LECTURE 22: MAPPING DEGREE, POINCARE DUALITY

Transcription

1 LECTURE 22: APPING DEGREE, POINCARE DUALITY 1. The mapping degree and its appliations Let, N be n-dimensional onneted oriented manifolds, and f : N a proper map. (If is ompat, then any smooth map f : N is proper.) Then the pull-bak map f : R = H n (N) H n () = R is a linear map, and thus is a map of the form λ λ. The onstant is alled the degree of f: Definition 1.1. The degree of a proper smooth map f : N is the number deg(f) so that f ω = deg(f) ω, ω Ω n (N). Remark. It is not hard to see If f : N and g : N P are both proper, then deg(g f) = deg(f)deg(g). If f and g are properly homotopi, then deg(f) = deg(g). Proposition 1.2. If a proper map f : N is not surjetive, then deg(f) = 0. N Proof. Sine any proper ontinuous map to a manifold is losed (.f. Lee, appendix A, theorem A.57), f() is a losed subset in N. If q f(), then one an find an open neighborhood Ũ of q so that Ũ f() =. We pik an n-form ω supported in Ũ so that ω = 1. Sine by our N onstrution, f ω = 0, we onlude deg(f) = 0. Reall (leture 18): if f is an orientation-preserving diffeomorphism, then f ω = ω. So N Lemma 1.3. If f : N is a diffeomorphism, then 1, f is orientation preserving, deg(f) = 1, f is orientation reversing. This fat implies the following remarkable property of the degree: deg(f) is always an integer! To see this, we may assume that f is surjetive. Aording to Sard s theorem, almost every point q N is a regular value of f. In what follows we fix a regular value q of f. Sine dim = dim N, f is loally a diffeomorphism at eah point in f 1 (q). It follows that f 1 (q) is disrete. On the other hand, by properness, f 1 (q) is ompat. It follows that f 1 (q) is a finite set. We will denote f 1 (q) = p 1,, p k }. We an take (why?) a neighborhood Ũ of q and neighborhoods U i of p i, 1 i k, so that 1

2 2 LECTURE 22: APPING DEGREE, POINCARE DUALITY We let Eah U i is the domain of an oriented hart ϕ, U i, V i } on. For i j, U i U j =. Ũ is the domain of an oriented hart ψ, Ũ, Ṽ } on N. f maps U i diffeomorphially to Ũ. f 1 (Ũ) = k U i. σ i = Theorem 1.4. deg(f) = k σ i. 1, if f : Ui Ũ is orientation preserving at p i, 1, if f : U i Ũ is orientation reversing at p i. Proof. We take ω Ω n (Ũ) so that ω = 1. Then f ω is supported in f 1 (Ũ) = N k U i, and k k k f ω = f ω = σ i ω = σ i. U i Ũ The theorem follows. As an appliation of degree theory, we an prove the following hairy ball theorem : Theorem 1.5. Even dimensional spheres do not admit non-vanishing smooth vetor fields. Proof. Let f : S 2n S 2n be the antipodal map, then by PSet 5 problem 6(a), f is orientationreversion and thus deg(f) = 1. On the other hand, suppose X is a non-vanishing smooth vetor field on S 2n R. By normalizing the vetors, we may assume X p = 1 for all p S 2n. We will think of p and X p as vetors in R, and onsider the map F (p, t) = p os(tπ) + X p sin(tπ). Then for eah t R, F (, t) is a map from S 2n to S 2n. So F is a homotopy between F (, 0) = Id S 2n and F (, 1) = f, the antipodal map. It follows deg(f) = deg(id S 2n) = 1, a ontradition. Another appliation is Theorem 1.6. Suppose is an n-dimensional oriented ompat manifold with smooth and onneted boundary, X a onneted oriented (n 1)-manifold, and f : X a smooth map that extends to a smooth map g : X. Then deg(f) = 0. Proof. Pik ω Ω n 1 (X) so that ω = 1. Then X deg(f) = deg(f) ω = f ω = g ω = d(g ω) = g dω = 0. X Remark. Aording to Whitney approximation theorem, one has (Theorem 6.26 in Lee s book) Any ontinuous map is homotopi to a smooth map.

3 LECTURE 22: APPING DEGREE, POINCARE DUALITY 3 (Theorem 6.29 in Lee s book) If a ontinuous map f is homotopi to smooth maps g 1 and g 2, then g 1 and g 2 are smoothly homotopi. So one an define the degree of a ontinuous map to be the degree of orresponding smooth maps. All the theorems we proved above apply to ontinuous maps. In fat, in algebrai topology, degree is developed for ontinuous maps: in that ase there is no smoothness at all. Corollary 1.7 (Brouwer Fixed Point Theorem). Every ontinuous map from B n (=the unit ball in R n ) to itself has a fixed point. Proof. Let F : B n B n be a ontinuous map without fixed point. Then G : B n S n 1, x x F (x) x F (x) is the extension of g = G S n 1. By theorem 1.6, deg(g) = 0. On the other hand, the map H : S n 1 [0, 1] S n 1, (x, t) x tf (x) x tf (x) is a homotopy between the identity map and g. So deg(g) = deg(id) = 1. Contradition. 2. The Poinaré duality and its appliations Let be an oriented manifold of dimension n. We have the following maps : H k () Hl () H k+l (), ([ω], [η]) [ω η]. : Hn () R, [ω] ω. For any 0 k n, onsider the bilinear pairing map The map P k P k : H k () H n k () R, P k ([ω], [η]) = indues the following Poinaré duality operator P k : H k () ( H n k () ), P k ([ω]) = ω η. } η ω η. For example, P 0 maps the element 1 R = H0 () to the linear map : H n () R, η η on H n (), so that one an think of as an element in (Hn ()). The major theorem we would like to disuss in this setion is Theorem 2.1 (Poinaré duality). For any oriented manifold and any k, the Poinaré duality map P k is a linear isomorphism from Hk () to ( H n k () ). Remark. If H n k () is finite dimensional, then ( H n k () ) is isomorphi to H n k ().

4 4 LECTURE 22: APPING DEGREE, POINCARE DUALITY Remark. Any losed submanifold S of o-dimension k defines an element in ( H n k S () ), and thus defines an element in H k (). For example, the element in H0 () orresponding to the (o-dimension zero sub-)manifold is 1. R, k = n, R, k = 0, Example. For = R n, we have H k (Rn ) = and H 0, k n, k (R n ) = 0, k 0. R, k = 0, n, Example. For = S n, we have H k (S n ) = H k (Sn ) = 0, k 0, n. Example. For any ompat onneted oriented manifold of dimension n, we have already seen H n () = R and H0 () = R. Example. Let = i N (i, i + 1) be a ountable union of disjoint open intervals. Then H 0 () = i N R = (a 1, a 2, ) a i R} and H 1 () = i N R = (a 1, a 2, ) a i R, all but finitely many a is are zero }, and the Poinaré duality follows from the well-known fat in algebra: ( i N R) = i N R. It is also well-known that ( i N R) i N R. So in general ( H k ()) H n k (). In what follows we will sketh a proof of Poinaré duality for oriented manifolds admitting a finite good over, although the theorem holds for any oriented manifold. We need Lemma 2.2. The following diagram ommutes: H k () α Hk (U) Hk (V ) β ( 1) k+1 δ.. Hk (U V ) Hk+1 ().... P k.. PU k Pk V.. PU V k.... H n k () H k (U) H k (V ) H n k (U V ) H n k 1 ().. α β.... δ where the bottom row is the dual of the ayer-vietoris sequene for ompatly supported de Rham ohomology groups. Sketh of proof of Poinaré duality for oriented manifolds admitting a finite good over. We proeed by indution. The theorem holds for admitting one good hart, in whih ase R n, and the isomorphism follows from the two versions of Poinaré lemma that we proved. Now suppose the theorem holds for manifolds admitting a good over of no more than k 1 open sets, and suppose admits a good over U 1,, U k }. We let U = U 1 U k 1 and V = U k. Then U, V and U V all admit a good over of no more than k 1 open sets. By indution hypothesis, PU k, Pk V and Pk U V are all isomorphisms. By the above lemma and the five lemma (see leture 20), P k is an isomorphism. P k+1..

5 LECTURE 22: APPING DEGREE, POINCARE DUALITY 5 Sine for any onneted nonompat manifold, H 0 () = 0, we get another proof of Corollary 2.3. For any n-dimensional onneted nonompat oriented manifold, H n () = 0. Another orollary is (the result atually holds for any manifold ) Corollary 2.4. For any oriented manifold whose ompat supported ohomology groups are finite dimensional, H k+l ( R l ) H k (). Proof. H k+l ( R l ) H n+l k l ( R l ) H n k () Hk (). Reall that the Betti number b k = dim H k (). So Corollary 2.5. If is a ompat oriented manifold of dimension n, then b k = b n k. Reall that the Euler harateristi χ() = k ( 1)k b k. Corollary 2.6. Let be a ompat oriented manifold. (1) If dim = 2n + 1, then χ() = 0. (2) If dim = 4n + 2, then χ() is even. Proof. (1) If dim = 2n + 1, we have n χ() = ( 1) k b k = ( 1) k b k + k=n+1 ( 1) k b k = (2) Suppose dim = 4n + 2, then the same argument yields 4n+2 χ() = ( 1) k b k = 2n n (( 1) k + ( 1) k )b k = 0. (( 1) k + ( 1) 4n+2 k )b k + b. Sine ( 1) k + ( 1) 4n+2 k = ±2, it remains to prove b is even. This follows from the nondegeneray of the pairing P If fat, for any [ω], [η] H (), we have P ([ω], [η]) = ω η = It follows that the matrix for the pairing P b b matrix. It follows that : H () H() R. ( 1) η ω = P ([η], [ω]). : H () H () R is an anti-symmetri det(p ) = det((p )T ) = ( 1) b det(p ). So b must be an even number, otherwise det(p ) = 0 and thus P is not non-degenerate. Note that in partiular, we proved that for a ompat oriented surfae Σ g = T 2 # #T 2 of genus g, dim H 1 (Σ g) is even.