A closed oriented 4-manifold M has H_0 = Z, H_1 = 0, H_2 = Z^3, H_3 = 0, H_4 = Z. What does Poincare duality tell us about its cohomology?
AH^k(M) = H_k(M) for all k
BH^0 = Z, H^1 = 0, H^2 = Z^3, H^3 = 0, H^4 = Z — the same as homology since all groups are free
CH^0 = Z, H^1 = 0, H^2 = Z^3, H^3 = Z^3, H^4 = Z
DPoincare duality does not apply since H_1 = 0
Poincare duality gives H^k(M) ≅ H_{4-k}(M). So H^0 ≅ H_4 = Z, H^1 ≅ H_3 = 0, H^2 ≅ H_2 = Z^3, H^3 ≅ H_1 = 0, H^4 ≅ H_0 = Z. Since all homology groups are free, the universal coefficient theorem gives H^k ≅ Hom(H_k, Z) ≅ H_k, which happens to agree with Poincare duality in this case. But Poincare duality is the deeper statement: it connects H^k to H_{n-k}, not to H_k.
Question 2 Multiple Choice
Poincare duality requires the manifold to be oriented. What goes wrong for non-orientable manifolds?
ANon-orientable manifolds do not have homology groups
BThe fundamental class [M] ∈ H_n(M; Z) does not exist for non-orientable manifolds, so the cap product isomorphism fails
CNon-orientable manifolds have H_n = 0 with Z coefficients, so there is no class to cap with
DBoth B and C are correct (they describe the same phenomenon)
For a closed connected non-orientable n-manifold, H_n(M; Z) = 0 (there is no fundamental class with integer coefficients, because the manifold cannot be consistently oriented). Without a fundamental class, the Poincare duality isomorphism α ↦ α ∩ [M] has no [M] to use. However, every closed manifold HAS a fundamental class with Z/2Z coefficients ([M] ∈ H_n(M; Z/2Z) ≅ Z/2Z), so Poincare duality holds with Z/2Z coefficients for all closed manifolds, orientable or not.
Question 3 True / False
The real projective plane RP^2 is a closed 2-manifold with H_0 = Z, H_1 = Z/2Z, H_2 = 0 (integer coefficients). This violates Poincare duality because H^0 ≅ Z ≇ H_2 = 0.
TTrue
FFalse
Answer: True
This does not 'violate' Poincare duality — it confirms that Poincare duality requires orientability. RP^2 is non-orientable, so the theorem does not apply with Z coefficients. With Z/2Z coefficients, RP^2 has H_0 = H_1 = H_2 = Z/2Z, and Poincare duality holds: H^k(RP^2; Z/2Z) ≅ H_{2-k}(RP^2; Z/2Z). The statement in the question is technically true — the integer homology does not display Poincare symmetry — but the reason is that the hypothesis (oriented) is not met.
Question 4 Short Answer
Explain what the fundamental class [M] ∈ H_n(M; Z) represents for a closed oriented n-manifold, and why it is essential for Poincare duality.
Think about your answer, then reveal below.
Model answer: The fundamental class [M] is the unique generator of H_n(M; Z) ≅ Z that is compatible with the orientation of M. It is represented by the sum of all n-simplices in any triangulation, oriented consistently with the global orientation. The cap product map α ↦ α ∩ [M] from H^k(M) to H_{n-k}(M) uses [M] as the 'bridge' between cohomology and homology: it takes a k-cocycle α, evaluates it on the front k-face of the fundamental cycle, and returns the remaining (n-k)-chain as the Poincare dual. Without [M], there is no bridge.
The fundamental class encodes the entire manifold as a single homology class. It exists if and only if M is orientable and closed. The Poincare duality isomorphism is not abstract nonsense — it is a concrete geometric operation (cap product with the fundamental class) that pairs each cohomology class with a dual homology class of complementary dimension.