From Hatcher, "Algebraic Topology," Chapter 2, "Singular Homology" section (p. 108-109 in my copy):

Cycles in singular homology are defined algebraically, but they can be given a somewhat more geometric interpretation in terms of maps from finite $\Delta$-complexes. To see this, note that a singular $n$-chain $\xi$ can always be written in the form $\sum_i \varepsilon_i \sigma_i$ with $\varepsilon_i = \pm 1$, allowing repetitions of the singular $n$-simplices $\sigma_i$. Given such an $n$-chain $\xi = \sum_i \varepsilon_i \sigma_i$, when we compute $\partial \xi$ as a sum of singular $(n-1)$-simplices with signs $\pm 1$, there may be some canceling pairs consisting of two identical singular $(n-1)$-simplices with opposite signs. Choosing a maximal collection of canceling pairs, construct an $n$-dimensional $\Delta$ complex $K_\xi$ from a disjoint union of $n$-simplices, one for each $\sigma_i$, by identifying the pairs of $(n-1)$-dimensional faces corresponding to the chosen canceling pairs. The $\sigma_i$'s then induce a map $K_\xi \rightarrow X$. If $\xi$ is a cycle, all the $(n-1)$ simplices of $K_\xi$ come from canceling pairs, hence are faces of exactly two $n$-simplices of $K_\xi$. Thus $K_\xi$ is a manifold, locally homeomorphic to $\mathbb{R}^n$, except at a subcomplex of dimension at most $n - 2$.

It's this last bit that has me confused, perhaps because I'm unable to visualize a nontrivial 3-manifold. We're building a $\Delta$-complex by identifying the $(n-1)$-faces of simplices; there are only finitely many such simplices; we're not required to identify the faces in any unusual way; and so forth. How, then, can we end up with something which is not a manifold? I'd really appreciate an explicit example.

I've been told that this is a general example of a "pseudomanifold," but all the examples of pseudomanifolds that I've been able to locate and follow wind up making a non-manifold by essentially identifying vertices to get pinched spaces. This can't happen under the present construction because we're always identifying the largest proper faces. So I'm quite confused by the situation.

EDIT: Looks like there are some more questions raised in the comments as to what the construction actually means. If there's a name for this construction, or any other source that discusses it, I'd appreciate a reference. And of course if anyone can shed any additional light on the topic that would be quite welcome too.

EDIT: I may have inadvertently overwritten someone else's edit just then, judging by a message that popped up. Apologies if so. (How do I tell / fix this?)

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    $\begingroup$ I don't want to deter you from asking questions at all, I just want to mention that I would not worry too much about this, unless you are particularly interested in triangulations and what not coming from elements of the homology groups of a space. $\endgroup$ – Sean Tilson Feb 14 '11 at 4:55
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    $\begingroup$ What concerns me is not so much that I can't come up with an example, but that it appears to be trivial to me to prove that such examples don't exist. This means that there's something seriously wrong with my understanding of either (1) the construction or (2) the subject matter (or most likely both). In the former case I'm of course willing to let it go; in the latter, not so much... $\endgroup$ – Daniel McLaury Feb 14 '11 at 5:00

First, a remark that I think is relevant: understanding the construction $K_{\xi}$ is related to the question of whether a homology class can be represented by a submanifold. This is not always possible; see e.g. this MO answer.

On the other hand, it is always possible for $n$-cycles for small values of $n$ (if we are considering the homology of a manifold), which may be one reason why it is hard to visualize non-manifold examples of $K_{\xi}$.

E.g. in the case $n = 2$, we are gluing triangles along their edges, with the rule that exactly two triangles meet along a common edge. This always gives a closed $2$-manifold. In general, unless I am confused, the fact that for $n = 2$ we always get a manifold implies in general that the non-manifold points are in codimension at least $n - 3$ (not just $n - 2$). So to find a counterexample we should take $n = 3$. I will give such a counterexample in a moment, but first let me describe a general approach to thinking about this situation.

When gluing simplices like this, the way to investigate whether the resulting object is a manifold or not is to consider the link of each vertex. Namely, if a bunch of $n$-simplices meet at a vertex $v$, take a transverse slice to each simplex just below the vertex (I am imagining that simplex is sitting on the face opposite to $v$, so that $v$ is at the top of the simplex), to get an $n-1$-simplex. Now these glue together to make a closed simplicial $n-1$-complex that surrounds $v$; this is the link of $v$. Note that these $n-1$-simplices meet along $n-2$-dimensional faces, and so the link is a $K_{\xi}$-type object, but of one dimension less.

In the case that $n = 2$, we get a bunch of segments being joined at their vertices, and hence a circle. No drama there.

But now suppose that $n = 3$. Then the link is formed by a bunch of triangles gluing together, and we have agreed that this gives a $2$-manifold. But which one?

If $K_{\xi}$ is locally Euclidean at $v$, then this surface would have to be a $2$-sphere. But a priori it doesn't have to be, and so in this case we can get a non-manifold example!

In practice, to make an example we should take a cone on a surface that is not a $2$-sphere, e.g. a cone on a $2$-torus.

Precisely: triangulate a $2$-torus, e.g. by choosing two triangles $\Delta_1$ and $\Delta_2$, and identifying the three edges of the first with the three edges of the second in the appropriate manner.

Now to form the cone on this, replace $\Delta_i$ by a $3$-simplex $\tilde{\Delta_i}$. Regard $\Delta_i$ as one of the faces of $\tilde{\Delta}_i$, and label the other three faces according to the edge along which they meet $\Delta_i$. Now glue the three faces of $\tilde{\Delta}_1$ other than $\Delta_1$ with the three faces of $\tilde{\Delta}_2$ other than $\Delta_2$ according to the same gluing scheme we used previously to construct the $2$-torus. What we end up with is a three dimensional simplicial complex whose boundary is equal to $\Delta_1 + \Delta_2$, i.e. a $2$-torus. But it is not a $3$-manifold; rather it is a cone on the $2$-torus.

If we let $\xi$ be the $3$-chain $\tilde{\Delta}_1 + \tilde{\Delta}_2$, then $K_{\xi}$ is the cone on the $2$-torus, and so is not a $3$-manifold.

Of course, this $\xi$ is not a cycle (so the boundary of $K_{\xi}$ is non-empty), but we could take two such cones and glue them along their common $2$-torus boundary to get a three-dimensional simplicial complex without boundary, and then take $\xi$ to be the sum of the four $3$-simplices involved; then $\xi$ would be a cycle, but $K_{\xi}$ would not be a manifold; it has two vertices whose links are $2$-tori rather than $2$-spheres.

Concluding remark: In dimension $2$, the only way to get pseudo-manifolds that are not manifolds is to explicitly glue together vertices, as you observed. But in dimension $3$ or higher, there are other examples, e.g. in dimension $3$ we can take cones on positive genus surfaces, as in the preceding construction.

  • $\begingroup$ You were right about my answer :) It's been deleted. Wonderful answer on your part! $\endgroup$ – Dylan Wilson Feb 14 '11 at 6:17
  • $\begingroup$ What about the pinched torus? Take a triangle and glue two of its edges together to get a cone (who says the two simplices touching an edge had to be different?). Glue two of these cones at the cone point. Now glue on a (triangulated) cylinder and you have a two-dimensional orientable pseudomanifold that's not a manifold. This is the example given on wikipedia; one could argue about whether the definition of pseudomanifold should be adjusted, but this example seems great for answering the question. $\endgroup$ – Dan Ramras Feb 14 '11 at 6:22
  • $\begingroup$ @Dan: Dear Dan, I agree that we can pinch a torus, but in this construction we identify two vertices, without identifying any edges on which they lie. So this cannot come up as an example of a $K_{\xi}$. My point was not to argue with the definition of pseudo-manifold, but just to address the OP's comment about pseudo-manifolds, namely: that the OP's observation that non-manifold pseudo-manifolds seem to arise only by pinching (i.e. identifying vertices) is correct in dimension $2$, but not in higher dimensions (where there are more subtle obstructions to being a manifold). Regards, $\endgroup$ – Matt E Feb 14 '11 at 6:27
  • $\begingroup$ To be a little clearer, taking $X$ as the pinched torus and $\xi$ as the fundamental class generating $H_2(X; \mathbb{Z})$, won't Hatcher's construction give $K_\xi = X$ and the identity map $X\to X$ realizing this cycle $\xi$? $\endgroup$ – Dan Ramras Feb 14 '11 at 6:29
  • $\begingroup$ I added my second comment too quickly; I think I finally see the point... $\endgroup$ – Dan Ramras Feb 14 '11 at 6:30

Edit: The example below doesn't give a cycle, so it's not really what we're looking for here.

Original post: The simplest example of what you're looking for is the dunce cap.

It's a compact 2-dimensional complex, and it's contractible, so it's definitely not a manifold w/o boundary.

  • $\begingroup$ Actually you seem to be aware of such examples (you mention pinching vertices); is there some reason you think this doesn't fit Hatcher's discussion? $\endgroup$ – Dan Ramras Feb 14 '11 at 0:50
  • $\begingroup$ Can you explain further how to get the dunce cap as a $K_\xi$? I thought you couldn't get anything of that sort, since it seems to require you to glue together three different edges, whereas we can only identify edges pairwise in a $K_\xi$. $\endgroup$ – Daniel McLaury Feb 14 '11 at 1:24
  • $\begingroup$ What if you start with the dunce cap itself as your delta complex X, and consider the 2-chain formed by taking one copy of the 2-simplex? I think you'll end up having the map $K_\xi \to X$ simply be the identity map on the dunce cap. It's true that you identify edges in pairs, but there's no guarantee in general that two of these identifications won't involve the same edges (as happens in this example). Thinking more carefully through this example, I have to wonder a bit about exactly what Hatcher means by a maximal collection of canceling pairs. $\endgroup$ – Dan Ramras Feb 14 '11 at 1:36
  • $\begingroup$ I think in the dunce cap example, the three faces of the singular 2-simplex I'm talking about are all the same as maps from [0,1] into the dunce cap. In the formula for the boundary, one of these (the "middle" one) appears with a minus sign, so I think Hatcher intends to have two "canceling pairs," with the middle edge appearing in both. Then when you build $K_\xi$ you'll get the dunce cap back (note that Hatcher isn't explicit about how you identify two canceling edges; you need to make sure you're using the correct convention... $\endgroup$ – Dan Ramras Feb 14 '11 at 1:41
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    $\begingroup$ @Dan- they are distinct in K_\xi, though they may map to the same thing in $X$, I think. $\endgroup$ – Dylan Wilson Feb 14 '11 at 4:42

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