Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
Quelle joie de vous retrouver ici, Tryphon! –  Georges Elencwajg Oct 30 '11 at 13:56
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. Nov 1 '11 at 1:58
@JacobSchlather: That's not quite how homology interacts with products. See my response below. –  Aaron Mazel-Gee Nov 1 '11 at 20:42
show 1 more comment

2 Answers

up vote 7 down vote accepted

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.

share|improve this answer
Neat! I guess you could do everything over, say $\mathbb{Z}/2$ - that would just kill the Tor term straight away –  Juan S Nov 1 '11 at 7:57
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 1 '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 1 '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 1 '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 1 '11 at 22:08
show 4 more comments

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.

share|improve this answer
You are assuming the manifolds are connected and oriented. –  Mariano Suárez-Alvarez Nov 1 '11 at 0:08
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 1 '11 at 2:34
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 1 '11 at 7:59
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 1 '11 at 17:59
@Sven: Integral (co)homology doesn't admit Kunneth isomorphisms; see my closely-related answer. –  Aaron Mazel-Gee Nov 1 '11 at 18:04
show 5 more comments

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.