Geometric difference between a homotopy equivalance and a homeomorphism

[mathwonk]

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.

 

[homeomorphic]

very cool. So just to get a feel for this, how do I construct a closed 4 manifold without boundary whose fundamental group is the fundamental group of the Klein bottle?

To make the CW complex that has the fundamental group of the Klein bottle, I guess you could use the Klein bottle itself, but the general construction is as follows. Start with a zero cell. Then, for each generator, attach a 1-cell to the 0-cell. So, two generators in this case. There is one relation, so that will be a word in the generators. Attach a 2-cell so that its boundary goes around each 1-cell according to that word. That will kill off that relation (and the least normal subgroup that contains it).

To do this in the case of manifolds, you can do something similar. You can think of a handle as a thickened up cell. If you want to build an n-manifold, you have to fatten up any k-cells you want to add, so that they are n-dimensional, so you cross them with an n-k-disk. That is a handle. Attaching maps get a little bit more complicated with handles than with CW complex because you need a framing of the core to figure out how to glue the thickened handle on. You won't run into any trouble getting any free group. Start with a 0-handle, which is a thickened 0-cell. Then, attach 1-handles to the boundary, one for each generator. Then, you want to add the 2-handles to kill the relations, as before. I guess you run into trouble here if you have a 3-manifold because the attaching spheres in that case are 1-dimensional and you trying to glue them along the boundary of the 3-manifold, which is a surface, so you don't have enough room to make it an embedding. I think that's the only problem you run into. So, this seems to work for 4-dimensions or higher. What you have at this point is a 4-manifold with boundary. I think you could take the dual handle-decomposition of the manifold and glue it back to itself, this time thinking of the handles as 3-handles and 4-handles. Just as adding a 3-cell will not touch the fundamental group (cellular approximation theorem), a 3-handle will not touch it. Same for 4-handles. Then, connect sum as many simply-connected guys as you want.

Also suppose you take a high dimensional lattice, say a 95 dimensional lattice (free abelian group). On the universal covering space of this manifold, this huge lattice must act on a simply connected 4 manifold. That would be interesting to see. Do you just add 95 copies of S^1xS^3 to the 4 sphere with 190 open balls removed?

The manifold itself will be 95 1-handles attached to a 4-ball, plus a bunch of relations that will make every thing commute. Then, dual handles, for each of those. I'm not sure about the universal cover. I'm getting tired.

 

[homeomorphic]

on these same lines I remember reading that Morse functions can be used to show that every smooth compact manifold without boundary is homotopy equivalent to a CW complex.

Why is this?

I would rephrase it this way. Morse functions can be used to show that any smooth manifold can be build out of handles. Just shrink all the handles down to their cores. Unthicken them. Then, you have a CW complex.

The idea is to look a Morse function on a cobordism with only one critical point, since you can reduce it to that case, by tweaking the Morse function slightly so that none of the critical points are at the same level, and then cutting everything out except one critical point at a time. What you get is a handle of index equal to the index of the critical point (which is the number of minus signs you get when you diagonalize the Hessian or the dimension of the stable manifold when you flow along a gradient-like vector field).

You could also appeal to Whitehead's theorem that smooth manifolds admit a triangulation. I think the idea there is that a smooth manifold admits a local triangulation because it looks locally like R^n, which has an obvious triangulation. Then, somehow, I think you can subdivide enough so that you can patch together all the triangulations. Something like that. I don't know exactly how it goes.

 

[homeomorphic]

Are these E-M spaces all CW-complexes spaces? manifolds?

I guess they have to be homotopy equivalent to CW complexes, since they can be realized as CW complexes, and they are unique up to homotopy equivalence. The way you construct them is what I have been talking about. Make a wedge of spheres to get generators of the non-zero homotopy group. By the Hurewicz theorem, it's isomorphic to the homology in that dimension because all the lower homotopy groups vanish. Then, just keep attaching cells to kill off all the homotopy groups above that dimension. This doesn't give you a very concrete construction, so in the end, you don't really know what you've built. But some Eilenberg-Maclane spaces occur in nature, so to speak, like ℝP^∞, ℂP^∞, or S^1. As far as I know, those are the only naturally occurring ones. Some asked for examples of them when I first encountered them in my algebraic topology class, and the prof didn't seem to know of any other than those few examples and the abstract construction of them. Of course, maybe the term "natural occurring" doesn't have too much meaning.

 

[WWGD]

I guess they have to be homotopy equivalent to CW complexes, since they can be realized as CW complexes, and they are unique up to homotopy equivalence. The way you construct them is what I have been talking about. Make a wedge of spheres to get generators of the non-zero homotopy group. By the Hurewicz theorem, it's isomorphic to the homology in that dimension because all the lower homotopy groups vanish. Then, just keep attaching cells to kill off all the homotopy groups above that dimension. This doesn't give you a very concrete construction, so in the end, you don't really know what you've built. But some Eilenberg-Maclane spaces occur in nature, so to speak, like ℝP^∞, ℂP^∞, or S^1. As far as I know, those are the only naturally occurring ones. Some asked for examples of them when I first encountered them in my algebraic topology class, and the prof didn't seem to know of any other than those few examples and the abstract construction of them. Of course, maybe the term "natural occurring" doesn't have too much meaning.

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.

 

[homeomorphic]

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).

 

[lavinia]

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,

\pi_{1}(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.

 

[lavinia]

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?

 

[homeomorphic]

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.

 

[lavinia]

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.

 

[lavinia]

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?

 

[mathwonk]

read the first page of this paper for some related results:

http://deepblue.lib.umich.edu/bitstream/2027.42/29969/1/0000331.pdf

but perhaps these examples are not compact.

 

[mathwonk]

here you go: (lens spaces)

http://en.wikipedia.org/wiki/Spherical_3-manifold

 

[mathwonk]

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."

 

[homeomorphic]

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.

 

[lavinia]

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?