LECTURE 21: COMPACTLY SUPPORTED DE RHAM COHOMOLOGY 1. Compactly Supported De Rham cohomology For any ! 2 M, we can define the support of ! to be supp(!) = fp 2 M j !p 6= 0g: As usual, we say ! is compactly supported if supp(!) is compact in M. We denote k k Ωc (M) = f! 2 Ω (M) j ! is compactly supportedg; k k Zc (M) = f! 2 Ωc (M) j d! = 0g; k k k−1 Bc (M) = f! 2 Ωc (M) j ! = dη for some η 2 Ωc (M)g: 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 − fpg. Then any 1 n n n ! 2 Zc (R ) defines a closed 1-form, still denoted by !, on S which is supported in S − U for some neighborhood U of p. Since H1(Sn) = 0, ! is exact, i.e. there exists η 2 Ω0(Sn) = C1(Sn) so that ! = 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 ~ 2 Ωc (S − fpg) = Ωc (R ) and dη~ = !. 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 − fpg, and for ! 2 Zc (R ), viewed as an element in Zk(Sn) that is supported in some Sn − U, one can find η 2 Ωk−1(Sn) such that ! = dη. The neighborhood U of p can be taken as a contractible one. Then the fact dη = ! = 0 in U implies that η is exact in U, i.e. one can find a µ 2 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 = !. 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); (!; η) 7! [! ^ 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 fpg, Hc (R ) 6= Hc (fpg). Now let ' : M ! N be a smooth map. Unfortunately, in general we cannot define the pull- back map '∗ 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 ' : M ! N is proper, then the pull-back '∗! of a compactly supported differential form k ∗ k ! 2 Ωc (N) is a compactly supported differential form on M. So the map ' : 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([!]) = ((1)∗[!]; −(2)∗[!]) and βk([!1]; [!2]) = (ι1)∗[!1] + (ι2)∗[!2]: 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;! 7! !: 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 2 C1( ), 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 2 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;! 7! !: M M k n−1 Note that by the Stokes' lemma, if ! 2 Bc (M), i.e. ! = dη for some η 2 Ωc (M), then Z Z Z Z ! = dη = dη = η = 0: M M supp(η) @(supp(η)) R So M induces a linear map Z Z n : Hc (M) ! R; [!] 7! !: M M This map is obviously surjective: One just fix a volume form !, then for any c, one can find a smooth function f that is compactly supported in a coordinate chart U such that R f! = c. The surjectivity of this map has an important consequence: n n R Proposition 2.1. If ! 2 Ω (S ) and Sn ! = 0, then ! 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 ! = 0, then [!] = 0, i.e. ! 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 ! = 0 for some ! 2 Ωc (R ), then n n n n R n n ! 2 Bc (R ). As before we will identify R = S − fpg. We denote by ι : R ! S the canonical n n n n embedding. Then for any ! 2 Ωc (R ), we have ι∗! 2 Ω (S ). The condition Z Z ι∗! = ! = 0 Sn Rn n−1 n implies ι∗! = dη for some η 2 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 ! vanishes. Then we adjust η to η~ = η − d(ρµ), where ρ is a bump function and µ 2 Ωk−2(U) is such that dµ = η on U. Thenη ~ is n a compactly supported (n − 1)-form on R and dη~ = !. As an immediate corollary, we get k Corollary 2.3. If M admits a finite good cover, then dim Hc (M) < 1 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) ! 2 Ωc (M) such that ! = 0 =) ! = dµ for some µ 2 Ωc (M): M To prove this, we can use induction on the number of open sets that is needed to cover the support of ! by a good cover. If we can cover the support of ! by a good cover which contains only one n chart, then the Poincar´elemma implies (1).