Poincaré duality
/ pwan-kah-RAY /
On a closed oriented manifold there is a striking symmetry: the homology in dimension k mirrors the homology in the complementary dimension n - k. A surface's loops pair with its loops, a 4-manifold's surfaces pair with its surfaces. Poincare duality is the theorem that makes this mirror exact, identifying cohomology in one degree with homology in the complementary degree, and it is the deepest structural fact about the topology of manifolds.
The precise statement: let M be a closed (compact, no boundary) oriented manifold of dimension n. Orientation provides a fundamental class [M] in H_n(M; Z). Then capping with [M] is an isomorphism H^k(M; Z) -> H_{n-k}(M; Z) for every k. Dually, and more symmetrically over a field, there is a perfect pairing H^k(M) cross H^{n-k}(M) -> H^n(M) = Z given by cup product followed by evaluation on [M]; 'perfect' means neither factor has a class that pairs to zero with everything. Concretely this is intersection: two cycles of complementary dimension in general position meet in finitely many points, and the signed count is exactly the duality pairing.
Among the consequences: the Betti numbers of a closed oriented n-manifold satisfy b_k = b_{n-k}, a palindrome; for a 2-manifold this gives b_0 = b_2 and recovers the genus from b_1. On a closed oriented manifold of dimension 4m the middle pairing H^{2m} cross H^{2m} -> Z is a symmetric bilinear form, the intersection form, whose signature is a powerful invariant feeding into the classification of 4-manifolds. Poincare duality is what makes manifolds so much more rigid than general spaces.
The hypotheses are not optional and are the usual trap. Closed: a manifold with boundary needs the modified Lefschetz duality pairing absolute against relative cohomology. Oriented: without orientation the integral statement fails — the fundamental class over Z does not exist; one falls back to Z/2 coefficients, where duality holds for all closed manifolds (Poincare duality mod 2). And it is about manifolds, not arbitrary spaces: a space that merely has the right Betti numbers is not thereby a manifold. State 'closed oriented' every time or the theorem is simply false.
On the closed oriented surface of genus g, Poincare duality pairs H^1 with H^1 via cup product into H^2 = Z. In the basis a_1, b_1, ..., a_g, b_g of H^1 the intersection form is the standard symplectic form: a_i pairs with b_i to 1 and everything else to 0. Geometrically the loop a_i meets b_i in exactly one point, which is the pairing value.
On a genus-g surface the intersection form on H^1 is the standard symplectic pairing.
Poincare duality needs the manifold closed and oriented; drop orientation and the integral statement fails (use Z/2 coefficients instead), and for manifolds with boundary you need Lefschetz duality between absolute and relative cohomology. Having the right Betti numbers does not make a space a manifold.