Lecture 15. de Rham cohomology In this lecture we will show how differential forms can be used to define topo- logical invariants of manifolds. This is closely related to other constructions in algebraic topology such as simplicial homology and cohomology, singular homology and cohomology, and Cechˇ cohomology. 15.1 Cocycles and coboundaries Let us first note some applications of Stokes’ theorem: Let ω be a k-form on a differentiable manifold M.For any oriented k-dimensional compact sub- manifold Σ of M, this gives us a real number by integration: " ω : Σ → ω. Σ (Here we really mean the integral over Σ of the form obtained by pulling back ω under the inclusion map). Now suppose we have two such submanifolds, Σ0 and Σ1, which are (smoothly) homotopic. That is, we have a smooth map F : Σ × [0, 1] → M with F |Σ×{i} an immersion describing Σi for i =0, 1. Then d(F∗ω)isa (k + 1)-form on the (k + 1)-dimensional oriented manifold with boundary Σ × [0, 1], and Stokes’ theorem gives " " " d(F∗ω)= ω − ω. Σ×[0,1] Σ1 Σ1 In particular, if dω =0,then d(F∗ω)=F∗(dω)=0, and we deduce that ω = ω. Σ1 Σ0 This says that k-forms with exterior derivative zero give a well-defined functional on homotopy classes of compact oriented k-dimensional submani- folds of M. We know some examples of k-forms with exterior derivative zero, namely those of the form ω = dη for some (k − 1)-form η. But Stokes’ theorem then gives that Σ ω = Σ dη =0,sointhese cases the functional we defined on homotopy classes of submanifolds is trivial. 138 Lecture 15. de Rham cohomology This leads us to consider the space of ‘non-trivial’ functionals on homotopy classes of submanifolds: Each of these is defined by a k-form ω with exterior derivative zero, but is unchanged if we add the exterior derivative of an arbitrary (k − 1)-form to ω. We call a k-form with exterior derivative zero a k-cocycle, and a k-form which is an exterior derivative of a form is called a k-coboundary. The space of k-cocycles on M is a vector space, denoted Zk(M), and the space of k- coboundaries is then dΩk−1(M), which is contained in Zk(M). 15.2 Cohomology groups and Betti numbers We define the k-th de Rham cohomology group of M, denoted Hk(M), to be Zk(M) Hk(M)= . dΩk−1(M) Thus an element of Hk(M)isdefined by any k-cocycle ω, but is unchanged by changing ω to ω + dη for any (k − 1)-form η, which agrees with the notion we produced before of a ‘nontrivial’ functional on homotopy classes of submanifolds. An element of Hk(M)iscalled a cohomology class, and the cohomology class containing a k-cocycle ω is denoted [ω]. Thus [ω]={ω + dη : η ∈ Ωk−1(M)}. Since the exterior derivative and Stokes’ theorem do not depend in any wayonthe presence of a Riemannian metric on M, the cohomology groups of M depend only on the differentiable structure on M.Itturns out that they in fact depend only on the topological structure of M, and not on the differentiable structure at all — any two homeomorphic manifolds have the same cohomology groups5. The groups Hk(M) are therefore topological invariants, which can be used to distinguish manifolds from each other: If two manifolds have different cohomology groups, they cannot be homeomorphic (let alone diffeomorphic). The k-the cohomology group Hk(M)isareal vector space. The dimension of this vector space is called the kth Betti number of M, and denoted bk(M). 5 The de Rham theorem states that the de Rham cohomology groups are isomorphic to the singular or Cechˇ cohomology groups with real coefficients, and these are defined in purely topological terms. It is also a consequence of this theorem that the cohomology groups are finite dimensional. 15.4 The group H1(M) 139 15.3 The group H0(M) The group H0(M)isrelatively easy to understand: The space Z0(M)isjust the space of functions on M with derivative zero, which is the space of locally constant functions. We interpret Ω−1 as the trivial vector space. Therefore H0(M) Z0(M)=RN where N is the number of connected components of M. Thefore b0(M)isequal to the number of connected components of M. 15.4 The group H1(M) 1 The group H (M)isclosely related to the fundamental group π1(M). We will examine some aspects of this relationship: Proposition 15.4.1 Suppose M is connected. If ω ∈ Z1(M) and [ω] =0 in H1(M), then there exists a smooth curve γ : S1 → M such that " ω(γ):= γ∗ω =0 . S1 Proof. We will prove that if ω(γ)=0for every smooth map γ : S1 → M, then [ω]=0inH1(M). On each connected component of M choose a ‘base point’ x0.Wedefine a ∞ 0 function f ∈ C (M) Ω (M)bysetting f(x0)=0and extending to other points of M according to " f(x)= γ∗ω [0,1] for any γ :[0, 1] → M with γ(0) = x0 and γ(1) = x. This is well-defined, since if γ1 and γ2 are two such curves, then the curve γ1#(−γ2) obtained by concatenating γ1 and −γ2 (i.e. γ2 with orientation reversed) gives a map from S1 to M,sobyassumption " " " − ∗ ∗ − ∗ 0= (γ1#( γ2)) ω = γ1 ω γ2 ω, S1 S1 S1 and so the value of f(x)isindependent of the choice of γ. Finally, df = ω, since in a chart ϕ about x, " 1+t d ∗ df (∂i)=∂if = γ ω = ω(∂i) dt 0 −1 where γ(1 + t)=ϕ (ϕ(x)+tei). This shows that ω = df ,sothat [ω]=0inH1(M). 140 Lecture 15. de Rham cohomology Corollary 15.4.2 If M has finite fundamental group then H1(M)=0.In particular if M is simply connected, then H1(M)=0. Proof. Let ω ∈ Ω1(M) with dω =0.Then for any closed loop γ : S1 → M,we have [γ]n =0inπ1(M) for some integer n. Therefore we have a homotopy F : S1 × [0, 1] → M from γ#γ...#γ to the constant loop c, and Stokes’ theorem gives " " " 0= F ∗dω = (γ# ...#γ)∗ω − c∗ω = nω(γ). S1×[0,1] S1 S1 Since this is true for all closed loops γ, Proposition 15.4.1 applies to show [ω]=0inH1(M), and so H1(M)=0. The same argument tells us something more: In fact H1(M)isasubspace of the dual space of the vector space G ⊗ R, where G is the abelianisation of π1(M), which is the abelian group given by taking π1(M) and imposing the −1 −1 extra relations aba b =1for all elements a and b.Inparticular, b1(M) is no greater than the smallest number of generators of π1(M). In fact it turns out (at least for compact manifolds) that H1(M)isisomorphic to the torsion-free part of the abelianisation of π1(M), as described above. We will not prove this here. 15.5 Homotopy invariance In this section we will prove a remarkable topological invariance property of cohomology groups: They do not change when the space is continuously deformed. More precisely, suppose M and N are two manifolds, and F is a smooth map from M to N. Then the pullback of forms induces a homomorphism of cohomology groups: If ω ∈ Ωk(N)isacocycle, then so is F ∗ω ∈ Ωk(M), since d(F ∗ω)=F ∗(dω). Also, is ω = dη then F ∗ω = F ∗dη = d(F ∗η), so this map is well-defined on cohomology. Proposition 15.5.1 Let F : M × [0, 1] → N be a smooth map, and set ∈ ∗ ft(x)=F (x, t) for each t [0, 1]. Then ft is independent of t. Proof. Let ω ∈ Ωk(N)beacocycle. Then we can write ∗ F ω = ω0 + dt ∧ ω1 ∈ k ∈ k−1 ∗ where ω0 Ω (M) and ω1 Ω (M) for each t. Then ft ω = ω0 for each t. Since F ∗ω is a cocycle, we have ∂ω 0=dF ∗ω = dt ∧ 0 − d ω + ... ∂t M 1 15.6 The Poincar´e Lemma 141 and therefore " " " 1 1 1 ∗ − ∗ − ∂ω0 f1 ω f0 ω = ω0(1) ω0(0) = dt = dM ω1dt = dM ω1dt. 0 ∂t 0 0 ∗ ∗ Therefore f1 ω and f0 ω represent the same cohomology class. Corollary 15.5.2 If the smooth map f : M → N is a homotopy equivalence (that is, there exists a continuous map g : N → M such that f ◦ g and g ◦ f are both homotopic to the identity) then f ∗ is an isomorphism. 15.6 The Poincar´e Lemma We will compute the cohomology groups for a simple example: A subset B in Rn is star-shaped (with respect to the origin) if for every point y ∈ B, the interval {ty : t ∈ [0, 1]} is in B. Proposition 15.6.1 (The Poincar´eLemma). Let B beastar-shaped open set in Rn. Then Hk(B)={0} for k =1,...,n. Proof. We need to show that for k>0every k-cocycle is a k-coboundary. In other words, given a k-form ω on B with dω =0,weneed to find a (k − 1)-form η such that ω = dη.Wewill do this with k replaced by k +1. i0 ∧ ∧ ik ∈ j Write ω = ωi0...ik dx ... dx .Fory B write y = y ej and define " 1 j k ηi1...ik = y t (ωty)ji1...ik dt.