Poincaré Duality

The homology of a closed orientable manifold reads the same forwards and backwards — a symmetry strong enough to rule spaces out.

The idea

Poincaré duality is the theorem that the homology of a closed orientable manifold is symmetric about its middle dimension.

Theorem (Poincaré duality).

Let $M$ be a closed connected orientable $n$-manifold — compact, without boundary, and locally homeomorphic to $\mathbb{R}^{n}$ — so that it carries a fundamental class $[M] \in H_{n}(M)$. Then capping with $[M]$ is an isomorphism $H^{k}(M) \longrightarrow H_{n-k}(M)$ for every $k$.

For computation, read the theorem through Betti numbers. Write $b_{k}$ for the rank of $H_{k}(M;\mathbb{Z})$, its number of $\mathbb{Z}$ summands.

Corollary.

For every closed connected orientable $n$-manifold, $b_{k} = b_{n-k}$ for all $k$.

So the list of Betti numbers of a closed orientable manifold reads the same forwards and backwards.

The symmetry is a constraint: half of a manifold's homology determines the other half, and a proposed list that is not a palindrome belongs to no closed orientable manifold at all. Since $b_{0} = 1$ for a connected space, $b_{n} = 1$ as well, which makes the top homology group a test for orientability. No relation of this kind holds for spaces in general — the symmetry comes from the manifold condition, from the space looking the same near each of its points.

Ways to work on it

• Walkthrough. The duality isomorphism of a closed orientable manifold and the Betti-number symmetry it produces.
• Proof. Prove that a closed odd-dimensional manifold has Euler characteristic zero, by pairing each degree with its complement.
• Practice. Use Poincaré duality to complete Betti-number lists and reject impossible ones.
• Hardest. Decide whether a proposed Euler characteristic is possible for a closed orientable 6-manifold.

Not sure where to start? Take the ten-question placement test.