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It seems obvious that, if the sphere $S^n$ is homeomorphic to a product $X\times Y$ of topological spaces, then either $X$ or $Y$ is a point. How can one prove that?

Tournesol
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    Quelle joie de vous retrouver ici, Tryphon! – Georges Elencwajg Oct 30 '11 at 13:56
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    I can't quite justify a few steps, so I'm posting this as a comment. Note that $H_n(S^n)\cong \mathbb Z$ so $H_n(X) \times H_n(Y) \cong \mathbb Z$. Since $\mathbb Z$ can't be written as a non-trivial direct product, either $X$ or $Y$ is $n$-simply connected, assume $Y$ is. In particular $X$ and $Y$ must both be manifolds of dim $m,k\leq n$ where $m+k=n$. Suppose that $\dim X =m < n$, then we have that $H_n(X)$ is nontrivial where $n>m$, which isn't possible. It follows that $X$ must be an $n$-dimensional manifold. So $Y$ must be a $0$ dimensional connected manifold, I.E. a point. – JSchlather Oct 30 '11 at 15:24
  • @Jacob: Thanks! I think your argument quite correctly shows that, assuming that $X$ and $Y$ are CW-complexes, $Y$ is indeed a point. – Tournesol Oct 31 '11 at 16:47
  • For those who didn't quite get Georges's reference: you may know Professor Tryphon Tournesol better as Professor Cuthbert Calculus... @Georges: I noticed only now that you and the author of that cartoon are namesakes! :) – J. M. ain't a mathematician Nov 01 '11 at 01:58
  • @JacobSchlather: That's not quite how homology interacts with products. See my response below. – Aaron Mazel-Gee Nov 01 '11 at 20:42
  • @Aaron ah, right I was pretty sure something was wrong with my argument. – JSchlather Nov 02 '11 at 00:45

2 Answers2

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Suppose $S^n = X \times Y$, where $X$ and $Y$ are arbitrary topological spaces (that are homotopy-equivalent to CW complexes). With integer coefficients, there is the (unnaturally) split Kunneth sexseq

$0 \rightarrow \bigoplus_i (H_i(X) \otimes H_{m-i}(Y)) \rightarrow H_m(X \times Y) \rightarrow \bigoplus_i \mbox{Tor}(H_i(X),H_{m-i-1}(Y)) \rightarrow 0$.

Clearly $X$ and $Y$ must be path-connected since $\pi_0$ takes products to products (and obviously we should assume $n \geq 1$), so $H_0(X) = H_0(Y)=\mathbb{Z}$. Since $-\otimes \mathbb{Z}$ does nothing to an abelian group, this means that we're getting a copy of $H_*(X)$ in $H_*(X\times Y)$ from the inclusion above when it's tensored against $H_0(Y)$, and similarly for $H_*(Y)$. Moreover, all the homology of $S^n$ must come from the inclusions in the above sexseq, since $\mbox{Tor}$ always consists entirely of torsion. So without loss of generality, $H_*(X) \cong H_*(S^n)$ and $H_*(Y) \cong H_*(\mbox{pt})$. Since $\pi_1$ takes products to products, both $X$ and $Y$ are simply-connected. So the projection $S^n =X \times Y \rightarrow X$ is a homology isomorphism of simply-connected spaces and hence is a (weak) homotopy equivalence, while $Y$ is a simply-connected space with trivial integral homology so it must be (weakly) contractible. Thus the factorization $S^n = X \times Y$ is trivial.

Aaron Mazel-Gee
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  • Neat! I guess you could do everything over, say $\mathbb{Z}/2$ - that would just kill the Tor term straight away – Juan S Nov 01 '11 at 07:57
  • Over any field, the Kunneth sexseq becomes a Kunneth isomorphism. However, there's no immediate Relative Hurewicz sort of relationship between homotopy and Z/2 homology, so this wouldn't suffice. I was considering doing it at Z/p for all p and also at $\mathbb{Q}$, which would suffice, but this seemed like more of a pain because there didn't seem to be any immediate reason why it couldn't be that among the various torsion you get, some comes from $X$ and some comes from $Y$. – Aaron Mazel-Gee Nov 01 '11 at 18:02
  • I should amend my second sentence: There is a framework called "Serre's C-theory" in which one ignores certain "classes of abelian groups", e.g. abelian groups consisting of odd torsion (or possibly just no even torsion, I'm forgetting which is the right thing). For example, a homomorphism is a "mod-C epimorphism" if the cokernel is in C. In this setting, there is a "mod C Hurewicz theorem", which says what you'd hope. (See the excellent book by Mosher & Tangora for details.)... – Aaron Mazel-Gee Nov 01 '11 at 20:39
  • This led to the first really serious bite (by Serre) that anyone took out of the homotopy groups of spheres, and it also set off a long string of generalizations of the idea of "working one prime at a time" in homotopy theory. – Aaron Mazel-Gee Nov 01 '11 at 20:41
  • Opps, you are right - I had kind of glossed over the fact that you are using the Hurewicz theorem to get that $S^n \to X$ is a weak homotopy equivalence. (I love Mosher & Tangora too. I guess you could ignore all odd torsion, since 2 is prime) – Juan S Nov 01 '11 at 22:08
  • If you do this at 2, then all you get is that $S^n \rightarrow X$ is an isomorphism in 2-localized homotopy. But then, when you do it at 3, you might get that $S^n \rightarrow Y$ is an isomorphism in 3-localized homotopy! This is what I was alluding to in my first comment. – Aaron Mazel-Gee Nov 02 '11 at 01:53
  • Sorry that was misleading - I was just referring to 'e.g. abelian groups consisting of odd torsion (or possibly just no even torsion, I'm forgetting which is the right thing).'! – Juan S Nov 02 '11 at 02:05
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    I'm missing something. How do you go from "Y is weakly contractible" to "therefore the factorization is trivial" ? For example, if one applies the same argument to $\mathbb{R}^n$ one would learn that all factorizations of $\mathbb{R}^n$ are trivial - but this is not true. I agree that this argument shows that if $X$ and $Y$ are manifolds, then the factorization is trivial. – Jason DeVito - on hiatus Nov 07 '11 at 03:06
  • Well I'm using the techniques of algebraic topology, so I could only hope to obtain a result up to homotopy equivalence in the first place. Certainly this has a different flavor than factorizations of Euclidean space. I guess since the question says "homeomorphism", this isn't truly a complete answer... and given the examples at the related question http://math.stackexchange.com/questions/78055/decomposition-of-a-manifold/78166#78166 , I'd be surprised if the sphere really is on-the-nose indecomposable anyways. – Aaron Mazel-Gee Nov 08 '11 at 07:34
  • @AaronMazel-Gee You are too careful in your final comment. One simply needs to know a) that factors of manifolds have a good notion of dimension, which is additive under products, b) that a 0-dimensional, path-connected manifold factor is a point, and c) that a k-dimensional manifold factor has no homology in dimension higher than k. The framework that gives you all of these (the first two quickly) is the fact that manifold factors are "homology manifolds". –  Jan 13 '19 at 06:13
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Assuming that everything is a connected manifold.

Let $X$, $Y$ such that $X \times Y = S^n$. Before we apply the Kunneth theorem lets note that the dimensions of $X$ and $Y$ must add up to to $n$. $$\bigoplus_{i+j=k} H^i(X)\otimes H^j(Y)=H^k(S^n) $$

Suppose $\dim X = i < n$. Then $H^i(X)=\mathbb{Z}=H^{n-i}(Y)$ as these are the only possible non-zero cohomology groups that add up to $n$ which is required by the Kunneth theorem. But therefore $H^i(S^n)=\bigoplus H^k(X)\oplus H^j(Y) \neq 0 $ as the summation contains $H^i(X) \oplus H^0(Y)=\mathbb{Z}$ which is a contradiction. Therefore, wlog, $i=n$ and $Y$ is a point.

Sven
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  • You are assuming the manifolds are connected and oriented. – Mariano Suárez-Álvarez Nov 01 '11 at 00:08
  • Well, a product of manifolds is orientable iff both factors are. My complaint is that we're assuming $X$ and $Y$ are manifolds! – Aaron Mazel-Gee Nov 01 '11 at 02:34
  • Is it possible for $X \times Y \simeq M$ (in the category of topological spaces) and for $M$ to be a manifold, and $X,Y$ not to be? – Juan S Nov 01 '11 at 07:59
  • Certainly! Take $X$ to be homotopy equivalent to a manifold but not homeomorphic to one, and take $Y$ to be a point. Being a manifold is a very strong condition; among other things, it means that you have Poincare duality (possibly twisted). Thus, it is a much more restrictive answer to show that $S^n$ is not the product of two manifolds. – Aaron Mazel-Gee Nov 01 '11 at 17:59
  • @Sven: Integral (co)homology doesn't admit Kunneth isomorphisms; see my closely-related answer. – Aaron Mazel-Gee Nov 01 '11 at 18:04
  • @Aaron - if $X$ is only homotopy equivalent to a manifold, then how is $X \times Y$ where $Y$ is a point a manifold? – Juan S Nov 01 '11 at 22:13
  • @JuanS: I took your $\simeq$ to mean homotopy equivalent; I guess that's not what you meant. If you meant homeomorphism, then I don't know. I wouldn't be surprised if there's such a pathological example, but point-set topology is very much not my forte. You should ask this separately. Of course, it's really a local question... – Aaron Mazel-Gee Nov 02 '11 at 01:35
  • Yes I thought that might be misleading (that was why I added the in the category of topological spaces). I wouldn't be surprised either - I was wondering if it is such a strong thing to assume in this question – Juan S Nov 02 '11 at 02:06
  • Unless you can answer your question affirmatively, then yes! Manifolds have lots of extra structure; most notably they have Poincare duality. Anyways, this is an essentially homotopical question, and as such it should be solved as one. There wouldn't be much to modify about the proof, except that "dimension" should be replaced with "homological dimension". – Aaron Mazel-Gee Nov 02 '11 at 05:54
  • At the risk of having the computer yell at me for too many comments - see this thread, where t.b. gives examples of non-manifold decompositions of a manifold – Juan S Nov 03 '11 at 01:58