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Suppose $X^n$ is an orientable compact orbifold (without boundary) with stabilisers in codimension 2, and $\bar X^n$ is the underlying topological space. We can assume, moreover, that $X^n$ is a quotient of a manifold $X'^n$ by an action of finite group $G$.

Is it true that, for simplicial homologies of $\bar X^n$, we have $H_{n-k}(\bar X^n, \mathbb R)$ is dual to $H_k(\bar X^n,\mathbb R)$?

If not, what is a simplest counterexample, and what is the correct statement? If yes, what would be a reference?

PS. It seems to me that this should be true in the case when $X^n$ is a global quotient of a manifold by a finite group, because I guess in this case the simplicial homology of $X^n$ should be equal to the invariants of the action of $G$ on $H_k(X'^n)$. At the same time, actions on $H_k(X'^n)$ and $H_{n-k}(X'^n)$ are dual. Is this reasoning correct?

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Your proof for global quotients is correct, although the homology is more naturally the co-invariants, though these are isomorphic to the invariants by the transfer. The dual of the co-invariants is naturally the invariants of the dual. So to say that the co-invariants of two dual spaces are again dual, you need to adjust the pairing by the the order of the group, though people don't usually worry about such factors.

For the general case, a useful tool is homology manifolds. This is a local property which the underlying (or "coarse") space of an orbifold has with $\mathbb Q$ or $\mathbb R$ coefficients. The same argument you applied to the homology in the global quotient case applies to the local homology. Homology manifolds have Poincare duality. Unfortunately, the definition I linked in wikipedia is not quite correct. Your user name suggests that you might like sheaf theory. The correct definition is that a certain sheaf is constant; the incorrect definition checks only the stalks. In fact, this sheaf is the dualizing sheaf, though sheaves are overkill to proving PD for homology manifolds.

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  • $\begingroup$ Dear Ben, thank you for the answer! Would you suggest some text available online (apart from wiki, preferably some lecture notes or an article) where I would be able to read a bit about Homology manifolds? $\endgroup$
    – aglearner
    Commented Aug 20, 2011 at 8:26
  • $\begingroup$ Hey Ben, actually the statement on stalks implies that the dualising sheaf is the orientation sheaf, and hence Poincaré duality applies. One way of seeing on, say, a stratified and orientable psuedomanifold is that the condition on stalks implies that the constant sheaf is isomorphic to the intersection cohomology complex (by the unicity of the IC sheaf). Hence constant = IC = dualising because the IC (of the trivial local system) is self-dual. $\endgroup$
    – Geordie Williamson
    Commented Aug 20, 2011 at 22:00
  • $\begingroup$ I think Bredon's book Sheaf Theory talks about homology manifolds, but, again, sheaf theory is overkill. If you just want Poincare duality, look at proofs you know and try to generalize them. eg, I think the one in May's Short Course generalizes. $\endgroup$
    – Ben Wieland
    Commented Aug 23, 2011 at 3:30
  • $\begingroup$ @Geordie an example of what can go wrong is the mapping torus of a degree 2 map of a sphere. That is, glue the ends of $S\times I$ by a degree 2 map. The local cohomology is at every point the cohomology of a sphere. But the sheaf of local cohomology is not the constant sheaf $\mathbb Z$. The right thing to do is to check whether the restriction from a small open set to a point is an isomorphism, but it's actually multiplication by 2. Maybe these groups associated with points aren't actually stalks of everything. I think the relevant sheaf is loc sys $\mathbb Z[1/2]$ with monodromy 2. $\endgroup$
    – Ben Wieland
    Commented Nov 18, 2013 at 20:26
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Chapter 5 of "Orbifolds and stringy topology" by Adem, Leida, Ruan have a version of Poincare duality in the groupoid setting. This is probably a very general result on this.

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