Introduction to de Rham cohomology. Pekka Pankka

Transcription

1 Introduction to de Rham cohomology Pekka Pankka December 11, 2013

2 Preface These are lecture notes for the course Johdatus de Rham kohomologiaan lectured fall 2013 at Department of Mathematics and Statistics at the University of Jyväskylä. The main purpose of these lecture notes is to organize the topics discussed on the lectures. They are not meant as a comprehensive material on the topic! These lectures follow closely the book of Madsen and Tornehave From Calculus to Cohomology [7] and the reader is strongly encouraged to consult it for more details. There are also several alternative sources e.g. [1, 9] on differential forms and de Rham theory and [4, 5, 3] on multilinear algebra. 1

3 Bibliography [1] H. Cartan. Differential forms. Translated from the French. Houghton Mifflin Co., Boston, Mass, [2] W. Greub. Linear algebra. Springer-Verlag, New York, fourth edition, Graduate Texts in Mathematics, No. 23. [3] W. Greub. Multilinear algebra. Springer-Verlag, New York, second edition, Universitext. [4] S. Helgason. Differential geometry and symmetric spaces. Pure and Applied Mathematics, Vol. XII. Academic Press, New York, [5] S. Helgason. Differential geometry, Lie groups, and symmetric spaces, volume 80 of Pure and Applied Mathematics. Academic Press Inc. [Harcourt Brace Jovanovich Publishers], New York, [6] M. W. Hirsch. Differential topology, volume 33 of Graduate Texts in Mathematics. Springer-Verlag, New York, Corrected reprint of the 1976 original. [7] I. Madsen and J. Tornehave. From calculus to cohomology. Cambridge University Press, Cambridge, de Rham cohomology and characteristic classes. [8] W. Rudin. Real and complex analysis. McGraw-Hill Book Co., New York, third edition, [9] L. W. Tu. An introduction to manifolds. Universitext. Springer, New York, second edition,

4 Contents 1 Alternating algebra Some linear algebra Multilinear maps Alternating multilinear maps Differential forms Exterior derivative Pull-back of forms De Rham cohomology Poincaré lemma Chain complexes Long exact sequence Homotopy Applications 52 5 Manifolds and bundles Topological manifolds and bundles Smooth manifolds Tangent bundle Cotangent bundle Exterior bundles Exterior derivative de Rham cohomology Orientable manifolds 78 7 Integration Euclidean case Manifold case Integration on domains with smooth boundary

5 8 Poincaré duality Special case Exact sequence for compactly supported cohomology Exact sequence for dual spaces Main steps

6 Chapter 1 Alternating algebra Let v 1,..., v n be vectors in R n. The volume of the parallelepiped P (v 1,..., v n ) = {t 1 v t n v n R n : 0 t i 1, i = 1,..., n} is given by det [v 1 v n ]. Whereas the absolute value of the determinant is independent on the order of vectors, the sign of det [v 1 v n ] depends on the order of (v 1,..., v n ). The quantity det [v 1 v n ] is therefore sometimes called signed volume. The role of the sign is to detect the so-called orientation of vectors (v 1,..., v n ). The notion of volume and signed volume based on the determinant allows us to consider n-dimensional objects in n-dimensional space. In this section, we discuss multilinear algebra which allows us to consider volumes (and signed volumes ) of k-dimensional (linear) objects in n-dimensional space when k < n. We discuss these geometric ideas in later sections and develop first the necessary linear theory. 1.1 Some linear algebra We begin by recalling some facts from linear algebra; see e.g. [2, Chapter I & II] for a detail treatment. Let V be a (real) vector space, that is, we have operations +: V V V, (u, v) u + v, : R V V, (a, v) av, for which the triple (V, +, ) satisfies the axioms of a vector space. 5

7 Linear maps and dual spaces Given vector spaces V and V, a function f : V V is a called a linear map if f(u + av) = f(u) + af(v) for all u, v V and a R. A bijective linear map is called a (linear) isomorphism. Given linear maps f, g : V V and a R, the mapping f + ag : V V, v f(v) + ag(v), is a linear map V V. Thus the set Hom(V, V ) of all linear maps V V is also a vector space under operations +: Hom(V, V ) Hom(V, V ) Hom(V, V ), (f, g) f + g, : R Hom(V, V ) Hom(V, V ), (a, f) af. An important special case is the dual space V = Hom(V, R) of V. Theorem (Dual basis). Let V be a finite dimensional vector space. Then V = V. Furthermore, if (e i ) n i=1 is a basis of V, then (ε i) n i=1, where ε i : V R is the map { 1, i = j ε i (e j ) = δ ij = 0, i j, is a basis of V. Proof. Exercise. The basis (ε i ) i of V in Theorem is called dual basis (of (e i ) i ). Note that bases of V and V do not have the same cardinality (i.e V and V do not have the same dimension ) if V is infinite dimensional! Induced maps Given f Hom(V, W ) and a linear map ϕ: U V, we have a composition f ϕ Hom(U, W ): U ϕ V f ϕ The composition with fixed map ϕ induces a linear map as formalized in the following lemma. W f 6

8 Lemma Let U, V, W be vector spaces. Let ϕ: U V be a linear map. Then ϕ : Hom(V, W ) Hom(U, W ), f f ϕ is a linear map. Moreover, if ϕ is an isomorphism then ϕ is an isomorphism. Similarly, we may also consider f Hom(U, V ) and a fixed linear map ψ : V W : U f V f ψ W ψ Lemma Let U, V, W be vector spaces. Let ψ : V W be a linear map. Then ϕ : Hom(U, V ) Hom(U, W ), f ψ f is a linear map. Moreover, if ψ is an isomorphism then ψ is an isomorphism. The mapping ϕ in Lemma is so-called pull-back (under ϕ). The mapping ψ is called as push-forward. As an immediate corollary of Lemma we have the following observation. Corollary Let U and V be vector spaces and ϕ: U V a linear map. Then ϕ : V U, f f ϕ is a linear map. Moreover, ϕ is an if ϕ is an isomorphism. Subspaces and quotients A subset W V of a vector space V is a subspace of V if u + aw W for all u, v W and a R. A coset of an element v V with respect to W is the subset v + W = {v + w V : w W }. The relation W on V, given by the formula u W v u v W, is an equivalence relation with equivalence classes {v + W : v V }. The set V/W of these equivalence classes has a natural structure of a vector space given by operations +: V/W V/W V/W, ((u + W ), (v + W )) (u + v) + W, : R V/W V/W, (a, v + V/W ) (av) + W, that is, (u + W ) + a(v + W ) = (u + av) + W 7

9 for all u, v W and a R. The space V/W is called the quotient space of V (with respect to W ). Note that the mapping π : V V/W, v v + W, is linear. The mapping p is called quotient (or canonical) map. A fundamental fact of on linear mappings is that the kernel and the image are vector spaces, that is, let f : V V be a linear map between vector spaces, then ker f = f 1 (0) = {v V : f(v) = 0} Imf = f[v ] = {f(v) V : v V } are subspaces of V and V, respectively. Theorem (Isomorphism theorem). Let f : V V be a linear map between vector spaces V and V. Then the mapping ϕ: V/ ker f Imf, v + W f(v) + W, is a linear isomorphism satisfying V p f V/ ker f = ϕ where p: V V/ ker f is the canonical map. The proof is left as a voluntary exercise. Imf Products and sums Given a set I (finite or infinite) and a vector space V i for each i I, the product space i I V i has a natural linear structure given by (v i ) i + a(v i) i = (v i + av i) i where v i, v i V i and a R. For I =, we declare i I V i = {0}. If V i = V j =: V for all i, j I, denote V I = i I V. Note that, V I is, in fact, the space of all functions I V. Note also that, for n N, we have V n = V V = V {1,...,n} = {all functions {1,..., n} V }. The sum of two vector spaces V W has (at least) three different meanings in the literature. Abstractly, V W is the product space V W. More concretely, if V and W are subspaces of a vector space U, then V + W is the subspace of U spanned by V and W, that is, V + W = {v + w U : v V, w W }. If V W = {0}, notation V W for V +W is commonly used. If U, however, is an inner product space, then notation V W is commonly reserved for the case that v w for all v V and w W. (Be warned!) 8

10 1.2 Multilinear maps Definition Let V 1,..., V k, and W be vector spaces. A function f : V 1 V k W is k-linear if, for all v j V j (j = 1,..., k), f(v 1,..., v i 1, v + aw, v i+1, v k ) = f(v 1,, v i 1, v, v i+1, v k ) + af(v 1,, v i 1, w, v i+1, v k ) for all i {1,..., k}, v, w V i, and a R. A mapping is multilinear if it is k-linear for some k 1. Example Let V be a (real) vector space. Then the duality pairing V V R defined by (ϕ, v) ϕ(v) is bilinear. Remark A function f : V 1 V k W is k-linear if and only if for every sequence (v 1,..., v k ) V 1 V k and each 1 i k functions are linear mappings V i W. v f(v 1,..., v i 1, v, v i+1, v k ) Example Let V be a (real) vector space. Then the duality pairing V V R defined by (ϕ, v) ϕ(v) is bilinear (i.e. 2-linear). Example An inner product, : V V R is bilinear. Values of multilinear maps V k R depend only on values on basis elements. This can be formalized as follows. Lemma Let V be an n-dimensional vector space and {e 1,..., e n } a basis of V. Let f : V k R be k-linear, and v i = n j=1 v ije j V for i = 1,..., k. Then f(v 1,..., v k ) = f(e j1,..., e jk )v 1ji v kjk (j 1,...,j k ) {1,...,n} k Proof. Since f is k-linear, f(v 1,..., v k ) = f n v 1j1 e ji,..., j 1 =1 = j 1 n v kjk e jk j k =1 j k f (v 1j1 e ji,..., v kjk e jk ) = j 1 j k (v 1j1 v kjk )f(e ji,..., e jk ) Corollary Let V be an n-dimensional vector space and (e 1,..., e n ) a basis of V. Suppose f, g : V k W satisfy f(e j1,..., e jk ) = g(e j1,... e jk ) for all multi-indices (j 1,..., j k ) {1,... n} k. Then f = g. 9

11 1.2.1 Vector space of multilinear maps Definition Let V 1,..., V k, V, W be vector spaces. We denote by Mult(V 1 V k, W ) the set of all k-linear maps V 1 V k W. We denote Mult k (V, W ) = Mult( k, W ). Lemma Let V 1,..., V k, W be vector spaces. Then Mult(V 1 V k, W ) is a (vector) subspace of Hom(V 1 V k, W ). Proof. Clearly f + ag : V 1 V k W is k-linear if f, g Mult(V 1 V k, W ) and a R. Definition Let U, V and W be vector spaces and ϕ: U V a linear map. Given a k-linear map f : V k W, the map ϕ f : U k W is defined by formula (1.2.1) (ϕ f)(u 1,..., u k ) = f(ϕ(u 1 ),..., ϕ(u k )) for u 1,..., u k U. Lemma Let U, V and W be vector spaces and ϕ: U V a linear map. Then ϕ f Mult k (U, W ) for every f Mult k (V, W ). Furthermore, the map ϕ : Mult k (V, W ) Mult k (U, W ) defined by formula (1.2.1), is linear Tensor product This section is added for completeness. We do not use tensor products in the following sections. Let f : V k R and g : U l R be k- and l-linear maps respectively. Define f g : V k U l R, (f g)(v 1,..., v k, u 1,..., u l ) = f(v 1,..., v k )g(u 1,..., v k ) Lemma Let f : V k R and g : U l R be multilinear. Then f g is multilinear. Moreover, if h: W m R is multilinear, then f (g h) = (f g) h. Proof. Exercise. Lemma Let U, V be vector space, f Mult k (V ), g Mult l (V ), and ϕ: U V a linear map. Then Proof. Exercise. ϕ (f g) = ϕ f ϕ g. 10

12 Lemma Let V be an n-dimensional vector space, (e 1,..., e n ) a basis of V and (ε 1,..., ε n ) the corresponding dual basis. Let f : V k R. Then there exists coefficients a J R, J = (j 1,... j k ) {1,..., n} k, for which f = J=(j 1,...,j k ) {1,...n} k a J ε j1 ε jk Moreover, a J = f(e j1,..., e jk ) for J = (j 1,..., j k ) {1,..., n} k. Proof. Let J = (j 1,... j k ) and v i = k j=1 v ije j for i = 1,..., k. Then ε j1 ε jk (v 1,..., v k ) = Thus, by Lemma 1.2.6, f(v 1,..., v k ) = = (j 1,...,v k ) = (j 1,...,v k ) (ε j1 ε jk )(e i1,..., e ik )v 1ii v kik (i 1,...,i k ) {1,...,n} k (i 1,...,i k ) {1,...,n} k ε j1 (e i1 ) ε jk (e ik )v 1ii v kik = v 1ji v kjk f(e j1,..., e jk )v 1j1 v kjk f(e j1,..., e jk )ε j1 ε jk (v 1j1,..., v kjk ). Lemma Let V be an n-dimensional vector space, (e 1,..., e n ) a basis of V and (ε 1,..., ε n ) the corresponding dual basis. Then is a basis of Mult k (V, R). Proof. Exercise. (ε j1 ε jk ) (j1,...,j k ) {1,...,n} k Remark Multilinear maps are discussed in many sources, see e.g. [4, 5] or [3]. Note, however, that for example in [3], the point of view is more abstract and the tensor product refers to a vector space which linearizes multilinear maps. This more abstarct approach can be viewed as the next step in this theory (which we do not take in these notes). 11

13 1.3 Alternating multilinear maps The material in this section is gathered from [7, Chapter 2]; alternatively see [3]. Throughout this section V and W are (real) vector spaces. Definition A k-linear map f : V V W is alternating if, for 1 i < j k, (1.3.1) f(v 1,..., v i 1, v j, v i+1,..., v j 1, v i, v j+1,..., v k ) = f(v 1,..., v k ) for all v 1,..., v k V. Example Let ϕ: R 2 R 2 R 2 2 be the mapping [ ] u1 v (u, v) 1, u 2 v 2 where u = (u 1, u 2 ) and v = (v 1, v 2 ). Then ϕ is a linear map and the mapping f : R 2 R 2 R, (u, v) det ϕ(u, v), is an alternating 2-linear map. In fact, in this sense, a determinant (R n ) n R is an alternating n-linear map for all n. Lemma A k-linear mapping f : V k W is alternating if and only if, for all (v 1,..., v k ) V k, (1.3.2) f(v 1,..., v k ) = 0 if v i = v j for i j. Example Let f, g V and define ω : V V R by formula ω(v, v ) = f(v)g(v ) f(v )g(v) for v, v V. Then ω is 2-linear and alternating, and hence ω Alt 2 (V ). It will turn out that ω is an alternating product (or wedge) of f and g. Example Let (ε 1, ε 2 ) be the dual basis of the standard basis (e 1, e 2 ) of R 2. If f = ε 1, g = ε 2, and ω are as in the previous example, then [ ] x1 x ω((x 1, y 1 ), (x 2, y 2 )) = x 1 y 2 x 2 y 1 = det 2 y 1 y 2 for all (x 1, y 1 ), (x 2, y 2 ) R 2. 12

14 1.3.1 Space Alt k (V ) Definition Let V be a vector space. We denote Alt k (V ) = {f : V k R: f is an alternating k linear}. Here V 0 = R and we identify Alt 0 (V ) = R = R. Lemma Let V be a vector space. Mult k (V, R), and hence of Hom( k V, R). Then Alt k (V ) is a subspace of Proof. Since Alt k (V ) Mult k (V, R), it suffices to observe that (clearly) f + ag : V k R is alternating for every f, g Alt k (V ) and a R. Remark Alt 1 (V ) = {f : V R: f is k linear} = Hom(V, R) = V. Lemma Let V, W be vector spaces and ϕ: V W a linear map. Then ϕ : Alt k (W ) Alt k (V ) defined by (v 1,..., v k ) ω(ϕ(v 1 ),..., ϕ(v k )) (as in Definition 1.2.1), is well-defined and linear. Proof. Since ϕ : Mult k (W, R) Mult k (V, R) is well-defined and linear, it suffices to observe that (clearly) ϕ ω is alternating for every ω Alt k (V ) Intermission: permutations Let k 1. A bijection {1,..., k} {1,..., k} is called a permutation. We denote by S k the set of all permutations {1,... k} {1,..., k}. A permutation τ S k is a transposition if there exists 1 i < j k for which τ(m) = m for m i, j and τ(i) = j (and τ(j) = i); we denote by τ ij = τ (k) ij the transposition satisfying τ ij (i) = j. A permutation is even if it can be written as an even number of transpositions. Otherwise, a permutation is called odd. A permutation is either even or odd. We formalize this as follows. Proposition A permutation has a sign, that is, there exists a function sign: S k {±1} satisfying { +1, σ is an even permutation, sign(σ) = 1, σ is an odd permutation. Proof. Induction on k. (Exercise.) 13

15 Corollary The sign: S k {±1} satisfies In particular, for transpositions τ 1,..., τ m. sign(σ σ ) = sign(σ)sign(σ ). sign(τ 1 τ m ) = ( 1) m Remark The set S k is a group under composition, that is, the product σσ is the composition σ σ. In Proposition and Corollary we implicitely state that S k is generated by transpostions and the function sign is, in fact, a group homomorphism (S k, ) ({±1}, ). We do not pursue these details here. The main observation on permutions and alternating maps is the following lemma. Lemma Let f : V k W be an alternating k-linear map and σ S k. Then, for (v 1,..., v k ) V k, f(v σ(1),..., v σ(k) ) = sign(σ)f(v 1,..., v k ). Proof. Let σ = τ 1 τ m S k and denote σ = τ 1 1 σ. Then σ = τ 1 σ and, by alternating, Thus, by induction, f(v σ(1),..., v σ(k) ) = f(v τ1 (σ (1)),..., v τ1 (σ (k))) = f(v σ (1),..., v σ (k)). f(v σ(1),..., v σ(k) ) = ( 1) m f(v 1,..., v k ) = sign(σ)f(v 1,..., v k ). Lemma gives an easy proof for an alternative characterization of alternate multilinear maps. Lemma Let V and W be vector spaces and f : V k W be a k- linear map. Then f is alternating if and only if f(v 1,..., v k ) = 0 whenever v i = v i+1 for some 1 i < k. Proof. Exercise. Exercise Let f : V k W be a map and σ S k. Define the map σ # ω : V k W by (σ # f)(v 1,..., v k ) = f(v σ(1),..., v σ(k) ). Show that (σ # f) is an alternating k-linear map if and only if f is an alternating k-linear map. 14

16 1.3.3 Exterior product Definition Let k, l 1. A permutation σ S k+l, is a (k, l)-shuffle if σ(1) < σ(2) < < σ(k) and σ(k + 1) < < σ(k + l). We denote by S(k, l) the set of all (k, l)-shuffles. Definition Let ω Alt k (V ) and τ Alt l (V ). The exterior (or the wedge) product ω τ Alt k+l (V ) is defined by ω τ(v 1,..., v k+l ) = sign(σ)ω(v σ(1),..., v σ(k) )τ(v σ(k+1),..., v σ(k+l) ) for v 1,..., v k+l V. σ S(k,l Lemma Let ω Alt k (V ) and τ Alt l (V ). Then ω τ Alt k+l (V ). Proof. (See also [7, Lemma 2.6].) By Lemma it suffices to show that if v i = v i+1 for some 1 i < k + l. Let ω τ(v 1,..., v k+1 ) = 0 S 0 = {σ S(k, l): {i, i + 1} σ({1,..., k}) or {i, i + 1} σ({k + 1,..., k + l})} S = {σ S(k, l): i σ({1,..., k}) and i + 1 σ({k + 1,..., k + l})} S + = {σ S(k, l): i + 1 σ({1,..., k}) and i σ({k + 1,..., k + l})} Since v i = v i+1, we have, for every σ S 0, either ω(v σ(1),..., v σ(k) ) = 0 or τ(v σ(k+1 ),..., v σ(k+l) ) = 0. Thus ω τ(v 1,..., v k+l ) = sign(σ)ω(v σ(1),..., v σ(k) )τ(v σ(k+1),..., v σ(k+l) ) σ S(k,l) = It suffices to show that σ S + sign(σ)ω(v σ(1),..., v σ(k) )τ(v σ(k+1),..., v σ(k+l) ) + σ S sign(σ)ω(v σ(1),..., v σ(k) )τ(v σ(k+1),..., v σ(k+l) ). σ S + sign(σ)ω(v σ(1),..., v σ(k) )τ(v σ(k+1),..., v σ(k+l) ) = σ S sign(σ)ω(v σ(1),..., v σ(k) )τ(v σ(k+1),..., v σ(k+l) ). 15

17 Let α = τ i,i+1 be the transposition interchanging i and i + 1. Step 1: We show first that S S +, σ α σ, is a well-defined bijection. The inverse of the map is σ α σ, so it a bijection. It suffices to show that the target is S +. Let σ S. Then there exists 1 j k for which σ(j) = i and k + 1 j k + l for which σ(j ) = i + 1. We observe that, for 1 j < k, i + 1 < σ(j + 1) since i + 1 σ({1,..., k}) and σ(j + 1) > σ(j) = i. Similarly, and σ(j 1) < i. Since α is a transposition, we have Thus (α(σ(1)),..., α(σ(k))) = (σ(1),..., σ(j 1), α(σ(j)), σ(j + 1),..., σ(k)) = (σ(1),..., σ(j 1), i + 1, σ(j + 1),..., σ(k)) (α(σ(k + 1)),..., α(σ(k + l))) = ( σ(k + l),..., σ(j 1), α(σ(j )), σ(j + 1),..., σ(k + l) ) = ( σ(k + 1),..., σ(j 1), i, σ(j + 1),..., σ(k + l) ). α(σ(1)) < < α(σ(k)) and α(σ(k + 1)) < < α(σ(k + l)). Hence α σ S(k, l) and α σ S +. This ends Step 1. Step 2: We show now that, for σ S, ω(v α(σ(1)),..., v α(σ(k)) )τ(v α(σ(k+1)),..., v α(σ(k+l)) ) = ω(v σ(1),..., v σ(k) )τ(v σ(k+1),..., v σ(k+l) ). Let 1 j k and k + 1 j k + l be as in the proof of Subclaim 1. Since v i = v i+1, we have ω(v α(σ(1)),..., v α(σ(k)) )τ(v α(σ(k+1)),..., v α(σ(k+l)) ) = ω(v σ(1),..., v σ(j 1), v i+1, v σ(j+1),..., v σ(k) ) τ(v σ(k+1),..., v σ(j 1), v i, v σ(j +1),..., v σ(k+l) ) = ω(v σ(1),..., v σ(j 1), v i, v σ(j+1),..., v σ(k) ) τ(v σ(k+1),..., v σ(j 1), v i+1, v σ(j +1),..., v σ(k+l) ) = ω(v σ(1),..., v σ(k) )τ(v σ(k+1),..., v σ(k+l) ). 16

18 This ends Step 2. Final step: Combining previous steps, we have ω τ(v 1,..., v k+l ) = σ S sign(α σ )ω(v α σ (1),..., v α σ (k))τ(v α σ (k+1),..., v α σ (k+l)) + σ S sign(σ)ω(v σ(1),..., v σ(k) )τ(v σ(k+1),..., v σ(k+l) ) = ( 1) σ S sign(σ )ω(v σ (1),..., v σ (k))τ(v σ (k+1),..., v σ (k+l)) + σ S sign(σ)ω(v σ(1),..., v σ(k) )τ(v σ(k+1),..., v σ(k+l) ) = 0. The exterior product is bilinear in the following sense. Lemma Let ω, ω Alt k (V ), τ, τ Alt l (V ), and a R. Then Proof. Exercise. (ω + aω ) τ = ω τ + aω τ ω (τ + aτ ) = ω τ + aω τ. Lemma Let ω Alt k (V ) and τ Alt l (V ). Then ω τ = ( 1) kl τ ω. The proof of Lemma ([7, Lemma 2.8]) is based on the following observation on shuffles. Lemma Let k, l 1 and α S k+l the permutation { k + i, i l α(i) = i l, i l + 1 Then S(l, k) S(k, l), σ σ α, is a bijection. Furthermore, sign(α) = ( 1) kl. Proof. Let 1 j < l. Then k < α(j) < α(j + 1) k + l. Thus, for α S(l, k), σ(α(j)) < σ(α(j + 1)). Hence σ α S(k, l). Thus σ σ α is well-defined. It is clearly a bijection (the inverse map is given by σ σ α 1 ). 17

19 We show now that sign(α) = ( 1) kl. For 1 i k, let β i S k+l be the permutation β i = τ i,i+1 τ i+1,i+2 τ l+i 1,l+i. Then i, j = l + i β i (j) = j + 1, for i j < l + i j, otherwise. Moreover, sign(β i ) = ( 1) l for every 1 i k. By induction, α = β k β 1. Thus sign(α) = sign(β k β 1 ) = sign(β k ) sign(β 1 ) = ( 1) kl. Proof of Lemma Let α S k+l be the permutation in Lemma Then τ(v σ(α(1)),..., v σ(α(l)) ) = τ(v σ(k+1),..., v σ(k+l) ) and Hence ω(v σ(α(l+1),..., v σ(α(k+l) ) = ω(v σ(1),..., v σ(k) ). τ ω(v 1,..., v k+l ) = sign(σ)τ(v σ(1),..., v σ(l) )ω(v σ(l+1),..., v σ(l+k) ) σ S(l,k) = sign(σ α)τ(v σ (α(1)),..., v σ (α(l))ω(v σ (α(l+1)),..., v σ (α(l+k))) σ S(k,l) = sign(α) sign(σ )τ(v σ (k+1)),..., v σ (l+k))ω(v σ (1),..., v σ (k)) σ S(k,α) = ( 1) kl ω τ(v 1,..., v k+l ). For completeness, we remark that the exterior product is associative. Lemma Let ω Alt k (V ), τ Alt l (V ), and ξ Alt m (V ). Then (ω τ) ξ = ω (τ ξ) Alt k+l+m (V ). Proof. [7, Lemma 2.9]. 18

20 1.3.4 Alternating multilinear maps and determinants Lemma Let V be a vector space and ω 1,..., ω k Alt 1 (V ). Then ω 1 (v 1 ) ω 1 (v k ) ω 1 ω k (v 1,..., v k ) = det..... ω k (v 1 ) ω k (v k ) for all v 1,..., v k V. Proof. The proof is by induction. The claim clearly holds for k = 1. Suppose it holds for k 1. Then, by the definition of the exterior product, we have, for k + 1, ω 1 ω k+1 (v 1,..., v k+1 ) = ω 1 (ω 2 ω k+1 ) (v 1,..., v k+1 ) = sign(σ)ω 1 (v σ(1) )ω 2 ω k+1 (v σ(2),..., v σ(k+1) ) σ S(1,k) k+1 = ( 1) i+1 ω 1 (v i )ω 2 ω k+1 (v 1,..., v i 1, v i+1,..., v k+1 ). i=1 For i = 1,..., k + 1, let A i1 be the matrix ω 2 (v 1 ) ω 2 (v i 1 ) ω 2 (v i+1 ) ω k+1 (v σ(k+1) ) A i1 = ω k+1 (v 1 ) ω k+1 (v i 1 ) ω k+1 (v i+1 ) ω k+1 (v σ(k+1) ) Then, by the induction assumption and the expansion of the determinant along the first row ω 1 ω k+1 (v 1,..., v k+1 ) = k+1 ( 1) i+1 ω 1 (v i ) det A i1 i=1 ω 1 (v 1 ) ω 1 (v k+1 ) = det..... ω k+1 (v 1 ) ω k+1 (v k+1 ) Corollary Let ω 1,..., ω k Alt 1 (V ). Then ω 1 ω k 0 if and only if ω 1,..., ω k are linearly independent in V. 19

21 Proof. Suppose ω 1 ω k 0. Suppose towards contradiction that ω 1,..., ω k are linearly dependent, that is, there exists a 1,..., a k R not all zeroes, so that a 1 ω a k ω k = 0. We may assume that a 1 0 and solve ω 1 to obtain Thus ω 1 = 1 a 1 (a 2 ω a k ω k ). ω 1 ω k = k i=2 a i a 1 ω i ω 2 ω k = 0, since ω i ω 2 ω k = 0 for all i = 1,..., k by Lemma This is a contradiction. Thus linear maps ω 1,..., ω k are linearly independent. Suppose now that ω 1 ω k = 0 and suppose towards contradiction that (ω 1,..., ω k ) is linearly independent. Then there exists vectors e 1,..., e k V for which ω i (e j ) = δ ij. Then ω 1 (e 1 ) ω k (e 1 ) ω 1 (e 1 ) ω 1 (e k ) det..... = det..... ω 1 (e k ) ω k (e k ) ω k (e 1 ) ω k (e k ) = ω 1 ω k (e 1,..., e k ) = 0. Thus columns of the determinant are linearly dependent and there exists coefficients a 1,..., a k R (not all zero) so that a 1 ω 1 (e 1 ). ω 1 (e k ) + + a k We may assume that a 1 0 and we obtain ω 1 (e 1 ). ω k (e k ) = 0. ω 1 (e i ) = 1 a 1 k ω j (e i ) j=1 for all i = 1,..., k. But this is contradiction, since ω 1 (e 1 ) = 1 and 1 a 1 k ω j (e 1 ) = 0. j=1 Thus (ω 1, ω 2,..., ω k ) is linearly dependent. 20

22 1.3.5 Basis of Alt k (V ) Let n k 1. Recall that σ S(k, n k) is a bijection {1,..., n} {1,..., n} satisfying 1 σ(1) < < σ(k) n and σ(k + 1) < < σ(n). Theorem Let V be n-dimensional (real) vector space, (e 1,..., e n ) a basis of V and (ε 1,..., ε n ) the corresponding dual basis. Then (1.3.3) (ε σ(1) ε σ(k) ) σ S(k,n k) is a basis of Alt k (V ). Moreover, for every ω Alt k (V ), (1.3.4) ω = ω(e σ(1),..., e σ(k) )ε σ(1) ε σ(k). σ S(k,n k) Corollary Let V be an n-dimensional vector space. For 0 k n, ( n ) dim Alt k (V ) = #S(k, n k) =. k In particular, Alt n (V ) = R. Moreover, for k > n, Alt k (V ) = {0}. Proof of Theorem We show first the linear independence. Suppose a σ R (σ S(k, n k)) are such coefficients that a σ ε σ(1) ε σ(k) = 0. Let τ S(k, n k). By Lemma , for every σ S(k, n k). Thus σ ε σ(1) ε σ(k) (e τ(1),..., e τ(k) ) = δ σ,τ a τ = a τ ε τ(1) ε τ(k) (e τ(1),..., e τ(k) ) = σ = 0. a σ ε σ(1) ε σ(k) (e τ(1),..., e τ(k) ) for every τ S(k, n k). Hence (1.3.3) is linearly independent. We show now (1.3.4). This proves that the sequence in (1.3.3) spans Alt k (V ). Let v i = k j=1 v ije j V for i = 1,..., k. Then ω(v 1,..., v k ) = v 1j1 v kjk ω(e j1,... e jk ) = = (j 1,...,j k ) σ S(n,k) σ S(n,k) v 1,τ(σ(1)) v k,τ(σ(k)) ω(e τ(σ(1)),... e τ(σ(k)) ) τ S k sign(τ)v 1,τ(σ(1)) v k,τ(σ(k)) ω(e σ(1),... e σ(k) ) τ S k 21

23 On the other hand, ε σ(1) ε σ(k) (e α(1),... e α(k) ) = { 1, α = σ 0, otherwise for all σ, α S(k, n k) (by Lemma or directly from definition). Thus (1.3.5) ε σ(1) ε σ(k) (v 1,..., v k ) = sign(τ)v 1,τ(α(1)) v k,τ(α(k)) ε σ(1) ε σ(k) (e α(1),... e α(k) ) α S(k,n k) τ S k = sign(τ)v 1,τ(σ(1)) v k,τ(σ(k)). τ S k Hence ω(v 1,..., v k ) = = = σ S(k,n k) σ S(k,n k) σ S(k,n k) τ S k sign(τ)v 1,τ(σ(1)) v k,τ(σ(k)) ω(e σ(1),... e σ(k) ) ε σ(1) ε σ(k) (v 1,..., v k )ω(e σ(1),... e σ(k) ) ω(e σ(1),... e σ(k) )ε σ(1) ε σ(k) (v 1,..., v k ). Remark Note that, for n = k and σ = id, we may combine Lemma and (1.3.5) to obtain the formula det [v 1 v n ] = ε 1 ε n (v 1,..., v n ) = τ S n sign(τ)v 1τ(1) v nτ(n). This representation formula is usually taken as a definition of the determinant. Taking the formula for the determinant in Remark and combining it with Lemma , we also get a (useful) formula which could be taken as the definition of exterior product of dual vectors. Lemma Let V be a vector space and let ω 1,..., ω k Alt k (V ). Then ω 1 ω k (v 1,..., v k ) = σ S k sign(σ)ω 1 (v σ(1) ) ω n (v σ(k) ). 22

24 1.3.6 Exterior ring Alt (V ) Let V be an n-dimensional vector space. Having the exterior product at our disposal, we may define the exterior ring Alt (V ) of the vector space V by setting Alt (V ) = k 0 Alt k (V ) = Alt 0 (V ) Alt 1 (V ) Alt n (V ). Thus the elements of Alt (V ) are sequences ω = (ω 0,..., ω n ). Since there is no confusion between k- and l-linear maps, we may also unambiguously write ω = n (0,..., 0, ω k, 0,..., 0) = ω 0 + ω ω n, k=0 where ω k Alt k (V ); note that components in this sum are uniquely determined. The exterior product : Alt k (V ) Alt l (V ) Alt k+l (V ) induces a multiplication : Alt (V ) Alt (V ) Alt (V ) with ( n ) ( n ) n ω k τ k = ω k τ l. k=0 k=0 k,l=0 Definition Vector space Alt (V ) with exterior product is called an exterior ring of V. Given a linear map ϕ: V W, there exists a ring homomorphism ϕ : Alt (W ) Alt (V ) given by the formula ϕ ( n ω k ) = k=0 n ϕ ω k. The map ϕ is clearly linear and, by Lemma (below), it satisfies for all ω, τ Alt (W ). k=0 ϕ (ω τ) = ϕ ω ϕ τ Lemma Let V, W be vector spaces, ω Alt k (W ), τ Alt l (W ), and ϕ: V W a linear map. Then Proof. Exercise. ϕ (ω τ) = ϕ ω ϕ τ. 23

25 Chapter 2 Differential forms Let n, k 0. Let (e 1,..., e n ) be a standard basis of R n and (ε 1,..., ε n ) the corresponding dual basis. By results in the previous section, we have that Alt k (R n ) is a vector space with basis (ε σ ) σ S(k,n k), where Thus we have linear isomorphism ε σ = ε σ(1) ε σ(k). R ( n k ) = R S(k,n k) = Alt k (R n ), where R S(k,n k) is the vector space of all functions S(k, n k) R. We give R S(k,n k) and Alt k (R n ) topologies which are induced by these linear isomorphisms. (A careful reader checks that the given topologies are independent on the chosen linear isomorphisms. We leave this to the careful reader, though.) Let U R n be an open set. Let ω : U Alt k (R n ) be a function. Then ω(x) Alt k (R n ) for every x U. Thus, for every x U, there exists coeffiecients f σ (x) R for which ω(x) = σ S(k,n k) f σ (x)ε σ. In particular, for every σ S(k, n k), we have a function f σ : U R. For every α S(k, n k) and x U, we also have the formula (ω(x)) (e α(1),..., e α(k) ) = f α (x). Based on this observation, we give the following definition for smoothness. (A careful reader again verifies that the smoothness does not depend on chosen basis.) 24

26 Definition Let U R n be an open set. Alt k (R n ) is C -smooth if the function U R, A function ω : U is smooth for every α S(k, n k). x (ω(x)) (e α(1),..., e α(k) ), Convention To avoid unneccessary parenthesis, it is common to write ω x = ω(x) Alt k (R n ) and for x Ω and v 1,..., v k R n. ω x (v 1,..., v k ) = (ω(x)) (v 1,..., v k ) Definition Let U R n be an open set. A differential k-form is a C -smooth function U Alt k (R n ). We denote by Ω k (U) the vector space of all differential k-forms U Alt k (R n ). Remark We also call elements of Ω k (U) differential forms, k- forms, or just forms, for short. Example (1) Let U R n be an open set. Then x ε 1 ε n is a differential form Ω k (U). We call this usually a volume form and denote it by vol U. (2) Let U = R n \ {0}. Then the map ρ: U Alt n 1 (U), ( ) x (ρ x )(v 1,..., v n 1 ) = ε 1 ε n x, v 1,..., v n 1 is an (n 1)-form in Ω n 1 (U) (Exercise). The exterior product Alt (R n ) extends to differential forms as follows. Definition Let U R n be an open set. Let ω Ω k (U) and τ Ω l (U). The map ω τ : U Alt k+l (R n ) defined by (2.0.1) (ω τ) x = ω x τ x for x Ω is the exterior product ω τ Ω k+l (U) of ω and τ. Abstractly, we have have obtained a bilinear operator : Ω k (U) Ω l (U) Ω k+l (U). 25

27 2.1 Exterior derivative Let U be an open set in R n and f : U R a smooth function. Definition The differential df of f is the 1-form df : U Alt 1 (R n ) defined by f(x + tv) f(x) (df) x (v) = lim. t 0 t Remark The differential df of f is nothing but the standard linear map given by the directional derivatives. By the multivariable calculus, df(v) = v, f, where f : U R n is the gradient of f f = f x 1 (x). f x n (x) Example Let x = (x 1,..., x n ): R n R n be the indentity map with coordinate functions x i : R n R (i.e. x i (v 1,..., v n ) = v i ). Then dx i = ε i,. where ε i is the element of the (standard) dual basis.(exercise.) Corollary Let U R n be an open set and ω Ω k (U). Then ω = ω σ dx σ(1) dx σ(k), σ S(k,n k) where ω σ is the function ω σ (x) = ω x (e σ(1),..., e σ(k) ). Proof. Exercise Exterior derivitive of a k-form Observation Let U be an open set and ω Ω k (U). Let v = (v 1,..., v k ) (R n ) k and consider the function ω v : U R defined by ω v (x) = ω x (v 1,..., v k ) = ω σ (x)ε σ(1) ε σ(k) (v 1,..., v k ). Then, for w R n, σ S(k,n k) (dω v ω v (x + tw) ω v (x) ) x (w) = lim t 0 t 1 = lim (ω σ (x + tw) ω σ (x))ε σ (v 1,..., v k ) t 0 t = σ σ (dω σ ) x (w)ε σ (v 1,..., v k ) 26

28 Remark (Alternative approach). Since Alt k (V ) is a vector space, we may also consider ω : U Alt k (U) as a vector valued function (or a map!) and define the differential (Dω) x at x U as a linear map R n Alt k (R n ) by formula 1 (Dω) x (w) = lim t 0 t (ω x+tw ω x ). Note that (Dω) x (w) is an alternating k-linear map and 1 ((Dω) x (w)) (v 1,..., v k ) = lim t 0 t (ω x+tw ω x ) (v 1,..., v k ) 1 = lim t 0 t (ω x+tw(v 1,..., v k ) ω x (v 1,..., v k )) for v 1,..., v k R n. Thus, for v = (v 1,..., v k ) (R n ) k and w R n, ((Dω) x (w)) (v 1,..., v k ) = (dω v ) x (w). Convention Given a k-form ω Ω k (U). In what follows we use the notation ω x (v 1,..., v i,..., v k ) = ω x (v 1,..., v i 1, v i+1,..., v k ) to indicate that we have removed the ith entry. Observation We make now the final observation before the definition of the exterior product. Let ω = fdx i1 dx ik Ω k (U). Then, for v 1,..., v k+1 R n, (2.1.1) df dx i1 dx ik (v 1,..., v k+1 ) = sign(σ)df(v σ(1) )dx i1 dx ik (v σ(2),..., v σ(k+1) ) σ S(1,k) k+1 = ( 1) i 1 df(v i )dx i1 dx ik (v 1,..., v i,..., v k+1 ) i=1 Definition Let ω Ω k (U). The exterior derivative dω Ω k+1 (U) of ω is the (k + 1)-form (dω) x (v 1,..., v k+1 ) = σ S(k,n k) i=1 k+1 ( 1) i 1 (dω σ ) x (v i )ε σ (v 1,..., v i,..., v k ). Remark (1) By (2.1.1), we have that dω = dω σ dx σ(1) dx σ(k). σ S(k,n k) 27

29 (2) Clearly for ω, τ Ω k (U) and a R. d(ω + aτ) = dω + adτ (3) Since dx i1 dx ik has constant coefficients, we have for every i 1,..., i k {1,..., n}. d (dx i1 dx ik ) = 0 The two most important properties of the exterior product are the Leibniz rule and the fact that the composition of exterior derivatives is zero. We begin with the latter. Lemma Let ω Ω k (U). Then d 2 ω = d(dω) = 0. Proof. Let f C (U) and dx I = dx i1 dx ik, where I = (i 1,..., i k ). Then ( n ) d 2 f n n ( 2 ) f (fdx I ) = d dx i dx I = dx j dx i dx I x i x j x i i=1 j=1 i=1 k 2 f = (dx j dx i ) dx I x j,i=1 j x i = ( 2 ) f 2 f (dx j dx i ) dx I = 0. x j<i j x i x i x j The claim now follows by linearity. The Leibniz rule for the exterior product has the following form. Lemma Let ω Ω k (U) and τ Ω l (U). Then d(ω τ) = dω τ + ( 1) k ω dτ. Proof. It suffices to consider cases ω = fdx I and τ = gdx J, where dx I = dx i1 dx ik and dx J = dx j1 dx jl for I = (i 1,..., i k ) and J = (j 1,..., j l ). Since d (ω τ) = d ((fg)dx I dx J ) = d(fg) dx I dx J = (gdf + fdg) dx I dx J = (df dx I ) gdx J + fdg dx I dx J = dω τ + ( 1) k fdx I dg dx J = dω τ + ( 1) k ω dτ 28

30 We record as an observation that the exterior product for functions, Leibniz rule and the composition property uniquely determine the exterior product. Corollary Suppose that, for every k 0, ˆd k : Ω k (U) Ω k+1 (U) is a linear operator satisfying (1) ˆd 0 f = df for every f Ω 0 (U), (2) ˆd k+1 ˆd k = 0, (3) ˆd k+l (ω τ) = d k ω τ + ( 1) k ω ˆd l τ for ω Ω k (U) and τ Ω l (U). Then ˆd k = d for every k 0. Proof. See [7, Theorem 3.7]. 2.2 Pull-back of forms We begin with an observation, which we record as a lemma. Lemma Let f : U U be a C -smooth map. Then f : Ω k (U ) Ω k (U) defined by formula (2.2.1) (f ω) x (v 1,..., v k ) = ω f(x) ((Df)v 1,..., (Df)v k ) is a well-defined linear operator. Remark Note that (2.2.1) is equivalent to for every x U. (f ω) x = ((Df) x ) ω f(x) Definition Let f : U U be a C -smooth map. The mapping f : Ω k (U ) Ω k (U) is the pull-back (of differential forms) defined by f. Having Lemma at our disposal, we immediately obtain the fact that pull-back and the wedge product commute. Lemma Let U R m and U R n be open sets, and let f : U U be a C -smooth map. Then, for ω Ω k (U ) and τ Ω l (U ), we have f (ω τ) = (f ω) (f τ). In particular, for u Ω 0 (U) = C (U), f (u) = u f We also have the following observation. 29

31 Lemma Let U R n and U R m be open sets, and let f = (f 1,..., f n ): U U be a C -smooth map. Then for every i = 1,..., n. Proof. Exercise. f (dx i ) = df i The exterior derivative and the pull-back commute. Lemma Let U R m and U R n be open sets, and let f : U U be a C -smooth map. Then for every ω Ω k (U ). df ω = f dω Proof. By linearity, it suffices to consider ω = udx i1 dx ik. By the chain rule, ( m ) f (du) = f u m ( ) u dx i = f f (dx i ) x i=1 i x i=1 i m ( ) u n = f f i n m ( ) dx j u fi = f dx j x i x j x i x j = i=1 n j=1 j=1 (u f) dx j = d(u f). x j The second observation is that d (df i1 df ik ) = 0 j=1 i=1 for all I = (i 1,..., i k ). Indeed, this follows from the Leibniz rule by indeuction in k as follows: d (df i1 df ik ) = d(df i 1 ) df i2 df ik + ( 1)df i1 d (df i2 df ik ) = 0, where we used the induction assumption on df i2 df ik and the fact that d 2 = 0. Having both of these observation at our disposal, we get df ω = d ((u f)df i1 df ik ) = d(u f) (df i1 df ik ) + (u f) d (df i1 df ik ) = f du f dx i1 f dx ik = f (du dx i1 dx ik ) = f (dω). 30

32 Chapter 3 De Rham cohomology Let U R n be an open set. Since d 2 = d d: Ω k 1 (U) Ω k+1 (U) is zero opeator for every k 1, we have ( ) ( ) (3.0.1) im d: Ω k 1 (U) Ω k (U) ker d: Ω k (U) Ω k+1 (U). Thus Imd is a subspace of ker d for all k 1. For convenience, we set Ω k (U) = {0} for all k < 0 and set d = 0: Ω k (U) Ω k+1 (U) for all k < 0. Then (3.0.1) holds for all k Z. Definition Let U R n be an open set. The quotient vector space H k (U) = ker(d: Ωk (U) Ω k+1 (U)) im(d: Ω k 1 (U) Ω k (U)) = {ω Ωk (U): dω = 0} {dτ : τ Ω k 1 (U)} is the kth de Rham cohomology (group) of U. Recall that elemtents of the quotient space H k (U) are equivalence classes of k-forms. Given ω ker(d: Ω k (U) Ω k+1 (U)), we denote the equivalence class by [ω] = {ω + dτ Ω k (U): τ Ω k 1 (U)}. Definition Let U R n be an open set. A form ω Ω k (U) is closed if dω = 0 and exact if there exists a (k 1)-form τ Ω k 1 (U) for which dτ = ω. Thus H k (U) = {closed k forms in U} {exact k forms in U} and de Rham cohomology H k (U) is a vector space which classifies the closed k-forms in U upto exact forms. The 0th cohomology H 0 (U) counts the number of components of the set U. This is typically formalized as follows. 31

33 Lemma Let U be an open set. Then H 0 (U). is the vector space of maps U R which are locally constant. Proof. By definition H 0 (U) = ker(d: Ω0 (U) Ω 1 (U)) Im(d: Ω 1 (U) Ω 0 (U)). Since d: Ω 1 (U) Ω 0 (U) is the zero map, we have H 0 (U) = ker(d: Ω 0 (U) Ω 1 (U)) On the other hand, given f C (U), df = 0 if and only if f is locally constant. The claim follows. Again there is a natural pull-back. Lemma Let U R n and U R m be open sets, and let f : U U be a smooth mapping. Then the mapping H k (f): H k (U ) H k (U) defined by the formula [ω] [f ω] is well-defined and linear. Proof. To show that the formula is well-defined, we have to show that it is independent on the representative. Let ω, ω Ω k (U ) be closed forms so that [ω] = [ω ]. Then, by definition of the quotient, ω ω = dτ where τ Ω k 1 (U ). Since f ω = f (ω + dτ) = f ω + df τ, we have f ω f ω = df τ. Thus [f ω] = [f ω ] and H k (f) is well-defined. To show the linearity, let again ω, ω Ω k (U ) and a R. Then f (ω + aω ) = f ω + af ω. Since [f ω + af ω ] = [f ω] + a[f ω ] by the definition of operations in the quotient space, linearity of H k (f) follows. Remark It is typical to denote the mapping H k (f) by f. Thus, on the cohomological level, the pull-back is defined by f [ω] = [f ω]. Lemma Let U, V, W be open sets in R n, R m, and R l, respectively, and let f : U V and g : V W be C -smooth maps. Then and f g = (g f) : H k (W ) H k (U) id = id: H k (U) H k (U) for each k. In paricular, f : H k (V ) H k (U) if f is a diffeomorphism (i.e. f is a homeomorphism so that f and f 1 are C -smooth). 32

34 Proof. Given ω Ω k (W ) and x U, we have by the chain rule (f g ) x ω = (f ) x (g ω) = (Df x ) (g ω) f(x) = (Df) x (Dg) f(x) ω g(f(x)) = ((Dg) f(x) (Df) x ) ω (g f)(x) = D(g f) x ω (g f)(x) = (g f) xω. Thus (f g )[ω] = [(f g )ω] = [(g f) (ω)] = (g f) [ω]. Since id ω = ω, the second claim follows immediately. If f is a diffeomorphism, we observe that Thus f is an isomorphism. f (f 1 ) = id = id and (f 1 ) f = id. The exterior product is naturally defined on the cohomological level. Lemma Let U R n be an open set. Then the mapping H k (U) H l (U) H k+l (U) defined by is well-defined and bilinear. [ξ] [ζ] = [ξ ζ] Proof. Let ξ, ξ Ω k (U) and ζ, ζ Ω k (U) be closed forms for which ξ ξ = dα and ζ ζ = dβ for some α Ω k 1 (U) and β Ω k 1 (U). Since ζ is closed, d(α ζ) = dα ζ + ( 1) l 1 α dζ = dα ζ. Similarly, Thus Thus d(ξ β) = dξ β + ( 1) k ξ dβ = ( 1) k ξ dβ. ξ ζ = (ξ + dα) (ζ + dβ) = ξ ζ + ξ dβ + dα ζ + dα dβ = ξ ζ + ( 1) k d(ξ β) + d(α ζ) + d(α dβ) ( ) = ξ ζ + d ( 1) k ξ β + α ζ + α β. [ξ ζ ] = [ξ ζ ] and the mapping is well-defined. Cheking the bilinearity is left to the interested reader. 33

35 3.1 Poincaré lemma A fundamental fact in de Rham theory is that { H k (B n ) R, k = 0 = 0, k > 0. We obtain this fact from slightly more general result. Recall that a set U R n is star-like if there exists x 0 U so that, {(1 t)x 0 + tx R n : 0 t 1} U for every x U. The point x 0 is a center of U. Theorem (Poincaré lemma). Let U R n be an open star-like set. Then { dim H k 1, k = 0 (U) = 0, k > 0. Proof. Every star-like set is connected. Thus every locally constant function on U is constant. Hence H 0 (U) = R by Lemma Let k > 0. We need to show that ker(d: Ω k (U) Ω k+1 (U)) = Im(d: Ω k 1 (U) Ω k (U)). Idea: Suppose we find, for every k > 0, a linear map S k : Ω k (U) Ω k 1 (U) satisfying ω = ds k ω + S k+1 dω for every ω Ω k (U). Then every closed k-form is exact. Indeed, suppose ω is a closed k-form. Then ω = ds k ω + S k+1 dω = ds k ω. (This operator S k is a chain homotopy operator.) Execution of the idea. We abuse the notation and denote by t: U R R the projection (x, s) s. We also denote by x k : U R R the projection x k (y 1,..., y n, s) = y k (as usual). Given ω Ω k (U R), we have (3.1.1) ω = I=(i 1,...,i k ) 1 i 1 < <i k n = I ω I dx I + J ω I dx i1 d ik + ω J dt dx J, J=(j 1,...,j k 1 ) 1 j 1 < <j k 1 n ω J dt dx j1 d jk 1 where functions ω I C (U) and ω J C (U) are uniquely determined. Thus the linear map Ŝk : Ω k (U R) Ω k 1 (U), defined by formula ( 1 ) (Ŝkω) y = ω J (y, s) ds dx J J=(j 1,...,j k 1 ) 1 j 1 < <j k 1 n 34 0

36 for y U, is well-defined; here we use the notation in (3.1.1). Let ω as in (3.1.1). Then ( Ŝ k+1 dω = Ŝk+1 dω I dx I + ) dω J dt dx J I J ( ω I = Ŝk+1 t dt dx I + n ω I dx l dx I + x l I I l=1 J = Ŝk+1 ω I t dt dx I ω J dt dx l dx J x l I J,l n l=1 ω J x l dx l dt dx J ) Thus ( 1 (dŝk + Ŝk+1d)ω = d J 0 = ( 1 J,l 0 + ( 1 I 0 ) ( ω J (, s) ds dx J + Ŝk+1 ω J x l (, s) ds ω I (, s) ds t ) dx l dx J ) dx I ( 1 J,l 0 I dω I dx I + J ) ω J (, s) ds dx l dx J x l dω J dt dx J ) = I (ω I (, 1) ω I (, 0)) dx I Use of star-likeness. Since U is star-like, we may fix a center x 0 U of U. Define F : U R U by (x, t) x 0 + λ(t)(x x 0 ), where λ C (R) is a function so that λ(t) = 0 for t 0, 0 λ(t) 1 for 0 t 1, and λ(t) = 1 for t 1. (Existence: exercise.) Then F is a smooth map satisfying F (x, 0) = x, and F (x, 1) = x 0 for every x U. Note that F : Ω k (U) Ω k (U R). For every ω Ω k (U), we have (Exercise!) (F ω) (,1) = ω and (F ω) (,0) = 0. Let S k = Ŝk F : Ω k (U) Ω k 1 (U). Then d S k + S k+1 d = d Ŝk F + Ŝk+1 F d = d Ŝk F + Ŝk+1 d F = Thus ( (Ŝk+1d + dŝk) F ) ω = (F ω) (,1) (F ω) (,0) = ω. The proof is complete. ( ) d Ŝk + Ŝk+1 d F. 35

37 3.2 Chain complexes Let U R n open. Then ( Ω k (U) ) is a sequence of vector spaces. Exterior k Z derivatives (d: Ω k (U) Ω k+1 (U) give a sequence of linear maps with the property that the composition Ω k (U) d Ω k+1 (U) d Ω k+2 (U) is always a zero map. Thus the sequence of pairs ( (Ω k (U), d) ) is a chain complex. In this section we consider abstract chain complexes and then apply the algebraic observations to de Rham cohomology. Definition (Chain complex). The sequence A = (A k, d k ) k Z vector spaces and linear maps d k : A k A k+1 is a chain complex if d k+1 d k = 0. Remark Note that we have chosen here so-called +1-grading (i.e. d k increases the index by 1). It would be also possible to choose 1-grading, which is used e.g. in singular homology (or in homological algebra in general). Both gradings lead to the same theory. Definition The kth homology the chain complex A = (A k, d k ) is H k (A ) = ker d k Imd k 1. The elements of ker d k are called k-cycles and the elements of Imd k 1 k-boundaries. The elements of H k (A ) are homology classes. Example Let U R n open. Denote by Ω k c (U) the compactly supported k-forms in R n, that is, forms ω Ω k (U) for which the closure of {x U : ω x 0} is compact. Since dω Ω k+1 c (U) for every ω Ω k c (U), we obtain a chain complex Ω c(u) = (Ω k c (U), d) k Z. Its homology H k (Ω c(u)) = ker(d: Ωk c (U) Ω k+1 c (U)) Im(d: Ω k 1 (U) Ω k c (U)) is called the compactly supported cohomology H k c (U) of U Exact sequences of vector spaces Definition Let A, B, C be vector spaces and f : A B and g : B C linear maps. The sequence c is exact if ker g = Imf. A f B g C 36

38 Similarly, a sequence f k 2 A k 1 f k 1 A k f k A k+1 f k+1 of vector spaces and linear maps is exact if A k 1 f k 1 A k f k A k+1 is exact whenever defined. Exactness of a sequence encodes familiar properties of linear maps. Lemma (1) If the sequence B g C 0 is exact, then g is surjective. (2) If the sequence 0 A f B is exact, then f is injective. Proof. Exercise. Definition An exact sequence is called a short exact sequence. 0 A f B g C 0 Example Let f : A B be an injective linear map. Then 0 A f B π B/ ker f 0 is an exact sequence; here π is the (canonical) linear map v v + ker f. A fundamental observation on short exact sequences of vector spaces is the following lemma on splitting. Lemma Let 0 A f B g C 0 be a short exact sequence. Then B = A C. In particularly, if A and C are finite dimensional then so is B. 37

39 Proof. (We give a (slightly) abstract proof which does not refer to a basis. See [7, Lemma 4.1] for another proof.) Since g is surjective, we have the diagram B g C π B/ ker g ĝ where π is the (canonical) quotient map and ĝ a linear isomorphism. Let V B be a subspace so that V ker g = B (i.e V + ker g = B and V ker g = {0}). Since Im(π V ) = B/ ker g and ker(π V ) = {0}, we have that π V : V B/ ker g is an isomorphism. Since f is injective, we have that f : A Imf is an isomorphism. Since Imf = ker g, we conclude that B = ker g V = Imf (B/ ker g) = A C. (Note that first is understood in the sense of subspaces and the latter abstractly.) Exact sequences of chain complexes Definition Let A = (A k, d A k ) and B = (B k, d B k ) be chain complexes. A chain map f : A B is a sequence (f k : A k B k ) of linear maps satisfying for each k. A k d A k A k+1 f k f k+1 d B k B k B k+1 Lemma Let f : A B be a chain map. Then f = H k (f): H k (A ) H k (B ), [c] [f(c)], is well-defined and linear. Proof. To show that f is well-defined, let c and c be k-cycles so that c c = dc for c A k 1, i.e. [c] = [c ]. Then f(c) f(c ) = f(dc ) = d f(c ). Thus [f(c)] = [f(c )] and f is well-defined. Linearity is left to the interested reader. Definition Let A = (A k, d A k ) and B = (B k, d B k ) are chain complexes. Chain maps f : A B and g : A B are chain homotopic if there exists a linear maps s k : A k B k 1 for which f g = d B k 1 s k s k+1 d A k. 38

40 Lemma For chain homotopic maps f, g : A B, f = g : H k (A ) H k (B ). Proof. Exercise. (Compare to the operator S k in the proof of Poincaré lemma.) Exercise Let A = (A k, d A k ) and B = (B k, d B k ) be chain complexes. Define A B = (A k B k, d A k db k ); here da k db k : A k B k A k+1 B k+1 is the map (c, c ) (d A k c, db k c ). Show that H k (A B ) = H k (A ) H k (B ). Terminology introduced for vector spaces can easily be extended to chain complexes. A sequence A f B g C of chain complexes and chain maps is exact if A k f k B k g k C k is exact for each k. Similarly, an exact f g sequence 0 A B C 0 (of chain complexes) is called a short exact sequence (of chain complexes). A non-trivial example Let U 1 and U 2 be open sets in R n. We have four inclusions: U 1 U 2 U 1 j 1 j 2 U 2 i 1 U 1 U 2 i 2 It is clear that I k : Ω k (U 1 U 2 ) Ω k (U 1 ) Ω k (U 2 ), ω (i 1ω, i 2ω), and J k : Ω k (U 1 ) Ω k (U 2 ) Ω k (U 1 U 2 ), (ω 1, ω 2 ) j 1ω 1 j 2ω 2, are chain maps. Theorem The sequence 0 Ω k (U 1 U 2 ) I k Ω k (U 1 ) Ω k (U 2 ) J k Ω k (U 1 U 2 ) 0 is a short exact sequence of chain complexes. 39

41 We use the following version on the partition of unity in the proof. Lemma Let U and V be open sets in R n. Then there exist C - smooth functions λ: U V : [0, 1] and µ: U V [0, 1] for which spt λ U, spt µ V, and λ + µ = 1; here spt f = cl U V {x U V : f(x) 0}. Proof. Exercise. Proof of Theorem We begin with a general remark. Let V W R n be open sets and let ι: V W be the inclusion. Then, for all ω = I ω Idx I Ω k (W ), ι (ω) = I (ω I ι)ι (dx I ) = I (ω I V )dx I. We begin now the actual proof. Let ι: U 1 U 2 U 1 U 2 be the inclusion. Then ι = i 1 j 1 = i 2 j 2. In particular, j 1 i 1 = (i 1 j 1 ) = (i 2 j 2 ) = j 2 i 2. We have to show the exactness at three places. Case I: To show that I k is injective, suppose ω = I ω Idx I Ω k (U 1 U 2 ) is in the kernel of I k. Then i 1 ω = 0 and i 2 ω = 0. Thus ω I U 1 = 0 and ω I U 2 = 0 for each I. Hence ω I = 0 for every I. Hence ω = 0 and I k is injective. Case II: Let ω = I ω Idx I Ω k (U 1 U 2 ). Then J k I k (ω) = J k (i 1ω, i 2ω) = j 1i 1ω j 2i 2ω = 0. Thus ImI k ker J k. To show the converse, let (ω 1, ω 2 ) ker J k, where ω 1 = I ω 1,Idx I and ω 2 = I ω 2,Idx I. Then 0 = j 1ω 1 j 2ω 2 = I (ω 1,I U 1 U 2 ω 2,I U 1 U 2 ) dx I and, in particular, ω 1,I U 1 U 2 = ω 2,I U 1 U 2. We extend ω 1,I and ω 2,I by zero to functions of U 1 U 2. Let (λ 1, λ 2 ) be a smooth partition of unity for (U 1, U 2 ) so that λ 1 U 1 \ U 2 = 1. Then ω 1,I (x), x U 1 \ U 2 (3.2.1) λ 2 (x)ω 1,I (x) + λ 1 (x)ω 2,I (x) = ω 1,I (x) = ω 2,I (x), x U 1 U 2 ω 2,I (x), x U 2 \ U 1 Let ω = I (λ 2 ω 1,I + λ 1 ω 2,I ) dx I Ω k (U 1 U 2 ). 40