Algebraic Topology II Notes Week 12

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1 Algebraic Topology II Notes Week 12 1 Cohomology Theory (Continued) 1.1 More Applications of Poincaré Duality Proposition 1.1. Any homotopy equivalence CP 2n f CP 2n preserves orientation (n 1). In other words, f : H 4n (CP 2n ) H 4n (CP 2n ) [CP 2n ] [CP 2n ] Proof. Say α generates H 2 (CP 2n ). Then, as f induces an isomorphism on H 2, we have: f (α) = ±α. We also know that α 2n generates H 4n (CP 2n ), i.e. α 2n, [CP 2n ] = 1, or If f ([CP 2n ]) = [CP 2n ], then which is a contradiction! H 4n (CP 2n ) [CP 2n ] H 0 (CP 2n ) α 2n 1 1 = α 2n, [CP 2n ] = α 2n, f ([CP 2n ]) = f (α 2n ), [CP 2n ] = (f (α)) 2n, [CP 2n ] (1) = (±α) 2n ), [CP 2n ] = α 2n, [CP 2n ] = 1, Proposition 1.2. Let M n be a closed, connected, oriented n-manifold and let f : S n M be a continuous map of non-zero degree, i.e., the morphism f : H n (S n ; Z) H n (M; Z) is non-trivial. Show that H (M; Q) = H (S n ; Q). Proof. Assume that 1 i n 1 such that H i (M; Q) = 0. Then by Universal Coefficient Theorem, H i (M; Q) = Hom Q (H i (M; Q), Q) 0 If 0 α H i (M; Q) is the generator, then by Poincaré Duality, β H n i (M; Q) such that α β generates H n (M; Q) = Q. Especially, α β 0. On one hand, f : H i (M; Q) H i (S n ; Q) = 0 is a zero map. Consequently, f (α β) = f (α) f (β) = 0 0 = 0. On the other hand, f (α β) = (deg f) generator of H n (S n ; Q) 0, a contradiction. 1

2 Definition 1.3 (Manifold with Boundary). M is a n-manifold with boundary if any x M has a neighbourhood U x homeomorphic to R n or R n +(x n 0). if U x = R n, H n (M, M x) = H n (U x, U x x) = Z if U x = R n +, H n (M, M x) = H n (U x, U x x) = H n (R n +, R n + {0}) = 0 And the boundary of M is defined to be M = {x M H n (M, M x) = 0}. Example 1.4. (D n ) = S n 1, (R n +) = R n 1. Remark 1. M is a manifold of dimenison n 1 with no boundary. Definition 1.5. (M, M) is orientable, if M \ M is orientable as a manifold with no boundary. Proposition 1.6. If (M, M) is compact, orientable n-manifold with boundary, then! µ M H n (M, M) inducing local orientations µ x H n (M, M x) at all x M \ M. Note 1. In the long exact sequence for the pair (M, M), we have if M is oriented. H n (M, M) [M] = µ M [ M] H n 1 ( M) Theorem 1.7 (Poincaré Duality). If (M, M) is a connected, oriented n-manifold with boundary, then Hc(M) i µ M = where Hc(M, i M) === def lim support. H n i (M, M) or Hc(M, i M) µ M H n i (M). = Kcompact M\ M H i (M, (M \ K) M) is the cohomology with compact Proposition 1.8. If M n = V n+1 is a connected manifold with V compact, then the Euler characteristic χ(m) is even. An immediate corollary is Corollary 1.9. RP 2n, CP 2n, HP 2n cannot be boundaries of compact manifolds. In order to prove Proposition 1.8, we need another proposition: 2

3 Proposition Assume V 2n+1 is an oriented, compact manifold with connected boundary V = M 2n. If R is a field (especially Z/2Z if M is non-orientable), then dim R H n (M; R) is even. Proof of Proposition Let s consider the long exact sequence for the pair (V, M): H n (V ; R) i H n (M; R) δ H n+1 (V, M; R) = [M] = [V ] i H n (M; R) H n (V ; R) where i, i are induced by the inclusion i : M = V V. By exactness, Im i = Ker δ P.D. = Ker i, so dim Im i = dim Ker i = dim H n (M; R) dim Im i. Since i, i are Hom-dual, dim Im i = dim Im i. Therefore, dim H n (M; R) = dim H n (M; R) = 2 dim Im i is even. Proof of Proposition 1.8. If n = dim M is odd, then χ(m) = 0 is even. If n = 2m is even, then work with Z/2Z-coefficients, χ(m) = 2m i=0 ( 1) i dim Z/2Z H i (M; Z/2Z) m 1 (1) = 2 ( 1) i dim Z/2Z H i (M; Z/2Z) + ( 1) m dim Z/2Z H m (M; Z/2Z) i=0 dim Z/2Z H m (M; Z/2Z) mod 2 (2) 0 mod 2 Equation (1) is due to Poincaré Duality, dim Z/2Z H n i (M; Z/2Z) = dim Z/2Z H i (M; Z/2Z), 0 i m 1. Congruence (2) is by Proposition Note 2. Im i H n (M 2n ) is self-annihilating with respect to cup product, i.e. if α, β Im i, then α β = 0. by Proposition 1.10, dim Im i = 1 2 dim Hn (M 2n ). Proof. For any α = i (α), β = i (β) B, where α, β H 2n (V ), we have δ(α β) = δ(i (α) i (β)) = δi (α β) = 0 3

4 Hence, α β Ker (δ : H 4n (M) H 4n+1 (V )) = 0 by the following commutative diagram H 4n (M) δ H 4n+1 (V ) = P.D. H 0 (M) = P.D. H 0 (V ) with the bottom arrow an injection. 1.2 Signature Let M n be a closed, oriented, n-manifold. σ(m) = 0 if 4 dim M = n. if dim M = 4k, then σ(m) is defined to be the signature of the non-degenerate cup product pairing (, ) : H 2k (M; R) H 2k (M; R) R (α, β) (α, β)[m] σ(m):=(the number of positive eignvalues)-(the number of negative eignvalues). ( ) 0 1 Example σ(s 2 S 2 ) = σ = 0, σ(cp 2n ) = 1, σ(cp 2 #CP 2 ) = Remark 2. Signature σ is a cobordism invariant, i.e. if W = M N, then σ(m) = σ(n). Theorem If in the above notations M 4k = V 4k+1 is connected with V compact and orientable, then σ(m) = 0. Remark 3. σ(cp 2 # CP 2 ) = 0. In fact, CP 2 # CP 2 is the boundary of a connected, oriented 5-manifold; σ(cp 2 #CP 2 ) = 2 0, so CP 2 #CP 2 is NOT the boundary of a connected, oriented 5-manifold, but as we will see later, it IS the boundary of a non-orientable 5-manifold. 4

5 Proof. Let A = H 2n (M) and we ll always work over R-coefficients. We have a non-degenerate and symmetric pairing ϕ : A A R. Let A + be the subspace on which the pairing is positive-definite, A be the subspace on which the pairing is negative-definite. Let r = dim A +, 2l = dim A (by Proposition 1.10). Then, automatically, dim A = 2l r and σ(m) = r (2l r) = 2r 2l. In order to prove that σ(m) = 0, we want to show that r = l! Let B A be the self-annihilating l-dimensional subspace given by Proposition 1.8 and Note 2. Therefore, A + B = {0}, A B = {0}. Hence, dim A + + dim B dim A = 2l, i.e. r + l 2l i.e. r l dim A + dim B dim A = 2l, i.e. 2l r + l 2l i.e. r l In conclusion, r = l and σ(m) = Connected Sums Definition 1.13 (Connected Sum). Let M n, N n be closed, connected, oriented n-manifolds, then their connected sum is defined to be M#N := (M \ D n 1 ) f (N \ D n 2 ) where f : D n 1 = S n 1 D n 2 = S n 1 is an orientation-reversing homeomorphism. Remark 4. M#N is still a closed, connected, oriented n-manifold, H 0 (M#N) = Z, H n (M#N) = Z and H k (M#N) = H k (M) H k (N), 0 < k < n. Remark 5. Cup product α β = 0 for any α H k (M) and β H l (N) with k, l > 0. Example S 2 S 2 and CP 2 #CP 2 have the same cohomology groups, but different cohomogy rings, since H 0 = Z, H 2 = Z Z = Zα Zβ, H 4 = Z, α β 0 in S 2 S 2 ; while α β = 0 in CP 2 #CP 2. 5

6 Example σ(cp 2 #CP 2 ) = 2, so CP 2 #CP 2 cannot be the boundary of a compact oriented 5-manifold. However, CP 2 #CP 2 = W 5, where W 5 is a compact non-orientable 5-manifold. W can be constructed as follows: 1. Start with (CP 2 I)#(RP 2 S 3 ). 2. Run an orientation reversing path γ from one CP 2 to the other, by traveling along an orientation reversing path in RP Enlarge the path to a tube and remove its interior. What is left is a 5-dimensional non-orientable manifold with W = CP 2 #CP 2. 2 Homotopy Theory Let X be a topological space, x 0 X. Earlier we defined the fundamental group π 1 (X, x 0 ) = {f : (I, I) (X, x 0 )}/ = {f : (S 1, s 0 ) (X, x 0 )}/ where refers to the homotopy of maps. And also some properties of π 1 : functorial, i.e. f : X Y induces f : π 1 (X, x 0 ) π 1 (Y, f(x 0 )); homotopy invariant; group structure; related to H Higher Homotopy Groups Definition 2.1 (Higher homotopy groups). π n (X, x 0 ) = {f : (I n, I n ) (X, x 0 )}/ = {f : (S n, s 0 ) (X, x 0 )}/ 6

7 where I n = [0, 1] n, I n = {(x 1,..., x n ) I n i, s.t. x i {0, 1}} and refers to homotopy of maps. When n = 0, I 0 =point, I 0 =. So Define for n 1, f, g π n (X, x 0 ), π 0 = set of connected components of X (f + g)(s 1,..., s n ) = { f(2s1, s 2,..., s n ) 0 s g(2s 1 1, s 2,..., s n ) 1 2 s 1 1 Lemma 2.2. If n 2, π n is an abelian group. Goal: Prove Whitehead s Theorem. Theorem 2.3 (Whitehead s Theorem). If a map f : X Y between connected CW complexes induces isomorphisms on π n for all n, then f is a homotopy equivalence. If moreover, f is the inclusion of a subcomplex X Y, then X is a deformation retract of Y. 7