# Poincare dual simplicial structure of complexes homotopy equivalent to manifolds

Given a closed $$n$$-manifold $$M$$, Poincare duality equips us with an isomorphism: $$H_k(M)\cong H^{n-k}(M)$$ Here I'm speaking of singular homology with coeffecient in $$\mathbb{Z}_2$$.

Suppose now $$M$$ has a triangulation $$K$$. Then clearly, we have a similar Poincare isomorphism as above for $$K$$ as well. In fact, even more can be said: there exists a polyhedral dual structure $$K^{\vee}$$ of $$K$$ which gives an isomorphism already at the level of (co)chain complexes: $$C_k(K^{\vee})\cong C^{n-k}(K)$$ (in order to be safe let's further assume we are considering smooth manifolds so stuff as https://mathoverflow.net/questions/194297/dual-cell-structures-on-manifolds do not occur)

My question is:

Is there an isomorphism of the form $$C_k(K^{\vee})\cong C^{n-k}(K)$$ even when $$K$$ is not necessarily a triangulation of $$M$$, but instead only homotopy equivalent $$M$$

I'll remark, that I'm in fact interested in Poincare-Lefschetz duality (i.e for manifolds with boundary) but thought it is better to start from here. Furthermore, I'm in particular interested in the case where $$K$$ is the nerve of some "good cover" of $$M$$ (and thus homotopy equivalent from the nerve lemma)

Your question is not very precise, but I would say the answer is no. Consider $$M$$ to be the $$0$$-manifold given by the singleton. Take the simplicial complex given by the simplest triangulation of $$[0,1]$$ (two vertices, one edge), which satisfies $$M \simeq [0,1]$$. Then the cellular complex of $$K$$ is $$C_0(K) = \mathbb{F}_2^2$$ and $$C_1(K) = \mathbb{F}_2$$. There is no triangulation $$K^\vee$$ such that $$C^0(K^\vee) = \mathbb{F}_2^2$$ and $$C^{-1}(K^\vee) = \mathbb{F}_2$$.
• I agree that my question is not very precise.. I wasn't aware of how problematic it is to construct a dual polyhedral structure to some abstract simplicial complex. Even for a triangulation of a manifold with boundary there are subtleties to deal with. I assume that's your point there right? But then I must ask, in your answer what do you mean by $K^\vee$? do you mean some standard construction of a dual complex for a triangulation of a manifold with boundary, one which gives a Lefschetz type duality? – Itamar Vigi Apr 2 at 13:49
• @ItamarVigi For a manifold with boundary, in general you would want an isomorphism $C_*(K) \cong C^{n-*}(K^\vee, \partial K^\vee)$ (this is Poincaré–Lefschetz duality). Here in my answer, $K^\vee$ is just a notation: what I am saying is that there is no cell complex $X$ such that $C^{-1}(X) = \mathbb{F}_2$. There is never anything in negative degrees. – Najib Idrissi Apr 2 at 13:51