36 to conclude that the union U U 2 is diffeomorphic to S. We also notice that in this case all the remaining neighborhoods U xj must contain in U U 2. Indeed, each U xi which intersects the circle U U 2 must be completely contained in it, because otherwise we would again get a contradiction with the Hausdorff property. Hence, we can eleiminate all neighborhoods which intersect U U 2. But then no other neighborhoods could be left because otherwise we would have M = (U U 2 ) U xj, i.e. the manifold U xj (U U 2 )= M could be presented as a union of two disjoint non-empty open sets which is impossible due to connectedness of M. Thus we conclude that in this case M = U U 2 is diffeomorphic to S. Finally in the case when G,2 consists of component, i.e. when it is connected, one can Use Lemma 9.37 to show that U U 2 is diffeomorphic to an open interval. Hence, we get a covering of M by k neighborhood diffeomorphic to S. Continuing inductively this process we will either find at some step two neighborhoods which intersect each other along two components, or continue to reduce the number of neighborhoods. However, at some moment the first situation should occur because otherwise we would get that M is diffeomorphic to an interval which is impossible because by assumption M is compact. b) Let M be a manifold, f : M M be a diffeomorphism. Suppose that f satisfies the following property: There exists a positive integer p such that for any point x M we have f p (x) = 36

137 f f f(x) = x, but the points x, f(x)..., f }{{} p (x) are all disjoint. The set {x, f(x)..., f p (x)} p M is called the trajectory of the point x under the action of f. It is clear that trajectory of two different points either coincide or disjoint. Then one can consider the quotient space X/f, whose points are trajectories of points of M under the action of f. Similarly to how it was done in a) one can define on M/f a structure of an n-dimensional manifold. c) Here is a version of the construction in a) for the case when trajectory of points are infinite. Let f : M M be a diffeomorphism which satisfies the following property: for each point x M there exists a neighborhood U x x such that all sets..., f 2 (U x ), f (U x ), U x, f(u x ), f 2 (U x ),... are mutually disjoint. In this case the trajectory {..., f 2 (x), f (x), x, f(x), f 2 (x),... } of each point is infinite. As in the case b), the trajectories of two different points either coincide or disjoint. The set M/f of all trajectories can again be endowed with a structure of a manifold of the same dimension as M Examples of manifolds. n-dimensional sphere S n. Consider the unit sphere S n n+ = { x = x 2 j = } Rn+. Let introduce on S n the structure of an n-dimensional manifold. Let N = (0,..., ) and S = (0,..., ) be the North and South poles of S n, respectively. Denote U = S n \ S, U + = S n \ N and consider the maps p ± : U ± R n given by the formula n x 2 j p ± (x,..., x n+ ) = (x,..., x n ). (9.7.) x n+ The maps p + : U + R n and p : U R n are called stereographic projections from the North and South poles, respectively. It is easy to see that stereographic projections are one-to one maps. Note that U + U = S n \ {S, N} and both images, p + (U + U ) and p (U + U ) coincide with R n \ 0. The map p p + : Rn \ 0 R n \ 0 is given by the formula p p + (x) = x x 2, (9.7.2) 37

138 and therefore it is a diffeomorphism R n \ 0 R n \ 0. Thus, the atlas which consists of two coordinate charts (U +, p + ) and (U, p ) defines on S n a structure of an n-dimensional manifold. One can equivalently defines the manifold S n as follows. Take two disjoint copies of R n, let call them R n and Rn 2. Denote M = Rn Rn 2, U = Rn \ 0 and U = R n 2 \ 0. Let f : U U be a difeomorphism defined by the formula f(x) = Then S n can be equivalently described as the quotient manifold M/f. x x 2, as in (9.7.2). Note that the -dimensional sphere is the circle S. It can be d as follows. Consider the map T : R R given by the formula T (x) = x +, x R. It satisfies the condition from and hence, one can define the manifold R/T. This manifold is diffeomorphic to S. 2. Real projective space. The real projective space RP n is the set of all lines in R n+ passing through the origin. One introduces on RP n a structure of an n-dimensional manifold as follows. For each j =,..., n + let us denote by U j the set of lines which are not parallel to the affine subspace Π j = {x j = }. Clearly, n+ U j = RP n. There is a natural one-to one map π j : U j Π j which associates with each line µ U j the unique intersection point of µ with Π j. Furthermore, each Π j can be identified with R n, and hence pairs (U j, π j ), j =,..., n + can be chosen as an atlas of coordinate charts. We leave it to the reader to check that this atlas indeed define on RP n a structure of a smooth manifold, i.e. that the transition maps between different coordinate charts are smooth. Exercise Let us view S n as the unit sphere in R n+. Consider a map p : S n RP n which associates to a point of S n the line passing through this point and the origin. Prove that this two-to-one map is smooth, and moreover a local diffeomorphism, i.e. that the restriction of p to a sufficiently small neighborhood of each point is a diffeomorphism. Use it to show that RP n is diffeomorphic to the quotient space S n /f, where f : S n S n is the antipodal map f(x) = x. 3. Products of manifolds and n-dimensional tori. Given two manifolds, M and N of dimension m and n, respectively, one can naturally endow the direct product M N = {(x, y); x M, y N} with a structure of a manifold of dimension m + n. Let {(U λ, Φ λ )} λ Λ and {(V γ, Ψ γ )} γ Γ be atlases for M and N, so that Φ λ : U λ U λ Rm, Ψ µ : V µ V µ R n are diffeomorphisms on open 38

139 subsets of R m and R n. Then the smooth structure on M N can be defined by an atlas {(U λ V γ, Φ λ Ψ γ )} λ Λ,γ Γ, where we denote by Φ λ Ψ γ : U λ V γ U λ V γ R m R n = R m+n are diffeomorphisms defined by the formula (x, y) (Φ λ (x)ψ µ (y)) for x U λ and y V µ. One can similarly define the direct product of any finite number of smooth manifolds. In particular the n-dimensional torus T n is defined as the product of n circles: T n = S } {{ S }. Let n us recall, that the circle S is diffeomorphic to R/T, i.e. a point of S is a real number up to adding any integer. Hence, the points of the torus T n can viewed as the points of R n up to adding any vector with all integer coordinates Submanifolds of an n-dimensional vector space Let V be an n-dimensional vector space. A subset A V is called a k-dimensional submanifold of V, or simply a k-submanifold of V, 0 k n, if for any points a A there exists a local coordinate chart (U a, u = (u,..., u n ) R n ) such that u(a) = 0 (i.e. the point a is the origin in this coordinate system) and A U a = {u = (u,..., u n ) U a ; u k+ = = u n = 0}. (9.7.3) We will always assume the local coordinates at least as smooth as necessary for our purposes (but at least C -smooth), but more precisely, one can talk of C m - submanifolds if the implied coordinate systems are at least C m -smooth. Note that in the above we can replace the vector space V by any n-dimensional manifold, and thus will get a notion of a k-dimesional submanifold of an n-dimensional manifold V. Example 9.4. Suppose a subset A U V is given by equations F = = F n k = 0 for some C m -smooth functions F,..., F n k on U. Suppose that for any point a A the differentials d a F,..., d a F n k V a are linearly independent. Then A U is a C m -smooth submanifold. Indeed, for each a A one can choose a linear functions l,..., l k V a such that together with d a F,..., d a F n k V a they form a basis of V. Consider functions L,..., L k : V R, defined by L j (x) = l j (x a) so that d a (L j ) = l j, j =,..., k. Then the Jacobian det D a F of the 39

140 map F : (L,..., L k, F,..., F n k ) : U R n does not vanish at a, and hence the inverse function theorem implies that this map is invertible in a smaller neighborhood U a U of the point a A. Hence, the functions u = L,..., u k = L k, u k+ = F,..., u n = F n k can be chosen as a local coordinate system in U a, and thus A U a = {u k+ = = u n = 0}. Note that the map u = (u,..., u k ) maps Ua A = U a A onto an open neighborhood Ũ = u(u a ) R k of the origin in R k R n, and therefore u = (u,... u k ) defines a local coordinates, so that the pair (Ua A, u ) is a coordinate chart. The restriction φ = φ eu of the parameterization map φ = u maps Ũ onto U a A. Thus φ a parameterization map for the neighborhood Ua A. The atlas {(Ua A, u )}a A defines on a a structure of a k-dimensional manifold. The complementary dimension n k is called the codimension of the submanifold A. We will denote dimension and codimension of A by dim A and codima, respectively. As we already mentioned above in Section 9.2 -dimensional submanifolds are usually called curves. We will also call 2-dimensional submanifolds surfaces and codimension submanifolds hypersurfaces. Sometimes k-dimensional submanifolds are called k-surfaces. Submanifolds of codimension 0 are open domains in V. An important class form graphical k-submanifolds. Let us recall that given a map f : B R n k, where B is a subset B R k, then graph is the set Γ f = {(x, y) R k R n k = R n ; x B, y = f(x)}. A (C m )-submanifold A V is called graphical with respect to a splitting Φ : R k R n k V, if there exist an open set U R k and a (C m )-smooth map f : U R n k such that A = Φ(Γ f ). In other words, A is graphical if there exists a coordinate system in V such that A = {x = (x,..., x n ); (x,... x k ) U, x j = f j (x,..., x k ), j = k +,..., n}. for some open set U R k and smooth functions, f k+,..., f n : U R. For a graphical submanifold there is a global coordinate system given by the projection of the submanifold to R k. It turns out that that any submanifold locally is graphical. 40

141 Proposition Let A V be a submanifold. Then for any point a A there is a neighborhood U a a such that U a A is graphical with respect to a splitting of V. (The splitting may depend on the point a A). We leave it to the reader to prove this proposition using the implicit function theorem. One can generalize the discussion in this section and define submanifolds of any manifold M, and not just the vector space V. In fact, the definition (9.7.3) can be used without any changes to define submanifolds of an arbitrary smooth manifold. A map f : M Q is called an embedding of a manifold M into another manifold Q if it is a diffeomorphism of M onto a submanifold A Q. In other words, f is an embedding if the image A = f(m) is a submanifold of Q and the map f viewed as a map M A is a diffeomorphism. One can prove that any n-dimensional manifold can be embedded into R N with a sufficiently large N (in fact N = 2n + is always sufficient). Hence, one can think of manifold as submanifold of some R n given up to a diffeomorphism, i.e. ignoring how this submanifold is embedded in the ambient space. In the exposition below we mostly restrict our discussion to submanifolds of R n rather than general abstract manifolds Submanifolds with boundary A slightly different notion is of a submanifold with boundary. A subset A V is called a k- dimensional submanifold with boundary, or simply a k-submanifold of V with boundary, 0 k < n, if for any points a A there is a neighborhood U a a in V and local (curvi-linear) coordinates (u,..., u n ) in U a with the origin at a if one of two conditions is satisfied: condition (9.7.3), or the following condition A U a = {u = (u,..., u n ) U a ; u 0, u k+ = = u n = 0}. (9.7.4) In the latter case the point a is called a boundary point of A, and the set of all boundary points is called the boundary of A and is denoted by A. As in the case of submanifolds without boundary, any submanifold with boundary has a structure of a manifold with boundary. 4

142 Figure 9.5: boundary point b. The parameterization φ introducing local coordinates near an interior point a and a Exercise Prove that if A is k-submanifold with boundary then A is a (k )-dimensional submanifold (without boundary). Remark As we already pointed out when we discussed manifolds with boundary, a submanifold with boundary is not a submanifold! 2. As it was already pointed out when we discussed -dimensional submanifolds with boundary, the boundary of a k-submanifold with boundary is not the same as its set-theoretic boundary, though traditionally the same notation A is used. Usually this should be clear from the context, what the notation A stands for in each concrete case. We will explicitly point this difference out when it could be confusing. A compact manifold (without boundary) is called closed. The boundary of any compact manifold with boundary is closed, i.e. ( A) =. Example An open ball Br n = BR n(0) = { n x 2 j < } Rn is a codimension 0 submanifold, 42

143 A closed ball D n r Its boundary D n R = D n R (0) = { n x 2 j } Rn ia codimension 0 submanifold with boundary. is an (n )-dimensional sphere Sn R = { n x 2 j = } Rn. It is a closed hypersurface. For k = 0,... n let us denote by L k the subspace L k = {x k+ = = x n = 0} R n. Then the intersections B k R = B n R L k, D k R = D n R L k, and S k R = S n R L k R n are, respectively a k-dimensional submanifold, a k-dimensional submanifold with boundary and a closed (k )-dimensional submanifold of R n. Among all above examples there is only one (which one?) for which the manifold boundary is the same as the set-theoretic boundary. A neighborhood of a boundary point a A can be always locally parameterized by the semiopen upper-half ball B + (0) = {x = (x,..., x k ) R k ; x 0, n x 2 j < }. We will finish this section by defining submanifolds with piece-wise smooth boundary. A subset A V is called a k-dimensional submanifold of V with piecewise smooth boundary or with boundary with corners, 0 k < n, if for any points a A there is a neighborhood U a a in V and local (curvi-linear) coordinates (u,..., u n ) in U a with the origin at a if one of three condittions satisfied: conditions (9.7.3), (9.7.4) or the following condition A U a = {u = (u,..., u n ) U a ; l (u) 0,..., l m (u) 0, u k+ = = u n = 0}, (9.7.5) where m > and l,..., l m (R k ) are linear functions In the latter case the point a is called a corner point of A. Note that the system of linear inequalities l (u) 0,..., l m (u) 0 defines a convex cone in R k. Hence, near a corner point of its boundary the manifold is diffeomorphic to a convex cone. Thus convex polyhedra and their diffeomorphic images are important examples of submanifolds with boundary with corners. 43

144 9.8 Tangent spaces and differential Suppose we are given two local parameterizations φ : G A and φ : G A. Suppose that 0 G G and φ(0) = φ(0) = a A. Then there exists a neighborhood U a in A such that U φ(g) φ( G). Denote G := φ (U), G := φ (Ũ). Then one has two coordinate charts on U: u = (u,..., u k ) = ) (φ G ) : U G, and ũ = (ũ,..., ũ k ) = ( φ eg : U G. Denote h := u φ eg = G G. We have φ = φ u φ = φ h, and hence the differentials dφ 0 and d φ 0 of parameterizations φ and φ at the origin map R k 0 isomorphically onto the same k-dimensional linear subspace T V a. Indeed, d 0 φ = d0 φ d 0 h. Thus the space T = d 0 φ(r k 0 ) V a is independent of the choice of parameterization. It is called the tangent space to the submanifold A at the point a A and will be denoted by T a A. If A is a submanifold with boundary and a A then there are defined both the k-dimensional tangent space T a A and its (k )-dimensional subspace T a ( A) T a A tangent to the boundary. Example Suppose a submanifold A V is globally parameterized by an embedding φ : G A V, G R k. Suppose the coordinates in R k are denoted by (u,..., u k ). Then the tangent space T a A at a point a = φ(b), b G is equal to the span ( φ Span (a),..., φ ) (a). u u k 2. In particular, suppose a submanifold A is graphical and given by equations x k+ = g (x,..., x k ),..., x n = g n k (x,..., x k ), (x,..., x k ) G R k. Take points b = (b,... b k ) G and a = (b,..., b k, g (b),..., g n k (b)) A. Then T a A = 44

145 Span (T,... T k ), where T =, 0,..., 0, g (b),..., g n k (b), }{{} x x k T = 0,,..., 0, g (b),..., g n k (b), }{{} x 2 x 2 k... T = 0, 0,...,, g (b),..., g n k (b). }{{} x k x k k 3. Suppose a hypersurface Σ R n is given by an equation Σ = {F = 0} for a smooth function F defined on an neighborhood of Σ and such that d a F 0 for any a Σ. In other words, the function F has no critical points on Σ. Take a point a Σ. Then T a Σ R n a is given by a linear equation n F x j (a)h j = 0, h = (h,..., h n ) R n a. Note that sometimes one is interested to define T a Σ as an affine subspace of R n = R n 0 and not as a linear subspace of R n a. We get the required equation by shifting the origin: T a Σ = {x = (x,..., x n ) R n ; n F x j (a)(x j a j ) = 0}. If for some parameterization φ : G A with φ(0) = a the composition f φ is differentiable at 0, and the linear map d 0 (f φ) (d 0 φ) : T a A W f(a) is called the differential of f at the point a and denoted, as usual, by d a f. Similarly one can define C m -smooth maps A W. Exercise Show that a map f : A W is differentiable at a point a A iff for some neighborhood U of a in V there exists a map F : U W that is differentiable at a and such that F U A = f U A, and we have df TaA = d a f. As it follows from the above discussion the map df TaA is independent of this extension. Similarly, any C m -smooth map of A locally extends to a C m -smooth map of a neighborhood of A in V. 45

146 Suppose that the image f(a) of a smooth map f : A W is contained in a submanifold B W. In this case the image d a f(t a A) is contained in T f(a) B. Hence, given a smooth map f : A B between two submanifolds A V and B W its differential at a point a can be viewed as a linear map d a f : T a A T f(a) B. Let us recall, that given two submanifolds A V and B W (with or without boundary), a smooth map f : A B is called a diffeomorphism if there exists a smooth inverse map : B A, i.e. f g : B B and g f : A A are both identity map. The submanifolds A and B are called in this case diffeomorphic. Exercise Let A, B be two diffeomorphic submanifolds. Prove that (a) if A is path-connected then so is B; (b) if A is compact then so is B; (c) if A = then B = ; (d) dim A = dim B; 5 2. Give an example of two diffeomorphic submanifolds, such that one is bounded and the other is not. 3. Prove that any closed connected -dimensional submanifold is diffeomorphic to the unit circle S = {x 2 + x2 2 = } R Vector bundles and their homomorphisms Let us put the above discussion in a bit more global and general setup. A collection of all tangent spaces {T a A} a A to a submanifold A is called its tangent bundle and denoted by T A or T (A). This is an example of a more general notion of a vector bundle of rank r over a set A V. One understands by this a family of r-dimensional vector subspaces L a V a, parameterized by points of A and continuously (or C m -smoothly) depending on a. More precisely one requires that each point a A has a neighborhood U A such that there exist linear independent vector fields v (a),..., v r (a) L a which continuously (smoothly, etc.) depend on a. 5 In fact, we will prove later that even homeomorphic manifolds should have the same dimension. 46

147 Besides the tangent bundle T (A) over a k-submanifold A an important example of a vector bundle over a submanifold A is its normal bundle NA = N(A), which is a vector bundle of rank n k formed by orthogonal complements N a A = Ta A V a of the tangent spaces T a A of A. We assume here that V is Euclidean space. A vector bundle L of rank k over A is called trivial if one can find k continuous linearly independent vector fields v (a),..., v k (a) L a defined for all a A. The set A is called the base of the bundle L. An important example of a trivial bundle is the bundle T V = {V a } a V. Exercise Prove that the tangent bundle to the unit circle S R 2 is trivial. Prove that the tangent bundle to S 2 R 3 is not trivial, but the tangent bundle to the unit sphere S 3 R 4 is trivial. (The case of S is easy, of S 3 is a bit more difficult, and of S 2 even more difficult. It turns out that the tangent bundle T S n to the unit sphere S n R n is trivial if and only if n = 2, 4 and 8. The only if part is a very deep topological fact which was proved by F. Adams in 960. Suppose we are given two vector bundles, L over A and L over à and a continuous (resp. smooth) map φ : A Ã. By a continuous (resp. smooth) homomorphism Φ : L L which covers the map φ : A à we understand a continuous (resp. smooth) family of linear maps Φ a : L a L φ(a). For instance, a C m -smooth map f : A B defines a C m -smooth homomorphism df : T A T B which covers f : A B. Here df = {d a f} a A is the family of linear maps d a f : T a A T f(a) B, a A. 9.0 Orientation By an orientation of a vector bundle L = {L a } a A over A we understand continuously depending on a orientation of all vector spaces L a. An orientation of a submanifold k is the same as an orientation of its tangent bundle T (A). A co-orientation of a k- submanifold A is an orientation of its normal bundle N(A) = T A in V. Note that not all bundles are orientable, i.e. some bundles admit no orientation. But if L is orientable and the base A is connected, then L admits exactly two orientations. Here is a simplest example of a non-orientable rank bundle of the circle S R 2. Let us identify a point a S with a complex number a = e iφ, and consider a line l a R 2 a directed 47

148 Figure 9.6: The orientation of the surface is induced by its co-orientation by the normal vector n. The orientation of the boundary us induced by the orientation of the surface. by the vector e iφ 2. Hence, when the point completes a turn around S the line l a rotates by the angle π. We leave it to the reader to make a precise argument why this bundle is not orientable. In fact, rank bundles are orientable if and only if they are trivial. If the ambient space V is oriented then co-orientation and orientation of a submanifold A determine each other according to the following rule. For each point a, let us choose any basis v,..., v k of T a (A) and any basis w,..., w n k of N a (A). Then w,..., w n k, v,..., v k is a basis of V a = V. Suppose one of the bundles, say N(A), is oriented. Let us assume that the basis w,..., w n k defines this orientation. Then we orient T a A by the basis v,..., v k if the basis w,..., w n k, v,..., v k defines the given orientation of V, and we pick the opposite orientation of T a A otherwise. Example (Induced orientation of the boundary of a submanifold.) Suppose A is an oriented manifold with boundary. Let us co-orient the boundary A by orienting the rank normal bundle to T ( A) in T (A) by the unit ourtward normal to T ( A) in T (A) vector field. Then the above rule determine an orientation of T ( A), and hence of A. 48

149 9. Integration of differential k-forms over k-dimensional submanifolds Let α be a differential k-form defined on an open set U V. Consider first a k-dimensional compact submanifold with boundary A U defined parametrically by an embedding φ : G A U, where G R k is possibly with boundary. Suppose that A is oriented by this embedding. Then we define α := A G φ α. Note that if we define A by a different embedding φ : G A, then we have φ = φ h, where h = φ φ : G G is a diffeomorphism. Hence, using Theorem 9.27 we get φ α = h (φ α) = φ α, eg eg G and hence α is independent of a choice of parameterization, provided that the orientation is A preserved. Let now A be any compact oriented submanifold with boundary. Let us choose a partition of unity = K θ j in a neighborhood of A such that each function is supported in some coordinate neighborhood of A. Denote α j = θ j α. Then α = K α j, where each form α j is supported in one of coordinate neighborhoods. Hence there exist orientation preserving embeddings φ j : G j A of domains with boundary G j R k, such that φ j (G j ) Supp(α j ), j =,..., K. Hence, we can define α j := φ jα j and A G j A K α := α j. A Lemma 9.5. The above definition of A α is independent of a choice of a partition of unity. Proof. Consider two different partitions of unity = K θ j and = U,..., U K ek θ j subordinated to coverings and Ũ,..., Ũ e K, respectively. Taking the product of two partitions we get another 49

150 partition = K ek i= j= θ ij, where θ ij := θ i θ j, which is subordinated to the covering by intersections U i U j, i =,..., K, j =,..., K. Denote α i := θ i α, α j := θ j α and α ij = θ ij α. Then K α ij = α j, ek j= α ij = α i and α = K α i = ek α j. Then, using the linearity of the integral we get i= K A α i = K ek i= j= A α ij = ek A α j. When k = the above definition of the integral coincides with the definition of the integral of a -form over an oriented curve which was given above in Section 9.. Let us extend the definition of integration of differential forms to an important case of integration of 0-form over oriented 0-dimensional submanifolds. Let us recall a compact oriented 0-dimensional submanifold of V is just a finite set of points a,..., a m V with assigned signs to every point. So in view of the additivity of the integral it is sufficient to define integration over point with a sign. On the other hand, a 0-form is just a function f : V R. So we define f := ±f(a). ±a A partition of unity is a convenient tool for studying integrals, but not so convenient for practical computations. The following proposition provides a more practical method for computations. Proposition Let A be a compact oriented submanifold of V and α a differential k-form given on a neighborhood of A. Suppose that A presented as a union A = N A j, where A j are codimension 0 submanifolds of A with boundary with corners. Suppose that A i and A j for any i j intersect only along pieces of their boundaries. Then α = A N A j α. In particular, if each A j is parameterized by an orientation preserving embedding φ j : G j A, 50

151 where G j R k is a domain with boundary with corners. Then A α = N G j φ jα. We leave the proof to the reader as an exercise. Exercise Compute the integral /3(x dx 2 dx 3 + x 2 dx 3 dx + x 3 dx dx 2 ), S where S is the sphere {x 2 + x x 2 3 = }, cooriented by its exterior normal vector. Solution. Let us present the sphere as the union of northern and southern hemispheres: S = S S +, where S = S {x 3 0}, S + = S {x 3 0}. Then ω = ω + ω. Let us first compute ω. S S + S S + We can parametrize S + by the map (u, v) (u, v, R 2 u 2 v 2 ), (u, v) {u 2 + v 2 R 2 } = D R. One can check that this parametrization agrees with the prescribed coorientation of S. Thus, we have ω = /3 S + D R ( udv d R 2 u 2 v 2 + vd R 2 u 2 v 2 du + ) R 2 u 2 v 2 du dv. Passing to polar coordinates (r, ϕ) in the plane (u, v) we get ω = /3 r cos ϕd(r sin ϕ) d R 2 r 2 + r sin ϕd R 2 r 2 d(r cos ϕ) S + P + R 2 r 2 d(r cos ϕ) d(r sin ϕ), 5

152 where P = {0 r R, 0 ϕ 2π}. Computing this integral we get S + ω = = 3 r3 cos 2 ϕdϕ dr + r3 sin 2 ϕdr dϕ + R 3 2 r 2 dr dϕ P R 2 r 2 R 2 r 2 P ( r 3 ) R 2 r + r R 2 r 2 dr dϕ 2 = 2π 3 R 0 rr 2 R dr = R 2 2πR2 R r2 3 2 r 2 = 2πR3 0 3 Similarly, one can compute that 2πR 3 ω =. 3 S Computing this last integral, one should notice the fact that the parametrization (u, v) (u, v, R 2 u 2 v 2 ) defines the wrong orientation of S. Thus one should use instead the parametrization (u, v) (v, u, R 2 u 2 v 2 ), and we get the answer S ω = 4πR3. 3 This is just the volume of the ball bounded by the sphere. The reason for such an answer will be clear below from Stokes theorem. 52

153 Part III Stokes theorem and its applications 53

154

155 Chapter 0 Stokes theorem 0. Statement of Stokes theorem Theorem 0.. Let A V be a compact oriented submanifold with boundary (and possibly with corners). Let ω be a C 2 -smooth differential form defined on a neighborhood U A. Then ω = dω. A A Here dω is the exterior differential of the form ω and A is the oriented boundary of A. We will discuss below what exactly Stokes theorem means for the case k 3 and n = dim V 3. Let us begin with the case k =, n = 2. Thus V = R 2. Let x, x 2 be coordinates in R 2 and U a domain in R 2 bounded by a smooth curve Γ = U. Let us co-orient Γ with the outward normal ν to the boundary of U. This defines a counter-clockwise orientation of Γ. Let ω = P (x, x 2 )dx + P 2 (x, x 2 )dx 2 be a differential -form. Then the above Stokes formula asserts or U U dω = Γ ω, ( P2 P ) dx dx 2 = P dx + P 2 dx 2. x x 2 Γ 55

156 Figure 0.: George Stokes (89-903) This is called Green s formula. In particular, when dω = dx dx 2, e.g. ω = xdy or ω = 2 (xdy ydx), the integral ω computes the area of the domain U. Γ Consider now the case n = 3, k = 2. Thus V = R 3, ω = P dx 2 dx 3 + P 2 dx 3 dx + P 3 dx dx 2. Let U R 3 be a domain bounded by a smooth surface S. We co-orient S with the exterior normal ν. Then dω = ( P + P 2 + P ) 3 dx dx 2 dx 3. x x 2 x 3 Thus, Stokes formula ω = dω S U gives in this case S P dx 2 dx 3 + P 2 dx 3 dx + P 3 dx dx 2 = U ( P + P 2 + P ) 3 dx dx 2 dx 3 x x 2 x 3 This is called the divergence theorem or Gauss-Ostrogradski s formula. 56

157 Figure 0.2: George Green (793-84) Consider the case k = 0, n =. Thus ω is just a function f on an interval I = [a, b]. The boundary I consists of 2 points: I = {a, b}. One should orient the point a with the sign and the point b with the sign +. Thus, Stokes formula in this case gives df = f, or b a [a,b] { a,+b} f (x)dx = f(b) f(a). This is Newton-Leibnitz formula. More generally, for a -dimensional oriented connected curve Γ R 3 with boundary Γ = B ( A) and any smooth function f we get the formula df = f = f(b) f(a), Γ B ( A) which we already proved earlier, see Theorem 9.8. Consider now the case n = 3, k =. Thus V = R 3 and ω = P dx +P 2 dx 2 +P 3 dx 3. Let S R 3 be an oriented surface with boundary 57

158 Figure 0.3: Carl Friedrich Gauss ( ) Mikhail Ostrogradski (80-862) Γ. We orient Γ in the same way, as in Green s theorem. Then Stokes formula dω = ω gives in this case S Γ S ( P3 P ) ( 2 P dx 2 dx 3 + P ) ( 3 P2 dx 3 dx + P ) dx dx 2 x 2 x 3 x 3 x x x 2 = This is the original Stokes theorem. Γ P dx + P 2 dx 2 + P 3 dx 3. Stokes theorem allows one to clarify the geometric meaning of the exterior differential. Lemma 0.2. Let β be a differential k-form in a domain U V. Take any point a U and vectors X,..., X k+ V a. Given ɛ > 0 let us consider the parallelepiped P (ɛx,..., ɛx k+ ) as a subset of V with vertices at points a i...i k+ = a + ɛ k+ i j X j, where each index i j takes values 0,. Then dβ a (X,... X k+ ) = lim ɛ 0 ɛ k+ P (ɛx,...,ɛx k+ ) β. 58