Hi all,

Today I was playing with the cohomology of manifolds and I noticed something intriguing. Although I might just have been caught out by a couple of enticing coincidences, it feels enough like there might be something going on that I thought I'd put it out here.

Let $M$ be an $n$-manifold with boundary $\partial M$. We write out the long exact homology sequence for the pair $(M, \partial M)$:

$$\cdots \to H_k(M) \to H_k(M, \partial M) \to H_{k - 1}(\partial M) \to \cdots$$

Let's apply Poincaré duality termwise, and keep the arrows where they were out of sheer faith. What we get is

$$\cdots \to H^{n - k}(M, \partial M) \to H^{n - k}(M) \to H^{n - k}(\partial M) \to \cdots$$

Surprisingly, this is the long exact cohomology sequence for the pair $(M, \partial M)$! To my mind, two things here are weird. The first is that intuitively, any functoriality properties Poincaré duality possesses should be arrow-reversing. The second is that we have a shift - but not a shift by a multiple of 3. So the boundary map in the homology sequence 'maps' to something that doesn't change degree in the cohomology sequence.

Let's play the same game with the Mayer-Vietoris sequence. For simplicity, suppose now $M$ is without boundary. Write $M = A \cup B$ where $A$ and $B$ are $n$-submanifolds-with-boundary of $M$ and $A \cap B$ is a submanifold of $M$ with boundary $\partial A \cup \partial B$. Then we have

$$\cdots \to H_k(A \cap B) \to H_k(A) \oplus H_k(B) \to H_k(M) \cdots$$

Hitting it termwise with Poincaré duality, and cruelly and unnaturally keeping the arrows where they are once again, we get

$$\cdots \to H^{n - k}(A \cap B, \partial A \cup \partial B) \to H^{n - k}(A, \partial A) \oplus H^{n - k}(B, \partial B) \to H^{n - k}(M) \to \cdots$$

This looks unfamiliar, but by looking at cochains it's not hard to see that there actually is a long exact sequence with these terms. However, this time we don't have the weird shift.

Now is there anything going on here, or just happenstance? Is there really a sense in which Poincaré duality is functorial with respect to long exact sequences? If so, what's the 'Poincaré dual' of the long exact sequence of the pair $(M, N)$ where $N$ is a tamely embedded submanifold of $M$?

Edit: Realised that in the final l.e.s. the arrows should actually go the other way, which is slightly less impressive. Even so...

  • $\begingroup$ I'm having trouble seeing why your first pair of sequences isn't arrow-reversing. It looks contravariant to me ... or was that the point? $\endgroup$ – S. Carnahan May 22 '10 at 18:55
  • $\begingroup$ What I meant was we have ... -> K -> L -> M -> ... mapping to ... -> PD(K) -> PD(L) -> PD(M) -> ... rather than ... -> PD(M) -> PD(L) -> PD(K) -> ... $\endgroup$ – Saul Glasman May 23 '10 at 10:12

One of the standard proofs of Poincaré duality, at least for those manifolds that have handle decompositions, provides a reason for some of these naturality properties. Every piecewise linear manifold, or every smooth manifold, has a handle decomposition, and many but not all topological manifolds also do. (Amazingly enough, the only exceptions are in 4 dimensions.) A handle decomposition gives rise to two different CW cellulations on the manifold, one using cores and the other using co-cores. Then this proof of Poincaré duality posits that the CW chain complex of one cellulation is identical to the CW cochain complex of the other cellulation.

You can extend this coincidence of chain complexes to both of your examples, the Mayer-Vietoris sequence and the exact sequence of a pair. Obtaining identical chain complexes also gives you other information, for instance that the Bockstein maps are the same.

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  • $\begingroup$ I'd never heard/read «cellulation». Is that your coinage? $\endgroup$ – Mariano Suárez-Álvarez May 22 '10 at 20:15
  • $\begingroup$ I heard that term from Mike Freedman. It sounded good to me. $\endgroup$ – Greg Kuperberg May 22 '10 at 20:19
  • $\begingroup$ This is very interesting; I have seen this proof of PD, but I didn't imagine it would lie behind the phenomena I asked about. Does this mean that similar things don't happen for other cohomology theories? $\endgroup$ – Saul Glasman May 23 '10 at 10:14
  • $\begingroup$ I don't know. I would suppose that similar things could happen for other cohomology theories, if you figured out how to compute them from a CW complex. The CW chain complex is really a spectral sequence in disguise. I suppose that that spectral sequence always exists, I just don't know what it does for generalized cohomology theories. It should also be said that this CW model is an explanation for your question, but probably not the explanation. $\endgroup$ – Greg Kuperberg May 23 '10 at 16:09

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