De Rham Cohomology

Home , De Rham cohomology, Real projective space

SUPPLEMENTARY NOTES FOR MAT4520 — DE RHAM COHOMOLOGY

BJØRN JAHREN

(Version: May 26, 2011)

1. The Mayer–Vietoris sequence for H∗. De Rham cohomology of spheres.

Let U and V be two open subsets of a manifold M. The diagram

jU U ∩ V / U

jV iU iV V / U ∪ V gives rise to two an exact sequence of de Rham complexes:

∗ ∗ ∗ ∗ (iU ,iV ) jU −jV 0 → Ω∗(U ∪ V ) / Ω∗(U) ⊕ Ω∗(V ) / Ω∗(U ∩ V ) → 0 ∗ ∗ The only non–obvious part is the surjectivity of jU − jV . Let {φU , φV } be a partition of unity subordinate to the covering {U, V }, ie Supp φU ⊂ U, k Supp φV ⊂ V and φU (p) + φV (p) = 1 for all p ∈ U ∪ V . If ω ∈ Ω (U ∩ V ), φU ω can be extended by 0 to V and φV ω can be extended by 0 to U, and ∗ ∗ jU (φV ω) − jV (−φV ω) = ω. As usual, the short exact sequence of chain complexes now gives rise to a long exact sequence of cohomology vector spaces · · · → Hk(U ∪ V ) → Hk(U) ⊕ Hk(V ) → Hk(U ∩ V ) → Hk+1(U ∪ V ) → · · · This is the Mayer–Vietoris exact sequence for de Rham cohomology. (Spi- vak, Theorem 11.3, p. 424).

Remark 1.1. Note that this also makes sense if U and V are disjoint, if we deﬁne Ωk(N) = 0 for all k if N = ∅. In this case the sequence reduces to the obvious isomorphisms Hk(U) ⊕ Hk(V ) = Hk(U ∪ V ).

As an application we shall use the Mayer–Vietoris exact sequence to com- n pute all cohomology of the spheres S . For n > 1 the result is: ( if k = 0 or k = n (1) Hk(Sn) =∼ R 0 otherwise 0 (The case n = 0 (two points) is trivial: again there are two R s, but they both lie in degree 0.) 1 2 BJØRN JAHREN

n n n Let U± = S −{(0,..., ±1)} ⊂ S . Then {U+,U−} is an open cover of S n n−1 of manifolds diﬀeomorphic to R , and U+ ∩ U−is diﬀeomorphic to S × k ∼ R. By the homotopy invariance of cohomology, we then have H (U) = k k ∼ k n−1 0 n ∼ H (point) and H (U+ ∩ U−) = H (S ). We also recall that H (S ) = R for n > 0, and Hk(Sn) =∼ 0 for k > n. We will prove (1) by induction on n, starting with n = 1. Then the Mayer–Vietoris sequence reduces to 2 1 1 0 → R → R ⊕ R → R → H (S ) → 0 1 1 ∼ This can only be exact if H (S ) = R. Now assume we have proved (1) for n − 1, n > 2. Then Mayer–Vietoris for Sn becomes

1 n 0 → R → R ⊕ R → R → H (S ) → 0 → · · · → 0 → Hk−1(Sn−1) → Hk(Sn) → 0 ··· Hence we see that H1(Sn) =∼ 0 and Hk(Sn) =∼ Hk−1(Sn−1) for k > 1. The induction step follows from this. Note that by homotopy invariance it now also follows that we have com- n n−1 puted the cohomology of R − {0} ≈ S × R. We give two classical applications of this result. The ﬁrst is

Theorem 1.2. (Brouwer’s ﬁxed point theorem.) Let f : Bn → Bn be a smooth map. Then f has a ﬁxed point, i. e. there is a point x ∈ Bn such that f(x) = x. Proof. Suppose not, i. e. f(x) 6= x for all x. Then there is a unique line `x through the two points x and f(x), and `x must intersect the boundary sphere Sn−1 of Bn in two points. Let g(x) be the intersection point closer to x than to f(x). To see that g is a smooth map Bn → Sn−1, observe that g(x) = x + t(x − f(x)) for a unique t > 0, which can be found by solving the quadratic equation (in t) |x + t(x − f(x)|2 = 1. ι g If x ∈ Sn−1 we get g(x) = x, hence the composition Sn−1 ⊂ Bn → Sn−1 is the identity map. Applying Hn−1 we get that the composition

∗ ∼ n−1 n−1 g n−1 n ι∗ n−1 n−1 ∼ R = H (S ) → H (B ) → H (S ) = R n−1 n is the identity map. But since H (B ) = 0, this is impossible. Before stating the next result we make an important observation concern- ing the antipodal map A : Sn−1 → Sn−1 deﬁned by A(x) = −x. Claim: The induced map A∗ : Hn−1(Sn−1) → Hn−1(Sn−1) is multiplication by (−1)n. To see this, recall that a generator of Hn−1(Sn−1) is represented by the  p   v1  0 0   0 form σ deﬁned by σp(v1, . . . , vn−1) = det  . , where p and the vis  .  vn−1 MAT4520 (May 26, 2011) 3

n ∗ are considered as vectors in R . The image of this generator under A is then represented by the form A∗σ0 given by  −p   −v1  (A∗σ0) (v , . . . , v ) = σ0 (A v ,...,A v ) = det   p 1 n−1 A(p) ∗ 1 ∗ n  .   .  −vn−1 n 0 = (−1) σp(v1, . . . , vn−1) The claim follows. We can now prove

Theorem 1.3. The sphere Sn admits a vector ﬁeld v such that v(p) 6= 0 for every p ∈ Sn if and only if n is odd. (Cf. Spivak, p. 276)

Proof. Assume Sn does admit such a vector field v. By normalizing v at every point, we may assume that |v(p)| = 1. Hence we may think of v as a function v : Sn → Sn such that p and v(p) are orthogonal for every p. Now define a homotopy F : Sn × I → Sn by F (p, t) = cos(πt) p + sin(πt) v(p) . Then F (p, 0) = p and F (p, 1) = −p, which means that F is a smooth homotopy between the identity map on Sn and A. Since homotopic maps induce the same map on cohomology, it follows that 1 = (−1)n+1. Hence n must be odd. Conversely, suppose n = 2k − 1. Then we can define a vector field v by the formula

v(p1, . . . , p2k) = (−p2, p1, −p4, p3,..., −p2k, p2k−1) n Clearly, v(p) 6= 0 for all p ∈ S .

2. More calculations: Projective spaces.

We start with the real projective spaces RP n, which we think of as obtained from Sn by identifying antipodal points. Then the quotient map π : Sn → RP n is a surjective local diﬀeomorphism, and such that π(p) = π(q) if and only if p = q or p = −q = A(q). Now we make use of the fact that A∗ :Ωk(Sn) → Ωk(Sn) has order two — i. e. (A∗)2 = (A2)∗ = I; the identity homomorphism. Here is an easy exercise in linear algebra: Assume P : V → V is a linear map such that P 2 = I. Then the vector space V splits as a direct sum of the +1 and −1 eigenspaces V+ and V− of 1 1 P . (Use the decomposition v = (v + P v) + (v − P v).) 2 2 ∗ k n k n k n Applying this to A we see that Ω (S ) splits as Ω (S )+ ⊕ Ω (S )−, and it is also easy to see that the exterior derivative d :Ωk(Sn) → Ωk+1(Sn) respects this decomposition. 4 BJØRN JAHREN

Taking cohomology, we get a similar decomposition of Hk(Sn), and our n n ∼ n+1 calculations show that H (S ) = R is equal to the (−1) –eigenspace of A∗ : Hn(Sn) → Hn(Sn). Now consider π∗ :Ωk(RP n) → Ωk(Sn). Since πA = π, we have A∗π∗ = ∗ ∗ k n k n π , so π (Ω (RP )) ⊆ Ω (S )+. ∗ k n k n Claim: π :Ω (RP ) → Ω (S )+ is an isomorphism. ( if k = 0 or k = n if n is odd Corollary. Hk(RP n) =∼ R 0 otherwise It remains to verify the Claim. Let η ∈ Ωk(Sn) be a k–form. Recall that π∗ is deﬁned by ∗ (π ω)p(v1,... ) = ωπ(p)((π∗)p v1,... ) . Hence π∗ω = η if and only if

(2) ωq(w1,... ) = ηp(v1,... ) n n for every p ∈ S such that π(p) = q and vi ∈ Sp such that (π∗)p vi = wi. But for every q, w1,... there are exactly two choices of p, v1,... ; if −1 −1 p, (π∗p) w1,... is one, then A(p),A∗(π∗p) w1,... is the other. Hence formula (2) determines ω uniquely if and only if ∗ ηp(v1,... ) = ηA(p)(A∗v1,... ) = (A η)p(v1,... ) k n For all p, v1,... — i. e. if and only if η ∈ Ω+(S ). We now turn to the complex projective spaces CP n. We will compute the de Rham cohomology of CP n by induction on n using the Mayer–Vietoris sequence. n n+1 ∗ ∗ Recall that CP = (C − {0})/C , where the group C = C − {0} acts n+1 on C − {0} by ζ · (z0, . . . , zn) = (ζz0, . . . , ζzn). The orbit (equivalence class) of (z0, . . . , zn) is denoted [z0 : ··· : zn](homogeneous coordinates.) Note that we can identify CP n−1 with the set of points in CP n having the last homogeneous coordinate equal to zero. Let U = CP n − CP n−1 — the open subset of points with the last homogeneous coordinate nonzero. Every such point has a unique representative n of the form (z0, . . . , zn−1, 1), and it follows that U is diﬀeomorphic to C . Let V ⊂ CP n be the complement of the point [0 : ··· : 0 : 1]. Clearly n n 2n U ∪ V = CP and U ∩ V ≈ C − {0} ≈ R − {0}, which has the same cohomology as S2n−1. A point with homogeneous coordinates [z0 : ··· : zn] lies in V if and only if (z0, . . . , zn−1) 6= (0,..., 0). Therefore the formula

g([z0 : ··· : zn]) = [z0 : ··· : zn−1 : 0] deﬁnes a retraction of V to CP n−1, i. e. a map g : V → CP n−1 such that n−1 g ◦ j = idCP n−1 , where j is the inclusion CP ⊂ V . Moreover,

G([z0 : ··· : zn], t) = [z0 : ··· : tzn] deﬁnes a homotopy G : V ×I → V such that G(p, 1) = p and G(p, 0) = jg(p). It follows that g and j induce inverse isomorphisms in cohomology, yielding isomorphisms Hk(V ) =∼ Hk(CP n−1) for all k. MAT4520 (May 26, 2011) 5

We are now ready to prove ( for k even and 0 k 2n Hk(CP n) =∼ R 6 6 0 otherwise Moreover, j∗ : Hk(CP n) → Hk(CP n−1) is an isomorphism for k < 2n. Proof. We start with n = 1. CP 1 ≈ S2, and the result is a special case of the calculation for spheres. Now assume we have proved the result for n − 1. We know already that 0 n ∼ k ∼ H (CP ) = R, so let k > 1. Since then H (U) = 0, the Mayer–Vietoris sequence reduces to a sequence isomorphic to j∗ · · · → Hk−1(S2n−1) → Hk(CP n) → Hk(CP n−1) → Hk(S2n−1) → · · · For 1 < k < 2n we see immediately that j∗ is an isomorphism. (Note that H2n−1(CP n−1) =∼ 0.) This is also true for k = 1, since e. g. the map H0(U) → H0(U ∩V ) is surjective. Finally, for k = 2n we get an isomorphism 2n−1 2n−1 ∼ 2n n H (S )=H (CP ).

3. Cohomology with compact support

If the diﬀerential form ω has compact support, then dω also has compact support. It follows that the de Rham complex (Ω∗(M), d) has a subcomplex

k d k+1 · · · → Ωc (M) → Ωc → · · · , k k where Ωc (M) = {ω ∈ Ω (M) | Supp ω compact}. (Spivak uses the notation k Cc (M) for this. Note that Supp (ω + η) ⊆ Supp ω ∪ Supp η and Supp rω = k k Supp ω if r 6= 0, hence Ωc (M) is also a vector subspace of Ω (M).) Definition 3.1. The de Rham cohomology groups with compact support are defined as the cohomology groups of this complex, i. e. k k k+1 k−1 k Hc (M) = ker(d :Ωc (M) → Ωc (M))/Im (d :Ωc (M) → Ωc (M)) If f : M → N is a smooth map which is proper, i. e. such f −1(K) is compact for every compact K ⊂ N, then it induces homomorphisms k k k k ∗ Ωc (N) → Ωc (M) and Hc (N) → Hc (M), both denoted f as before. But a new and more interesting ingredient is the map induced by inclusion k j : U ⊂ V of open subsets of a manifold M: given ω ∈ Ωc (U) we can extend k by 0 to the rest of V and obtain a form j∗ω ∈ Ωc (V ). (This is denoted j0(ω) in Spivak (p. 430).) It commutes with the exterior differentials, i. e. dj∗ = j∗d. Hence it induces a homomorphism k k j∗ : Hc (U) → Hc (V ). ∗ It is easy to see that this makes Hc a covariant functor on open subsets of M and inclusions. 6 BJØRN JAHREN

Easy observations. 0 0 1 • Hc (M) = ker(Ωc (M) → Ωc (M)) is the set of locally constant func- tions with compact support. In follows that if M is connected and n 0 non–compact (e. g. M = R ), then Hc (M) = 0. k k • If M is compact, there is no diﬀerence between Hc (M) and H (M). k k In general there is only a forgetful map Hc (M) → H (M). Note that if j : U ⊂ V is as above, the diagram j k ∗ k Hc (U) / Hc (V )

Hk(U) o Hk(V ) j∗ n commutes. For example, it follows that if U ⊂ R is an open subset, k k then the map Hc (U) → H (U) is trivial for all k. • The usual proof of homotopy invariance works here, as well, provided the homotopy is proper as a map M × I → N. The crucial observation is that the operator I (Spivak, p. 224) preserves forms of compact support. R Lemma 3.2. If M is oriented and of dimension n, then integration M n induces a nontrivial homomorphism Hc (M) → R. n R Proof. We first need to prove that if ω ∈ Ωc (M) is exact, then M ω = 0. So, let ω = dη, where η has compact support. n Consider first the case M = R . (Cf. p. 268.) Then there is a ball Br of some radius r such that Supp (ω) ⊂ int Br, and Stokes’ theorem gives Z Z Z ω = dη = η = 0 n R Br ∂Br In the general case, cover the compact set Supp (ω) with finitely many n coordinate neighborhoods U1,...,Ul diffeomorphic to R , and use a partition of unity argument to write η as a sum η1 + ··· + ηl of forms such that ηi has compact support contained in Ui. Then Z Z Z ω = Σidηi = Σi dηi = 0 , M M Ui by the first case. To show that the homomorphism is nontrivial, choose an orientation n preserving diffeomorphism g from an open subset U of M to R , and let n f : R → [0, ∞) be a nonzero function with compact support. Let j : U ⊂ M be the inclusion map. Then Z Z ∗ 1 n 1 n j∗g (f dx ∧ · · · ∧ d x ) = f dx . . . d x > 0 n M R Remark. If j : U ⊂ V is an inclusion map of open sets in M with orientations R R n induced from M, we have V j∗ω = U ω for any ω ∈ Ωc (U). Here is our first nontrivial complete calculation: MAT4520 (May 26, 2011) 7

( k n ∼ R if k = n Hc (R ) = 0 otherwise

1 n+1 n+1 n Proof. For a ∈ (−1, 1) let Ua = {(p , . . . , p ) ∈ S | p < a} ⊂ S and 1 n+1 n+1 n Va = {(p , . . . , p ) ∈ S | p > a} ⊂ S . Both subsets are diffeomorphic n 0 n n n to R . We know that Hc (R ) = 0 and Hc (R ) 6= 0 , hence it suffices to prove that the map k k n k n j∗ : Hc (U0) → Hc (S ) = H (S ) n is injective for all k, where j is the inclusion map U0 ⊂ S . This is trivial for k = 0, so we assume k > 1. 0 Let ω be a closed k–form with compact support on U0 such that j∗ω = dη n 0 for some (k − 1)–form on S . η may not have support in U0, but we will arrange that by subtracting a suitable exact form dτ. Since ω has compact support contained in U0, we can find a b < 0 such 0 0 that dη = j∗ω vanishes on Vb. Therefore we can find a (k − 2)–form τ 0 0 n on Vb such that dτ = η there. Let φ : S → [0, 1] be a smooth function which is constant equal to 1 in a neighborhood of V0 and 0 on Ub. Define k−2 n 0 0 τ ∈ Ω (S ) to be equal to φτ on Vb and 0 on Ub. Then η = η − dτ has 0 support in U0 and satisfies dη = dη = ω.

∗ 4. Mayer–Vietoris for Hc

If U and V are open sybsets of M, the diagram

jU U ∩ V / U

jV iU iV V / U ∪ V now gives rise to another short exact sequence of de Rham complexes, namely

(j ,−j ) ∗ U ∗ V ∗ ∗ ∗ iU ∗+iV ∗ ∗ 0 → Ωc (U ∩ V ) / Ωc (U) ⊕ Ωc (V ) / Ωc (U ∪ V ) → 0 Again, the only non–obvious fact is the surjectivity of the last map, which now is iU ∗ + iV ∗. To verify this, let (φU , φV ) be a partition of unity as at the k k beginning of section 1. Then, if ω ∈ Ωc (U ∪V ), we see that φU ω ∈ Ω (U ∪V ) k has compact support contained in U, and φV ω ∈ Ω (U ∪ V ) has compact support contained in V , and ω = iU ∗(φU ω) + iV ∗(φV ω). From the short exact sequence above we get the long, exact Mayer– Vietoris sequence for cohomology with compact support: k k k k k+1 · · · → Hc (U ∩ V ) → Hc (U) ⊕ Hc (V ) → Hc (U ∪ V ) → Hc (U ∩ V ) → · · ·

As an illustration, let us apply it to the same open cover {U+,U−} of n k n the sphere S as we used to compute H (S ). Then U+ and U− are both 8 BJØRN JAHREN

n diﬀeomorphic to R , and the Mayer–Vietoris sequence reduces to 0 n 1 k 0 → H (S ) → Hc (U+ ∩ U−) → 0 → · · · → 0 → Hc (U+ ∩ U−) → 0 n n n n n · · · → 0 → Hc (U+ ∩ U−) → Hc (U+) ⊕ Hc (U−) → H (S ) → 0 n−1 But U+ ∩ U− ≈ S × R, so this means that ( k n−1 ∼ R if k = 1 or k = n Hc (S × R) = 0 otherwise Note that this is isomorphic to the cohomology of Sn−1, only shifted one degree.

5. Products and the Poincare´ duality map

It is an easy exercise to show that wedge product of forms induces a bilinear pairing on de Rham cohomology: (3) ∪ : Hk(M) × Hl(M) → Hk+l(M) (“Cup product”.) In fact, the formula d(α ∧ β) = dα ∧ β + (−1)kα ∧ dβ immediately implies that the product of two cocycles is a cocycle, and if dα = 0 and dβ = 0 we have (α + dγ) ∧ (β + dη) = α ∧ β + d(γ ∧ β + (−1)kα ∧ η + γ ∧ dη) . Moreover, since Supp (α ∧ β) ⊂ Supp α ∩ Supp β, the wedge product of two forms has compact support if at least one of them has compact support. Hence we also have a pairing k l k+l (4) ∪ : H (M) × Hc(M) → Hc (M) The following properties of the cup product are easy to prove: Lemma ∗ k (i) The product in (3) gives H (M) = ⊕kH (M) the structure of a graded ring. A smooth map f : M → N induces a ring homomorphism f ∗ : H∗(N) → H∗(M). ∗ k (ii) (4) gives Hc (M) = ⊕kHc (M) the structure of a graded module over the graded ring H∗(M). (iii) Let j : U ⊂ V is the inclusion of open sets in M. If x ∈ Hk(V ) and l ∗ y ∈ Hc(U), then j∗(j x ∪ y) = x ∪ j∗y. ∗ ∗ ∗ Remark. Composition with j makes Hc (U) a module also over H (V ), and ∗ property (iii) says that j∗ is a homomorphism of H (V )–modules. Now assume that M is an oriented n–manifold. Then integration of n– R n forms defines a homomorphism U : Hc (U) → R for every open U ⊆ M, and we can define a new pairing k n−k H (U) × Hc (U) → R R n−k induced by (α, β) 7→ U α ∧ β. For each α this defines a map Hc (U) → R, and bilinearity means that we get a linear map k n−k ∗ n−k µU : H (U) → Hc (U) = HomR(Hc (U), R) MAT4520 (May 26, 2011) 9

n Example. If U = R , these groups are nontrivial only for k = 0. Then they 0 n n n ∗ n are both isomorphic to R, and µR : H (R ) → Hc (R ) is an isomorphism. 0 n To see this, it is enough to verify that it is non–zero. But H (R ) is generated by the class of the constant map f(p) = 1, hence the image under R µ n is represented by the map β 7→ n β, which we know takes non–zero R R values. Our main result is a vast generalization of this:

Theorem (“Poincar´eduality”) µM is an isomorphism for every oriented manifold M and every k. The proof will be based on a comparison of Mayer-Vietoris sequences: Given a diagram

jU U ∩ V / U

jV iU iV V / U ∪ V of open subsets of M, we consider the Mayer–Vietoris sequences for both H∗ ∗ and Hc . If we dualize all vector spaces and maps in the compact support sequence we get a new long exact sequence, and since the spaces U, V, U ∩V and U ∪ V are all n–manifolds with orientations coming from M, we can draw the following diagram:

(5) δ / Hk(U ∪ V ) / Hk(U) ⊕ Hk(V ) / Hk(U ∩ V ) / Hk+1(U ∪ V ) /

µU∪V µU ⊕µV µU∩V µU∪V ∗ l ∗ l ∗ l ∗ l ∗ δ l−1 ∗ / Hc(U ∪ V ) / Hc(U) ⊕ Hc(V ) / Hc(U ∩ V ) / Hc (U ∪ V ) / where l = n − k. (Note that then l − 1 = n − (k + 1).) Lemma This diagram is commutative, provided the map labeled δ is replaced by (−1)kδ.

Proof. Observe ﬁrst that the lower sequence could also have been obtained from the short exact sequence of chain complexes gotten by dualizing

(j ,−j ) ∗ U ∗ V ∗ ∗ ∗ iU ∗+iV ∗ ∗ 0 → Ωc (U ∩ V ) / Ωc (U) ⊕ Ωc (V ) / Ωc (U ∪ V ) → 0

The maps µU etc. are also deﬁned on the level of forms. Hence we can construct a diagram of cochain complexes (6) ∗ ∗ ∗ ∗ (iU ,iV ) jU −jV 0 → Ω∗(U ∪ V ) / Ω∗(U) ⊕ Ω∗(V ) / Ω∗(U ∩ V ) → 0

µU∪V µU ⊕µV µU∩V

∗ ∗ ∗ ∗ ∗ ∗ ∗ ∗ 0 → Ωc (U ∪ V ) ∗ ∗/ Ωc (U) ⊕ Ωc (V ) ∗ / ∗ Ωc (U ∩ V ) → 0 ((iU ∗) ,(iV ∗) ) (jU ∗) −(jV ∗) 10 BJØRN JAHREN

(Note that the dual of a linear map of the form f + g : A ⊕ B → C is (f ∗, g∗): C∗ → A∗ ⊕ B∗, and the dual of (f, g): A → B ⊕ C is the map f ∗ + g∗ : B∗ ⊕ C∗ → A∗.) Observe that this is a three–dimensional diagram, with the third direc- tions given by the exterior diﬀerential and its dual. We claim that it is commutative in each degree. This reduces to the following observation: Let j : W1 ⊆ W2 be an inclusion of open subsets of M. Then the diagram

j∗ k k Ω (W2) / Ω (W1)

µW2 µW1 j ∗ n−k ∗ ∗ n−k ∗ Ωc (W2) / Ωc (W1) k is commutative. In fact, starting with ω ∈ Ω (W2) and going around the n−k diagram in two ways results in the two linear maps Ωc (W1) → R given by η 7→ R j∗ω ∧ η and η 7→ R (ω ∧ j η). But these are equal, since W1 W2 ∗ ∗ j∗(j ω ∧ η) = ω ∧ j∗η and this form has support inside W1. However, the maps µW do not commute with exterior differentials, instead we have the formula k+1 ∗ µW (dω) = (−1) d (µW (ω)) , k n−k+1 for ω ∈ Ω (W ). To see this, we evaluate both sides on a form η ∈ Ωc (W ): Z Lhs : µW (dω)(η) = dω ∧ η W Z ∗ Rhs : d (µW (ω))(η) = ω ∧ dη W R k R R But W dω ∧ η + (−1) W ω ∧ dη) = W d(ω ∧ η) = 0. Going back to diagram (6), we now see that if we change the sign of the exterior differentials in one of the exact sequences — e. g. if we replace dk :Ωk(W ) → Ωk+1(W ) by (−1)k+1dk — then all possible squares in the diagram will commute. This will not change any cohomology, and the maps induced by inclusions will remain the same as before. Therefore we will still get a Mayer–Vietoris sequence, and the only difference will be that the map δ : Hk(U ∩ V ) → Hk+1(U ∪ V ) will be changed by the same sign (−1)k+1. But since diagram (6) now is completely commutative, it is easily seen that (5) also will be commutative. Using the 5–lemma, we now get the following easy, but important Corollary If Poincaréduality holds for U, V and U ∩ V , it also holds for U ∪ V . The following extension of this corollary provides one of our main tools for proving Poincaréduality (henceforth abbreviated to PD) for more and more complicated subsets of M. Lemma A. Suppose O is a set of open subsets which is closed under finite intersections. If PD holds for all sets in O, then it holds for all finite unions of sets in O. MAT4520 (May 26, 2011) 11

Proof. We prove that PD holds for sets of the form U1 ∪ · · · ∪ Uk, Ui ∈ O, by induction on k, starting with k = 1 which holds by assumption. So, assume PD holds for unions of k − 1 > 1 sets in O. Then it holds for Uk and U1 ∪ · · · ∪ Uk−1, and by the assumption on O and the induction hypothesis it also holds for Uk∩(U1∪· · ·∪Uk−1) = (Uk∩U1)∪· · ·∪(Uk∩Uk−1). Therefore it also holds for U1 ∪ · · · ∪ Uk by the Corollary above.

Let us call a set of subsets an fc–set if it is closed under finite intersections. Observe that if O is an fc–set, then the set Of of finite unions of sets in O is again fc. An example to keep in mind is the set of subsets of open boxes n of R — the sets of the form (a1, b1) × · · · × (an, bn) for real numbers ai, bi. This set satisfies both of the conditions of Lemma A. Note that in this case n O is in fact a basis for the topology, and every open subset of R can be written as a countable union of sets in O. We would like to have a version pf Lemma A for infinite unions, but this can not be established the same way. However, one special case is easy: ` Lemma B. Let U = i Ui be the union of disjoint open sets such that PD holds for every Ui. Then PD holds for U.

Proof. A form on U consists of a form on each Ui, and it follows easily k Q k that H (U) ≈ i H (Ui). However, a form with compact support is non- n−k L n−k trivial only on finitely many Ui. Therefore Hc (U) ≈ i Hc (Ui). But dualizing takes direct sums to direct products. It follows from Lemma A and Lemma B that if we start from a collection O of open subsets of M as in Lemma A, then PD will hold for all sets that can be obtained by finitely many times applying the processes of taking finite unions and arbitrary disjoint unions. The general Poincaréduality theorem follows if we can prove that M itself can be so obtained. The following Lemma is the crucial topological result we need: Lemma C. Let O be an fc–basis for the topology of a connected manifold M such that the closure of each element of O is compact. Then M can be written M = V1 ∪V2, where V1 and V2 are disjoint unions of countably many sets in Of . Before proving this, let us see how it leads to a Proof of Poincaréduality. First, observe that PD follows for all open n n subsets of R . Indeed, an open U ∈ R is a disjoint, countable union of connected open set, and if U is connected, it has a basis (e. g. the basis discussed above consisting of open boxes) satisfying the conditions of both Lemma A and Lemma B. If M is an arbitrary connected manifold, it has a basis consisting of sets n W such that W is diffeomorphic to an open subset of R , and such that the closure of W is compact. This basis is easily seen to be fc, and we have just seen that all its elements satisfy Poincaréduality. It now only remains to prove Lemma C. We start by choosing an increasing sequence of open sets U1 ⊂ · · · ⊂ Ui ⊂ Ui+1 ⊂ · · · such that S (1) i Ui = M 12 BJØRN JAHREN

(2) Every Ui has compact closure U i, and U i ⊂ Ui+1. (See e. g. the proof of Theorem 13, page 50 in Spivak.) We will deﬁne a new sequence of open sets Wi, i = 1, 2,... such that S S (i) U i ⊂ Wj ⊂ Ui+1 (hence Wi = M) j6i i (ii) Each Wi ∈ Of (iii) Wi ∩ Wi+2 = ∅ for all i.

The sets Wi will be defined by induction on i. Since U2 is open, it is a union of open sets from the basis O. Choose a finite number of these sets which cover the compact set U 1 and let W1 be their union. Now suppose Wj has been defined for j < i and consider the inclusion S of the compact set U i − j

Finally, property (iii) is true because Wi ⊆ Ui+1 − U i−1 and Wi+2 ⊆ Ui+3 − U i+1, and the two sets Ui+1 − U i−1 and Ui+3 − U i+1 are disjoint. S S Now V1 = i W2i−1 and V2 = i W2i are easily seen to be as in Lemma C. Poincar´eduality is the most important result in de Rham cohomology. Here are some immediate consequences. • Suppose M is a connected, orientable n–manifold. Then ( if M is compact Hn(M) =∼ R 0 if M is non–compact

n ∼ In contrast, for cohomology with compact support we get Hc (M) = R both in the compact and non–compact cases, and an isomorphim is given by R ω 7→ M ω. Note that any nonzero element will be a generator. In particular it follows that we can represent a generator by a form with support in an arbitrarily small neighborhood of any point. n ∼ • The calculation of Hc (M) = R gives also rise to an important invariant of proper maps f : M → N between connected, oriented n–manifolds. In ∗ n n fact, the induced map f : Hc (N) → Hc (M) must be multiplication by a real number, which we call the degree of f — deg f. If ω represents n ∼ a generator of Hc (N) = R, the degree of f can be calculated from the equation Z Z f ∗ω = deg f ω . M N For example, if f is a diﬀeomorphism, deg f = 1 if f is orientation preserving and deg f = −1 if f is orientation reversing. Also, if f is an n–fold, orientation preserving covering map, then deg f = n. It is an amazing fact that the degree is always an integer! (See Theorem 12, page 275 in Spivak.) MAT4520 (May 26, 2011) 13

k k ∼ k • From the homotopy invariance of H (M) we get H (M) = H (M × R) for all k. It follows that if M is orientable we have isomorphisms k ∗ ∼ n−k ∼ n−k ∼ k+1 ∗ Hc (M) = H (M) = H (M × R) = Hc (M × R) , or k ∼ k+1 Hc (M) = Hc (M × R) . n n This generalizes the cases M = S and M = R established earlier in terms of explicit calculations. It is a nice exercise to show that the isomorphism above can be represented by the map on the level of forms given by ω 7→ ω ∧ f dt, where t is the coordinate on R and f : R → R is a function with compact support satisfying R f dt = 1. R This observation is vastly generalized in exercise 2.

• One can show that all smooth manifolds can be covered by open sets Ui n diffeomorphic to R such that all finite intersections Ui1 ∩· · ·∩Uil are either n diffeomorhic to R or empty. If M is compact, this simplifies the proof of Poincaréduality greatly, since we in this case only need the induction procedure of Lemma A. But then it also follows that the cohomology (Hk = k Hc in this case) is finitely dimensional. Hence, if M is a compact, orientable n–manifold there are isomorphisms Hk(M) =∼ Hn−k(M) for all k. k n • Recall that H (CP ) is isomorphic to R if k is even 6 2n and 0 otherwise. We will now use Poincaréduality to determine the ring structure on ∗ n L k n H (CP ) = k H (CP ). ∗ n ∼ n+1 Claim: H (CP ) = R[u]/(u ), where u corresponds to an element in degree 2. What this says is that if u is a generator (i. e. a non–zero element) of 2 n k H (CP ), then u is non–zero for all k 6 n. We prove this by induction on n, starting trivially with n = 1. ∗ n−1 ∼ n Assume that H (CP ) = R[v]/(v ). By the computation earlier we know that the inclusion ι : CP n−1 ⊂ CP n induces an isomorphism on Hj for j < 2n. Hence there is an element u ∈ H2(CP n) such that ι∗(u) = v, and since ι∗(uk) = (ι∗(u))k = vk 6= 0 for k < n, it follows that uk 6= 0 for k < n. Now use that the Poincaréduality isomorphism H2(n−1)(CP n) → 2 n ∗ H (CP ) is given by µCP n (w)(u) = w u. This must be non–zero for all w 6= 0, in particular for w = un−1. Thus un 6= 0. We end with some remarks on non–orientable manifolds. If M is non- orientable, we can construct a double covering space π : Mf → M such that Mf is orientable, in a natural way. Let

Mf = {(p, op)| p ∈ M and op an orientation of the tangent space Mp} , and define π : Mf → M by π(p, op) = p. Then π is a surjective map such that −1 π (p) consists of two points — the two orientations of Mp. If (p, op) ∈ Mf, n let x : U → R be a local coordinate map such that p ∈ U. We can clearly choose x such that it is orientation preserving at p with respect to the given n orientation op and the standard orientation on R . Requiring that x be orientation preserving in all of U defines an orientation oq of Mq for every 14 BJØRN JAHREN q ∈ U, hence a bijection between U and the subset {(q, oq)|q ∈ U}. This defines a differentiable atlas on Mf such that π becomes a smooth covering ∼ map. Moreover, π∗q is an isomorphism Mf(q,oq) = Mp. Giving Mf(q,oq) the orientation corresponding to oq via this isomorphism defines an orientation of Mf. π : Mf → M is called the orientation covering of M. Example. If M is an even–dimensional real projective space RP 2m, the orientation covering can be identified with the covering S2m → RP 2m which we used to compute the cohomology of RP 2m. In that computation a crucial rôle was played by the antipodal map, which we now generalize to an arbitrary orientation covering. Let A : Mf → Mf be the map that interchanges the two points of π−1(p) for every p. Then A shares the following formal properties with the antipodal map on even spheres: – A is an orientation reversing diffeomorphism – π ◦ A = π – A ◦ A = id . Mf Just as in the sphere case we now get splittings of the vector spaces of forms (with and without compact support) into the +1 and −1 eigenspaces of A∗, and these splittings induce the corresponding splittings in cohomology. Moreover, just as before we can prove that we have isomorphisms Ωk(M) =∼ k k ∼ k Ω (Mf)+ and Ωc (M) = Ωc (Mf)+. In particular we obtain isomorphisms n ∼ n n ∼ n H (M) = H (Mf)+ and Hc (M) = Hc (Mf)+ But these groups both vanish! This is trivial in the case Hn for M non– compact, since then Hn(Mf) = 0. The remaining cases follow from the n n fact that A is orientation reversing, since then A∗ : Hc (Mf) → Hc (Mf) is multiplication by −1. n n Hence we have proved that H (M) = Hc (M) = 0 for all connected non–orientable manifolds M. Taken in combination with our earlier com- putations, we obtain the following characterization of orientable manifolds: n Corollary. A connected n–manifold M is orientable if and only if Hc (M) is non–vanishing. n If M is orientable, a choice of generator of Hc (M) defines an orientation. Two generators g1 and g2 determine the same orientation if and only if g2 = ag2 with a > 0. It follows that if M is oriented, a generator represented by a form ω R such that M ω = 1 is uniquely determined. This generator is called the orientation class of M.

Here are a couple of slightly challenging exercises for the ambitious stu- dent: Exercise 1. Poincar´eduality as discussed here breaks down in the non– orientable case, but there are still interesting relations between cohomology with and without compact support. As an example, show that k n−k ∗ ∼ k H (M) ⊕ Hc (M) = H (Mf) . MAT4520 (May 26, 2011) 15

(Hint: show ﬁrst that the diagram

A∗ Hk(Mf) / Hk(Mf) µ µ Mf Mf (A∗)∗ n−k ∗ n−k ∗ Hc (Mf) o Hc (Mf) commutes up to multiplication by −1.) Exercise 2. This is an exercise to test if you really understand the techniques used to prove Poincaréduality. Let M and N be two manifolds, and let πM , πN be the two projections from M × N to M and N. Then we can define maps × :Ωi(M) × Ωj(N) → Ωi+j(M × N) ∗ ∗ by (ω, η) 7→ ω × η = πM ω ∧ πN η (a) Show that × also restricts to a bilinear map i j i+j × :Ωc(M) × Ωc(N) → Ωc (M × N) , which can be used to define a linear map i j i+j Hc(M) ⊗ Hc (N) → Hc (M × N) . Doing this for all i and j such that i + j = k defines a linear map M j k−j k κM,N : Hc (M) ⊗ Hc (N) → Hc (M × N) j

(b) Show that κM,N is an isomorphism for all M and N and all k.(The Künneththeorem.) (Hint: Fix N and consider the source and target of κM,N as functors of M and open inclusions. Start by showing that these functors have Mayer– Vietoris sequences.) (c) What goes wrong if we try to prove the Künneth theorem for Hk? Prove that this theorem also holds if either M or N has finite dimensional cohomology in every dimension. (d) Prove that S2 × S4 and CP 3 are two manifolds which have isomorphic cohomology groups in every dimension but are not smoothly homotopy equivalent. ∗ ∗ Remark. The tensor product of the graded rings Hc (M) and Hc (N) can be organized into a graded ring by setting ∗ ∗ k M j k−j (Hc (M) ⊗ Hc (N)) = Hc (M) ⊗ Hc (N) j rs r s and defining (a ⊗ b)(c ⊗ d) = (−1) ac ⊗ bd if b ∈ Hc (N) and c ∈ Hc (M). ∗ ∗ ∗ Then κM,N : Hc (M) ⊗ Hc (N) → Hc (M × N) becomes an isomorphism of graded rings. Similar remarks apply to H∗.