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De Rham Cohomology 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 ob- tained 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 homo- geneous 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].
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