LECTURE 21: COMPACTLY SUPPORTED DE RHAM COHOMOLOGY

1. Compactly Supported De Rham cohomology For any ω ∈ M, we can define the support of ω to be

supp(ω) = {p ∈ M | ωp 6= 0}. As usual, we say ω is compactly supported if supp(ω) is compact in M. We denote k k Ωc (M) = {ω ∈ Ω (M) | ω is compactly supported}, k k Zc (M) = {ω ∈ Ωc (M) | dω = 0}, k k k−1 Bc (M) = {ω ∈ Ωc (M) | ω = dη for some η ∈ Ωc (M)}. k k Again Bc (M) is additive subgroup of Zc (M), and we call k k Zc (M) Hc (M) = k Bc (M) the k-th de Rham cohomology group with compact supports of M. n 0 n Example. Consider M = R (n ≥ 1). We have Hc (R ) = 0 since a closed 0-form must be a constant function, but there are no compactly-supported non-trivial constant functions on Rn. 1 n n n If n > 1, we also have Hc (R ) = 0. To see this we will identify R with S − {p}. Then any 1 n n n ω ∈ Zc (R ) defines a closed 1-form, still denoted by ω, on S which is supported in S − U for some neighborhood U of p. Since H1(Sn) = 0, ω is exact, i.e. there exists η ∈ Ω0(Sn) = C∞(Sn) so that ω = dη. Moreover, the fact dη = 0 on U implies that η equals some constant c on U. It 0 n 0 n follows that if we takeη ˜ = η − c, thenη ˜ ∈ Ωc (S − {p}) = Ωc (R ) and dη˜ = ω. k n More generally, for any k < n we have Hc (R ) = 0. The argument for k > 1 is a modification n n k n of the argument for k = 1 above: Again one identify R with S − {p}, and for ω ∈ Zc (R ), viewed as an element in Zk(Sn) that is supported in some Sn − U, one can find η ∈ Ωk−1(Sn) such that ω = dη. The neighborhood U of p can be taken as a contractible one. Then the fact dη = ω = 0 in U implies that η is exact in U, i.e. one can find a µ ∈ Ωk−2(U) such that η = dµ. Now one pick a bump function ρ on Sn which vanishes on Sn − U and equals 1 near p. Then η˜ = η − d(ρµ) is a well-defined (k − 1)-form on Sn that vanishes near p, i.e. defined a compactly supported (k − 1)-form on Rn, such that dη˜ = dη = ω. Remarks. It is easy to see k k (1) If M is compact, then Hc (M) = H (M). (2) As before, we can define a cup product k l k+l ∪ : Hc (M) × Hc(M) → Hc (M), (ω, η) 7→ [ω ∧ η] ∗ n k which makes Hc (M) = ⊕k=1Hc (M) a graded ring.

1 2 LECTURE 21: COMPACTLY SUPPORTED DE RHAM COHOMOLOGY

(3) For k = 0, we have 0 m Hc (M) ' R , where m is the number of compact connected components of M. In particular, if M is 0 noncompact and connected, then Hc (M) = 0. k (4) Hc (M) are no longer a homotopy invariant. For example, as we have seen above, although n 0 n 0 R is homotopy equivalent to the one point set {p}, Hc (R ) 6= Hc ({p}).

Now let ϕ : M → N be a smooth map. Unfortunately, in general we cannot define the pull- back map ϕ∗ between compactly supported de Rham cohomology groups as we did for the regular de Rham cohomology groups, since the pull-back of a compactly supported differential form need not have compact support. However, there are still two cases we can define a pull-back map: (1) If ϕ : M → N is proper, then the pull-back ϕ∗ω of a compactly supported differential form k ∗ k ω ∈ Ωc (N) is a compactly supported differential form on M. So the map ϕ : Hc (N) → k Hc (M) is still well-defined. In this case one can prove

Theorem 1.1. If f0, f1 : M → N are proper maps that are properly homotopic, then the ∗ ∗ k k induced maps f1 = f2 : Hc (N) → Hc (M). Note that any diffeomorphism is proper. So in particular we see

k k Corollary 1.2. If M is diffeomorphic to N, then Hc (M) = Hc (N). (2) If ι : U → M is an open submanifold, then we can define a “push-forward” map

k k ι∗ :Ωc (U) → Ωc (M) which extend a compactly supported k-form on U to a compactly supported k-form on M by “zero-extension”. It is easy to see dι∗ = ι∗d, so ι∗ descend to a linear map

k k ι∗ : Hc (U) → Hc (M). One also has the MV sequence for de Rham cohomology with compact supports:

Theorem 1.3. Suppose U, V are open subsets in M so that M = U ∪ V . Then there k k+1 exists linear maps δk : Hc (M) → Hc (U ∩ V ) so that the following sequence is exact

δk−1 k αk k k βk k δk k+1 αk+1 ··· −→ Hc (U ∩ V ) −→ Hc (U) ⊕ Hc (V ) −→ Hc (M) −→ Hc (U ∩ V ) −→ · · · ,

where αk and βk are given by

αk([ω]) = ((1)∗[ω], −(2)∗[ω]) and

βk([ω1], [ω2]) = (ι1)∗[ω1] + (ι2)∗[ω2]. LECTURE 21: COMPACTLY SUPPORTED DE RHAM COHOMOLOGY 3

2. The de Rham cohomology groups of top degree We start by the following example: 1 Example. Let’s try to find Hc (R). To do so we consider the integration map Z Z 1 1 : Zc (R) = Ωc (R) → R, ω 7→ ω. R R 1 This map is clearly linear, surjective and vanishes on Bc (R), so it induces a surjective linear map Z 1 : Hc (R) → R. R Moreover, if R f(t)dt = 0, where f ∈ C∞( ), then the function g(t) = R t f(τ)dτ is smooth R c R −∞ 1 and compactly supported and dg = f(t)dt. In other words, f(t)dt ∈ Bc (R), i.e. [f(t)dt] = 0. It follows that R is an isomorphism between H1( ) and . So H1( ) ' . R c R R c R R Now let M be an n-dimensional connected oriented manifold. Consider the map Z Z n :Ωc (M) → R, ω 7→ ω. M M k n−1 Note that by the Stokes’ lemma, if ω ∈ Bc (M), i.e. ω = dη for some η ∈ Ωc (M), then Z Z Z Z ω = dη = dη = η = 0. M M supp(η) ∂(supp(η)) R So M induces a linear map Z Z n : Hc (M) → R, [ω] 7→ ω. M M This map is obviously surjective: One just fix a volume form ω, then for any c, one can find a smooth function f that is compactly supported in a coordinate chart U such that R fω = c. The surjectivity of this map has an important consequence: n n R Proposition 2.1. If ω ∈ Ω (S ) and Sn ω = 0, then ω is exact. R n n n n Proof. We have seen that the map is linear and surjective. But Hc (S ) = H (S ) ' R. So R M R M must be a linear isometry. In other words, if Sn ω = 0, then [ω] = 0, i.e. ω is exact.  Using this, we can prove the following Poincar´elemma for compactly supported de Rham cohomology:  , k = n, Theorem 2.2 (Poincar´elemma). Hk( n) = R c R 0, k 6= n. Proof. The only cases that we have not proved is the case n = k ≥ 2. In other words we need to show that the surjective linear map Z n n : Hc (R ) → R. Rn 4 LECTURE 21: COMPACTLY SUPPORTED DE RHAM COHOMOLOGY

R n n is an isomorphism. More precisely, we need to prove that if n ω = 0 for some ω ∈ Ωc (R ), then n n n n R n n ω ∈ Bc (R ). As before we will identify R = S − {p}. We denote by ι : R → S the canonical n n n n embedding. Then for any ω ∈ Ωc (R ), we have ι∗ω ∈ Ω (S ). The condition Z Z ι∗ω = ω = 0 Sn Rn n−1 n implies ι∗ω = dη for some η ∈ Ω (S ). The rest of proof is identically the same as before: We take an open contractible neighborhood U of p in Sn on which ω vanishes. Then we adjust η to η˜ = η − d(ρµ), where ρ is a bump function and µ ∈ Ωk−2(U) is such that dµ = η on U. Thenη ˜ is n a compactly supported (n − 1)-form on R and dη˜ = ω.  As an immediate corollary, we get k Corollary 2.3. If M admits a finite good cover, then dim Hc (M) < ∞ for all k. Proof. Use Mayer-Vertoris sequence and induction and the number of open sets in a good cover. The same as the proof for the ordinary de Rham cohomology that we did last time.  More generally, we have R n Theorem 2.4. For any n-dimensional connected oriented manifold M, the map M : Hc (M) → R n is an isomorphism. In particular, Hc (M) ' R. R R Proof. Since M is linear and surjective, it remains to prove M is injective, i.e. Z n k−1 (1) ω ∈ Ωc (M) such that ω = 0 =⇒ ω = dµ for some µ ∈ Ωc (M). M To prove this, we can use induction on the number of open sets that is needed to cover the support of ω by a good cover. If we can cover the support of ω by a good cover which contains only one n chart, then the Poincar´elemma implies (1). Now suppose (1) holds for all elements in Ωc (M) n whose support can be covered by k − 1 “good charts”, and suppose ω ∈ Ωc (M) satisfies the fact that supp(ω) admits a good cover {U1, ··· ,Uk}. We let U = U1 ∪ · · · ∪ Uk−1 and V = Uk. We pick a partition of unity {ρU , ρV } of U ∪ V subordinate to the cover {U, V }, and let ωU = ρU ω, ωV = ρV ω. We pick an n-form ω0 supported in U ∩ V so that Z Z ω0 = ωU . M M R Then ωU −ω0 is supported in U which admits a good cover of k−1 good charts, and M ωU −ω0 = 0. So by induction hypothesis, ωU − ω0 = dηU k−1 R R R for some ηU ∈ Ωc (M). Similarly M ωV + ω0 = − M ωU + M ω0 = 0 implies

ωV + ω0 = dηV k−1 for some ηV ∈ Ωc (M). It follows that

ω = ωU + ωV = d(ηU + ηV ), k−1 where ηU + ηV ∈ Ωc (M).  LECTURE 21: COMPACTLY SUPPORTED DE RHAM COHOMOLOGY 5

As a corollary, we get n Corollary 2.5. Let M be compact, connected, orientable and dim M = n. Then HdR(M) ' R. For oriented non-compact manifolds, one has Theorem 2.6. For any non-compact connected orientable manifold M of dimension n, we have n HdR(M) = 0 Proof. In lecture 3 we showed that there exists an positive exhaustion function f for M. In other words, f is a smooth function on M so that for any c, the sub-level set f −1((−∞, c]) is compact in M. By adding a constant, we may assume inf f = 0. Since M is connected and non-compact, −1 we must have f(M) = [0, ∞). Now for each integer k we let Vk = f ((k − 2, k)). Then {Vk} is a locally finite open covering of M. Let {ρk} be a partition of unity subordinate to this covering. n n R For each k, we choose some ηk ∈ Ωc (Vk ∩ Vk+1) ⊂ Ωc (M) so that M ηk = 1. n n R n For any ω ∈ Ω (M), we let ωk = ρkω ∈ Ω (Vk). Let c1 = ω1. Then ω1 − c1η1 ∈ Ω (V1) and c V1 c Z (ω1 − c1η1) = 0. V1 n−1 So by theorem 2.4, one can find µ1 ∈ Ωc (V1) so that

dµ1 = ω1 − c1η1. R Next we choose c2 ∈ so that (ω2 + c1η1 − c2η2) = 0, and conclude that there exists µ2 ∈ R V2 n−1 Ωc (V2) such that dµ2 = ω2 + c1η1 − c2η2. n−1 Continuing this process, we can find a sequence ck ∈ R and µk ∈ ωc (Vk) so that

dµk = ωk + ck−1ηk−1 − ckηk. P n−1 Let µ = µk. Note that this is a locally finite sum, and thus defines an element in Ω (M). Moreover, by construction we have X X dµ = d µk = ωk = ω.  n Remark. If M is a non-orientable connected smooth manifold of dimension n, then HdR(M) = 0 n and Hc (M) = 0. For a proof, c.f. Lee, page 456-457.