GEOMETRY HW 12 CLAY SHONKWILER

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1 GEOMETRY HW 12 CLAY SHONKWILER 1 Let M 3 be a compact 3-manifold with no boundary, and let H 1 (M, Z) = Z r T where T is torsion. Show that H 2 (M, Z) = Z r if M is orientable, and H 2 (M, Z) = Z r 1 Z/2 if M is non-orientable. Proof. If M is orientable, then, since it s compact, we know, by Poincaré duality, that Z r T H 1 (M, Z) H 2 (M, Z). Hence, the free part of H 2 (M, Z) is Z r ; by universal coefficients, H 2 (M, Z) Z r T 2, where T 2 is torsion. On the other hand, again by Poincaré duality, H 2 (M, Z) H 1 (M, Z), which is torsion-free, so we see that H 2 (M, Z) = Z r (we also could have seen this by noting that, since M is orientable, H 3 (M, Z) has no torsion and using universal coefficients). In the non-orientable case, we argue by Euler characteristics. Note, first, that since M is non-orientable, H 3 (M, Z) = Z/2. By universal coefficients, this means that H 2 (M, Z) = F Z/2, where F is free and H 3 (M, Z) = 0. Now, by Homework 7 #1, the Euler characteristic 3 χ(m) = ( 1) i rk H i (M, G) i=0 where rk denotes the free rank and G is any coefficient group (what we showed in HW 7 #1 was that the Euler characteristic is independent of coefficients). Therefore, the result proved in HW 11 #1, that a compact orientable manifold of odd dimension has Euler characteristic 0 holds for Z/2- orientable compact manifolds as well; since M is Z/2-orientable, χ(m) = 0. Therefore, 3 0 = χ(m) = ( 1) i rk H i (M, Z) = 1 r+rk H 2 (M, Z) 0 = 1+rk H 2 (M, Z) r, i=0 so rk H 2 (M, Z) = r 1, meaning that F = Z r 1. Therefore, we conclude that H 2 (M, Z) = Z r 1 Z/2. 1

2 2 CLAY SHONKWILER 2 Let M 3 be a compact 3-manifold. Show that H 1 (M, Z) is infinite if M is orientable with non-empty, connected boundary M which is not diffeomorphic to S 2. Proof. Since M is non-empty, connected, orientable (orientability comes from Problem 3 below) and not diffeomorphic to S 2, M is a surface of genus g for g 1. Therefore, H 1 ( M, R) = R 2g. Now, we have the following piece of the long exact sequence for the pair: (1) H 1 (M, R) H 1 ( M, R) H 2 (M, M, R) Also, since M is compact and orientable, we know, by Lefshetz duality, that H 1 (M, R) H 2 (M, M, R). Since there s no torsion in R coefficients, H 1 (M, R) H 1 (M, R) H 2 (M, M, R). Therefore, if H 1 (M, R) = 0, (1) reduces to 0 R 2g 0 which is impossible, since the sequence is exact. Therefore, we see that H 1 (M, R) = R m for some m 1. By universal coefficients, this implies that H 1 (M, Z) = Z m T where T is torsion, and so H 1 (M, Z) is infinite. Show that π 1 (M) is infinite if M 3 is non-orientable and has no boundary. Proof. Note that, for any manifold, H 1 (M, Z) = Z r T for some r 0 and T is torsion. Therefore, by Problem 1 above, in the case where M 3 is non-orientable and has no boundary, H 2 (M, Z) = Z r 1 Z/2; since we can t have Z 1, we see that r 1 0, so r 1. Hence, we see that H 1 (M, Z) contains at least one copy of Z and, therefore, is infinite. Now, since H 1 (M, Z) is the abelianization of π 1 (M), the cardinality of π 1 (M) is at least as great as the cardinality of H 1 (M, Z), so we see that π 1 (M) is infinite. 3 Let M n be a compact manifold with non-empty boundary M, such that M is R-orientable. Show that M is R-orientable also, and that the boundary homomorphism H n (M, M, R) H n 1 ( M, R) takes fundamental class to fundamental class. Proof. Let φ be an orientation on M. Let p M and let U be a coordinate neighborhood of p. Let V = U M and let q U := U V. Then there exists a neighborhood U of q such that φ : H n (M, M U, R) H n (M, M {q }, R) maps α U to α q, a generator of H n(m, M {q }, R),

3 GEOMETRY HW 12 3 for all q U. Since we can consider U U throughout the below, we may as well just assume U = U. Then φ restricts to an isomorphism H n (M, M U, R) H n (M, M {q}, R). Since we can contract U down to the point q, we see that (M, M {q}) is a deformation retract of (M, M U ), so we also have an isomorphism H n (M, M {q}, R) H n (M, M U, R). Now, consider the triple (M, M U, M U); we have this piece of the associated long exact sequence: H n (M, M U, R) H n (M, M U, R) H n 1 (M U, M U, R) Since we can take a collar neighborhood W of the boundary and may assume U is entirely contained in W, M W is a deformation retract of both M and M U, so we see that H (M, M U, R) H (M, M) = 0; hence, Then, by excision, H n (M, M U, R) H n 1 (M U, M U, R). H n 1 (M U, M U, R) H n 1 ((M U ) (M U ), (M U) (M U ), R) (2) = H n 1 ( M, M V, R) On the other hand, since V is homeomorphic to an open ball in R n 1, it is contractible to a point, so (M U, (M U ) {p}) is a deformation retract of (M U, (M U ) V ) = (M U, M U), so we see that H n 1 (M U, M U, R) H n 1 (M U, (M U ) {p}), R). Now, by excision, for any p V, H n 1 (M U, (M U ) {p }, R) H n 1 ((M U ) (M U ), ((M U ) {p }) (M U ), R) (3) = H n 1 ( M, M {p }, R) Therefore, putting (2) and (3) together, we see that H n 1 ( M, M V, R) H n 1 ( M, M {p }, R) for all p V. Furthermore, if we follow this chain of isomorphisms up, H n 1 ( M, M V, R) H n (M, M U, R); the orientation φ on M restricts to an orientation on M, so φ consistently maps the generator of α U H n (M, M U, R) to a generator α q H n (M, M {q }, R). Thus, following the isomorphsims down, we see that the pre-image α V H n 1 ( M, M V, R) of α U under the isomorphism H n 1 ( M, M V, R) H n (M, M U, R) is consistently mapped to a generator α p H n 1 ( M, M {p }, R) for all p V. Since our original choice of the point p was arbitrary, we see that, therefore, we have an induced orientation on M. Now, since M is not compact, H n (M, R) = 0; since M is homotopy equivalent to M, H n (M, R) H n (M, R) = 0. Therefore, from the long

4 4 CLAY SHONKWILER exact sequence for the pair (M, M): H n ( M, R) H n (M, R) H n (M, M, R) H n 1 ( M, R) we see that : H n (M, M, R) H n 1 ( M, R) is injective. Now, if W is a collar and N = M W, then, since N is a deformation retract of M and M is homotopy equivalent to M, H n (M, M N, R) H n (M, M M, R) = H n (M, M, R) Since M is compact, so is N, and so the R-orientation of M determines a fundamental class α N H n (M, M N, R); let [M, M] be its image in H n (M, M, R). Then [M, M] α q H n (M, M {q}, R) H n (M, M {q}, R) for all q M. Therefore, following the chain of isomorphisms elucidated above, we see that [M, M] is consistently mapped to a generator α p H n 1 ( M, M {p}, R) for all p M. Hence, if we define [ M] := [M, M] (which is unique, since is injective, we see that [ M] α p for all p M, so [ M] defines a fundamental class on M. Let M n be a compact manifold with non-empty boundary M. that M cannot be a retract of M. 4 Show Proof. Not that, since M n is compact, M is a compact (n 1) manifold. Since every manifold is orientable in Z/2 coefficients, we know, by Poincaré duality, that H k ( M, Z/2) H n k 1 ( M, Z/2) for all k. Now, since M is compact, we know, by Lefshetz duality, that and H k (M, M, Z/2) H n k (M, Z/2) H k (M, M, Z/2) H n k (M, Z/2) for all k (this second is actually from the generalization of Lefshetz duality where we break M into the pieces M and ). Also, since M and M are connected, H 0 (M, Z/2) Z/2 H 0 ( M, Z/2). Since the long exact sequences in homology and cohomology of the pair (M, M) is natural with respect to maps, all of this implies that we have the following commutative diagram: H n (M, Z/2) H n (M, M, Z/2) H n 1 ( M, Z/2) H 0 (M, M, Z/2) H 0 (M, Z/2) H 0 ( M, Z/2) f H n 1 (M, Z/2) H 1 (M, M, Z/2) Since the diagram commutes, we see that H n (M, M, Z/2) H n 1 ( M, Z/2), so ker f = H n 1 ( M, Z/2), meaning f 0. On the other hand, if M is a retract of M, then f : H n 1 ( M) H n 1 (M) is injective, which would

5 GEOMETRY HW 12 5 imply that H n 1 ( M, Z/2) = 0. However, this is clearly impossible, since M is a compact, Z/2-orientable manifold and so H n 1 ( M, Z/2) must be isomorphic to Z/2. From this contradiction, then, we conclude that M cannot be a retract of M. 5 Let M 2n be a compact orientable manifold. If H n 1 (M, Z) has no torsion, show that H n (M, Z) has no torsion also. Proof. Since H n 1 (M, Z) has no torsion, it is free, so H n 1 (M, Z) Z r for some r 0. By Poincaré duality, H n 1 (M, Z) H n+1 (M, Z), so H n+1 (M, Z) Z r. By universal coefficients, the torsion of H n+1 (M, Z) is equal to the torsion in H n (M, Z). Since H n+1 (M, Z) has no torsion, this implies that H n (M, Z) is torsion-free. 6 Define the degree of a map between compact orientable manifolds. Show that a n-fold cover between compact orientable manifolds has degree ±n. Definition 6.1. Let M n and N n be compact, orientable manifolds. Then H n (M, Z) H n (N, Z) Z. If f : M N is continuous, then f : H n (N, Z) H n (M, Z) is, under these isomorphisms, a map Z Z. Hence, for a Z, f (a) = da for some d a; define the degree of f to be d. Proof. Let π : M n M n be an m-fold cover, with M and M both compact and orientable. Then, for each p M, there exists an evenly covered neighborhood U p with preimage π 1 (U p ) = 1 i m where π 1 (p) = {q p,1,..., q p,m }. By excision, V p,i H n (V p,i, V p,i {q p,i }, Z) = H n (N ((N {q p,i }) (V p,i {q p,i })), (N {q p,i }) (V p,i {q p,i }), Z) and H n (N, N {q p,i }, Z) H n (U, U {p}, Z) = H n (M ((M {p}) (U {p})), (M {p}) (U {p}), Z) H n (M, M {p}) for each p and each i. From the exact sequence of relative cohomology, H n 1 (N {q p,i }, Z) H n (N, N {q p,i }, Z) H n (N, Z) H n (N {q p,i }, Z) 0

6 6 CLAY SHONKWILER Now, since N {q p,i } has the homology of the n-sphere and N {q p,i } is not compact, the left and right terms are zero, so H n (N, N {q p,i }, Z) H n (N, Z) is an isomorphism for each p and i. A similar argument shows that we have an isomorphism H n (M, M {p}, Z) H n (M, Z). We also have maps φ p,i : Hn (N, N {q p,i }, Z) H n (N, N π 1 (p), Z) and ψp,i : H n (N, N π 1 (p), Z) H n (V p,i, V p,i {q p,i }, Z) induced by the inclusions and the map j : H n (N, N π 1 (p), Z) H n (N, Z) from the long exact sequence. Since everything is natural, this all gives rise to the following commutative diagram: (4) H n (U, U {p}, Z) H n (M, M {p}, Z) H n (M, Z) π H n (V p,i, V p,i {q p,i }, Z) ψ p,i π H n (N, N π 1 φ p,i (p), Z) H n (N, N {q p,i }, Z) j π H n (N, Z) Via the isomorphisms around the outside and since M and N are compact and orientable, we can identify H n (U, U {p}, Z) and H n (V p,i, V p,i {q p,i }, Z) with Z, and so the induced map π between them is multiplication by some integer d p,i. Since π is a local homeomorphism, we see that each d p,i = ±1. Now, we claim that deg f = i d p,i. By excision, n H n (V p,i, V p,i {q p,i }Z) = H n ( i V p,i, V p,i {q p,i }, Z) i=1 = H n (N ((N π 1 (p)) ( i U {q p,i })), (N π 1 (p)) ( i U {q p,i }), Z) H n (N, N π 1 (p), Z), where ψp,i is projection onto the ith summand. Since the upper triangle in (4) commutes, the inclusions into the ith summand are given by φ p,i. Via the identification with Z, then, and since the lower triange commutes, j φ p,i(1) = 1 for all i, so j (0,..., 0, 1, 0,..., 0) = 1 for each such term. Hence, j (a 1,..., a m ) = m i=1 a i. The commutativity of the upper square says that ψp,i π (1) = d p,i for each i, so π (1) = (d p,1,..., d pm ). Thus, the commutativity of the lower square shows that m deg π = d p,i. Furthermore, we saw above that each d p,i = ±1; whether it is +1 or 1 will be consistent for all i, depending on whether π is orientation-preserving or -reversing, so we conclude that deg π = ±m. i=1

7 DRL 3E3A, University of Pennsylvania address: GEOMETRY HW 12 7