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Can every topological (not necessarily smooth or PL) manifold be given the structure of a CW complex?

I'm pretty sure that the answer is yes. However, I have not managed to find a reference for this.

flag	asked Aug 27 2010 at 3:48 A grad student 81●3●4

@algori : I thought you had posted an (important sounding) comment? Why did you delete it? – A grad student Aug 27 2010 at 4:48

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It turns out that my first comment was a bit wrong. Here are the slides of A. Ranicki's talk in Orsay. www.maths.ed.ac.uk/~aar/slides/orsay.pdf It says on p. 5 there that a compact manifold of dimension other than 4 is a CW complex. There is a related conjecture that says that each closed manifold of dimension $\geq 5$ is homeomorphic to a polyhedron (there are 4-manifolds for which this is false). See arxiv.org/pdf/math/0212297. I'm not sure what if anything is known about the noncompact case. – algori Aug 27 2010 at 4:50

Update: recent work of Davis, Fowler, and Lafont front.math.ucdavis.edu/1304.3730 shows that in every dimension ≥6 there exists a closed aspherical manifold that is not homeomorphic to a simplicial complex. – Lee Mosher May 1 at 16:10

Ryan Budney · Answer 1 · 2010-08-27 04:04:25Z UTC

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Kirby and Siebenmann's paper "On the triangulation of manifolds and the Hauptvermutung" Bull AMS 75 (1969) is the standard reference for this, I believe.

The result is that compact topological manifolds have the homotopy-type of CW-complexes, to be precise.

answered Aug 27 2010 at 4:04

Ryan Budney
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I think the fact that they have the homotopy type of a CW complex is due to Milnor (it is in his paper about spaces homotopy equivalent to CW complexes). Do Kirby-Siebenmann just prove this, or do they prove that all compact manifolds are homeomorphic to CW complexes? Also, how about the noncompact case? – A grad student Aug 27 2010 at 4:08

But I thought the question was whether each has the "homeomorphism type" of a CW complex. – Dev Sinha Aug 27 2010 at 4:22

It's been a while since I've looked at that Milnor paper -- I suspect maybe he's arguing that manifolds have the homotopy-type of countable CWs, while Kirby-Siebenmann deal with compact manifolds and finite CWs. ? – Ryan Budney Aug 27 2010 at 4:23

@Ryan : Yes, I think that is what Milnor proved (it's also been a long time since I looked at it). – A grad student Aug 27 2010 at 4:27

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@Ryan, the open problem is not whether any compact manifold is homeomorphic to a CW complex (this was proved by Kirby-Siebenmann). The open problem is whether it has a (non-combinatorial) triangulation. @grad student, whatever is known in the noncompact case must be Kirby-Siebenmann's book. – Igor Belegradek Aug 27 2010 at 13:15

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Mariano Suárez-Alvarez · Answer 2 · 2010-08-27 03:53:01Z UTC

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See http://arxiv.org/abs/math/0609665

answered Aug 27 2010 at 3:53

Mariano Suárez-Alvarez
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That manifold isn't 2nd countable. Like most mathematicians, I only care about manifolds that are Hausdorff and 2nd countable. – A grad student Aug 27 2010 at 3:56

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I hope that the fact that you only care about those does not preclude you from enjoying learning about the rest. – Mariano Suárez-Alvarez Aug 27 2010 at 4:12

Are non-PL manifolds CW-complexes?

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