Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
    
@algori : I thought you had posted an (important sounding) comment? Why did you delete it? –  A grad student Aug 27 '10 at 4:48
3  
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 '10 at 4:50
2  
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 '13 at 16:10
    
Hatcher's Algebraic Topology p. 529 has a paragraph answering this question very clearly for compact manifolds (not including results in 2013 of course). However his references are to two long dense books, without page specification. –  hsp Sep 3 '13 at 15:47

2 Answers 2

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.

share|improve this answer
    
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 '10 at 4:08
    
But I thought the question was whether each has the "homeomorphism type" of a CW complex. –  Dev Sinha Aug 27 '10 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 '10 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 '10 at 4:27
2  
@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 '10 at 13:15

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

share|improve this answer
5  
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 '10 at 3:56
14  
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 '10 at 4:12

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.