- #1

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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?

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- Thread starter chhan92
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- #1

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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

Bacle2

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- #3

lavinia

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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?

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

- #4

Bacle2

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What I was thinking was something like this, which is not an actual solution, but

hopefully may restrict the possibilities:

i)The only S

ii) S

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 ℝ

So, I am kind of torn here.

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

- #5

lavinia

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What I was thinking was something like this, which is not an actual solution, but

hopefully may restrict the possibilities:

i)The only S^{n}'s that asmit a complect structure are S^{2}and S^{6}. And the product of complex manifolds is a complex manifold, with the given product complex structure.

ii) S^{n}is compact , for all n≥1 . If S^{n}where the product of spheres, it could not be a product of S^{2}'s and S^{6}.

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 ℝ^{n}then, by the stereographic projection, the 1-pt compactification of ℝ^{n}is S^{n}

So, I am kind of torn here.

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

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.

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But, lavinia, can you think of a counterexample?

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

Bacle2

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" 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 ℝ

- #8

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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?

This is not what the exercise asks of you. The exercise asks to prove that every compact manifold can be

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

- #9

Bacle2

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This is not what the exercise asks of you. The exercise asks to prove that every compact manifold can beembeddedin 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 [itex]\mathbb{R}^n[/itex].

Why not use a variant of Whitney (if the manifold is not smooth) , embed the manifold in ℝ

- #10

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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

Bacle2

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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)?

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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.

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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)?

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).

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- #14

Bacle2

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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.

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.

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- #15

Bacle2

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

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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

Bacle2

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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.

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.

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- #18

mathwonk

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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.

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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.

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.

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- #21

mathwonk

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"connected" sums?

- #22

Bacle2

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"connected" sums?

Yes, take any two manifolds M,N you identify disks in each and join the two manifolds along them, e.g.,: http://en.wikipedia.org/wiki/Connected_sum.

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e.g. the n-holed torus is just the connected sum of n tori.

- #24

Bacle2

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Right, I forgot to remove the disks first before doing the gluing; story of my life*. Yes, precisely; I was hopelessly trying to patch the original statement into being that every compact manifold is either a product of spheres and projective spaces, or the connected sum of finitely-many products as above.

BTW: I need to change my previous statement: every compact manifold is the compactification; not necessarily 1-pt compactification of R^n; projective space

is a clear counter; the embedded R^n (looking at RP^n --which I always confuse with NPR-- as the quotient S^n/~ , the copy of R^n is everything except the equator, but the

equator is not all collapsed to a single point.)

* Bizarre , meaningless comment.

BTW: I need to change my previous statement: every compact manifold is the compactification; not necessarily 1-pt compactification of R^n; projective space

is a clear counter; the embedded R^n (looking at RP^n --which I always confuse with NPR-- as the quotient S^n/~ , the copy of R^n is everything except the equator, but the

equator is not all collapsed to a single point.)

* Bizarre , meaningless comment.

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- #25

Bacle2

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Maybe to relate it more to Mathwonk's Algebraic Geometric world (please

correct this, Mathwonk, as most surely I missed something or said something wrong):

Connected sums are used to resolve singularities by blowing up

and blowing down curves to allow a tangent space to be well-defined,

i.e., we connect sum a number of copies of ℂP

candidates for tangent spaces at a point to be well-defined, since a tangent space of dimension n has a unique representative in ℂP

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