When is a finite cw-complex a compact topological manifold? - MathOverflow

When is a finite cw-complex a compact topological manifold?

2012-02-18T17:01:06Z

I think the statement of the question is pretty straightforward. Given a finite $n$-dimensional CW complex, are there necessary and sufficient conditions for determining that it is also a compact $n$-dimensional topological manifold (possibly with boundary)? Most of the questions I have found are about the converse, "When is a topological manifold a CW-complex?" so I thought it would be useful to consider the other side of the picture.

First of all, we already have 2nd countable and Hausdorff so we only need to determine when it is locally Euclidean. One necessary condition is that every point must be contained in the closure of at least one $n$-cell. One the other hand, a point can't be in too many $n$-cells, as the wedge of spheres is not a manifold. But these criteria are clearly not sufficient (or very precise: what does "too many" mean exactly?).

Another necessary condition that I think of right away is that the space must satisfy Poincare duality (with $\mathbb{Z}/2$ coefficients). I'm not sure if this is practical at all, but maybe it is useful if you are working with an explicit cell-structure and a concrete description of the cellular chain complex.

After that, I don't know how to proceed. I am assuming this is a difficult problem, since Google searches haven't answered my question yet. One idea that I have is to try and come up with an "obstruction" to this, where cell-by-cell we determine if every point in that cell admits an $n$-dimensional Euclidean nhd, and we have that the CW-complex is a manifold iff this "obstruction," computed at the homology/cohomology level, vanishes.

Any suggestions (or references to a solution) will be appreciated.

Answer by Adam for When is a finite cw-complex a compact topological manifold?

2012-02-18T19:42:35Z

Here is one necessary condition for the space being locally Euclidean: If $X$ is an $n$-dimensional manifold, then for every $p\in X$ the relative homology groups $H_*(X,X\setminus p;\mathbb{Z})$ are trivial except in dimension $n$ where they are equal to $\mathbb{Z}$. It's an easy check that this detects some of the obvious types of singularities. For example, if $p$ is the singular point in $X$, the wedge of two $S^n$s, then $H_n(X,X\setminus p;\mathbb{Z})=\mathbb{Z}^2$.

I'm pretty sure it isn't a sufficient condition, however, since one should be able to cook up a singular space where a neighborhood of some point looks is a cone on a homology sphere.

Answer by JeffE for When is a finite cw-complex a compact topological manifold?

2012-02-18T20:08:46Z

Classical results of Andrei Markov [The insolubility of the problem of homeomorphy. Dokl. Akad. Nauk SSSR 123:978–980, 1958] imply that it is formally undecidable whether a given finite simplicial complex is a 6-manifold. So there are no simple necessary and sufficient conditions for manifoldness of CW complexes.

Answer by Greg Friedman for When is a finite cw-complex a compact topological manifold?

2012-02-18T20:09:57Z

In a certain sense, this is one of the starting questions for surgery theory, which aims to classify all manifolds, to the extent possible. You can get some idea of the complexity of this problem (including your problem) by reading some surveys about surgery theory. A good place to start might be Surveys on Surgery Theory, edited by Cappell, Ranicki, and Rosenberg, or even have a look at the Wikipedia article on surgery theory. In general, even when you know your CW complex is homotopy equivalent to a manifold it requires non-trivial work to show that it's homeomorphic to a manifold. For example, every finite CW complex can be embedded in Euclidean space and stably this yields a unique spherical fibration over the space, called the Spivak spherical fibration. If your space is a manifold, this should be the stable normal bundle (or its appropriate PL or topological analogues). So things like this start coming into play and it gets complicated pretty quickly.