1) A $n$-dimensional homology manifold is a topological space $X$ such that for any $x\in X$, the homology groups $$H_p(X,X-x,\mathbb{Z})$$ are trivial unless $p=n$ where $$H_n(X,X-x,\mathbb{Z})\cong \mathbb{Z}$$

2) A $n$-dimensional pseudomanifold is a topological space together with a triangulation such that

  • each simplex is the face of a $n$-simplex,
  • every $(n – 1)$-simplex is a face of exactly two $n$-simplices for $n > 1$,
  • any two -dimensional simplices can be joined by a "chain" of n-dimensional simplices in which each pair of neighbouring simplices have a common $(n-1)$-dimensional face.

I am searching for examples of pseudomanifolds that satisfy Poincaré duality but that are not homeomorphic to homology manifolds. I would like to know if there exist examples of complex projective varieties that satisfy these properties: satisfy poincaré duality but are not homology manifolds.

Edit (Ben Wieland's example): Let $M^3$ be a manifold and let us consider $[0,1]\subset M$, glue the interval to itself by an involution $f$ that reverses the endpoints. Then we get a pseudomanifold $M'$ whose homotopy type is the same as $M$ then it satisfies Poincaré duality but it is not a homology manifold, you can look at the local homology of the equivalence class $[0]\in M'$ of $0\in [0,1]\subset M$. This is due to the fact that the link of $[0]$ in $M'$ is not connected ($M'$ is not normal).

The normalization of $M'$ is $M$ itself, thus I would like to know an example where the pseudomanifold is normal.

  • $\begingroup$ Your question is difficult to read. You say "and by classical I mean" before making any reference to classical things. You have undefined terminology, too. You have Poincare Duality Spaces in the title but not in the body, etc. It's probably best to assume people haven't seen the classical terminology as it's rarely used. $\endgroup$ Mar 20, 2015 at 18:05
  • $\begingroup$ For the definition of a homology manifold don't you also need the condition that when $p = n$ the homology group is $\mathbb{Z}$? Otherwise the "kissing banana" ($S^2$ with two points identified) would be a homology manifold. $\endgroup$ Mar 20, 2015 at 19:18
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    $\begingroup$ Dear Chris the Kissing banana does not satisfy Poincaré duality. Let us consider the cap product with a non-trivial top class $B\in H_2$ it induces the zero map between $H^1$ and $H_1$. $\endgroup$
    – David C
    Mar 20, 2015 at 19:57
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    $\begingroup$ How much do you care about the complex hypothesis? Here is a non-complex example: take a manifold of dimension at least 3, embed an interval, and glue the interval to itself by an involution that reverses the endpoints. $\endgroup$ Mar 22, 2015 at 18:29
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    $\begingroup$ My guess is that the semisimplicity of perverse sheaves shows that if the dualizing sheaf of a variety is not a local system (ie, if it is not a homology manifold), then it does not satisfy duality. At least with char zero coefficients: $\mathbb Q$-PD $\implies$ $\mathbb Q$-homology manifold. . . . Normal: good question. $\endgroup$ Mar 22, 2015 at 21:27

1 Answer 1


1. Introduction
2. Easy example
3. Normality, Duality
4. Normal example


The pseudomanifold and homology manifold conditions are both local conditions, while Poincaré duality is a global condition. It is possible for a pseudomanifold to fail the homology manifold conditions in several places, but so that the local deviations globally cancel, so it retains duality.

Simplicial complexes are "locally cone-like": every point $x$ has a neighborhood that is the cone on another space $L$, the link, with $x$ the cone point. Then $$H_*(X,X-\{x\})=H_*(CL,L\times I)=\tilde H_*(SL)=\tilde H_{*-1}(L)$$ So the homology manifold condition is that the links have the homology of spheres.

Easy example

The interval admits an involution reversing the endpoints. The quotient by this involution is again contractible. Think of this quotient operation as gluing one half of the interval to the other half. Embed the interval in an $n$-manifold. Form a quotient of the manifold by gluing the one half of the interval to the other half. If $n\geqslant 3$, this leaves the cells in degree $n$ and $n-1$ unchanged, so does not affect the pseudomanifold condition. Since we have replaced one contractible subspace by another, the homotopy type of the total space is unchanged and satisfies duality (with the pseudomanifold fundamental class, etc). But the the links are no longer spheres at any point along the interval. At the image of the ends of the interval, the link is two spheres glued in one point. At general points along the interval, it is two spheres glued in two points. At the image of the midpoint, the fixed point of the involution, the link is a single sphere with two of its points glued to each other.

Normality, Duality

That example involved changing a manifold in low dimensions, which doesn't violate the high dimensional pseudomanifold condition. To rule out such examples, one has the concept of a normal pseudomanifold (cf normal variety). The normalization of a pseudomanifold is produced by taking a disjoint union of $n$-simplices parameterized by those of the original space and gluing them along their $n-1$-faces, according to how the original was glued. Thus the normalization of the above example is the unmodified manifold. A normal pseudomanifold is one isomorphic to its normalization.

It may be useful to use the Verdier dualizing sheaf. The cohomology with coefficients in the dualizing sheaf matches the homology: $H^*(X; D)\cong H_*(X)$. An oriented pseudomanifold yields a map of sheaves $\mathbb Z\to D$. We seek Poincaré duality, that is, $H^*(X; \mathbb Z)\cong H_*(X)$. So we seek spaces where the map of sheaves induces an isomorphism on cohomology $H^*(X; \mathbb Z)\cong H^*(X; D)$, even though it is not an isomorphism of sheaves. That is, we seek the cone to be a nontrivial sheaf with no cohomology in any dimension. There are such sheaves, such as a local system supported on a circle with appropriate monodromy. That motivates the construction. It also gives immediate proofs of the claims, but it should not be hard to check them without using sheaves.

Normal example

Take an $n$-manifold manifold $M$ and an automorphism $\phi$ so that the action of on the homology in intermediate degrees is sufficiently mixing so that if we consider it an action of the group $\mathbb Z$ and take group homology with those coefficients, the homology is trivial. For example, $M=T^2$ and $\phi=\left(\begin{matrix}2&1\\1&1\\ \end{matrix}\right)$. Then form the mapping torus ($T_\phi=M\times I/(x,0)\sim(\phi(x),1)$). This is an $M$-bundle over the circle, homology computed by a spectral sequence $H_*(S^1;H_*(M))\Rightarrow H_*(T_\phi)$. By assumption on $\phi$, the spectral sequence degenerates and $H_*(T_\phi)=H_*(S^1\times S^n)$. Then cone off each copy of $M$, forming a circle of cone points. To put it another way: form the mapping cylinder of the map to the circle $T_\phi\to S^1$. Yet another way: the mapping torus of the self-map of the cone $\tilde\phi\colon CM\to CM$. This space is a normal pseudomanifold (with boundary) because those properties are preserved by cones and products. Its set of singular points is a circle, and the dualizing sheaf twists about it with the prescribed monodromy, so it contributes nothing to the sheaf cohomology. If you prefer a closed example, double the space along the boundary (or equivalently take the double mapping cylinder of the map to the circle; or the mapping torus of the automorphism of the suspension). This gives a pseudomanifold that is not a homology manifold, but which satisfies Poincaré duality. Just as the mapping torus had the homology of a manifold $S^1\times S^n$, this space has the homology of $S^1\times S^{n+1}$. Actually, that is only correct if we restrict to trivial coefficients. If we define Poincaré duality to be for all local systems, then this space fails, for some unwind the twist around the singular circles. However, we can eliminate them by killing the fundamental group by surgery. That is, cut out a neighborhood of a circle, $S^1\times D^{n+1}$, leaving a boundary $S^1\times S^n$ and fill it in with $D^2\times S^n$. Now local systems are trivial, so it satisfies full Poincaré duality.

  • $\begingroup$ I continue to doubt that there are any algebraic examples, but this one has a nice dualizing sheaf, so my earlier suggestion was inadequate to eliminate them. $\endgroup$ Mar 27, 2015 at 3:36
  • $\begingroup$ Also, the third bullet point in David's definition of pseudomanifold is common, but I think it's a silly axiom. It amounts to the normalization being connected. I guess the point is to reduce the number of orientations, just as a connected manifold has at most one. Connectedness hypotheses are usually a bad idea, but this one is also unwieldy. $\endgroup$ Mar 27, 2015 at 3:52
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    $\begingroup$ @DavidC maybe this example can be made algebraic. Replace the the hyberbolic automorphism with the order 6 (not 2,3,4) automorphism of the right elliptic curve. And replace the cone on an elliptic curve with an affine cone. The link of its singularity is not the elliptic curve, but the $S^1$ bundle over it. And replace the circle with a complete variety, say, another elliptic curve. But the surgery is hopeless, so twisted coefficients are out. $\endgroup$ Mar 28, 2015 at 17:17
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    $\begingroup$ (but if you are only interested in homology with $\mathbb Q$ coefficients, 2,3,4 are fine) $\endgroup$ Mar 28, 2015 at 20:58
  • $\begingroup$ Here's a modification of the algebraic example that feels less "finite." You can interpret the elliptic curve example saying that elliptic curves have a canonical polarization, so the canonical bundle of curves on $M_{1,1}$ has a projective structure, thus there is an associated bundle of affine cones. Same for $M_g$. But there are more interesting complete curves in $M_g$, so those support a bundle of affine cones, so the singularities twist more. $\endgroup$ Mar 31, 2015 at 0:32

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