## Geometric difference between a homotopy equivalance and a homeomorphism

 I think that the closed manifolds must have the same dimension since homotopy equivalence implies isomorphic homology.
Right, the one with bigger dimension will have higher up homology groups, so, yeah, it's not possible to pull off that idea if you're talking about closed manifolds.

 My question was a little different. Take a group - say the fundamental group of the Klein bottle or of the two ringed torus. How many closed manifolds without boundary can you make that have these group as fundamental groups? Is there an easy construction that shows that there are infinitely many?
It's easy to construct a CW complex having any given fundamental group by taking a wedge of circles and then gluing on disks to kill the relations. For 4-manifolds, the same type of trick will work if you use handles instead of cells, but there's some subtlety there because for 3-manifolds, not every group can be realized as a fundamental group. So, you can show that there exists a 4-manifold with each fundamental group. And you can attach higher handles until it gets capped off, I think, so that it's closed without affecting the fundamental group. Then, just connect sum with a bunch of simply-connected guys, and that should give you infinitely many with the same fundamental group.

Recognitions:
 Quote by homeomorphic It's easy to construct a CW complex having any given fundamental group by taking a wedge of circles and then gluing on disks to kill the relations. For 4-manifolds, the same type of trick will work if you use handles instead of cells, but there's some subtlety there because for 3-manifolds, not every group can be realized as a fundamental group. So, you can show that there exists a 4-manifold with each fundamental group. And you can attach higher handles until it gets capped off, I think, so that it's closed without affecting the fundamental group. Then, just connect sum with a bunch of simply-connected guys, and that should give you infinitely many with the same fundamental group.
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?

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?
 Recognitions: Homework Help Science Advisor I thought the example I gave you was the simplest possible, i.e. take as fundamental group the zero group, then there are already infinitely many 4 manifolds having that fundamental group, namely all smooth surfaces in CP^3. Once you have that it should be easy to construct infinitely many with any fundamental group, maybe by taking connected sums.

Recognitions:
 Quote by homeomorphic Freedmans theorem says the equivalence of intersection forms, plus the Kirby Siebenmann invariant implies homeomorphism. It's very important to say homeomorphism, not diffeomorphism. Actually, Freedman's theorem is a source of many exotic 4-manifolds because we have diffeomorphism invariants like Seiberg-Witten invariants that can distinguish 2-manifolds that have the same intersection form and Kirby-Siebenmann invariant. S^1 × S^3 is not simply connected (fundamental group is Z), so Freedman's theorem doesn't apply here.
Indeed. Just wanted to make the point of how homotopy-equiv. would seem to give you a homeo., by preserving homology--therefore intersection/intersection form, fund. groups, but, one needs the strong restriction of having closedness and compactness to get up to homeomorphism.

To show how weird things can get in 4-D without strong conditions, see e.g:

http://www.intlpress.com/JDG/archive/1994/39-3-491.pdf
 Recognitions: Science Advisor 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?
 Recognitions: Homework Help Science Advisor Here is a nice article on obstructions to finding a smooth complex projective algebraic variety with given fundamental group. For example the integers do not occur, nor does any free group on n generators, nor does the fundamental group of the klein bottle. http://library.msri.org/books/Book28/files/arapura.pdf

Recognitions:
Homework Help
 Quote by lavinia 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?

Layman's version:

a morse function with one min and one max say on a sphere, says you can construct the sphere up to homotopy by starting from a point (for the min) and adding one disc (for the max).

A torus with a morse function with one min, one max, and two flex points, says it has the homotopy type of a point (the min) with a circle added (first flex), then another circle (second flex), then a disc (max).

in general, the fundamental theorem of differential equation says that between consecutive critical points, the manifold looks like a cylinder was added, so no change in homotopy (the gradient flow sweeps out the cylinder). I.e. the homotopy type is determined by the critical points.

So if you have a morse function with a finite number of critical points, you get the homotopy type of a finite cell complex, and in fact the index of the critical point tells you the dimension of the cell to add.
 Recognitions: Homework Help Science Advisor 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.

 Quote by lavinia Here is a wilder example as an exercise. Remove the z-axis and the circle of radius 1 in the xy-plane from Euclidean 3 space. Show that this is homotopy equivalent to a torus.
Let B be the solid sphere of radius 2 in R^3. We can continuously deform everything outside of the sphere to the boundary of B via 2x/||x||. Now, if we make the open z-axis thicker then it will become a hole in the centre of the B. Similarly if we make the unit circle thicker than it will make the interior of B empty. The resulted space is obviously homotopic equivalent to a torus.

Thanks for your examples

Recognitions:
 Quote by mathwonk Layman's version: a morse function with one min and one max say on a sphere, says you can construct the sphere up to homotopy by starting from a point (for the min) and adding one disc (for the max). A torus with a morse function with one min, one max, and two flex points, says it has the homotopy type of a point (the min) with a circle added (first flex), then another circle (second flex), then a disc (max). in general, the fundamental theorem of differential equation says that between consecutive critical points, the manifold looks like a cylinder was added, so no change in homotopy (the gradient flow sweeps out the cylinder). I.e. the homotopy type is determined by the critical points. So if you have a morse function with a finite number of critical points, you get the homotopy type of a finite cell complex, and in fact the index of the critical point tells you the dimension of the cell to add.
yes but it seems like there are piece wise homotopy equivalences as one passes from on critical point to the next. But a global homotopy equivalence requires a map of the manifold into a CW complex and another from the CW complex into the manifold that are homotopy inverses of each other.

Recognitions:
 Quote by mathwonk Here is a nice article on obstructions to finding a smooth complex projective algebraic variety with given fundamental group. For example the integers do not occur, nor does any free group on n generators, nor does the fundamental group of the klein bottle. http://library.msri.org/books/Book28/files/arapura.pdf

So the Eilenberg-McLane spaces cannot be made into varieties? Are these E-M spaces all CW-complexes spaces? manifolds?

I mean, I know the trivial examples from, e.g., Wikipedia, on S^1 being a K(Z,1), etc., but I wonder if there are more general results.

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 Quote by Bacle2 So the Eilenberg-McLane spaces cannot be made into varieties? Are these E-M spaces all CW-complexes spaces? manifolds? I mean, I know the trivial examples from, e.g., Wikipedia, on S^1 being a K(Z,1), etc., but I wonder if there are more general results.
A manifold that is covered by Euclidean space is EM. The homotopy sequence of the fibration verifies this since Euclidean space is contractible and the fiber is discrete.

So all Riemann surfaces except the sphere I guess. all flat Riemannian manifolds such as tori and the Kelin bottle.

For finite groups, the EM's are probably all infinite dimensional CW complexes e.g the classifying space for Z/2Z is the infinite real projective space.
 Recognitions: Homework Help Science Advisor 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.

 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.

 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.

 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.