Proofs of Poincaré duality

Question

I know several proofs of Poincaré duality:

1. The original proof using dual cell complexes. Probably the nicest version of this uses a handle decomposition.

2. The argument (in Hatcher and many other places) using Mayer–Vietoris to glue together local charts one at a time.

3. There is a strange-looking proof in Chapter 11 of Milnor–Stasheff of the cohomology version (over a field) that uses the Thom isomorphism and a careful study of the diagonal map $M \rightarrow M \times M$.

4. Over the real numbers, you can prove it in de Rham cohomology using the Hodge-$\ast$ operator.

Does anyone know any other proofs of this result?

Answer 1

1. Alexander duality + Thom isomorphism + tubular neighborhood theorem = Poincaré Duality. Such an argument does not have to be circular, Dold-Puppe prove Alexander duality directly without reference to either Poincaré duality or the Thom isomorphism.

2. Spivak showed in his thesis that if some space $X$ has a tubular neighborhood up to homotopy, then it satisfies Poincaré Duality. The argument is different than that of (1) and uses the Serre spectral sequence.

3. A modern take on the previous proof is that Poincaré Duality can be proven using the "Six functor formalism". A particularly recent account by Volpe is The six operations in topology

4. There are various "scanning arguments" to construct Poincaré Duality which rely on using exponential maps to compare neighborhoods around $x$ to $T_xM$. These basically have two flavors: one which comes from applying homology to recover a Poincaré Duality map and one which comes from applying homotopy to recover a Poincaré Duality map. Both can be seen as special cases of "nonabelian Poincaré Duality" of Lurie, Ayala-Francis.

5. On the category of tame, oriented $n$-manifolds, $C_\ast(M)$, $ \bar{C}^{n-\ast}(M^+)$ provide two cosheaves "up to homotopy". By work of Ayala--Francis and Lurie, to prove the $C_\ast(M) \simeq \bar{C}^{n-\ast}(M^+)$ it suffices to show a quasiisomorphism exists (sufficiently naturally) for $M = \bigsqcup_i \mathbb{R}^n$. However, in this case Poincaré duality is trivial to prove. Note, this is basically a beefed up version of Hatcher's argument.

Finally, I should point out that Klein's work on The dualizing spectrum of a topological group is relevant to most of these examples and if one has a proof that the dualizing spectrum of a manifold is its Spivak normal fibration, then this would also yield a proof. Klein provides such an argument, but I have not checked carefully to see if it makes use of Poincaré Duality.

Answer 2

1. On a closed, oriented manifold both the sheaf of singular cochains and the sheaf of singular chains (with appropriate indexings) are resolutions of the constant sheaf of coefficients. (The hard part here is showing, if one wants to, that the resulting isomorphism on (co)homology is equivalent to the cap product.)
2. Dev Sinha, Anibal Medina, and I have been developing "geometric homology and cohomology" on smooth manifolds, originally due to Lipyanskiy. In this case, due to the definitions of geometric chains and cochains, Poincare duality is tautological and it's tautological that it comes from the cap product. The hard part here is showing that geometric homology and cohomology are the usual homology and cohomology theories.

Answer 3

In two and three dimensions one can (in principle) give proofs based on a geometric realization of homology classes by submanifolds. I’ll discuss the 3-D case since the 2D one follows from the classification of surfaces (or following the 3D argument below).

Moise proved that every 3-manifold is triangulable (and smooth), so if connected and orientable it has a fundamental class, and hence if closed $H_3(M;\mathbb{Z}) \cong \mathbb{Z} \cong H^0(M)$ (an upper bound may be obtained by taking a cell structure with one 3-cell and using the isomorphism with cellular homology).

For $\alpha \in H^1(M)$, there is a map $a:M\to S^1$ with pullback of the fundamental class $1\in H^1(S^1)$ $a^*(1)= \alpha$ by Brown representability. By Sard there is a point in the image of $a$ for which the map is transverse, and hence the preimage is a closed surface. One may show that this is well-defined up to homology (preimages of different transverse points bound submanifolds ), hence one gets a well-defined map $H^1(M) \to H_2(M)$. To see that this is onto, one may take an arbitrary 2-cycle and create a map of a surface into the manifold whose image is this cycle, then resolve singularities (see this answer for how to resolve branch point singularities). Orientability implies that this surface is 2-sided and homologically nontrivial implies it is has a non-separating component from which one may derive a map to $S^1$ realizing the surface as the preimage of a point. This shows that $H^1(M)\cong H_2(M)$.

For $\beta \in H^2(M)$, realize by a map $b:M\to K(Z,2) = CP^\infty $, and use cellular approximation to realize by a map factoring through $M\to CP^2$. Homotope this map to be transverse to $CP^1 \subset CP^2$, then the preimage will be a 1-submanifold. This gives a map $H^2(M) \to H_1(M)$, and one may show that this is well-defined and onto as in the previous paragraph.

For $H^3(M)$ one may similarly take maps to a $K(Z,3)$ which has 4-skeleton $S^3$, so this is the cohomotopy group of maps to $S^3$, then do similar arguments as before.

This sort of proof is overkill in many ways and not a practical proof. On the other hand, this is actually the way I that I think about Poincaré duality and homology in three dimensions.