So far the sources I read on Poincare duality applies to manifold without boundary (eg https://en.m.wikipedia.org/wiki/Poincar%C3%A9_duality).

Does Poincare duality always hold for simplicial complexes? I read that “a simplicial complex is a manifold if the links of all vertices are simplicial spheres.” What if this condition is not met, will Poincare duality fail?



1 Answer 1


Yes it will fail. Indeed, let $X$ a simplicial complex homeomorphic to the wedge of two spheres (for example two pyramids with a commun vertex). We have $H_2(X) \cong \Bbb Z^2$ and $H_0(X) \cong \Bbb Z$. Notice that the link at the singular point is $S^1 \sqcup S^1$.

Poincaré duality also fails for a non-orientable space (or you have to take coefficient in $\Bbb Z/2 \Bbb Z$). For compact orientable simplicial complex you can still have a duality but with a different homology theory, called intersection homology theory.

Roughly speaking, Poincaré duality fails because of the non-transversal intersection, so if you want Poincaré duality to holds you should allows only cycles which intersect "transversally enough".

  • $\begingroup$ Thanks. Can you explain slightly more how it fails? What is the cohomology for your first example? $\endgroup$
    – yoyostein
    Dec 11, 2017 at 12:57
  • 1
    $\begingroup$ @yoyostein : sure ! By universal coefficient theorem we have e.g $H^0(X, \Bbb Q) \cong \text{Hom}(H_0(X, \Bbb Q), \Bbb Q) \cong \Bbb Q$. So $H^0(X, \Bbb Q)$ can't be isomorphic to $H_2(X, \Bbb Q)$. An excellent reference for Poincaré Duality is the book by Kirk and Davis (which shows a more general Poincaré duality, working for non-compact and orientable manifolds). If you are interested in singularities an excellent reference is "An Introduction to Intersection Homology Theory" by Kirwan and Woolf, or "Intersection homology" by McPherson and Goresky (the original paper). $\endgroup$ Dec 11, 2017 at 16:52

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