# Manifold Question

1. Nov 21, 2011

### chhan92

I came across this problem in Willard, General Topology 18H.2,
where it says

"For compact n-manifold X, prove that X is a product of spheres"

I have no idea how to rigorously do it...

I searched internet for this, but has no hope... any helps?

2. Nov 22, 2011

### Bacle2

I don't know if this helps, but there is a result that says that every compact manifold is a compactification of ℝn , tho I don't have a nice proof (I remember the trick was to find the copy of ℝn within the compact manifold, like, e.g., ℝ1 itself is embedded in S1, etc.) I imagine it must be a simple type of compactification e.g., Alexandroff 1-pt. , otherwise , your not likely to end up with a manifold.

3. Nov 22, 2011

### lavinia

This is false. Very few manifolds are products of spheres.

4. Nov 22, 2011

### Bacle2

But, lavinia, can you think of a counterexample?

What I was thinking was something like this, which is not an actual solution, but
hopefully may restrict the possibilities:

i)The only Sn's that asmit a complect structure are S2 and S6. And the product of complex manifolds is a complex manifold, with the given product complex structure.

ii) Sn is compact , for all n≥1 . If Sn where the product of spheres, it could not be a product of S2's and S6.

BUT....

iii) OTOH, I think my previous statement about every manifold being the compactification -- which I think is the 1-pt. compactification , since the resulting space is a manifold, and compactifications can be very bizarre, e.g., Stone-Cech)-- of ℝn then, by the stereographic projection, the 1-pt compactification of ℝn is Sn

So, I am kind of torn here.

BTW : OP: do you mean real manifolds, right?

5. Nov 22, 2011

### lavinia

Maybe I jumped to a conclusion and the question did not mean any manifold. what manifold was supposed to be a product of spheres?

i am not sure about this but give the manifold a Riemannian metric. Then the exponential map at a point maps an open neighborhood of the origin in the tangent space at the point diffeomorphically onto a dense subset of the manifold. Add the complement and you compactify the ball to make the manifold. The complement is a set of measure zero> I think is it call the cut locus - or maybe the conjugate locus. For the sphere, it is the south pole if one exponentiates from the north pole. In this case it is a one point compactification.

6. Nov 22, 2011

### homeomorphic

You can only build simply-connected spaces using products of S^n for n>1.

If you include S^1, you get copies of Z in the fundamental group. So, for example, any manifold that has any torsion in its fundamental group would be a counter-example. And this even works if you consider things only up to homotopy type.

7. Nov 23, 2011

### Bacle2

Homeomorphic wrote:

" You can only build simply-connected spaces using products of S^n for n>1.

If you include S^1, you get copies of Z in the fundamental group. So, for example, any manifold that has any torsion in its fundamental group would be a counter-example. And this even works if you consider things only up to homotopy type. "

I see, so, say a projective space would be then a counterexample.

But, how about the claim that every compact manifold is the 1-pt compactification of ℝn?

8. Nov 23, 2011

### micromass

This is not what the exercise asks of you. The exercise asks to prove that every compact manifold can be embedded in a product of spheres.

Construct the embedding by using (1) and by using that a compact manifold is covered by a finite number of open sets homeomorphic to $\mathbb{R}^n$.

9. Nov 23, 2011

### Bacle2

Why not use a variant of Whitney (if the manifold is not smooth) , embed the manifold in ℝ(2n+1) , then do the stereo projection on ℝ(2n+1)?

10. Nov 23, 2011

### Jamma

Indeed- or, once it's embedded in a Euclidean space, you can easily embed that into a sphere of high enough dimension since it will locally look like Euclidean space.

This makes the question trivial. On the other hand, it is easy to prove that any sphere embeds, so if you could prove the theorem as stated for a suitable class of manifolds then you can easily show that a suitable choice of manifolds embeds into some Euclidean space, which seems rather deep (i.e. being a weaker version of the Whitney embedding theorem).

Clearly your manifolds need to be at least orientable if they are homeomorphic to a product of spheres- perhaps the question was asking you to prove this for only orientable surfaces?

11. Nov 24, 2011

### Bacle2

Still, I wonder if we can still save the OP question with a small change: is every compact manifold the product of a combination of spheres and projective spaces?

I have thought about the sort of reverse issue of a bundle p:E-->B with fiber F over a space being trivial, so that E=BxF: when/how can we tell if a space is a product space? How could we tell, e.g., that S^n is not a product space (I think S^n is not the top space of a trivial bundle)?

12. Nov 24, 2011

### Jamma

I highly doubt it.

What about 2d surfaces? The torus and sphere are fine, but what about genus n surfaces? We will be unable to obtain the "n-holed torus" for n not equal to one.

It also seems that there are issues with you not being able to classify surfaces of dimension greater than 3, and clearly there wouldn't be all that many different manifolds if they all were just products of spheres/projective spaces.

It seems, as someone else pointed out, that the question as stated is just simply wrong. I'm also assuming that the "repaired version" has a nice trick, since just using the Whitney embedding theorem makes the theorem pretty trivial. On the other hand, as I said, this proves the Whitney embedding theorem (with, possibly, a slightly higher minimum bound on the dimension), so I'm still a bit confused.

13. Nov 24, 2011

### Jamma

So you just mean the manifold is a product of manifolds? Well, the (co)homology should help, I'd think - if the manfold was a product of lower dimensional ones, then the fundamental classes of the lower dimensional ones will make an appearance somewhere in the product (take field coefficients Z/2, if you want). This is impossible for S^n because its (co)homology is concentrated in dimensions 0 and n (note that this argument does not work for general bundles - S^3, for example, is a fibre bundle of S^2 and S^1, the groups in dimensions 1 and 2 are killed off in the spectral sequence).

Last edited: Nov 24, 2011
14. Nov 24, 2011

### Bacle2

Yes, I am a bit confused myself, Jamma. And, you're right that the genus-n surface is not likely to be a product; we need connected sums for that, I would guess. It would be nice to see a counterexample if there is one.

But, re S^n, remember that I am referring to trivial bundles, not just any bundle, so S^3 does not apply there (Hopf ). But I'll look into the (co)homology idea.

Last edited: Nov 24, 2011
15. Nov 24, 2011

### Bacle2

And, while you're right about there not being that many manifolds--tho remember we are talking about compact manifolds, and that narrows it down a bit--I am more interested in seeing what may be wrong with the arguments, and for some counterexamples, and how to show a manifold is not a product; it seems like we use a sort of equivalent of primality version using cohomology.

I suspect, for n=4, things will breakdown more quickly, because of all the weirdness going on there.

16. Nov 24, 2011

### Jamma

Yeah, I know we're only looking at compact manifolds- it is impossible to classify all 4 dimensional manifolds up to homeomorphism due to a really nice argument incorporating the word problem on groups (if you aren't familiar with the argument, it's worth looking up).

And I know you only meant trivial bundles- what I said shows that non of the S^n are trivial bundles of lower dimensional manifolds because their cohomology is concentrated in dimensions 0 and n- if they were a product then the fundamental classes of the lower dimensional manifolds would make a contribution to some of the intermediate cohomology groups. I simply noted that this argument doesn't work for general fibre bundles, the Hopf fibration giving a counter example, and the reason it doesn't work is that the intermediate cohomology is killed off in the spectral sequence.

17. Nov 24, 2011

### Bacle2

Ah, I see, nice points. I'll review my spectral sequences (which I never fully got the 1st time around)

Still, Jamma, if we weaken bundles to fibrations, there are nice general results re classification: e.g., symplectic 4-manifolds are precisely those that admit a Lefschetz fibration. Hope I'm not drifting of too far from the OP; sorry otherwise.

Last edited: Nov 24, 2011
18. Nov 24, 2011

### mathwonk

this is a good discussion. consider this problem just in (real) dimension 2.

there are only two possibilities for such products, either S^2, or S1xS1.

Thus there would be only two compact surfaces if these were all possibilities.

But as has been pointed out by Jamma, there are infinitely many non homeomorphic real surfaces,

one for every genus g ≥ 0 even if we only look at orientable ones, and only the ones of genus 0 and 1 occur as such products.

The same argument shows there are only a finite number of products of spheres in any dimension n, but it seems you can

easily construct infinitely many non homeomorphic manifolds of dimension n by using the surfaces already mentioned.

19. Nov 24, 2011

### Jamma

They're not easy- I'm still not comfortable with them (I'm sure that not many are!), I'm not sure how you could obviously see that the differentials kill the cohomology in the spectral sequence, for example, I just know that they will because I know the cohomology of S^3.

I'll look up those classification results, sounds interesting.

20. Nov 24, 2011

### Jamma

To add a little in the relevant direction of the thread, it's interesting to note that in 2d, all surfaces are direct sums of tori or projective spaces. This can't be the case for higher dimensions, but it's hard to visualise why.