Geometric difference between a homotopy equivalance and a homeomorphism

In summary: Two homotopy equivalent compact manifolds without boundary that are not homeomorphic to each other.There are many such examples. The simplest is probably the sphere and the torus (which is a product of two circles). They are both compact, without boundary, and have the same homotopy type (they are both simply connected), but they are not homeomorphic.Another example is the Klein bottle and the real projective plane. They are both non-orientable compact manifolds without boundary and have the same homotopy type, but they are not homeomorphic. In summary, a homotopy equivalence between two spaces means that they can be continuously deformed into each other, but it does not necessarily mean they are
  • #36
Not to go too-far off-topic, but maybe a reasonable meaning for " x being naturally-occuring" is that one is somewhat likely to either run into x or hear about it while doing research that is not too wildly unusual.

Yeah, and I forgot to say the closed (orientable?) surfaces are Eilenberg Maclane. And aspherical manifolds of all sorts, since those are manifolds with vanishing higher homotopy groups (which came up in my 3-manifold readings).
 
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  • #37
The classifying spaces for flat bundles,bundles with discrete structure group, are all EMs. For finite groups these are all probably all infinite dimensional CW complexes. For instance the classifying space for Z2 bundles is the infinite real projective space.

An example of a Z2 bundle is the tangent bundle of the Klein bottle (itself an EM space).
It follows that the classifying map into the infinite Grassmann of 2 planes in Euclidean space can be factored through the infinite projective space. (I wonder though whether it can actually be factored through the two dimensional projective plane by following a ramified cover of the sphere by a torus with the antipodal map.)

In terms of group cohomology this corresponds to the projection map,

[itex]\pi_{1}[/itex](K) -> Z2 obtained by modding out the maximal two dimensional lattice.

Group cohomology is the same as the cohomology of the universal classifying space for vector bundles with that structure group - I think

In the case of the flat Klein bottle, this shows that its holonomy group is Z2.
 
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  • #38
mathwonk said:
note that even an exotic sphere has a morse function with just two critical points. so when you attach that disc you cannot always attach it differentiably in the usual way. In fact this is apparently how one produces exotic spheres. you produce a manifold that is not an ordinary smooth sphere somehow, but that does have a morse function with only two critical points. then it is homeomorphic to a sphere.

reference?
 
  • #39
The classifying spaces for flat bundles,bundles with discrete structure group, are all EMs.

Well, what I was saying is that most of those aren't some familiar space that has a name, like S^1.

For finite groups these are all probably all infinite dimensional CW complexes. For instance the classifying space for Z2 bundles is the infinite real projective space.

Not always. For example, for surface groups, as we just mentioned. And there are some aspherical manifolds that are finite-dimensional. This includes, for example, all hyperbolic 3-manifolds.

http://en.wikipedia.org/wiki/Aspherical_space
Group cohomology is the same as the cohomology of the universal classifying space for vector bundles with that structure group - I think

I would say principal bundles, rather than vector bundles. I would prefer to say the cohomology of G is the cohomology of a K(G,1). That's my favorite definition, and of course, it's equivalent to other definitions of group cohomology, like the one as a derived functor or from the bar resolution of Z over ZG.
 
  • #40
homeomorphic said:
Not always. For example, for surface groups, as we just mentioned. And there are some aspherical manifolds that are finite-dimensional. This includes, for example, all hyperbolic 3-manifolds.

http://en.wikipedia.org/wiki/Aspherical_space


I think that the fundamental groups of ashperical manifolds are infinite e.g. tori. The fundamental groups of closed orientable surfaces are infinite except for the sphere.

I would say principal bundles, rather than vector bundles. I would prefer to say the cohomology of G is the cohomology of a K(G,1). That's my favorite definition, and of course, it's equivalent to other definitions of group cohomology, like the one as a derived functor or from the bar resolution of Z over ZG.

Same thing I think.

For aspherical manifolds the fundamental domain in the universal covering space generates a free resolution of the integers over the fundamental group. e.g. for a 2 dimensional torus one has four vertices and edjes and one rectange as a basis over ZxZ.
 
  • #41
mathwonk said:
Lavinia, have you read the classical reference, Milnor's Morse Theory? This is Thm. 3.5 proved in roughly the first 24 pages. He proves that the region on the manifold "below" c+e where c is a critical value, has the homotopy type of the region below c-e with the attachment of a single cell determined by the index of the critical point with value c.

The passage from local to global you ask about may be Lemma 3.7. In it he proves that a homotopy equivalence between two spaces extends to one between the spaces obtained from them by attaching a cell.

He makes use of deformation retractions and ultimately uses Whitehead's theorem that a map is a homotopy equivalence if it induces isomorphism on homotopy groups, at least for spaces dominated by CW complexes.

Thanks again Mathwonk. I just browsed through the first chapter. The lemmas you mention do the trick.

So what are two homotopy equivalent compact manifold without boundary that are not homeomorphic?
 
  • #44
to be explicit:

"In particular, the lens spaces L(7,1) and L(7,2) give examples of two 3-manifolds that are homotopy equivalent but not homeomorphic."
 
  • #45
I think that the fundamental groups of ashperical manifolds are infinite e.g. tori. The fundamental groups of closed orientable surfaces are infinite except for the sphere.

True, I sometimes confuse finite with finitely generated. That's why I was confused. Yeah, I think if you even have a subgroup of finite order, you have to have an infinite-dimensional complex. The proof was really cool, but I'll have to try and remember it. There was a covering-spaces proof.
 
  • #46
mathwonk said:
to be explicit:

"In particular, the lens spaces L(7,1) and L(7,2) give examples of two 3-manifolds that are homotopy equivalent but not homeomorphic."

pretty cool. So how do their Morse functions distinguish them?
 

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